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Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2013, Article ID 367086, 13 pages http://dx.doi.org/10.1155/2013/367086 Research Article Implicit Active Contour Model with Local and Global Intensity Fitting Energies Xiaozeng Xu 1,2 and Chuanjiang He 1 1 College of Mathematics and Statistics, Chongqing University, Chongqing 401331, China 2 School of Mathematics and Statistics, Chongqing University of Technology, Chongqing 400054, China Correspondence should be addressed to Xiaozeng Xu; [email protected] Received 19 December 2012; Accepted 28 March 2013 Academic Editor: Oluwole Daniel Makinde Copyright © 2013 X. Xu and C. He. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We propose a new active contour model which integrates a local intensity fitting (LIF) energy with an auxiliary global intensity fitting (GIF) energy. e LIF energy is responsible for attracting the contour toward object boundaries and is dominant near object boundaries, while the GIF energy incorporates global image information to improve the robustness to initialization of the contours. e proposed model not only can provide desirable segmentation results in the presence of intensity inhomogeneity but also allows for more flexible initialization of the contour compared to the RSF and LIF models, and we give a theoretical proof to compute a unique steady state regardless of the initialization; that is, the convergence of the zero-level line is irrespective of the initial function. is means that we can obtain the same zero-level line in the steady state, if we choose the initial function as a bounded function. In particular, our proposed model has the capability of detecting multiple objects or objects with interior holes or blurred edges. 1. Introduction Implicit active contour models have been extensively studied and successfully used in image segmentation [13]. e basic idea is to evolve a contour under some constraints to extract the desired object. According to the nature of constraints, the existing active contour models can be categorised into two classes: edge-based models [47], and region-based models [817]. Each of them has its own pros and cons; the choice of them in applications depends on different characteristics of images. In this study, we focus on region-based models and consider images with intensity inhomogeneity. Unlike edge-based models that utilize typically an edge indicator depending on image gradient to perform contour extraction, region-based models usually use global and/or local statistics inside regions rather than gradient on edges to find a partition of image domain. ey generally have bet- ter performances in the presence of weak or discontinuous boundaries than edge-based models. Early popular region- based models tend to rely on intensity homogeneous (roughly constant or smooth) statistics in each of the regions to be segmented. erefore, they either lack the ability to deal with intensity inhomogeneity like the PC (piecewise constant) model [8] or are excessively expensive in computation like the PS (piecewise smooth) model [9]. To handle intensity inhomogeneity efficiently, some localized region-based models [1116] have been proposed recently. For example, Li et al. [12] recently proposed a region- scalable fitting (RSF) active contour model (originally termed as local binary fitting (LBF) model [11]). e RSF model draws upon intensity information in spatially varying local regions depending on a scale parameter, so it is able to deal with intensity inhomogeneity efficiently. Very recently, Zhang et al. [15] proposed a novel active contour model driven by local image fitting energy, which also can handle intensity inhomogeneity efficiently. e experimental results in [15] show that this model is more efficient than the RSF model, while yielding similar results. However, these two models easily get stuck in local minimums for the most of contour initializations. is makes it need user intervention to define the initial contours professionally. In this study, based on the PC model [8] and RSF model [12], we propose a new active contour model, which integrates a local intensity fitting (LIF) energy with an auxiliary global

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Page 1: Research Article Implicit Active Contour Model with Local ...downloads.hindawi.com/journals/mpe/2013/367086.pdfAcademic Editor: Oluwole Daniel Makinde ... (CV)proposedaregion-basedactivecontourmodel

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 367086 13 pageshttpdxdoiorg1011552013367086

Research ArticleImplicit Active Contour Model with Local andGlobal Intensity Fitting Energies

Xiaozeng Xu12 and Chuanjiang He1

1 College of Mathematics and Statistics Chongqing University Chongqing 401331 China2 School of Mathematics and Statistics Chongqing University of Technology Chongqing 400054 China

Correspondence should be addressed to Xiaozeng Xu xuxzcquteducn

Received 19 December 2012 Accepted 28 March 2013

Academic Editor Oluwole Daniel Makinde

Copyright copy 2013 X Xu and C He This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We propose a new active contour model which integrates a local intensity fitting (LIF) energy with an auxiliary global intensityfitting (GIF) energyThe LIF energy is responsible for attracting the contour toward object boundaries and is dominant near objectboundaries while the GIF energy incorporates global image information to improve the robustness to initialization of the contoursThe proposedmodel not only can provide desirable segmentation results in the presence of intensity inhomogeneity but also allowsfor more flexible initialization of the contour compared to the RSF and LIF models and we give a theoretical proof to compute aunique steady state regardless of the initialization that is the convergence of the zero-level line is irrespective of the initial functionThis means that we can obtain the same zero-level line in the steady state if we choose the initial function as a bounded functionIn particular our proposed model has the capability of detecting multiple objects or objects with interior holes or blurred edges

1 Introduction

Implicit active contour models have been extensively studiedand successfully used in image segmentation [1ndash3] The basicidea is to evolve a contour under some constraints to extractthe desired object According to the nature of constraints theexisting active contour models can be categorised into twoclasses edge-based models [4ndash7] and region-based models[8ndash17] Each of them has its own pros and cons the choice ofthem in applications depends on different characteristics ofimages In this study we focus on region-based models andconsider images with intensity inhomogeneity

Unlike edge-based models that utilize typically an edgeindicator depending on image gradient to perform contourextraction region-based models usually use global andorlocal statistics inside regions rather than gradient on edgesto find a partition of image domain They generally have bet-ter performances in the presence of weak or discontinuousboundaries than edge-based models Early popular region-basedmodels tend to rely on intensity homogeneous (roughlyconstant or smooth) statistics in each of the regions to besegmentedTherefore they either lack the ability to deal with

intensity inhomogeneity like the PC (piecewise constant)model [8] or are excessively expensive in computation like thePS (piecewise smooth) model [9]

To handle intensity inhomogeneity efficiently somelocalized region-based models [11ndash16] have been proposedrecently For example Li et al [12] recently proposed a region-scalable fitting (RSF) active contourmodel (originally termedas local binary fitting (LBF) model [11]) The RSF modeldraws upon intensity information in spatially varying localregions depending on a scale parameter so it is able to dealwith intensity inhomogeneity efficiently Very recently Zhanget al [15] proposed a novel active contour model drivenby local image fitting energy which also can handle intensityinhomogeneity efficiently The experimental results in [15]show that this model is more efficient than the RSF modelwhile yielding similar results However these two modelseasily get stuck in local minimums for the most of contourinitializations This makes it need user intervention to definethe initial contours professionally

In this study based on the PC model [8] and RSF model[12] we propose a new active contourmodel which integratesa local intensity fitting (LIF) energy with an auxiliary global

2 Mathematical Problems in Engineering

intensity fitting (GIF) energy The LIF energy is responsiblefor attracting the contour toward object boundaries andis dominant near object boundaries while the GIF energyincorporates global image information to improve the robust-ness to initialization of the contoursThe proposedmodel canefficiently handle intensity inhomogeneity while allowingfor more flexible initialization and maintaining the subpixelaccuracy

The remainder of this paper is organized as followsSection 2 briefly reviews the PC model [8] and RSF model[12] Section 3 introduces the proposed model Section 4presents experimental results using a set of synthetic and realimages This paper is summarized in Section 5

2 Related Works

21 Piecewise ConstantModel by Chan and Vese In [8] ChanandVese (CV) proposed a region-based active contourmodelthat relies on intensity homogeneous (roughly constant)statistics in each of the regions to be segmented In the CVmodel a contour 119862 is represented implicitly by the zero-levelset of a Lipschitz function 120601 Ω rarr R which is called a levelset function In what follows we let the level set function120601 take positive and negative values inside and outside thecontour 119862 respectively

Let 119868 Ω sub R2rarr R be an input image and let119867

120576be the

regularized Heaviside function the energy functional of theCV model is defined as

119864CV

(1198881 1198882 120601) = 120582

1intΩ

1003816100381610038161003816119868 minus 11988811003816100381610038161003816

2

119867120576(120601) 119889x

+1205821intΩ

1003816100381610038161003816119868 minus 11988821003816100381610038161003816

2

(1 minus 119867120576(120601)) 119889x

+ ]intΩ

1003816100381610038161003816nabla119867120576 (120601)1003816100381610038161003816 119889x

(1)

where 1205821 1205822gt 0 ] gt 0 are constantsThe regularized version

of119867(119911) is chosen as

119867120576(119911) =

1

2(1 +

2

120587arctan(119911

120576)) (2)

1198881and 1198882are the global averages of the image intensities in the

region x 120601(x) gt 0 and x 120601(x) lt 0 respectively that is

1198881(120601) =

intΩ

119868 (x)119867120576(120601 (x)) 119889x

intΩ

119867120576(120601 (x)) 119889x

1198882(120601) =

intΩ

119868 (x) (1 minus 119867120576(120601 (x))) 119889x

intΩ

(1 minus 119867120576(120601 (x))) 119889x

(3)

The solution of the CV model in fact leads to a piecewiseconstant segmentation of the original image 119868(x)

119868119866

(x) = 1198881119867120576(120601 (x)) + 119888

2(1 minus 119867

120576(120601 (x))) (4)

where 1198881and 1198882are the averages of the image intensities in the

region x 120601(x) gt 0 and x 120601(x) lt 0 respectively Suchconstants may be far away from the original image data if theintensities outside or inside the contour119862 = x 120601(x) = 0 arenot homogeneous As a result the CVmodel generally fails tosegment images with intensity inhomogeneity

22 Region-Scalable Fitting Model In order to improve theperformance of the global CVmodel [8] and the PSmodel [9]on images with inhomogeneity Li et al [11 12] recently pro-posed a novel region-based active contour model They in-troduced a kernel function and defined the following energyfunctional

119864RSF

(1198911 1198912 120601) = 120582

1int[int119870

120590(x minus y) 1003816100381610038161003816119868 (y) minus 1198911 (x)

1003816100381610038161003816

2

times119867120576(120601 (y)) 119889y] 119889x

+ 1205822int[int119870

120590(x minus y) 1003816100381610038161003816119868 (y) minus 1198911 (x)

1003816100381610038161003816

2

times (1 minus 119867120576(120601 (y))) 119889y] 119889x

+ ]intΩ

1003816100381610038161003816nabla119867120576 (120601 (x))1003816100381610038161003816 119889x

+ 120583intΩ

1

2(1003816100381610038161003816nabla120601 (x)

1003816100381610038161003816 minus 1)2

119889x(5)

where119870120590is a Gaussian kernel with standard deviation 120590 and

1198911(x) and 119891

2(x) are two smooth functions that approximate

the local image intensities inside and outside the contourrespectively They are computed by

1198911(x) =

intΩ

119870120590(x minus y) 119868 (y)119867

120576(120601 (y)) 119889y

intΩ

119870120590(x minus y)119867

120576(120601 (y)) 119889y

1198912(x) =

intΩ

119870120590(x minus y) 119868 (y) (1 minus 119867

120576(120601 (y))) 119889y

intΩ

119870120590(x minus y) (1 minus 119867

120576(120601 (y))) 119889y

(6)

The solution of the RSF model leads to a piecewisesmooth approximation of the original image 119868(x)

119868119871

(x) = 1198911(x)119867120576(120601 (x)) + 119891

2(x) (1 minus 119867

120576(120601 (x))) (7)

where the smooth functions 1198911(x) and 119891

2(x) are computed

by (6) 1198911(x) and 119891

2(x) are the averages of local intensities

on the two sides of the contour which are different from theconstants 119888

1and 1198882in the CVmodel the averages of the image

intensities on the two sides of the contour Therefore theRSF model can deal with intensity inhomogeneity efficientlyHowever it is sensitive to contour initialization (initial loca-tions sizes and shapes)

3 The Proposed Model

31 Description of Our Model Given a positive constant 120596(0 le 120596 le 1) from (4) and (7) we define the following energyfunctional

119864 (120601 1198881 1198882 1198911 1198912) = 120596119864

119866

(120601) + (1 minus 120596) 119864119871

(120601)

= 120596(1

2intΩ

10038161003816100381610038161003816119868(x) minus 119868119866(x)10038161003816100381610038161003816

2

119889x)

+ (1 minus 120596) (1

2intΩ

10038161003816100381610038161003816119868 (x) minus 119868119871 (119909)10038161003816100381610038161003816

2

119889x) (8)

Mathematical Problems in Engineering 3

where

119868119866

(x) = 1198881119867120576(120601 (x)) + 119888

2(1 minus 119867

120576(120601 (x)))

119868119871

(x) = 1198911(x)119867120576(120601 (x)) + 119891

2(x) (1 minus 119867

120576(120601 (x)))

(9)

Keeping 120601 fixed andminimizing the functional119864(120601 1198881 1198882

1198911 1198912) with respect to 119888

1 1198882 1198911 and 119891

2 we have

1198881=

intΩ

119868 (x)119867120576(120601 (x)) 119889x

intΩ

119867120576(120601 (x)) 119889x

1198882=

intΩ

119868 (x) (1 minus 119867120576(120601 (x))) 119889x

intΩ

(1 minus 119867120576(120601 (x))) 119889x

1198911(x) =

intΩ

119870120590(x minus 119910) 119868 (119910)119867

120576(120601 (119910)) 119889119910

intΩ

119870120590(x minus 119910)119867

120576(120601 (119910)) 119889119910

1198912(x) =

intΩ

119870120590(x minus 119910) 119868 (119910) (1 minus 119867

120576(120601 (119910))) 119889119910

intΩ

119870120590(x minus 119910) (1 minus 119867

120576(120601 (119910))) 119889119910

(10)

Keeping 1198881 1198882 1198911 and 119891

2fixed and minimizing the func-

tional 119864(120601 1198881 1198882 1198911 1198912) with respect to 120601 we obtain the cor-

responding gradient descent flow equation

120597120601

120597119905= 120575120576(120601) [120596 ((119868 minus 119868

119866

) (1198881minus 1198882))

+ (1 minus 120596) ((119868 minus 119868119871

) (1198911minus 1198912))]

= 120575120576(120601) [120596119865

119866

(120601) + (1 minus 120596) 119865119871

(120601)] = 119865 (120601)

(11)

where

120575120576(119911) = 119867

1015840

120576(119911) =

1

120587

120576

1205762 + 1199112 (12)

Like the CV and RSF models our model is also imple-mented using an alternative procedure for each iteration andthe corresponding level set function 120601119899 we first compute thefitting values 119888

119894(120601119899

) and 119891119894(120601119899

) and then obtain 120601119899+1 by min-imizing 119864(120601 119888

1(120601119899

) 1198882(120601119899

) 1198911(120601119899

) 1198912(120601119899

)) with respect to 120601This process is repeated until the zero-level set of 120601119899+1 isexactly on the object boundary

In the following we first discuss the properties of 119865119866(120601)and 119865119871(120601) and then analyze the behavior of (11)

32 Properties of 119865119866(120601) and 119865119871(120601) For the sake of simplicitywe state and prove the properties of 119865119866(120601) and 119865119871(120601) only foran image consisting of only two distinct gray levels

119868 (x) = 1198921 x isin 120596

1198922 x isin Ω 120596

(13)

where 1198921 1198922ge 0 with 119892

1= 1198922 120596 and Ω 120596 represent the ob-

jects of interest and the background respectively

Theorem 1 Let 119868(x) be an image by (13) Then one has

119865119866

(120601) =

(119872119899 minus 119873119898) (1198921minus 1198922)2

times [(119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

times(1198732

(119872 minus 119873)2

)minus1

in 120596

minus (119872119899 minus 119873119898) (1198921minus 1198922)2

times [119899 (119872 minus 119873)119867120576

+119873 (119898 minus 119899) (1 minus 119867120576)]

times(1198732

(119872 minus 119873)2

)minus1

in Ω 120596

(14)

and so

sign (119865119866 (120601)) =

+ sign (119872119899 minus 119873119898) in 120596

minus sign (119872119899 minus 119873119898) in Ω 120596

(15)

where

119872 = |Ω| 119898 = |120596| 119873 =1003816100381610038161003816120601 gt 0

1003816100381610038161003816

119899 =1003816100381610038161003816120596 cap 120601 gt 0

1003816100381610038161003816

(16)

in which |Ω| is the area of the regionΩ and similarly for others

Remarks (i) Due to the discrete nature of image |Ω| is in factthe number of pixels in the image 119868(x) and similarly to others

(ii) The cases of 119899 = 119898 119899 = 119873 and 119899 = 0 correspondto the zero-level lines of 120601(x) which are encircling the object(120596) inside the object and within the background (Ω

120596) respectively The cases of 119873 = 119899 =119898 and 119873 = 119899 =

119898 correspond to the zero-level line of 120601(x) that are partiallyinside the object and exactly on the object edge respectively

(iii) The significance of (15) is that the function 119865119866

(120601)

has the opposite sign in 120596 (object) and Ω 120596 (background)respectively

The proof of Theorem 1 is provided in Appendix A Thefollowing result will be used in the proof ofTheorem 7 whichguarantees that the evolution by (10) converges to a uniquestable value after finite time

Corollary 2 Let 119868(x) be an image by (13) Then one has thefollowing

(i) If119872119899 minus119873119898 ge 0 then

119865119866

(120601) ge (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

119865119866

(120601) le minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(17)

4 Mathematical Problems in Engineering

(ii) If119872119899 minus119873119898 le 0 then

119865119866

(120601) le (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

119865119866

(120601) ge minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(18)

The proof of Corollary 2 is given in Appendix AWe call the property in Theorem 1 an adaptive sign-

changing property of 119865119866(120601) Such property also holds for119865119871

(120601) which can be expressed by the following theorem

Theorem 3 Let 119868(x) be an image given by (13) Then one has

119865119871

(120601) =

(119875119902 minus 119876119901) (1198921minus 1198922)2

times [(119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times(1198762

(119875 minus 119876)2

)minus1

in 120596

minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times(1198762

(119875 minus 119876)2

)minus1

in Ω 120596

(19)

and so

sign (119865119871 (120601)) =

+ sign (119875119902 minus 119876119901) in 120596

minus sign (119875119902 minus 119876119901) in Ω 120596

(20)

where

119875 = intΩ

119870120590(x minus 120585) 119889120585 119901 = int

120596

119870120590(x minus 120585) 119889120585

119876 = int120601gt0

119870120590(x minus 120585) 119889120585

119902 = int120596cap120601gt0

119870120590(x minus 120585) 119889120585

(21)

This theorem shows that the local force 119865119871(120601) has exactlyopposite sign in 120596 (object) and in Ω 120596 (background)

The following result together with Corollary 2 will beused in the proof of Theorem 7 mentioned later

Corollary 4 Under the assumption ofTheorem 3 one has thefollowing

(i) If 119875119902 minus 119876119901 ge 0 then

119865119871

(120601) ge 119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

119865119871

(120601) le minus119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(22)

(ii) If 119875119902 minus 119876119901 le 0 then

119865119871

(120601) le 119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

119865119871

(120601) ge minus119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(23)

where

119878 = min 119870120590(x minus 120585) 120585 isin Ω) (24)

The proofs of Theorem 3 and Corollary 4 are similar toTheorem 1 and Corollary 2 respectively see Appendix B fordetails

33 Behavior of Our Model In this section we analyze thebehavior of our model (10) for image segmentation We willshow that the zero-level line of the evolution function startingwith a bounded function finally comes to a unique steadystate which separates the object from the background

Due to the discrete nature of image the continuous equa-tion (10) can be discretized in space with space step 1 (impliedpixel spacing) but also in time with a time step Δ119905 as follows

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

0 le 119894 le 119901 minus 1 0 le 119895 le 119902 minus 1

(25)

where 119901 times 119902 corresponds to the image size 119868119894119895= 119868(119894 119895) and

120601119896

119894119895= 120601(119896Δ119905 119894 119895) with 119896 ge 0 and 1206010

119894119895= 1206010(119894 119895)

Corollary 5 Let 1198721198990minus 1198730119898 = |Ω||120596 cap 120601

0

gt 0| minus |1206010

gt

0||120596| = 0

(i) If1198721198990minus 1198730119898 gt 0 then

119865119866

(120601119896

) gt (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119866

(120601119896

) lt minus (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= minus119860 lt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(26)

Mathematical Problems in Engineering 5

(ii) If1198721198990minus 1198730119898 lt 0 then

119865119866

(120601119896

) lt (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= 119860 lt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119866

(120601119896

) gt minus (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= minus119860 gt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(27)

where

119860 = (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987248 (28)

We provide the proof of Corollary 5 in Appendix C Sim-ilar analysis can prove the following corollary Corollary 6

Corollary 6 Let 1198751199020minus 1198760119901 = 0

(i) If 1198751199020minus 1198760119901 gt 0 then

119865119871

(120601119896

) gt (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= 119861 gt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119871

(120601119896

) lt minus (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= minus119861 lt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(29)

(ii) If 1198751199020minus 1198760119901 lt 0 then

119865119871

(120601119896

) lt (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= 119861 lt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119871

(120601119896

) gt minus (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= minus119861 gt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(30)

where

119861 = (1198751199020minus 1198760119901)

(1198921minus 1198922)2

11987548 (31)

Theorem 7 If 1206010(x) is a bounded function with (1198721198990minus

1198730119898)(119875119902

0minus 1198760119901) gt 0 then there exists a positive integer 119870

such that for 119896 ge 119870

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

if 1198721198990minus 1198730119898 gt 0 119875119902

0minus 1198760119901 gt 0

(32)

120601119896+1

119894119895lt 0 in 120596

120601119896+1

119894119895gt 0 in Ω 120596

if 1198721198990minus 1198730119898 lt 0 119875119902

0minus 1198760119901 lt 0

(33)

We provide the proof of Theorem 7 in Appendix DThe significance of Theorem 7 is that if we choose 1206010(x)

such that (1198721198990minus 1198730119898)(119875119902

0minus 1198760119901) gt 0 then the zero-level

line of120601(x) startingwith such initial function1206010(x) can finallycome to a unique steady state which separates object from thebackground

Remark Mathematically there does exist 1206010(x) such that(1198721198990minus1198730119898)(119875119902

0minus1198760119901) le 0 in practice however we always

can guarantee that

(1198721198990minus 1198730119898) (119875119902

0minus 1198760119901) gt 0 (34)

34 Discussion of Initial Function Theorem 7 guarantees thatthe proposed model computes a unique steady state regard-less of the initialization that is the convergence of the zero-level line 120601 = 0 is irrespective of the initial functionThis means that we can obtain the same zero-level line inthe steady state if we choose the initial function as a boundedfunction 1206010 with (119872119899

0minus 1198730119898)(119875119902

0minus 1198760119901) gt 0

In applications the initial function 1206010 can be defined

via a simple curve (closed curve or line segment) in imagedomain For example we can choose the initial function asa signed distance to a circle which is widely used in mostof image segmentation models with level set methods Forthe proposed model however we prefer to define the initialfunction 120601

0

(x) as a piecewise constant or constant functionas follows

(1) If the curve 119862 is a closed curve (eg circle or square)then 1206010(x) is defined by

1206010

(x) = +120588 (x) isin inside (119862) minus120588 (x) isin outside (119862)

(35)

where 120588 = 0 is a constant(2) If the curve 119862 is a line segment that partitions the

image domainΩ into two disjoint regionsΩ1andΩ

2

(eg the left and right half regions of image domainΩ respectively) withΩ = Ω

1cup Ω2andΩ

1cap Ω2= Φ

then 1206010(119909 119910) is defined by

1206010

(x) = +120588 (x) isin Ω

1

minus120588 (x) isin Ω2

(36)

where 120588 = 0 is a constant

6 Mathematical Problems in Engineering

(3) We can also define a zero function as follows

1206010

(x) = 0 (37)

Next we prove the fact that with 1206010(x) = 0 120601(x) becomesa sign-changing function and satisfy the condition of (119872119899

1minus

1198731119898)(119875119902

1minus 1198761119901) gt 0 after the first iteration

Theorem 8 If the initial function 1206010(x) = 0 in Ω and

1198880

1(x) = 119891

0

1(x) = 2 sdot mean

xisinΩ119868 (x)

1198880

2(x) = 119891

0

2(x) = 0

(38)

then after the first iteration one has

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(39)

whereΩ1cap Ω2= Φ Ω

1= Φ Ω2

= Φ

We provide the proof of Theorem 8 in Appendix EBy (25) andTheorem 8 we have

1206011

(x) = 120575120576(0) mean

xisinΩ119868 (x)

times (119868 minus meanxisinΩ

119868 (x))

lt 0 x isin Ω1

gt 0 x isin Ω2

(40)

Therefore 1206011(x) became a sign-changing function Thenusing two distinct gray levels of (13) and the above demon-stration we have

(1198721198991minus 1198731119898) (119875119902

1minus 1198761119901)

= (119898 (119872 minus1198731)) (119901 (119875 minus 119876

1))

= 119898119901 (119872 minus1198731) (119875 minus 119876

1) gt 0

(41)

4 Implementation and Experimental Results

41 Implementation In tradition PDE-based methods a cer-tain diffusion term is usually included in the evolution equa-tion to regularize the evolving function but which increasesthe computational cost Recently [16] proposed a novelscheme to regularize the evolving function that is Gaussianfiltering the evolving function after each of iterations Weadopt this scheme to regularize the evolving function 119906 ateach of iteration that is 120601119899 = 119866

120588lowast 120601119899 where 119901 controls the

regularization strength Such regularization procedure hassome advantages over the traditional regularization termsuch as computational efficiency and better smoothing effectsee [16] for more explanations

The main procedures of the proposed algorithm can besummarized as follows

(1) Initialize the evolving function 1206010(x)(2) Compute 119888

119894(120601) and 119891

119894(120601) (119894 = 1 2) using (10)

(3) Evolve the function 120601 according to (11)(4) Regularize the function 120601 with a Gaussian filter that

is 120601119899 = 119866120588lowast 120601119899

(5) Check whether the evolution is finished If not returnto step 2

42 Experimental Results In this section we show experi-mentally that the proposedmodel not only can provide desir-able segmentation results in the presence of intensity inho-mogeneity but also allows for more flexible initialization ofthe contour compared to the RSF and LIF models

In our numerical experiments for our model we choosethe parameters as follows 120576 = 15 for the regularized Diracfunction Δ119905=0025 (time step) ℎ = 1 (space step) 120596 = 01and 119899 = 5 The MATLAB source code of the LIF modelalgorithm is downloaded at httpwww4comppolyueduhksimcslzhangcodeLIFzip and the RSF model algorithmis downloaded at httpwwwengruconnedusimcmlicodeRSF v0 v01rar All experiments were run under MATLAB2010b on a PC with Dual 27 GHz processor

We will utilize two region overlap metrics to evaluatethe performances of the three models quantitatively The twometrics are the ratio of segmentation error (RSE) [18] and thedice similarity coefficient (DSC) [14 18 19] If 119878

1and 1198782rep-

resent a given baseline foreground region (eg true object)and the foreground region found by the model respectivelythen the two metrics are defined as follows

RSE =119873 (1198781 1198782) + 119873 (119878

2 1198781)

119873 (Ω)

DSC =2119873 (119878

1cap 1198782)

119873 (1198781) + 119873 (119878

2)

(42)

where 119873(sdot) indicates the number of pixels in the enclosedregion andΩ is image domainThe closer the DSC value to 1the better the segmentation Since119873(119878

11198782)+119873(119878

21198781) is the

number of the pixelsmistakenly classified by themodel lowerRSE means that there are fewer pixels misclassified that isthe image can be segmented more accurately Thus a perfectsegmentation will give DSC = 1 RSE = 0

In the first example (Figures 1ndash3) we mainly verify thecomputation of a unique steady state of the zero-level line of120601 starting with three types of representative initial functionsthat is signed distance function piecewise constant functionsby (26)-(27) and zero function respectively The top rowof Figure 1 shows the evolution of an active contour (iezero-level line 120601 = 0) with the function 120601 initialized toa signed distance function piecewise constant functions by(26)-(27) with 120588 = 1 and zero function respectively whilethe bottom row shows the corresponding evolution of120601Withsuch initializations the zero-level line of 120601 converges to thesame steady state

In Figure 2 we show that our model has the capabilityof detecting multiple objects or objects with interior holesor blurred edges only starting with a zero function Thecontours (zero-level lines) evolution processes are shown inthe second column to the forth column

Mathematical Problems in Engineering 7

Figure 1 Segmentations of our model for two real images with 120601 initialized as different functions Columns 1 to 4 signed distance functionpiecewise constant functions by (26)-(27) with 120588 = 1 and zero function

Figure 2 Applications of our model to four images (slug cell ventriculus sinister MR and real plane images) The curve evolution processfrom the initial contour (in the first column) to the final contour (in the fourth column) is shown in every row for the corresponding image

8 Mathematical Problems in Engineering

(a) (b) (c) (d)

Figure 3 Comparisons of three models Rows 1ndash3 RSF model LIF model and our model with 1206010= 0 For the RSF and LIF models initial

and final contours are shown in green and red color respectively Columns 1ndash4 the 4 pictures indicate (a) (b) (c) (d) respectively

Figure 4 Segmentation results of both models for a hand phantomUpper row LIF model Lower row our model with 120601

0= 0 Column

2 zoomed view of the narrow parts in hand fingers

We choose the segmentation results of the RSF and LIFmodels as baseline foreground regions and then computeDSC values for the corresponding images The RSE and DSCvalues for the four real images are given inTable 1 fromwhich

Table 1 RSE and DSC for RSF and LIF models and our model

RSF model LIF modelRSE () DSC () RSE () DSC ()

Figure 3(a) 027 9717 299 9891Figure 3(b) 271 9637 012 9913Figure 3(c) 049 8795 045 8806Figure 3(d) 016 9766 076 9891

we can see that the proposed algorithm works as well as theRSF and LIFmodels for imageswith intensity inhomogeneity

The experimental results shown in Figure 4 validate thatour method can also achieve subpixel segmentation accuracyas in [15] As can be seen from Figures 4(b) and 4(d) bothmodels achieve subpixel segmentation accuracy of the fingerboundaries The final contour accurately reflects the truehand shape

5 Conclusion

In this study we propose a new active contour model inte-grating a local intensity fitting (LIF) energy with an auxiliaryglobal intensity fitting (GIF) energy The LIF energy isresponsible for attracting the contour toward object bound-aries and is dominant near object boundaries while the GIFenergy incorporates global image information to improve therobustness to initialization of the contours The proposedmodel can efficiently handle intensity inhomogeneity while

Mathematical Problems in Engineering 9

allowing for more flexible initialization and maintaining thesubpixel accuracy The utility model has the advantages ofallowing for more flexible initialization of the contour andthe capability of detecting multiple objects or objects withinterior holes or blurred edges But [14] has not been imple-mented

Appendices

A Proofs of Theorem 1 and Corollary 2

Proof of Theorem 1 Clearly 120601 gt 0 =Φ and 120601 lt 0 =Φowing to the sign-changing property of 120601(119909 119910) Thus by (3)we have

1198881(120601) =

1198921119899 + 1198922(119873 minus 119899)

119873=(1198921minus 1198922) 119899 + 119892

2119873

119873

1198882(120601) =

1198921(119898 minus 119899) + 119892

2[(119872 minus 119873) minus (119898 minus 119899)]

119872 minus 119873

=(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)

119872 minus119873

(A1)

and so

1198881(120601) minus 119888

2(120601) = ((119872 minus 119873) [(119892

1minus 1198922) 119899 + 119892

2119873]

minus119873 [(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)])

times (119873 (119872 minus119873))minus1

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

(A2)

Therefore in 120596 we get

119903119897119865119866

= (1198881(120601) minus 119888

2(120601)) ((119892

1minus 1198881(120601))119867

120576(120601)

+ (1198921minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)sdot(1198921minus 1198922)

119873 (119872 minus119873)

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

=(119872119899 minus 119873119898) times (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

(A3)

and in Ω 120596 we have

119865119866

(120601) = (1198881(120601) minus 119888

2(120601)) ((119892

2minus 1198881(120601))119867

120576(120601)

+ (1198922minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

times [ minus (1198921minus 1198922)

times (119899 (119872 minus 119873)119867120576(120601)

+119873 (119898 minus 119899) (1 minus 119867120576(120601)))

times (119873 (119872 minus119873))minus1

]

=minus (119872119899 minus 119873119898) (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [(119899 (119872 minus 119873)119867120576(120601)119873 (119898 minus 119899) (1 minus 119867

120576(120601)))]

(A4)

This completes the proof of (14)The last assertion (15) followsclearly from the following fact

(119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576(120601))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A5)

(119899 (119872 minus 119873)119867120576(120601) + 119873 (119898 minus 119899) (1 minus 119867

120576(120601)))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A6)

The proof of Theorem 1 is completed

Now we give the proof of Corollary 2 (i) Since

1198722

= (119873 + (119872 minus119873))2

= 1198732

+ (119872 minus119873)2

+ 2119873 (119872 minus119873)

ge 4119873 (119872 minus119873)

(A7)

by (14) (A5) and (A7) we have

119865119866

(120601) = (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

times (1198732

(119872 minus 119873)2

)minus1

ge (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

(A8)

10 Mathematical Problems in Engineering

and by (14) and (A6)

119865119866

(120601) = minus (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119899 (119872 minus 119873)119867120576

+119873 (119898 minus 119899) (1 minus 119867120576))]

times (1198732

(119872 minus 119873)2

)minus1

le minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(A9)

This completes the proof of (17) and similarly for (18) Theproof is completed

B Proofs of Theorem 3 and Corollary 4

Proof of Theorem 3 By (6) we have

1198911(x) =

int120601gt0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601gt0

119870120590(xminus120585) 119889120585

= (int

120601gt0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

119906gt119868

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601gt0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

times (int

119906gt119868

119870120590(119909minus120585) 119889120585)

minus1

=1198921119902 + 1198922(119876 minus 119902)

119876

(B1)

1198912(x) =

int120601le0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601le0

119870120590(xminus120585) 119889120585

= (int

120601le0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601le0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

=1198921(119901 minus 119902) + 119892

2[(119875 minus 119876) minus (119901 minus 119902)]

119875 minus 119876

(B2)

and so

1198911(x) minus 119891

2(x) =

(1198921minus 1198922) (119875119902 minus 119876119901)

119876 (119875 minus 119876) (B3)

Therefore in 120596 we get

119865119871

(120601) = (1198981(120601) minus 119898

2(120601))

times ((1198921minus 1198981(120601))119867

120576(120601)

+ (1198921minus 1198982(120601)) (1 minus 119867

120576(120601)))

= (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B4)

and in Ω 120596 we have

119865119871

= (1198981(120601) minus 119898

2(120601))

times ((1198922minus 1198981(120601))119867

120576(120601)

+ (1198922minus 1198982(120601)) (1 minus 119867

120576(120601)))

= minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B5)

Mathematical Problems in Engineering 11

This completes the proof of (19) The last assertion (20) fol-lows clearly from the following fact(119875 minus 119876) (119876 minus 119902)119867

120576(120601)

+ 119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))

= 119867120576(120601) (int

Ω

minusint

120601gt0

)119870120590(xminus120585) 119889120585

sdot (int

120601gt0

minusint

120601gt0cap120596)119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601)) int

120601gt0

119870120590(xminus120585) 119889120585

sdot ((intΩ

+int

120601gt0cap120596) minus (int

120596

+int

120601gt0

))119870120590(xminus120585) 119889120585

= 119867120576(120601) int

120601le0

119870120590(xminus120585) 119889120585

sdot (int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot (int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

ge 119867120576(120601) (int

120601le0

119878119889120585) sdot (int

120601gt0(120601gt0cap120596)

119878119889120585)

+ (1 minus 119867120576(120601)) (int

120601gt0

119878119889120585) sdot (int

120601le0(120601le0cap120596)

119878119889120585)

= 119878 [ (119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 minus119873) minus (119898 minus 119899)) (1 minus 119867120576(120601))]

ge 119878

(B6)119902 (119875 minus 119876)119867

120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))

= 119867120576(120601) int

120601gt0cap120596119870120590(xminus120585) 119889120585 sdot int

120601le0

119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot int

120601le0cap120596119870120590(xminus120585) 119889120585

ge 119867120576(120601) (int

119906gt119868cap120596119878119889120585) sdot (int

119906le119868

119878119889120585)

+ (1 minus 119867120576(120601)) (int

119906gt119868

119878119889120585) sdot (int

119906le119868cap120596119878119889120585)

= 119878 [119899 (119872 minus 119873)119867120576(120601) + (1 minus 119867

120576(120601))119873 (119898 minus 119899)]

ge 119878

(B7)

where 119878 = min119870120590(xminus120585) 120585 isin Ω)

Now we give the proof of Corollary 4 (i) Since

1198752

ge 4119875 (119875 minus 119876) (B8)

we have by (19) and (B6)

119865119871

(120601) = (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

ge (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

(B9)

and by (19) and (B7)

119865119871

(120601) = minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

le minus (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(B10)

This completes the proof of (19) and similarly to (20) Theproof is completed

C Proof of Corollary 5

Proof (i) By Corollary 2 (i) we have

119865119866

(1206010

) ge (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0

in 120596

119865119866

(1206010

) le minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

(C1)

At the first iteration by (25) and (C1) we have

1206011

119894119895= 1206010

119894119895+ Δ119905

sdot 119865119866

(1206010

119894119895)

ge 1206010

119894119895+ Δ119905 sdot 120575

120576(1206010

)119860 in 120596

le 1206010

119894119895minus Δ119905 sdot 120575

120576(1206010

)119860 in Ω 120596

(C2)

12 Mathematical Problems in Engineering

or equivalently

1206011

(x) ge 1206010

(x) + Δ119905 sdot 120575120576(1206010

)119860 gt 1206010

(x) in 120596

1206011

(x) le 1206010

(x) minus Δ119905 sdot 120575120576(1206010

)119860 lt 1206010

(x) in Ω 120596

(C3)

which implies that

1198721198991minus 1198731119898 = |Ω|

10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816minus100381610038161003816100381610038161206011

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816) |120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816|120596|

gt |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816) |120596|

= |Ω|1003816100381610038161003816120596 cap 120601

0gt 0

1003816100381610038161003816 minus10038161003816100381610038161206010 gt 0

1003816100381610038161003816 |120596|

= 1198721198990minus 1198730119898 gt 0

(C4)

In a manner similar to the steps used to obtain (26)-(27) and(C1)ndash (C4) we obtain

119865119866

(119868 1206011

) ge (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

gt (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

119865119866

(119868 1206011

) le minus (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

lt minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

1206012

(x) ge 1206011

(x) + Δ119905 sdot 120575120576(1206011

)119860 gt 1206011

(x) in 120596

1206012

(x) le 1206011

(x) minus Δ119905 sdot 120575120576(1206011

)119860 lt 1206011

(x) in Ω 120596

1198721198992minus 1198732119898 gt 119872119899

1minus 1198731119898 gt 119872119899

0minus 1198730119898 gt 0

(C5)

D Proof of Theorem 7

Proof We prove (32) and similarly to (33) By (25)Corollary 5(i) and Corollary 6(i) we get

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

= 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119865119871

(120601119896

119894119895) + 120596119865

119866

(120601119896

119894119895))

ge 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

) + 120596119860 sdot 120575120576(120601119896+1

))

= 120601119896

119894119895+ ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in 120596

(119896 ge 1 119896 isin 119911+

)

le 120601119896

119894119895+ Δ119905

sdot (minus ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

)

+120596119860 sdot 120575120576(120601119896+1

)))

= 120601119896

119894119895minus ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(D1)

which implies that

120601119896+1

119894119895ge 1206010

119894+ ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in 120596

120601119896+1

119894119895le 1206010

119894minus (119896 + 1)119860 ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in Ω 120596

(D2)

Because 1206010(x) is bounded in domain Ω there clearly exists a

positive integer 119870 such that

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

for 119896 ge 119870 (D3)

The proof is completed

Mathematical Problems in Engineering 13

E Proof of Theorem 8

Proof From (11) we have

119865 (0) = 120575120576(0) [120596 ((119868 minus 119868

119866

) sdot 2meanxisinΩ

119868 (x))

+ (1 minus 120596) ((119868 minus 119868119871

) sdot 2meanxisinΩ

119868 (x))]

=2120575120576(0)mean

xisinΩ119868 (x) (119868 minus 2119867

120576(0)mean

xisinΩ119868 (x))

=120575120576(0)mean

xisinΩ119868 (x) (119868 minusmean

xisinΩ119868 (x))

(E1)

Clearly

minxisinΩ

119868 (x) lt meanxisinΩ

119868 (x) lt maxxisinΩ

119868 (x) (E2)

which implies that

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(E3)

where

Ω1= 119868 (x) lt mean

xisinΩ119868 (x)

Ω2= 119868 (x) gt mean

xisinΩ119868 (x)

(E4)

The proof of Theorem 8 is completed

Acknowledgments

The authors would like to thank the anonymous reviewersfor their valuable comments and suggestions to improve thispaper Besides this work was supported by the Fundamen-tal Research Funds for the Central University Grant no(CDJXS10100006)

References

[1] V Caselles F Catte T Coll and F Dibos ldquoA geometricmodel for active contours in image processingrdquo NumerischeMathematik vol 66 no 1 pp 1ndash31 1993

[2] T Chan M Moelich and B Sandberg ldquoSome recent develop-ments in variational image segmentationrdquo UCLA CAM Reportcam 06-52 2006

[3] L He Z Peng B Everding et al ldquoA comparative study ofdeformable contour methods on medical image segmentationrdquoImage and Vision Computing vol 26 no 2 pp 141ndash163 2008

[4] C Li C Xu C Gui andM D Fox ldquoLevel set evolution withoutre-initialization a new variational formulationrdquo in Proceedingsof the IEEE Computer Society Conference on Computer Visionand Pattern Recognition (CVPR rsquo05) pp 430ndash436 San DiegoCalif USA June 2005

[5] C Li C Xu C Gui and M D Fox ldquoDistance regularized levelset evolution and its application to image segmentationrdquo IEEETransactions on Image Processing vol 19 no 12 pp 3243ndash32542010

[6] B Zhou and C LMu ldquoLevel set evolution for boundary extrac-tion based on a p-Laplace equationrdquo Applied MathematicalModelling vol 34 no 12 pp 3910ndash3916 2010

[7] Y Wang and C He ldquoAdaptive level set evolution starting with aconstant functionrdquoAppliedMathematical Modelling vol 36 no7 pp 3217ndash3228 2012

[8] T F Chan and L A Vese ldquoActive contours without edgesrdquo IEEETransactions on Image Processing vol 10 no 2 pp 266ndash2772001

[9] A Tsai A Yezzi andA SWillsky ldquoCurve evolution implemen-tation of theMumford-Shah functional for image segmentationdenoising interpolation andmagnificationrdquo IEEETransactionson Image Processing vol 10 no 8 pp 1169ndash1186 2001

[10] T Brox and D Cremers ldquoOn the statistical interpretation of thepiecewise smooth Mumford-Shah functionalrdquo in Proceedingsof the International Conference on Scale Space and VariationalMethods in Computer Vision (SSVM rsquo07) pp 203ndash213 IschiaItaly 2007

[11] C Li C Y Kao J CGore andZDing ldquoImplicit active contoursdriven by local binary fitting energyrdquo in Proceedings of the IEEEComputer Society Conference on Computer Vision and PatternRecognition (CVPR rsquo07) pp 1ndash7 Washington DC USA June2007

[12] C Li C Y Kao J C Gore and Z Ding ldquoMinimization ofregion-scalable fitting energy for image segmentationrdquo IEEETransactions on Image Processing vol 17 no 10 pp 1940ndash19492008

[13] S Lankton and A Tannenbaum ldquoLocalizing region-basedactive contoursrdquo IEEE Transactions on Image Processing vol 17no 11 pp 2029ndash2039 2008

[14] L Wang C Li Q Sun D Xia and C Y Kao ldquoActive contoursdriven by local and global intensity fitting energy with applica-tion to brain MR image segmentationrdquo Computerized MedicalImaging and Graphics vol 33 no 7 pp 520ndash531 2009

[15] K Zhang H Song and L Zhang ldquoActive contours driven bylocal image fitting energyrdquo Pattern Recognition vol 43 no 4pp 1199ndash1206 2010

[16] K Zhang L Zhang H Song and W Zhou ldquoActive contourswith selective local or global segmentation a new formulationand level set methodrdquo Image and Vision Computing vol 28 no4 pp 668ndash676 2010

[17] Y Yuan and C He ldquoVariational level set methods for image seg-mentation based on both 1198712 and Sobolev gradientsrdquo NonlinearAnalysis Real World Applications vol 13 no 2 pp 959ndash9662012

[18] B Liu H D Cheng J Huang J Tian X Tang and J LiuldquoProbability density difference-based active contour for ultra-sound image segmentationrdquo Pattern Recognition vol 43 no 6pp 2028ndash2042 2010

[19] D W Shattuck S R Sandor-Leahy K A Schaper D ARottenberg and RM Leahy ldquoMagnetic resonance image tissueclassification using a partial volume modelrdquo NeuroImage vol13 no 5 pp 856ndash876 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 2: Research Article Implicit Active Contour Model with Local ...downloads.hindawi.com/journals/mpe/2013/367086.pdfAcademic Editor: Oluwole Daniel Makinde ... (CV)proposedaregion-basedactivecontourmodel

2 Mathematical Problems in Engineering

intensity fitting (GIF) energy The LIF energy is responsiblefor attracting the contour toward object boundaries andis dominant near object boundaries while the GIF energyincorporates global image information to improve the robust-ness to initialization of the contoursThe proposedmodel canefficiently handle intensity inhomogeneity while allowingfor more flexible initialization and maintaining the subpixelaccuracy

The remainder of this paper is organized as followsSection 2 briefly reviews the PC model [8] and RSF model[12] Section 3 introduces the proposed model Section 4presents experimental results using a set of synthetic and realimages This paper is summarized in Section 5

2 Related Works

21 Piecewise ConstantModel by Chan and Vese In [8] ChanandVese (CV) proposed a region-based active contourmodelthat relies on intensity homogeneous (roughly constant)statistics in each of the regions to be segmented In the CVmodel a contour 119862 is represented implicitly by the zero-levelset of a Lipschitz function 120601 Ω rarr R which is called a levelset function In what follows we let the level set function120601 take positive and negative values inside and outside thecontour 119862 respectively

Let 119868 Ω sub R2rarr R be an input image and let119867

120576be the

regularized Heaviside function the energy functional of theCV model is defined as

119864CV

(1198881 1198882 120601) = 120582

1intΩ

1003816100381610038161003816119868 minus 11988811003816100381610038161003816

2

119867120576(120601) 119889x

+1205821intΩ

1003816100381610038161003816119868 minus 11988821003816100381610038161003816

2

(1 minus 119867120576(120601)) 119889x

+ ]intΩ

1003816100381610038161003816nabla119867120576 (120601)1003816100381610038161003816 119889x

(1)

where 1205821 1205822gt 0 ] gt 0 are constantsThe regularized version

of119867(119911) is chosen as

119867120576(119911) =

1

2(1 +

2

120587arctan(119911

120576)) (2)

1198881and 1198882are the global averages of the image intensities in the

region x 120601(x) gt 0 and x 120601(x) lt 0 respectively that is

1198881(120601) =

intΩ

119868 (x)119867120576(120601 (x)) 119889x

intΩ

119867120576(120601 (x)) 119889x

1198882(120601) =

intΩ

119868 (x) (1 minus 119867120576(120601 (x))) 119889x

intΩ

(1 minus 119867120576(120601 (x))) 119889x

(3)

The solution of the CV model in fact leads to a piecewiseconstant segmentation of the original image 119868(x)

119868119866

(x) = 1198881119867120576(120601 (x)) + 119888

2(1 minus 119867

120576(120601 (x))) (4)

where 1198881and 1198882are the averages of the image intensities in the

region x 120601(x) gt 0 and x 120601(x) lt 0 respectively Suchconstants may be far away from the original image data if theintensities outside or inside the contour119862 = x 120601(x) = 0 arenot homogeneous As a result the CVmodel generally fails tosegment images with intensity inhomogeneity

22 Region-Scalable Fitting Model In order to improve theperformance of the global CVmodel [8] and the PSmodel [9]on images with inhomogeneity Li et al [11 12] recently pro-posed a novel region-based active contour model They in-troduced a kernel function and defined the following energyfunctional

119864RSF

(1198911 1198912 120601) = 120582

1int[int119870

120590(x minus y) 1003816100381610038161003816119868 (y) minus 1198911 (x)

1003816100381610038161003816

2

times119867120576(120601 (y)) 119889y] 119889x

+ 1205822int[int119870

120590(x minus y) 1003816100381610038161003816119868 (y) minus 1198911 (x)

1003816100381610038161003816

2

times (1 minus 119867120576(120601 (y))) 119889y] 119889x

+ ]intΩ

1003816100381610038161003816nabla119867120576 (120601 (x))1003816100381610038161003816 119889x

+ 120583intΩ

1

2(1003816100381610038161003816nabla120601 (x)

1003816100381610038161003816 minus 1)2

119889x(5)

where119870120590is a Gaussian kernel with standard deviation 120590 and

1198911(x) and 119891

2(x) are two smooth functions that approximate

the local image intensities inside and outside the contourrespectively They are computed by

1198911(x) =

intΩ

119870120590(x minus y) 119868 (y)119867

120576(120601 (y)) 119889y

intΩ

119870120590(x minus y)119867

120576(120601 (y)) 119889y

1198912(x) =

intΩ

119870120590(x minus y) 119868 (y) (1 minus 119867

120576(120601 (y))) 119889y

intΩ

119870120590(x minus y) (1 minus 119867

120576(120601 (y))) 119889y

(6)

The solution of the RSF model leads to a piecewisesmooth approximation of the original image 119868(x)

119868119871

(x) = 1198911(x)119867120576(120601 (x)) + 119891

2(x) (1 minus 119867

120576(120601 (x))) (7)

where the smooth functions 1198911(x) and 119891

2(x) are computed

by (6) 1198911(x) and 119891

2(x) are the averages of local intensities

on the two sides of the contour which are different from theconstants 119888

1and 1198882in the CVmodel the averages of the image

intensities on the two sides of the contour Therefore theRSF model can deal with intensity inhomogeneity efficientlyHowever it is sensitive to contour initialization (initial loca-tions sizes and shapes)

3 The Proposed Model

31 Description of Our Model Given a positive constant 120596(0 le 120596 le 1) from (4) and (7) we define the following energyfunctional

119864 (120601 1198881 1198882 1198911 1198912) = 120596119864

119866

(120601) + (1 minus 120596) 119864119871

(120601)

= 120596(1

2intΩ

10038161003816100381610038161003816119868(x) minus 119868119866(x)10038161003816100381610038161003816

2

119889x)

+ (1 minus 120596) (1

2intΩ

10038161003816100381610038161003816119868 (x) minus 119868119871 (119909)10038161003816100381610038161003816

2

119889x) (8)

Mathematical Problems in Engineering 3

where

119868119866

(x) = 1198881119867120576(120601 (x)) + 119888

2(1 minus 119867

120576(120601 (x)))

119868119871

(x) = 1198911(x)119867120576(120601 (x)) + 119891

2(x) (1 minus 119867

120576(120601 (x)))

(9)

Keeping 120601 fixed andminimizing the functional119864(120601 1198881 1198882

1198911 1198912) with respect to 119888

1 1198882 1198911 and 119891

2 we have

1198881=

intΩ

119868 (x)119867120576(120601 (x)) 119889x

intΩ

119867120576(120601 (x)) 119889x

1198882=

intΩ

119868 (x) (1 minus 119867120576(120601 (x))) 119889x

intΩ

(1 minus 119867120576(120601 (x))) 119889x

1198911(x) =

intΩ

119870120590(x minus 119910) 119868 (119910)119867

120576(120601 (119910)) 119889119910

intΩ

119870120590(x minus 119910)119867

120576(120601 (119910)) 119889119910

1198912(x) =

intΩ

119870120590(x minus 119910) 119868 (119910) (1 minus 119867

120576(120601 (119910))) 119889119910

intΩ

119870120590(x minus 119910) (1 minus 119867

120576(120601 (119910))) 119889119910

(10)

Keeping 1198881 1198882 1198911 and 119891

2fixed and minimizing the func-

tional 119864(120601 1198881 1198882 1198911 1198912) with respect to 120601 we obtain the cor-

responding gradient descent flow equation

120597120601

120597119905= 120575120576(120601) [120596 ((119868 minus 119868

119866

) (1198881minus 1198882))

+ (1 minus 120596) ((119868 minus 119868119871

) (1198911minus 1198912))]

= 120575120576(120601) [120596119865

119866

(120601) + (1 minus 120596) 119865119871

(120601)] = 119865 (120601)

(11)

where

120575120576(119911) = 119867

1015840

120576(119911) =

1

120587

120576

1205762 + 1199112 (12)

Like the CV and RSF models our model is also imple-mented using an alternative procedure for each iteration andthe corresponding level set function 120601119899 we first compute thefitting values 119888

119894(120601119899

) and 119891119894(120601119899

) and then obtain 120601119899+1 by min-imizing 119864(120601 119888

1(120601119899

) 1198882(120601119899

) 1198911(120601119899

) 1198912(120601119899

)) with respect to 120601This process is repeated until the zero-level set of 120601119899+1 isexactly on the object boundary

In the following we first discuss the properties of 119865119866(120601)and 119865119871(120601) and then analyze the behavior of (11)

32 Properties of 119865119866(120601) and 119865119871(120601) For the sake of simplicitywe state and prove the properties of 119865119866(120601) and 119865119871(120601) only foran image consisting of only two distinct gray levels

119868 (x) = 1198921 x isin 120596

1198922 x isin Ω 120596

(13)

where 1198921 1198922ge 0 with 119892

1= 1198922 120596 and Ω 120596 represent the ob-

jects of interest and the background respectively

Theorem 1 Let 119868(x) be an image by (13) Then one has

119865119866

(120601) =

(119872119899 minus 119873119898) (1198921minus 1198922)2

times [(119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

times(1198732

(119872 minus 119873)2

)minus1

in 120596

minus (119872119899 minus 119873119898) (1198921minus 1198922)2

times [119899 (119872 minus 119873)119867120576

+119873 (119898 minus 119899) (1 minus 119867120576)]

times(1198732

(119872 minus 119873)2

)minus1

in Ω 120596

(14)

and so

sign (119865119866 (120601)) =

+ sign (119872119899 minus 119873119898) in 120596

minus sign (119872119899 minus 119873119898) in Ω 120596

(15)

where

119872 = |Ω| 119898 = |120596| 119873 =1003816100381610038161003816120601 gt 0

1003816100381610038161003816

119899 =1003816100381610038161003816120596 cap 120601 gt 0

1003816100381610038161003816

(16)

in which |Ω| is the area of the regionΩ and similarly for others

Remarks (i) Due to the discrete nature of image |Ω| is in factthe number of pixels in the image 119868(x) and similarly to others

(ii) The cases of 119899 = 119898 119899 = 119873 and 119899 = 0 correspondto the zero-level lines of 120601(x) which are encircling the object(120596) inside the object and within the background (Ω

120596) respectively The cases of 119873 = 119899 =119898 and 119873 = 119899 =

119898 correspond to the zero-level line of 120601(x) that are partiallyinside the object and exactly on the object edge respectively

(iii) The significance of (15) is that the function 119865119866

(120601)

has the opposite sign in 120596 (object) and Ω 120596 (background)respectively

The proof of Theorem 1 is provided in Appendix A Thefollowing result will be used in the proof ofTheorem 7 whichguarantees that the evolution by (10) converges to a uniquestable value after finite time

Corollary 2 Let 119868(x) be an image by (13) Then one has thefollowing

(i) If119872119899 minus119873119898 ge 0 then

119865119866

(120601) ge (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

119865119866

(120601) le minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(17)

4 Mathematical Problems in Engineering

(ii) If119872119899 minus119873119898 le 0 then

119865119866

(120601) le (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

119865119866

(120601) ge minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(18)

The proof of Corollary 2 is given in Appendix AWe call the property in Theorem 1 an adaptive sign-

changing property of 119865119866(120601) Such property also holds for119865119871

(120601) which can be expressed by the following theorem

Theorem 3 Let 119868(x) be an image given by (13) Then one has

119865119871

(120601) =

(119875119902 minus 119876119901) (1198921minus 1198922)2

times [(119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times(1198762

(119875 minus 119876)2

)minus1

in 120596

minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times(1198762

(119875 minus 119876)2

)minus1

in Ω 120596

(19)

and so

sign (119865119871 (120601)) =

+ sign (119875119902 minus 119876119901) in 120596

minus sign (119875119902 minus 119876119901) in Ω 120596

(20)

where

119875 = intΩ

119870120590(x minus 120585) 119889120585 119901 = int

120596

119870120590(x minus 120585) 119889120585

119876 = int120601gt0

119870120590(x minus 120585) 119889120585

119902 = int120596cap120601gt0

119870120590(x minus 120585) 119889120585

(21)

This theorem shows that the local force 119865119871(120601) has exactlyopposite sign in 120596 (object) and in Ω 120596 (background)

The following result together with Corollary 2 will beused in the proof of Theorem 7 mentioned later

Corollary 4 Under the assumption ofTheorem 3 one has thefollowing

(i) If 119875119902 minus 119876119901 ge 0 then

119865119871

(120601) ge 119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

119865119871

(120601) le minus119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(22)

(ii) If 119875119902 minus 119876119901 le 0 then

119865119871

(120601) le 119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

119865119871

(120601) ge minus119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(23)

where

119878 = min 119870120590(x minus 120585) 120585 isin Ω) (24)

The proofs of Theorem 3 and Corollary 4 are similar toTheorem 1 and Corollary 2 respectively see Appendix B fordetails

33 Behavior of Our Model In this section we analyze thebehavior of our model (10) for image segmentation We willshow that the zero-level line of the evolution function startingwith a bounded function finally comes to a unique steadystate which separates the object from the background

Due to the discrete nature of image the continuous equa-tion (10) can be discretized in space with space step 1 (impliedpixel spacing) but also in time with a time step Δ119905 as follows

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

0 le 119894 le 119901 minus 1 0 le 119895 le 119902 minus 1

(25)

where 119901 times 119902 corresponds to the image size 119868119894119895= 119868(119894 119895) and

120601119896

119894119895= 120601(119896Δ119905 119894 119895) with 119896 ge 0 and 1206010

119894119895= 1206010(119894 119895)

Corollary 5 Let 1198721198990minus 1198730119898 = |Ω||120596 cap 120601

0

gt 0| minus |1206010

gt

0||120596| = 0

(i) If1198721198990minus 1198730119898 gt 0 then

119865119866

(120601119896

) gt (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119866

(120601119896

) lt minus (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= minus119860 lt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(26)

Mathematical Problems in Engineering 5

(ii) If1198721198990minus 1198730119898 lt 0 then

119865119866

(120601119896

) lt (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= 119860 lt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119866

(120601119896

) gt minus (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= minus119860 gt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(27)

where

119860 = (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987248 (28)

We provide the proof of Corollary 5 in Appendix C Sim-ilar analysis can prove the following corollary Corollary 6

Corollary 6 Let 1198751199020minus 1198760119901 = 0

(i) If 1198751199020minus 1198760119901 gt 0 then

119865119871

(120601119896

) gt (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= 119861 gt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119871

(120601119896

) lt minus (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= minus119861 lt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(29)

(ii) If 1198751199020minus 1198760119901 lt 0 then

119865119871

(120601119896

) lt (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= 119861 lt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119871

(120601119896

) gt minus (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= minus119861 gt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(30)

where

119861 = (1198751199020minus 1198760119901)

(1198921minus 1198922)2

11987548 (31)

Theorem 7 If 1206010(x) is a bounded function with (1198721198990minus

1198730119898)(119875119902

0minus 1198760119901) gt 0 then there exists a positive integer 119870

such that for 119896 ge 119870

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

if 1198721198990minus 1198730119898 gt 0 119875119902

0minus 1198760119901 gt 0

(32)

120601119896+1

119894119895lt 0 in 120596

120601119896+1

119894119895gt 0 in Ω 120596

if 1198721198990minus 1198730119898 lt 0 119875119902

0minus 1198760119901 lt 0

(33)

We provide the proof of Theorem 7 in Appendix DThe significance of Theorem 7 is that if we choose 1206010(x)

such that (1198721198990minus 1198730119898)(119875119902

0minus 1198760119901) gt 0 then the zero-level

line of120601(x) startingwith such initial function1206010(x) can finallycome to a unique steady state which separates object from thebackground

Remark Mathematically there does exist 1206010(x) such that(1198721198990minus1198730119898)(119875119902

0minus1198760119901) le 0 in practice however we always

can guarantee that

(1198721198990minus 1198730119898) (119875119902

0minus 1198760119901) gt 0 (34)

34 Discussion of Initial Function Theorem 7 guarantees thatthe proposed model computes a unique steady state regard-less of the initialization that is the convergence of the zero-level line 120601 = 0 is irrespective of the initial functionThis means that we can obtain the same zero-level line inthe steady state if we choose the initial function as a boundedfunction 1206010 with (119872119899

0minus 1198730119898)(119875119902

0minus 1198760119901) gt 0

In applications the initial function 1206010 can be defined

via a simple curve (closed curve or line segment) in imagedomain For example we can choose the initial function asa signed distance to a circle which is widely used in mostof image segmentation models with level set methods Forthe proposed model however we prefer to define the initialfunction 120601

0

(x) as a piecewise constant or constant functionas follows

(1) If the curve 119862 is a closed curve (eg circle or square)then 1206010(x) is defined by

1206010

(x) = +120588 (x) isin inside (119862) minus120588 (x) isin outside (119862)

(35)

where 120588 = 0 is a constant(2) If the curve 119862 is a line segment that partitions the

image domainΩ into two disjoint regionsΩ1andΩ

2

(eg the left and right half regions of image domainΩ respectively) withΩ = Ω

1cup Ω2andΩ

1cap Ω2= Φ

then 1206010(119909 119910) is defined by

1206010

(x) = +120588 (x) isin Ω

1

minus120588 (x) isin Ω2

(36)

where 120588 = 0 is a constant

6 Mathematical Problems in Engineering

(3) We can also define a zero function as follows

1206010

(x) = 0 (37)

Next we prove the fact that with 1206010(x) = 0 120601(x) becomesa sign-changing function and satisfy the condition of (119872119899

1minus

1198731119898)(119875119902

1minus 1198761119901) gt 0 after the first iteration

Theorem 8 If the initial function 1206010(x) = 0 in Ω and

1198880

1(x) = 119891

0

1(x) = 2 sdot mean

xisinΩ119868 (x)

1198880

2(x) = 119891

0

2(x) = 0

(38)

then after the first iteration one has

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(39)

whereΩ1cap Ω2= Φ Ω

1= Φ Ω2

= Φ

We provide the proof of Theorem 8 in Appendix EBy (25) andTheorem 8 we have

1206011

(x) = 120575120576(0) mean

xisinΩ119868 (x)

times (119868 minus meanxisinΩ

119868 (x))

lt 0 x isin Ω1

gt 0 x isin Ω2

(40)

Therefore 1206011(x) became a sign-changing function Thenusing two distinct gray levels of (13) and the above demon-stration we have

(1198721198991minus 1198731119898) (119875119902

1minus 1198761119901)

= (119898 (119872 minus1198731)) (119901 (119875 minus 119876

1))

= 119898119901 (119872 minus1198731) (119875 minus 119876

1) gt 0

(41)

4 Implementation and Experimental Results

41 Implementation In tradition PDE-based methods a cer-tain diffusion term is usually included in the evolution equa-tion to regularize the evolving function but which increasesthe computational cost Recently [16] proposed a novelscheme to regularize the evolving function that is Gaussianfiltering the evolving function after each of iterations Weadopt this scheme to regularize the evolving function 119906 ateach of iteration that is 120601119899 = 119866

120588lowast 120601119899 where 119901 controls the

regularization strength Such regularization procedure hassome advantages over the traditional regularization termsuch as computational efficiency and better smoothing effectsee [16] for more explanations

The main procedures of the proposed algorithm can besummarized as follows

(1) Initialize the evolving function 1206010(x)(2) Compute 119888

119894(120601) and 119891

119894(120601) (119894 = 1 2) using (10)

(3) Evolve the function 120601 according to (11)(4) Regularize the function 120601 with a Gaussian filter that

is 120601119899 = 119866120588lowast 120601119899

(5) Check whether the evolution is finished If not returnto step 2

42 Experimental Results In this section we show experi-mentally that the proposedmodel not only can provide desir-able segmentation results in the presence of intensity inho-mogeneity but also allows for more flexible initialization ofthe contour compared to the RSF and LIF models

In our numerical experiments for our model we choosethe parameters as follows 120576 = 15 for the regularized Diracfunction Δ119905=0025 (time step) ℎ = 1 (space step) 120596 = 01and 119899 = 5 The MATLAB source code of the LIF modelalgorithm is downloaded at httpwww4comppolyueduhksimcslzhangcodeLIFzip and the RSF model algorithmis downloaded at httpwwwengruconnedusimcmlicodeRSF v0 v01rar All experiments were run under MATLAB2010b on a PC with Dual 27 GHz processor

We will utilize two region overlap metrics to evaluatethe performances of the three models quantitatively The twometrics are the ratio of segmentation error (RSE) [18] and thedice similarity coefficient (DSC) [14 18 19] If 119878

1and 1198782rep-

resent a given baseline foreground region (eg true object)and the foreground region found by the model respectivelythen the two metrics are defined as follows

RSE =119873 (1198781 1198782) + 119873 (119878

2 1198781)

119873 (Ω)

DSC =2119873 (119878

1cap 1198782)

119873 (1198781) + 119873 (119878

2)

(42)

where 119873(sdot) indicates the number of pixels in the enclosedregion andΩ is image domainThe closer the DSC value to 1the better the segmentation Since119873(119878

11198782)+119873(119878

21198781) is the

number of the pixelsmistakenly classified by themodel lowerRSE means that there are fewer pixels misclassified that isthe image can be segmented more accurately Thus a perfectsegmentation will give DSC = 1 RSE = 0

In the first example (Figures 1ndash3) we mainly verify thecomputation of a unique steady state of the zero-level line of120601 starting with three types of representative initial functionsthat is signed distance function piecewise constant functionsby (26)-(27) and zero function respectively The top rowof Figure 1 shows the evolution of an active contour (iezero-level line 120601 = 0) with the function 120601 initialized toa signed distance function piecewise constant functions by(26)-(27) with 120588 = 1 and zero function respectively whilethe bottom row shows the corresponding evolution of120601Withsuch initializations the zero-level line of 120601 converges to thesame steady state

In Figure 2 we show that our model has the capabilityof detecting multiple objects or objects with interior holesor blurred edges only starting with a zero function Thecontours (zero-level lines) evolution processes are shown inthe second column to the forth column

Mathematical Problems in Engineering 7

Figure 1 Segmentations of our model for two real images with 120601 initialized as different functions Columns 1 to 4 signed distance functionpiecewise constant functions by (26)-(27) with 120588 = 1 and zero function

Figure 2 Applications of our model to four images (slug cell ventriculus sinister MR and real plane images) The curve evolution processfrom the initial contour (in the first column) to the final contour (in the fourth column) is shown in every row for the corresponding image

8 Mathematical Problems in Engineering

(a) (b) (c) (d)

Figure 3 Comparisons of three models Rows 1ndash3 RSF model LIF model and our model with 1206010= 0 For the RSF and LIF models initial

and final contours are shown in green and red color respectively Columns 1ndash4 the 4 pictures indicate (a) (b) (c) (d) respectively

Figure 4 Segmentation results of both models for a hand phantomUpper row LIF model Lower row our model with 120601

0= 0 Column

2 zoomed view of the narrow parts in hand fingers

We choose the segmentation results of the RSF and LIFmodels as baseline foreground regions and then computeDSC values for the corresponding images The RSE and DSCvalues for the four real images are given inTable 1 fromwhich

Table 1 RSE and DSC for RSF and LIF models and our model

RSF model LIF modelRSE () DSC () RSE () DSC ()

Figure 3(a) 027 9717 299 9891Figure 3(b) 271 9637 012 9913Figure 3(c) 049 8795 045 8806Figure 3(d) 016 9766 076 9891

we can see that the proposed algorithm works as well as theRSF and LIFmodels for imageswith intensity inhomogeneity

The experimental results shown in Figure 4 validate thatour method can also achieve subpixel segmentation accuracyas in [15] As can be seen from Figures 4(b) and 4(d) bothmodels achieve subpixel segmentation accuracy of the fingerboundaries The final contour accurately reflects the truehand shape

5 Conclusion

In this study we propose a new active contour model inte-grating a local intensity fitting (LIF) energy with an auxiliaryglobal intensity fitting (GIF) energy The LIF energy isresponsible for attracting the contour toward object bound-aries and is dominant near object boundaries while the GIFenergy incorporates global image information to improve therobustness to initialization of the contours The proposedmodel can efficiently handle intensity inhomogeneity while

Mathematical Problems in Engineering 9

allowing for more flexible initialization and maintaining thesubpixel accuracy The utility model has the advantages ofallowing for more flexible initialization of the contour andthe capability of detecting multiple objects or objects withinterior holes or blurred edges But [14] has not been imple-mented

Appendices

A Proofs of Theorem 1 and Corollary 2

Proof of Theorem 1 Clearly 120601 gt 0 =Φ and 120601 lt 0 =Φowing to the sign-changing property of 120601(119909 119910) Thus by (3)we have

1198881(120601) =

1198921119899 + 1198922(119873 minus 119899)

119873=(1198921minus 1198922) 119899 + 119892

2119873

119873

1198882(120601) =

1198921(119898 minus 119899) + 119892

2[(119872 minus 119873) minus (119898 minus 119899)]

119872 minus 119873

=(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)

119872 minus119873

(A1)

and so

1198881(120601) minus 119888

2(120601) = ((119872 minus 119873) [(119892

1minus 1198922) 119899 + 119892

2119873]

minus119873 [(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)])

times (119873 (119872 minus119873))minus1

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

(A2)

Therefore in 120596 we get

119903119897119865119866

= (1198881(120601) minus 119888

2(120601)) ((119892

1minus 1198881(120601))119867

120576(120601)

+ (1198921minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)sdot(1198921minus 1198922)

119873 (119872 minus119873)

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

=(119872119899 minus 119873119898) times (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

(A3)

and in Ω 120596 we have

119865119866

(120601) = (1198881(120601) minus 119888

2(120601)) ((119892

2minus 1198881(120601))119867

120576(120601)

+ (1198922minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

times [ minus (1198921minus 1198922)

times (119899 (119872 minus 119873)119867120576(120601)

+119873 (119898 minus 119899) (1 minus 119867120576(120601)))

times (119873 (119872 minus119873))minus1

]

=minus (119872119899 minus 119873119898) (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [(119899 (119872 minus 119873)119867120576(120601)119873 (119898 minus 119899) (1 minus 119867

120576(120601)))]

(A4)

This completes the proof of (14)The last assertion (15) followsclearly from the following fact

(119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576(120601))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A5)

(119899 (119872 minus 119873)119867120576(120601) + 119873 (119898 minus 119899) (1 minus 119867

120576(120601)))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A6)

The proof of Theorem 1 is completed

Now we give the proof of Corollary 2 (i) Since

1198722

= (119873 + (119872 minus119873))2

= 1198732

+ (119872 minus119873)2

+ 2119873 (119872 minus119873)

ge 4119873 (119872 minus119873)

(A7)

by (14) (A5) and (A7) we have

119865119866

(120601) = (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

times (1198732

(119872 minus 119873)2

)minus1

ge (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

(A8)

10 Mathematical Problems in Engineering

and by (14) and (A6)

119865119866

(120601) = minus (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119899 (119872 minus 119873)119867120576

+119873 (119898 minus 119899) (1 minus 119867120576))]

times (1198732

(119872 minus 119873)2

)minus1

le minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(A9)

This completes the proof of (17) and similarly for (18) Theproof is completed

B Proofs of Theorem 3 and Corollary 4

Proof of Theorem 3 By (6) we have

1198911(x) =

int120601gt0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601gt0

119870120590(xminus120585) 119889120585

= (int

120601gt0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

119906gt119868

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601gt0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

times (int

119906gt119868

119870120590(119909minus120585) 119889120585)

minus1

=1198921119902 + 1198922(119876 minus 119902)

119876

(B1)

1198912(x) =

int120601le0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601le0

119870120590(xminus120585) 119889120585

= (int

120601le0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601le0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

=1198921(119901 minus 119902) + 119892

2[(119875 minus 119876) minus (119901 minus 119902)]

119875 minus 119876

(B2)

and so

1198911(x) minus 119891

2(x) =

(1198921minus 1198922) (119875119902 minus 119876119901)

119876 (119875 minus 119876) (B3)

Therefore in 120596 we get

119865119871

(120601) = (1198981(120601) minus 119898

2(120601))

times ((1198921minus 1198981(120601))119867

120576(120601)

+ (1198921minus 1198982(120601)) (1 minus 119867

120576(120601)))

= (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B4)

and in Ω 120596 we have

119865119871

= (1198981(120601) minus 119898

2(120601))

times ((1198922minus 1198981(120601))119867

120576(120601)

+ (1198922minus 1198982(120601)) (1 minus 119867

120576(120601)))

= minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B5)

Mathematical Problems in Engineering 11

This completes the proof of (19) The last assertion (20) fol-lows clearly from the following fact(119875 minus 119876) (119876 minus 119902)119867

120576(120601)

+ 119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))

= 119867120576(120601) (int

Ω

minusint

120601gt0

)119870120590(xminus120585) 119889120585

sdot (int

120601gt0

minusint

120601gt0cap120596)119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601)) int

120601gt0

119870120590(xminus120585) 119889120585

sdot ((intΩ

+int

120601gt0cap120596) minus (int

120596

+int

120601gt0

))119870120590(xminus120585) 119889120585

= 119867120576(120601) int

120601le0

119870120590(xminus120585) 119889120585

sdot (int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot (int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

ge 119867120576(120601) (int

120601le0

119878119889120585) sdot (int

120601gt0(120601gt0cap120596)

119878119889120585)

+ (1 minus 119867120576(120601)) (int

120601gt0

119878119889120585) sdot (int

120601le0(120601le0cap120596)

119878119889120585)

= 119878 [ (119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 minus119873) minus (119898 minus 119899)) (1 minus 119867120576(120601))]

ge 119878

(B6)119902 (119875 minus 119876)119867

120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))

= 119867120576(120601) int

120601gt0cap120596119870120590(xminus120585) 119889120585 sdot int

120601le0

119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot int

120601le0cap120596119870120590(xminus120585) 119889120585

ge 119867120576(120601) (int

119906gt119868cap120596119878119889120585) sdot (int

119906le119868

119878119889120585)

+ (1 minus 119867120576(120601)) (int

119906gt119868

119878119889120585) sdot (int

119906le119868cap120596119878119889120585)

= 119878 [119899 (119872 minus 119873)119867120576(120601) + (1 minus 119867

120576(120601))119873 (119898 minus 119899)]

ge 119878

(B7)

where 119878 = min119870120590(xminus120585) 120585 isin Ω)

Now we give the proof of Corollary 4 (i) Since

1198752

ge 4119875 (119875 minus 119876) (B8)

we have by (19) and (B6)

119865119871

(120601) = (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

ge (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

(B9)

and by (19) and (B7)

119865119871

(120601) = minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

le minus (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(B10)

This completes the proof of (19) and similarly to (20) Theproof is completed

C Proof of Corollary 5

Proof (i) By Corollary 2 (i) we have

119865119866

(1206010

) ge (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0

in 120596

119865119866

(1206010

) le minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

(C1)

At the first iteration by (25) and (C1) we have

1206011

119894119895= 1206010

119894119895+ Δ119905

sdot 119865119866

(1206010

119894119895)

ge 1206010

119894119895+ Δ119905 sdot 120575

120576(1206010

)119860 in 120596

le 1206010

119894119895minus Δ119905 sdot 120575

120576(1206010

)119860 in Ω 120596

(C2)

12 Mathematical Problems in Engineering

or equivalently

1206011

(x) ge 1206010

(x) + Δ119905 sdot 120575120576(1206010

)119860 gt 1206010

(x) in 120596

1206011

(x) le 1206010

(x) minus Δ119905 sdot 120575120576(1206010

)119860 lt 1206010

(x) in Ω 120596

(C3)

which implies that

1198721198991minus 1198731119898 = |Ω|

10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816minus100381610038161003816100381610038161206011

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816) |120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816|120596|

gt |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816) |120596|

= |Ω|1003816100381610038161003816120596 cap 120601

0gt 0

1003816100381610038161003816 minus10038161003816100381610038161206010 gt 0

1003816100381610038161003816 |120596|

= 1198721198990minus 1198730119898 gt 0

(C4)

In a manner similar to the steps used to obtain (26)-(27) and(C1)ndash (C4) we obtain

119865119866

(119868 1206011

) ge (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

gt (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

119865119866

(119868 1206011

) le minus (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

lt minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

1206012

(x) ge 1206011

(x) + Δ119905 sdot 120575120576(1206011

)119860 gt 1206011

(x) in 120596

1206012

(x) le 1206011

(x) minus Δ119905 sdot 120575120576(1206011

)119860 lt 1206011

(x) in Ω 120596

1198721198992minus 1198732119898 gt 119872119899

1minus 1198731119898 gt 119872119899

0minus 1198730119898 gt 0

(C5)

D Proof of Theorem 7

Proof We prove (32) and similarly to (33) By (25)Corollary 5(i) and Corollary 6(i) we get

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

= 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119865119871

(120601119896

119894119895) + 120596119865

119866

(120601119896

119894119895))

ge 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

) + 120596119860 sdot 120575120576(120601119896+1

))

= 120601119896

119894119895+ ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in 120596

(119896 ge 1 119896 isin 119911+

)

le 120601119896

119894119895+ Δ119905

sdot (minus ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

)

+120596119860 sdot 120575120576(120601119896+1

)))

= 120601119896

119894119895minus ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(D1)

which implies that

120601119896+1

119894119895ge 1206010

119894+ ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in 120596

120601119896+1

119894119895le 1206010

119894minus (119896 + 1)119860 ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in Ω 120596

(D2)

Because 1206010(x) is bounded in domain Ω there clearly exists a

positive integer 119870 such that

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

for 119896 ge 119870 (D3)

The proof is completed

Mathematical Problems in Engineering 13

E Proof of Theorem 8

Proof From (11) we have

119865 (0) = 120575120576(0) [120596 ((119868 minus 119868

119866

) sdot 2meanxisinΩ

119868 (x))

+ (1 minus 120596) ((119868 minus 119868119871

) sdot 2meanxisinΩ

119868 (x))]

=2120575120576(0)mean

xisinΩ119868 (x) (119868 minus 2119867

120576(0)mean

xisinΩ119868 (x))

=120575120576(0)mean

xisinΩ119868 (x) (119868 minusmean

xisinΩ119868 (x))

(E1)

Clearly

minxisinΩ

119868 (x) lt meanxisinΩ

119868 (x) lt maxxisinΩ

119868 (x) (E2)

which implies that

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(E3)

where

Ω1= 119868 (x) lt mean

xisinΩ119868 (x)

Ω2= 119868 (x) gt mean

xisinΩ119868 (x)

(E4)

The proof of Theorem 8 is completed

Acknowledgments

The authors would like to thank the anonymous reviewersfor their valuable comments and suggestions to improve thispaper Besides this work was supported by the Fundamen-tal Research Funds for the Central University Grant no(CDJXS10100006)

References

[1] V Caselles F Catte T Coll and F Dibos ldquoA geometricmodel for active contours in image processingrdquo NumerischeMathematik vol 66 no 1 pp 1ndash31 1993

[2] T Chan M Moelich and B Sandberg ldquoSome recent develop-ments in variational image segmentationrdquo UCLA CAM Reportcam 06-52 2006

[3] L He Z Peng B Everding et al ldquoA comparative study ofdeformable contour methods on medical image segmentationrdquoImage and Vision Computing vol 26 no 2 pp 141ndash163 2008

[4] C Li C Xu C Gui andM D Fox ldquoLevel set evolution withoutre-initialization a new variational formulationrdquo in Proceedingsof the IEEE Computer Society Conference on Computer Visionand Pattern Recognition (CVPR rsquo05) pp 430ndash436 San DiegoCalif USA June 2005

[5] C Li C Xu C Gui and M D Fox ldquoDistance regularized levelset evolution and its application to image segmentationrdquo IEEETransactions on Image Processing vol 19 no 12 pp 3243ndash32542010

[6] B Zhou and C LMu ldquoLevel set evolution for boundary extrac-tion based on a p-Laplace equationrdquo Applied MathematicalModelling vol 34 no 12 pp 3910ndash3916 2010

[7] Y Wang and C He ldquoAdaptive level set evolution starting with aconstant functionrdquoAppliedMathematical Modelling vol 36 no7 pp 3217ndash3228 2012

[8] T F Chan and L A Vese ldquoActive contours without edgesrdquo IEEETransactions on Image Processing vol 10 no 2 pp 266ndash2772001

[9] A Tsai A Yezzi andA SWillsky ldquoCurve evolution implemen-tation of theMumford-Shah functional for image segmentationdenoising interpolation andmagnificationrdquo IEEETransactionson Image Processing vol 10 no 8 pp 1169ndash1186 2001

[10] T Brox and D Cremers ldquoOn the statistical interpretation of thepiecewise smooth Mumford-Shah functionalrdquo in Proceedingsof the International Conference on Scale Space and VariationalMethods in Computer Vision (SSVM rsquo07) pp 203ndash213 IschiaItaly 2007

[11] C Li C Y Kao J CGore andZDing ldquoImplicit active contoursdriven by local binary fitting energyrdquo in Proceedings of the IEEEComputer Society Conference on Computer Vision and PatternRecognition (CVPR rsquo07) pp 1ndash7 Washington DC USA June2007

[12] C Li C Y Kao J C Gore and Z Ding ldquoMinimization ofregion-scalable fitting energy for image segmentationrdquo IEEETransactions on Image Processing vol 17 no 10 pp 1940ndash19492008

[13] S Lankton and A Tannenbaum ldquoLocalizing region-basedactive contoursrdquo IEEE Transactions on Image Processing vol 17no 11 pp 2029ndash2039 2008

[14] L Wang C Li Q Sun D Xia and C Y Kao ldquoActive contoursdriven by local and global intensity fitting energy with applica-tion to brain MR image segmentationrdquo Computerized MedicalImaging and Graphics vol 33 no 7 pp 520ndash531 2009

[15] K Zhang H Song and L Zhang ldquoActive contours driven bylocal image fitting energyrdquo Pattern Recognition vol 43 no 4pp 1199ndash1206 2010

[16] K Zhang L Zhang H Song and W Zhou ldquoActive contourswith selective local or global segmentation a new formulationand level set methodrdquo Image and Vision Computing vol 28 no4 pp 668ndash676 2010

[17] Y Yuan and C He ldquoVariational level set methods for image seg-mentation based on both 1198712 and Sobolev gradientsrdquo NonlinearAnalysis Real World Applications vol 13 no 2 pp 959ndash9662012

[18] B Liu H D Cheng J Huang J Tian X Tang and J LiuldquoProbability density difference-based active contour for ultra-sound image segmentationrdquo Pattern Recognition vol 43 no 6pp 2028ndash2042 2010

[19] D W Shattuck S R Sandor-Leahy K A Schaper D ARottenberg and RM Leahy ldquoMagnetic resonance image tissueclassification using a partial volume modelrdquo NeuroImage vol13 no 5 pp 856ndash876 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 3: Research Article Implicit Active Contour Model with Local ...downloads.hindawi.com/journals/mpe/2013/367086.pdfAcademic Editor: Oluwole Daniel Makinde ... (CV)proposedaregion-basedactivecontourmodel

Mathematical Problems in Engineering 3

where

119868119866

(x) = 1198881119867120576(120601 (x)) + 119888

2(1 minus 119867

120576(120601 (x)))

119868119871

(x) = 1198911(x)119867120576(120601 (x)) + 119891

2(x) (1 minus 119867

120576(120601 (x)))

(9)

Keeping 120601 fixed andminimizing the functional119864(120601 1198881 1198882

1198911 1198912) with respect to 119888

1 1198882 1198911 and 119891

2 we have

1198881=

intΩ

119868 (x)119867120576(120601 (x)) 119889x

intΩ

119867120576(120601 (x)) 119889x

1198882=

intΩ

119868 (x) (1 minus 119867120576(120601 (x))) 119889x

intΩ

(1 minus 119867120576(120601 (x))) 119889x

1198911(x) =

intΩ

119870120590(x minus 119910) 119868 (119910)119867

120576(120601 (119910)) 119889119910

intΩ

119870120590(x minus 119910)119867

120576(120601 (119910)) 119889119910

1198912(x) =

intΩ

119870120590(x minus 119910) 119868 (119910) (1 minus 119867

120576(120601 (119910))) 119889119910

intΩ

119870120590(x minus 119910) (1 minus 119867

120576(120601 (119910))) 119889119910

(10)

Keeping 1198881 1198882 1198911 and 119891

2fixed and minimizing the func-

tional 119864(120601 1198881 1198882 1198911 1198912) with respect to 120601 we obtain the cor-

responding gradient descent flow equation

120597120601

120597119905= 120575120576(120601) [120596 ((119868 minus 119868

119866

) (1198881minus 1198882))

+ (1 minus 120596) ((119868 minus 119868119871

) (1198911minus 1198912))]

= 120575120576(120601) [120596119865

119866

(120601) + (1 minus 120596) 119865119871

(120601)] = 119865 (120601)

(11)

where

120575120576(119911) = 119867

1015840

120576(119911) =

1

120587

120576

1205762 + 1199112 (12)

Like the CV and RSF models our model is also imple-mented using an alternative procedure for each iteration andthe corresponding level set function 120601119899 we first compute thefitting values 119888

119894(120601119899

) and 119891119894(120601119899

) and then obtain 120601119899+1 by min-imizing 119864(120601 119888

1(120601119899

) 1198882(120601119899

) 1198911(120601119899

) 1198912(120601119899

)) with respect to 120601This process is repeated until the zero-level set of 120601119899+1 isexactly on the object boundary

In the following we first discuss the properties of 119865119866(120601)and 119865119871(120601) and then analyze the behavior of (11)

32 Properties of 119865119866(120601) and 119865119871(120601) For the sake of simplicitywe state and prove the properties of 119865119866(120601) and 119865119871(120601) only foran image consisting of only two distinct gray levels

119868 (x) = 1198921 x isin 120596

1198922 x isin Ω 120596

(13)

where 1198921 1198922ge 0 with 119892

1= 1198922 120596 and Ω 120596 represent the ob-

jects of interest and the background respectively

Theorem 1 Let 119868(x) be an image by (13) Then one has

119865119866

(120601) =

(119872119899 minus 119873119898) (1198921minus 1198922)2

times [(119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

times(1198732

(119872 minus 119873)2

)minus1

in 120596

minus (119872119899 minus 119873119898) (1198921minus 1198922)2

times [119899 (119872 minus 119873)119867120576

+119873 (119898 minus 119899) (1 minus 119867120576)]

times(1198732

(119872 minus 119873)2

)minus1

in Ω 120596

(14)

and so

sign (119865119866 (120601)) =

+ sign (119872119899 minus 119873119898) in 120596

minus sign (119872119899 minus 119873119898) in Ω 120596

(15)

where

119872 = |Ω| 119898 = |120596| 119873 =1003816100381610038161003816120601 gt 0

1003816100381610038161003816

119899 =1003816100381610038161003816120596 cap 120601 gt 0

1003816100381610038161003816

(16)

in which |Ω| is the area of the regionΩ and similarly for others

Remarks (i) Due to the discrete nature of image |Ω| is in factthe number of pixels in the image 119868(x) and similarly to others

(ii) The cases of 119899 = 119898 119899 = 119873 and 119899 = 0 correspondto the zero-level lines of 120601(x) which are encircling the object(120596) inside the object and within the background (Ω

120596) respectively The cases of 119873 = 119899 =119898 and 119873 = 119899 =

119898 correspond to the zero-level line of 120601(x) that are partiallyinside the object and exactly on the object edge respectively

(iii) The significance of (15) is that the function 119865119866

(120601)

has the opposite sign in 120596 (object) and Ω 120596 (background)respectively

The proof of Theorem 1 is provided in Appendix A Thefollowing result will be used in the proof ofTheorem 7 whichguarantees that the evolution by (10) converges to a uniquestable value after finite time

Corollary 2 Let 119868(x) be an image by (13) Then one has thefollowing

(i) If119872119899 minus119873119898 ge 0 then

119865119866

(120601) ge (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

119865119866

(120601) le minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(17)

4 Mathematical Problems in Engineering

(ii) If119872119899 minus119873119898 le 0 then

119865119866

(120601) le (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

119865119866

(120601) ge minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(18)

The proof of Corollary 2 is given in Appendix AWe call the property in Theorem 1 an adaptive sign-

changing property of 119865119866(120601) Such property also holds for119865119871

(120601) which can be expressed by the following theorem

Theorem 3 Let 119868(x) be an image given by (13) Then one has

119865119871

(120601) =

(119875119902 minus 119876119901) (1198921minus 1198922)2

times [(119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times(1198762

(119875 minus 119876)2

)minus1

in 120596

minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times(1198762

(119875 minus 119876)2

)minus1

in Ω 120596

(19)

and so

sign (119865119871 (120601)) =

+ sign (119875119902 minus 119876119901) in 120596

minus sign (119875119902 minus 119876119901) in Ω 120596

(20)

where

119875 = intΩ

119870120590(x minus 120585) 119889120585 119901 = int

120596

119870120590(x minus 120585) 119889120585

119876 = int120601gt0

119870120590(x minus 120585) 119889120585

119902 = int120596cap120601gt0

119870120590(x minus 120585) 119889120585

(21)

This theorem shows that the local force 119865119871(120601) has exactlyopposite sign in 120596 (object) and in Ω 120596 (background)

The following result together with Corollary 2 will beused in the proof of Theorem 7 mentioned later

Corollary 4 Under the assumption ofTheorem 3 one has thefollowing

(i) If 119875119902 minus 119876119901 ge 0 then

119865119871

(120601) ge 119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

119865119871

(120601) le minus119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(22)

(ii) If 119875119902 minus 119876119901 le 0 then

119865119871

(120601) le 119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

119865119871

(120601) ge minus119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(23)

where

119878 = min 119870120590(x minus 120585) 120585 isin Ω) (24)

The proofs of Theorem 3 and Corollary 4 are similar toTheorem 1 and Corollary 2 respectively see Appendix B fordetails

33 Behavior of Our Model In this section we analyze thebehavior of our model (10) for image segmentation We willshow that the zero-level line of the evolution function startingwith a bounded function finally comes to a unique steadystate which separates the object from the background

Due to the discrete nature of image the continuous equa-tion (10) can be discretized in space with space step 1 (impliedpixel spacing) but also in time with a time step Δ119905 as follows

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

0 le 119894 le 119901 minus 1 0 le 119895 le 119902 minus 1

(25)

where 119901 times 119902 corresponds to the image size 119868119894119895= 119868(119894 119895) and

120601119896

119894119895= 120601(119896Δ119905 119894 119895) with 119896 ge 0 and 1206010

119894119895= 1206010(119894 119895)

Corollary 5 Let 1198721198990minus 1198730119898 = |Ω||120596 cap 120601

0

gt 0| minus |1206010

gt

0||120596| = 0

(i) If1198721198990minus 1198730119898 gt 0 then

119865119866

(120601119896

) gt (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119866

(120601119896

) lt minus (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= minus119860 lt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(26)

Mathematical Problems in Engineering 5

(ii) If1198721198990minus 1198730119898 lt 0 then

119865119866

(120601119896

) lt (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= 119860 lt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119866

(120601119896

) gt minus (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= minus119860 gt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(27)

where

119860 = (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987248 (28)

We provide the proof of Corollary 5 in Appendix C Sim-ilar analysis can prove the following corollary Corollary 6

Corollary 6 Let 1198751199020minus 1198760119901 = 0

(i) If 1198751199020minus 1198760119901 gt 0 then

119865119871

(120601119896

) gt (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= 119861 gt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119871

(120601119896

) lt minus (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= minus119861 lt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(29)

(ii) If 1198751199020minus 1198760119901 lt 0 then

119865119871

(120601119896

) lt (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= 119861 lt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119871

(120601119896

) gt minus (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= minus119861 gt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(30)

where

119861 = (1198751199020minus 1198760119901)

(1198921minus 1198922)2

11987548 (31)

Theorem 7 If 1206010(x) is a bounded function with (1198721198990minus

1198730119898)(119875119902

0minus 1198760119901) gt 0 then there exists a positive integer 119870

such that for 119896 ge 119870

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

if 1198721198990minus 1198730119898 gt 0 119875119902

0minus 1198760119901 gt 0

(32)

120601119896+1

119894119895lt 0 in 120596

120601119896+1

119894119895gt 0 in Ω 120596

if 1198721198990minus 1198730119898 lt 0 119875119902

0minus 1198760119901 lt 0

(33)

We provide the proof of Theorem 7 in Appendix DThe significance of Theorem 7 is that if we choose 1206010(x)

such that (1198721198990minus 1198730119898)(119875119902

0minus 1198760119901) gt 0 then the zero-level

line of120601(x) startingwith such initial function1206010(x) can finallycome to a unique steady state which separates object from thebackground

Remark Mathematically there does exist 1206010(x) such that(1198721198990minus1198730119898)(119875119902

0minus1198760119901) le 0 in practice however we always

can guarantee that

(1198721198990minus 1198730119898) (119875119902

0minus 1198760119901) gt 0 (34)

34 Discussion of Initial Function Theorem 7 guarantees thatthe proposed model computes a unique steady state regard-less of the initialization that is the convergence of the zero-level line 120601 = 0 is irrespective of the initial functionThis means that we can obtain the same zero-level line inthe steady state if we choose the initial function as a boundedfunction 1206010 with (119872119899

0minus 1198730119898)(119875119902

0minus 1198760119901) gt 0

In applications the initial function 1206010 can be defined

via a simple curve (closed curve or line segment) in imagedomain For example we can choose the initial function asa signed distance to a circle which is widely used in mostof image segmentation models with level set methods Forthe proposed model however we prefer to define the initialfunction 120601

0

(x) as a piecewise constant or constant functionas follows

(1) If the curve 119862 is a closed curve (eg circle or square)then 1206010(x) is defined by

1206010

(x) = +120588 (x) isin inside (119862) minus120588 (x) isin outside (119862)

(35)

where 120588 = 0 is a constant(2) If the curve 119862 is a line segment that partitions the

image domainΩ into two disjoint regionsΩ1andΩ

2

(eg the left and right half regions of image domainΩ respectively) withΩ = Ω

1cup Ω2andΩ

1cap Ω2= Φ

then 1206010(119909 119910) is defined by

1206010

(x) = +120588 (x) isin Ω

1

minus120588 (x) isin Ω2

(36)

where 120588 = 0 is a constant

6 Mathematical Problems in Engineering

(3) We can also define a zero function as follows

1206010

(x) = 0 (37)

Next we prove the fact that with 1206010(x) = 0 120601(x) becomesa sign-changing function and satisfy the condition of (119872119899

1minus

1198731119898)(119875119902

1minus 1198761119901) gt 0 after the first iteration

Theorem 8 If the initial function 1206010(x) = 0 in Ω and

1198880

1(x) = 119891

0

1(x) = 2 sdot mean

xisinΩ119868 (x)

1198880

2(x) = 119891

0

2(x) = 0

(38)

then after the first iteration one has

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(39)

whereΩ1cap Ω2= Φ Ω

1= Φ Ω2

= Φ

We provide the proof of Theorem 8 in Appendix EBy (25) andTheorem 8 we have

1206011

(x) = 120575120576(0) mean

xisinΩ119868 (x)

times (119868 minus meanxisinΩ

119868 (x))

lt 0 x isin Ω1

gt 0 x isin Ω2

(40)

Therefore 1206011(x) became a sign-changing function Thenusing two distinct gray levels of (13) and the above demon-stration we have

(1198721198991minus 1198731119898) (119875119902

1minus 1198761119901)

= (119898 (119872 minus1198731)) (119901 (119875 minus 119876

1))

= 119898119901 (119872 minus1198731) (119875 minus 119876

1) gt 0

(41)

4 Implementation and Experimental Results

41 Implementation In tradition PDE-based methods a cer-tain diffusion term is usually included in the evolution equa-tion to regularize the evolving function but which increasesthe computational cost Recently [16] proposed a novelscheme to regularize the evolving function that is Gaussianfiltering the evolving function after each of iterations Weadopt this scheme to regularize the evolving function 119906 ateach of iteration that is 120601119899 = 119866

120588lowast 120601119899 where 119901 controls the

regularization strength Such regularization procedure hassome advantages over the traditional regularization termsuch as computational efficiency and better smoothing effectsee [16] for more explanations

The main procedures of the proposed algorithm can besummarized as follows

(1) Initialize the evolving function 1206010(x)(2) Compute 119888

119894(120601) and 119891

119894(120601) (119894 = 1 2) using (10)

(3) Evolve the function 120601 according to (11)(4) Regularize the function 120601 with a Gaussian filter that

is 120601119899 = 119866120588lowast 120601119899

(5) Check whether the evolution is finished If not returnto step 2

42 Experimental Results In this section we show experi-mentally that the proposedmodel not only can provide desir-able segmentation results in the presence of intensity inho-mogeneity but also allows for more flexible initialization ofthe contour compared to the RSF and LIF models

In our numerical experiments for our model we choosethe parameters as follows 120576 = 15 for the regularized Diracfunction Δ119905=0025 (time step) ℎ = 1 (space step) 120596 = 01and 119899 = 5 The MATLAB source code of the LIF modelalgorithm is downloaded at httpwww4comppolyueduhksimcslzhangcodeLIFzip and the RSF model algorithmis downloaded at httpwwwengruconnedusimcmlicodeRSF v0 v01rar All experiments were run under MATLAB2010b on a PC with Dual 27 GHz processor

We will utilize two region overlap metrics to evaluatethe performances of the three models quantitatively The twometrics are the ratio of segmentation error (RSE) [18] and thedice similarity coefficient (DSC) [14 18 19] If 119878

1and 1198782rep-

resent a given baseline foreground region (eg true object)and the foreground region found by the model respectivelythen the two metrics are defined as follows

RSE =119873 (1198781 1198782) + 119873 (119878

2 1198781)

119873 (Ω)

DSC =2119873 (119878

1cap 1198782)

119873 (1198781) + 119873 (119878

2)

(42)

where 119873(sdot) indicates the number of pixels in the enclosedregion andΩ is image domainThe closer the DSC value to 1the better the segmentation Since119873(119878

11198782)+119873(119878

21198781) is the

number of the pixelsmistakenly classified by themodel lowerRSE means that there are fewer pixels misclassified that isthe image can be segmented more accurately Thus a perfectsegmentation will give DSC = 1 RSE = 0

In the first example (Figures 1ndash3) we mainly verify thecomputation of a unique steady state of the zero-level line of120601 starting with three types of representative initial functionsthat is signed distance function piecewise constant functionsby (26)-(27) and zero function respectively The top rowof Figure 1 shows the evolution of an active contour (iezero-level line 120601 = 0) with the function 120601 initialized toa signed distance function piecewise constant functions by(26)-(27) with 120588 = 1 and zero function respectively whilethe bottom row shows the corresponding evolution of120601Withsuch initializations the zero-level line of 120601 converges to thesame steady state

In Figure 2 we show that our model has the capabilityof detecting multiple objects or objects with interior holesor blurred edges only starting with a zero function Thecontours (zero-level lines) evolution processes are shown inthe second column to the forth column

Mathematical Problems in Engineering 7

Figure 1 Segmentations of our model for two real images with 120601 initialized as different functions Columns 1 to 4 signed distance functionpiecewise constant functions by (26)-(27) with 120588 = 1 and zero function

Figure 2 Applications of our model to four images (slug cell ventriculus sinister MR and real plane images) The curve evolution processfrom the initial contour (in the first column) to the final contour (in the fourth column) is shown in every row for the corresponding image

8 Mathematical Problems in Engineering

(a) (b) (c) (d)

Figure 3 Comparisons of three models Rows 1ndash3 RSF model LIF model and our model with 1206010= 0 For the RSF and LIF models initial

and final contours are shown in green and red color respectively Columns 1ndash4 the 4 pictures indicate (a) (b) (c) (d) respectively

Figure 4 Segmentation results of both models for a hand phantomUpper row LIF model Lower row our model with 120601

0= 0 Column

2 zoomed view of the narrow parts in hand fingers

We choose the segmentation results of the RSF and LIFmodels as baseline foreground regions and then computeDSC values for the corresponding images The RSE and DSCvalues for the four real images are given inTable 1 fromwhich

Table 1 RSE and DSC for RSF and LIF models and our model

RSF model LIF modelRSE () DSC () RSE () DSC ()

Figure 3(a) 027 9717 299 9891Figure 3(b) 271 9637 012 9913Figure 3(c) 049 8795 045 8806Figure 3(d) 016 9766 076 9891

we can see that the proposed algorithm works as well as theRSF and LIFmodels for imageswith intensity inhomogeneity

The experimental results shown in Figure 4 validate thatour method can also achieve subpixel segmentation accuracyas in [15] As can be seen from Figures 4(b) and 4(d) bothmodels achieve subpixel segmentation accuracy of the fingerboundaries The final contour accurately reflects the truehand shape

5 Conclusion

In this study we propose a new active contour model inte-grating a local intensity fitting (LIF) energy with an auxiliaryglobal intensity fitting (GIF) energy The LIF energy isresponsible for attracting the contour toward object bound-aries and is dominant near object boundaries while the GIFenergy incorporates global image information to improve therobustness to initialization of the contours The proposedmodel can efficiently handle intensity inhomogeneity while

Mathematical Problems in Engineering 9

allowing for more flexible initialization and maintaining thesubpixel accuracy The utility model has the advantages ofallowing for more flexible initialization of the contour andthe capability of detecting multiple objects or objects withinterior holes or blurred edges But [14] has not been imple-mented

Appendices

A Proofs of Theorem 1 and Corollary 2

Proof of Theorem 1 Clearly 120601 gt 0 =Φ and 120601 lt 0 =Φowing to the sign-changing property of 120601(119909 119910) Thus by (3)we have

1198881(120601) =

1198921119899 + 1198922(119873 minus 119899)

119873=(1198921minus 1198922) 119899 + 119892

2119873

119873

1198882(120601) =

1198921(119898 minus 119899) + 119892

2[(119872 minus 119873) minus (119898 minus 119899)]

119872 minus 119873

=(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)

119872 minus119873

(A1)

and so

1198881(120601) minus 119888

2(120601) = ((119872 minus 119873) [(119892

1minus 1198922) 119899 + 119892

2119873]

minus119873 [(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)])

times (119873 (119872 minus119873))minus1

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

(A2)

Therefore in 120596 we get

119903119897119865119866

= (1198881(120601) minus 119888

2(120601)) ((119892

1minus 1198881(120601))119867

120576(120601)

+ (1198921minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)sdot(1198921minus 1198922)

119873 (119872 minus119873)

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

=(119872119899 minus 119873119898) times (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

(A3)

and in Ω 120596 we have

119865119866

(120601) = (1198881(120601) minus 119888

2(120601)) ((119892

2minus 1198881(120601))119867

120576(120601)

+ (1198922minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

times [ minus (1198921minus 1198922)

times (119899 (119872 minus 119873)119867120576(120601)

+119873 (119898 minus 119899) (1 minus 119867120576(120601)))

times (119873 (119872 minus119873))minus1

]

=minus (119872119899 minus 119873119898) (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [(119899 (119872 minus 119873)119867120576(120601)119873 (119898 minus 119899) (1 minus 119867

120576(120601)))]

(A4)

This completes the proof of (14)The last assertion (15) followsclearly from the following fact

(119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576(120601))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A5)

(119899 (119872 minus 119873)119867120576(120601) + 119873 (119898 minus 119899) (1 minus 119867

120576(120601)))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A6)

The proof of Theorem 1 is completed

Now we give the proof of Corollary 2 (i) Since

1198722

= (119873 + (119872 minus119873))2

= 1198732

+ (119872 minus119873)2

+ 2119873 (119872 minus119873)

ge 4119873 (119872 minus119873)

(A7)

by (14) (A5) and (A7) we have

119865119866

(120601) = (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

times (1198732

(119872 minus 119873)2

)minus1

ge (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

(A8)

10 Mathematical Problems in Engineering

and by (14) and (A6)

119865119866

(120601) = minus (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119899 (119872 minus 119873)119867120576

+119873 (119898 minus 119899) (1 minus 119867120576))]

times (1198732

(119872 minus 119873)2

)minus1

le minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(A9)

This completes the proof of (17) and similarly for (18) Theproof is completed

B Proofs of Theorem 3 and Corollary 4

Proof of Theorem 3 By (6) we have

1198911(x) =

int120601gt0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601gt0

119870120590(xminus120585) 119889120585

= (int

120601gt0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

119906gt119868

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601gt0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

times (int

119906gt119868

119870120590(119909minus120585) 119889120585)

minus1

=1198921119902 + 1198922(119876 minus 119902)

119876

(B1)

1198912(x) =

int120601le0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601le0

119870120590(xminus120585) 119889120585

= (int

120601le0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601le0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

=1198921(119901 minus 119902) + 119892

2[(119875 minus 119876) minus (119901 minus 119902)]

119875 minus 119876

(B2)

and so

1198911(x) minus 119891

2(x) =

(1198921minus 1198922) (119875119902 minus 119876119901)

119876 (119875 minus 119876) (B3)

Therefore in 120596 we get

119865119871

(120601) = (1198981(120601) minus 119898

2(120601))

times ((1198921minus 1198981(120601))119867

120576(120601)

+ (1198921minus 1198982(120601)) (1 minus 119867

120576(120601)))

= (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B4)

and in Ω 120596 we have

119865119871

= (1198981(120601) minus 119898

2(120601))

times ((1198922minus 1198981(120601))119867

120576(120601)

+ (1198922minus 1198982(120601)) (1 minus 119867

120576(120601)))

= minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B5)

Mathematical Problems in Engineering 11

This completes the proof of (19) The last assertion (20) fol-lows clearly from the following fact(119875 minus 119876) (119876 minus 119902)119867

120576(120601)

+ 119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))

= 119867120576(120601) (int

Ω

minusint

120601gt0

)119870120590(xminus120585) 119889120585

sdot (int

120601gt0

minusint

120601gt0cap120596)119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601)) int

120601gt0

119870120590(xminus120585) 119889120585

sdot ((intΩ

+int

120601gt0cap120596) minus (int

120596

+int

120601gt0

))119870120590(xminus120585) 119889120585

= 119867120576(120601) int

120601le0

119870120590(xminus120585) 119889120585

sdot (int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot (int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

ge 119867120576(120601) (int

120601le0

119878119889120585) sdot (int

120601gt0(120601gt0cap120596)

119878119889120585)

+ (1 minus 119867120576(120601)) (int

120601gt0

119878119889120585) sdot (int

120601le0(120601le0cap120596)

119878119889120585)

= 119878 [ (119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 minus119873) minus (119898 minus 119899)) (1 minus 119867120576(120601))]

ge 119878

(B6)119902 (119875 minus 119876)119867

120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))

= 119867120576(120601) int

120601gt0cap120596119870120590(xminus120585) 119889120585 sdot int

120601le0

119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot int

120601le0cap120596119870120590(xminus120585) 119889120585

ge 119867120576(120601) (int

119906gt119868cap120596119878119889120585) sdot (int

119906le119868

119878119889120585)

+ (1 minus 119867120576(120601)) (int

119906gt119868

119878119889120585) sdot (int

119906le119868cap120596119878119889120585)

= 119878 [119899 (119872 minus 119873)119867120576(120601) + (1 minus 119867

120576(120601))119873 (119898 minus 119899)]

ge 119878

(B7)

where 119878 = min119870120590(xminus120585) 120585 isin Ω)

Now we give the proof of Corollary 4 (i) Since

1198752

ge 4119875 (119875 minus 119876) (B8)

we have by (19) and (B6)

119865119871

(120601) = (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

ge (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

(B9)

and by (19) and (B7)

119865119871

(120601) = minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

le minus (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(B10)

This completes the proof of (19) and similarly to (20) Theproof is completed

C Proof of Corollary 5

Proof (i) By Corollary 2 (i) we have

119865119866

(1206010

) ge (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0

in 120596

119865119866

(1206010

) le minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

(C1)

At the first iteration by (25) and (C1) we have

1206011

119894119895= 1206010

119894119895+ Δ119905

sdot 119865119866

(1206010

119894119895)

ge 1206010

119894119895+ Δ119905 sdot 120575

120576(1206010

)119860 in 120596

le 1206010

119894119895minus Δ119905 sdot 120575

120576(1206010

)119860 in Ω 120596

(C2)

12 Mathematical Problems in Engineering

or equivalently

1206011

(x) ge 1206010

(x) + Δ119905 sdot 120575120576(1206010

)119860 gt 1206010

(x) in 120596

1206011

(x) le 1206010

(x) minus Δ119905 sdot 120575120576(1206010

)119860 lt 1206010

(x) in Ω 120596

(C3)

which implies that

1198721198991minus 1198731119898 = |Ω|

10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816minus100381610038161003816100381610038161206011

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816) |120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816|120596|

gt |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816) |120596|

= |Ω|1003816100381610038161003816120596 cap 120601

0gt 0

1003816100381610038161003816 minus10038161003816100381610038161206010 gt 0

1003816100381610038161003816 |120596|

= 1198721198990minus 1198730119898 gt 0

(C4)

In a manner similar to the steps used to obtain (26)-(27) and(C1)ndash (C4) we obtain

119865119866

(119868 1206011

) ge (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

gt (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

119865119866

(119868 1206011

) le minus (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

lt minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

1206012

(x) ge 1206011

(x) + Δ119905 sdot 120575120576(1206011

)119860 gt 1206011

(x) in 120596

1206012

(x) le 1206011

(x) minus Δ119905 sdot 120575120576(1206011

)119860 lt 1206011

(x) in Ω 120596

1198721198992minus 1198732119898 gt 119872119899

1minus 1198731119898 gt 119872119899

0minus 1198730119898 gt 0

(C5)

D Proof of Theorem 7

Proof We prove (32) and similarly to (33) By (25)Corollary 5(i) and Corollary 6(i) we get

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

= 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119865119871

(120601119896

119894119895) + 120596119865

119866

(120601119896

119894119895))

ge 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

) + 120596119860 sdot 120575120576(120601119896+1

))

= 120601119896

119894119895+ ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in 120596

(119896 ge 1 119896 isin 119911+

)

le 120601119896

119894119895+ Δ119905

sdot (minus ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

)

+120596119860 sdot 120575120576(120601119896+1

)))

= 120601119896

119894119895minus ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(D1)

which implies that

120601119896+1

119894119895ge 1206010

119894+ ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in 120596

120601119896+1

119894119895le 1206010

119894minus (119896 + 1)119860 ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in Ω 120596

(D2)

Because 1206010(x) is bounded in domain Ω there clearly exists a

positive integer 119870 such that

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

for 119896 ge 119870 (D3)

The proof is completed

Mathematical Problems in Engineering 13

E Proof of Theorem 8

Proof From (11) we have

119865 (0) = 120575120576(0) [120596 ((119868 minus 119868

119866

) sdot 2meanxisinΩ

119868 (x))

+ (1 minus 120596) ((119868 minus 119868119871

) sdot 2meanxisinΩ

119868 (x))]

=2120575120576(0)mean

xisinΩ119868 (x) (119868 minus 2119867

120576(0)mean

xisinΩ119868 (x))

=120575120576(0)mean

xisinΩ119868 (x) (119868 minusmean

xisinΩ119868 (x))

(E1)

Clearly

minxisinΩ

119868 (x) lt meanxisinΩ

119868 (x) lt maxxisinΩ

119868 (x) (E2)

which implies that

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(E3)

where

Ω1= 119868 (x) lt mean

xisinΩ119868 (x)

Ω2= 119868 (x) gt mean

xisinΩ119868 (x)

(E4)

The proof of Theorem 8 is completed

Acknowledgments

The authors would like to thank the anonymous reviewersfor their valuable comments and suggestions to improve thispaper Besides this work was supported by the Fundamen-tal Research Funds for the Central University Grant no(CDJXS10100006)

References

[1] V Caselles F Catte T Coll and F Dibos ldquoA geometricmodel for active contours in image processingrdquo NumerischeMathematik vol 66 no 1 pp 1ndash31 1993

[2] T Chan M Moelich and B Sandberg ldquoSome recent develop-ments in variational image segmentationrdquo UCLA CAM Reportcam 06-52 2006

[3] L He Z Peng B Everding et al ldquoA comparative study ofdeformable contour methods on medical image segmentationrdquoImage and Vision Computing vol 26 no 2 pp 141ndash163 2008

[4] C Li C Xu C Gui andM D Fox ldquoLevel set evolution withoutre-initialization a new variational formulationrdquo in Proceedingsof the IEEE Computer Society Conference on Computer Visionand Pattern Recognition (CVPR rsquo05) pp 430ndash436 San DiegoCalif USA June 2005

[5] C Li C Xu C Gui and M D Fox ldquoDistance regularized levelset evolution and its application to image segmentationrdquo IEEETransactions on Image Processing vol 19 no 12 pp 3243ndash32542010

[6] B Zhou and C LMu ldquoLevel set evolution for boundary extrac-tion based on a p-Laplace equationrdquo Applied MathematicalModelling vol 34 no 12 pp 3910ndash3916 2010

[7] Y Wang and C He ldquoAdaptive level set evolution starting with aconstant functionrdquoAppliedMathematical Modelling vol 36 no7 pp 3217ndash3228 2012

[8] T F Chan and L A Vese ldquoActive contours without edgesrdquo IEEETransactions on Image Processing vol 10 no 2 pp 266ndash2772001

[9] A Tsai A Yezzi andA SWillsky ldquoCurve evolution implemen-tation of theMumford-Shah functional for image segmentationdenoising interpolation andmagnificationrdquo IEEETransactionson Image Processing vol 10 no 8 pp 1169ndash1186 2001

[10] T Brox and D Cremers ldquoOn the statistical interpretation of thepiecewise smooth Mumford-Shah functionalrdquo in Proceedingsof the International Conference on Scale Space and VariationalMethods in Computer Vision (SSVM rsquo07) pp 203ndash213 IschiaItaly 2007

[11] C Li C Y Kao J CGore andZDing ldquoImplicit active contoursdriven by local binary fitting energyrdquo in Proceedings of the IEEEComputer Society Conference on Computer Vision and PatternRecognition (CVPR rsquo07) pp 1ndash7 Washington DC USA June2007

[12] C Li C Y Kao J C Gore and Z Ding ldquoMinimization ofregion-scalable fitting energy for image segmentationrdquo IEEETransactions on Image Processing vol 17 no 10 pp 1940ndash19492008

[13] S Lankton and A Tannenbaum ldquoLocalizing region-basedactive contoursrdquo IEEE Transactions on Image Processing vol 17no 11 pp 2029ndash2039 2008

[14] L Wang C Li Q Sun D Xia and C Y Kao ldquoActive contoursdriven by local and global intensity fitting energy with applica-tion to brain MR image segmentationrdquo Computerized MedicalImaging and Graphics vol 33 no 7 pp 520ndash531 2009

[15] K Zhang H Song and L Zhang ldquoActive contours driven bylocal image fitting energyrdquo Pattern Recognition vol 43 no 4pp 1199ndash1206 2010

[16] K Zhang L Zhang H Song and W Zhou ldquoActive contourswith selective local or global segmentation a new formulationand level set methodrdquo Image and Vision Computing vol 28 no4 pp 668ndash676 2010

[17] Y Yuan and C He ldquoVariational level set methods for image seg-mentation based on both 1198712 and Sobolev gradientsrdquo NonlinearAnalysis Real World Applications vol 13 no 2 pp 959ndash9662012

[18] B Liu H D Cheng J Huang J Tian X Tang and J LiuldquoProbability density difference-based active contour for ultra-sound image segmentationrdquo Pattern Recognition vol 43 no 6pp 2028ndash2042 2010

[19] D W Shattuck S R Sandor-Leahy K A Schaper D ARottenberg and RM Leahy ldquoMagnetic resonance image tissueclassification using a partial volume modelrdquo NeuroImage vol13 no 5 pp 856ndash876 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 4: Research Article Implicit Active Contour Model with Local ...downloads.hindawi.com/journals/mpe/2013/367086.pdfAcademic Editor: Oluwole Daniel Makinde ... (CV)proposedaregion-basedactivecontourmodel

4 Mathematical Problems in Engineering

(ii) If119872119899 minus119873119898 le 0 then

119865119866

(120601) le (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

119865119866

(120601) ge minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(18)

The proof of Corollary 2 is given in Appendix AWe call the property in Theorem 1 an adaptive sign-

changing property of 119865119866(120601) Such property also holds for119865119871

(120601) which can be expressed by the following theorem

Theorem 3 Let 119868(x) be an image given by (13) Then one has

119865119871

(120601) =

(119875119902 minus 119876119901) (1198921minus 1198922)2

times [(119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times(1198762

(119875 minus 119876)2

)minus1

in 120596

minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times(1198762

(119875 minus 119876)2

)minus1

in Ω 120596

(19)

and so

sign (119865119871 (120601)) =

+ sign (119875119902 minus 119876119901) in 120596

minus sign (119875119902 minus 119876119901) in Ω 120596

(20)

where

119875 = intΩ

119870120590(x minus 120585) 119889120585 119901 = int

120596

119870120590(x minus 120585) 119889120585

119876 = int120601gt0

119870120590(x minus 120585) 119889120585

119902 = int120596cap120601gt0

119870120590(x minus 120585) 119889120585

(21)

This theorem shows that the local force 119865119871(120601) has exactlyopposite sign in 120596 (object) and in Ω 120596 (background)

The following result together with Corollary 2 will beused in the proof of Theorem 7 mentioned later

Corollary 4 Under the assumption ofTheorem 3 one has thefollowing

(i) If 119875119902 minus 119876119901 ge 0 then

119865119871

(120601) ge 119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

119865119871

(120601) le minus119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(22)

(ii) If 119875119902 minus 119876119901 le 0 then

119865119871

(120601) le 119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

119865119871

(120601) ge minus119878 (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(23)

where

119878 = min 119870120590(x minus 120585) 120585 isin Ω) (24)

The proofs of Theorem 3 and Corollary 4 are similar toTheorem 1 and Corollary 2 respectively see Appendix B fordetails

33 Behavior of Our Model In this section we analyze thebehavior of our model (10) for image segmentation We willshow that the zero-level line of the evolution function startingwith a bounded function finally comes to a unique steadystate which separates the object from the background

Due to the discrete nature of image the continuous equa-tion (10) can be discretized in space with space step 1 (impliedpixel spacing) but also in time with a time step Δ119905 as follows

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

0 le 119894 le 119901 minus 1 0 le 119895 le 119902 minus 1

(25)

where 119901 times 119902 corresponds to the image size 119868119894119895= 119868(119894 119895) and

120601119896

119894119895= 120601(119896Δ119905 119894 119895) with 119896 ge 0 and 1206010

119894119895= 1206010(119894 119895)

Corollary 5 Let 1198721198990minus 1198730119898 = |Ω||120596 cap 120601

0

gt 0| minus |1206010

gt

0||120596| = 0

(i) If1198721198990minus 1198730119898 gt 0 then

119865119866

(120601119896

) gt (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119866

(120601119896

) lt minus (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= minus119860 lt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(26)

Mathematical Problems in Engineering 5

(ii) If1198721198990minus 1198730119898 lt 0 then

119865119866

(120601119896

) lt (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= 119860 lt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119866

(120601119896

) gt minus (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= minus119860 gt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(27)

where

119860 = (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987248 (28)

We provide the proof of Corollary 5 in Appendix C Sim-ilar analysis can prove the following corollary Corollary 6

Corollary 6 Let 1198751199020minus 1198760119901 = 0

(i) If 1198751199020minus 1198760119901 gt 0 then

119865119871

(120601119896

) gt (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= 119861 gt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119871

(120601119896

) lt minus (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= minus119861 lt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(29)

(ii) If 1198751199020minus 1198760119901 lt 0 then

119865119871

(120601119896

) lt (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= 119861 lt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119871

(120601119896

) gt minus (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= minus119861 gt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(30)

where

119861 = (1198751199020minus 1198760119901)

(1198921minus 1198922)2

11987548 (31)

Theorem 7 If 1206010(x) is a bounded function with (1198721198990minus

1198730119898)(119875119902

0minus 1198760119901) gt 0 then there exists a positive integer 119870

such that for 119896 ge 119870

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

if 1198721198990minus 1198730119898 gt 0 119875119902

0minus 1198760119901 gt 0

(32)

120601119896+1

119894119895lt 0 in 120596

120601119896+1

119894119895gt 0 in Ω 120596

if 1198721198990minus 1198730119898 lt 0 119875119902

0minus 1198760119901 lt 0

(33)

We provide the proof of Theorem 7 in Appendix DThe significance of Theorem 7 is that if we choose 1206010(x)

such that (1198721198990minus 1198730119898)(119875119902

0minus 1198760119901) gt 0 then the zero-level

line of120601(x) startingwith such initial function1206010(x) can finallycome to a unique steady state which separates object from thebackground

Remark Mathematically there does exist 1206010(x) such that(1198721198990minus1198730119898)(119875119902

0minus1198760119901) le 0 in practice however we always

can guarantee that

(1198721198990minus 1198730119898) (119875119902

0minus 1198760119901) gt 0 (34)

34 Discussion of Initial Function Theorem 7 guarantees thatthe proposed model computes a unique steady state regard-less of the initialization that is the convergence of the zero-level line 120601 = 0 is irrespective of the initial functionThis means that we can obtain the same zero-level line inthe steady state if we choose the initial function as a boundedfunction 1206010 with (119872119899

0minus 1198730119898)(119875119902

0minus 1198760119901) gt 0

In applications the initial function 1206010 can be defined

via a simple curve (closed curve or line segment) in imagedomain For example we can choose the initial function asa signed distance to a circle which is widely used in mostof image segmentation models with level set methods Forthe proposed model however we prefer to define the initialfunction 120601

0

(x) as a piecewise constant or constant functionas follows

(1) If the curve 119862 is a closed curve (eg circle or square)then 1206010(x) is defined by

1206010

(x) = +120588 (x) isin inside (119862) minus120588 (x) isin outside (119862)

(35)

where 120588 = 0 is a constant(2) If the curve 119862 is a line segment that partitions the

image domainΩ into two disjoint regionsΩ1andΩ

2

(eg the left and right half regions of image domainΩ respectively) withΩ = Ω

1cup Ω2andΩ

1cap Ω2= Φ

then 1206010(119909 119910) is defined by

1206010

(x) = +120588 (x) isin Ω

1

minus120588 (x) isin Ω2

(36)

where 120588 = 0 is a constant

6 Mathematical Problems in Engineering

(3) We can also define a zero function as follows

1206010

(x) = 0 (37)

Next we prove the fact that with 1206010(x) = 0 120601(x) becomesa sign-changing function and satisfy the condition of (119872119899

1minus

1198731119898)(119875119902

1minus 1198761119901) gt 0 after the first iteration

Theorem 8 If the initial function 1206010(x) = 0 in Ω and

1198880

1(x) = 119891

0

1(x) = 2 sdot mean

xisinΩ119868 (x)

1198880

2(x) = 119891

0

2(x) = 0

(38)

then after the first iteration one has

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(39)

whereΩ1cap Ω2= Φ Ω

1= Φ Ω2

= Φ

We provide the proof of Theorem 8 in Appendix EBy (25) andTheorem 8 we have

1206011

(x) = 120575120576(0) mean

xisinΩ119868 (x)

times (119868 minus meanxisinΩ

119868 (x))

lt 0 x isin Ω1

gt 0 x isin Ω2

(40)

Therefore 1206011(x) became a sign-changing function Thenusing two distinct gray levels of (13) and the above demon-stration we have

(1198721198991minus 1198731119898) (119875119902

1minus 1198761119901)

= (119898 (119872 minus1198731)) (119901 (119875 minus 119876

1))

= 119898119901 (119872 minus1198731) (119875 minus 119876

1) gt 0

(41)

4 Implementation and Experimental Results

41 Implementation In tradition PDE-based methods a cer-tain diffusion term is usually included in the evolution equa-tion to regularize the evolving function but which increasesthe computational cost Recently [16] proposed a novelscheme to regularize the evolving function that is Gaussianfiltering the evolving function after each of iterations Weadopt this scheme to regularize the evolving function 119906 ateach of iteration that is 120601119899 = 119866

120588lowast 120601119899 where 119901 controls the

regularization strength Such regularization procedure hassome advantages over the traditional regularization termsuch as computational efficiency and better smoothing effectsee [16] for more explanations

The main procedures of the proposed algorithm can besummarized as follows

(1) Initialize the evolving function 1206010(x)(2) Compute 119888

119894(120601) and 119891

119894(120601) (119894 = 1 2) using (10)

(3) Evolve the function 120601 according to (11)(4) Regularize the function 120601 with a Gaussian filter that

is 120601119899 = 119866120588lowast 120601119899

(5) Check whether the evolution is finished If not returnto step 2

42 Experimental Results In this section we show experi-mentally that the proposedmodel not only can provide desir-able segmentation results in the presence of intensity inho-mogeneity but also allows for more flexible initialization ofthe contour compared to the RSF and LIF models

In our numerical experiments for our model we choosethe parameters as follows 120576 = 15 for the regularized Diracfunction Δ119905=0025 (time step) ℎ = 1 (space step) 120596 = 01and 119899 = 5 The MATLAB source code of the LIF modelalgorithm is downloaded at httpwww4comppolyueduhksimcslzhangcodeLIFzip and the RSF model algorithmis downloaded at httpwwwengruconnedusimcmlicodeRSF v0 v01rar All experiments were run under MATLAB2010b on a PC with Dual 27 GHz processor

We will utilize two region overlap metrics to evaluatethe performances of the three models quantitatively The twometrics are the ratio of segmentation error (RSE) [18] and thedice similarity coefficient (DSC) [14 18 19] If 119878

1and 1198782rep-

resent a given baseline foreground region (eg true object)and the foreground region found by the model respectivelythen the two metrics are defined as follows

RSE =119873 (1198781 1198782) + 119873 (119878

2 1198781)

119873 (Ω)

DSC =2119873 (119878

1cap 1198782)

119873 (1198781) + 119873 (119878

2)

(42)

where 119873(sdot) indicates the number of pixels in the enclosedregion andΩ is image domainThe closer the DSC value to 1the better the segmentation Since119873(119878

11198782)+119873(119878

21198781) is the

number of the pixelsmistakenly classified by themodel lowerRSE means that there are fewer pixels misclassified that isthe image can be segmented more accurately Thus a perfectsegmentation will give DSC = 1 RSE = 0

In the first example (Figures 1ndash3) we mainly verify thecomputation of a unique steady state of the zero-level line of120601 starting with three types of representative initial functionsthat is signed distance function piecewise constant functionsby (26)-(27) and zero function respectively The top rowof Figure 1 shows the evolution of an active contour (iezero-level line 120601 = 0) with the function 120601 initialized toa signed distance function piecewise constant functions by(26)-(27) with 120588 = 1 and zero function respectively whilethe bottom row shows the corresponding evolution of120601Withsuch initializations the zero-level line of 120601 converges to thesame steady state

In Figure 2 we show that our model has the capabilityof detecting multiple objects or objects with interior holesor blurred edges only starting with a zero function Thecontours (zero-level lines) evolution processes are shown inthe second column to the forth column

Mathematical Problems in Engineering 7

Figure 1 Segmentations of our model for two real images with 120601 initialized as different functions Columns 1 to 4 signed distance functionpiecewise constant functions by (26)-(27) with 120588 = 1 and zero function

Figure 2 Applications of our model to four images (slug cell ventriculus sinister MR and real plane images) The curve evolution processfrom the initial contour (in the first column) to the final contour (in the fourth column) is shown in every row for the corresponding image

8 Mathematical Problems in Engineering

(a) (b) (c) (d)

Figure 3 Comparisons of three models Rows 1ndash3 RSF model LIF model and our model with 1206010= 0 For the RSF and LIF models initial

and final contours are shown in green and red color respectively Columns 1ndash4 the 4 pictures indicate (a) (b) (c) (d) respectively

Figure 4 Segmentation results of both models for a hand phantomUpper row LIF model Lower row our model with 120601

0= 0 Column

2 zoomed view of the narrow parts in hand fingers

We choose the segmentation results of the RSF and LIFmodels as baseline foreground regions and then computeDSC values for the corresponding images The RSE and DSCvalues for the four real images are given inTable 1 fromwhich

Table 1 RSE and DSC for RSF and LIF models and our model

RSF model LIF modelRSE () DSC () RSE () DSC ()

Figure 3(a) 027 9717 299 9891Figure 3(b) 271 9637 012 9913Figure 3(c) 049 8795 045 8806Figure 3(d) 016 9766 076 9891

we can see that the proposed algorithm works as well as theRSF and LIFmodels for imageswith intensity inhomogeneity

The experimental results shown in Figure 4 validate thatour method can also achieve subpixel segmentation accuracyas in [15] As can be seen from Figures 4(b) and 4(d) bothmodels achieve subpixel segmentation accuracy of the fingerboundaries The final contour accurately reflects the truehand shape

5 Conclusion

In this study we propose a new active contour model inte-grating a local intensity fitting (LIF) energy with an auxiliaryglobal intensity fitting (GIF) energy The LIF energy isresponsible for attracting the contour toward object bound-aries and is dominant near object boundaries while the GIFenergy incorporates global image information to improve therobustness to initialization of the contours The proposedmodel can efficiently handle intensity inhomogeneity while

Mathematical Problems in Engineering 9

allowing for more flexible initialization and maintaining thesubpixel accuracy The utility model has the advantages ofallowing for more flexible initialization of the contour andthe capability of detecting multiple objects or objects withinterior holes or blurred edges But [14] has not been imple-mented

Appendices

A Proofs of Theorem 1 and Corollary 2

Proof of Theorem 1 Clearly 120601 gt 0 =Φ and 120601 lt 0 =Φowing to the sign-changing property of 120601(119909 119910) Thus by (3)we have

1198881(120601) =

1198921119899 + 1198922(119873 minus 119899)

119873=(1198921minus 1198922) 119899 + 119892

2119873

119873

1198882(120601) =

1198921(119898 minus 119899) + 119892

2[(119872 minus 119873) minus (119898 minus 119899)]

119872 minus 119873

=(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)

119872 minus119873

(A1)

and so

1198881(120601) minus 119888

2(120601) = ((119872 minus 119873) [(119892

1minus 1198922) 119899 + 119892

2119873]

minus119873 [(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)])

times (119873 (119872 minus119873))minus1

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

(A2)

Therefore in 120596 we get

119903119897119865119866

= (1198881(120601) minus 119888

2(120601)) ((119892

1minus 1198881(120601))119867

120576(120601)

+ (1198921minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)sdot(1198921minus 1198922)

119873 (119872 minus119873)

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

=(119872119899 minus 119873119898) times (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

(A3)

and in Ω 120596 we have

119865119866

(120601) = (1198881(120601) minus 119888

2(120601)) ((119892

2minus 1198881(120601))119867

120576(120601)

+ (1198922minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

times [ minus (1198921minus 1198922)

times (119899 (119872 minus 119873)119867120576(120601)

+119873 (119898 minus 119899) (1 minus 119867120576(120601)))

times (119873 (119872 minus119873))minus1

]

=minus (119872119899 minus 119873119898) (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [(119899 (119872 minus 119873)119867120576(120601)119873 (119898 minus 119899) (1 minus 119867

120576(120601)))]

(A4)

This completes the proof of (14)The last assertion (15) followsclearly from the following fact

(119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576(120601))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A5)

(119899 (119872 minus 119873)119867120576(120601) + 119873 (119898 minus 119899) (1 minus 119867

120576(120601)))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A6)

The proof of Theorem 1 is completed

Now we give the proof of Corollary 2 (i) Since

1198722

= (119873 + (119872 minus119873))2

= 1198732

+ (119872 minus119873)2

+ 2119873 (119872 minus119873)

ge 4119873 (119872 minus119873)

(A7)

by (14) (A5) and (A7) we have

119865119866

(120601) = (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

times (1198732

(119872 minus 119873)2

)minus1

ge (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

(A8)

10 Mathematical Problems in Engineering

and by (14) and (A6)

119865119866

(120601) = minus (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119899 (119872 minus 119873)119867120576

+119873 (119898 minus 119899) (1 minus 119867120576))]

times (1198732

(119872 minus 119873)2

)minus1

le minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(A9)

This completes the proof of (17) and similarly for (18) Theproof is completed

B Proofs of Theorem 3 and Corollary 4

Proof of Theorem 3 By (6) we have

1198911(x) =

int120601gt0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601gt0

119870120590(xminus120585) 119889120585

= (int

120601gt0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

119906gt119868

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601gt0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

times (int

119906gt119868

119870120590(119909minus120585) 119889120585)

minus1

=1198921119902 + 1198922(119876 minus 119902)

119876

(B1)

1198912(x) =

int120601le0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601le0

119870120590(xminus120585) 119889120585

= (int

120601le0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601le0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

=1198921(119901 minus 119902) + 119892

2[(119875 minus 119876) minus (119901 minus 119902)]

119875 minus 119876

(B2)

and so

1198911(x) minus 119891

2(x) =

(1198921minus 1198922) (119875119902 minus 119876119901)

119876 (119875 minus 119876) (B3)

Therefore in 120596 we get

119865119871

(120601) = (1198981(120601) minus 119898

2(120601))

times ((1198921minus 1198981(120601))119867

120576(120601)

+ (1198921minus 1198982(120601)) (1 minus 119867

120576(120601)))

= (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B4)

and in Ω 120596 we have

119865119871

= (1198981(120601) minus 119898

2(120601))

times ((1198922minus 1198981(120601))119867

120576(120601)

+ (1198922minus 1198982(120601)) (1 minus 119867

120576(120601)))

= minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B5)

Mathematical Problems in Engineering 11

This completes the proof of (19) The last assertion (20) fol-lows clearly from the following fact(119875 minus 119876) (119876 minus 119902)119867

120576(120601)

+ 119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))

= 119867120576(120601) (int

Ω

minusint

120601gt0

)119870120590(xminus120585) 119889120585

sdot (int

120601gt0

minusint

120601gt0cap120596)119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601)) int

120601gt0

119870120590(xminus120585) 119889120585

sdot ((intΩ

+int

120601gt0cap120596) minus (int

120596

+int

120601gt0

))119870120590(xminus120585) 119889120585

= 119867120576(120601) int

120601le0

119870120590(xminus120585) 119889120585

sdot (int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot (int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

ge 119867120576(120601) (int

120601le0

119878119889120585) sdot (int

120601gt0(120601gt0cap120596)

119878119889120585)

+ (1 minus 119867120576(120601)) (int

120601gt0

119878119889120585) sdot (int

120601le0(120601le0cap120596)

119878119889120585)

= 119878 [ (119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 minus119873) minus (119898 minus 119899)) (1 minus 119867120576(120601))]

ge 119878

(B6)119902 (119875 minus 119876)119867

120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))

= 119867120576(120601) int

120601gt0cap120596119870120590(xminus120585) 119889120585 sdot int

120601le0

119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot int

120601le0cap120596119870120590(xminus120585) 119889120585

ge 119867120576(120601) (int

119906gt119868cap120596119878119889120585) sdot (int

119906le119868

119878119889120585)

+ (1 minus 119867120576(120601)) (int

119906gt119868

119878119889120585) sdot (int

119906le119868cap120596119878119889120585)

= 119878 [119899 (119872 minus 119873)119867120576(120601) + (1 minus 119867

120576(120601))119873 (119898 minus 119899)]

ge 119878

(B7)

where 119878 = min119870120590(xminus120585) 120585 isin Ω)

Now we give the proof of Corollary 4 (i) Since

1198752

ge 4119875 (119875 minus 119876) (B8)

we have by (19) and (B6)

119865119871

(120601) = (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

ge (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

(B9)

and by (19) and (B7)

119865119871

(120601) = minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

le minus (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(B10)

This completes the proof of (19) and similarly to (20) Theproof is completed

C Proof of Corollary 5

Proof (i) By Corollary 2 (i) we have

119865119866

(1206010

) ge (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0

in 120596

119865119866

(1206010

) le minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

(C1)

At the first iteration by (25) and (C1) we have

1206011

119894119895= 1206010

119894119895+ Δ119905

sdot 119865119866

(1206010

119894119895)

ge 1206010

119894119895+ Δ119905 sdot 120575

120576(1206010

)119860 in 120596

le 1206010

119894119895minus Δ119905 sdot 120575

120576(1206010

)119860 in Ω 120596

(C2)

12 Mathematical Problems in Engineering

or equivalently

1206011

(x) ge 1206010

(x) + Δ119905 sdot 120575120576(1206010

)119860 gt 1206010

(x) in 120596

1206011

(x) le 1206010

(x) minus Δ119905 sdot 120575120576(1206010

)119860 lt 1206010

(x) in Ω 120596

(C3)

which implies that

1198721198991minus 1198731119898 = |Ω|

10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816minus100381610038161003816100381610038161206011

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816) |120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816|120596|

gt |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816) |120596|

= |Ω|1003816100381610038161003816120596 cap 120601

0gt 0

1003816100381610038161003816 minus10038161003816100381610038161206010 gt 0

1003816100381610038161003816 |120596|

= 1198721198990minus 1198730119898 gt 0

(C4)

In a manner similar to the steps used to obtain (26)-(27) and(C1)ndash (C4) we obtain

119865119866

(119868 1206011

) ge (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

gt (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

119865119866

(119868 1206011

) le minus (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

lt minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

1206012

(x) ge 1206011

(x) + Δ119905 sdot 120575120576(1206011

)119860 gt 1206011

(x) in 120596

1206012

(x) le 1206011

(x) minus Δ119905 sdot 120575120576(1206011

)119860 lt 1206011

(x) in Ω 120596

1198721198992minus 1198732119898 gt 119872119899

1minus 1198731119898 gt 119872119899

0minus 1198730119898 gt 0

(C5)

D Proof of Theorem 7

Proof We prove (32) and similarly to (33) By (25)Corollary 5(i) and Corollary 6(i) we get

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

= 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119865119871

(120601119896

119894119895) + 120596119865

119866

(120601119896

119894119895))

ge 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

) + 120596119860 sdot 120575120576(120601119896+1

))

= 120601119896

119894119895+ ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in 120596

(119896 ge 1 119896 isin 119911+

)

le 120601119896

119894119895+ Δ119905

sdot (minus ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

)

+120596119860 sdot 120575120576(120601119896+1

)))

= 120601119896

119894119895minus ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(D1)

which implies that

120601119896+1

119894119895ge 1206010

119894+ ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in 120596

120601119896+1

119894119895le 1206010

119894minus (119896 + 1)119860 ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in Ω 120596

(D2)

Because 1206010(x) is bounded in domain Ω there clearly exists a

positive integer 119870 such that

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

for 119896 ge 119870 (D3)

The proof is completed

Mathematical Problems in Engineering 13

E Proof of Theorem 8

Proof From (11) we have

119865 (0) = 120575120576(0) [120596 ((119868 minus 119868

119866

) sdot 2meanxisinΩ

119868 (x))

+ (1 minus 120596) ((119868 minus 119868119871

) sdot 2meanxisinΩ

119868 (x))]

=2120575120576(0)mean

xisinΩ119868 (x) (119868 minus 2119867

120576(0)mean

xisinΩ119868 (x))

=120575120576(0)mean

xisinΩ119868 (x) (119868 minusmean

xisinΩ119868 (x))

(E1)

Clearly

minxisinΩ

119868 (x) lt meanxisinΩ

119868 (x) lt maxxisinΩ

119868 (x) (E2)

which implies that

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(E3)

where

Ω1= 119868 (x) lt mean

xisinΩ119868 (x)

Ω2= 119868 (x) gt mean

xisinΩ119868 (x)

(E4)

The proof of Theorem 8 is completed

Acknowledgments

The authors would like to thank the anonymous reviewersfor their valuable comments and suggestions to improve thispaper Besides this work was supported by the Fundamen-tal Research Funds for the Central University Grant no(CDJXS10100006)

References

[1] V Caselles F Catte T Coll and F Dibos ldquoA geometricmodel for active contours in image processingrdquo NumerischeMathematik vol 66 no 1 pp 1ndash31 1993

[2] T Chan M Moelich and B Sandberg ldquoSome recent develop-ments in variational image segmentationrdquo UCLA CAM Reportcam 06-52 2006

[3] L He Z Peng B Everding et al ldquoA comparative study ofdeformable contour methods on medical image segmentationrdquoImage and Vision Computing vol 26 no 2 pp 141ndash163 2008

[4] C Li C Xu C Gui andM D Fox ldquoLevel set evolution withoutre-initialization a new variational formulationrdquo in Proceedingsof the IEEE Computer Society Conference on Computer Visionand Pattern Recognition (CVPR rsquo05) pp 430ndash436 San DiegoCalif USA June 2005

[5] C Li C Xu C Gui and M D Fox ldquoDistance regularized levelset evolution and its application to image segmentationrdquo IEEETransactions on Image Processing vol 19 no 12 pp 3243ndash32542010

[6] B Zhou and C LMu ldquoLevel set evolution for boundary extrac-tion based on a p-Laplace equationrdquo Applied MathematicalModelling vol 34 no 12 pp 3910ndash3916 2010

[7] Y Wang and C He ldquoAdaptive level set evolution starting with aconstant functionrdquoAppliedMathematical Modelling vol 36 no7 pp 3217ndash3228 2012

[8] T F Chan and L A Vese ldquoActive contours without edgesrdquo IEEETransactions on Image Processing vol 10 no 2 pp 266ndash2772001

[9] A Tsai A Yezzi andA SWillsky ldquoCurve evolution implemen-tation of theMumford-Shah functional for image segmentationdenoising interpolation andmagnificationrdquo IEEETransactionson Image Processing vol 10 no 8 pp 1169ndash1186 2001

[10] T Brox and D Cremers ldquoOn the statistical interpretation of thepiecewise smooth Mumford-Shah functionalrdquo in Proceedingsof the International Conference on Scale Space and VariationalMethods in Computer Vision (SSVM rsquo07) pp 203ndash213 IschiaItaly 2007

[11] C Li C Y Kao J CGore andZDing ldquoImplicit active contoursdriven by local binary fitting energyrdquo in Proceedings of the IEEEComputer Society Conference on Computer Vision and PatternRecognition (CVPR rsquo07) pp 1ndash7 Washington DC USA June2007

[12] C Li C Y Kao J C Gore and Z Ding ldquoMinimization ofregion-scalable fitting energy for image segmentationrdquo IEEETransactions on Image Processing vol 17 no 10 pp 1940ndash19492008

[13] S Lankton and A Tannenbaum ldquoLocalizing region-basedactive contoursrdquo IEEE Transactions on Image Processing vol 17no 11 pp 2029ndash2039 2008

[14] L Wang C Li Q Sun D Xia and C Y Kao ldquoActive contoursdriven by local and global intensity fitting energy with applica-tion to brain MR image segmentationrdquo Computerized MedicalImaging and Graphics vol 33 no 7 pp 520ndash531 2009

[15] K Zhang H Song and L Zhang ldquoActive contours driven bylocal image fitting energyrdquo Pattern Recognition vol 43 no 4pp 1199ndash1206 2010

[16] K Zhang L Zhang H Song and W Zhou ldquoActive contourswith selective local or global segmentation a new formulationand level set methodrdquo Image and Vision Computing vol 28 no4 pp 668ndash676 2010

[17] Y Yuan and C He ldquoVariational level set methods for image seg-mentation based on both 1198712 and Sobolev gradientsrdquo NonlinearAnalysis Real World Applications vol 13 no 2 pp 959ndash9662012

[18] B Liu H D Cheng J Huang J Tian X Tang and J LiuldquoProbability density difference-based active contour for ultra-sound image segmentationrdquo Pattern Recognition vol 43 no 6pp 2028ndash2042 2010

[19] D W Shattuck S R Sandor-Leahy K A Schaper D ARottenberg and RM Leahy ldquoMagnetic resonance image tissueclassification using a partial volume modelrdquo NeuroImage vol13 no 5 pp 856ndash876 2001

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 5: Research Article Implicit Active Contour Model with Local ...downloads.hindawi.com/journals/mpe/2013/367086.pdfAcademic Editor: Oluwole Daniel Makinde ... (CV)proposedaregion-basedactivecontourmodel

Mathematical Problems in Engineering 5

(ii) If1198721198990minus 1198730119898 lt 0 then

119865119866

(120601119896

) lt (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= 119860 lt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119866

(120601119896

) gt minus (1198721198990minus 1198730119898)

times(1198921minus 1198922)2

11987244= minus119860 gt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(27)

where

119860 = (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987248 (28)

We provide the proof of Corollary 5 in Appendix C Sim-ilar analysis can prove the following corollary Corollary 6

Corollary 6 Let 1198751199020minus 1198760119901 = 0

(i) If 1198751199020minus 1198760119901 gt 0 then

119865119871

(120601119896

) gt (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= 119861 gt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119871

(120601119896

) lt minus (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= minus119861 lt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(29)

(ii) If 1198751199020minus 1198760119901 lt 0 then

119865119871

(120601119896

) lt (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= 119861 lt 0 in 120596

(119896 ge 1 119896 isin 119911+

)

119865119871

(120601119896

) gt minus (1198751199020minus 1198760119901)

times(1198921minus 1198922)2

11987544= minus119861 gt 0 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(30)

where

119861 = (1198751199020minus 1198760119901)

(1198921minus 1198922)2

11987548 (31)

Theorem 7 If 1206010(x) is a bounded function with (1198721198990minus

1198730119898)(119875119902

0minus 1198760119901) gt 0 then there exists a positive integer 119870

such that for 119896 ge 119870

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

if 1198721198990minus 1198730119898 gt 0 119875119902

0minus 1198760119901 gt 0

(32)

120601119896+1

119894119895lt 0 in 120596

120601119896+1

119894119895gt 0 in Ω 120596

if 1198721198990minus 1198730119898 lt 0 119875119902

0minus 1198760119901 lt 0

(33)

We provide the proof of Theorem 7 in Appendix DThe significance of Theorem 7 is that if we choose 1206010(x)

such that (1198721198990minus 1198730119898)(119875119902

0minus 1198760119901) gt 0 then the zero-level

line of120601(x) startingwith such initial function1206010(x) can finallycome to a unique steady state which separates object from thebackground

Remark Mathematically there does exist 1206010(x) such that(1198721198990minus1198730119898)(119875119902

0minus1198760119901) le 0 in practice however we always

can guarantee that

(1198721198990minus 1198730119898) (119875119902

0minus 1198760119901) gt 0 (34)

34 Discussion of Initial Function Theorem 7 guarantees thatthe proposed model computes a unique steady state regard-less of the initialization that is the convergence of the zero-level line 120601 = 0 is irrespective of the initial functionThis means that we can obtain the same zero-level line inthe steady state if we choose the initial function as a boundedfunction 1206010 with (119872119899

0minus 1198730119898)(119875119902

0minus 1198760119901) gt 0

In applications the initial function 1206010 can be defined

via a simple curve (closed curve or line segment) in imagedomain For example we can choose the initial function asa signed distance to a circle which is widely used in mostof image segmentation models with level set methods Forthe proposed model however we prefer to define the initialfunction 120601

0

(x) as a piecewise constant or constant functionas follows

(1) If the curve 119862 is a closed curve (eg circle or square)then 1206010(x) is defined by

1206010

(x) = +120588 (x) isin inside (119862) minus120588 (x) isin outside (119862)

(35)

where 120588 = 0 is a constant(2) If the curve 119862 is a line segment that partitions the

image domainΩ into two disjoint regionsΩ1andΩ

2

(eg the left and right half regions of image domainΩ respectively) withΩ = Ω

1cup Ω2andΩ

1cap Ω2= Φ

then 1206010(119909 119910) is defined by

1206010

(x) = +120588 (x) isin Ω

1

minus120588 (x) isin Ω2

(36)

where 120588 = 0 is a constant

6 Mathematical Problems in Engineering

(3) We can also define a zero function as follows

1206010

(x) = 0 (37)

Next we prove the fact that with 1206010(x) = 0 120601(x) becomesa sign-changing function and satisfy the condition of (119872119899

1minus

1198731119898)(119875119902

1minus 1198761119901) gt 0 after the first iteration

Theorem 8 If the initial function 1206010(x) = 0 in Ω and

1198880

1(x) = 119891

0

1(x) = 2 sdot mean

xisinΩ119868 (x)

1198880

2(x) = 119891

0

2(x) = 0

(38)

then after the first iteration one has

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(39)

whereΩ1cap Ω2= Φ Ω

1= Φ Ω2

= Φ

We provide the proof of Theorem 8 in Appendix EBy (25) andTheorem 8 we have

1206011

(x) = 120575120576(0) mean

xisinΩ119868 (x)

times (119868 minus meanxisinΩ

119868 (x))

lt 0 x isin Ω1

gt 0 x isin Ω2

(40)

Therefore 1206011(x) became a sign-changing function Thenusing two distinct gray levels of (13) and the above demon-stration we have

(1198721198991minus 1198731119898) (119875119902

1minus 1198761119901)

= (119898 (119872 minus1198731)) (119901 (119875 minus 119876

1))

= 119898119901 (119872 minus1198731) (119875 minus 119876

1) gt 0

(41)

4 Implementation and Experimental Results

41 Implementation In tradition PDE-based methods a cer-tain diffusion term is usually included in the evolution equa-tion to regularize the evolving function but which increasesthe computational cost Recently [16] proposed a novelscheme to regularize the evolving function that is Gaussianfiltering the evolving function after each of iterations Weadopt this scheme to regularize the evolving function 119906 ateach of iteration that is 120601119899 = 119866

120588lowast 120601119899 where 119901 controls the

regularization strength Such regularization procedure hassome advantages over the traditional regularization termsuch as computational efficiency and better smoothing effectsee [16] for more explanations

The main procedures of the proposed algorithm can besummarized as follows

(1) Initialize the evolving function 1206010(x)(2) Compute 119888

119894(120601) and 119891

119894(120601) (119894 = 1 2) using (10)

(3) Evolve the function 120601 according to (11)(4) Regularize the function 120601 with a Gaussian filter that

is 120601119899 = 119866120588lowast 120601119899

(5) Check whether the evolution is finished If not returnto step 2

42 Experimental Results In this section we show experi-mentally that the proposedmodel not only can provide desir-able segmentation results in the presence of intensity inho-mogeneity but also allows for more flexible initialization ofthe contour compared to the RSF and LIF models

In our numerical experiments for our model we choosethe parameters as follows 120576 = 15 for the regularized Diracfunction Δ119905=0025 (time step) ℎ = 1 (space step) 120596 = 01and 119899 = 5 The MATLAB source code of the LIF modelalgorithm is downloaded at httpwww4comppolyueduhksimcslzhangcodeLIFzip and the RSF model algorithmis downloaded at httpwwwengruconnedusimcmlicodeRSF v0 v01rar All experiments were run under MATLAB2010b on a PC with Dual 27 GHz processor

We will utilize two region overlap metrics to evaluatethe performances of the three models quantitatively The twometrics are the ratio of segmentation error (RSE) [18] and thedice similarity coefficient (DSC) [14 18 19] If 119878

1and 1198782rep-

resent a given baseline foreground region (eg true object)and the foreground region found by the model respectivelythen the two metrics are defined as follows

RSE =119873 (1198781 1198782) + 119873 (119878

2 1198781)

119873 (Ω)

DSC =2119873 (119878

1cap 1198782)

119873 (1198781) + 119873 (119878

2)

(42)

where 119873(sdot) indicates the number of pixels in the enclosedregion andΩ is image domainThe closer the DSC value to 1the better the segmentation Since119873(119878

11198782)+119873(119878

21198781) is the

number of the pixelsmistakenly classified by themodel lowerRSE means that there are fewer pixels misclassified that isthe image can be segmented more accurately Thus a perfectsegmentation will give DSC = 1 RSE = 0

In the first example (Figures 1ndash3) we mainly verify thecomputation of a unique steady state of the zero-level line of120601 starting with three types of representative initial functionsthat is signed distance function piecewise constant functionsby (26)-(27) and zero function respectively The top rowof Figure 1 shows the evolution of an active contour (iezero-level line 120601 = 0) with the function 120601 initialized toa signed distance function piecewise constant functions by(26)-(27) with 120588 = 1 and zero function respectively whilethe bottom row shows the corresponding evolution of120601Withsuch initializations the zero-level line of 120601 converges to thesame steady state

In Figure 2 we show that our model has the capabilityof detecting multiple objects or objects with interior holesor blurred edges only starting with a zero function Thecontours (zero-level lines) evolution processes are shown inthe second column to the forth column

Mathematical Problems in Engineering 7

Figure 1 Segmentations of our model for two real images with 120601 initialized as different functions Columns 1 to 4 signed distance functionpiecewise constant functions by (26)-(27) with 120588 = 1 and zero function

Figure 2 Applications of our model to four images (slug cell ventriculus sinister MR and real plane images) The curve evolution processfrom the initial contour (in the first column) to the final contour (in the fourth column) is shown in every row for the corresponding image

8 Mathematical Problems in Engineering

(a) (b) (c) (d)

Figure 3 Comparisons of three models Rows 1ndash3 RSF model LIF model and our model with 1206010= 0 For the RSF and LIF models initial

and final contours are shown in green and red color respectively Columns 1ndash4 the 4 pictures indicate (a) (b) (c) (d) respectively

Figure 4 Segmentation results of both models for a hand phantomUpper row LIF model Lower row our model with 120601

0= 0 Column

2 zoomed view of the narrow parts in hand fingers

We choose the segmentation results of the RSF and LIFmodels as baseline foreground regions and then computeDSC values for the corresponding images The RSE and DSCvalues for the four real images are given inTable 1 fromwhich

Table 1 RSE and DSC for RSF and LIF models and our model

RSF model LIF modelRSE () DSC () RSE () DSC ()

Figure 3(a) 027 9717 299 9891Figure 3(b) 271 9637 012 9913Figure 3(c) 049 8795 045 8806Figure 3(d) 016 9766 076 9891

we can see that the proposed algorithm works as well as theRSF and LIFmodels for imageswith intensity inhomogeneity

The experimental results shown in Figure 4 validate thatour method can also achieve subpixel segmentation accuracyas in [15] As can be seen from Figures 4(b) and 4(d) bothmodels achieve subpixel segmentation accuracy of the fingerboundaries The final contour accurately reflects the truehand shape

5 Conclusion

In this study we propose a new active contour model inte-grating a local intensity fitting (LIF) energy with an auxiliaryglobal intensity fitting (GIF) energy The LIF energy isresponsible for attracting the contour toward object bound-aries and is dominant near object boundaries while the GIFenergy incorporates global image information to improve therobustness to initialization of the contours The proposedmodel can efficiently handle intensity inhomogeneity while

Mathematical Problems in Engineering 9

allowing for more flexible initialization and maintaining thesubpixel accuracy The utility model has the advantages ofallowing for more flexible initialization of the contour andthe capability of detecting multiple objects or objects withinterior holes or blurred edges But [14] has not been imple-mented

Appendices

A Proofs of Theorem 1 and Corollary 2

Proof of Theorem 1 Clearly 120601 gt 0 =Φ and 120601 lt 0 =Φowing to the sign-changing property of 120601(119909 119910) Thus by (3)we have

1198881(120601) =

1198921119899 + 1198922(119873 minus 119899)

119873=(1198921minus 1198922) 119899 + 119892

2119873

119873

1198882(120601) =

1198921(119898 minus 119899) + 119892

2[(119872 minus 119873) minus (119898 minus 119899)]

119872 minus 119873

=(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)

119872 minus119873

(A1)

and so

1198881(120601) minus 119888

2(120601) = ((119872 minus 119873) [(119892

1minus 1198922) 119899 + 119892

2119873]

minus119873 [(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)])

times (119873 (119872 minus119873))minus1

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

(A2)

Therefore in 120596 we get

119903119897119865119866

= (1198881(120601) minus 119888

2(120601)) ((119892

1minus 1198881(120601))119867

120576(120601)

+ (1198921minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)sdot(1198921minus 1198922)

119873 (119872 minus119873)

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

=(119872119899 minus 119873119898) times (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

(A3)

and in Ω 120596 we have

119865119866

(120601) = (1198881(120601) minus 119888

2(120601)) ((119892

2minus 1198881(120601))119867

120576(120601)

+ (1198922minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

times [ minus (1198921minus 1198922)

times (119899 (119872 minus 119873)119867120576(120601)

+119873 (119898 minus 119899) (1 minus 119867120576(120601)))

times (119873 (119872 minus119873))minus1

]

=minus (119872119899 minus 119873119898) (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [(119899 (119872 minus 119873)119867120576(120601)119873 (119898 minus 119899) (1 minus 119867

120576(120601)))]

(A4)

This completes the proof of (14)The last assertion (15) followsclearly from the following fact

(119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576(120601))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A5)

(119899 (119872 minus 119873)119867120576(120601) + 119873 (119898 minus 119899) (1 minus 119867

120576(120601)))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A6)

The proof of Theorem 1 is completed

Now we give the proof of Corollary 2 (i) Since

1198722

= (119873 + (119872 minus119873))2

= 1198732

+ (119872 minus119873)2

+ 2119873 (119872 minus119873)

ge 4119873 (119872 minus119873)

(A7)

by (14) (A5) and (A7) we have

119865119866

(120601) = (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

times (1198732

(119872 minus 119873)2

)minus1

ge (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

(A8)

10 Mathematical Problems in Engineering

and by (14) and (A6)

119865119866

(120601) = minus (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119899 (119872 minus 119873)119867120576

+119873 (119898 minus 119899) (1 minus 119867120576))]

times (1198732

(119872 minus 119873)2

)minus1

le minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(A9)

This completes the proof of (17) and similarly for (18) Theproof is completed

B Proofs of Theorem 3 and Corollary 4

Proof of Theorem 3 By (6) we have

1198911(x) =

int120601gt0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601gt0

119870120590(xminus120585) 119889120585

= (int

120601gt0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

119906gt119868

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601gt0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

times (int

119906gt119868

119870120590(119909minus120585) 119889120585)

minus1

=1198921119902 + 1198922(119876 minus 119902)

119876

(B1)

1198912(x) =

int120601le0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601le0

119870120590(xminus120585) 119889120585

= (int

120601le0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601le0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

=1198921(119901 minus 119902) + 119892

2[(119875 minus 119876) minus (119901 minus 119902)]

119875 minus 119876

(B2)

and so

1198911(x) minus 119891

2(x) =

(1198921minus 1198922) (119875119902 minus 119876119901)

119876 (119875 minus 119876) (B3)

Therefore in 120596 we get

119865119871

(120601) = (1198981(120601) minus 119898

2(120601))

times ((1198921minus 1198981(120601))119867

120576(120601)

+ (1198921minus 1198982(120601)) (1 minus 119867

120576(120601)))

= (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B4)

and in Ω 120596 we have

119865119871

= (1198981(120601) minus 119898

2(120601))

times ((1198922minus 1198981(120601))119867

120576(120601)

+ (1198922minus 1198982(120601)) (1 minus 119867

120576(120601)))

= minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B5)

Mathematical Problems in Engineering 11

This completes the proof of (19) The last assertion (20) fol-lows clearly from the following fact(119875 minus 119876) (119876 minus 119902)119867

120576(120601)

+ 119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))

= 119867120576(120601) (int

Ω

minusint

120601gt0

)119870120590(xminus120585) 119889120585

sdot (int

120601gt0

minusint

120601gt0cap120596)119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601)) int

120601gt0

119870120590(xminus120585) 119889120585

sdot ((intΩ

+int

120601gt0cap120596) minus (int

120596

+int

120601gt0

))119870120590(xminus120585) 119889120585

= 119867120576(120601) int

120601le0

119870120590(xminus120585) 119889120585

sdot (int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot (int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

ge 119867120576(120601) (int

120601le0

119878119889120585) sdot (int

120601gt0(120601gt0cap120596)

119878119889120585)

+ (1 minus 119867120576(120601)) (int

120601gt0

119878119889120585) sdot (int

120601le0(120601le0cap120596)

119878119889120585)

= 119878 [ (119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 minus119873) minus (119898 minus 119899)) (1 minus 119867120576(120601))]

ge 119878

(B6)119902 (119875 minus 119876)119867

120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))

= 119867120576(120601) int

120601gt0cap120596119870120590(xminus120585) 119889120585 sdot int

120601le0

119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot int

120601le0cap120596119870120590(xminus120585) 119889120585

ge 119867120576(120601) (int

119906gt119868cap120596119878119889120585) sdot (int

119906le119868

119878119889120585)

+ (1 minus 119867120576(120601)) (int

119906gt119868

119878119889120585) sdot (int

119906le119868cap120596119878119889120585)

= 119878 [119899 (119872 minus 119873)119867120576(120601) + (1 minus 119867

120576(120601))119873 (119898 minus 119899)]

ge 119878

(B7)

where 119878 = min119870120590(xminus120585) 120585 isin Ω)

Now we give the proof of Corollary 4 (i) Since

1198752

ge 4119875 (119875 minus 119876) (B8)

we have by (19) and (B6)

119865119871

(120601) = (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

ge (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

(B9)

and by (19) and (B7)

119865119871

(120601) = minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

le minus (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(B10)

This completes the proof of (19) and similarly to (20) Theproof is completed

C Proof of Corollary 5

Proof (i) By Corollary 2 (i) we have

119865119866

(1206010

) ge (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0

in 120596

119865119866

(1206010

) le minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

(C1)

At the first iteration by (25) and (C1) we have

1206011

119894119895= 1206010

119894119895+ Δ119905

sdot 119865119866

(1206010

119894119895)

ge 1206010

119894119895+ Δ119905 sdot 120575

120576(1206010

)119860 in 120596

le 1206010

119894119895minus Δ119905 sdot 120575

120576(1206010

)119860 in Ω 120596

(C2)

12 Mathematical Problems in Engineering

or equivalently

1206011

(x) ge 1206010

(x) + Δ119905 sdot 120575120576(1206010

)119860 gt 1206010

(x) in 120596

1206011

(x) le 1206010

(x) minus Δ119905 sdot 120575120576(1206010

)119860 lt 1206010

(x) in Ω 120596

(C3)

which implies that

1198721198991minus 1198731119898 = |Ω|

10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816minus100381610038161003816100381610038161206011

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816) |120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816|120596|

gt |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816) |120596|

= |Ω|1003816100381610038161003816120596 cap 120601

0gt 0

1003816100381610038161003816 minus10038161003816100381610038161206010 gt 0

1003816100381610038161003816 |120596|

= 1198721198990minus 1198730119898 gt 0

(C4)

In a manner similar to the steps used to obtain (26)-(27) and(C1)ndash (C4) we obtain

119865119866

(119868 1206011

) ge (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

gt (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

119865119866

(119868 1206011

) le minus (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

lt minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

1206012

(x) ge 1206011

(x) + Δ119905 sdot 120575120576(1206011

)119860 gt 1206011

(x) in 120596

1206012

(x) le 1206011

(x) minus Δ119905 sdot 120575120576(1206011

)119860 lt 1206011

(x) in Ω 120596

1198721198992minus 1198732119898 gt 119872119899

1minus 1198731119898 gt 119872119899

0minus 1198730119898 gt 0

(C5)

D Proof of Theorem 7

Proof We prove (32) and similarly to (33) By (25)Corollary 5(i) and Corollary 6(i) we get

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

= 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119865119871

(120601119896

119894119895) + 120596119865

119866

(120601119896

119894119895))

ge 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

) + 120596119860 sdot 120575120576(120601119896+1

))

= 120601119896

119894119895+ ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in 120596

(119896 ge 1 119896 isin 119911+

)

le 120601119896

119894119895+ Δ119905

sdot (minus ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

)

+120596119860 sdot 120575120576(120601119896+1

)))

= 120601119896

119894119895minus ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(D1)

which implies that

120601119896+1

119894119895ge 1206010

119894+ ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in 120596

120601119896+1

119894119895le 1206010

119894minus (119896 + 1)119860 ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in Ω 120596

(D2)

Because 1206010(x) is bounded in domain Ω there clearly exists a

positive integer 119870 such that

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

for 119896 ge 119870 (D3)

The proof is completed

Mathematical Problems in Engineering 13

E Proof of Theorem 8

Proof From (11) we have

119865 (0) = 120575120576(0) [120596 ((119868 minus 119868

119866

) sdot 2meanxisinΩ

119868 (x))

+ (1 minus 120596) ((119868 minus 119868119871

) sdot 2meanxisinΩ

119868 (x))]

=2120575120576(0)mean

xisinΩ119868 (x) (119868 minus 2119867

120576(0)mean

xisinΩ119868 (x))

=120575120576(0)mean

xisinΩ119868 (x) (119868 minusmean

xisinΩ119868 (x))

(E1)

Clearly

minxisinΩ

119868 (x) lt meanxisinΩ

119868 (x) lt maxxisinΩ

119868 (x) (E2)

which implies that

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(E3)

where

Ω1= 119868 (x) lt mean

xisinΩ119868 (x)

Ω2= 119868 (x) gt mean

xisinΩ119868 (x)

(E4)

The proof of Theorem 8 is completed

Acknowledgments

The authors would like to thank the anonymous reviewersfor their valuable comments and suggestions to improve thispaper Besides this work was supported by the Fundamen-tal Research Funds for the Central University Grant no(CDJXS10100006)

References

[1] V Caselles F Catte T Coll and F Dibos ldquoA geometricmodel for active contours in image processingrdquo NumerischeMathematik vol 66 no 1 pp 1ndash31 1993

[2] T Chan M Moelich and B Sandberg ldquoSome recent develop-ments in variational image segmentationrdquo UCLA CAM Reportcam 06-52 2006

[3] L He Z Peng B Everding et al ldquoA comparative study ofdeformable contour methods on medical image segmentationrdquoImage and Vision Computing vol 26 no 2 pp 141ndash163 2008

[4] C Li C Xu C Gui andM D Fox ldquoLevel set evolution withoutre-initialization a new variational formulationrdquo in Proceedingsof the IEEE Computer Society Conference on Computer Visionand Pattern Recognition (CVPR rsquo05) pp 430ndash436 San DiegoCalif USA June 2005

[5] C Li C Xu C Gui and M D Fox ldquoDistance regularized levelset evolution and its application to image segmentationrdquo IEEETransactions on Image Processing vol 19 no 12 pp 3243ndash32542010

[6] B Zhou and C LMu ldquoLevel set evolution for boundary extrac-tion based on a p-Laplace equationrdquo Applied MathematicalModelling vol 34 no 12 pp 3910ndash3916 2010

[7] Y Wang and C He ldquoAdaptive level set evolution starting with aconstant functionrdquoAppliedMathematical Modelling vol 36 no7 pp 3217ndash3228 2012

[8] T F Chan and L A Vese ldquoActive contours without edgesrdquo IEEETransactions on Image Processing vol 10 no 2 pp 266ndash2772001

[9] A Tsai A Yezzi andA SWillsky ldquoCurve evolution implemen-tation of theMumford-Shah functional for image segmentationdenoising interpolation andmagnificationrdquo IEEETransactionson Image Processing vol 10 no 8 pp 1169ndash1186 2001

[10] T Brox and D Cremers ldquoOn the statistical interpretation of thepiecewise smooth Mumford-Shah functionalrdquo in Proceedingsof the International Conference on Scale Space and VariationalMethods in Computer Vision (SSVM rsquo07) pp 203ndash213 IschiaItaly 2007

[11] C Li C Y Kao J CGore andZDing ldquoImplicit active contoursdriven by local binary fitting energyrdquo in Proceedings of the IEEEComputer Society Conference on Computer Vision and PatternRecognition (CVPR rsquo07) pp 1ndash7 Washington DC USA June2007

[12] C Li C Y Kao J C Gore and Z Ding ldquoMinimization ofregion-scalable fitting energy for image segmentationrdquo IEEETransactions on Image Processing vol 17 no 10 pp 1940ndash19492008

[13] S Lankton and A Tannenbaum ldquoLocalizing region-basedactive contoursrdquo IEEE Transactions on Image Processing vol 17no 11 pp 2029ndash2039 2008

[14] L Wang C Li Q Sun D Xia and C Y Kao ldquoActive contoursdriven by local and global intensity fitting energy with applica-tion to brain MR image segmentationrdquo Computerized MedicalImaging and Graphics vol 33 no 7 pp 520ndash531 2009

[15] K Zhang H Song and L Zhang ldquoActive contours driven bylocal image fitting energyrdquo Pattern Recognition vol 43 no 4pp 1199ndash1206 2010

[16] K Zhang L Zhang H Song and W Zhou ldquoActive contourswith selective local or global segmentation a new formulationand level set methodrdquo Image and Vision Computing vol 28 no4 pp 668ndash676 2010

[17] Y Yuan and C He ldquoVariational level set methods for image seg-mentation based on both 1198712 and Sobolev gradientsrdquo NonlinearAnalysis Real World Applications vol 13 no 2 pp 959ndash9662012

[18] B Liu H D Cheng J Huang J Tian X Tang and J LiuldquoProbability density difference-based active contour for ultra-sound image segmentationrdquo Pattern Recognition vol 43 no 6pp 2028ndash2042 2010

[19] D W Shattuck S R Sandor-Leahy K A Schaper D ARottenberg and RM Leahy ldquoMagnetic resonance image tissueclassification using a partial volume modelrdquo NeuroImage vol13 no 5 pp 856ndash876 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 6: Research Article Implicit Active Contour Model with Local ...downloads.hindawi.com/journals/mpe/2013/367086.pdfAcademic Editor: Oluwole Daniel Makinde ... (CV)proposedaregion-basedactivecontourmodel

6 Mathematical Problems in Engineering

(3) We can also define a zero function as follows

1206010

(x) = 0 (37)

Next we prove the fact that with 1206010(x) = 0 120601(x) becomesa sign-changing function and satisfy the condition of (119872119899

1minus

1198731119898)(119875119902

1minus 1198761119901) gt 0 after the first iteration

Theorem 8 If the initial function 1206010(x) = 0 in Ω and

1198880

1(x) = 119891

0

1(x) = 2 sdot mean

xisinΩ119868 (x)

1198880

2(x) = 119891

0

2(x) = 0

(38)

then after the first iteration one has

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(39)

whereΩ1cap Ω2= Φ Ω

1= Φ Ω2

= Φ

We provide the proof of Theorem 8 in Appendix EBy (25) andTheorem 8 we have

1206011

(x) = 120575120576(0) mean

xisinΩ119868 (x)

times (119868 minus meanxisinΩ

119868 (x))

lt 0 x isin Ω1

gt 0 x isin Ω2

(40)

Therefore 1206011(x) became a sign-changing function Thenusing two distinct gray levels of (13) and the above demon-stration we have

(1198721198991minus 1198731119898) (119875119902

1minus 1198761119901)

= (119898 (119872 minus1198731)) (119901 (119875 minus 119876

1))

= 119898119901 (119872 minus1198731) (119875 minus 119876

1) gt 0

(41)

4 Implementation and Experimental Results

41 Implementation In tradition PDE-based methods a cer-tain diffusion term is usually included in the evolution equa-tion to regularize the evolving function but which increasesthe computational cost Recently [16] proposed a novelscheme to regularize the evolving function that is Gaussianfiltering the evolving function after each of iterations Weadopt this scheme to regularize the evolving function 119906 ateach of iteration that is 120601119899 = 119866

120588lowast 120601119899 where 119901 controls the

regularization strength Such regularization procedure hassome advantages over the traditional regularization termsuch as computational efficiency and better smoothing effectsee [16] for more explanations

The main procedures of the proposed algorithm can besummarized as follows

(1) Initialize the evolving function 1206010(x)(2) Compute 119888

119894(120601) and 119891

119894(120601) (119894 = 1 2) using (10)

(3) Evolve the function 120601 according to (11)(4) Regularize the function 120601 with a Gaussian filter that

is 120601119899 = 119866120588lowast 120601119899

(5) Check whether the evolution is finished If not returnto step 2

42 Experimental Results In this section we show experi-mentally that the proposedmodel not only can provide desir-able segmentation results in the presence of intensity inho-mogeneity but also allows for more flexible initialization ofthe contour compared to the RSF and LIF models

In our numerical experiments for our model we choosethe parameters as follows 120576 = 15 for the regularized Diracfunction Δ119905=0025 (time step) ℎ = 1 (space step) 120596 = 01and 119899 = 5 The MATLAB source code of the LIF modelalgorithm is downloaded at httpwww4comppolyueduhksimcslzhangcodeLIFzip and the RSF model algorithmis downloaded at httpwwwengruconnedusimcmlicodeRSF v0 v01rar All experiments were run under MATLAB2010b on a PC with Dual 27 GHz processor

We will utilize two region overlap metrics to evaluatethe performances of the three models quantitatively The twometrics are the ratio of segmentation error (RSE) [18] and thedice similarity coefficient (DSC) [14 18 19] If 119878

1and 1198782rep-

resent a given baseline foreground region (eg true object)and the foreground region found by the model respectivelythen the two metrics are defined as follows

RSE =119873 (1198781 1198782) + 119873 (119878

2 1198781)

119873 (Ω)

DSC =2119873 (119878

1cap 1198782)

119873 (1198781) + 119873 (119878

2)

(42)

where 119873(sdot) indicates the number of pixels in the enclosedregion andΩ is image domainThe closer the DSC value to 1the better the segmentation Since119873(119878

11198782)+119873(119878

21198781) is the

number of the pixelsmistakenly classified by themodel lowerRSE means that there are fewer pixels misclassified that isthe image can be segmented more accurately Thus a perfectsegmentation will give DSC = 1 RSE = 0

In the first example (Figures 1ndash3) we mainly verify thecomputation of a unique steady state of the zero-level line of120601 starting with three types of representative initial functionsthat is signed distance function piecewise constant functionsby (26)-(27) and zero function respectively The top rowof Figure 1 shows the evolution of an active contour (iezero-level line 120601 = 0) with the function 120601 initialized toa signed distance function piecewise constant functions by(26)-(27) with 120588 = 1 and zero function respectively whilethe bottom row shows the corresponding evolution of120601Withsuch initializations the zero-level line of 120601 converges to thesame steady state

In Figure 2 we show that our model has the capabilityof detecting multiple objects or objects with interior holesor blurred edges only starting with a zero function Thecontours (zero-level lines) evolution processes are shown inthe second column to the forth column

Mathematical Problems in Engineering 7

Figure 1 Segmentations of our model for two real images with 120601 initialized as different functions Columns 1 to 4 signed distance functionpiecewise constant functions by (26)-(27) with 120588 = 1 and zero function

Figure 2 Applications of our model to four images (slug cell ventriculus sinister MR and real plane images) The curve evolution processfrom the initial contour (in the first column) to the final contour (in the fourth column) is shown in every row for the corresponding image

8 Mathematical Problems in Engineering

(a) (b) (c) (d)

Figure 3 Comparisons of three models Rows 1ndash3 RSF model LIF model and our model with 1206010= 0 For the RSF and LIF models initial

and final contours are shown in green and red color respectively Columns 1ndash4 the 4 pictures indicate (a) (b) (c) (d) respectively

Figure 4 Segmentation results of both models for a hand phantomUpper row LIF model Lower row our model with 120601

0= 0 Column

2 zoomed view of the narrow parts in hand fingers

We choose the segmentation results of the RSF and LIFmodels as baseline foreground regions and then computeDSC values for the corresponding images The RSE and DSCvalues for the four real images are given inTable 1 fromwhich

Table 1 RSE and DSC for RSF and LIF models and our model

RSF model LIF modelRSE () DSC () RSE () DSC ()

Figure 3(a) 027 9717 299 9891Figure 3(b) 271 9637 012 9913Figure 3(c) 049 8795 045 8806Figure 3(d) 016 9766 076 9891

we can see that the proposed algorithm works as well as theRSF and LIFmodels for imageswith intensity inhomogeneity

The experimental results shown in Figure 4 validate thatour method can also achieve subpixel segmentation accuracyas in [15] As can be seen from Figures 4(b) and 4(d) bothmodels achieve subpixel segmentation accuracy of the fingerboundaries The final contour accurately reflects the truehand shape

5 Conclusion

In this study we propose a new active contour model inte-grating a local intensity fitting (LIF) energy with an auxiliaryglobal intensity fitting (GIF) energy The LIF energy isresponsible for attracting the contour toward object bound-aries and is dominant near object boundaries while the GIFenergy incorporates global image information to improve therobustness to initialization of the contours The proposedmodel can efficiently handle intensity inhomogeneity while

Mathematical Problems in Engineering 9

allowing for more flexible initialization and maintaining thesubpixel accuracy The utility model has the advantages ofallowing for more flexible initialization of the contour andthe capability of detecting multiple objects or objects withinterior holes or blurred edges But [14] has not been imple-mented

Appendices

A Proofs of Theorem 1 and Corollary 2

Proof of Theorem 1 Clearly 120601 gt 0 =Φ and 120601 lt 0 =Φowing to the sign-changing property of 120601(119909 119910) Thus by (3)we have

1198881(120601) =

1198921119899 + 1198922(119873 minus 119899)

119873=(1198921minus 1198922) 119899 + 119892

2119873

119873

1198882(120601) =

1198921(119898 minus 119899) + 119892

2[(119872 minus 119873) minus (119898 minus 119899)]

119872 minus 119873

=(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)

119872 minus119873

(A1)

and so

1198881(120601) minus 119888

2(120601) = ((119872 minus 119873) [(119892

1minus 1198922) 119899 + 119892

2119873]

minus119873 [(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)])

times (119873 (119872 minus119873))minus1

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

(A2)

Therefore in 120596 we get

119903119897119865119866

= (1198881(120601) minus 119888

2(120601)) ((119892

1minus 1198881(120601))119867

120576(120601)

+ (1198921minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)sdot(1198921minus 1198922)

119873 (119872 minus119873)

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

=(119872119899 minus 119873119898) times (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

(A3)

and in Ω 120596 we have

119865119866

(120601) = (1198881(120601) minus 119888

2(120601)) ((119892

2minus 1198881(120601))119867

120576(120601)

+ (1198922minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

times [ minus (1198921minus 1198922)

times (119899 (119872 minus 119873)119867120576(120601)

+119873 (119898 minus 119899) (1 minus 119867120576(120601)))

times (119873 (119872 minus119873))minus1

]

=minus (119872119899 minus 119873119898) (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [(119899 (119872 minus 119873)119867120576(120601)119873 (119898 minus 119899) (1 minus 119867

120576(120601)))]

(A4)

This completes the proof of (14)The last assertion (15) followsclearly from the following fact

(119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576(120601))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A5)

(119899 (119872 minus 119873)119867120576(120601) + 119873 (119898 minus 119899) (1 minus 119867

120576(120601)))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A6)

The proof of Theorem 1 is completed

Now we give the proof of Corollary 2 (i) Since

1198722

= (119873 + (119872 minus119873))2

= 1198732

+ (119872 minus119873)2

+ 2119873 (119872 minus119873)

ge 4119873 (119872 minus119873)

(A7)

by (14) (A5) and (A7) we have

119865119866

(120601) = (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

times (1198732

(119872 minus 119873)2

)minus1

ge (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

(A8)

10 Mathematical Problems in Engineering

and by (14) and (A6)

119865119866

(120601) = minus (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119899 (119872 minus 119873)119867120576

+119873 (119898 minus 119899) (1 minus 119867120576))]

times (1198732

(119872 minus 119873)2

)minus1

le minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(A9)

This completes the proof of (17) and similarly for (18) Theproof is completed

B Proofs of Theorem 3 and Corollary 4

Proof of Theorem 3 By (6) we have

1198911(x) =

int120601gt0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601gt0

119870120590(xminus120585) 119889120585

= (int

120601gt0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

119906gt119868

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601gt0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

times (int

119906gt119868

119870120590(119909minus120585) 119889120585)

minus1

=1198921119902 + 1198922(119876 minus 119902)

119876

(B1)

1198912(x) =

int120601le0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601le0

119870120590(xminus120585) 119889120585

= (int

120601le0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601le0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

=1198921(119901 minus 119902) + 119892

2[(119875 minus 119876) minus (119901 minus 119902)]

119875 minus 119876

(B2)

and so

1198911(x) minus 119891

2(x) =

(1198921minus 1198922) (119875119902 minus 119876119901)

119876 (119875 minus 119876) (B3)

Therefore in 120596 we get

119865119871

(120601) = (1198981(120601) minus 119898

2(120601))

times ((1198921minus 1198981(120601))119867

120576(120601)

+ (1198921minus 1198982(120601)) (1 minus 119867

120576(120601)))

= (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B4)

and in Ω 120596 we have

119865119871

= (1198981(120601) minus 119898

2(120601))

times ((1198922minus 1198981(120601))119867

120576(120601)

+ (1198922minus 1198982(120601)) (1 minus 119867

120576(120601)))

= minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B5)

Mathematical Problems in Engineering 11

This completes the proof of (19) The last assertion (20) fol-lows clearly from the following fact(119875 minus 119876) (119876 minus 119902)119867

120576(120601)

+ 119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))

= 119867120576(120601) (int

Ω

minusint

120601gt0

)119870120590(xminus120585) 119889120585

sdot (int

120601gt0

minusint

120601gt0cap120596)119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601)) int

120601gt0

119870120590(xminus120585) 119889120585

sdot ((intΩ

+int

120601gt0cap120596) minus (int

120596

+int

120601gt0

))119870120590(xminus120585) 119889120585

= 119867120576(120601) int

120601le0

119870120590(xminus120585) 119889120585

sdot (int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot (int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

ge 119867120576(120601) (int

120601le0

119878119889120585) sdot (int

120601gt0(120601gt0cap120596)

119878119889120585)

+ (1 minus 119867120576(120601)) (int

120601gt0

119878119889120585) sdot (int

120601le0(120601le0cap120596)

119878119889120585)

= 119878 [ (119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 minus119873) minus (119898 minus 119899)) (1 minus 119867120576(120601))]

ge 119878

(B6)119902 (119875 minus 119876)119867

120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))

= 119867120576(120601) int

120601gt0cap120596119870120590(xminus120585) 119889120585 sdot int

120601le0

119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot int

120601le0cap120596119870120590(xminus120585) 119889120585

ge 119867120576(120601) (int

119906gt119868cap120596119878119889120585) sdot (int

119906le119868

119878119889120585)

+ (1 minus 119867120576(120601)) (int

119906gt119868

119878119889120585) sdot (int

119906le119868cap120596119878119889120585)

= 119878 [119899 (119872 minus 119873)119867120576(120601) + (1 minus 119867

120576(120601))119873 (119898 minus 119899)]

ge 119878

(B7)

where 119878 = min119870120590(xminus120585) 120585 isin Ω)

Now we give the proof of Corollary 4 (i) Since

1198752

ge 4119875 (119875 minus 119876) (B8)

we have by (19) and (B6)

119865119871

(120601) = (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

ge (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

(B9)

and by (19) and (B7)

119865119871

(120601) = minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

le minus (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(B10)

This completes the proof of (19) and similarly to (20) Theproof is completed

C Proof of Corollary 5

Proof (i) By Corollary 2 (i) we have

119865119866

(1206010

) ge (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0

in 120596

119865119866

(1206010

) le minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

(C1)

At the first iteration by (25) and (C1) we have

1206011

119894119895= 1206010

119894119895+ Δ119905

sdot 119865119866

(1206010

119894119895)

ge 1206010

119894119895+ Δ119905 sdot 120575

120576(1206010

)119860 in 120596

le 1206010

119894119895minus Δ119905 sdot 120575

120576(1206010

)119860 in Ω 120596

(C2)

12 Mathematical Problems in Engineering

or equivalently

1206011

(x) ge 1206010

(x) + Δ119905 sdot 120575120576(1206010

)119860 gt 1206010

(x) in 120596

1206011

(x) le 1206010

(x) minus Δ119905 sdot 120575120576(1206010

)119860 lt 1206010

(x) in Ω 120596

(C3)

which implies that

1198721198991minus 1198731119898 = |Ω|

10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816minus100381610038161003816100381610038161206011

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816) |120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816|120596|

gt |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816) |120596|

= |Ω|1003816100381610038161003816120596 cap 120601

0gt 0

1003816100381610038161003816 minus10038161003816100381610038161206010 gt 0

1003816100381610038161003816 |120596|

= 1198721198990minus 1198730119898 gt 0

(C4)

In a manner similar to the steps used to obtain (26)-(27) and(C1)ndash (C4) we obtain

119865119866

(119868 1206011

) ge (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

gt (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

119865119866

(119868 1206011

) le minus (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

lt minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

1206012

(x) ge 1206011

(x) + Δ119905 sdot 120575120576(1206011

)119860 gt 1206011

(x) in 120596

1206012

(x) le 1206011

(x) minus Δ119905 sdot 120575120576(1206011

)119860 lt 1206011

(x) in Ω 120596

1198721198992minus 1198732119898 gt 119872119899

1minus 1198731119898 gt 119872119899

0minus 1198730119898 gt 0

(C5)

D Proof of Theorem 7

Proof We prove (32) and similarly to (33) By (25)Corollary 5(i) and Corollary 6(i) we get

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

= 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119865119871

(120601119896

119894119895) + 120596119865

119866

(120601119896

119894119895))

ge 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

) + 120596119860 sdot 120575120576(120601119896+1

))

= 120601119896

119894119895+ ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in 120596

(119896 ge 1 119896 isin 119911+

)

le 120601119896

119894119895+ Δ119905

sdot (minus ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

)

+120596119860 sdot 120575120576(120601119896+1

)))

= 120601119896

119894119895minus ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(D1)

which implies that

120601119896+1

119894119895ge 1206010

119894+ ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in 120596

120601119896+1

119894119895le 1206010

119894minus (119896 + 1)119860 ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in Ω 120596

(D2)

Because 1206010(x) is bounded in domain Ω there clearly exists a

positive integer 119870 such that

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

for 119896 ge 119870 (D3)

The proof is completed

Mathematical Problems in Engineering 13

E Proof of Theorem 8

Proof From (11) we have

119865 (0) = 120575120576(0) [120596 ((119868 minus 119868

119866

) sdot 2meanxisinΩ

119868 (x))

+ (1 minus 120596) ((119868 minus 119868119871

) sdot 2meanxisinΩ

119868 (x))]

=2120575120576(0)mean

xisinΩ119868 (x) (119868 minus 2119867

120576(0)mean

xisinΩ119868 (x))

=120575120576(0)mean

xisinΩ119868 (x) (119868 minusmean

xisinΩ119868 (x))

(E1)

Clearly

minxisinΩ

119868 (x) lt meanxisinΩ

119868 (x) lt maxxisinΩ

119868 (x) (E2)

which implies that

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(E3)

where

Ω1= 119868 (x) lt mean

xisinΩ119868 (x)

Ω2= 119868 (x) gt mean

xisinΩ119868 (x)

(E4)

The proof of Theorem 8 is completed

Acknowledgments

The authors would like to thank the anonymous reviewersfor their valuable comments and suggestions to improve thispaper Besides this work was supported by the Fundamen-tal Research Funds for the Central University Grant no(CDJXS10100006)

References

[1] V Caselles F Catte T Coll and F Dibos ldquoA geometricmodel for active contours in image processingrdquo NumerischeMathematik vol 66 no 1 pp 1ndash31 1993

[2] T Chan M Moelich and B Sandberg ldquoSome recent develop-ments in variational image segmentationrdquo UCLA CAM Reportcam 06-52 2006

[3] L He Z Peng B Everding et al ldquoA comparative study ofdeformable contour methods on medical image segmentationrdquoImage and Vision Computing vol 26 no 2 pp 141ndash163 2008

[4] C Li C Xu C Gui andM D Fox ldquoLevel set evolution withoutre-initialization a new variational formulationrdquo in Proceedingsof the IEEE Computer Society Conference on Computer Visionand Pattern Recognition (CVPR rsquo05) pp 430ndash436 San DiegoCalif USA June 2005

[5] C Li C Xu C Gui and M D Fox ldquoDistance regularized levelset evolution and its application to image segmentationrdquo IEEETransactions on Image Processing vol 19 no 12 pp 3243ndash32542010

[6] B Zhou and C LMu ldquoLevel set evolution for boundary extrac-tion based on a p-Laplace equationrdquo Applied MathematicalModelling vol 34 no 12 pp 3910ndash3916 2010

[7] Y Wang and C He ldquoAdaptive level set evolution starting with aconstant functionrdquoAppliedMathematical Modelling vol 36 no7 pp 3217ndash3228 2012

[8] T F Chan and L A Vese ldquoActive contours without edgesrdquo IEEETransactions on Image Processing vol 10 no 2 pp 266ndash2772001

[9] A Tsai A Yezzi andA SWillsky ldquoCurve evolution implemen-tation of theMumford-Shah functional for image segmentationdenoising interpolation andmagnificationrdquo IEEETransactionson Image Processing vol 10 no 8 pp 1169ndash1186 2001

[10] T Brox and D Cremers ldquoOn the statistical interpretation of thepiecewise smooth Mumford-Shah functionalrdquo in Proceedingsof the International Conference on Scale Space and VariationalMethods in Computer Vision (SSVM rsquo07) pp 203ndash213 IschiaItaly 2007

[11] C Li C Y Kao J CGore andZDing ldquoImplicit active contoursdriven by local binary fitting energyrdquo in Proceedings of the IEEEComputer Society Conference on Computer Vision and PatternRecognition (CVPR rsquo07) pp 1ndash7 Washington DC USA June2007

[12] C Li C Y Kao J C Gore and Z Ding ldquoMinimization ofregion-scalable fitting energy for image segmentationrdquo IEEETransactions on Image Processing vol 17 no 10 pp 1940ndash19492008

[13] S Lankton and A Tannenbaum ldquoLocalizing region-basedactive contoursrdquo IEEE Transactions on Image Processing vol 17no 11 pp 2029ndash2039 2008

[14] L Wang C Li Q Sun D Xia and C Y Kao ldquoActive contoursdriven by local and global intensity fitting energy with applica-tion to brain MR image segmentationrdquo Computerized MedicalImaging and Graphics vol 33 no 7 pp 520ndash531 2009

[15] K Zhang H Song and L Zhang ldquoActive contours driven bylocal image fitting energyrdquo Pattern Recognition vol 43 no 4pp 1199ndash1206 2010

[16] K Zhang L Zhang H Song and W Zhou ldquoActive contourswith selective local or global segmentation a new formulationand level set methodrdquo Image and Vision Computing vol 28 no4 pp 668ndash676 2010

[17] Y Yuan and C He ldquoVariational level set methods for image seg-mentation based on both 1198712 and Sobolev gradientsrdquo NonlinearAnalysis Real World Applications vol 13 no 2 pp 959ndash9662012

[18] B Liu H D Cheng J Huang J Tian X Tang and J LiuldquoProbability density difference-based active contour for ultra-sound image segmentationrdquo Pattern Recognition vol 43 no 6pp 2028ndash2042 2010

[19] D W Shattuck S R Sandor-Leahy K A Schaper D ARottenberg and RM Leahy ldquoMagnetic resonance image tissueclassification using a partial volume modelrdquo NeuroImage vol13 no 5 pp 856ndash876 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 7: Research Article Implicit Active Contour Model with Local ...downloads.hindawi.com/journals/mpe/2013/367086.pdfAcademic Editor: Oluwole Daniel Makinde ... (CV)proposedaregion-basedactivecontourmodel

Mathematical Problems in Engineering 7

Figure 1 Segmentations of our model for two real images with 120601 initialized as different functions Columns 1 to 4 signed distance functionpiecewise constant functions by (26)-(27) with 120588 = 1 and zero function

Figure 2 Applications of our model to four images (slug cell ventriculus sinister MR and real plane images) The curve evolution processfrom the initial contour (in the first column) to the final contour (in the fourth column) is shown in every row for the corresponding image

8 Mathematical Problems in Engineering

(a) (b) (c) (d)

Figure 3 Comparisons of three models Rows 1ndash3 RSF model LIF model and our model with 1206010= 0 For the RSF and LIF models initial

and final contours are shown in green and red color respectively Columns 1ndash4 the 4 pictures indicate (a) (b) (c) (d) respectively

Figure 4 Segmentation results of both models for a hand phantomUpper row LIF model Lower row our model with 120601

0= 0 Column

2 zoomed view of the narrow parts in hand fingers

We choose the segmentation results of the RSF and LIFmodels as baseline foreground regions and then computeDSC values for the corresponding images The RSE and DSCvalues for the four real images are given inTable 1 fromwhich

Table 1 RSE and DSC for RSF and LIF models and our model

RSF model LIF modelRSE () DSC () RSE () DSC ()

Figure 3(a) 027 9717 299 9891Figure 3(b) 271 9637 012 9913Figure 3(c) 049 8795 045 8806Figure 3(d) 016 9766 076 9891

we can see that the proposed algorithm works as well as theRSF and LIFmodels for imageswith intensity inhomogeneity

The experimental results shown in Figure 4 validate thatour method can also achieve subpixel segmentation accuracyas in [15] As can be seen from Figures 4(b) and 4(d) bothmodels achieve subpixel segmentation accuracy of the fingerboundaries The final contour accurately reflects the truehand shape

5 Conclusion

In this study we propose a new active contour model inte-grating a local intensity fitting (LIF) energy with an auxiliaryglobal intensity fitting (GIF) energy The LIF energy isresponsible for attracting the contour toward object bound-aries and is dominant near object boundaries while the GIFenergy incorporates global image information to improve therobustness to initialization of the contours The proposedmodel can efficiently handle intensity inhomogeneity while

Mathematical Problems in Engineering 9

allowing for more flexible initialization and maintaining thesubpixel accuracy The utility model has the advantages ofallowing for more flexible initialization of the contour andthe capability of detecting multiple objects or objects withinterior holes or blurred edges But [14] has not been imple-mented

Appendices

A Proofs of Theorem 1 and Corollary 2

Proof of Theorem 1 Clearly 120601 gt 0 =Φ and 120601 lt 0 =Φowing to the sign-changing property of 120601(119909 119910) Thus by (3)we have

1198881(120601) =

1198921119899 + 1198922(119873 minus 119899)

119873=(1198921minus 1198922) 119899 + 119892

2119873

119873

1198882(120601) =

1198921(119898 minus 119899) + 119892

2[(119872 minus 119873) minus (119898 minus 119899)]

119872 minus 119873

=(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)

119872 minus119873

(A1)

and so

1198881(120601) minus 119888

2(120601) = ((119872 minus 119873) [(119892

1minus 1198922) 119899 + 119892

2119873]

minus119873 [(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)])

times (119873 (119872 minus119873))minus1

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

(A2)

Therefore in 120596 we get

119903119897119865119866

= (1198881(120601) minus 119888

2(120601)) ((119892

1minus 1198881(120601))119867

120576(120601)

+ (1198921minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)sdot(1198921minus 1198922)

119873 (119872 minus119873)

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

=(119872119899 minus 119873119898) times (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

(A3)

and in Ω 120596 we have

119865119866

(120601) = (1198881(120601) minus 119888

2(120601)) ((119892

2minus 1198881(120601))119867

120576(120601)

+ (1198922minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

times [ minus (1198921minus 1198922)

times (119899 (119872 minus 119873)119867120576(120601)

+119873 (119898 minus 119899) (1 minus 119867120576(120601)))

times (119873 (119872 minus119873))minus1

]

=minus (119872119899 minus 119873119898) (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [(119899 (119872 minus 119873)119867120576(120601)119873 (119898 minus 119899) (1 minus 119867

120576(120601)))]

(A4)

This completes the proof of (14)The last assertion (15) followsclearly from the following fact

(119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576(120601))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A5)

(119899 (119872 minus 119873)119867120576(120601) + 119873 (119898 minus 119899) (1 minus 119867

120576(120601)))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A6)

The proof of Theorem 1 is completed

Now we give the proof of Corollary 2 (i) Since

1198722

= (119873 + (119872 minus119873))2

= 1198732

+ (119872 minus119873)2

+ 2119873 (119872 minus119873)

ge 4119873 (119872 minus119873)

(A7)

by (14) (A5) and (A7) we have

119865119866

(120601) = (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

times (1198732

(119872 minus 119873)2

)minus1

ge (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

(A8)

10 Mathematical Problems in Engineering

and by (14) and (A6)

119865119866

(120601) = minus (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119899 (119872 minus 119873)119867120576

+119873 (119898 minus 119899) (1 minus 119867120576))]

times (1198732

(119872 minus 119873)2

)minus1

le minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(A9)

This completes the proof of (17) and similarly for (18) Theproof is completed

B Proofs of Theorem 3 and Corollary 4

Proof of Theorem 3 By (6) we have

1198911(x) =

int120601gt0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601gt0

119870120590(xminus120585) 119889120585

= (int

120601gt0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

119906gt119868

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601gt0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

times (int

119906gt119868

119870120590(119909minus120585) 119889120585)

minus1

=1198921119902 + 1198922(119876 minus 119902)

119876

(B1)

1198912(x) =

int120601le0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601le0

119870120590(xminus120585) 119889120585

= (int

120601le0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601le0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

=1198921(119901 minus 119902) + 119892

2[(119875 minus 119876) minus (119901 minus 119902)]

119875 minus 119876

(B2)

and so

1198911(x) minus 119891

2(x) =

(1198921minus 1198922) (119875119902 minus 119876119901)

119876 (119875 minus 119876) (B3)

Therefore in 120596 we get

119865119871

(120601) = (1198981(120601) minus 119898

2(120601))

times ((1198921minus 1198981(120601))119867

120576(120601)

+ (1198921minus 1198982(120601)) (1 minus 119867

120576(120601)))

= (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B4)

and in Ω 120596 we have

119865119871

= (1198981(120601) minus 119898

2(120601))

times ((1198922minus 1198981(120601))119867

120576(120601)

+ (1198922minus 1198982(120601)) (1 minus 119867

120576(120601)))

= minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B5)

Mathematical Problems in Engineering 11

This completes the proof of (19) The last assertion (20) fol-lows clearly from the following fact(119875 minus 119876) (119876 minus 119902)119867

120576(120601)

+ 119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))

= 119867120576(120601) (int

Ω

minusint

120601gt0

)119870120590(xminus120585) 119889120585

sdot (int

120601gt0

minusint

120601gt0cap120596)119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601)) int

120601gt0

119870120590(xminus120585) 119889120585

sdot ((intΩ

+int

120601gt0cap120596) minus (int

120596

+int

120601gt0

))119870120590(xminus120585) 119889120585

= 119867120576(120601) int

120601le0

119870120590(xminus120585) 119889120585

sdot (int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot (int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

ge 119867120576(120601) (int

120601le0

119878119889120585) sdot (int

120601gt0(120601gt0cap120596)

119878119889120585)

+ (1 minus 119867120576(120601)) (int

120601gt0

119878119889120585) sdot (int

120601le0(120601le0cap120596)

119878119889120585)

= 119878 [ (119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 minus119873) minus (119898 minus 119899)) (1 minus 119867120576(120601))]

ge 119878

(B6)119902 (119875 minus 119876)119867

120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))

= 119867120576(120601) int

120601gt0cap120596119870120590(xminus120585) 119889120585 sdot int

120601le0

119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot int

120601le0cap120596119870120590(xminus120585) 119889120585

ge 119867120576(120601) (int

119906gt119868cap120596119878119889120585) sdot (int

119906le119868

119878119889120585)

+ (1 minus 119867120576(120601)) (int

119906gt119868

119878119889120585) sdot (int

119906le119868cap120596119878119889120585)

= 119878 [119899 (119872 minus 119873)119867120576(120601) + (1 minus 119867

120576(120601))119873 (119898 minus 119899)]

ge 119878

(B7)

where 119878 = min119870120590(xminus120585) 120585 isin Ω)

Now we give the proof of Corollary 4 (i) Since

1198752

ge 4119875 (119875 minus 119876) (B8)

we have by (19) and (B6)

119865119871

(120601) = (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

ge (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

(B9)

and by (19) and (B7)

119865119871

(120601) = minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

le minus (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(B10)

This completes the proof of (19) and similarly to (20) Theproof is completed

C Proof of Corollary 5

Proof (i) By Corollary 2 (i) we have

119865119866

(1206010

) ge (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0

in 120596

119865119866

(1206010

) le minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

(C1)

At the first iteration by (25) and (C1) we have

1206011

119894119895= 1206010

119894119895+ Δ119905

sdot 119865119866

(1206010

119894119895)

ge 1206010

119894119895+ Δ119905 sdot 120575

120576(1206010

)119860 in 120596

le 1206010

119894119895minus Δ119905 sdot 120575

120576(1206010

)119860 in Ω 120596

(C2)

12 Mathematical Problems in Engineering

or equivalently

1206011

(x) ge 1206010

(x) + Δ119905 sdot 120575120576(1206010

)119860 gt 1206010

(x) in 120596

1206011

(x) le 1206010

(x) minus Δ119905 sdot 120575120576(1206010

)119860 lt 1206010

(x) in Ω 120596

(C3)

which implies that

1198721198991minus 1198731119898 = |Ω|

10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816minus100381610038161003816100381610038161206011

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816) |120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816|120596|

gt |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816) |120596|

= |Ω|1003816100381610038161003816120596 cap 120601

0gt 0

1003816100381610038161003816 minus10038161003816100381610038161206010 gt 0

1003816100381610038161003816 |120596|

= 1198721198990minus 1198730119898 gt 0

(C4)

In a manner similar to the steps used to obtain (26)-(27) and(C1)ndash (C4) we obtain

119865119866

(119868 1206011

) ge (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

gt (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

119865119866

(119868 1206011

) le minus (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

lt minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

1206012

(x) ge 1206011

(x) + Δ119905 sdot 120575120576(1206011

)119860 gt 1206011

(x) in 120596

1206012

(x) le 1206011

(x) minus Δ119905 sdot 120575120576(1206011

)119860 lt 1206011

(x) in Ω 120596

1198721198992minus 1198732119898 gt 119872119899

1minus 1198731119898 gt 119872119899

0minus 1198730119898 gt 0

(C5)

D Proof of Theorem 7

Proof We prove (32) and similarly to (33) By (25)Corollary 5(i) and Corollary 6(i) we get

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

= 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119865119871

(120601119896

119894119895) + 120596119865

119866

(120601119896

119894119895))

ge 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

) + 120596119860 sdot 120575120576(120601119896+1

))

= 120601119896

119894119895+ ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in 120596

(119896 ge 1 119896 isin 119911+

)

le 120601119896

119894119895+ Δ119905

sdot (minus ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

)

+120596119860 sdot 120575120576(120601119896+1

)))

= 120601119896

119894119895minus ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(D1)

which implies that

120601119896+1

119894119895ge 1206010

119894+ ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in 120596

120601119896+1

119894119895le 1206010

119894minus (119896 + 1)119860 ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in Ω 120596

(D2)

Because 1206010(x) is bounded in domain Ω there clearly exists a

positive integer 119870 such that

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

for 119896 ge 119870 (D3)

The proof is completed

Mathematical Problems in Engineering 13

E Proof of Theorem 8

Proof From (11) we have

119865 (0) = 120575120576(0) [120596 ((119868 minus 119868

119866

) sdot 2meanxisinΩ

119868 (x))

+ (1 minus 120596) ((119868 minus 119868119871

) sdot 2meanxisinΩ

119868 (x))]

=2120575120576(0)mean

xisinΩ119868 (x) (119868 minus 2119867

120576(0)mean

xisinΩ119868 (x))

=120575120576(0)mean

xisinΩ119868 (x) (119868 minusmean

xisinΩ119868 (x))

(E1)

Clearly

minxisinΩ

119868 (x) lt meanxisinΩ

119868 (x) lt maxxisinΩ

119868 (x) (E2)

which implies that

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(E3)

where

Ω1= 119868 (x) lt mean

xisinΩ119868 (x)

Ω2= 119868 (x) gt mean

xisinΩ119868 (x)

(E4)

The proof of Theorem 8 is completed

Acknowledgments

The authors would like to thank the anonymous reviewersfor their valuable comments and suggestions to improve thispaper Besides this work was supported by the Fundamen-tal Research Funds for the Central University Grant no(CDJXS10100006)

References

[1] V Caselles F Catte T Coll and F Dibos ldquoA geometricmodel for active contours in image processingrdquo NumerischeMathematik vol 66 no 1 pp 1ndash31 1993

[2] T Chan M Moelich and B Sandberg ldquoSome recent develop-ments in variational image segmentationrdquo UCLA CAM Reportcam 06-52 2006

[3] L He Z Peng B Everding et al ldquoA comparative study ofdeformable contour methods on medical image segmentationrdquoImage and Vision Computing vol 26 no 2 pp 141ndash163 2008

[4] C Li C Xu C Gui andM D Fox ldquoLevel set evolution withoutre-initialization a new variational formulationrdquo in Proceedingsof the IEEE Computer Society Conference on Computer Visionand Pattern Recognition (CVPR rsquo05) pp 430ndash436 San DiegoCalif USA June 2005

[5] C Li C Xu C Gui and M D Fox ldquoDistance regularized levelset evolution and its application to image segmentationrdquo IEEETransactions on Image Processing vol 19 no 12 pp 3243ndash32542010

[6] B Zhou and C LMu ldquoLevel set evolution for boundary extrac-tion based on a p-Laplace equationrdquo Applied MathematicalModelling vol 34 no 12 pp 3910ndash3916 2010

[7] Y Wang and C He ldquoAdaptive level set evolution starting with aconstant functionrdquoAppliedMathematical Modelling vol 36 no7 pp 3217ndash3228 2012

[8] T F Chan and L A Vese ldquoActive contours without edgesrdquo IEEETransactions on Image Processing vol 10 no 2 pp 266ndash2772001

[9] A Tsai A Yezzi andA SWillsky ldquoCurve evolution implemen-tation of theMumford-Shah functional for image segmentationdenoising interpolation andmagnificationrdquo IEEETransactionson Image Processing vol 10 no 8 pp 1169ndash1186 2001

[10] T Brox and D Cremers ldquoOn the statistical interpretation of thepiecewise smooth Mumford-Shah functionalrdquo in Proceedingsof the International Conference on Scale Space and VariationalMethods in Computer Vision (SSVM rsquo07) pp 203ndash213 IschiaItaly 2007

[11] C Li C Y Kao J CGore andZDing ldquoImplicit active contoursdriven by local binary fitting energyrdquo in Proceedings of the IEEEComputer Society Conference on Computer Vision and PatternRecognition (CVPR rsquo07) pp 1ndash7 Washington DC USA June2007

[12] C Li C Y Kao J C Gore and Z Ding ldquoMinimization ofregion-scalable fitting energy for image segmentationrdquo IEEETransactions on Image Processing vol 17 no 10 pp 1940ndash19492008

[13] S Lankton and A Tannenbaum ldquoLocalizing region-basedactive contoursrdquo IEEE Transactions on Image Processing vol 17no 11 pp 2029ndash2039 2008

[14] L Wang C Li Q Sun D Xia and C Y Kao ldquoActive contoursdriven by local and global intensity fitting energy with applica-tion to brain MR image segmentationrdquo Computerized MedicalImaging and Graphics vol 33 no 7 pp 520ndash531 2009

[15] K Zhang H Song and L Zhang ldquoActive contours driven bylocal image fitting energyrdquo Pattern Recognition vol 43 no 4pp 1199ndash1206 2010

[16] K Zhang L Zhang H Song and W Zhou ldquoActive contourswith selective local or global segmentation a new formulationand level set methodrdquo Image and Vision Computing vol 28 no4 pp 668ndash676 2010

[17] Y Yuan and C He ldquoVariational level set methods for image seg-mentation based on both 1198712 and Sobolev gradientsrdquo NonlinearAnalysis Real World Applications vol 13 no 2 pp 959ndash9662012

[18] B Liu H D Cheng J Huang J Tian X Tang and J LiuldquoProbability density difference-based active contour for ultra-sound image segmentationrdquo Pattern Recognition vol 43 no 6pp 2028ndash2042 2010

[19] D W Shattuck S R Sandor-Leahy K A Schaper D ARottenberg and RM Leahy ldquoMagnetic resonance image tissueclassification using a partial volume modelrdquo NeuroImage vol13 no 5 pp 856ndash876 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 8: Research Article Implicit Active Contour Model with Local ...downloads.hindawi.com/journals/mpe/2013/367086.pdfAcademic Editor: Oluwole Daniel Makinde ... (CV)proposedaregion-basedactivecontourmodel

8 Mathematical Problems in Engineering

(a) (b) (c) (d)

Figure 3 Comparisons of three models Rows 1ndash3 RSF model LIF model and our model with 1206010= 0 For the RSF and LIF models initial

and final contours are shown in green and red color respectively Columns 1ndash4 the 4 pictures indicate (a) (b) (c) (d) respectively

Figure 4 Segmentation results of both models for a hand phantomUpper row LIF model Lower row our model with 120601

0= 0 Column

2 zoomed view of the narrow parts in hand fingers

We choose the segmentation results of the RSF and LIFmodels as baseline foreground regions and then computeDSC values for the corresponding images The RSE and DSCvalues for the four real images are given inTable 1 fromwhich

Table 1 RSE and DSC for RSF and LIF models and our model

RSF model LIF modelRSE () DSC () RSE () DSC ()

Figure 3(a) 027 9717 299 9891Figure 3(b) 271 9637 012 9913Figure 3(c) 049 8795 045 8806Figure 3(d) 016 9766 076 9891

we can see that the proposed algorithm works as well as theRSF and LIFmodels for imageswith intensity inhomogeneity

The experimental results shown in Figure 4 validate thatour method can also achieve subpixel segmentation accuracyas in [15] As can be seen from Figures 4(b) and 4(d) bothmodels achieve subpixel segmentation accuracy of the fingerboundaries The final contour accurately reflects the truehand shape

5 Conclusion

In this study we propose a new active contour model inte-grating a local intensity fitting (LIF) energy with an auxiliaryglobal intensity fitting (GIF) energy The LIF energy isresponsible for attracting the contour toward object bound-aries and is dominant near object boundaries while the GIFenergy incorporates global image information to improve therobustness to initialization of the contours The proposedmodel can efficiently handle intensity inhomogeneity while

Mathematical Problems in Engineering 9

allowing for more flexible initialization and maintaining thesubpixel accuracy The utility model has the advantages ofallowing for more flexible initialization of the contour andthe capability of detecting multiple objects or objects withinterior holes or blurred edges But [14] has not been imple-mented

Appendices

A Proofs of Theorem 1 and Corollary 2

Proof of Theorem 1 Clearly 120601 gt 0 =Φ and 120601 lt 0 =Φowing to the sign-changing property of 120601(119909 119910) Thus by (3)we have

1198881(120601) =

1198921119899 + 1198922(119873 minus 119899)

119873=(1198921minus 1198922) 119899 + 119892

2119873

119873

1198882(120601) =

1198921(119898 minus 119899) + 119892

2[(119872 minus 119873) minus (119898 minus 119899)]

119872 minus 119873

=(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)

119872 minus119873

(A1)

and so

1198881(120601) minus 119888

2(120601) = ((119872 minus 119873) [(119892

1minus 1198922) 119899 + 119892

2119873]

minus119873 [(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)])

times (119873 (119872 minus119873))minus1

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

(A2)

Therefore in 120596 we get

119903119897119865119866

= (1198881(120601) minus 119888

2(120601)) ((119892

1minus 1198881(120601))119867

120576(120601)

+ (1198921minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)sdot(1198921minus 1198922)

119873 (119872 minus119873)

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

=(119872119899 minus 119873119898) times (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

(A3)

and in Ω 120596 we have

119865119866

(120601) = (1198881(120601) minus 119888

2(120601)) ((119892

2minus 1198881(120601))119867

120576(120601)

+ (1198922minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

times [ minus (1198921minus 1198922)

times (119899 (119872 minus 119873)119867120576(120601)

+119873 (119898 minus 119899) (1 minus 119867120576(120601)))

times (119873 (119872 minus119873))minus1

]

=minus (119872119899 minus 119873119898) (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [(119899 (119872 minus 119873)119867120576(120601)119873 (119898 minus 119899) (1 minus 119867

120576(120601)))]

(A4)

This completes the proof of (14)The last assertion (15) followsclearly from the following fact

(119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576(120601))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A5)

(119899 (119872 minus 119873)119867120576(120601) + 119873 (119898 minus 119899) (1 minus 119867

120576(120601)))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A6)

The proof of Theorem 1 is completed

Now we give the proof of Corollary 2 (i) Since

1198722

= (119873 + (119872 minus119873))2

= 1198732

+ (119872 minus119873)2

+ 2119873 (119872 minus119873)

ge 4119873 (119872 minus119873)

(A7)

by (14) (A5) and (A7) we have

119865119866

(120601) = (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

times (1198732

(119872 minus 119873)2

)minus1

ge (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

(A8)

10 Mathematical Problems in Engineering

and by (14) and (A6)

119865119866

(120601) = minus (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119899 (119872 minus 119873)119867120576

+119873 (119898 minus 119899) (1 minus 119867120576))]

times (1198732

(119872 minus 119873)2

)minus1

le minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(A9)

This completes the proof of (17) and similarly for (18) Theproof is completed

B Proofs of Theorem 3 and Corollary 4

Proof of Theorem 3 By (6) we have

1198911(x) =

int120601gt0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601gt0

119870120590(xminus120585) 119889120585

= (int

120601gt0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

119906gt119868

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601gt0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

times (int

119906gt119868

119870120590(119909minus120585) 119889120585)

minus1

=1198921119902 + 1198922(119876 minus 119902)

119876

(B1)

1198912(x) =

int120601le0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601le0

119870120590(xminus120585) 119889120585

= (int

120601le0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601le0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

=1198921(119901 minus 119902) + 119892

2[(119875 minus 119876) minus (119901 minus 119902)]

119875 minus 119876

(B2)

and so

1198911(x) minus 119891

2(x) =

(1198921minus 1198922) (119875119902 minus 119876119901)

119876 (119875 minus 119876) (B3)

Therefore in 120596 we get

119865119871

(120601) = (1198981(120601) minus 119898

2(120601))

times ((1198921minus 1198981(120601))119867

120576(120601)

+ (1198921minus 1198982(120601)) (1 minus 119867

120576(120601)))

= (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B4)

and in Ω 120596 we have

119865119871

= (1198981(120601) minus 119898

2(120601))

times ((1198922minus 1198981(120601))119867

120576(120601)

+ (1198922minus 1198982(120601)) (1 minus 119867

120576(120601)))

= minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B5)

Mathematical Problems in Engineering 11

This completes the proof of (19) The last assertion (20) fol-lows clearly from the following fact(119875 minus 119876) (119876 minus 119902)119867

120576(120601)

+ 119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))

= 119867120576(120601) (int

Ω

minusint

120601gt0

)119870120590(xminus120585) 119889120585

sdot (int

120601gt0

minusint

120601gt0cap120596)119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601)) int

120601gt0

119870120590(xminus120585) 119889120585

sdot ((intΩ

+int

120601gt0cap120596) minus (int

120596

+int

120601gt0

))119870120590(xminus120585) 119889120585

= 119867120576(120601) int

120601le0

119870120590(xminus120585) 119889120585

sdot (int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot (int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

ge 119867120576(120601) (int

120601le0

119878119889120585) sdot (int

120601gt0(120601gt0cap120596)

119878119889120585)

+ (1 minus 119867120576(120601)) (int

120601gt0

119878119889120585) sdot (int

120601le0(120601le0cap120596)

119878119889120585)

= 119878 [ (119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 minus119873) minus (119898 minus 119899)) (1 minus 119867120576(120601))]

ge 119878

(B6)119902 (119875 minus 119876)119867

120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))

= 119867120576(120601) int

120601gt0cap120596119870120590(xminus120585) 119889120585 sdot int

120601le0

119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot int

120601le0cap120596119870120590(xminus120585) 119889120585

ge 119867120576(120601) (int

119906gt119868cap120596119878119889120585) sdot (int

119906le119868

119878119889120585)

+ (1 minus 119867120576(120601)) (int

119906gt119868

119878119889120585) sdot (int

119906le119868cap120596119878119889120585)

= 119878 [119899 (119872 minus 119873)119867120576(120601) + (1 minus 119867

120576(120601))119873 (119898 minus 119899)]

ge 119878

(B7)

where 119878 = min119870120590(xminus120585) 120585 isin Ω)

Now we give the proof of Corollary 4 (i) Since

1198752

ge 4119875 (119875 minus 119876) (B8)

we have by (19) and (B6)

119865119871

(120601) = (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

ge (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

(B9)

and by (19) and (B7)

119865119871

(120601) = minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

le minus (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(B10)

This completes the proof of (19) and similarly to (20) Theproof is completed

C Proof of Corollary 5

Proof (i) By Corollary 2 (i) we have

119865119866

(1206010

) ge (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0

in 120596

119865119866

(1206010

) le minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

(C1)

At the first iteration by (25) and (C1) we have

1206011

119894119895= 1206010

119894119895+ Δ119905

sdot 119865119866

(1206010

119894119895)

ge 1206010

119894119895+ Δ119905 sdot 120575

120576(1206010

)119860 in 120596

le 1206010

119894119895minus Δ119905 sdot 120575

120576(1206010

)119860 in Ω 120596

(C2)

12 Mathematical Problems in Engineering

or equivalently

1206011

(x) ge 1206010

(x) + Δ119905 sdot 120575120576(1206010

)119860 gt 1206010

(x) in 120596

1206011

(x) le 1206010

(x) minus Δ119905 sdot 120575120576(1206010

)119860 lt 1206010

(x) in Ω 120596

(C3)

which implies that

1198721198991minus 1198731119898 = |Ω|

10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816minus100381610038161003816100381610038161206011

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816) |120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816|120596|

gt |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816) |120596|

= |Ω|1003816100381610038161003816120596 cap 120601

0gt 0

1003816100381610038161003816 minus10038161003816100381610038161206010 gt 0

1003816100381610038161003816 |120596|

= 1198721198990minus 1198730119898 gt 0

(C4)

In a manner similar to the steps used to obtain (26)-(27) and(C1)ndash (C4) we obtain

119865119866

(119868 1206011

) ge (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

gt (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

119865119866

(119868 1206011

) le minus (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

lt minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

1206012

(x) ge 1206011

(x) + Δ119905 sdot 120575120576(1206011

)119860 gt 1206011

(x) in 120596

1206012

(x) le 1206011

(x) minus Δ119905 sdot 120575120576(1206011

)119860 lt 1206011

(x) in Ω 120596

1198721198992minus 1198732119898 gt 119872119899

1minus 1198731119898 gt 119872119899

0minus 1198730119898 gt 0

(C5)

D Proof of Theorem 7

Proof We prove (32) and similarly to (33) By (25)Corollary 5(i) and Corollary 6(i) we get

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

= 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119865119871

(120601119896

119894119895) + 120596119865

119866

(120601119896

119894119895))

ge 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

) + 120596119860 sdot 120575120576(120601119896+1

))

= 120601119896

119894119895+ ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in 120596

(119896 ge 1 119896 isin 119911+

)

le 120601119896

119894119895+ Δ119905

sdot (minus ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

)

+120596119860 sdot 120575120576(120601119896+1

)))

= 120601119896

119894119895minus ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(D1)

which implies that

120601119896+1

119894119895ge 1206010

119894+ ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in 120596

120601119896+1

119894119895le 1206010

119894minus (119896 + 1)119860 ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in Ω 120596

(D2)

Because 1206010(x) is bounded in domain Ω there clearly exists a

positive integer 119870 such that

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

for 119896 ge 119870 (D3)

The proof is completed

Mathematical Problems in Engineering 13

E Proof of Theorem 8

Proof From (11) we have

119865 (0) = 120575120576(0) [120596 ((119868 minus 119868

119866

) sdot 2meanxisinΩ

119868 (x))

+ (1 minus 120596) ((119868 minus 119868119871

) sdot 2meanxisinΩ

119868 (x))]

=2120575120576(0)mean

xisinΩ119868 (x) (119868 minus 2119867

120576(0)mean

xisinΩ119868 (x))

=120575120576(0)mean

xisinΩ119868 (x) (119868 minusmean

xisinΩ119868 (x))

(E1)

Clearly

minxisinΩ

119868 (x) lt meanxisinΩ

119868 (x) lt maxxisinΩ

119868 (x) (E2)

which implies that

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(E3)

where

Ω1= 119868 (x) lt mean

xisinΩ119868 (x)

Ω2= 119868 (x) gt mean

xisinΩ119868 (x)

(E4)

The proof of Theorem 8 is completed

Acknowledgments

The authors would like to thank the anonymous reviewersfor their valuable comments and suggestions to improve thispaper Besides this work was supported by the Fundamen-tal Research Funds for the Central University Grant no(CDJXS10100006)

References

[1] V Caselles F Catte T Coll and F Dibos ldquoA geometricmodel for active contours in image processingrdquo NumerischeMathematik vol 66 no 1 pp 1ndash31 1993

[2] T Chan M Moelich and B Sandberg ldquoSome recent develop-ments in variational image segmentationrdquo UCLA CAM Reportcam 06-52 2006

[3] L He Z Peng B Everding et al ldquoA comparative study ofdeformable contour methods on medical image segmentationrdquoImage and Vision Computing vol 26 no 2 pp 141ndash163 2008

[4] C Li C Xu C Gui andM D Fox ldquoLevel set evolution withoutre-initialization a new variational formulationrdquo in Proceedingsof the IEEE Computer Society Conference on Computer Visionand Pattern Recognition (CVPR rsquo05) pp 430ndash436 San DiegoCalif USA June 2005

[5] C Li C Xu C Gui and M D Fox ldquoDistance regularized levelset evolution and its application to image segmentationrdquo IEEETransactions on Image Processing vol 19 no 12 pp 3243ndash32542010

[6] B Zhou and C LMu ldquoLevel set evolution for boundary extrac-tion based on a p-Laplace equationrdquo Applied MathematicalModelling vol 34 no 12 pp 3910ndash3916 2010

[7] Y Wang and C He ldquoAdaptive level set evolution starting with aconstant functionrdquoAppliedMathematical Modelling vol 36 no7 pp 3217ndash3228 2012

[8] T F Chan and L A Vese ldquoActive contours without edgesrdquo IEEETransactions on Image Processing vol 10 no 2 pp 266ndash2772001

[9] A Tsai A Yezzi andA SWillsky ldquoCurve evolution implemen-tation of theMumford-Shah functional for image segmentationdenoising interpolation andmagnificationrdquo IEEETransactionson Image Processing vol 10 no 8 pp 1169ndash1186 2001

[10] T Brox and D Cremers ldquoOn the statistical interpretation of thepiecewise smooth Mumford-Shah functionalrdquo in Proceedingsof the International Conference on Scale Space and VariationalMethods in Computer Vision (SSVM rsquo07) pp 203ndash213 IschiaItaly 2007

[11] C Li C Y Kao J CGore andZDing ldquoImplicit active contoursdriven by local binary fitting energyrdquo in Proceedings of the IEEEComputer Society Conference on Computer Vision and PatternRecognition (CVPR rsquo07) pp 1ndash7 Washington DC USA June2007

[12] C Li C Y Kao J C Gore and Z Ding ldquoMinimization ofregion-scalable fitting energy for image segmentationrdquo IEEETransactions on Image Processing vol 17 no 10 pp 1940ndash19492008

[13] S Lankton and A Tannenbaum ldquoLocalizing region-basedactive contoursrdquo IEEE Transactions on Image Processing vol 17no 11 pp 2029ndash2039 2008

[14] L Wang C Li Q Sun D Xia and C Y Kao ldquoActive contoursdriven by local and global intensity fitting energy with applica-tion to brain MR image segmentationrdquo Computerized MedicalImaging and Graphics vol 33 no 7 pp 520ndash531 2009

[15] K Zhang H Song and L Zhang ldquoActive contours driven bylocal image fitting energyrdquo Pattern Recognition vol 43 no 4pp 1199ndash1206 2010

[16] K Zhang L Zhang H Song and W Zhou ldquoActive contourswith selective local or global segmentation a new formulationand level set methodrdquo Image and Vision Computing vol 28 no4 pp 668ndash676 2010

[17] Y Yuan and C He ldquoVariational level set methods for image seg-mentation based on both 1198712 and Sobolev gradientsrdquo NonlinearAnalysis Real World Applications vol 13 no 2 pp 959ndash9662012

[18] B Liu H D Cheng J Huang J Tian X Tang and J LiuldquoProbability density difference-based active contour for ultra-sound image segmentationrdquo Pattern Recognition vol 43 no 6pp 2028ndash2042 2010

[19] D W Shattuck S R Sandor-Leahy K A Schaper D ARottenberg and RM Leahy ldquoMagnetic resonance image tissueclassification using a partial volume modelrdquo NeuroImage vol13 no 5 pp 856ndash876 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 9: Research Article Implicit Active Contour Model with Local ...downloads.hindawi.com/journals/mpe/2013/367086.pdfAcademic Editor: Oluwole Daniel Makinde ... (CV)proposedaregion-basedactivecontourmodel

Mathematical Problems in Engineering 9

allowing for more flexible initialization and maintaining thesubpixel accuracy The utility model has the advantages ofallowing for more flexible initialization of the contour andthe capability of detecting multiple objects or objects withinterior holes or blurred edges But [14] has not been imple-mented

Appendices

A Proofs of Theorem 1 and Corollary 2

Proof of Theorem 1 Clearly 120601 gt 0 =Φ and 120601 lt 0 =Φowing to the sign-changing property of 120601(119909 119910) Thus by (3)we have

1198881(120601) =

1198921119899 + 1198922(119873 minus 119899)

119873=(1198921minus 1198922) 119899 + 119892

2119873

119873

1198882(120601) =

1198921(119898 minus 119899) + 119892

2[(119872 minus 119873) minus (119898 minus 119899)]

119872 minus 119873

=(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)

119872 minus119873

(A1)

and so

1198881(120601) minus 119888

2(120601) = ((119872 minus 119873) [(119892

1minus 1198922) 119899 + 119892

2119873]

minus119873 [(1198921minus 1198922) (119898 minus 119899) + 119892

2(119872 minus 119873)])

times (119873 (119872 minus119873))minus1

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

(A2)

Therefore in 120596 we get

119903119897119865119866

= (1198881(120601) minus 119888

2(120601)) ((119892

1minus 1198881(120601))119867

120576(120601)

+ (1198921minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)sdot(1198921minus 1198922)

119873 (119872 minus119873)

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

=(119872119899 minus 119873119898) times (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

(A3)

and in Ω 120596 we have

119865119866

(120601) = (1198881(120601) minus 119888

2(120601)) ((119892

2minus 1198881(120601))119867

120576(120601)

+ (1198922minus 1198882(120601)) (1 minus 119867

120576(120601)))

=(1198921minus 1198922) (119872119899 minus 119873119898)

119873 (119872 minus119873)

times [ minus (1198921minus 1198922)

times (119899 (119872 minus 119873)119867120576(120601)

+119873 (119898 minus 119899) (1 minus 119867120576(120601)))

times (119873 (119872 minus119873))minus1

]

=minus (119872119899 minus 119873119898) (119892

1minus 1198922)2

1198732(119872 minus 119873)2

times [(119899 (119872 minus 119873)119867120576(120601)119873 (119898 minus 119899) (1 minus 119867

120576(120601)))]

(A4)

This completes the proof of (14)The last assertion (15) followsclearly from the following fact

(119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576(120601))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A5)

(119899 (119872 minus 119873)119867120576(120601) + 119873 (119898 minus 119899) (1 minus 119867

120576(120601)))

ge 119867120576(120601) + (1 minus 119867

120576(120601)) ge 1

(A6)

The proof of Theorem 1 is completed

Now we give the proof of Corollary 2 (i) Since

1198722

= (119873 + (119872 minus119873))2

= 1198732

+ (119872 minus119873)2

+ 2119873 (119872 minus119873)

ge 4119873 (119872 minus119873)

(A7)

by (14) (A5) and (A7) we have

119865119866

(120601) = (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119872 minus 119873) (119873 minus 119899)119867120576

+ 119873 ((119872 + 119899) minus (119898 + 119873)) (1 minus 119867120576)]

times (1198732

(119872 minus 119873)2

)minus1

ge (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in 120596

(A8)

10 Mathematical Problems in Engineering

and by (14) and (A6)

119865119866

(120601) = minus (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119899 (119872 minus 119873)119867120576

+119873 (119898 minus 119899) (1 minus 119867120576))]

times (1198732

(119872 minus 119873)2

)minus1

le minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(A9)

This completes the proof of (17) and similarly for (18) Theproof is completed

B Proofs of Theorem 3 and Corollary 4

Proof of Theorem 3 By (6) we have

1198911(x) =

int120601gt0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601gt0

119870120590(xminus120585) 119889120585

= (int

120601gt0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

119906gt119868

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601gt0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

times (int

119906gt119868

119870120590(119909minus120585) 119889120585)

minus1

=1198921119902 + 1198922(119876 minus 119902)

119876

(B1)

1198912(x) =

int120601le0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601le0

119870120590(xminus120585) 119889120585

= (int

120601le0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601le0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

=1198921(119901 minus 119902) + 119892

2[(119875 minus 119876) minus (119901 minus 119902)]

119875 minus 119876

(B2)

and so

1198911(x) minus 119891

2(x) =

(1198921minus 1198922) (119875119902 minus 119876119901)

119876 (119875 minus 119876) (B3)

Therefore in 120596 we get

119865119871

(120601) = (1198981(120601) minus 119898

2(120601))

times ((1198921minus 1198981(120601))119867

120576(120601)

+ (1198921minus 1198982(120601)) (1 minus 119867

120576(120601)))

= (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B4)

and in Ω 120596 we have

119865119871

= (1198981(120601) minus 119898

2(120601))

times ((1198922minus 1198981(120601))119867

120576(120601)

+ (1198922minus 1198982(120601)) (1 minus 119867

120576(120601)))

= minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B5)

Mathematical Problems in Engineering 11

This completes the proof of (19) The last assertion (20) fol-lows clearly from the following fact(119875 minus 119876) (119876 minus 119902)119867

120576(120601)

+ 119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))

= 119867120576(120601) (int

Ω

minusint

120601gt0

)119870120590(xminus120585) 119889120585

sdot (int

120601gt0

minusint

120601gt0cap120596)119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601)) int

120601gt0

119870120590(xminus120585) 119889120585

sdot ((intΩ

+int

120601gt0cap120596) minus (int

120596

+int

120601gt0

))119870120590(xminus120585) 119889120585

= 119867120576(120601) int

120601le0

119870120590(xminus120585) 119889120585

sdot (int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot (int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

ge 119867120576(120601) (int

120601le0

119878119889120585) sdot (int

120601gt0(120601gt0cap120596)

119878119889120585)

+ (1 minus 119867120576(120601)) (int

120601gt0

119878119889120585) sdot (int

120601le0(120601le0cap120596)

119878119889120585)

= 119878 [ (119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 minus119873) minus (119898 minus 119899)) (1 minus 119867120576(120601))]

ge 119878

(B6)119902 (119875 minus 119876)119867

120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))

= 119867120576(120601) int

120601gt0cap120596119870120590(xminus120585) 119889120585 sdot int

120601le0

119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot int

120601le0cap120596119870120590(xminus120585) 119889120585

ge 119867120576(120601) (int

119906gt119868cap120596119878119889120585) sdot (int

119906le119868

119878119889120585)

+ (1 minus 119867120576(120601)) (int

119906gt119868

119878119889120585) sdot (int

119906le119868cap120596119878119889120585)

= 119878 [119899 (119872 minus 119873)119867120576(120601) + (1 minus 119867

120576(120601))119873 (119898 minus 119899)]

ge 119878

(B7)

where 119878 = min119870120590(xminus120585) 120585 isin Ω)

Now we give the proof of Corollary 4 (i) Since

1198752

ge 4119875 (119875 minus 119876) (B8)

we have by (19) and (B6)

119865119871

(120601) = (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

ge (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

(B9)

and by (19) and (B7)

119865119871

(120601) = minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

le minus (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(B10)

This completes the proof of (19) and similarly to (20) Theproof is completed

C Proof of Corollary 5

Proof (i) By Corollary 2 (i) we have

119865119866

(1206010

) ge (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0

in 120596

119865119866

(1206010

) le minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

(C1)

At the first iteration by (25) and (C1) we have

1206011

119894119895= 1206010

119894119895+ Δ119905

sdot 119865119866

(1206010

119894119895)

ge 1206010

119894119895+ Δ119905 sdot 120575

120576(1206010

)119860 in 120596

le 1206010

119894119895minus Δ119905 sdot 120575

120576(1206010

)119860 in Ω 120596

(C2)

12 Mathematical Problems in Engineering

or equivalently

1206011

(x) ge 1206010

(x) + Δ119905 sdot 120575120576(1206010

)119860 gt 1206010

(x) in 120596

1206011

(x) le 1206010

(x) minus Δ119905 sdot 120575120576(1206010

)119860 lt 1206010

(x) in Ω 120596

(C3)

which implies that

1198721198991minus 1198731119898 = |Ω|

10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816minus100381610038161003816100381610038161206011

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816) |120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816|120596|

gt |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816) |120596|

= |Ω|1003816100381610038161003816120596 cap 120601

0gt 0

1003816100381610038161003816 minus10038161003816100381610038161206010 gt 0

1003816100381610038161003816 |120596|

= 1198721198990minus 1198730119898 gt 0

(C4)

In a manner similar to the steps used to obtain (26)-(27) and(C1)ndash (C4) we obtain

119865119866

(119868 1206011

) ge (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

gt (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

119865119866

(119868 1206011

) le minus (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

lt minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

1206012

(x) ge 1206011

(x) + Δ119905 sdot 120575120576(1206011

)119860 gt 1206011

(x) in 120596

1206012

(x) le 1206011

(x) minus Δ119905 sdot 120575120576(1206011

)119860 lt 1206011

(x) in Ω 120596

1198721198992minus 1198732119898 gt 119872119899

1minus 1198731119898 gt 119872119899

0minus 1198730119898 gt 0

(C5)

D Proof of Theorem 7

Proof We prove (32) and similarly to (33) By (25)Corollary 5(i) and Corollary 6(i) we get

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

= 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119865119871

(120601119896

119894119895) + 120596119865

119866

(120601119896

119894119895))

ge 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

) + 120596119860 sdot 120575120576(120601119896+1

))

= 120601119896

119894119895+ ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in 120596

(119896 ge 1 119896 isin 119911+

)

le 120601119896

119894119895+ Δ119905

sdot (minus ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

)

+120596119860 sdot 120575120576(120601119896+1

)))

= 120601119896

119894119895minus ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(D1)

which implies that

120601119896+1

119894119895ge 1206010

119894+ ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in 120596

120601119896+1

119894119895le 1206010

119894minus (119896 + 1)119860 ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in Ω 120596

(D2)

Because 1206010(x) is bounded in domain Ω there clearly exists a

positive integer 119870 such that

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

for 119896 ge 119870 (D3)

The proof is completed

Mathematical Problems in Engineering 13

E Proof of Theorem 8

Proof From (11) we have

119865 (0) = 120575120576(0) [120596 ((119868 minus 119868

119866

) sdot 2meanxisinΩ

119868 (x))

+ (1 minus 120596) ((119868 minus 119868119871

) sdot 2meanxisinΩ

119868 (x))]

=2120575120576(0)mean

xisinΩ119868 (x) (119868 minus 2119867

120576(0)mean

xisinΩ119868 (x))

=120575120576(0)mean

xisinΩ119868 (x) (119868 minusmean

xisinΩ119868 (x))

(E1)

Clearly

minxisinΩ

119868 (x) lt meanxisinΩ

119868 (x) lt maxxisinΩ

119868 (x) (E2)

which implies that

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(E3)

where

Ω1= 119868 (x) lt mean

xisinΩ119868 (x)

Ω2= 119868 (x) gt mean

xisinΩ119868 (x)

(E4)

The proof of Theorem 8 is completed

Acknowledgments

The authors would like to thank the anonymous reviewersfor their valuable comments and suggestions to improve thispaper Besides this work was supported by the Fundamen-tal Research Funds for the Central University Grant no(CDJXS10100006)

References

[1] V Caselles F Catte T Coll and F Dibos ldquoA geometricmodel for active contours in image processingrdquo NumerischeMathematik vol 66 no 1 pp 1ndash31 1993

[2] T Chan M Moelich and B Sandberg ldquoSome recent develop-ments in variational image segmentationrdquo UCLA CAM Reportcam 06-52 2006

[3] L He Z Peng B Everding et al ldquoA comparative study ofdeformable contour methods on medical image segmentationrdquoImage and Vision Computing vol 26 no 2 pp 141ndash163 2008

[4] C Li C Xu C Gui andM D Fox ldquoLevel set evolution withoutre-initialization a new variational formulationrdquo in Proceedingsof the IEEE Computer Society Conference on Computer Visionand Pattern Recognition (CVPR rsquo05) pp 430ndash436 San DiegoCalif USA June 2005

[5] C Li C Xu C Gui and M D Fox ldquoDistance regularized levelset evolution and its application to image segmentationrdquo IEEETransactions on Image Processing vol 19 no 12 pp 3243ndash32542010

[6] B Zhou and C LMu ldquoLevel set evolution for boundary extrac-tion based on a p-Laplace equationrdquo Applied MathematicalModelling vol 34 no 12 pp 3910ndash3916 2010

[7] Y Wang and C He ldquoAdaptive level set evolution starting with aconstant functionrdquoAppliedMathematical Modelling vol 36 no7 pp 3217ndash3228 2012

[8] T F Chan and L A Vese ldquoActive contours without edgesrdquo IEEETransactions on Image Processing vol 10 no 2 pp 266ndash2772001

[9] A Tsai A Yezzi andA SWillsky ldquoCurve evolution implemen-tation of theMumford-Shah functional for image segmentationdenoising interpolation andmagnificationrdquo IEEETransactionson Image Processing vol 10 no 8 pp 1169ndash1186 2001

[10] T Brox and D Cremers ldquoOn the statistical interpretation of thepiecewise smooth Mumford-Shah functionalrdquo in Proceedingsof the International Conference on Scale Space and VariationalMethods in Computer Vision (SSVM rsquo07) pp 203ndash213 IschiaItaly 2007

[11] C Li C Y Kao J CGore andZDing ldquoImplicit active contoursdriven by local binary fitting energyrdquo in Proceedings of the IEEEComputer Society Conference on Computer Vision and PatternRecognition (CVPR rsquo07) pp 1ndash7 Washington DC USA June2007

[12] C Li C Y Kao J C Gore and Z Ding ldquoMinimization ofregion-scalable fitting energy for image segmentationrdquo IEEETransactions on Image Processing vol 17 no 10 pp 1940ndash19492008

[13] S Lankton and A Tannenbaum ldquoLocalizing region-basedactive contoursrdquo IEEE Transactions on Image Processing vol 17no 11 pp 2029ndash2039 2008

[14] L Wang C Li Q Sun D Xia and C Y Kao ldquoActive contoursdriven by local and global intensity fitting energy with applica-tion to brain MR image segmentationrdquo Computerized MedicalImaging and Graphics vol 33 no 7 pp 520ndash531 2009

[15] K Zhang H Song and L Zhang ldquoActive contours driven bylocal image fitting energyrdquo Pattern Recognition vol 43 no 4pp 1199ndash1206 2010

[16] K Zhang L Zhang H Song and W Zhou ldquoActive contourswith selective local or global segmentation a new formulationand level set methodrdquo Image and Vision Computing vol 28 no4 pp 668ndash676 2010

[17] Y Yuan and C He ldquoVariational level set methods for image seg-mentation based on both 1198712 and Sobolev gradientsrdquo NonlinearAnalysis Real World Applications vol 13 no 2 pp 959ndash9662012

[18] B Liu H D Cheng J Huang J Tian X Tang and J LiuldquoProbability density difference-based active contour for ultra-sound image segmentationrdquo Pattern Recognition vol 43 no 6pp 2028ndash2042 2010

[19] D W Shattuck S R Sandor-Leahy K A Schaper D ARottenberg and RM Leahy ldquoMagnetic resonance image tissueclassification using a partial volume modelrdquo NeuroImage vol13 no 5 pp 856ndash876 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 10: Research Article Implicit Active Contour Model with Local ...downloads.hindawi.com/journals/mpe/2013/367086.pdfAcademic Editor: Oluwole Daniel Makinde ... (CV)proposedaregion-basedactivecontourmodel

10 Mathematical Problems in Engineering

and by (14) and (A6)

119865119866

(120601) = minus (119872119899 minus 119873119898) (1198921minus 1198922)2

times [ (119899 (119872 minus 119873)119867120576

+119873 (119898 minus 119899) (1 minus 119867120576))]

times (1198732

(119872 minus 119873)2

)minus1

le minus (119872119899 minus 119873119898)(1198921minus 1198922)2

11987244 in Ω 120596

(A9)

This completes the proof of (17) and similarly for (18) Theproof is completed

B Proofs of Theorem 3 and Corollary 4

Proof of Theorem 3 By (6) we have

1198911(x) =

int120601gt0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601gt0

119870120590(xminus120585) 119889120585

= (int

120601gt0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

119906gt119868

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601gt0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

times (int

119906gt119868

119870120590(119909minus120585) 119889120585)

minus1

=1198921119902 + 1198922(119876 minus 119902)

119876

(B1)

1198912(x) =

int120601le0

119870120590(xminus120585) 119868 (120585) 119889120585

int120601le0

119870120590(xminus120585) 119889120585

= (int

120601le0cap120596119870120590(xminus120585) 119868 (120585) 119889120585

+int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119868 (120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

= (1198921int

120601le0cap120596119870120590(xminus120585) 119889120585

+1198922int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

times (int

120601le0

119870120590(xminus120585) 119889120585)

minus1

=1198921(119901 minus 119902) + 119892

2[(119875 minus 119876) minus (119901 minus 119902)]

119875 minus 119876

(B2)

and so

1198911(x) minus 119891

2(x) =

(1198921minus 1198922) (119875119902 minus 119876119901)

119876 (119875 minus 119876) (B3)

Therefore in 120596 we get

119865119871

(120601) = (1198981(120601) minus 119898

2(120601))

times ((1198921minus 1198981(120601))119867

120576(120601)

+ (1198921minus 1198982(120601)) (1 minus 119867

120576(120601)))

= (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B4)

and in Ω 120596 we have

119865119871

= (1198981(120601) minus 119898

2(120601))

times ((1198922minus 1198981(120601))119867

120576(120601)

+ (1198922minus 1198982(120601)) (1 minus 119867

120576(120601)))

= minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

(B5)

Mathematical Problems in Engineering 11

This completes the proof of (19) The last assertion (20) fol-lows clearly from the following fact(119875 minus 119876) (119876 minus 119902)119867

120576(120601)

+ 119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))

= 119867120576(120601) (int

Ω

minusint

120601gt0

)119870120590(xminus120585) 119889120585

sdot (int

120601gt0

minusint

120601gt0cap120596)119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601)) int

120601gt0

119870120590(xminus120585) 119889120585

sdot ((intΩ

+int

120601gt0cap120596) minus (int

120596

+int

120601gt0

))119870120590(xminus120585) 119889120585

= 119867120576(120601) int

120601le0

119870120590(xminus120585) 119889120585

sdot (int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot (int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

ge 119867120576(120601) (int

120601le0

119878119889120585) sdot (int

120601gt0(120601gt0cap120596)

119878119889120585)

+ (1 minus 119867120576(120601)) (int

120601gt0

119878119889120585) sdot (int

120601le0(120601le0cap120596)

119878119889120585)

= 119878 [ (119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 minus119873) minus (119898 minus 119899)) (1 minus 119867120576(120601))]

ge 119878

(B6)119902 (119875 minus 119876)119867

120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))

= 119867120576(120601) int

120601gt0cap120596119870120590(xminus120585) 119889120585 sdot int

120601le0

119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot int

120601le0cap120596119870120590(xminus120585) 119889120585

ge 119867120576(120601) (int

119906gt119868cap120596119878119889120585) sdot (int

119906le119868

119878119889120585)

+ (1 minus 119867120576(120601)) (int

119906gt119868

119878119889120585) sdot (int

119906le119868cap120596119878119889120585)

= 119878 [119899 (119872 minus 119873)119867120576(120601) + (1 minus 119867

120576(120601))119873 (119898 minus 119899)]

ge 119878

(B7)

where 119878 = min119870120590(xminus120585) 120585 isin Ω)

Now we give the proof of Corollary 4 (i) Since

1198752

ge 4119875 (119875 minus 119876) (B8)

we have by (19) and (B6)

119865119871

(120601) = (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

ge (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

(B9)

and by (19) and (B7)

119865119871

(120601) = minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

le minus (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(B10)

This completes the proof of (19) and similarly to (20) Theproof is completed

C Proof of Corollary 5

Proof (i) By Corollary 2 (i) we have

119865119866

(1206010

) ge (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0

in 120596

119865119866

(1206010

) le minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

(C1)

At the first iteration by (25) and (C1) we have

1206011

119894119895= 1206010

119894119895+ Δ119905

sdot 119865119866

(1206010

119894119895)

ge 1206010

119894119895+ Δ119905 sdot 120575

120576(1206010

)119860 in 120596

le 1206010

119894119895minus Δ119905 sdot 120575

120576(1206010

)119860 in Ω 120596

(C2)

12 Mathematical Problems in Engineering

or equivalently

1206011

(x) ge 1206010

(x) + Δ119905 sdot 120575120576(1206010

)119860 gt 1206010

(x) in 120596

1206011

(x) le 1206010

(x) minus Δ119905 sdot 120575120576(1206010

)119860 lt 1206010

(x) in Ω 120596

(C3)

which implies that

1198721198991minus 1198731119898 = |Ω|

10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816minus100381610038161003816100381610038161206011

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816) |120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816|120596|

gt |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816) |120596|

= |Ω|1003816100381610038161003816120596 cap 120601

0gt 0

1003816100381610038161003816 minus10038161003816100381610038161206010 gt 0

1003816100381610038161003816 |120596|

= 1198721198990minus 1198730119898 gt 0

(C4)

In a manner similar to the steps used to obtain (26)-(27) and(C1)ndash (C4) we obtain

119865119866

(119868 1206011

) ge (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

gt (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

119865119866

(119868 1206011

) le minus (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

lt minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

1206012

(x) ge 1206011

(x) + Δ119905 sdot 120575120576(1206011

)119860 gt 1206011

(x) in 120596

1206012

(x) le 1206011

(x) minus Δ119905 sdot 120575120576(1206011

)119860 lt 1206011

(x) in Ω 120596

1198721198992minus 1198732119898 gt 119872119899

1minus 1198731119898 gt 119872119899

0minus 1198730119898 gt 0

(C5)

D Proof of Theorem 7

Proof We prove (32) and similarly to (33) By (25)Corollary 5(i) and Corollary 6(i) we get

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

= 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119865119871

(120601119896

119894119895) + 120596119865

119866

(120601119896

119894119895))

ge 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

) + 120596119860 sdot 120575120576(120601119896+1

))

= 120601119896

119894119895+ ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in 120596

(119896 ge 1 119896 isin 119911+

)

le 120601119896

119894119895+ Δ119905

sdot (minus ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

)

+120596119860 sdot 120575120576(120601119896+1

)))

= 120601119896

119894119895minus ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(D1)

which implies that

120601119896+1

119894119895ge 1206010

119894+ ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in 120596

120601119896+1

119894119895le 1206010

119894minus (119896 + 1)119860 ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in Ω 120596

(D2)

Because 1206010(x) is bounded in domain Ω there clearly exists a

positive integer 119870 such that

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

for 119896 ge 119870 (D3)

The proof is completed

Mathematical Problems in Engineering 13

E Proof of Theorem 8

Proof From (11) we have

119865 (0) = 120575120576(0) [120596 ((119868 minus 119868

119866

) sdot 2meanxisinΩ

119868 (x))

+ (1 minus 120596) ((119868 minus 119868119871

) sdot 2meanxisinΩ

119868 (x))]

=2120575120576(0)mean

xisinΩ119868 (x) (119868 minus 2119867

120576(0)mean

xisinΩ119868 (x))

=120575120576(0)mean

xisinΩ119868 (x) (119868 minusmean

xisinΩ119868 (x))

(E1)

Clearly

minxisinΩ

119868 (x) lt meanxisinΩ

119868 (x) lt maxxisinΩ

119868 (x) (E2)

which implies that

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(E3)

where

Ω1= 119868 (x) lt mean

xisinΩ119868 (x)

Ω2= 119868 (x) gt mean

xisinΩ119868 (x)

(E4)

The proof of Theorem 8 is completed

Acknowledgments

The authors would like to thank the anonymous reviewersfor their valuable comments and suggestions to improve thispaper Besides this work was supported by the Fundamen-tal Research Funds for the Central University Grant no(CDJXS10100006)

References

[1] V Caselles F Catte T Coll and F Dibos ldquoA geometricmodel for active contours in image processingrdquo NumerischeMathematik vol 66 no 1 pp 1ndash31 1993

[2] T Chan M Moelich and B Sandberg ldquoSome recent develop-ments in variational image segmentationrdquo UCLA CAM Reportcam 06-52 2006

[3] L He Z Peng B Everding et al ldquoA comparative study ofdeformable contour methods on medical image segmentationrdquoImage and Vision Computing vol 26 no 2 pp 141ndash163 2008

[4] C Li C Xu C Gui andM D Fox ldquoLevel set evolution withoutre-initialization a new variational formulationrdquo in Proceedingsof the IEEE Computer Society Conference on Computer Visionand Pattern Recognition (CVPR rsquo05) pp 430ndash436 San DiegoCalif USA June 2005

[5] C Li C Xu C Gui and M D Fox ldquoDistance regularized levelset evolution and its application to image segmentationrdquo IEEETransactions on Image Processing vol 19 no 12 pp 3243ndash32542010

[6] B Zhou and C LMu ldquoLevel set evolution for boundary extrac-tion based on a p-Laplace equationrdquo Applied MathematicalModelling vol 34 no 12 pp 3910ndash3916 2010

[7] Y Wang and C He ldquoAdaptive level set evolution starting with aconstant functionrdquoAppliedMathematical Modelling vol 36 no7 pp 3217ndash3228 2012

[8] T F Chan and L A Vese ldquoActive contours without edgesrdquo IEEETransactions on Image Processing vol 10 no 2 pp 266ndash2772001

[9] A Tsai A Yezzi andA SWillsky ldquoCurve evolution implemen-tation of theMumford-Shah functional for image segmentationdenoising interpolation andmagnificationrdquo IEEETransactionson Image Processing vol 10 no 8 pp 1169ndash1186 2001

[10] T Brox and D Cremers ldquoOn the statistical interpretation of thepiecewise smooth Mumford-Shah functionalrdquo in Proceedingsof the International Conference on Scale Space and VariationalMethods in Computer Vision (SSVM rsquo07) pp 203ndash213 IschiaItaly 2007

[11] C Li C Y Kao J CGore andZDing ldquoImplicit active contoursdriven by local binary fitting energyrdquo in Proceedings of the IEEEComputer Society Conference on Computer Vision and PatternRecognition (CVPR rsquo07) pp 1ndash7 Washington DC USA June2007

[12] C Li C Y Kao J C Gore and Z Ding ldquoMinimization ofregion-scalable fitting energy for image segmentationrdquo IEEETransactions on Image Processing vol 17 no 10 pp 1940ndash19492008

[13] S Lankton and A Tannenbaum ldquoLocalizing region-basedactive contoursrdquo IEEE Transactions on Image Processing vol 17no 11 pp 2029ndash2039 2008

[14] L Wang C Li Q Sun D Xia and C Y Kao ldquoActive contoursdriven by local and global intensity fitting energy with applica-tion to brain MR image segmentationrdquo Computerized MedicalImaging and Graphics vol 33 no 7 pp 520ndash531 2009

[15] K Zhang H Song and L Zhang ldquoActive contours driven bylocal image fitting energyrdquo Pattern Recognition vol 43 no 4pp 1199ndash1206 2010

[16] K Zhang L Zhang H Song and W Zhou ldquoActive contourswith selective local or global segmentation a new formulationand level set methodrdquo Image and Vision Computing vol 28 no4 pp 668ndash676 2010

[17] Y Yuan and C He ldquoVariational level set methods for image seg-mentation based on both 1198712 and Sobolev gradientsrdquo NonlinearAnalysis Real World Applications vol 13 no 2 pp 959ndash9662012

[18] B Liu H D Cheng J Huang J Tian X Tang and J LiuldquoProbability density difference-based active contour for ultra-sound image segmentationrdquo Pattern Recognition vol 43 no 6pp 2028ndash2042 2010

[19] D W Shattuck S R Sandor-Leahy K A Schaper D ARottenberg and RM Leahy ldquoMagnetic resonance image tissueclassification using a partial volume modelrdquo NeuroImage vol13 no 5 pp 856ndash876 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 11: Research Article Implicit Active Contour Model with Local ...downloads.hindawi.com/journals/mpe/2013/367086.pdfAcademic Editor: Oluwole Daniel Makinde ... (CV)proposedaregion-basedactivecontourmodel

Mathematical Problems in Engineering 11

This completes the proof of (19) The last assertion (20) fol-lows clearly from the following fact(119875 minus 119876) (119876 minus 119902)119867

120576(120601)

+ 119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))

= 119867120576(120601) (int

Ω

minusint

120601gt0

)119870120590(xminus120585) 119889120585

sdot (int

120601gt0

minusint

120601gt0cap120596)119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601)) int

120601gt0

119870120590(xminus120585) 119889120585

sdot ((intΩ

+int

120601gt0cap120596) minus (int

120596

+int

120601gt0

))119870120590(xminus120585) 119889120585

= 119867120576(120601) int

120601le0

119870120590(xminus120585) 119889120585

sdot (int

120601gt0(120601gt0cap120596)

119870120590(xminus120585) 119889120585)

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot (int

120601le0(120601le0cap120596)

119870120590(xminus120585) 119889120585)

ge 119867120576(120601) (int

120601le0

119878119889120585) sdot (int

120601gt0(120601gt0cap120596)

119878119889120585)

+ (1 minus 119867120576(120601)) (int

120601gt0

119878119889120585) sdot (int

120601le0(120601le0cap120596)

119878119889120585)

= 119878 [ (119872 minus 119873) (119873 minus 119899)119867120576(120601)

+ 119873 ((119872 minus119873) minus (119898 minus 119899)) (1 minus 119867120576(120601))]

ge 119878

(B6)119902 (119875 minus 119876)119867

120576(120601) + 119876 (119901 minus 119902) (1 minus 119867

120576(120601))

= 119867120576(120601) int

120601gt0cap120596119870120590(xminus120585) 119889120585 sdot int

120601le0

119870120590(xminus120585) 119889120585

+ (1 minus 119867120576(120601))

times int

120601gt0

119870120590(xminus120585) 119889120585 sdot int

120601le0cap120596119870120590(xminus120585) 119889120585

ge 119867120576(120601) (int

119906gt119868cap120596119878119889120585) sdot (int

119906le119868

119878119889120585)

+ (1 minus 119867120576(120601)) (int

119906gt119868

119878119889120585) sdot (int

119906le119868cap120596119878119889120585)

= 119878 [119899 (119872 minus 119873)119867120576(120601) + (1 minus 119867

120576(120601))119873 (119898 minus 119899)]

ge 119878

(B7)

where 119878 = min119870120590(xminus120585) 120585 isin Ω)

Now we give the proof of Corollary 4 (i) Since

1198752

ge 4119875 (119875 minus 119876) (B8)

we have by (19) and (B6)

119865119871

(120601) = (119875119902 minus 119876119901) (1198921minus 1198922)2

times [ (119875 minus 119876) (119876 minus 119902)119867120576(120601)

+119876 ((119875 + 119902) minus (119901 + 119876)) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

ge (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in 120596

(B9)

and by (19) and (B7)

119865119871

(120601) = minus (119875119902 minus 119876119901) (1198921minus 1198922)2

times [119902 (119875 minus 119876)119867120576(120601)

+119876 (119901 minus 119902) (1 minus 119867120576(120601))]

times (1198762

(119875 minus 119876)2

)minus1

le minus (119875119902 minus 119876119901)(1198921minus 1198922)2

11987544 in Ω 120596

(B10)

This completes the proof of (19) and similarly to (20) Theproof is completed

C Proof of Corollary 5

Proof (i) By Corollary 2 (i) we have

119865119866

(1206010

) ge (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0

in 120596

119865119866

(1206010

) le minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

(C1)

At the first iteration by (25) and (C1) we have

1206011

119894119895= 1206010

119894119895+ Δ119905

sdot 119865119866

(1206010

119894119895)

ge 1206010

119894119895+ Δ119905 sdot 120575

120576(1206010

)119860 in 120596

le 1206010

119894119895minus Δ119905 sdot 120575

120576(1206010

)119860 in Ω 120596

(C2)

12 Mathematical Problems in Engineering

or equivalently

1206011

(x) ge 1206010

(x) + Δ119905 sdot 120575120576(1206010

)119860 gt 1206010

(x) in 120596

1206011

(x) le 1206010

(x) minus Δ119905 sdot 120575120576(1206010

)119860 lt 1206010

(x) in Ω 120596

(C3)

which implies that

1198721198991minus 1198731119898 = |Ω|

10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816minus100381610038161003816100381610038161206011

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816) |120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816|120596|

gt |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816) |120596|

= |Ω|1003816100381610038161003816120596 cap 120601

0gt 0

1003816100381610038161003816 minus10038161003816100381610038161206010 gt 0

1003816100381610038161003816 |120596|

= 1198721198990minus 1198730119898 gt 0

(C4)

In a manner similar to the steps used to obtain (26)-(27) and(C1)ndash (C4) we obtain

119865119866

(119868 1206011

) ge (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

gt (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

119865119866

(119868 1206011

) le minus (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

lt minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

1206012

(x) ge 1206011

(x) + Δ119905 sdot 120575120576(1206011

)119860 gt 1206011

(x) in 120596

1206012

(x) le 1206011

(x) minus Δ119905 sdot 120575120576(1206011

)119860 lt 1206011

(x) in Ω 120596

1198721198992minus 1198732119898 gt 119872119899

1minus 1198731119898 gt 119872119899

0minus 1198730119898 gt 0

(C5)

D Proof of Theorem 7

Proof We prove (32) and similarly to (33) By (25)Corollary 5(i) and Corollary 6(i) we get

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

= 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119865119871

(120601119896

119894119895) + 120596119865

119866

(120601119896

119894119895))

ge 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

) + 120596119860 sdot 120575120576(120601119896+1

))

= 120601119896

119894119895+ ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in 120596

(119896 ge 1 119896 isin 119911+

)

le 120601119896

119894119895+ Δ119905

sdot (minus ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

)

+120596119860 sdot 120575120576(120601119896+1

)))

= 120601119896

119894119895minus ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(D1)

which implies that

120601119896+1

119894119895ge 1206010

119894+ ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in 120596

120601119896+1

119894119895le 1206010

119894minus (119896 + 1)119860 ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in Ω 120596

(D2)

Because 1206010(x) is bounded in domain Ω there clearly exists a

positive integer 119870 such that

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

for 119896 ge 119870 (D3)

The proof is completed

Mathematical Problems in Engineering 13

E Proof of Theorem 8

Proof From (11) we have

119865 (0) = 120575120576(0) [120596 ((119868 minus 119868

119866

) sdot 2meanxisinΩ

119868 (x))

+ (1 minus 120596) ((119868 minus 119868119871

) sdot 2meanxisinΩ

119868 (x))]

=2120575120576(0)mean

xisinΩ119868 (x) (119868 minus 2119867

120576(0)mean

xisinΩ119868 (x))

=120575120576(0)mean

xisinΩ119868 (x) (119868 minusmean

xisinΩ119868 (x))

(E1)

Clearly

minxisinΩ

119868 (x) lt meanxisinΩ

119868 (x) lt maxxisinΩ

119868 (x) (E2)

which implies that

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(E3)

where

Ω1= 119868 (x) lt mean

xisinΩ119868 (x)

Ω2= 119868 (x) gt mean

xisinΩ119868 (x)

(E4)

The proof of Theorem 8 is completed

Acknowledgments

The authors would like to thank the anonymous reviewersfor their valuable comments and suggestions to improve thispaper Besides this work was supported by the Fundamen-tal Research Funds for the Central University Grant no(CDJXS10100006)

References

[1] V Caselles F Catte T Coll and F Dibos ldquoA geometricmodel for active contours in image processingrdquo NumerischeMathematik vol 66 no 1 pp 1ndash31 1993

[2] T Chan M Moelich and B Sandberg ldquoSome recent develop-ments in variational image segmentationrdquo UCLA CAM Reportcam 06-52 2006

[3] L He Z Peng B Everding et al ldquoA comparative study ofdeformable contour methods on medical image segmentationrdquoImage and Vision Computing vol 26 no 2 pp 141ndash163 2008

[4] C Li C Xu C Gui andM D Fox ldquoLevel set evolution withoutre-initialization a new variational formulationrdquo in Proceedingsof the IEEE Computer Society Conference on Computer Visionand Pattern Recognition (CVPR rsquo05) pp 430ndash436 San DiegoCalif USA June 2005

[5] C Li C Xu C Gui and M D Fox ldquoDistance regularized levelset evolution and its application to image segmentationrdquo IEEETransactions on Image Processing vol 19 no 12 pp 3243ndash32542010

[6] B Zhou and C LMu ldquoLevel set evolution for boundary extrac-tion based on a p-Laplace equationrdquo Applied MathematicalModelling vol 34 no 12 pp 3910ndash3916 2010

[7] Y Wang and C He ldquoAdaptive level set evolution starting with aconstant functionrdquoAppliedMathematical Modelling vol 36 no7 pp 3217ndash3228 2012

[8] T F Chan and L A Vese ldquoActive contours without edgesrdquo IEEETransactions on Image Processing vol 10 no 2 pp 266ndash2772001

[9] A Tsai A Yezzi andA SWillsky ldquoCurve evolution implemen-tation of theMumford-Shah functional for image segmentationdenoising interpolation andmagnificationrdquo IEEETransactionson Image Processing vol 10 no 8 pp 1169ndash1186 2001

[10] T Brox and D Cremers ldquoOn the statistical interpretation of thepiecewise smooth Mumford-Shah functionalrdquo in Proceedingsof the International Conference on Scale Space and VariationalMethods in Computer Vision (SSVM rsquo07) pp 203ndash213 IschiaItaly 2007

[11] C Li C Y Kao J CGore andZDing ldquoImplicit active contoursdriven by local binary fitting energyrdquo in Proceedings of the IEEEComputer Society Conference on Computer Vision and PatternRecognition (CVPR rsquo07) pp 1ndash7 Washington DC USA June2007

[12] C Li C Y Kao J C Gore and Z Ding ldquoMinimization ofregion-scalable fitting energy for image segmentationrdquo IEEETransactions on Image Processing vol 17 no 10 pp 1940ndash19492008

[13] S Lankton and A Tannenbaum ldquoLocalizing region-basedactive contoursrdquo IEEE Transactions on Image Processing vol 17no 11 pp 2029ndash2039 2008

[14] L Wang C Li Q Sun D Xia and C Y Kao ldquoActive contoursdriven by local and global intensity fitting energy with applica-tion to brain MR image segmentationrdquo Computerized MedicalImaging and Graphics vol 33 no 7 pp 520ndash531 2009

[15] K Zhang H Song and L Zhang ldquoActive contours driven bylocal image fitting energyrdquo Pattern Recognition vol 43 no 4pp 1199ndash1206 2010

[16] K Zhang L Zhang H Song and W Zhou ldquoActive contourswith selective local or global segmentation a new formulationand level set methodrdquo Image and Vision Computing vol 28 no4 pp 668ndash676 2010

[17] Y Yuan and C He ldquoVariational level set methods for image seg-mentation based on both 1198712 and Sobolev gradientsrdquo NonlinearAnalysis Real World Applications vol 13 no 2 pp 959ndash9662012

[18] B Liu H D Cheng J Huang J Tian X Tang and J LiuldquoProbability density difference-based active contour for ultra-sound image segmentationrdquo Pattern Recognition vol 43 no 6pp 2028ndash2042 2010

[19] D W Shattuck S R Sandor-Leahy K A Schaper D ARottenberg and RM Leahy ldquoMagnetic resonance image tissueclassification using a partial volume modelrdquo NeuroImage vol13 no 5 pp 856ndash876 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 12: Research Article Implicit Active Contour Model with Local ...downloads.hindawi.com/journals/mpe/2013/367086.pdfAcademic Editor: Oluwole Daniel Makinde ... (CV)proposedaregion-basedactivecontourmodel

12 Mathematical Problems in Engineering

or equivalently

1206011

(x) ge 1206010

(x) + Δ119905 sdot 120575120576(1206010

)119860 gt 1206010

(x) in 120596

1206011

(x) le 1206010

(x) minus Δ119905 sdot 120575120576(1206010

)119860 lt 1206010

(x) in Ω 120596

(C3)

which implies that

1198721198991minus 1198731119898 = |Ω|

10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816minus100381610038161003816100381610038161206011

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816) |120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

1

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

1

gt 010038161003816100381610038161003816|120596|

gt |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= |Ω 120596|10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816|120596|

minus10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816|120596|

= (|Ω 120596| + |120596|)10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

minus (10038161003816100381610038161003816120596 cap 120601

0

gt 010038161003816100381610038161003816

+10038161003816100381610038161003816(Ω 120596) cap 120601

0

gt 010038161003816100381610038161003816) |120596|

= |Ω|1003816100381610038161003816120596 cap 120601

0gt 0

1003816100381610038161003816 minus10038161003816100381610038161206010 gt 0

1003816100381610038161003816 |120596|

= 1198721198990minus 1198730119898 gt 0

(C4)

In a manner similar to the steps used to obtain (26)-(27) and(C1)ndash (C4) we obtain

119865119866

(119868 1206011

) ge (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

gt (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= 119860 gt 0 in 120596

119865119866

(119868 1206011

) le minus (1198721198991minus 1198731119898)

(1198921minus 1198922)2

11987244

lt minus (1198721198990minus 1198730119898)

(1198921minus 1198922)2

11987244= minus119860 lt 0

in Ω 120596

1206012

(x) ge 1206011

(x) + Δ119905 sdot 120575120576(1206011

)119860 gt 1206011

(x) in 120596

1206012

(x) le 1206011

(x) minus Δ119905 sdot 120575120576(1206011

)119860 lt 1206011

(x) in Ω 120596

1198721198992minus 1198732119898 gt 119872119899

1minus 1198731119898 gt 119872119899

0minus 1198730119898 gt 0

(C5)

D Proof of Theorem 7

Proof We prove (32) and similarly to (33) By (25)Corollary 5(i) and Corollary 6(i) we get

120601119896+1

119894119895= 120601119896

119894119895+ Δ119905 sdot 119865 (120601

119896

119894119895)

= 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119865119871

(120601119896

119894119895) + 120596119865

119866

(120601119896

119894119895))

ge 120601119896

119894119895+ Δ119905

sdot ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

) + 120596119860 sdot 120575120576(120601119896+1

))

= 120601119896

119894119895+ ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in 120596

(119896 ge 1 119896 isin 119911+

)

le 120601119896

119894119895+ Δ119905

sdot (minus ((1 minus 120596) 119861 sdot 120575120576(120601119896+1

)

+120596119860 sdot 120575120576(120601119896+1

)))

= 120601119896

119894119895minus ((1 minus 120596) 119861 + 120596119860)

sdot 120575120576(120601119896+1

) sdot Δ119905 in Ω 120596

(119896 ge 1 119896 isin 119911+

)

(D1)

which implies that

120601119896+1

119894119895ge 1206010

119894+ ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in 120596

120601119896+1

119894119895le 1206010

119894minus (119896 + 1)119860 ((1 minus 120596) 119861 + 120596119860sdot) sdot

119896+1

prod

119894=0

120575120576(120601119894

) sdot Δ119905

in Ω 120596

(D2)

Because 1206010(x) is bounded in domain Ω there clearly exists a

positive integer 119870 such that

120601119896+1

119894119895gt 0 in 120596

120601119896+1

119894119895lt 0 in Ω 120596

for 119896 ge 119870 (D3)

The proof is completed

Mathematical Problems in Engineering 13

E Proof of Theorem 8

Proof From (11) we have

119865 (0) = 120575120576(0) [120596 ((119868 minus 119868

119866

) sdot 2meanxisinΩ

119868 (x))

+ (1 minus 120596) ((119868 minus 119868119871

) sdot 2meanxisinΩ

119868 (x))]

=2120575120576(0)mean

xisinΩ119868 (x) (119868 minus 2119867

120576(0)mean

xisinΩ119868 (x))

=120575120576(0)mean

xisinΩ119868 (x) (119868 minusmean

xisinΩ119868 (x))

(E1)

Clearly

minxisinΩ

119868 (x) lt meanxisinΩ

119868 (x) lt maxxisinΩ

119868 (x) (E2)

which implies that

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(E3)

where

Ω1= 119868 (x) lt mean

xisinΩ119868 (x)

Ω2= 119868 (x) gt mean

xisinΩ119868 (x)

(E4)

The proof of Theorem 8 is completed

Acknowledgments

The authors would like to thank the anonymous reviewersfor their valuable comments and suggestions to improve thispaper Besides this work was supported by the Fundamen-tal Research Funds for the Central University Grant no(CDJXS10100006)

References

[1] V Caselles F Catte T Coll and F Dibos ldquoA geometricmodel for active contours in image processingrdquo NumerischeMathematik vol 66 no 1 pp 1ndash31 1993

[2] T Chan M Moelich and B Sandberg ldquoSome recent develop-ments in variational image segmentationrdquo UCLA CAM Reportcam 06-52 2006

[3] L He Z Peng B Everding et al ldquoA comparative study ofdeformable contour methods on medical image segmentationrdquoImage and Vision Computing vol 26 no 2 pp 141ndash163 2008

[4] C Li C Xu C Gui andM D Fox ldquoLevel set evolution withoutre-initialization a new variational formulationrdquo in Proceedingsof the IEEE Computer Society Conference on Computer Visionand Pattern Recognition (CVPR rsquo05) pp 430ndash436 San DiegoCalif USA June 2005

[5] C Li C Xu C Gui and M D Fox ldquoDistance regularized levelset evolution and its application to image segmentationrdquo IEEETransactions on Image Processing vol 19 no 12 pp 3243ndash32542010

[6] B Zhou and C LMu ldquoLevel set evolution for boundary extrac-tion based on a p-Laplace equationrdquo Applied MathematicalModelling vol 34 no 12 pp 3910ndash3916 2010

[7] Y Wang and C He ldquoAdaptive level set evolution starting with aconstant functionrdquoAppliedMathematical Modelling vol 36 no7 pp 3217ndash3228 2012

[8] T F Chan and L A Vese ldquoActive contours without edgesrdquo IEEETransactions on Image Processing vol 10 no 2 pp 266ndash2772001

[9] A Tsai A Yezzi andA SWillsky ldquoCurve evolution implemen-tation of theMumford-Shah functional for image segmentationdenoising interpolation andmagnificationrdquo IEEETransactionson Image Processing vol 10 no 8 pp 1169ndash1186 2001

[10] T Brox and D Cremers ldquoOn the statistical interpretation of thepiecewise smooth Mumford-Shah functionalrdquo in Proceedingsof the International Conference on Scale Space and VariationalMethods in Computer Vision (SSVM rsquo07) pp 203ndash213 IschiaItaly 2007

[11] C Li C Y Kao J CGore andZDing ldquoImplicit active contoursdriven by local binary fitting energyrdquo in Proceedings of the IEEEComputer Society Conference on Computer Vision and PatternRecognition (CVPR rsquo07) pp 1ndash7 Washington DC USA June2007

[12] C Li C Y Kao J C Gore and Z Ding ldquoMinimization ofregion-scalable fitting energy for image segmentationrdquo IEEETransactions on Image Processing vol 17 no 10 pp 1940ndash19492008

[13] S Lankton and A Tannenbaum ldquoLocalizing region-basedactive contoursrdquo IEEE Transactions on Image Processing vol 17no 11 pp 2029ndash2039 2008

[14] L Wang C Li Q Sun D Xia and C Y Kao ldquoActive contoursdriven by local and global intensity fitting energy with applica-tion to brain MR image segmentationrdquo Computerized MedicalImaging and Graphics vol 33 no 7 pp 520ndash531 2009

[15] K Zhang H Song and L Zhang ldquoActive contours driven bylocal image fitting energyrdquo Pattern Recognition vol 43 no 4pp 1199ndash1206 2010

[16] K Zhang L Zhang H Song and W Zhou ldquoActive contourswith selective local or global segmentation a new formulationand level set methodrdquo Image and Vision Computing vol 28 no4 pp 668ndash676 2010

[17] Y Yuan and C He ldquoVariational level set methods for image seg-mentation based on both 1198712 and Sobolev gradientsrdquo NonlinearAnalysis Real World Applications vol 13 no 2 pp 959ndash9662012

[18] B Liu H D Cheng J Huang J Tian X Tang and J LiuldquoProbability density difference-based active contour for ultra-sound image segmentationrdquo Pattern Recognition vol 43 no 6pp 2028ndash2042 2010

[19] D W Shattuck S R Sandor-Leahy K A Schaper D ARottenberg and RM Leahy ldquoMagnetic resonance image tissueclassification using a partial volume modelrdquo NeuroImage vol13 no 5 pp 856ndash876 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 13: Research Article Implicit Active Contour Model with Local ...downloads.hindawi.com/journals/mpe/2013/367086.pdfAcademic Editor: Oluwole Daniel Makinde ... (CV)proposedaregion-basedactivecontourmodel

Mathematical Problems in Engineering 13

E Proof of Theorem 8

Proof From (11) we have

119865 (0) = 120575120576(0) [120596 ((119868 minus 119868

119866

) sdot 2meanxisinΩ

119868 (x))

+ (1 minus 120596) ((119868 minus 119868119871

) sdot 2meanxisinΩ

119868 (x))]

=2120575120576(0)mean

xisinΩ119868 (x) (119868 minus 2119867

120576(0)mean

xisinΩ119868 (x))

=120575120576(0)mean

xisinΩ119868 (x) (119868 minusmean

xisinΩ119868 (x))

(E1)

Clearly

minxisinΩ

119868 (x) lt meanxisinΩ

119868 (x) lt maxxisinΩ

119868 (x) (E2)

which implies that

sign (119865 (0)) = minus1 x isin Ω

1

1 x isin Ω2

(E3)

where

Ω1= 119868 (x) lt mean

xisinΩ119868 (x)

Ω2= 119868 (x) gt mean

xisinΩ119868 (x)

(E4)

The proof of Theorem 8 is completed

Acknowledgments

The authors would like to thank the anonymous reviewersfor their valuable comments and suggestions to improve thispaper Besides this work was supported by the Fundamen-tal Research Funds for the Central University Grant no(CDJXS10100006)

References

[1] V Caselles F Catte T Coll and F Dibos ldquoA geometricmodel for active contours in image processingrdquo NumerischeMathematik vol 66 no 1 pp 1ndash31 1993

[2] T Chan M Moelich and B Sandberg ldquoSome recent develop-ments in variational image segmentationrdquo UCLA CAM Reportcam 06-52 2006

[3] L He Z Peng B Everding et al ldquoA comparative study ofdeformable contour methods on medical image segmentationrdquoImage and Vision Computing vol 26 no 2 pp 141ndash163 2008

[4] C Li C Xu C Gui andM D Fox ldquoLevel set evolution withoutre-initialization a new variational formulationrdquo in Proceedingsof the IEEE Computer Society Conference on Computer Visionand Pattern Recognition (CVPR rsquo05) pp 430ndash436 San DiegoCalif USA June 2005

[5] C Li C Xu C Gui and M D Fox ldquoDistance regularized levelset evolution and its application to image segmentationrdquo IEEETransactions on Image Processing vol 19 no 12 pp 3243ndash32542010

[6] B Zhou and C LMu ldquoLevel set evolution for boundary extrac-tion based on a p-Laplace equationrdquo Applied MathematicalModelling vol 34 no 12 pp 3910ndash3916 2010

[7] Y Wang and C He ldquoAdaptive level set evolution starting with aconstant functionrdquoAppliedMathematical Modelling vol 36 no7 pp 3217ndash3228 2012

[8] T F Chan and L A Vese ldquoActive contours without edgesrdquo IEEETransactions on Image Processing vol 10 no 2 pp 266ndash2772001

[9] A Tsai A Yezzi andA SWillsky ldquoCurve evolution implemen-tation of theMumford-Shah functional for image segmentationdenoising interpolation andmagnificationrdquo IEEETransactionson Image Processing vol 10 no 8 pp 1169ndash1186 2001

[10] T Brox and D Cremers ldquoOn the statistical interpretation of thepiecewise smooth Mumford-Shah functionalrdquo in Proceedingsof the International Conference on Scale Space and VariationalMethods in Computer Vision (SSVM rsquo07) pp 203ndash213 IschiaItaly 2007

[11] C Li C Y Kao J CGore andZDing ldquoImplicit active contoursdriven by local binary fitting energyrdquo in Proceedings of the IEEEComputer Society Conference on Computer Vision and PatternRecognition (CVPR rsquo07) pp 1ndash7 Washington DC USA June2007

[12] C Li C Y Kao J C Gore and Z Ding ldquoMinimization ofregion-scalable fitting energy for image segmentationrdquo IEEETransactions on Image Processing vol 17 no 10 pp 1940ndash19492008

[13] S Lankton and A Tannenbaum ldquoLocalizing region-basedactive contoursrdquo IEEE Transactions on Image Processing vol 17no 11 pp 2029ndash2039 2008

[14] L Wang C Li Q Sun D Xia and C Y Kao ldquoActive contoursdriven by local and global intensity fitting energy with applica-tion to brain MR image segmentationrdquo Computerized MedicalImaging and Graphics vol 33 no 7 pp 520ndash531 2009

[15] K Zhang H Song and L Zhang ldquoActive contours driven bylocal image fitting energyrdquo Pattern Recognition vol 43 no 4pp 1199ndash1206 2010

[16] K Zhang L Zhang H Song and W Zhou ldquoActive contourswith selective local or global segmentation a new formulationand level set methodrdquo Image and Vision Computing vol 28 no4 pp 668ndash676 2010

[17] Y Yuan and C He ldquoVariational level set methods for image seg-mentation based on both 1198712 and Sobolev gradientsrdquo NonlinearAnalysis Real World Applications vol 13 no 2 pp 959ndash9662012

[18] B Liu H D Cheng J Huang J Tian X Tang and J LiuldquoProbability density difference-based active contour for ultra-sound image segmentationrdquo Pattern Recognition vol 43 no 6pp 2028ndash2042 2010

[19] D W Shattuck S R Sandor-Leahy K A Schaper D ARottenberg and RM Leahy ldquoMagnetic resonance image tissueclassification using a partial volume modelrdquo NeuroImage vol13 no 5 pp 856ndash876 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 14: Research Article Implicit Active Contour Model with Local ...downloads.hindawi.com/journals/mpe/2013/367086.pdfAcademic Editor: Oluwole Daniel Makinde ... (CV)proposedaregion-basedactivecontourmodel

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of