research article dynamical analysis of a pest management...

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 204642 18 pageshttpdxdoiorg1011552013204642

Research ArticleDynamical Analysis of a Pest Management Model withSaturated Growth Rate and State Dependent Impulsive Effects

Wencai Zhao1 Tongqian Zhang1 Xinzhu Meng2 and Yang Yang2

1 College of Science Shandong University of Science and Technology Qingdao 266590 China2 College of Information Science and Engineering Shandong University of Science and Technology Qingdao 266590 China

Correspondence should be addressed to Tongqian Zhang zhangtongqiansdusteducn

Received 11 March 2013 Accepted 20 May 2013

Academic Editor Tonghua Zhang

Copyright copy 2013 Wencai Zhao et al 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

Anewpestmanagementmathematicalmodelwith saturated growth is proposedThe integrated pestmanagement (IPM) strategy byintroducing two state dependent pulses into the model is considered Firstly we analyze singular points of the model qualitativelyand get the condition for focus point Secondly by using geometry theory of impulsive differential equation the existence andstability of periodic solution of the system are discussed Lastly some examples and numerical simulations are given to illustrateour results

1 Introduction

Forest diseases and insect pests are the most destructivenatural disasters to forestry they are always an importantconstraint to forestry production worldwide and cause majoreconomic losses in forestry For example spruce budwormmountain pine beetle and gypsy moth have led to significantlosses in Canadian forests [1] Therefore how to preventthe occurrence and spreading of forest diseases and insectpests and reduce the economic loss is always an importantsubject in theory and practice Mathematical models havebeen established to study change rule of the number of forestpests these models include ordinary differential equation(ODE) system difference equation system partial differentialequation (PDE) system and so forth In order to controlpests integrated pestmanagement (IPM) [2ndash4] including twomeasures will be taken one is chemical control (spraying pes-ticide) and the other is biological control (releasing naturalenemy) In the process of pest management if the amountof pests is small biological control will be implemented andif the amount of pests is large integrated pest control willbe applied Usually system with sudden disturbance alwaysleads to impulsive system In the forest pest managementsystem humanrsquos interference can cause a sudden change ofthe number or density of pests So impulsive differential equa-tion (IDE) can be established to explain this phenomenon

Many biological models based on time impulse [5ndash13] or onestate impulse [14 15] have been studied by many authorsThe model with two state impulses is very rare In [16 17]two models with two state impulses are established to studythe existence and the stability of periodic solution of themodels respectively by Poincare map and the Lambert 119872function In [18] a model with two state impulses is builtup and the geometrical analysis of the model is made byBendixson theorem of impulsive differential equations andthe successor function Motivated by [15 17 18] we getthe following reasonable model by introducing two statedependent impulses

1199091015840(119905) =

119903119909

1 + 119886119909minus 119887119909119910

1199101015840(119905) = 119888119909119910 minus 119889119910

119909 = ℎ1 ℎ2 or 119909 = ℎ

1 119910 gt 119910

lowast

Δ119909 = 0

Δ119910 = 120572119909 = ℎ1 119910 le 119910

lowast

Δ119909 = minus119901119909

Δ119910 = minus119902119910 + 120591119909 = ℎ2

(1)

where Δ119909(119905) = 119909(119905+) minus 119909(119905) Δ119910(119905) = 119910(119905

+) minus 119910(119905) 119909(119905)

is the density of pests at time 119905 and 119910(119905) is the density ofnatural enemies 119903119909(1 + 119886119909) is saturated growth rate 119887 is

2 Abstract and Applied Analysis

O

x

N M

Y

X

x+

Figure 1 The successor function of system (2)

predation rate of natural enemies and 119888 is the growth rateof natural enemies after digestion and absorption 119889 is thedeath rate of natural enemies ℎ

1and ℎ2are the thresholdwith

slight damage and serious damage to the forest respectively119910lowast is the ordinate of intersection of the line 119909 = ℎ

1and

(119903(1 + 119886119909)) minus 119887119910 = 0 120572 120591 are the amount of natural enemies119910(119905) released one time 119901 and 119902 are death rates of pests andnatural enemies due to pesticide respectively According tothe biological significance we only discuss 1198772

+= (119909 119910) | 119909 ge

0 119910 ge 0The paper is organized as follows In Section 2 some

definitions and theorems are given for the later use InSection 3 the stability of system (1) without impulsive effectis investigated qualitatively In Section 4 the existence andstability of periodic solution to system (1) are investigatedusing the successor function Bendixson theorem and sta-bility theorem of periodic solution of impulsive differentialequations In Section 5 an example and some simulations areexerted to prove the results

2 Preliminaries

We consider the state impulsive differential equation

1199091015840(119905) = 119875 (119909 119910)

1199101015840(119905) = 119876 (119909 119910)

(119909 119910) notin 119872119909 119910

Δ119909 = 120572 (119909 119910)

Δ119910 = 120573 (119909 119910) (119909 119910) isin 119872119909 119910

(2)

where119872 is one-dimensional line or curve in 1198772119872 is calledimpulsive set and the function 119868 is continuous mapping119868(119872) = 119873 119868 is called the impulse function and 119873 is calledthe corresponding image set

Definition 1 If there exists a point 119875 in the image set suchthat 119891(119875 119879) = 119876 isin 119872119909 119910 and the impulsive mapping119868(119876) = 119868(119891(119875 119879)) = 119875 isin 119873 then one calls 119891(119875 119879) theperiodic solution of system (2)

Definition 2 Function

119891 (119909) = 119891 (119909+) minus 119891 (119909) (3)

is called a successor function of the point 119909 (see Figure 1)

Theorem 3 (Bendixson theorem of impulsive differentialequations) Assume119866 is a Bendixson region of system (2) if119866does not contain any critical points of system (2) then system(2) has a closed orbit in 119866

Theorem 4 Assume that in continuous dynamic system (119883

Π) there exist two points 1199091 1199092in the pulse phase concentra-

tion such that the successor function 119891(1199091) gt 0 and 119891(119909

2) lt 0

then there exists a point 119875 between the two points 1199091 1199092that

makes 119891(119875) = 0 that is the system has periodic solution

Theorem 5 (see [19 20]) Consider the state impulsive differ-ential equation

1199091015840(119905) = 119891 (119909 119910)

1199101015840(119905) = 119892 (119909 119910)

Φ (119909 119910) = 0

Δ119909 = 120572 (119909 119910)

Δ119910 = 120573 (119909 119910) Φ (119909 119910) = 0

(4)

where the function Φ(119909 119910) = 0 is corresponding to the impul-sive set119872 The 119879-periodic solution 119909 = 120601(119905) 119910 = 120593(119905) of themodel is orbitally asymptotically stable if the multiplier 120583

2lt 1

is calculated by the following formula

1205832=

119902

prod

119896=1

Δ119896exp [int

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905))

+120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905]

(5)

where

Δ119896= (119891+(120597120573

120597119910sdot120597Φ

120597119909minus120597120573

120597119909sdot120597Φ

120597119910+120597Φ

120597119909)

+119892+(120597120572

120597119909sdot120597Φ

120597119910minus120597120572

120597119910sdot120597Φ

120597119909+120597Φ

120597119910))

times (119891120597Φ

120597119909+ 119892

120597Φ

120597119910)

minus1

(6)

and 119891 119892 (120597120572120597119909) (120597120572120597119910) (120597120573120597119909) (120597120573120597119910) (120597Φ120597119909)(120597Φ120597119910) are calculated at the point (120601(119905

119896) 120593(119905119896)) and 119891

+=

119891(120601(119905+

119896) 120593(119905+

119896)) 119892+= 119892(120601(119905

+

119896) 120593(119905+

119896))

3 The Stability of System (1) withoutImpulsive Effect

In this section we consider system (1) without impulsiveeffect Let 120572 120591 119901 119902 = 0 then we get the following system

1199091015840

(119905) =119903119909

1 + 119886119909minus 119887119909119910

1199101015840

(119905) = 119910 (minus119889 + 119888119909)

(7)

Abstract and Applied Analysis 3

The system has two equilibrium points 1198640(0 0) and 119864(119909alefsym

119910alefsym) where 119909alefsym = 119889119888 and 119910alefsym = 119903119888(119887(119886119889 + 119888)) Assume that

the variational matrix of the system (7) is as follows

119869 = (

119903

(1 + 119886119909)2minus 119887119910 minus119887119909

119888119910 119888119909 minus 119889

) (8)

At 1198640(0 0)

119869 (1198640) = (

119903 0

0 minus119889) (9)

Obviously 1198640(0 0) is saddle point At 119864(119909alefsym 119910alefsym)

119869 (119864 (119909alefsym 119910alefsym)) = (

minus119903119886119888119889

(119888 + 119886119889)2minus119887119889

119888

1199031198882

119887 (119888 + 119886119889)0

) (10)

The characteristic equation of 119869(119864) satisfies 119891(120582) = 1205822 + 119901120582 +119902 = 0 where 119901 = 119903119886119888119889(119888 + 119886119889)2 119902 = (119903119888119889(119886119889 + 119888)) gt 0 andΔ = 119901

2minus 4119902 = 119903119888119889[119903119886

2119888119889 minus 4(119886119889 + 119888)

3](119886119889 + 119888)

4 Analyse Δwe have the following conclusion

(i) if Δ gt 0 that is 1199031198862119888119889 gt 4(119886119889 + 119888)3 then 119864(119909alefsym 119910alefsym) isstable node

(ii) if Δ = 0 that is 1199031198862119888119889 = 4(119886119889 + 119888)3 then 119864(119909alefsym 119910alefsym) isstable critical node

(iii) if Δ lt 0 that is 1199031198862119888119889 lt 4(119886119889 + 119888)3 then 119864(119909alefsym 119910alefsym) isstable focus

Denote (1198671) 1199031198862119888119889 lt 4(119886119889 + 119888)3 we have the following the-

orem

Theorem 6 The positive equilibrium point of system (7) isstable focus provided (119867

1) is true

Theorem 7 The solution of system (7) is bounded

Proof Let (119909(119905) 119910(119905)) be solution of system (7) satisfied initialconditions 119909(119905

0) = 1199090gt 0 119910(119905

0) = 1199100gt 0 From the second

equation of system (7) it follows that 119889119910119889119905 gt 0 for 119909 gt 119889119888And there exists 119879

0gt 1199050 such that orbit (119909(119905) 119910(119905)) intersects

with isoclinic line 119910 = 119903(119887(1 + 119886119909)) at point (119909(1198790) 119910(1198790))

Thus we have 119909(119905) lt 119909(1198790) for 119905 gt 119879

0 Let119870 gt 119909(119879

0) we have

119909(119905) lt 119870 for 119905 gt 1198790

On the other hand making three lines in the first quad-rant Γ

1119909minus119870 = 0 Γ

2119910+(119888119887)119909minus119872 = 0 and Γ

3119910minus119872+(119889119887) =

0 we can acquire that

119889Γ2

119889119905= [119910 (119888119909 minus 119889) +

119888

119887119909 (

119903

1 + 119886119909minus 119887119910)]

Γ2

= [119888119903119909

119887 (1 + 119886119909)+119889119888119909

119887] minus119872119889

(11)

Then we have 119889Γ2119889119905 lt 0 for 119889119888 lt 119909 lt 119870 and 119872 large

enough Similarly

119889Γ3

119889119905= 119910 (119888119909 minus 119889) (12)

5

4

3

2

1

0

543210

y

x

Figure 2 Vector graph of system (7) with a stable focus where 119903 =06 119886 = 1 119887 = 05 119888 = 08 and 119889 = 04

Then we have 119889Γ3119889119905 lt 0 for 0 lt 119909 lt 119889119888 According to the

previous discussion there exists an areaΩwhich is composedof 119909 = 0 119910 = 0 Γ

1 Γ2 Γ3 For all initial points (119909(119905

0) 119910(1199050))

exist119879 gt 0 for 119905 gt 119879 we can acquire (119909(119905) 119910(119905)) isin Ω Thus thesolution of system (7) satisfied initial conditions (119909(119905

0) 119910(1199050))

is bounded

Theorem 8 The positive equilibrium point of system (7) isglobally asymptotically stable provided condition (119867

1) is true

Proof Firstly we claim that the positive equilibrium point ofsystem (7) is stable focus point and around it there does notexist closed rail line provided condition (119867

1) is true In fact

let Dulac function be 119861 = 1119909119910 in regard to the system (7) itfollows that

119863 =120597 (119875119861)

120597119909+(119876119861)

120597119910=

minus119903119886

119910(1 + 119886119909)2lt 0 (13)

Due to the Bendixson-Dulac theorem around 119864 there doesnot exist closed rail line SynthesizeTheorems 6 and 7 there-fore we have that the positive equilibrium point of system (7)is globally asymptotically stable provided condition (119867

1) is

true

Vector graph of system (7) with a stable focus can be seenin Figure 2

4 The Analysis of System (1) with StateDependent on Pulse

41 The Existence of Periodic Solutions of the System In thissection we will discuss the existence of periodic solution ofsystem (1) by using the successor function defined in this


Q

R

P

E

LY

XO h1 (1 minus p)h2 h2 d

c

L998400

ylowast

N1 N2 M2

M1

Figure 3 The structure graph of system (1)

Q

R

P

E

O

LY

Xh1 (1 minus p)h2 h2

d

c

L998400

ylowast

N1 N2 M2

M1

C1

C+1

C0

Figure 4 The orbit starting from the point 1198620above 119875 (Case 1 in

Section 411)

paper and qualitative analysis Let = 0 119910 = 0 we knowthe curve 1198711015840 119910 = 119903(119887(1 + 119886119909)) and 119910-axis are 119883-isoline andthe line119871119909 = 119889119888 and119909-axis are119884-isoline From the previousdiscussion we know119864(119909alefsym 119910alefsym) is a stable focus provided (119867

1)

is true Denote the first impulsive set as1198721= (119909 119910) | 119909 =

ℎ1 0 le 119910 le 119910

lowast the second impulsive set as119872

2= (119909 119910) |

119909 = ℎ2 119910 ge 0 the image set corresponding to set 119872

1as

1198731= (119909 119910) | 119909 = ℎ

1 0 lt 119910 le 119910

lowast+ 120572 and the image

set corresponding to set 1198722as 1198732= (119909 119910) | 119909 = (1 minus

119901)ℎ2 119910 ge 120591 Let points 119875119876 and 119877 be the intersection of line

1198711015840with lines119873

11198732 and119872

2 respectivelyDue to the practical

significance in this paper we assume the set always lies in theleft side of stable focus 119864 that is ℎ

1lt (1 minus 119901)ℎ

2lt ℎ2lt 119889119888

We will start our discussion about the existence of periodic

Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

ylowast

N1 N2 M2

M1

C1

C+1

C0

d

c

D1

D+1

D0

Figure 5 The orbit starting from the point 1198620(Case 2 in

Section 411)

solution for (1)with the initial point1198620(1199091198880

1199101198880

)The structuregraph of system (1) can be seen in Figure 3

411 The Orbit Starting from the Point 1198620on 1198731above 119875

Starting from 1198620 the orbit intersects with 119872

1at point

1198621(1199091198881

1199101198881

) and then produces pulse to the point 11986210158401(1199091198881015840

1

1199101198881015840

1

)

on1198721 According to (1) the following is got 119910

1198621015840

1

= 1199101198621

+ 120572After 119899 time impulse the point 119862+

1(119909119888+

1

119910119888+

1

) should satisfy119910lowastlt 119910119862+

1

le 119910lowast+ 120572 Thus we only need to consider case

119910lowastlt 1199101198620

le 119910lowast+ 120572 and three cases should be discussed

Case 1 If the impulsive point 119862+1corresponding to 119862

1is

exactly1198620 thus the curve119862

01198621119862+

1shall constitute a periodic

orbit of (1) (see Figure 4)


1is

below 1198620on 119873

1 obviously the successor function of 119862

0

satisfies 119891(1198620) = 119910119862+

1

minus 1199101198620

lt 0 On the other hand choosea point 119863

0next to 119875 on 119873

1(ie |119863

0119875| lt 120576) The orbit of (1)

starting from point 1198630intersects with 119872

1at point 119863

1and

then jumps to point 119863+1in the line 119873

1 Because 119863

0is next

to 119875 1198631is next to 119875 and 119910

119863+

1

= 1199101198631

+ 120572 thus the successorfunction of119863

0satisfies119891(119863

0) = 119910119863+

1

minus1199101198630

gt 0 ByTheorem 4there exists a period solution (see Figure 5)


1is

above 1198620on 1198731 obviously 119891(119862

0) gt 0 On the other hand

the orbit starting from point119860(ℎ1 119910lowast+120572) intersects with119872

1

at point 1198601and then jumps to point 119860+

1(ℎ1 119910119860+

1

) in the line1198731 According to system (1) the following is got 119910lowast lt 119910

119860+

1

lt

119910lowast+ 120572 then the successor function of 119860 satisfies 119891(119860) lt 0

Based on the previous discussion according to Theorem 4there exists a periodic orbit of (1) (see Figure 6)

412 The Orbit Starting from the Point 119876 Starting from 119876the orbit intersects with 119872

2at point 119862

1and then produces


A

Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

ylowast

N1 N2 M2

M1

C1

C+1

C0

d

c

A1

A+1


Section 411)

P

R

E

O X

Q

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c

Figure 7 The orbit starting from the point 119876 (Case 1 inSection 412)

pulse to the point 119862+1on 1198732 According to (1) the following

is got

119909119862+

1

= (1 minus 119901) ℎ2

119910119862+

1

= (1 minus 119902) 1199101198621

+ 120591

(14)

By selecting 120591 three cases should be discussed


1is

exactly 119876 thus the curve 1198761198621119862+




1is

above 119876 on 1198732 thus 119891(119876) = 119910

119862+

1

minus 119910119876

gt 0 Choose

E

R

Q

P

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c

D1

D2

D0

D+2


a point 1198630above 119862+

1in the straight line 119873

2 Only when it

intersects with 1198731at 119875 can the orbit of (1) starting from

point 1198630be vertical After intersecting with119872

2at point 119863

2

through1198732from left to right it jumps to point119863+

2in the line

1198732 According to the existence and uniqueness of impulsive

differential equations1198632must be definitely below 119862

1on1198722

and 119863+2must be below 119862

+

1on 1198732 therefore the successor

function of 1198630is 119891(119863

0) = 119910

119863+

2

minus 1199101198630

lt 0 Based on theprevious discussion we can see that the region 120581 surroundedby the closed curve 119863

0119875119863111986321198621119862+

11198630is a positive invariant

set of (1) and it contains no equilibrium point Accordingto Theorem 3 there exists a periodic orbit of (1) in 120581 (seeFigure 8)


1is

below 119876 on1198732 we shall have the successor function 119891(119876) =

119910119862+

1

minus 119910119876lt 0

In the meantime take a point 1198630next to 119861((1 minus 119901)ℎ

2 0)

on1198732(|1198630119861| ≪ 120576) Starting from119863

0 the orbit intersects with

1198722at point 119863

1and then produces pulse to the point 119863+

1on

1198732 According to (1) the following is got

119909119863+

1

= (1 minus 119901) ℎ2

119910119863+

1

= (1 minus 119902) 1199101198631

+ 120591

(15)

Thus119863+1must be above119863

0 and the successor function of119863

0

is 119891(1198630) = 119910119863+

1

minus 1199101198630

gt 0As a result the region 120581 surrounded by the closed curve

119876119863011986311198621119876 involves a periodic solution of (1) (see Figure 9)

413 The Orbit Starting from the Point 1198620between the Second

Impulsive Set 1198722and Its Image Set 119873

2 The orbit starting

from point 1198620 between line 119873

2and line119872

2 intersects with


B

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c

D1

D0

D+1


Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

ylowast

C2

C0

d

c

Figure 10 The orbit starting from the point 1198620(Case 1(a) in

Section 413)

line1198722at 1198621and jumps onto point 119862+

1on1198732 There are the

following cases with changing 120591

Case 1 If 119862+1on 1198732is exactly 119876 the orbit from point 119862+

1

moves to the point 1198622on1198722 For 119862

2 there are the following

cases

(a) If 1198622is exactly 119862

1 then the curve 119862

1119862+

11198622forms a

periodic orbit of (1) (see Figure 10)(b) If 119862

2on1198722is below 119862

1 the successor function of 119862

1

is 119891(1198621) = 1199101198622

minus 1199101198621

lt 0In the meantime take a point119863

0between119873

2and119872

2

next to 119909-axis After starting from 1198630 the orbit hits

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2M2

M1

C1

C0

C2

C+1

D1

D2D0

D+1

d

c

Figure 11 The orbit starting from the point 1198620(Case 1(b) in

Section 413)

P

Q

R

E

XO

LY

h1(1 minus p)h2 h2

L998400

N1N2 M2

M1

C1C0

C2

C+1

d

c


Section 413)

the point1198631on1198722 then jumps onto the point119863+

1on

1198732 and thenmoves to the point119863

2on1198722 Obviously

1198632is above 119863

1 thus the successor function of 119863

1is

119891(1198631) = 1199101198632

minus 1199101198631

gt 0

As a result we get a periodic orbit in the region120581 encircled by the closed curve 119862

1119862+

1119863+

111986311198621(see

Figure 11)

(c) If1198622on1198722is above119862

1 which does not exist periodic

orbit otherwise the curve 119862+11198622and 119862

01198621must be

crossed which leads to the conflict with the existenceand uniqueness of solution to impulsive differentialequations


P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1C2

D1

D2L998400

C+1

D+1

C0

D0


Section 413)

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1

C2

L998400

C+1

C0

ylowast

C3


Section 413)

Case 2 If 119862+1on 1198732is below 119876 the same conclusion can

be got in the similar discussion previously mentioned (seeFigures 12 and 13)

Case 3 If119862+1on1198732is above119876 for the orbit starting from119862

+

1

there exist the following cases

(a) If the orbit starting from119862+

1crosses119873

1from the right

to the left and becomes vertical only when it goesthrough the line 1198711015840 the point 119862

2is the intersection

of the orbit and 1198731 The same conclusion can be got

as the discussion in Section 411 (see Figure 14)

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1C2 L998400

C+1

C0


Section 413)

P

Q

RE

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C0

C2

C+1

D1

D2

D0

D+1

d

c


Section 413)

(b) The orbit starting from 119862+

1becomes vertical only

when it crosses 119875119876 and then it moves on automati-cally to119862

2on1198722 which is exactly119862

1 or below119862

1 or

above 1198621

If 1198622on 119872

2is exactly 119862


1119862+

11198622

forms a periodic orbit (see Figure 15)

If 1198622on1198722is below 119862

1 then the successor function

of 1198621is 119891(119862

1) = 1199101198622

minus 1199101198621

lt 0

On the other hand take a point 1198630between 119873

2and

1198722near to 119909-axis and the orbit starting from 119863

0

hits the point 1198631on 119872

2 then jumps onto the point

119863+

1on 1198732 and then returns to the point 119863

2on 119872

2


18

16

14

12

1

08

06

014 015 016 017 018 019 02

y(t)

x(t)

(a) Phase diagram of system (24)

014

015

016

017

018

019

02

x(t

t)

10 20 30 40 50 60

(b) Time series of system (24)

18

16

14

12

1

08

06

y(t)

t

10 20 30 40 50 60

(c) Time series of system (24)

Figure 17 Illustration of basic behavior of system (24)

Obviously1198632on1198722is above119863

1 Thus the successor

function of1198631is 119891(119863

1) = 1199101198632

minus 1199101198631

gt 0Based on the previous discussion above there existsa periodic orbit in the region 120581 encircled 119863

11198621119862+

11198622

and 119862+1119863+

111986321198622(see Figure 16)

(c) If1198622on1198722is above119862


orbit the reason is the same as in Case 1(c) inSection 413

42 The Stability Analysis of Periodic Solutions ofthe System (1)

421 The Stability of Periodic Solution about Impulsive Set1198721

Theorem 9 Let 119909 = 120601(119905) 119910 = 120593(119905) be the periodic solution ofsystem (1) and 120601(0) = ℎ

1 120593(0) = 120593

0 if 1205930lt (119903119887(1+119886ℎ

1))+120572

the periodic solution is stable


014 015 016 017 018 019 02

18

16

14

12

1

08

04

02

06

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t

)

10 20 30 40 50 60


18

16

14

12

1

08

04

02

06

y(t)

t

10 20 30 40 50 60



Proof Let

119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = 0 120573 (119909 119910) = 120572

Φ (119909 119910) = 119909 minus ℎ1

(16)

and (120601(0) 120593(0)) = (ℎ1 1205930) (120601(119879) 120593(119879)) = (119909

1 1199101) (120601(119879+)

120593(119879+)) = (119909

1 1199101+ 120572) = (ℎ

1 1205930) one gets

120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889

120597120572

120597119909= 0

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= 0

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(17)

10 Abstract and Applied Analysisy(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05



According toTheorem 5 we have

Δ1=119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))=

119891 (ℎ1 1205930)

119891 (ℎ1 1205930minus 120572)

=(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

11205930+ 119887ℎ1120572 minus 119887ℎ

1120572

(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

1(1205930minus 120572)

= 1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

1205832= Δ1expint

119879

0

[120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))] 119889119905

= Δ1expint

119879

0

[119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889]119889119905

lt Δ1expint

119879

0

[119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889] 119889119905

Abstract and Applied Analysis 11y(t)

12

11

1

09

08

07

06

05

0450403503025x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

12

11

1

09

08

07

06

05



= Δ1exp [int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905)]

= Δ1

1205930minus 120572

1205930

= (1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

)1205930minus 120572

1205930

(18)

Obviously if (119903(1 + 119886ℎ1)) minus 119887(120593

0minus 120572) gt 0 that is 120593

0lt

(119903119887(1+ 119886ℎ1)) +120572 we have 120583

2lt 1 Thus the periodic solution

to the system (1) is stable This completes the proof


Theorem 10 Let 119909 = 120601(119905) 119910 = 120593(119905) be the periodic solutionof system (1) and (120601(0) 120593(0)) = ((1 minus 119901)ℎ

2 1205930) if 120593

0lt

min(119903(1 minus 119902)119887(1 + 119886ℎ2)) + 120591 (1119887119902)(119887120591 minus (119886119901(1 minus 119902)119903ℎ

2(1 +


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05



119886ℎ2)(1 + 119886(1 minus 119901)ℎ

2))) or 120593

0gt max(119903(1 minus 119902)119887(1 + 119886ℎ

2)) +

120591 (1119887119902)(119887120591 minus (119886119901(1 minus 119902)119903ℎ2(1 + 119886ℎ

2)(1 + 119886(1 minus 119901)ℎ

2))) the

periodic solution to the system (1) is stable

Proof Let 119909 = 120601(119905) 119910 = 120593(119905) be the 119879 periodic solution ofsystem (1) and

119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = minus119901119909 120573 (119909 119910) = minus119902119910 + 120591

Φ (119909 119910) = 119909 minus ℎ2

(19)

(120601(119879+) 120593(119879

+)) = ((1minus119901)120601(119879) (1minus119902)120593(119879)+120591) (120601(119879) 120593(119879)) =

(ℎ2 (1205930minus 120591)(1 minus 119902)) one gets

120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

09

08

07

06

05



120597120572

120597119909= minus119901

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= minus119902

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(20)

Thus

Δ1=(1 minus 119902) 119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))

= ((1 minus 119902) [119903120601 (119879+)

1 + 119886120601 (119879+)minus 119887120601 (119879

+) 120593 (119879

+)])


y(t)

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

08

07

06

05



times (119903120601 (119879)

1 + 119886120601 (119879)minus 119887120601 (119879) 120593 (119879))

minus1

= ((1 minus 119902) [119903 (1 minus 119901) ℎ

2

1 + 119886 (1 minus 119901) ℎ2

minus 119887 (1 minus 119901) ℎ21205930])

times (119903ℎ2

1 + 119886ℎ2

minus 119887ℎ2

1205930minus 120591

1 minus 119902)

minus1

= ((1 minus 119901) (1 minus 119902) [119903

1 + 119886 (1 minus 119901) ℎ2

minus 1198871205930])

times (119903

1 + 119886ℎ2

minus 1198871205930minus 120591

1 minus 119902)

minus1

1205832= Δ1expint

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905


1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06



=(1 minus 119901) (1 minus 119902) [(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)

we have 1205832lt 1 then the periodic solution to the system (1) is

stable This completes the proof

5 Example and Numerical Simulation

To verify the validity of our results we give an example Let119903 = 06 119886 = 1 119887 = 05 119888 = 08 119889 = 04 119901 = 05 119902 = 02ℎ1= 02 and ℎ

2= 045 By calculation we obtain 119910lowast = 1

119875(02 1) 119876(0225 09796) and 119864(119909alefsym 119910alefsym) = (05 08) Thenwe have

1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)

Numerical analysis of system (24) is being done using Maple140 We have the following cases

Case 1 Initial point 1198620is in the line119873

1 where 120572 = 05

(a) Let 1198620be (02 12) above 119875(02 1) (see Figure 17)

(b) Let 1198620be (02 06) below 119875(02 1) (see Figure 18)

Case 2 Initial point 1198620is in the line 119873

2 Let 119862

0be

119876(0225 09796)

(a) The impulsive point119862+1corresponding to119862

1is exactly

119876 where 120591 = 062 (see Figure 19)(b) The impulsive point 119862+

1corresponding to 119862

1is above

119876 on1198732 where 120591 = 08 (see Figure 20)

(c) The impulsive point 119862+1corresponding to 119862

1is below


Case 3 Initial point 1198620is between the second impulsive set

1198722and its image set119873

2 Let 119862

0be (025 09)


1is exactly



1is below


(c) (1) The impulsive point 119862+1corresponding to 119862

1is

above 119876 on1198732 where 120591 = 08 (see Figure 24)

(2) The impulsive point 119862+1corresponding to 119862

1is

above 119876 on 1198732 where 120591 = 1 and 120572 = 05 (see

Figure 25)

All the simulations show agreement with the results inthis paper

6 Conclusion

Certainly the actual pest management will be very compli-cated such biological models may be described by more statedependent pulses and time impulse but they are difficultto study by the use of successor functions In this papera pest management model with saturated growth rate isconstructed We proposed two state dependent impulsiveeffects in the model according to quantity of pests andnatural enemies Our primary results are to explain the effectsof human interference in the actual pest management ondynamics of the system If the quantity of pests is less thanℎ1 but the quantity of natural enemies is enough relatively

(gt119910lowast) human interference is unnecessary If the quantity ofpests reaches a certain amount ℎ

1 but the quantity of natural

enemies is less than or equal to 119910lowast only biological control(releasing natural enemy) is implemented and if the quantityof pests is large (reach ℎ

2) integrated pest control (spraying

pesticide and releasing natural enemy) is applied We thinkour results will offer help to the actual pest management

Acknowledgments

This work is supported by the Shandong Provincial NaturalSciences Funds of China (Grant no ZR2012AM012) anda Project of Shandong Province Higher Educational Scienceand Technology Program of China (Grant no J13LI05)

References

[1] ldquoCanadian Forest Servicerdquo httpcfsnrcangccahome[2] S M Cook Z R Khan and J A Pickett ldquoThe use of push-

pull strategies in integrated pest managementrdquo Annual Reviewof Entomology vol 52 pp 375ndash400 2007

[3] N C Elliott J A Farrell A P Gutierrez et al Integrated PestManagement Springer New York NY USA 1995

[4] MKogan ldquoIntegrated pestmanagement historical perspectivesand contemporary developmentsrdquo Annual Review of Entomol-ogy vol 43 pp 243ndash270 1998

[5] C Dai M Zhao and L Chen ldquoHomoclinic bifurcation insemi-continuous dynamic systemsrdquo International Journal ofBiomathematics vol 5 no 6 Article ID 1250059 2012

[6] Z Li L Chen and J Huang ldquoPermanence and periodicity ofa delayed ratio-dependent predator-prey model with Hollingtype functional response and stage structurerdquo Journal of Com-putational and Applied Mathematics vol 233 no 2 pp 173ndash1872009

[7] B Liu L Chen and Y Zhang ldquoThe dynamics of a prey-dependent consumption model concerning impulsive controlstrategyrdquo Applied Mathematics and Computation vol 169 no 1pp 305ndash320 2005

[8] X Meng J Jiao and L Chen ldquoThe dynamics of an age struc-tured predator-prey model with disturbing pulse and time


delaysrdquo Nonlinear Analysis Real World Applications An Inter-national Multidisciplinary Journal vol 9 no 2 pp 547ndash5612008

[9] X Song M Hao and X Meng ldquoA stage-structured predator-prey model with disturbing pulse and time delaysrdquo AppliedMathematical Modelling vol 33 no 1 pp 211ndash223 2009

[10] S Tang and L Chen ldquoModelling and analysis of integratedpest management strategyrdquoDiscrete and Continuous DynamicalSystems B vol 4 no 3 pp 759ndash768 2004

[11] H Zhang L Chen and J J Nieto ldquoA delayed epidemic modelwith stage-structure and pulses for pest management strategyrdquoNonlinear Analysis Real World Applications An InternationalMultidisciplinary Journal vol 9 no 4 pp 1714ndash1726 2008

[12] T Zhang X Meng and Y Song ldquoThe dynamics of a high-dimensional delayed pest management model with impulsivepesticide input and harvesting prey at different fixedmomentsrdquoNonlinear Dynamics vol 64 no 1-2 pp 1ndash12 2011

[13] T Zhang X Meng Y Song et al ldquoDynamical analysis ofdelayed plant disease models with continuous or impulsivecultural control strategiesrdquo Abstract and Applied Analysis vol2012 Article ID 428453 25 pages 2012

[14] L Nie Z Teng L Hu and J Peng ldquoExistence and stabilityof periodic solution of a predator-prey model with state-dependent impulsive effectsrdquo Mathematics and Computers inSimulation vol 79 no 7 pp 2122ndash2134 2009

[15] S Tang and R A Cheke ldquoState-dependent impulsive models ofintegrated pestmanagement (IPM) strategies and their dynamicconsequencesrdquo Journal of Mathematical Biology vol 50 no 3pp 257ndash292 2005

[16] L Nie J Peng Z Teng and L Hu ldquoExistence and stability ofperiodic solution of a Lotka-Volterra predator-prey model withstate dependent impulsive effectsrdquo Journal of Computationaland Applied Mathematics vol 224 no 2 pp 544ndash555 2009

[17] L Nie Z Teng L Hu and J Peng ldquoQualitative analysis ofa modified Leslie-Gower and Holling-type II predator-preymodel with state dependent impulsive effectsrdquo Nonlinear Anal-ysis RealWorld Applications An International MultidisciplinaryJournal vol 11 no 3 pp 1364ndash1373 2010

[18] L Zhao L Chen and Q Zhang ldquoThe geometrical analysis ofa predator-prey model with two state impulsesrdquo MathematicalBiosciences vol 238 no 2 pp 55ndash64 2012

[19] D Bainov and P S Simeonov Impulsive Differential EquationsPeriodic Solutions and Applications Longman Scientific ampTechnical Harlow UK 1993

[20] V LakshmikanthamD D Bainov and P S SimeonovTheory ofImpulsive Differential Equations World Scientific Teaneck NJUSA 1989

Submit your manuscripts athttpwwwhindawicom

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Hindawi Publishing Corporationhttpwwwhindawicom

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


O

x

N M

Y

X

x+

Figure 1 The successor function of system (2)

predation rate of natural enemies and 119888 is the growth rateof natural enemies after digestion and absorption 119889 is thedeath rate of natural enemies ℎ

1and ℎ2are the thresholdwith

slight damage and serious damage to the forest respectively119910lowast is the ordinate of intersection of the line 119909 = ℎ

1and

(119903(1 + 119886119909)) minus 119887119910 = 0 120572 120591 are the amount of natural enemies119910(119905) released one time 119901 and 119902 are death rates of pests andnatural enemies due to pesticide respectively According tothe biological significance we only discuss 1198772

+= (119909 119910) | 119909 ge

0 119910 ge 0The paper is organized as follows In Section 2 some

definitions and theorems are given for the later use InSection 3 the stability of system (1) without impulsive effectis investigated qualitatively In Section 4 the existence andstability of periodic solution to system (1) are investigatedusing the successor function Bendixson theorem and sta-bility theorem of periodic solution of impulsive differentialequations In Section 5 an example and some simulations areexerted to prove the results

2 Preliminaries

We consider the state impulsive differential equation

1199091015840(119905) = 119875 (119909 119910)

1199101015840(119905) = 119876 (119909 119910)

(119909 119910) notin 119872119909 119910

Δ119909 = 120572 (119909 119910)

Δ119910 = 120573 (119909 119910) (119909 119910) isin 119872119909 119910

(2)

where119872 is one-dimensional line or curve in 1198772119872 is calledimpulsive set and the function 119868 is continuous mapping119868(119872) = 119873 119868 is called the impulse function and 119873 is calledthe corresponding image set

Definition 1 If there exists a point 119875 in the image set suchthat 119891(119875 119879) = 119876 isin 119872119909 119910 and the impulsive mapping119868(119876) = 119868(119891(119875 119879)) = 119875 isin 119873 then one calls 119891(119875 119879) theperiodic solution of system (2)

Definition 2 Function

119891 (119909) = 119891 (119909+) minus 119891 (119909) (3)

is called a successor function of the point 119909 (see Figure 1)

Theorem 3 (Bendixson theorem of impulsive differentialequations) Assume119866 is a Bendixson region of system (2) if119866does not contain any critical points of system (2) then system(2) has a closed orbit in 119866

Theorem 4 Assume that in continuous dynamic system (119883

Π) there exist two points 1199091 1199092in the pulse phase concentra-

tion such that the successor function 119891(1199091) gt 0 and 119891(119909

2) lt 0

then there exists a point 119875 between the two points 1199091 1199092that

makes 119891(119875) = 0 that is the system has periodic solution

Theorem 5 (see [19 20]) Consider the state impulsive differ-ential equation

1199091015840(119905) = 119891 (119909 119910)

1199101015840(119905) = 119892 (119909 119910)

Φ (119909 119910) = 0

Δ119909 = 120572 (119909 119910)

Δ119910 = 120573 (119909 119910) Φ (119909 119910) = 0

(4)

where the function Φ(119909 119910) = 0 is corresponding to the impul-sive set119872 The 119879-periodic solution 119909 = 120601(119905) 119910 = 120593(119905) of themodel is orbitally asymptotically stable if the multiplier 120583

2lt 1

is calculated by the following formula

1205832=

119902

prod

119896=1

Δ119896exp [int

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905))

+120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905]

(5)

where

Δ119896= (119891+(120597120573

120597119910sdot120597Φ

120597119909minus120597120573

120597119909sdot120597Φ

120597119910+120597Φ

120597119909)

+119892+(120597120572

120597119909sdot120597Φ

120597119910minus120597120572

120597119910sdot120597Φ

120597119909+120597Φ

120597119910))

times (119891120597Φ

120597119909+ 119892

120597Φ

120597119910)

minus1

(6)

and 119891 119892 (120597120572120597119909) (120597120572120597119910) (120597120573120597119909) (120597120573120597119910) (120597Φ120597119909)(120597Φ120597119910) are calculated at the point (120601(119905

119896) 120593(119905119896)) and 119891

+=

119891(120601(119905+

119896) 120593(119905+

119896)) 119892+= 119892(120601(119905

+

119896) 120593(119905+

119896))

3 The Stability of System (1) withoutImpulsive Effect

In this section we consider system (1) without impulsiveeffect Let 120572 120591 119901 119902 = 0 then we get the following system

1199091015840

(119905) =119903119909

1 + 119886119909minus 119887119909119910

1199101015840

(119905) = 119910 (minus119889 + 119888119909)

(7)





119869 = (

119903

(1 + 119886119909)2minus 119887119910 minus119887119909

119888119910 119888119909 minus 119889

) (8)

At 1198640(0 0)

119869 (1198640) = (

119903 0

0 minus119889) (9)


119869 (119864 (119909alefsym 119910alefsym)) = (

minus119903119886119888119889

(119888 + 119886119889)2minus119887119889

119888

1199031198882

119887 (119888 + 119886119889)0

) (10)


2minus 4119902 = 119903119888119889[119903119886

2119888119889 minus 4(119886119889 + 119888)

3](119886119889 + 119888)






orem


1) is true



0) = 1199090gt 0 119910(119905






0 Let119870 gt 119909(119879

0) we have

119909(119905) lt 119870 for 119905 gt 1198790


1119909minus119870 = 0 Γ

2119910+(119888119887)119909minus119872 = 0 and Γ

3119910minus119872+(119889119887) =


119889Γ2

119889119905= [119910 (119888119909 minus 119889) +

119888

119887119909 (

119903

1 + 119886119909minus 119887119910)]

Γ2

= [119888119903119909

119887 (1 + 119886119909)+119889119888119909

119887] minus119872119889

(11)


enough Similarly

119889Γ3

119889119905= 119910 (119888119909 minus 119889) (12)

5

4

3

2

1

0

543210

y

x





0) 119910(1199050))


0) 119910(1199050))

is bounded


1) is true


1) is true In fact


119863 =120597 (119875119861)

120597119909+(119876119861)

120597119910=

minus119903119886

119910(1 + 119886119909)2lt 0 (13)


1) is

true





Q

R

P

E

LY


c

L998400

ylowast

N1 N2 M2

M1


Q

R

P

E

O

LY


d

c

L998400

ylowast

N1 N2 M2

M1

C1

C+1

C0


Section 411)


1)


ℎ1 0 le 119910 le 119910


2= (119909 119910) |


1as

1198731= (119909 119910) | 119909 = ℎ

1 0 lt 119910 le 119910




1198711015840with lines119873

11198732 and119872



1lt (1 minus 119901)ℎ

2lt ℎ2lt 119889119888


Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

ylowast

N1 N2 M2

M1

C1

C+1

C0

d

c

D1

D+1

D0


Section 411)


1199101198880




1at point

1198621(1199091198881

1199101198881


1

1199101198881015840

1

)


1198621015840

1

= 1199101198621


1(119909119888+

1

119910119888+

1


1


119910lowastlt 1199101198620



1is


01198621119862+




1is

below 1198620on 119873


0

satisfies 119891(1198620) = 119910119862+

1

minus 1199101198620


0next to 119875 on 119873

1(ie |119863

0119875| lt 120576) The orbit of (1)


1at point 119863

1and


1 Because 119863

0is next

to 119875 1198631is next to 119875 and 119910

119863+

1

= 1199101198631


0satisfies119891(119863

0) = 119910119863+

1

minus1199101198630



1is




1


1(ℎ1 119910119860+

1


119860+

1

lt




2at point 119862

1and then produces


A

Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

ylowast

N1 N2 M2

M1

C1

C+1

C0

d

c

A1

A+1


Section 411)

P

R

E

O X

Q

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c



is got

119909119862+

1

= (1 minus 119901) ℎ2

119910119862+

1

= (1 minus 119902) 1199101198621

+ 120591

(14)



1is





1is

above 119876 on 1198732 thus 119891(119876) = 119910

119862+

1

minus 119910119876

gt 0 Choose

E

R

Q

P

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c

D1

D2

D0

D+2


a point 1198630above 119862+


2 Only when it



2at point 119863

2


2in the line



1on1198722


+


function of 1198630is 119891(119863

0) = 119910

119863+

2

minus 1199101198630


0119875119863111986321198621119862+




1is


119910119862+

1

minus 119910119876lt 0


2 0)

on1198732(|1198630119861| ≪ 120576) Starting from119863


1198722at point 119863


1on


119909119863+

1

= (1 minus 119901) ℎ2

119910119863+

1

= (1 minus 119902) 1199101198631

+ 120591

(15)



0

is 119891(1198630) = 119910119863+

1

minus 1199101198630







2and line119872

2 intersects with


B

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c

D1

D0

D+1


Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

ylowast

C2

C0

d

c


Section 413)





1



cases

(a) If 1198622is exactly 119862


1119862+

11198622forms a


2on1198722is below 119862


1

is 119891(1198621) = 1199101198622

minus 1199101198621


0between119873

2and119872

2


P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2M2

M1

C1

C0

C2

C+1

D1

D2D0

D+1

d

c


Section 413)

P

Q

R

E

XO

LY

h1(1 minus p)h2 h2

L998400

N1N2 M2

M1

C1C0

C2

C+1

d

c


Section 413)


1on



1198632is above 119863


1is

119891(1198631) = 1199101198632

minus 1199101198631

gt 0


1119862+

1119863+

111986311198621(see

Figure 11)

(c) If1198622on1198722is above119862



01198621must be



P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1C2

D1

D2L998400

C+1

D+1

C0

D0


Section 413)

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1

C2

L998400

C+1

C0

ylowast

C3


Section 413)




+

1



1crosses119873

1from the right





P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1C2 L998400

C+1

C0


Section 413)

P

Q

RE

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C0

C2

C+1

D1

D2

D0

D+1

d

c


Section 413)





1 or below119862

1 or

above 1198621

If 1198622on 119872

2is exactly 119862


1119862+

11198622


If 1198622on1198722is below 119862


of 1198621is 119891(119862

1) = 1199101198622

minus 1199101198621

lt 0


2and


0



119863+


2on 119872

2


18

16

14

12

1

08

06

014 015 016 017 018 019 02

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t)

10 20 30 40 50 60


18

16

14

12

1

08

06

y(t)

t

10 20 30 40 50 60





function of1198631is 119891(119863

1) = 1199101198632

minus 1199101198631


11198621119862+

11198622

and 119862+1119863+

111986321198622(see Figure 16)

(c) If1198622on1198722is above119862






1 120593(0) = 120593

0 if 1205930lt (119903119887(1+119886ℎ

1))+120572



014 015 016 017 018 019 02

18

16

14

12

1

08

04

02

06

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t

)

10 20 30 40 50 60


18

16

14

12

1

08

04

02

06

y(t)

t

10 20 30 40 50 60



Proof Let

119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = 0 120573 (119909 119910) = 120572

Φ (119909 119910) = 119909 minus ℎ1

(16)

and (120601(0) 120593(0)) = (ℎ1 1205930) (120601(119879) 120593(119879)) = (119909

1 1199101) (120601(119879+)

120593(119879+)) = (119909

1 1199101+ 120572) = (ℎ

1 1205930) one gets

120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889

120597120572

120597119909= 0

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= 0

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(17)


09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05




Δ1=119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))=

119891 (ℎ1 1205930)

119891 (ℎ1 1205930minus 120572)

=(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

11205930+ 119887ℎ1120572 minus 119887ℎ

1120572

(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

1(1205930minus 120572)

= 1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

1205832= Δ1expint

119879

0

[120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))] 119889119905

= Δ1expint

119879

0

[119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889]119889119905

lt Δ1expint

119879

0

[119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889] 119889119905


12

11

1

09

08

07

06

05

0450403503025x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

12

11

1

09

08

07

06

05



= Δ1exp [int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905)]

= Δ1

1205930minus 120572

1205930

= (1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

)1205930minus 120572

1205930

(18)


0minus 120572) gt 0 that is 120593

0lt

(119903119887(1+ 119886ℎ1)) +120572 we have 120583





2 1205930) if 120593

0lt


2(1 +


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05



119886ℎ2)(1 + 119886(1 minus 119901)ℎ

2))) or 120593

0gt max(119903(1 minus 119902)119887(1 + 119886ℎ

2)) +

120591 (1119887119902)(119887120591 minus (119886119901(1 minus 119902)119903ℎ2(1 + 119886ℎ

2)(1 + 119886(1 minus 119901)ℎ

2))) the



119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = minus119901119909 120573 (119909 119910) = minus119902119910 + 120591

Φ (119909 119910) = 119909 minus ℎ2

(19)

(120601(119879+) 120593(119879

+)) = ((1minus119901)120601(119879) (1minus119902)120593(119879)+120591) (120601(119879) 120593(119879)) =


120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

09

08

07

06

05



120597120572

120597119909= minus119901

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= minus119902

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(20)

Thus

Δ1=(1 minus 119902) 119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))

= ((1 minus 119902) [119903120601 (119879+)

1 + 119886120601 (119879+)minus 119887120601 (119879

+) 120593 (119879

+)])


y(t)

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

08

07

06

05



times (119903120601 (119879)

1 + 119886120601 (119879)minus 119887120601 (119879) 120593 (119879))

minus1

= ((1 minus 119902) [119903 (1 minus 119901) ℎ

2

1 + 119886 (1 minus 119901) ℎ2

minus 119887 (1 minus 119901) ℎ21205930])

times (119903ℎ2

1 + 119886ℎ2

minus 119887ℎ2

1205930minus 120591

1 minus 119902)

minus1

= ((1 minus 119901) (1 minus 119902) [119903

1 + 119886 (1 minus 119901) ℎ2

minus 1198871205930])

times (119903

1 + 119886ℎ2

minus 1198871205930minus 120591

1 minus 119902)

minus1

1205832= Δ1expint

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905


1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra













119869 = (

119903

(1 + 119886119909)2minus 119887119910 minus119887119909

119888119910 119888119909 minus 119889

) (8)

At 1198640(0 0)

119869 (1198640) = (

119903 0

0 minus119889) (9)


119869 (119864 (119909alefsym 119910alefsym)) = (

minus119903119886119888119889

(119888 + 119886119889)2minus119887119889

119888

1199031198882

119887 (119888 + 119886119889)0

) (10)


2minus 4119902 = 119903119888119889[119903119886

2119888119889 minus 4(119886119889 + 119888)

3](119886119889 + 119888)






orem


1) is true



0) = 1199090gt 0 119910(119905






0 Let119870 gt 119909(119879

0) we have

119909(119905) lt 119870 for 119905 gt 1198790


1119909minus119870 = 0 Γ

2119910+(119888119887)119909minus119872 = 0 and Γ

3119910minus119872+(119889119887) =


119889Γ2

119889119905= [119910 (119888119909 minus 119889) +

119888

119887119909 (

119903

1 + 119886119909minus 119887119910)]

Γ2

= [119888119903119909

119887 (1 + 119886119909)+119889119888119909

119887] minus119872119889

(11)


enough Similarly

119889Γ3

119889119905= 119910 (119888119909 minus 119889) (12)

5

4

3

2

1

0

543210

y

x





0) 119910(1199050))


0) 119910(1199050))

is bounded


1) is true


1) is true In fact


119863 =120597 (119875119861)

120597119909+(119876119861)

120597119910=

minus119903119886

119910(1 + 119886119909)2lt 0 (13)


1) is

true





Q

R

P

E

LY


c

L998400

ylowast

N1 N2 M2

M1


Q

R

P

E

O

LY


d

c

L998400

ylowast

N1 N2 M2

M1

C1

C+1

C0


Section 411)


1)


ℎ1 0 le 119910 le 119910


2= (119909 119910) |


1as

1198731= (119909 119910) | 119909 = ℎ

1 0 lt 119910 le 119910




1198711015840with lines119873

11198732 and119872



1lt (1 minus 119901)ℎ

2lt ℎ2lt 119889119888


Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

ylowast

N1 N2 M2

M1

C1

C+1

C0

d

c

D1

D+1

D0


Section 411)


1199101198880




1at point

1198621(1199091198881

1199101198881


1

1199101198881015840

1

)


1198621015840

1

= 1199101198621


1(119909119888+

1

119910119888+

1


1


119910lowastlt 1199101198620



1is


01198621119862+




1is

below 1198620on 119873


0

satisfies 119891(1198620) = 119910119862+

1

minus 1199101198620


0next to 119875 on 119873

1(ie |119863

0119875| lt 120576) The orbit of (1)


1at point 119863

1and


1 Because 119863

0is next

to 119875 1198631is next to 119875 and 119910

119863+

1

= 1199101198631


0satisfies119891(119863

0) = 119910119863+

1

minus1199101198630



1is




1


1(ℎ1 119910119860+

1


119860+

1

lt




2at point 119862

1and then produces


A

Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

ylowast

N1 N2 M2

M1

C1

C+1

C0

d

c

A1

A+1


Section 411)

P

R

E

O X

Q

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c



is got

119909119862+

1

= (1 minus 119901) ℎ2

119910119862+

1

= (1 minus 119902) 1199101198621

+ 120591

(14)



1is





1is

above 119876 on 1198732 thus 119891(119876) = 119910

119862+

1

minus 119910119876

gt 0 Choose

E

R

Q

P

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c

D1

D2

D0

D+2


a point 1198630above 119862+


2 Only when it



2at point 119863

2


2in the line



1on1198722


+


function of 1198630is 119891(119863

0) = 119910

119863+

2

minus 1199101198630


0119875119863111986321198621119862+




1is


119910119862+

1

minus 119910119876lt 0


2 0)

on1198732(|1198630119861| ≪ 120576) Starting from119863


1198722at point 119863


1on


119909119863+

1

= (1 minus 119901) ℎ2

119910119863+

1

= (1 minus 119902) 1199101198631

+ 120591

(15)



0

is 119891(1198630) = 119910119863+

1

minus 1199101198630







2and line119872

2 intersects with


B

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c

D1

D0

D+1


Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

ylowast

C2

C0

d

c


Section 413)





1



cases

(a) If 1198622is exactly 119862


1119862+

11198622forms a


2on1198722is below 119862


1

is 119891(1198621) = 1199101198622

minus 1199101198621


0between119873

2and119872

2


P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2M2

M1

C1

C0

C2

C+1

D1

D2D0

D+1

d

c


Section 413)

P

Q

R

E

XO

LY

h1(1 minus p)h2 h2

L998400

N1N2 M2

M1

C1C0

C2

C+1

d

c


Section 413)


1on



1198632is above 119863


1is

119891(1198631) = 1199101198632

minus 1199101198631

gt 0


1119862+

1119863+

111986311198621(see

Figure 11)

(c) If1198622on1198722is above119862



01198621must be



P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1C2

D1

D2L998400

C+1

D+1

C0

D0


Section 413)

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1

C2

L998400

C+1

C0

ylowast

C3


Section 413)




+

1



1crosses119873

1from the right





P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1C2 L998400

C+1

C0


Section 413)

P

Q

RE

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C0

C2

C+1

D1

D2

D0

D+1

d

c


Section 413)





1 or below119862

1 or

above 1198621

If 1198622on 119872

2is exactly 119862


1119862+

11198622


If 1198622on1198722is below 119862


of 1198621is 119891(119862

1) = 1199101198622

minus 1199101198621

lt 0


2and


0



119863+


2on 119872

2


18

16

14

12

1

08

06

014 015 016 017 018 019 02

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t)

10 20 30 40 50 60


18

16

14

12

1

08

06

y(t)

t

10 20 30 40 50 60





function of1198631is 119891(119863

1) = 1199101198632

minus 1199101198631


11198621119862+

11198622

and 119862+1119863+

111986321198622(see Figure 16)

(c) If1198622on1198722is above119862






1 120593(0) = 120593

0 if 1205930lt (119903119887(1+119886ℎ

1))+120572



014 015 016 017 018 019 02

18

16

14

12

1

08

04

02

06

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t

)

10 20 30 40 50 60


18

16

14

12

1

08

04

02

06

y(t)

t

10 20 30 40 50 60



Proof Let

119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = 0 120573 (119909 119910) = 120572

Φ (119909 119910) = 119909 minus ℎ1

(16)

and (120601(0) 120593(0)) = (ℎ1 1205930) (120601(119879) 120593(119879)) = (119909

1 1199101) (120601(119879+)

120593(119879+)) = (119909

1 1199101+ 120572) = (ℎ

1 1205930) one gets

120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889

120597120572

120597119909= 0

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= 0

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(17)


09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05




Δ1=119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))=

119891 (ℎ1 1205930)

119891 (ℎ1 1205930minus 120572)

=(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

11205930+ 119887ℎ1120572 minus 119887ℎ

1120572

(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

1(1205930minus 120572)

= 1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

1205832= Δ1expint

119879

0

[120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))] 119889119905

= Δ1expint

119879

0

[119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889]119889119905

lt Δ1expint

119879

0

[119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889] 119889119905


12

11

1

09

08

07

06

05

0450403503025x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

12

11

1

09

08

07

06

05



= Δ1exp [int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905)]

= Δ1

1205930minus 120572

1205930

= (1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

)1205930minus 120572

1205930

(18)


0minus 120572) gt 0 that is 120593

0lt

(119903119887(1+ 119886ℎ1)) +120572 we have 120583





2 1205930) if 120593

0lt


2(1 +


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05



119886ℎ2)(1 + 119886(1 minus 119901)ℎ

2))) or 120593

0gt max(119903(1 minus 119902)119887(1 + 119886ℎ

2)) +

120591 (1119887119902)(119887120591 minus (119886119901(1 minus 119902)119903ℎ2(1 + 119886ℎ

2)(1 + 119886(1 minus 119901)ℎ

2))) the



119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = minus119901119909 120573 (119909 119910) = minus119902119910 + 120591

Φ (119909 119910) = 119909 minus ℎ2

(19)

(120601(119879+) 120593(119879

+)) = ((1minus119901)120601(119879) (1minus119902)120593(119879)+120591) (120601(119879) 120593(119879)) =


120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

09

08

07

06

05



120597120572

120597119909= minus119901

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= minus119902

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(20)

Thus

Δ1=(1 minus 119902) 119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))

= ((1 minus 119902) [119903120601 (119879+)

1 + 119886120601 (119879+)minus 119887120601 (119879

+) 120593 (119879

+)])


y(t)

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

08

07

06

05



times (119903120601 (119879)

1 + 119886120601 (119879)minus 119887120601 (119879) 120593 (119879))

minus1

= ((1 minus 119902) [119903 (1 minus 119901) ℎ

2

1 + 119886 (1 minus 119901) ℎ2

minus 119887 (1 minus 119901) ℎ21205930])

times (119903ℎ2

1 + 119886ℎ2

minus 119887ℎ2

1205930minus 120591

1 minus 119902)

minus1

= ((1 minus 119901) (1 minus 119902) [119903

1 + 119886 (1 minus 119901) ℎ2

minus 1198871205930])

times (119903

1 + 119886ℎ2

minus 1198871205930minus 120591

1 minus 119902)

minus1

1205832= Δ1expint

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905


1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Q

R

P

E

LY


c

L998400

ylowast

N1 N2 M2

M1


Q

R

P

E

O

LY


d

c

L998400

ylowast

N1 N2 M2

M1

C1

C+1

C0


Section 411)


1)


ℎ1 0 le 119910 le 119910


2= (119909 119910) |


1as

1198731= (119909 119910) | 119909 = ℎ

1 0 lt 119910 le 119910




1198711015840with lines119873

11198732 and119872



1lt (1 minus 119901)ℎ

2lt ℎ2lt 119889119888


Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

ylowast

N1 N2 M2

M1

C1

C+1

C0

d

c

D1

D+1

D0


Section 411)


1199101198880




1at point

1198621(1199091198881

1199101198881


1

1199101198881015840

1

)


1198621015840

1

= 1199101198621


1(119909119888+

1

119910119888+

1


1


119910lowastlt 1199101198620



1is


01198621119862+




1is

below 1198620on 119873


0

satisfies 119891(1198620) = 119910119862+

1

minus 1199101198620


0next to 119875 on 119873

1(ie |119863

0119875| lt 120576) The orbit of (1)


1at point 119863

1and


1 Because 119863

0is next

to 119875 1198631is next to 119875 and 119910

119863+

1

= 1199101198631


0satisfies119891(119863

0) = 119910119863+

1

minus1199101198630



1is




1


1(ℎ1 119910119860+

1


119860+

1

lt




2at point 119862

1and then produces


A

Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

ylowast

N1 N2 M2

M1

C1

C+1

C0

d

c

A1

A+1


Section 411)

P

R

E

O X

Q

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c



is got

119909119862+

1

= (1 minus 119901) ℎ2

119910119862+

1

= (1 minus 119902) 1199101198621

+ 120591

(14)



1is





1is

above 119876 on 1198732 thus 119891(119876) = 119910

119862+

1

minus 119910119876

gt 0 Choose

E

R

Q

P

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c

D1

D2

D0

D+2


a point 1198630above 119862+


2 Only when it



2at point 119863

2


2in the line



1on1198722


+


function of 1198630is 119891(119863

0) = 119910

119863+

2

minus 1199101198630


0119875119863111986321198621119862+




1is


119910119862+

1

minus 119910119876lt 0


2 0)

on1198732(|1198630119861| ≪ 120576) Starting from119863


1198722at point 119863


1on


119909119863+

1

= (1 minus 119901) ℎ2

119910119863+

1

= (1 minus 119902) 1199101198631

+ 120591

(15)



0

is 119891(1198630) = 119910119863+

1

minus 1199101198630







2and line119872

2 intersects with


B

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c

D1

D0

D+1


Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

ylowast

C2

C0

d

c


Section 413)





1



cases

(a) If 1198622is exactly 119862


1119862+

11198622forms a


2on1198722is below 119862


1

is 119891(1198621) = 1199101198622

minus 1199101198621


0between119873

2and119872

2


P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2M2

M1

C1

C0

C2

C+1

D1

D2D0

D+1

d

c


Section 413)

P

Q

R

E

XO

LY

h1(1 minus p)h2 h2

L998400

N1N2 M2

M1

C1C0

C2

C+1

d

c


Section 413)


1on



1198632is above 119863


1is

119891(1198631) = 1199101198632

minus 1199101198631

gt 0


1119862+

1119863+

111986311198621(see

Figure 11)

(c) If1198622on1198722is above119862



01198621must be



P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1C2

D1

D2L998400

C+1

D+1

C0

D0


Section 413)

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1

C2

L998400

C+1

C0

ylowast

C3


Section 413)




+

1



1crosses119873

1from the right





P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1C2 L998400

C+1

C0


Section 413)

P

Q

RE

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C0

C2

C+1

D1

D2

D0

D+1

d

c


Section 413)





1 or below119862

1 or

above 1198621

If 1198622on 119872

2is exactly 119862


1119862+

11198622


If 1198622on1198722is below 119862


of 1198621is 119891(119862

1) = 1199101198622

minus 1199101198621

lt 0


2and


0



119863+


2on 119872

2


18

16

14

12

1

08

06

014 015 016 017 018 019 02

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t)

10 20 30 40 50 60


18

16

14

12

1

08

06

y(t)

t

10 20 30 40 50 60





function of1198631is 119891(119863

1) = 1199101198632

minus 1199101198631


11198621119862+

11198622

and 119862+1119863+

111986321198622(see Figure 16)

(c) If1198622on1198722is above119862






1 120593(0) = 120593

0 if 1205930lt (119903119887(1+119886ℎ

1))+120572



014 015 016 017 018 019 02

18

16

14

12

1

08

04

02

06

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t

)

10 20 30 40 50 60


18

16

14

12

1

08

04

02

06

y(t)

t

10 20 30 40 50 60



Proof Let

119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = 0 120573 (119909 119910) = 120572

Φ (119909 119910) = 119909 minus ℎ1

(16)

and (120601(0) 120593(0)) = (ℎ1 1205930) (120601(119879) 120593(119879)) = (119909

1 1199101) (120601(119879+)

120593(119879+)) = (119909

1 1199101+ 120572) = (ℎ

1 1205930) one gets

120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889

120597120572

120597119909= 0

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= 0

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(17)


09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05




Δ1=119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))=

119891 (ℎ1 1205930)

119891 (ℎ1 1205930minus 120572)

=(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

11205930+ 119887ℎ1120572 minus 119887ℎ

1120572

(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

1(1205930minus 120572)

= 1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

1205832= Δ1expint

119879

0

[120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))] 119889119905

= Δ1expint

119879

0

[119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889]119889119905

lt Δ1expint

119879

0

[119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889] 119889119905


12

11

1

09

08

07

06

05

0450403503025x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

12

11

1

09

08

07

06

05



= Δ1exp [int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905)]

= Δ1

1205930minus 120572

1205930

= (1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

)1205930minus 120572

1205930

(18)


0minus 120572) gt 0 that is 120593

0lt

(119903119887(1+ 119886ℎ1)) +120572 we have 120583





2 1205930) if 120593

0lt


2(1 +


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05



119886ℎ2)(1 + 119886(1 minus 119901)ℎ

2))) or 120593

0gt max(119903(1 minus 119902)119887(1 + 119886ℎ

2)) +

120591 (1119887119902)(119887120591 minus (119886119901(1 minus 119902)119903ℎ2(1 + 119886ℎ

2)(1 + 119886(1 minus 119901)ℎ

2))) the



119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = minus119901119909 120573 (119909 119910) = minus119902119910 + 120591

Φ (119909 119910) = 119909 minus ℎ2

(19)

(120601(119879+) 120593(119879

+)) = ((1minus119901)120601(119879) (1minus119902)120593(119879)+120591) (120601(119879) 120593(119879)) =


120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

09

08

07

06

05



120597120572

120597119909= minus119901

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= minus119902

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(20)

Thus

Δ1=(1 minus 119902) 119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))

= ((1 minus 119902) [119903120601 (119879+)

1 + 119886120601 (119879+)minus 119887120601 (119879

+) 120593 (119879

+)])


y(t)

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

08

07

06

05



times (119903120601 (119879)

1 + 119886120601 (119879)minus 119887120601 (119879) 120593 (119879))

minus1

= ((1 minus 119902) [119903 (1 minus 119901) ℎ

2

1 + 119886 (1 minus 119901) ℎ2

minus 119887 (1 minus 119901) ℎ21205930])

times (119903ℎ2

1 + 119886ℎ2

minus 119887ℎ2

1205930minus 120591

1 minus 119902)

minus1

= ((1 minus 119901) (1 minus 119902) [119903

1 + 119886 (1 minus 119901) ℎ2

minus 1198871205930])

times (119903

1 + 119886ℎ2

minus 1198871205930minus 120591

1 minus 119902)

minus1

1205832= Δ1expint

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905


1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










A

Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

ylowast

N1 N2 M2

M1

C1

C+1

C0

d

c

A1

A+1


Section 411)

P

R

E

O X

Q

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c



is got

119909119862+

1

= (1 minus 119901) ℎ2

119910119862+

1

= (1 minus 119902) 1199101198621

+ 120591

(14)



1is





1is

above 119876 on 1198732 thus 119891(119876) = 119910

119862+

1

minus 119910119876

gt 0 Choose

E

R

Q

P

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c

D1

D2

D0

D+2


a point 1198630above 119862+


2 Only when it



2at point 119863

2


2in the line



1on1198722


+


function of 1198630is 119891(119863

0) = 119910

119863+

2

minus 1199101198630


0119875119863111986321198621119862+




1is


119910119862+

1

minus 119910119876lt 0


2 0)

on1198732(|1198630119861| ≪ 120576) Starting from119863


1198722at point 119863


1on


119909119863+

1

= (1 minus 119901) ℎ2

119910119863+

1

= (1 minus 119902) 1199101198631

+ 120591

(15)



0

is 119891(1198630) = 119910119863+

1

minus 1199101198630







2and line119872

2 intersects with


B

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c

D1

D0

D+1


Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

ylowast

C2

C0

d

c


Section 413)





1



cases

(a) If 1198622is exactly 119862


1119862+

11198622forms a


2on1198722is below 119862


1

is 119891(1198621) = 1199101198622

minus 1199101198621


0between119873

2and119872

2


P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2M2

M1

C1

C0

C2

C+1

D1

D2D0

D+1

d

c


Section 413)

P

Q

R

E

XO

LY

h1(1 minus p)h2 h2

L998400

N1N2 M2

M1

C1C0

C2

C+1

d

c


Section 413)


1on



1198632is above 119863


1is

119891(1198631) = 1199101198632

minus 1199101198631

gt 0


1119862+

1119863+

111986311198621(see

Figure 11)

(c) If1198622on1198722is above119862



01198621must be



P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1C2

D1

D2L998400

C+1

D+1

C0

D0


Section 413)

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1

C2

L998400

C+1

C0

ylowast

C3


Section 413)




+

1



1crosses119873

1from the right





P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1C2 L998400

C+1

C0


Section 413)

P

Q

RE

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C0

C2

C+1

D1

D2

D0

D+1

d

c


Section 413)





1 or below119862

1 or

above 1198621

If 1198622on 119872

2is exactly 119862


1119862+

11198622


If 1198622on1198722is below 119862


of 1198621is 119891(119862

1) = 1199101198622

minus 1199101198621

lt 0


2and


0



119863+


2on 119872

2


18

16

14

12

1

08

06

014 015 016 017 018 019 02

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t)

10 20 30 40 50 60


18

16

14

12

1

08

06

y(t)

t

10 20 30 40 50 60





function of1198631is 119891(119863

1) = 1199101198632

minus 1199101198631


11198621119862+

11198622

and 119862+1119863+

111986321198622(see Figure 16)

(c) If1198622on1198722is above119862






1 120593(0) = 120593

0 if 1205930lt (119903119887(1+119886ℎ

1))+120572



014 015 016 017 018 019 02

18

16

14

12

1

08

04

02

06

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t

)

10 20 30 40 50 60


18

16

14

12

1

08

04

02

06

y(t)

t

10 20 30 40 50 60



Proof Let

119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = 0 120573 (119909 119910) = 120572

Φ (119909 119910) = 119909 minus ℎ1

(16)

and (120601(0) 120593(0)) = (ℎ1 1205930) (120601(119879) 120593(119879)) = (119909

1 1199101) (120601(119879+)

120593(119879+)) = (119909

1 1199101+ 120572) = (ℎ

1 1205930) one gets

120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889

120597120572

120597119909= 0

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= 0

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(17)


09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05




Δ1=119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))=

119891 (ℎ1 1205930)

119891 (ℎ1 1205930minus 120572)

=(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

11205930+ 119887ℎ1120572 minus 119887ℎ

1120572

(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

1(1205930minus 120572)

= 1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

1205832= Δ1expint

119879

0

[120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))] 119889119905

= Δ1expint

119879

0

[119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889]119889119905

lt Δ1expint

119879

0

[119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889] 119889119905


12

11

1

09

08

07

06

05

0450403503025x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

12

11

1

09

08

07

06

05



= Δ1exp [int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905)]

= Δ1

1205930minus 120572

1205930

= (1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

)1205930minus 120572

1205930

(18)


0minus 120572) gt 0 that is 120593

0lt

(119903119887(1+ 119886ℎ1)) +120572 we have 120583





2 1205930) if 120593

0lt


2(1 +


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05



119886ℎ2)(1 + 119886(1 minus 119901)ℎ

2))) or 120593

0gt max(119903(1 minus 119902)119887(1 + 119886ℎ

2)) +

120591 (1119887119902)(119887120591 minus (119886119901(1 minus 119902)119903ℎ2(1 + 119886ℎ

2)(1 + 119886(1 minus 119901)ℎ

2))) the



119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = minus119901119909 120573 (119909 119910) = minus119902119910 + 120591

Φ (119909 119910) = 119909 minus ℎ2

(19)

(120601(119879+) 120593(119879

+)) = ((1minus119901)120601(119879) (1minus119902)120593(119879)+120591) (120601(119879) 120593(119879)) =


120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

09

08

07

06

05



120597120572

120597119909= minus119901

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= minus119902

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(20)

Thus

Δ1=(1 minus 119902) 119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))

= ((1 minus 119902) [119903120601 (119879+)

1 + 119886120601 (119879+)minus 119887120601 (119879

+) 120593 (119879

+)])


y(t)

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

08

07

06

05



times (119903120601 (119879)

1 + 119886120601 (119879)minus 119887120601 (119879) 120593 (119879))

minus1

= ((1 minus 119902) [119903 (1 minus 119901) ℎ

2

1 + 119886 (1 minus 119901) ℎ2

minus 119887 (1 minus 119901) ℎ21205930])

times (119903ℎ2

1 + 119886ℎ2

minus 119887ℎ2

1205930minus 120591

1 minus 119902)

minus1

= ((1 minus 119901) (1 minus 119902) [119903

1 + 119886 (1 minus 119901) ℎ2

minus 1198871205930])

times (119903

1 + 119886ℎ2

minus 1198871205930minus 120591

1 minus 119902)

minus1

1205832= Δ1expint

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905


1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










B

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

d

c

D1

D0

D+1


Q

R

P

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C+1

ylowast

C2

C0

d

c


Section 413)





1



cases

(a) If 1198622is exactly 119862


1119862+

11198622forms a


2on1198722is below 119862


1

is 119891(1198621) = 1199101198622

minus 1199101198621


0between119873

2and119872

2


P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2M2

M1

C1

C0

C2

C+1

D1

D2D0

D+1

d

c


Section 413)

P

Q

R

E

XO

LY

h1(1 minus p)h2 h2

L998400

N1N2 M2

M1

C1C0

C2

C+1

d

c


Section 413)


1on



1198632is above 119863


1is

119891(1198631) = 1199101198632

minus 1199101198631

gt 0


1119862+

1119863+

111986311198621(see

Figure 11)

(c) If1198622on1198722is above119862



01198621must be



P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1C2

D1

D2L998400

C+1

D+1

C0

D0


Section 413)

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1

C2

L998400

C+1

C0

ylowast

C3


Section 413)




+

1



1crosses119873

1from the right





P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1C2 L998400

C+1

C0


Section 413)

P

Q

RE

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C0

C2

C+1

D1

D2

D0

D+1

d

c


Section 413)





1 or below119862

1 or

above 1198621

If 1198622on 119872

2is exactly 119862


1119862+

11198622


If 1198622on1198722is below 119862


of 1198621is 119891(119862

1) = 1199101198622

minus 1199101198621

lt 0


2and


0



119863+


2on 119872

2


18

16

14

12

1

08

06

014 015 016 017 018 019 02

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t)

10 20 30 40 50 60


18

16

14

12

1

08

06

y(t)

t

10 20 30 40 50 60





function of1198631is 119891(119863

1) = 1199101198632

minus 1199101198631


11198621119862+

11198622

and 119862+1119863+

111986321198622(see Figure 16)

(c) If1198622on1198722is above119862






1 120593(0) = 120593

0 if 1205930lt (119903119887(1+119886ℎ

1))+120572



014 015 016 017 018 019 02

18

16

14

12

1

08

04

02

06

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t

)

10 20 30 40 50 60


18

16

14

12

1

08

04

02

06

y(t)

t

10 20 30 40 50 60



Proof Let

119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = 0 120573 (119909 119910) = 120572

Φ (119909 119910) = 119909 minus ℎ1

(16)

and (120601(0) 120593(0)) = (ℎ1 1205930) (120601(119879) 120593(119879)) = (119909

1 1199101) (120601(119879+)

120593(119879+)) = (119909

1 1199101+ 120572) = (ℎ

1 1205930) one gets

120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889

120597120572

120597119909= 0

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= 0

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(17)


09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05




Δ1=119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))=

119891 (ℎ1 1205930)

119891 (ℎ1 1205930minus 120572)

=(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

11205930+ 119887ℎ1120572 minus 119887ℎ

1120572

(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

1(1205930minus 120572)

= 1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

1205832= Δ1expint

119879

0

[120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))] 119889119905

= Δ1expint

119879

0

[119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889]119889119905

lt Δ1expint

119879

0

[119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889] 119889119905


12

11

1

09

08

07

06

05

0450403503025x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

12

11

1

09

08

07

06

05



= Δ1exp [int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905)]

= Δ1

1205930minus 120572

1205930

= (1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

)1205930minus 120572

1205930

(18)


0minus 120572) gt 0 that is 120593

0lt

(119903119887(1+ 119886ℎ1)) +120572 we have 120583





2 1205930) if 120593

0lt


2(1 +


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05



119886ℎ2)(1 + 119886(1 minus 119901)ℎ

2))) or 120593

0gt max(119903(1 minus 119902)119887(1 + 119886ℎ

2)) +

120591 (1119887119902)(119887120591 minus (119886119901(1 minus 119902)119903ℎ2(1 + 119886ℎ

2)(1 + 119886(1 minus 119901)ℎ

2))) the



119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = minus119901119909 120573 (119909 119910) = minus119902119910 + 120591

Φ (119909 119910) = 119909 minus ℎ2

(19)

(120601(119879+) 120593(119879

+)) = ((1minus119901)120601(119879) (1minus119902)120593(119879)+120591) (120601(119879) 120593(119879)) =


120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

09

08

07

06

05



120597120572

120597119909= minus119901

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= minus119902

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(20)

Thus

Δ1=(1 minus 119902) 119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))

= ((1 minus 119902) [119903120601 (119879+)

1 + 119886120601 (119879+)minus 119887120601 (119879

+) 120593 (119879

+)])


y(t)

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

08

07

06

05



times (119903120601 (119879)

1 + 119886120601 (119879)minus 119887120601 (119879) 120593 (119879))

minus1

= ((1 minus 119902) [119903 (1 minus 119901) ℎ

2

1 + 119886 (1 minus 119901) ℎ2

minus 119887 (1 minus 119901) ℎ21205930])

times (119903ℎ2

1 + 119886ℎ2

minus 119887ℎ2

1205930minus 120591

1 minus 119902)

minus1

= ((1 minus 119901) (1 minus 119902) [119903

1 + 119886 (1 minus 119901) ℎ2

minus 1198871205930])

times (119903

1 + 119886ℎ2

minus 1198871205930minus 120591

1 minus 119902)

minus1

1205832= Δ1expint

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905


1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1C2

D1

D2L998400

C+1

D+1

C0

D0


Section 413)

P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1

C2

L998400

C+1

C0

ylowast

C3


Section 413)




+

1



1crosses119873

1from the right





P

Q

R

E

XO

LY

h1 (1 minus p)h2 h2

N1 N2 M2

M1

d

c

C1C2 L998400

C+1

C0


Section 413)

P

Q

RE

XO

LY

h1 (1 minus p)h2 h2

L998400

N1 N2 M2

M1

C1

C0

C2

C+1

D1

D2

D0

D+1

d

c


Section 413)





1 or below119862

1 or

above 1198621

If 1198622on 119872

2is exactly 119862


1119862+

11198622


If 1198622on1198722is below 119862


of 1198621is 119891(119862

1) = 1199101198622

minus 1199101198621

lt 0


2and


0



119863+


2on 119872

2


18

16

14

12

1

08

06

014 015 016 017 018 019 02

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t)

10 20 30 40 50 60


18

16

14

12

1

08

06

y(t)

t

10 20 30 40 50 60





function of1198631is 119891(119863

1) = 1199101198632

minus 1199101198631


11198621119862+

11198622

and 119862+1119863+

111986321198622(see Figure 16)

(c) If1198622on1198722is above119862






1 120593(0) = 120593

0 if 1205930lt (119903119887(1+119886ℎ

1))+120572



014 015 016 017 018 019 02

18

16

14

12

1

08

04

02

06

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t

)

10 20 30 40 50 60


18

16

14

12

1

08

04

02

06

y(t)

t

10 20 30 40 50 60



Proof Let

119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = 0 120573 (119909 119910) = 120572

Φ (119909 119910) = 119909 minus ℎ1

(16)

and (120601(0) 120593(0)) = (ℎ1 1205930) (120601(119879) 120593(119879)) = (119909

1 1199101) (120601(119879+)

120593(119879+)) = (119909

1 1199101+ 120572) = (ℎ

1 1205930) one gets

120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889

120597120572

120597119909= 0

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= 0

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(17)


09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05




Δ1=119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))=

119891 (ℎ1 1205930)

119891 (ℎ1 1205930minus 120572)

=(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

11205930+ 119887ℎ1120572 minus 119887ℎ

1120572

(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

1(1205930minus 120572)

= 1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

1205832= Δ1expint

119879

0

[120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))] 119889119905

= Δ1expint

119879

0

[119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889]119889119905

lt Δ1expint

119879

0

[119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889] 119889119905


12

11

1

09

08

07

06

05

0450403503025x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

12

11

1

09

08

07

06

05



= Δ1exp [int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905)]

= Δ1

1205930minus 120572

1205930

= (1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

)1205930minus 120572

1205930

(18)


0minus 120572) gt 0 that is 120593

0lt

(119903119887(1+ 119886ℎ1)) +120572 we have 120583





2 1205930) if 120593

0lt


2(1 +


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05



119886ℎ2)(1 + 119886(1 minus 119901)ℎ

2))) or 120593

0gt max(119903(1 minus 119902)119887(1 + 119886ℎ

2)) +

120591 (1119887119902)(119887120591 minus (119886119901(1 minus 119902)119903ℎ2(1 + 119886ℎ

2)(1 + 119886(1 minus 119901)ℎ

2))) the



119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = minus119901119909 120573 (119909 119910) = minus119902119910 + 120591

Φ (119909 119910) = 119909 minus ℎ2

(19)

(120601(119879+) 120593(119879

+)) = ((1minus119901)120601(119879) (1minus119902)120593(119879)+120591) (120601(119879) 120593(119879)) =


120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

09

08

07

06

05



120597120572

120597119909= minus119901

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= minus119902

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(20)

Thus

Δ1=(1 minus 119902) 119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))

= ((1 minus 119902) [119903120601 (119879+)

1 + 119886120601 (119879+)minus 119887120601 (119879

+) 120593 (119879

+)])


y(t)

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

08

07

06

05



times (119903120601 (119879)

1 + 119886120601 (119879)minus 119887120601 (119879) 120593 (119879))

minus1

= ((1 minus 119902) [119903 (1 minus 119901) ℎ

2

1 + 119886 (1 minus 119901) ℎ2

minus 119887 (1 minus 119901) ℎ21205930])

times (119903ℎ2

1 + 119886ℎ2

minus 119887ℎ2

1205930minus 120591

1 minus 119902)

minus1

= ((1 minus 119901) (1 minus 119902) [119903

1 + 119886 (1 minus 119901) ℎ2

minus 1198871205930])

times (119903

1 + 119886ℎ2

minus 1198871205930minus 120591

1 minus 119902)

minus1

1205832= Δ1expint

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905


1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










18

16

14

12

1

08

06

014 015 016 017 018 019 02

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t)

10 20 30 40 50 60


18

16

14

12

1

08

06

y(t)

t

10 20 30 40 50 60





function of1198631is 119891(119863

1) = 1199101198632

minus 1199101198631


11198621119862+

11198622

and 119862+1119863+

111986321198622(see Figure 16)

(c) If1198622on1198722is above119862






1 120593(0) = 120593

0 if 1205930lt (119903119887(1+119886ℎ

1))+120572



014 015 016 017 018 019 02

18

16

14

12

1

08

04

02

06

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t

)

10 20 30 40 50 60


18

16

14

12

1

08

04

02

06

y(t)

t

10 20 30 40 50 60



Proof Let

119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = 0 120573 (119909 119910) = 120572

Φ (119909 119910) = 119909 minus ℎ1

(16)

and (120601(0) 120593(0)) = (ℎ1 1205930) (120601(119879) 120593(119879)) = (119909

1 1199101) (120601(119879+)

120593(119879+)) = (119909

1 1199101+ 120572) = (ℎ

1 1205930) one gets

120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889

120597120572

120597119909= 0

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= 0

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(17)


09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05




Δ1=119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))=

119891 (ℎ1 1205930)

119891 (ℎ1 1205930minus 120572)

=(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

11205930+ 119887ℎ1120572 minus 119887ℎ

1120572

(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

1(1205930minus 120572)

= 1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

1205832= Δ1expint

119879

0

[120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))] 119889119905

= Δ1expint

119879

0

[119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889]119889119905

lt Δ1expint

119879

0

[119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889] 119889119905


12

11

1

09

08

07

06

05

0450403503025x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

12

11

1

09

08

07

06

05



= Δ1exp [int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905)]

= Δ1

1205930minus 120572

1205930

= (1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

)1205930minus 120572

1205930

(18)


0minus 120572) gt 0 that is 120593

0lt

(119903119887(1+ 119886ℎ1)) +120572 we have 120583





2 1205930) if 120593

0lt


2(1 +


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05



119886ℎ2)(1 + 119886(1 minus 119901)ℎ

2))) or 120593

0gt max(119903(1 minus 119902)119887(1 + 119886ℎ

2)) +

120591 (1119887119902)(119887120591 minus (119886119901(1 minus 119902)119903ℎ2(1 + 119886ℎ

2)(1 + 119886(1 minus 119901)ℎ

2))) the



119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = minus119901119909 120573 (119909 119910) = minus119902119910 + 120591

Φ (119909 119910) = 119909 minus ℎ2

(19)

(120601(119879+) 120593(119879

+)) = ((1minus119901)120601(119879) (1minus119902)120593(119879)+120591) (120601(119879) 120593(119879)) =


120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

09

08

07

06

05



120597120572

120597119909= minus119901

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= minus119902

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(20)

Thus

Δ1=(1 minus 119902) 119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))

= ((1 minus 119902) [119903120601 (119879+)

1 + 119886120601 (119879+)minus 119887120601 (119879

+) 120593 (119879

+)])


y(t)

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

08

07

06

05



times (119903120601 (119879)

1 + 119886120601 (119879)minus 119887120601 (119879) 120593 (119879))

minus1

= ((1 minus 119902) [119903 (1 minus 119901) ℎ

2

1 + 119886 (1 minus 119901) ℎ2

minus 119887 (1 minus 119901) ℎ21205930])

times (119903ℎ2

1 + 119886ℎ2

minus 119887ℎ2

1205930minus 120591

1 minus 119902)

minus1

= ((1 minus 119901) (1 minus 119902) [119903

1 + 119886 (1 minus 119901) ℎ2

minus 1198871205930])

times (119903

1 + 119886ℎ2

minus 1198871205930minus 120591

1 minus 119902)

minus1

1205832= Δ1expint

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905


1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










014 015 016 017 018 019 02

18

16

14

12

1

08

04

02

06

y(t)

x(t)


014

015

016

017

018

019

02

x(t

t

)

10 20 30 40 50 60


18

16

14

12

1

08

04

02

06

y(t)

t

10 20 30 40 50 60



Proof Let

119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = 0 120573 (119909 119910) = 120572

Φ (119909 119910) = 119909 minus ℎ1

(16)

and (120601(0) 120593(0)) = (ℎ1 1205930) (120601(119879) 120593(119879)) = (119909

1 1199101) (120601(119879+)

120593(119879+)) = (119909

1 1199101+ 120572) = (ℎ

1 1205930) one gets

120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889

120597120572

120597119909= 0

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= 0

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(17)


09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05




Δ1=119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))=

119891 (ℎ1 1205930)

119891 (ℎ1 1205930minus 120572)

=(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

11205930+ 119887ℎ1120572 minus 119887ℎ

1120572

(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

1(1205930minus 120572)

= 1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

1205832= Δ1expint

119879

0

[120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))] 119889119905

= Δ1expint

119879

0

[119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889]119889119905

lt Δ1expint

119879

0

[119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889] 119889119905


12

11

1

09

08

07

06

05

0450403503025x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

12

11

1

09

08

07

06

05



= Δ1exp [int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905)]

= Δ1

1205930minus 120572

1205930

= (1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

)1205930minus 120572

1205930

(18)


0minus 120572) gt 0 that is 120593

0lt

(119903119887(1+ 119886ℎ1)) +120572 we have 120583





2 1205930) if 120593

0lt


2(1 +


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05



119886ℎ2)(1 + 119886(1 minus 119901)ℎ

2))) or 120593

0gt max(119903(1 minus 119902)119887(1 + 119886ℎ

2)) +

120591 (1119887119902)(119887120591 minus (119886119901(1 minus 119902)119903ℎ2(1 + 119886ℎ

2)(1 + 119886(1 minus 119901)ℎ

2))) the



119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = minus119901119909 120573 (119909 119910) = minus119902119910 + 120591

Φ (119909 119910) = 119909 minus ℎ2

(19)

(120601(119879+) 120593(119879

+)) = ((1minus119901)120601(119879) (1minus119902)120593(119879)+120591) (120601(119879) 120593(119879)) =


120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

09

08

07

06

05



120597120572

120597119909= minus119901

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= minus119902

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(20)

Thus

Δ1=(1 minus 119902) 119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))

= ((1 minus 119902) [119903120601 (119879+)

1 + 119886120601 (119879+)minus 119887120601 (119879

+) 120593 (119879

+)])


y(t)

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

08

07

06

05



times (119903120601 (119879)

1 + 119886120601 (119879)minus 119887120601 (119879) 120593 (119879))

minus1

= ((1 minus 119902) [119903 (1 minus 119901) ℎ

2

1 + 119886 (1 minus 119901) ℎ2

minus 119887 (1 minus 119901) ℎ21205930])

times (119903ℎ2

1 + 119886ℎ2

minus 119887ℎ2

1205930minus 120591

1 minus 119902)

minus1

= ((1 minus 119901) (1 minus 119902) [119903

1 + 119886 (1 minus 119901) ℎ2

minus 1198871205930])

times (119903

1 + 119886ℎ2

minus 1198871205930minus 120591

1 minus 119902)

minus1

1205832= Δ1expint

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905


1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05




Δ1=119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))=

119891 (ℎ1 1205930)

119891 (ℎ1 1205930minus 120572)

=(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

11205930+ 119887ℎ1120572 minus 119887ℎ

1120572

(119903ℎ1 (1 + 119886ℎ

1)) minus 119887ℎ

1(1205930minus 120572)

= 1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

1205832= Δ1expint

119879

0

[120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))] 119889119905

= Δ1expint

119879

0

[119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889]119889119905

lt Δ1expint

119879

0

[119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889] 119889119905


12

11

1

09

08

07

06

05

0450403503025x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

12

11

1

09

08

07

06

05



= Δ1exp [int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905)]

= Δ1

1205930minus 120572

1205930

= (1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

)1205930minus 120572

1205930

(18)


0minus 120572) gt 0 that is 120593

0lt

(119903119887(1+ 119886ℎ1)) +120572 we have 120583





2 1205930) if 120593

0lt


2(1 +


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05



119886ℎ2)(1 + 119886(1 minus 119901)ℎ

2))) or 120593

0gt max(119903(1 minus 119902)119887(1 + 119886ℎ

2)) +

120591 (1119887119902)(119887120591 minus (119886119901(1 minus 119902)119903ℎ2(1 + 119886ℎ

2)(1 + 119886(1 minus 119901)ℎ

2))) the



119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = minus119901119909 120573 (119909 119910) = minus119902119910 + 120591

Φ (119909 119910) = 119909 minus ℎ2

(19)

(120601(119879+) 120593(119879

+)) = ((1minus119901)120601(119879) (1minus119902)120593(119879)+120591) (120601(119879) 120593(119879)) =


120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

09

08

07

06

05



120597120572

120597119909= minus119901

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= minus119902

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(20)

Thus

Δ1=(1 minus 119902) 119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))

= ((1 minus 119902) [119903120601 (119879+)

1 + 119886120601 (119879+)minus 119887120601 (119879

+) 120593 (119879

+)])


y(t)

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

08

07

06

05



times (119903120601 (119879)

1 + 119886120601 (119879)minus 119887120601 (119879) 120593 (119879))

minus1

= ((1 minus 119902) [119903 (1 minus 119901) ℎ

2

1 + 119886 (1 minus 119901) ℎ2

minus 119887 (1 minus 119901) ℎ21205930])

times (119903ℎ2

1 + 119886ℎ2

minus 119887ℎ2

1205930minus 120591

1 minus 119902)

minus1

= ((1 minus 119901) (1 minus 119902) [119903

1 + 119886 (1 minus 119901) ℎ2

minus 1198871205930])

times (119903

1 + 119886ℎ2

minus 1198871205930minus 120591

1 minus 119902)

minus1

1205832= Δ1expint

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905


1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










12

11

1

09

08

07

06

05

0450403503025x(t)


t

10 20 30 40 50 60

045

04

035

03

025

x(t)


y(t)

t

10 20 30 40 50 60

12

11

1

09

08

07

06

05



= Δ1exp [int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905)]

= Δ1

1205930minus 120572

1205930

= (1 minus119887120572

(119903 (1 + 119886ℎ1)) minus 119887 (120593

0minus 120572)

)1205930minus 120572

1205930

(18)


0minus 120572) gt 0 that is 120593

0lt

(119903119887(1+ 119886ℎ1)) +120572 we have 120583





2 1205930) if 120593

0lt


2(1 +


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05



119886ℎ2)(1 + 119886(1 minus 119901)ℎ

2))) or 120593

0gt max(119903(1 minus 119902)119887(1 + 119886ℎ

2)) +

120591 (1119887119902)(119887120591 minus (119886119901(1 minus 119902)119903ℎ2(1 + 119886ℎ

2)(1 + 119886(1 minus 119901)ℎ

2))) the



119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = minus119901119909 120573 (119909 119910) = minus119902119910 + 120591

Φ (119909 119910) = 119909 minus ℎ2

(19)

(120601(119879+) 120593(119879

+)) = ((1minus119901)120601(119879) (1minus119902)120593(119879)+120591) (120601(119879) 120593(119879)) =


120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

09

08

07

06

05



120597120572

120597119909= minus119901

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= minus119902

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(20)

Thus

Δ1=(1 minus 119902) 119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))

= ((1 minus 119902) [119903120601 (119879+)

1 + 119886120601 (119879+)minus 119887120601 (119879

+) 120593 (119879

+)])


y(t)

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

08

07

06

05



times (119903120601 (119879)

1 + 119886120601 (119879)minus 119887120601 (119879) 120593 (119879))

minus1

= ((1 minus 119902) [119903 (1 minus 119901) ℎ

2

1 + 119886 (1 minus 119901) ℎ2

minus 119887 (1 minus 119901) ℎ21205930])

times (119903ℎ2

1 + 119886ℎ2

minus 119887ℎ2

1205930minus 120591

1 minus 119902)

minus1

= ((1 minus 119901) (1 minus 119902) [119903

1 + 119886 (1 minus 119901) ℎ2

minus 1198871205930])

times (119903

1 + 119886ℎ2

minus 1198871205930minus 120591

1 minus 119902)

minus1

1205832= Δ1expint

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905


1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

09

08

07

06

05



119886ℎ2)(1 + 119886(1 minus 119901)ℎ

2))) or 120593

0gt max(119903(1 minus 119902)119887(1 + 119886ℎ

2)) +

120591 (1119887119902)(119887120591 minus (119886119901(1 minus 119902)119903ℎ2(1 + 119886ℎ

2)(1 + 119886(1 minus 119901)ℎ

2))) the



119891 (119909 119910) =119903119909

1 + 119886119909minus 119887119909119910

119892 (119909 119910) = 119910 (minus119889 + 119888119909)

120572 (119909 119910) = minus119901119909 120573 (119909 119910) = minus119902119910 + 120591

Φ (119909 119910) = 119909 minus ℎ2

(19)

(120601(119879+) 120593(119879

+)) = ((1minus119901)120601(119879) (1minus119902)120593(119879)+120591) (120601(119879) 120593(119879)) =


120597119891

120597119909=

119903

(1 + 119886119909)2minus 119887119910

120597119892

120597119910= 119888119909 minus 119889


y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

09

08

07

06

05



120597120572

120597119909= minus119901

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= minus119902

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(20)

Thus

Δ1=(1 minus 119902) 119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))

= ((1 minus 119902) [119903120601 (119879+)

1 + 119886120601 (119879+)minus 119887120601 (119879

+) 120593 (119879

+)])


y(t)

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

08

07

06

05



times (119903120601 (119879)

1 + 119886120601 (119879)minus 119887120601 (119879) 120593 (119879))

minus1

= ((1 minus 119902) [119903 (1 minus 119901) ℎ

2

1 + 119886 (1 minus 119901) ℎ2

minus 119887 (1 minus 119901) ℎ21205930])

times (119903ℎ2

1 + 119886ℎ2

minus 119887ℎ2

1205930minus 120591

1 minus 119902)

minus1

= ((1 minus 119901) (1 minus 119902) [119903

1 + 119886 (1 minus 119901) ℎ2

minus 1198871205930])

times (119903

1 + 119886ℎ2

minus 1198871205930minus 120591

1 minus 119902)

minus1

1205832= Δ1expint

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905


1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










y(t)

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

09

08

07

06

05



120597120572

120597119909= minus119901

120597120572

120597119910= 0

120597120573

120597119909= 0

120597120573

120597119910= minus119902

120597Φ

120597119909= 1

120597Φ

120597119910= 0

(20)

Thus

Δ1=(1 minus 119902) 119891 (120601 (119879

+) 120593 (119879

+))

119891 (120601 (119879) 120593 (119879))

= ((1 minus 119902) [119903120601 (119879+)

1 + 119886120601 (119879+)minus 119887120601 (119879

+) 120593 (119879

+)])


y(t)

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

08

07

06

05



times (119903120601 (119879)

1 + 119886120601 (119879)minus 119887120601 (119879) 120593 (119879))

minus1

= ((1 minus 119902) [119903 (1 minus 119901) ℎ

2

1 + 119886 (1 minus 119901) ℎ2

minus 119887 (1 minus 119901) ℎ21205930])

times (119903ℎ2

1 + 119886ℎ2

minus 119887ℎ2

1205930minus 120591

1 minus 119902)

minus1

= ((1 minus 119901) (1 minus 119902) [119903

1 + 119886 (1 minus 119901) ℎ2

minus 1198871205930])

times (119903

1 + 119886ℎ2

minus 1198871205930minus 120591

1 minus 119902)

minus1

1205832= Δ1expint

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905


1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










y(t)

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

08

07

06

05



times (119903120601 (119879)

1 + 119886120601 (119879)minus 119887120601 (119879) 120593 (119879))

minus1

= ((1 minus 119902) [119903 (1 minus 119901) ℎ

2

1 + 119886 (1 minus 119901) ℎ2

minus 119887 (1 minus 119901) ℎ21205930])

times (119903ℎ2

1 + 119886ℎ2

minus 119887ℎ2

1205930minus 120591

1 minus 119902)

minus1

= ((1 minus 119901) (1 minus 119902) [119903

1 + 119886 (1 minus 119901) ℎ2

minus 1198871205930])

times (119903

1 + 119886ℎ2

minus 1198871205930minus 120591

1 minus 119902)

minus1

1205832= Δ1expint

119879

0

(120597119891

120597119909(120601 (119905) 120593 (119905)) +

120597119892

120597119910(120601 (119905) 120593 (119905))) 119889119905


1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










1

11

09

08

07

06

05

025 03 035 04 045x(t)


t

10 20 30 40 50 60

025

03

035

04

045

x(t)


y(t)

t

10 20 30 40 50 60

1

11

09

08

07

06

05



= Δ1expint

119879

0

(119903

(1 + 119886120601)2minus 119887120593 + 119888120601 minus 119889)119889119905

lt Δ1expint

119879

0

(119903

1 + 119886120601minus 119887120593 + 119888120601 minus 119889)119889119905

= Δ1exp(int

119879

0

1

120601 (119905)119889120601 (119905) + int

119879

0

1

120593 (119905)119889120593 (119905))

= Δ1

1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930


02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










02502015 03 035 04 045x(t)

y(t)

18

16

14

12

1

08

06


t

10 20 30 40 50 60

015

02

025

03

035

04

045

x(t)


t

10 20 30 40 50 60

y(t)

18

16

14

12

1

08

06




2)) minus 119887120593

0]

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

times1

(1 minus 119902) (1 minus 119901)

1205930minus 120591

1205930

le(119903 (1 + 119886 (1 minus 119901) ℎ

2)) minus 119887120593

0

(119903 (1 + 119886ℎ2)) minus 119887 ((120593

0minus 120591) (1 minus 119902))

(21)

If

1205930lt min

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(22)


or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










or

1205930gt max

119903 (1 minus 119902)

119887 (1 + 119886ℎ2)+ 120591

1

119887119902(119887120591 minus

119886119901 (1 minus 119902) 119903ℎ2

(1 + 119886ℎ2) (1 + 119886 (1 minus 119901) ℎ

2))

(23)







1199091015840(119905) =

06119909

1 + 119909minus 05119909119910

1199101015840(119905) = 08119909119910 minus 04119910

119909 = 02 045 or 119909 = 02 119910 gt 1

Δ119909 = 0

Δ119910 = 120572119909 = 02 119910 le 1

Δ119909 = minus05119909

Δ119910 = minus02119910 + 120591119909 = 045

(24)



1 where 120572 = 05




2 Let 119862

0be

119876(0225 09796)


1is exactly



1is above



1is below




2 Let 119862

0be (025 09)


1is exactly



1is below



1is



1is


Figure 25)


6 Conclusion







Acknowledgments


References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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