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International Journal of Differential Equations

Recent Advances in Numerical Methods and Analysis for Nonlinear Differential Equations

Lead Guest Editor: Dongfang LiGuest Editors: Jinming Wen and Jiwei Zhang

Recent Advances in Numerical Methodsand Analysis for Nonlinear Differential Equations

International Journal of Differential Equations

Recent Advances in Numerical Methodsand Analysis for Nonlinear Differential Equations

Lead Guest Editor: Dongfang LiGuest Editors: Jinming Wen and Jiwei Zhang

Copyright © 2019 Hindawi. All rights reserved.

This is a special issue published in “International Journal of Differential Equations.” All articles are open access articles distributed underthe Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, providedthe original work is properly cited.

Editorial Board

Om P. Agrawal, USAKartik Ariyur, USAElena Braverman, CanadaDer-Chen Chang, USAToka Diagana, USAM. El-Gebeily, Saudi ArabiaMostafa Eslami, IranKhalil Ezzinbi, MoroccoGiovanni P. Galdi, USADavood D. Ganji, IranYoshikazu Giga, JapanJaume Giné, SpainJerome A. Goldstein, USASaid R. Grace, EgyptEmmanuel Hebey, FranceMayer Humi, USAElena Kaikina, Mexico

Qingkai Kong, USAP. A. Krutitskii, RussiaAlexander Kurganov, USAJosé A Langa, SpainDaniel Franco Leis, SpainXiaodi Li, ChinaFawang Liu, AustraliaYuji Liu, ChinaWen-Xiu Ma, USATimothy R. Marchant, AustraliaSalim Messaoudi, Saudi ArabiaShaher Momani, JordanG. Mandata N’guérékata, USAKanishka Perera, USARamón Quintanilla, SpainJulio D. Rossi, ArgentinaSamir H. Saker, Egypt

Martin Schechter, USAIoannis G. Stratis, GreeceJian-Ping Sun, ChinaGuido Sweers, GermanyDan Tiba, RomaniaA. Vatsala, USAPeiguang Wang, ChinaGershon Wolansky, IsraelPatricia J. Y. Wong, SingaporeJen-Chih Yao, TaiwanJingxue Yin, ChinaJianshe Yu, ChinaVjacheslav Yurko, RussiaSining Zheng, ChinaWenming Zou, ChinaXingfu Zou, Canada

Contents

Recent Advances in Numerical Methods and Analysis for Nonlinear Differential EquationsDongfang Li , Jinming Wen, and Jiwei ZhangEditorial (1 page), Article ID 3243510, Volume 2019 (2019)

Application of Optimal Homotopy Asymptotic Method to SomeWell-Known Linear and NonlinearTwo-Point Boundary Value ProblemsMuhammad Asim Khan , Shafiq Ullah, and Norhashidah Hj. Mohd AliResearch Article (11 pages), Article ID 8725014, Volume 2018 (2019)

Global Stability of an Economic Model with a Continuous Delay of Kaldor Type ModifiedAka Fulgence Nindjin, Tetchi Albin N’guessan , Sahoua Hypolithe Okou A Kpetihi ,and Kessé Tiban TiaResearch Article (18 pages), Article ID 7212795, Volume 2018 (2019)

Linear 𝜃-Method and Compact 𝜃-Method for Generalised Reaction-Diffusion Equation with DelayFengyan Wu, Qiong Wang, Xiujun Cheng , and Xiaoli ChenResearch Article (13 pages), Article ID 6402576, Volume 2018 (2019)

Stability Analysis of Additive Runge-Kutta Methods for Delay-Integro-Differential EquationsHongyu Qin , Zhiyong Wang, Fumin Zhu , and Jinming WenResearch Article (5 pages), Article ID 8241784, Volume 2018 (2019)

Well-Posedness and Numerical Study for Solutions of a Parabolic Equation with Variable-ExponentNonlinearitiesJamal H. Al-Smail, Salim A. Messaoudi , and Ala A. TalahmehResearch Article (9 pages), Article ID 9754567, Volume 2018 (2019)

http://orcid.org/0000-0002-2901-3919http://orcid.org/0000-0001-9046-9264http://orcid.org/0000-0002-9544-7221http://orcid.org/0000-0001-5504-9387http://orcid.org/0000-0002-0466-3923http://orcid.org/0000-0002-8001-7966http://orcid.org/0000-0001-8385-3858http://orcid.org/0000-0003-1061-0075

EditorialRecent Advances in Numerical Methods and Analysis forNonlinear Differential Equations

Dongfang Li ,1 JinmingWen,2 and Jiwei Zhang3

1School of Mathematics and Statistics, Huazhong University of Science and Technology, Wuhan 430074, China2Department of Electrical and Computer Engineering, University of Toronto, Toronto, ON M5S3G4, Canada3Beijing Computational Science Research Center, Beijing 100193, China

Correspondence should be addressed to Dongfang Li; [email protected]

Received 26 November 2018; Accepted 26 November 2018; Published 1 January 2019

Copyright © 2019 Dongfang Li et al. is is an open access article distributed under the Creative Commons Attribution License,which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

e construction and analysis of numerical schemes fornonlinear differential equations are very important. A well-designed numerical scheme not only may save computationalcosts for long-time simulations in the real-world problems,but also can give a better performance in the prediction ofthe mathematical models. erefore, it is highly desirable todevelop effective and efficient numerical schemes and theirnumerical analysis for nonlinear differential equations.

e paper authored by J. H. Al-Smail et al. studies theexistence and uniqueness of weak solutions for the nonlinearparabolic equations and then presents a linearized finiteelement Galerkin method to numerically solve the modelequations. e decay property of solutions is confirmed bythe numerical results.

e paper by H. Qin et al. deals with the stability analysisof additive Runge-Kutta methods for delay-integrodifferen-tial equations arising from some spatially discretized time-dependent partial differential equations. Such equations oenown different stiff terms. In such cases, it is more effective inapplying some split numerical schemes to approximate thesystems. e given results imply that if the additive Runge-Kutta methods are algebraically stable, the perturbationsof the numerical solutions are controlled by the initialperturbations from the system and the methods.

e paper of F. Wu et al. is concerned with the analysisof the two fully discrete numerical schemes for solvingdelay reaction-diffusion equation. Solvability and conver-gence of the fully discrete numerical schemes are studied.Besides, asymptotical stability of the fully discrete schemes

is investigated extensively. Several examples are presented atlast to confirm the theoretical results.

e paper by A. F. Nindjin et al. studies continuousnonlinear economic dynamics with a continuous delay. It isshown that themodel is bounded and admits an attractor unit(set). In addition, the delay term can justify the bifurcation ofan economic model of the stationary growth towards cyclicgrowth. Several numerical results are given to illustrate thegiven results.

e paper of M. A. Khan et al. presents an approximatesolution for somewell-known linear and nonlinear two-pointboundary value problems by using the optimal homotopyasymptotic method. Numerical results are provided to showthe effectiveness and reliability of the proposed method, andsome comparisons are also given to show the advantages ofthe methods.

Conflicts of Interest

e authors declare that they have no conflicts of interest.

Acknowledgments

We are grateful to all the authors who have made a contribu-tion to this special issue.

Dongfang LiJinmingWenJiwei Zhang

HindawiInternational Journal of Differential EquationsVolume 2019, Article ID 3243510, 1 pagehttps://doi.org/10.1155/2019/3243510

http://orcid.org/0000-0002-2901-3919https://creativecommons.org/licenses/by/4.0/https://doi.org/10.1155/2019/3243510

Research ArticleApplication of Optimal Homotopy AsymptoticMethod to Some Well-Known Linear and Nonlinear Two-PointBoundary Value Problems

Muhammad Asim Khan ,1 Shafiq Ullah,2 and Norhashidah Hj. Mohd Ali1

1School of Mathematical Sciences, Universiti Sains Malaysia, 11800 Penang, Malaysia2Department of Mathematics, University of Peshawar, Pakistan

Correspondence should be addressed to Muhammad Asim Khan; [email protected]

Received 20 May 2018; Accepted 21 October 2018; Published 3 December 2018

Guest Editor: Dongfang Li

Copyright © 2018 MuhammadAsimKhan et al.This is an open access article distributed under the Creative Commons AttributionLicense,whichpermits unrestricteduse, distribution, and reproduction in anymedium, provided the original work is properly cited.

The objective of this paper is to obtain an approximate solution for some well-known linear and nonlinear two-point boundaryvalue problems. For this purpose, a semianalytical method known as optimal homotopy asymptotic method (OHAM) is used.Moreover, optimal homotopy asymptotic method does not involve any discretization, linearization, or small perturbations and thatis why it reduces the computations a lot. OHAM results show the effectiveness and reliability of OHAM for application to two-pointboundary value problems. The obtained results are compared to the exact solutions and homotopy perturbation method (HPM).

1. Introduction

Two-point boundary value problems (TPBVP) have manyapplications in the field of science and engineering [1, 2].These problems arise in many physical situations like mod-eling of chemical reactions, heat transfer, viscous fluids, dif-fusions, deflection of beams, the solution of optimal controlproblems, etc. Due to thewide applications and importance ofboundary value problems (BVP) in science and engineeringwe need solutions to these problems.

There are many techniques available for the solution of-of BVP like Adomian Decomposition Method (ADM) [3–7], Extended Adomian Decomposition Method (EADM)[8],Differential Transformation Method (DTM) [9], VariationalIteration Method (VIM) [10], Perturbation methods(PMs)[1, 11–13], and so on. Perturbation methods are easy to solvebut they require small parameters which are sometimes notan easy task. Recently V. Marinca et al. presented optimalhomotopy asymptotic method (OHAM) [14] for the solutionof BVP, which did not require small parameters. The methodcan also be applied to solve the stationary solution of somepartial differential equations, e.g., gKdv equation, nonlinearparabolic problems, and so on [15–20]. In OHAM, the

concept of homotopy is used together with the perturbationtechniques. Here, OHAM is applied to TPBVP to check theapplicability of OHAM for TPBVP.

2. Basics of OHAM

Let us take the BVP whose general form is the following:

L (𝑤 (𝜉)) + N (𝑤 (𝜉)) + ϝ (𝜉) = 0,𝐵 (𝑤, 𝑑𝑤𝑑𝜉 )

(1)

whereL is a linear operator, 𝜉 is independent variable, N isthe nonlinear operator, ϝ(𝜉) is a known function, and 𝐵 is aboundary operator.

Homotopy on OHAM can be constructed as

(1 − þ) (L (𝜑 (𝜉, þ) + ϝ (𝜉))= 𝐻 (þ) (L (𝜑 (𝜉, þ) + ϝ (𝜉) + N (𝜑 (𝜉, þ))) ,

𝐵(𝜑 (𝜉, þ) , 𝜕𝜑 (𝜉, þ)𝜕𝜉 )(2)

HindawiInternational Journal of Differential EquationsVolume 2018, Article ID 8725014, 11 pageshttps://doi.org/10.1155/2018/8725014

http://orcid.org/0000-0001-9046-9264https://creativecommons.org/licenses/by/4.0/https://creativecommons.org/licenses/by/4.0/https://doi.org/10.1155/2018/8725014

2 International Journal of Differential Equations

where þ ∈ [0, 1] is an embedding parameter, 𝜑(𝜉, þ) is anunknown function, 𝐻(þ) is a nonzero auxiliary function forþ ̸= 0, and𝐻(þ) is of the form

𝐻(þ) = þC1 + þ2C2 + þ3C3 + ⋅ ⋅ ⋅ (3)Clearly when þ = 0 then 𝐻(0) = 0. And obviously, whenþ = 0 then 𝜑(𝜉, 0) = 𝑤0(𝜉).When þ = 1 then 𝜑(𝜉, 1) = 𝑤(𝜉).So as þ increases from 0 to 1, the solution 𝜑(𝜉, þ) varies from𝑤0(𝜉) to the exact solution𝑤(𝜉), where𝑤0 (𝜉) is obtained from(2) for þ = 0

L (𝑤0 (𝜉) + ϝ (𝜉) = 0 (4)The proposed solution of (1) will be of the form

𝜑 (𝜉, þ,C𝑖) = 𝑤0 (𝜉) + ∑𝑘≥1

𝑤𝑘 (𝜉,C𝑖) þ𝑘, 𝑖 = 1, 2, 3, . . . (5)Substituting this value of 𝜑(𝜉, þ,C𝑖) into (1), after somecalculations, we can obtain the governing equations of 𝑤0(𝜉)by using (4) and 𝑤𝑘(𝜉), that is,L (𝑤1 (𝜉)) = C1N0 (𝑤0 (𝜉)) ,𝐵 (𝑤1, 𝑑𝑤1𝑑𝜉 ) = 0

(6)

L (𝑤𝑘 (𝜉) − 𝑤𝑘−1 (𝜉))= C𝑘N0 (𝑤0 (𝜉)) +

𝑘−1∑𝑖=1

C𝑖L (𝑤𝑘−1 (𝜉)+ N𝑘−1 (𝑤0 (𝜉) , 𝑤1 (𝜉) , 𝑤2 (𝜉) , . . . , 𝑤𝑘−1 (𝜉)) ,

𝑘 = 2, 3, 4, . . . ,𝐵 (𝑤𝑘, 𝑑𝑤𝑘𝑑𝜉 ) = 0,

(7)

whereN𝑚(𝑤0(𝜉), 𝑤1(𝜉), 𝑤2(𝜉), . . . , 𝑤𝑚(𝜉)) is the coefficient ofþ𝑚 in the series expansion ofN(𝜑(𝜉, þ,C𝑖))with respect to theembedding parameter þ. AndN (𝜑 (𝜉, þ,C𝑖)) = N0 (𝑤0 (𝜉))

+ ∑𝑚≥1

N𝑚 (𝑤0, 𝑤1, 𝑤2, . . . , 𝑤𝑚) þ𝑚,𝑖 = 1, 2, 3, . . . , 𝑚

(8)

where 𝜑(𝜉, þ,C𝑖) is given by (5). The convergence of series(5) depends on the convergence of the constants C𝑖𝑠, if theseconstants are convergent at þ = 1, then the solution becomes

𝑤 (𝜉,C𝑖) = 𝑤0 (𝜉) + ∑𝑘≥1

(𝑤𝑘 (𝜉,C𝑖)) . (9)Generally, the 𝑚𝑡ℎ order solution of the problem can beobtained in the form

𝑤(𝑚) (𝜉,C𝑖) = 𝑤0 (𝜉) +𝑚∑𝑘=1

(𝑤𝑘 (𝜉,C𝑖)) ,𝑖 = 1, 2, 3, . . . , 𝑚

(10)

Putting this solution in (1) we get the following residual:

R (𝜉,C𝑖) = L (𝑤(𝑚) (𝜉,C𝑖) + ϝ (𝜉))+ N (𝑤(𝑚) (𝜉,C𝑖)) , 𝑖 = 1, 2, 3, . . . , 𝑚

(11)

If R(𝜉,C𝑖) = 0, then the solution is going to be exact,but generally, such a situation does not arise in nonlinearproblems but the functional defined below can be minimized

𝐽 (𝜉,C𝑖) = ∫𝑥1

𝑥0

R2 (𝜉,C𝑖) 𝑑𝜉, (12)

where 𝑥0 and 𝑥1 are two constants depending on the givenproblem. The values of C𝑖𝑠 can be optimally found by thecondition

𝜕𝐽𝜕C1 =

𝜕𝐽𝜕C2 = ⋅ ⋅ ⋅ =

𝜕𝐽𝜕C𝑚 = 0 (13)

After knowing these constants, the solution (10) is welldetermined.

3. Examples

To check the applicability of OHAM for TPBVP, in thissection four examples of TPBVP are presented in which oneexample is linear and the remaining are nonlinear.

3.1. Example 1. Let us consider the linear problem [1] ofsecond order

𝑤 (𝜉) = 𝑤 (𝜉) − 𝑒(𝜉−1) − 1, 0 < 𝜉 < 1,𝑤 (0) = 0,𝑤 (1) = 1

(14)

The exact solution of problem (14) is 𝜉(1 − 𝑒𝜉−1). Nowaccording toOHAML(𝑤0(𝜉)) = 𝑤(𝜉)−𝑤(𝜉), the nonlinearpart N(𝑤(𝜉)) = 0 and ϝ(𝜉) = 𝑒(𝜉−1) + 1.

The zeroth-order problem is

𝑤0 (𝜉) − 𝑤0 (𝜉) = 1 + 𝑒𝜉−1,𝑤0 (0) = 0,𝑤0 (1) = 1

(15)

The solution of (15) is

𝑤0 (𝜉) = (𝑒 − 𝑒𝜉) 𝜉

𝜉 (16)The first-order problem is

𝑤1 (𝜉) − 𝑤1 (𝜉) = −1 − 𝑒𝜉−1 − C1 − C1𝑒𝜉−1 + 𝑤0 (𝜉)+ C1𝑤0 (𝜉) − 𝑤0 (𝜉)− C1𝑤0 (𝜉) ,

𝑤1 (0) = 0,𝑤1 (1) = 0.

(17)

International Journal of Differential Equations 3

Table 1: Comparison of the third-order OHAM solution with the exact solution and HPM.

𝜉 OHAM Solution (𝑤(2)) Exact HPM [1] |𝑤(2) − 𝐸𝑥𝑎𝑐𝑡|0.1 0.059343 0.059343 0.05934820 1.38778 × 10−170.3 0.151024 0.151024 0.15103441 2.77556 × 10−170.5 0.196735 0.196735 0.19673826 2.77556 × 10−170.7 0.181427 0.181427 0.18142196 5.55112 × 10−170.9 0.0856463 0.0856463 0.08564186 5.55112 × 10−17

Ｑ()

0.20

0.15

0.10

0.05

0.00

0.0 0.2 0.4 0.6 0.8 1.0

OHAM

Exact

Figure 1: Comparison between exact solution (dashed line) andapproximate solution (dotted line) for example 1.


𝑤1 (𝜉) = 0 (18)The second-order problem is

𝑤2 (𝜉) = −C2 − exp (𝜉 − 1)C2 + C2𝑤0 (𝜉) + 𝑤1 (𝜉)+ C1𝑤1 (𝜉) − 𝑤2 (𝜉) − C2𝑤0 (𝜉) − 𝑤1 (𝜉)− C1𝑤1 (𝜉) ,

𝑤2 (0) = 0,𝑤2 (1) = 0

(19)


𝑤2 (𝜉) = 0 (20)And the third-order approximate solution of the bvp (14) isas follows:

𝑤(2) (𝜉) = 𝑤0 (𝜉) + 𝑤1 (𝜉) + 𝑤2 (𝜉) (21)𝑤(2) (𝜉) = (𝑒 − 𝑒

𝜉) 𝜉𝜉 (22)

Table 1 shows the comparison between the exact solution andthe approximate solution obtained by OHAM. Figure 1 of thesolution also shows well agreement with the exact solution.

3.2. Example 2. Consider the nonlinear two-point boundaryvalue problem [1] of the type

𝑤 (𝜉) = 𝑤3 (𝜉) − 𝑤 (𝜉) 𝑤 (𝜉) , 𝜉 ∈ [1, 2] ,𝑤 (1) = 12 ,𝑤 (2) = 13

(23)

According to OHAM L(𝑤(𝑥)) = 𝑤(𝜉) and N(𝑤(𝜉)) =u(𝜉)u(𝜉) − 𝑤3(𝜉), while ϝ(𝜉) = 0. The exact solution of (23)is 1/(𝜉 + 1). Now proceeding with the same lines as above wehave the following zeroth-order problem:

𝑤0 (𝜉) = 0,𝑤0 (1) = 12 ,𝑤0 (2) = 13

(24)


𝑤0 (𝜉) = 4 − 𝜉6 (25)Now the first-order problem is

𝑤1 (𝜉) = C1𝑤30 (𝜉) − C1𝑤0 (𝜉) 𝑤0 (𝜉) − 𝑤0 (𝜉)− C1𝑤0 (𝜉)

(26)

𝑤1 (1) = 0,𝑤1 (2) = 0 (27)

The solution of (26) is𝑤1 (𝜉)= −930C1 + 1649𝜉C1 − 880𝜉21C + 180𝜉31C − 20𝜉41C + 𝜉5C14320

(28)

The second-order problem is

𝑤2 (𝜉) = C2𝑤30 (𝜉) + 3C1𝑤20 (𝜉)𝑤1 (𝜉)− C2𝑤0 (𝜉)𝑤0 (𝜉) − C1𝑤1 (𝜉) 𝑤0 (𝜉)− C1𝑤0 (𝜉)𝑤1 (𝜉) − C2𝑤0 (𝜉) − 𝑤1 (𝜉)− C1𝑤1 (𝜉) ,

𝑤2 (1) = 0,𝑤2 (2) = 0.

(29)


Table 2: Comparison of second-order OHAM solution with the exact solution for example 2.

𝜉 OHAM Solution (𝑤(3)) Exact |𝑤(3) − 𝐸𝑥𝑎𝑐𝑡|1.1 0.47619 0.47619 2.0597 × 10−71.3 0.434783 0.434783 5.38284 × 10−71.5 0.400001 0.4 1.39261 × 10−61.7 0.370371 0.37037 6.7426 × 10−71.9 0.344828 0.344828 8.10196 × 10−8

0.50

0.45

0.40

0.35

1.0 1.2 1.4 1.6 1.8 2.0

OHAM

Exact

Ｑ()

Figure 2: Comparison between exact solution (dashed line) and approximate solution (dotted line) for example 2.


𝑤2 (𝜉) = 1130636800 (𝜉 − 2) (𝜉 − 1) {30240 (−465 + 𝜉 (127 + (𝜉 − 17) 𝜉)C1− (8985375 + 𝜉 (4253423 + 𝜉 (−3664113 + 𝜉 (1100519 + 5𝜉 (−36795 + 𝜉 (3709 + 7 (−33 + 𝜉) 𝜉))))))C21+ 30240 (−465 + 𝜉 (127 + (−17 + 𝜉) 𝜉))C2)

(30)

The third-order problem is

C3𝑤30 (𝜉) + 3C2𝑤20 (𝜉)𝑤1 (𝜉) + 3C1𝑤0 (𝜉) 𝑤21 (𝜉)+ 3C1𝑤20 (𝜉)𝑤2 (𝜉) − 3C3𝑤0 (𝜉) 𝑤0 (𝜉)− C2𝑤1 (𝜉) 𝑤0 (𝜉) − C1𝑤2 (𝜉) 𝑤0 (𝜉)− C2𝑤0 (𝜉) 𝑤1 (𝜉) − C1𝑤1 (𝜉) 𝑤1 (𝜉)− C1𝑤0 (𝜉) 𝑤2 (𝜉) − C3𝑤0 (𝜉) − C2𝑤1 (𝜉)− 𝑤2 (𝜉) − C1𝑤2 (𝜉) + 𝑤3 (𝜉) = 0

(31)

The solution of the third-order problem results a large output,therefore not included here.

Now the third-order approximate solution is

𝑤(3) (𝜉) = 𝑤0 (𝜉) + 𝑤1 (𝜉) + 𝑤2 (𝜉) + 𝑤3 (𝜉) (32)

C𝑖𝑠 has the following values and then substituting in theabove solution we will get the approximate solution. 𝑤(3)(𝜉)is given in Appendix (A.1).

C1 = −0.9637924142971654,C2 = −0.0002296939939480446,C3 = −0.000014314891134337846,

(33)

The solution at the points given in Table 2 and the graph of thesolution is shown in Figure 2. Here it is third-order OHAMsolution while the HPM[1] gives the accuracy up to 9 decimalplaces in 7th order.

3.3. Example 3. Now we consider higher order TPBVP oforder four. The problem is

𝑑4𝑤 (𝜉)𝑑𝜉4 = 𝑤2 (𝜉) + ϝ (𝜉) , 0 ≤ 𝜉 ≤ 1 (34)

with the boundary conditions 𝑤(0) = 0, 𝑤(0) = 0, 𝑤(1) = 1,and 𝑤(1) = 1.


Table 3: Comparison of second-order OHAM solution with the exact solution for example 3.

𝜉 OHAM Solution (𝑤(2)) Exact |𝑤(2) − 𝐸𝑥𝑎𝑐𝑡|0.2 0.077119 0.0771200 6.152306 × 10−110.4 0.279039 0.2790400 1.314346 × 10−100.6 0.538559 0.5385600 1.001054 × 10−100.8 0.788479 0.7884800 2.415356 × 10−11

Where L(𝑤(𝜉)) = 𝑑4𝑤(𝜉)/𝑑𝑥4, N(𝑤(𝜉)) = 𝑤2(𝜉), andϝ(𝜉) = −𝜉10 + 4𝜉9 − 4𝜉8 − 4𝜉7 + 8𝜉6 − 4𝜉4 + 120𝜉 − 48, theexact solution of problem (34) is𝑤𝑒𝑥𝑎𝑐𝑡 = 𝜉5 −2𝜉4 +2𝜉2. Aftersolving this by themethod described in Section 2, we have thefollowing zeroth-order problem:

48 − 120𝜉 + 4𝜉4 − 8𝜉6 + 4𝜉7 + 4𝜉8 − 4𝜉9 + 𝜉10

+ 𝑑4𝑤0 (𝜉)𝑑𝜉4 = 0(35)

𝑤0 (0) = 0,𝑤0 (0) = 0,𝑤0 (1) = 1,𝑤0 (1) = 1

(36)

The solution to (35) is

𝑤0 (𝜉) = 11081080 (2155683𝜉2 + 8038𝜉3 − 2162160𝜉4

+ 1081080𝜉5 − 2574𝜉8 + 1716𝜉10 − 546𝜉11− 364𝜉12 + 252𝜉13 − 45𝜉14)

(37)

The first-order problem is

− 48 + 120𝜉 − 4𝜉4 + 8𝜉6 − 4𝜉7 − 4𝜉8 + 4𝜉9 − 𝜉10− 48C1 + 120𝜉C1 − 4𝜉41C + 8𝜉6C1 − 4𝜉7C1− 4𝜉8C1 + 4𝜉9C1 − 𝜉10C1 + C1𝑤20 (𝜉)− (1 + C1) 𝑑

4𝑤0 (𝜉)𝑑𝜉4 +𝑑4𝑤1 (𝜉)𝑑𝜉4 = 0

(38)

𝑤1 (0) = 0,𝑤1 (0) = 0,𝑤1 (1) = 0,𝑤1 (1) = 0.

(39)

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0OHAM

Exact

Ｑ()


The second-order problem is

− 48C2 + 120𝜉C2 − 4𝜉4C2 + 8𝜉6C2 − 4𝜉7C2 − 4𝜉8C2+ 4𝜉9C2 − 𝜉10C2 + C2𝑤20 (𝜉) + 2C1𝑤0 (𝜉) 𝑤1 (𝜉)− C2𝑤(4)0 (𝜉) − 𝑤(4)1 (𝜉) − C1𝑤(4)1 (𝜉) + 𝑤(4)2 (𝜉) = 0

(40)

𝑤2 (0) = 0,𝑤2 (0) = 0,𝑤2 (1) = 0,𝑤2 (1) = 0.

(41)

The solutions of problem (38) and (40) are very large; there-fore we did not write it here. The constants C1 and C2 havethe values −1.0011320722175725 and −1.079468959963785×10−6, respectively. Table 3 and Figure 3 show a good agree-ment with the exact values. The approximate solution 𝑤2(𝜉)is given in Appendix (A.2).

3.4. Example 4. At last, consider the second-order nonlinearTPBVP[1]

𝑤 (𝜉) = 𝑤2 (𝜉) + 2𝜋2cos (2𝜋𝜉) − sin2 (2𝜋𝜉) ,0 ≤ 𝜉 ≤ 1 (42)

𝑤 (0) = 0,𝑤 (1) = 0. (43)


Table 4: Comparison of third-order OHAM solution with the exact solution.

𝜉 OHAM Solution (𝑤(3)) Exact |𝑤(3) − 𝐸𝑥𝑎𝑐𝑡|0.1 0.0954915 0.0954915 5.59262 × 10−90.3 0.654508 0.654508 1.23779 × 10−80.5 0.999999 1. 1.51565 × 10−80.7 0.654508 0.654508 1.23779 × 10−80.9 0.0954915 0.0954915 5.59262 × 10−9

The exact solution of (42) is sin2(𝜋𝜉). Solving (42) by themethod depicted in Section 2, we have the following zerothorder problem:

−2𝜋2cos (2𝜋𝜉) + sin4 (𝜋𝜉) + 𝑤0 (𝜉) = 0,𝑤0 (0) = 0,𝑤0 (1) = 0

(44)

The solution of (44) is given by

𝑤0 (𝜉) = 1128𝜋2 (15 − 64𝜋2 + 24𝜋2𝜉 − 24𝜋2𝜉2

− 16 cos (2𝜋𝜉) − 64𝜋2cos (2𝜋𝜉) + cos (4𝜋𝜉)) .(45)

The first-, second-, and third-order problems are given in(46), (47), and (48) respectively.

2𝜋2cos (2𝜋𝜉) (1 + C1) − sin4 (𝜋𝜉) (1 + C1)+ C1𝑤20 (𝜉) − 𝑤0 (𝜉) − C1𝑤0 (𝜉) + 𝑤1 (𝜉) = 0,

𝑤1 (0) = 0,𝑤1 (1) = 0

(46)

{2𝜋2cos (2𝜋𝜉) − sin4 (𝜋𝜉)}C2 + C2𝑤20 (𝜉)+ 2C1𝑤0 (𝜉)𝑤1 (𝜉) − C2𝑤0 (𝜉) − 𝑤1 (𝜉)− C1𝑤1 (𝜉) + 𝑤2 (𝜉) = 0,

𝑤2 (0) = 0,𝑤2 (1) = 0

(47)

{2𝜋2cos (2𝜋𝜉) − sin4 (𝜋𝜉)}C3 + C3𝑤20 (𝜉)+ 2C2𝑤0 (𝜉)𝑤1 (𝜉) + C1𝑤21 (𝜉) + 2C1𝑤0 (𝜉) 𝑤2 (𝜉)+ C3𝑤0 (𝜉) − C2𝑤1 (𝜉) − 𝑤2 (𝜉) − C1𝑤2 (𝜉)+ 𝑤3 (𝜉) = 0

(48)

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0OHAM

Exact

Ｑ()


𝑤3 (0) = 0,𝑤3 (1) = 0 (49)

The solutions of problem (46), (47), and (48) are very largeand therefore cannot be written here but the table of valuesand the graph are shown in Table 4 and Figure 4, respectively.The approximate solution 𝑤(3)(𝜉) is written in Appendix(A.3). The values of the constants C𝑖𝑠 can be found by (13)which are given as follows:

C1 = −0.9030981665320986,C2 = −0.00569345107292796,C3 = 0.00021560218552318884

(50)

4. Conclusion

This paper reveals that OHAM is a very strong methodfor solving TPBVP and gives us a more accurate solutionas compared to other methods. In these examples onlysecond- and third-order solution gives us the accuracy upto 8 or 10 decimal places; therefore it is concluded that thismethod converges very fast to the exact solution and insome problems like example 1 it gives us the exact solution.The plots and tables show well agreement with the exactsolution.


Appendix

(3)𝑤 (𝜉) = 4 − 𝜉6 +1

4320 (−930C1 + 1649𝜉C1 − 880𝜉2C1 + 180𝜉3C1 − 20𝜉4C1 + 𝜉5C1) +1

130636800 (−28123200C1+ 49865760𝜉C1 − 26611200𝜉2C1 + 5443200𝜉3C1 − 604800𝜉4C1 + 30240𝜉5C1 − 17970750C21 + 18449279𝜉C21+ 11103120𝜉2C21 − 17446800𝜉3C21 + 7333620𝜉4C21 − 1689534𝜉5C21 + 241920𝜉6C21 − 22080𝜉7C21 + 1260𝜉8C21− 35𝜉8C21 − 28123200C2 + 49865760𝜉C2 − 26611200𝜉2C2 + 5443200𝜉3C2 − 604800𝜉4C2 + 30240𝜉5C2)+ 113450364928000 (−2895564672000C1 + 5134178649600𝜉C1 − 2739889152000𝜉2C1 + 560431872000𝜉3C1− 62270208000𝜉4C1 + 3113510400𝜉5C1 − 3700536840000C21 + 3799075531680𝜉C21 + 2286354470400𝜉2C21− 3592645056000𝜉3C21 + 1510139030400𝜉4C21 − 347908841280𝜉5C21 + 49816166400𝜉6C21 − 4546713600𝜉7C21+ 259459200𝜉8C21 − 7207200𝜉9C21 − 1161287826270C31 + 270053629823𝜉C31 + 1988069792280𝜉2C31− 963479454080𝜉3C31 − 507564676190𝜉4C31 + 549609764847𝜉5C31 − 221549328000𝜉6C31 + 54219285120𝜉7C31− 9069733530𝜉8C31 + 1086825025𝜉9C31 − 93716480𝜉10C31 + 5653440𝜉11C31 − 220220𝜉12C31 + 4235𝜉13C31− 2895564672000C2 + 5134178649600𝜉C2 − 2739889152000𝜉2C2 + 560431872000𝜉3C2 − 62270208000𝜉4C2+ 3113510400𝜉5C2 − 3700536840000C1C2 + 3799075531680𝜉C1C2 + 2286354470400𝜉2C1C2− 3592645056000𝜉3C1C2 + 1510139030400𝜉4C1C2 − 347908841280𝜉5C1C2 + 49816166400𝜉6C1C2− 4546713600𝜉7C1C2 + 259459200𝜉8C1C2 − 7207200𝜉9C1C2 − 2895564672000C3 + 5134178649600𝜉C3− 2739889152000𝜉2C3 + 560431872000𝜉3C3 − 62270208000𝜉4C3 + 3113510400𝜉5C3) .

(A.1)

𝑤(2) (𝜉) = (2155683𝜉2 + 8038𝜉3 − 2162160𝜉4 + 1081080𝜉5 − 2574𝜉8 + 1716𝜉10 − 546𝜉11 − 364𝜉12 + 252𝜉13 − 45𝜉14) /1081080 + 1/2360410309588661890560000(−14113828503813453911359𝜉2C1+ 17512915766704666962322𝜉3C1 − 5586404113887189501420𝜉8C1 − 23144774000952226800𝜉9C1+ 3735433519034813810160𝜉10C1 − 1179691619002973294400𝜉11C1 − 797705495192034374400𝜉12C1+ 550212193377310464000𝜉13C1 − 97319245468388853600𝜉14C1 + 2551023644304960𝜉15C1− 856729299998872080𝜉16C1 + 279036341094724200𝜉17C1 + 247466505045614400𝜉18C1− 155275084044250800𝜉19C1 − 3650519253533040𝜉20C1 + 26398020790188000𝜉21C1− 8405056667479680𝜉22C1 + 897927547083840𝜉23C1 − 38159719228800𝜉24C1 + 21095475553152𝜉25C1+ 4049797421376𝜉26C1 − 6052906241984𝜉27C1 + 1221181635008𝜉28C1 + 475889853600𝜉29C1− 295593372480𝜉30C1 + 60656299200𝜉31C1 − 4738773375𝜉32C1)+ ((−32669682535166038177800562717403451774787200𝜉2C1+ 40537647046565498276872580694218893070457600𝜉3C1− 12931009390154359168924266696920562467136000𝜉8C1


+ 40537647046565498276872580694218893070457600𝜉3C1− 12931009390154359168924266696920562467136000𝜉8C1− 53573870389240769630688977156787421440000𝜉9C1+ 8646514810996349122418420307721339216128000𝜉10C1− 2730665933188143522672522412342587755520000𝜉11C1− 1846471726465979561236202970676720619520000𝜉12C1+ 1273591901712376237386288026597835571200000𝜉13C1− 225267641104971069200162200032882362880000𝜉14C1 + 5904927396321076366168727485651968000𝜉15C1− 1983095815708383589080913455027281664000𝜉16C1 + 645893399999572047403581539117439360000𝜉17C1+ 572817797505678403219571120181611520000𝜉18C1+ 8620466886846280020965176767572231570851512𝜉10C21− 2701942454241603867437283593123940978268560𝜉11C21− 1853296246170788567680120913554535247983040𝜉12C21+ 1273591901712376237386288026597835571200000𝜉13C21− 223127551098503747806099303888254244271880𝜉14C21 + 24188753111272816253620159558815649776𝜉15C21− 3954294999119289026707499780027151742404𝜉16C21 + 1280476731768992936554600505857043216034𝜉17C21+ 1146299848262837493979094855639474837488𝜉18C21 − 716583570253989121237950089315154553020𝜉19C21− 18153410829102066767079156365109506436𝜉20C21 + 122325062475997255309755518897728337880𝜉21C21− 38583478912638559610674295035606805760𝜉22C21 + 4064708510722834491402346035781177920𝜉23C21− 304944482768990447155003185921595200𝜉24C21 + 172131824316102518718368542082822400𝜉25C21+ 31225989769228784108742227282106504𝜉26C21 − 48696064947262435404352469277878496𝜉27C21+ 9955954110432674815496294613892416𝜉28C21 + 3768051722420250956380935649060800𝜉29C21− 2341220030385245677826756247386880𝜉30C21 + 474526294591146486759733985232768𝜉31C21− 40917013297075779490615187391192𝜉32C21 + 3867571077895433794192598216016𝜉33C21− 521615945593005771502089327960𝜉34C21 − 578651890632287750071052046372𝜉35C21+ 265019543551924982873921756160𝜉36C21 − 7980251045802826679782141248𝜉37C21− 25268923385823287170976713452𝜉38C21 + 8700984701390689923286015560𝜉39C21− 1340309891030782820607242784𝜉40C21 + 135045166608803637462708432𝜉41C21− 18683464135506877900207872𝜉42C21 − 3279443942818588507522560𝜉43C21+ 3820823675617325912753400𝜉44C21 − 805194420243411190696704𝜉45C21 − 143582548822550108696880𝜉46C21+ 115436320975091250290520𝜉47C21 − 27795688158510545773500𝜉48C21 + 3304269870772182097500𝜉49C21


− 165213493538609104875𝜉50C21 − 32669682535166038177800562717403451774787200𝜉2C2+ 40537647046565498276872580694218893070457600𝜉3C2− 12931009390154359168924266696920562467136000𝜉8C2− 53573870389240769630688977156787421440000𝜉9C2+ 8646514810996349122418420307721339216128000𝜉10C2− 2730665933188143522672522412342587755520000𝜉11C2− 1846471726465979561236202970676720619520000𝜉12C2+ 1273591901712376237386288026597835571200000𝜉13C2− 225267641104971069200162200032882362880000𝜉14C2 + 5904927396321076366168727485651968000𝜉15C2− 1983095815708383589080913455027281664000𝜉16C2 + 645893399999572047403581539117439360000𝜉17C2+ 572817797505678403219571120181611520000𝜉18C2 − 359419678365531276056662133636240640000𝜉19C2− 8449961331839350421935750500018432000𝜉20C2 + 61104253784799540864465470704550400000𝜉21C2− 19455425077784486158148965087647744000𝜉22C2 + 2078458576628113491233720925699072000𝜉23C2− 88329393580142122870344465623040000𝜉24C2 + 48830300656099081258580171252121600𝜉25C2+ 9374181927486220561295860381900800𝜉26C2 − 14010835209405620008882949076787200𝜉27C2+ 2826704059972590662056252066406400𝜉28C2 + 1101555855990308389571183162880000𝜉29C2− 684218434127313503766165801984000𝜉30C2 + 140402870708442139347389199360000𝜉31C2− 10968974274097042136514781200000𝜉32C2)) /5463709258346094058387175634104714600448000000+ ((−6381240944360971702818350529525900150881006436955309839999659827200000𝜉2C1)) /1067204898300215968287533458623769753803390805870787878857474048000000000

(A.2)

𝑤(3) (𝜉) = (15 + 64𝜋2 + 24𝜋2𝜉 − 24𝜋2𝜉2 − 16 cos [2𝜋𝜉] − 64𝜋2 cos [2𝜋𝜉] + cos [4𝜋𝜉]) / (128𝜋2) + 1/ (94371840𝜋6)⋅ (2312275C1 + 12037120𝜋2C1 + 11059200𝜋4C1 + 1018080𝜋2𝜉C1 + 8824320𝜋4𝜉C1 + 19224576𝜋6𝜉C1− 1018080𝜋2𝜉2C1 − 8478720𝜋4𝜉2C1 − 17694720𝜋6𝜉2C1 − 691200𝜋4𝜉3C1 − 2949120𝜋6𝜉3C1 + 345600𝜋4𝜉4C1+ 1198080𝜋6𝜉4C1 + 331776𝜋6𝜉5C1 − 110592𝜋6𝜉6C1 − 2373120 cos [2𝜋𝜉]C1 − 12441600𝜋2 cos [2𝜋𝜉]C1− 11796480𝜋4 cos [2𝜋𝜉]C1 − 1105920𝜋2𝜉 cos [2𝜋𝜉]C1 − 4423680𝜋4𝜉 cos [2𝜋𝜉]C1 + 1105920𝜋2𝜉2 cos [2𝜋𝜉]C1+ 4423680𝜋4𝜉2 cos [2𝜋𝜉]C1 + 63360 cos [4𝜋𝜉]C1 + 414720𝜋2 cos [4𝜋𝜉]C1 + 737280𝜋4 cos [4𝜋𝜉]C1+ 17280𝜋2𝜉 cos [4𝜋𝜉]C1 − 17280𝜋2𝜉2 cos [4𝜋𝜉]C1 − 2560 cos [6𝜋𝜉]C1 − 10240𝜋2 cos [6𝜋𝜉]C1 + 45 cos [8𝜋𝜉]C1+ 1105920𝜋 sin [2𝜋𝜉]C1 + 4423680𝜋3 sin [2𝜋𝜉]C1 − 2211840𝜋𝜉 sin [2𝜋𝜉]C1 − 8847360𝜋3𝜉 sin [2𝜋𝜉]C1− 8640𝜋 sin [4𝜋𝜉]C1 + 17280𝜋𝜉 sin [4𝜋𝜉]C1) + 1/ (487049291366400𝜋10) (11933558784000𝜋4C1) + 1/(35659800916682342400𝜋12) 𝜉 (384696643800268800𝜋8C1 + 3334400329855795200𝜋10C1+ 7264291475800719360𝜋12C1 + 317056740153753600𝜋4C21 + 3055348438990848000𝜋6C21


+ 9761324410664386560𝜋8C21 + 14367361192521891840𝜋10C21 + 15851242925787709440𝜋12C21+ 222124734074698455C31 + 2171404189900081136𝜋2C31 + 7100075693557310464𝜋4C31+ 10842199196917637120𝜋6C31 + 14250836120871567360𝜋8C31 + 11649501141852487680𝜋10C31+ 8684066594856370176𝜋12C31 + 384696643800268800𝜋8C2 + 3334400329855795200𝜋10C2+ 7264291475800719360𝜋12C2 + 317056740153753600𝜋4C1C2 + 3055348438990848000𝜋6C1C2+ 9761324410664386560𝜋8C1C2 + 14367361192521891840𝜋10C1C2 + 15851242925787709440𝜋12C1C2+ 384696643800268800𝜋8C3 + 3334400329855795200𝜋10C3 + 7264291475800719360𝜋12C3) + 1/(3770462866155503616000000𝜋14) (92382929312808960000000𝜋8C1 + 480922211281010688000000𝜋10C1+ 441851117127598080000000𝜋12C1 + 241156678171145748480000𝜋4C21 + 1273402703047211089920000𝜋6C21+ 1659580957335748608000000𝜋8C21 + 1923688845124042752000000𝜋10C21+ 883702234255196160000000𝜋12C21 + 594085555282615418420429C31 + 2862436440052784429431616𝜋2C31+ 2597952006766286551449600𝜋4C31 + 3066666484153535692800000𝜋6C31 + 2139343103453036544000000𝜋8C31+ 1442766633843032064000000𝜋10C31 + 441851117127598080000000𝜋12C31 + 92382929312808960000000𝜋8C2+ 480922211281010688000000𝜋10C2 + 441851117127598080000000𝜋12C2+ 241156678171145748480000𝜋4C1C2 + 1273402703047211089920000𝜋6C1C2+ 1659580957335748608000000𝜋8C1C2 + 1923688845124042752000000𝜋10C1C2+ 883702234255196160000000𝜋12C1C2 + 92382929312808960000000𝜋8C3 + 480922211281010688000000𝜋10C3+ 441851117127598080000000𝜋12C3) + 1/ (124684618589798400𝜋12) (−1345093160140800𝜋8𝜉2C1)

(A.3)

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

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Research ArticleGlobal Stability of an Economic Model with a Continuous Delayof Kaldor Type Modified

Aka Fulgence Nindjin, Tetchi Albin N’guessan , Sahoua Hypolithe Okou A Kpetihi ,and Kessé Tiban TiaUFR de Mathématiques et Informatique Université Félix Houphouët Boigny d’Abidjan Cocody 22 BP 582 Abidjan 22, Côte d’Ivoire

Correspondence should be addressed to Tetchi Albin N’guessan; [email protected] andSahoua Hypolithe Okou A Kpetihi; [email protected]

Received 7 March 2018; Revised 13 May 2018; Accepted 17 May 2018; Published 2 August 2018

Academic Editor: Dongfang Li

Copyright © 2018 Aka Fulgence Nindjin et al. This is an open access article distributed under the Creative Commons AttributionLicense, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properlycited.

This paper studies continuous nonlinear economic dynamics with a continuous delay of a Kaldor type modified in dimension two.The important results are, on the one hand, the boundedness of solutions, the existence of an attractive set, and the permanence ofthe system and, on the other hand, the local and global stability of equilibrium points.

1. Model

The complex nonlinear dynamic has been introduced intothe analysis of economic phenomena to explain not only thefluctuations observed in the series studies but also the eco-nomic crisis in the capitalist system. Thus, economists suchas Goodwin (1967) and Kaldor (1955-1956) have employeddynamicmodels to explain that the cyclic and chaotic growthcurves are the economic phenomena endogenous to thecapitalist system itself. Within the framework of structuralreforms for economic dynamics, NINDJIN et al. in [1] suggesta proven model originating from ecology which could have acertain range in analysis and regulation of financial systems.This model of dimension two describes how the GDP andan economic capital interact in accordance with the modelof Kaldor modified in order to increase the resilience ofthese dynamics against possible disruptions. Mathematicalanalysis of thismodel (see [1]) demonstrates that it is boundedand permanent and admits under certain circumstances anattractive set. On the one hand, this permanence manifestsitself in the form of stationary growth of capital stock andproduct (stable interior equilibrium point). On the otherhand, it appears in the capital cyclic growth and the product(limit cyclic). So, a capital stock rupture or a long-termproduction is prevented. It is also shown that the financialsystem stability (relative to the capital and the product) is

overall; i.e., it does not depend on either stock level orproduction at the initial time. Facing a possible disruptionof one of the control parameters of the economic system, wehave analyzed, in [2], the model bifurcations. We have shownthat the model admits a transcritical bifurcation, a pitchforkbifurcation or a Hopf bifurcation. In the last one, when themodel is disrupted, it changes from a stationary balancedgrowth to a cyclic growth by preventing a GDP or capitalcrisis. Interactions between the GDP and the economy’scapital could not be clearly and definitively understood orexplained regardless of past situations which may affect thepresent or the future. In this paper, we are going to focuson the way in which investments are evolved and savingsare established. Indeed, economies finance their investmentsthrough their own savings or those of other economiesvia financial structures with an interest rate. Regarding thesavings, it is known that they are made up of a portion ofprofits or wages. Thus, when the economy is able to be self-funding by its savings, then the net profit increases 𝜋𝑛𝑒𝑡 =𝜋 − 𝑟𝐷 (i.e., profit, 𝜋 deprived of the portion set aside to payinterest, 𝑟 debt, 𝐷). So, saving at a time 𝑡 depends on the netprofit; consequently, the GDP from a time 𝑡0 = 𝑡 − 𝑇 where𝑇 > 0. Let us suppose that 𝑇 is the deadline needed by thissaving to reach a certain threshold likely to ensure the self-financing of the investments of the economy.

HindawiInternational Journal of Differential EquationsVolume 2018, Article ID 7212795, 18 pageshttps://doi.org/10.1155/2018/7212795

http://orcid.org/0000-0002-9544-7221http://orcid.org/0000-0001-5504-9387https://doi.org/10.1155/2018/7212795


Let us consider the dynamics with no delay of themodified Kaldor type and the following assumption:

�̇� = [𝑎0𝑌(1 − 𝑌𝐶0)]+ 𝛼 (𝑎1𝐾 + 𝑏1) 𝑌(1 − 𝑚2𝐾𝑌 + 𝐶2)− [𝛼 𝑚1𝑌𝑌 + 𝐶1 ]𝐾,

�̇� = [(𝑎2𝑌 + 𝑏2) (1 − 𝑚2𝐾𝑌 + 𝐶2)]𝐾 − 𝐾,𝑌 (0) > 0, 𝐾 (0) > 0, 𝑌,𝐾 ∈ 𝐶1 ([0; +∞ [ ;R+) ,

(1.1)

with (𝑎0, 𝐶0, 𝐶1, 𝐶2, 𝑚1, 𝑚2, 𝛼) ∈ (R∗+)7 and (𝑎1, 𝑎2, 𝑏1,𝑏2, ) ∈ (R+)5.𝑌 denotes the product,𝐾 denotes the stock of capital, and�̇� and �̇� indicate, respectively, the growth rate of the productand the stock of capital depending on the following economicparameters:

(i) 𝑎0 the trend of rate of increase in GDP for a given(future) period in absence (or in neglect) of the losses,

(ii) 𝐶0 themaximum (monetary) value ofGDPwe can getfrom this economy for that given period,

(iii) 𝛼 the currency adjustment factor,(iv) 𝑚1 the maximum (monetary) value of the genuine

saving of this economy for the given period,

(v) 𝐶1 the maximum (monetary) value of the savingsupported by the economy in the given period,

(vi) 𝑚2 the maximum value of the investment rates lossesfor the given period,

(vii) 𝐶2 the maximum capital stock for the given period,(viii) 𝑎1 the derivative relative to the capital of the invest-

ment rate in the absence (or in the neglect) of thelosses (when𝑚2 = 0) for the given period,

(ix) 𝑏1 the investment rate when the capital is null (𝐾 =0) and this rate has suffered no loss (𝑚2 = 0) for thegiven period,

(x) 𝑎2 the share of the GDP converted into stock of capitalfor the given period,

(xi) 𝑏2 the accumulation rate of capital when the product isnull (𝑌 = 0) and that the investment rate has sufferedno loss (𝑚2 = 0) for the given period.

(see [1], page 4).

Assumption 1. The ratio saving-capital 𝑔(𝑌,𝐾) = 𝑚1𝑌/(𝑌 +𝐶1) at a given time 𝑡 depends on the GDP produced since 𝑡0 =𝑡 −𝑇 with a probability of exponential lag (see, [3], appendix,page 272).

So, the ratio saving-capital with delay is 𝑔(𝑡, 𝑇) =𝑚1𝑥(𝑡)/(𝑥(𝑡) + 𝐶1) with 𝑥(𝑡) = ∫𝑡𝑡−𝑇 𝜇𝑒−𝜇(𝑡−𝜉)𝑌(𝜉)𝑑𝜉 so that∫𝑡−∞

𝜇𝑒−𝜇(𝑡−𝜉)𝑑𝜉 = 1, 𝜇 ∈ R∗+. Then,�̇� = −𝜇 (𝑥 − 𝑌) − 𝜇𝑒−𝜇𝑇𝑌 (𝑡 − 𝑇) . (1)

By replacing 𝑔(𝑌,𝐾) by 𝑔(𝑡, 𝑇), one obtains continuousnonlinear economic dynamics with a continuous delay of aKaldor type modified in dimension two. Enclosing (1) to thesystemwith a continuous delay, one gets the following systemwith discrete delay in dimension three:

�̇� = 𝑎0 [1 − 𝑌𝐶0 ]𝑌+ 𝛼 (𝑎1𝐾 + 𝑏1) [(1 − 𝑚2𝐾𝑌 + 𝐶2)]𝑌− 𝛼 [ 𝑚1𝑥𝑥 + 𝐶1 ]𝐾,

�̇� = (𝑎2𝑌 + 𝑏2) [(1 − 𝑚2𝐾𝑌 + 𝐶2)]𝐾 − 𝐾,�̇� = −𝜇 (𝑥 − 𝑌) − 𝜇𝑒−𝜇𝑇𝑌 (𝑡 − 𝑇) ,𝑌,𝐾, 𝑥 ∈ 𝐶1 ([0; +∞ [; R+) , 𝑌 (0) > 0, 𝐾 (0) > 0,

(1.2)

with (𝑎0, 𝑎1, 𝑎2, 𝑏1, 𝑏2, 𝑐0, 𝑐1, 𝑐2, 𝑚1, 𝑚2, , 𝛼, 𝑇, 𝜇) ∈ (R∗+)14,𝑌𝑡(𝜃) = 𝑌(𝑡 + 𝜃), ∀𝜃 ∈ [−𝑇; 0], and the feedback function𝑌0 = 𝜑1 ∈ 𝐶([−𝑇; 0];R+) is consequently 𝑌0(0) = 𝑌(0).To facilitate the qualitative study of system (1.2)

which possesses 14 parameters (𝑎0, 𝑎1, 𝑎2, 𝑏1, 𝑏2, 𝑐0, 𝑐1, 𝑐2,𝑚1, 𝑚2, , 𝛼, 𝑇, 𝜇) ∈ (R∗+)14, let us change the variables byreducing the number of parameters.

Let us define the new variables:

𝜏 = 𝑎0𝑡,𝑢 (𝜏) = 𝑌 (𝑡)𝑐0 ,V (𝜏) = 𝑚2𝑐0 K (t) ,𝑤 (𝑡) = 𝑥𝑐0 = ∫

𝑡

𝑡−𝑇𝜆𝑒−𝜆(𝑡−𝜉)𝑢 (𝜉) 𝑑𝜉.

(2)

Let us define the new parameters of control:

𝛽1 = 𝛼𝑎1𝐶0𝑎0𝑚2 ,𝛽2 = 𝑎2𝐶0𝑎0 ,𝛼1 = 𝛼𝑏1𝑎0 ,


𝛼2 = 𝑏2𝑎0 ,𝑑1 = 𝐶1𝐶0 ,𝑑2 = 𝐶2𝐶0 ,𝛾 = 𝛼 𝑚1𝑎0𝑚2 ,𝛿 = 𝑎0 ,

(3)

𝑇 = 𝑎0𝑇, and 𝜆 = 𝜇/𝑎0.Then, system (1.2) becomes�̇� (𝜏) = [1 − 𝑢 (𝜏)] 𝑢 (𝜏)

+ (𝛽1V (𝜏) + 𝛼1) (1 − V (𝜏)𝑢 (𝜏) + 𝑑2)𝑢 (𝜏)− 𝛾𝑤 (𝜏)𝑤 (𝜏) + 𝑑1 V (𝜏) ,

V̇ (𝜏) = (𝛽2𝑢 (𝜏) + 𝛼2) (1 − V (𝜏)𝑢 (𝜏) + 𝑑2) V (𝜏) − 𝛿V (𝜏) ,

�̇� (𝜏) = −𝜆 (𝑤 (𝜏) − 𝑢 (𝜏)) − 𝜆𝑒−𝜆𝑇𝑢 (𝜏 − 𝑇) ,𝑢, V, 𝑤 ∈ 𝐶1 ([0; +∞ [; R+) 𝑢 (0) > 0, V (0) > 0, 𝑤 (0) > 0, (1.3)with (𝑑1, 𝑑2, 𝛾, 𝛼1, 𝛼2, 𝛽1, 𝛽2, 𝛿, 𝑇, 𝜆) ∈ (R∗+)10 and 𝑢0 =𝜑 ∈ 𝐶([−𝑇; 0];R+) so that 𝑢0(0) = 𝑢(0).In the long run, we shall adopt the following notations:

𝜑 = sup𝜃∈[−𝑇;0]

{𝜑 (𝜃)} ,𝑘 = 1 − 𝑒−𝜆𝑇,

and 𝑤 (0) = ∫0−𝑇

𝜆𝜑 (𝜃) 𝑒𝜆𝜃𝑑𝜃.(4)

2. Boundedness and Equilibria Points of (1.3)2.1. Boundedness of the Solutions of Model (1.3)Lemma 2. The interior int(R3+) and the boundary 𝜕(R3+) of apositive cone R3+ are invariant for model (1.3).Proof. Given 0 ≤ 𝑇0 < +∞. Let 𝜏 be in [−𝑇; 𝑇0[. One knowsthat

𝑢 (𝜏) = 𝑢 (0) exp{∫𝜏0[1 − 𝑢 (𝑡) + (𝛽1V (𝑡) + 𝛼1) (1 − V (𝑡)𝑢 (𝑡) + 𝑑2) −

𝛾V (𝑡) 𝑤 (𝑡)𝑢 (𝑡) (𝑤 (𝑡) + 𝑑1)] 𝑑𝑡} ,V (𝜏) = V (0) exp{∫𝜏

0[(𝛽2𝑢 (𝑡) + 𝛼2) (1 − V (𝑡)𝑢 (𝑡) + 𝑑2) − 𝛿] 𝑑𝑡} ,

𝑤 (𝜏) = 𝑤 (0) exp{∫𝜏0

−𝜆 (𝑤 (𝑡) − 𝑢 (𝑡)) − 𝜆𝑒−𝜆𝑇𝑢 (𝑡 − 𝑇)𝑤 (𝑡) 𝑑𝑡} .(5)

Thus, on the one hand, (𝑢(0), V(0), 𝑤(0)) ∈ 𝜕(R3+) ⇒(𝑢(𝜏), V(𝜏), 𝑤(𝜏)) ∈ 𝜕(R3+), and, on the other hand, (𝑢(0),V(0), 𝑤(0)) ∈ int(R3+) ⇒ (𝑢(𝜏), V(𝜏), 𝑤(𝜏)) ∈ int(R3+).Lemma 3 (see [4]). Given (𝐴, 𝐵) ∈ R2+ and 𝜙 a continuousand derivable function as there is 𝑡0 ≥ 0 verifying 𝜙(𝑡0) > 0.Then, ∀𝑡 ≥ 𝑡0,

𝑑𝜙𝑑𝑡 ≤ 𝜙 (𝐵 − 𝐴𝜙) ⇒lim sup𝑡→+∞

𝜙 (𝑡) ≤ 𝐵𝐴,(6)

𝑑𝜙𝑑𝑡 ≥ 𝜙 (𝐵 − 𝐴𝜙) ⇒lim inf𝑡→+∞

𝜙 (𝑡) ≥ 𝐵𝐴.(7)

Theorem 4. Let us assume that 𝛼2 > 𝛿. Let us pose𝑀𝑢 = 4𝑑2𝛽1 + (𝛽1𝑑2 + 2𝛼1) 𝛽1𝑑2 + 𝛼21𝛽1𝑑2 (4 − 𝛽1) if 0 < 𝛽1 < 4and 𝑀𝑢 = 1 + 𝛼1 if 𝛽1 = 0,𝑀V = (𝑀𝑢 + 𝑑2) (𝛽2𝑀𝑢 + 𝛼2 − 𝛿)𝛼2and 𝑚V = (𝛼2 − 𝛿) 𝑑2𝛽2𝑑2 + 𝛼2 ,𝑚(0)𝑢 = 𝑚𝑢 + 𝛽1𝑚V,𝑚𝑢 = −𝛽1𝑑2𝑀2V −

𝛼1𝑑2𝑀V + 1 + 𝛼1,

𝑚𝜀𝑢 = 𝑚(0)𝑢 + √[𝑚(0)𝑢 − 𝜀]2 − 4𝛾𝑘𝑀𝑢𝑀V/𝑑1

2 ,


0 ≤ 𝜀 ≤ 𝑚(0)𝑢 − 2√𝛾𝑘𝑀𝑢𝑀V𝑑1or 𝑚(0)𝑢 + 2√𝛾𝑘𝑀𝑢𝑀V𝑑1 ≤ 𝜀,𝑀𝑤 = 𝑘𝑀𝑢and 𝑚𝑤 = 𝑘𝑚𝜀𝑢.

(8)

If 𝑚(0)𝑢 ≥ 2√𝛾𝑘𝑀𝑢𝑀V𝑑1 , (9)

and, then, model (1.3) is permanent. Otherwise, the set definedby

A+ = {(𝑢, V, 𝑤) ∈ R3+ : 𝑚𝜀𝑢 ≤ 𝑢 ≤ 𝑀𝑢, 𝑚V ≤ V≤ 𝑀V and 𝑚𝑤 ≤ 𝑤 ≤ 𝑀𝑤} (10)

Which is a bounded set, positively invariant for model (1.3).Proof. Let us consider system (1.3) and Lemma 3.

(1) One has the following: V̇(𝜏) does not depend on thevariable 𝑤. So, one obtains the same results like thoseof the model with no delay, i.e.,

lim inf 𝑡→+∞[V(𝑡)] ≥ 𝑚V and lim sup𝑡→+∞[V(𝑡)] ≤𝑀V.(2) One has the following:𝑑𝑢/𝑑𝑡 ≤ (1−𝑢)𝑢+(𝛽1V+𝛼1)(1−

V/(𝑢 + 𝑑2))𝑢 because −𝛾𝑤V/(𝑤 + 𝑑1) < 0.Then, 𝑑𝑢/𝑑𝑡 ≤ (𝐵1−𝐴1𝑢)𝑢with (𝐴1, 𝐵1) = (1; 1+𝛼1)if 𝛽1 = 0and (𝐴1, 𝐵1)= ((4−𝛽1)/4; (4𝑑2𝛽1 + (𝛽1𝑑2+2𝛼1)𝛽1𝑑2 +𝛼21)/4𝛽1𝑑2) if 0 < 𝛽1 < 4 and, in that case,maxV{(𝛽1V+𝛼1)(1−V/(𝑢+𝑑2))𝑢} = 𝑢[𝛽1(𝑢+𝑑2)+𝛼1]2/4𝛽1(𝑢+𝑑2).By property (6) of Lemma 3, one obtainslim sup𝑡→+∞[𝑢(𝑡)] ≤ 𝑀𝑢.

(3) One knows that 0 ≤ 𝑢(𝜏). Let us suppose that 𝑢(0) > 0and, then, there exist 𝑡0 so that, ∀𝜏 ≥ 𝑡0, 0 < 𝑢(𝜏) ≤𝑀𝑢. Let us suppose that 𝑙 = liminf 𝑡→+∞[𝑢(𝑡)]. Thus,for any 𝜀 very tiny, there is 𝑡1 > 0 so that, ∀𝜏 > 𝑡1,𝑢(𝜏) ≥ 𝑙 − 𝜀 > 0.

Let us pose 𝑡2 = max(𝑡0, 𝑡1) and𝑚 = 𝑙 − 𝜀.Then, ∀𝜏 >𝑡2, 0 < 𝑚 < 𝑢(𝜏) ≤ 𝑀𝑢. Thus∀𝜏 > 𝑡2,0 < 𝑤 (𝜏) < 𝑢 (𝜏)𝑤 (𝜏)𝑚 ≤ 𝑀𝑢𝑤 (𝜏)𝑚 .

(11)

Hence, ∀𝜏 > 𝑡2, −𝛾𝑤(𝜏)/(𝑤(𝜏) + 𝑑1) >−𝛾(𝑢(𝜏)𝑤(𝜏)/𝑚)/(𝑢(𝜏)𝑤(𝜏)/𝑚 + 𝑑1) for 𝑔(𝑥) =−𝛾𝑥/(𝑥 + 𝑑1) is decreasing upon R+. On the onehand,

it is known that 𝑤(𝑡) = ∫𝑡𝑡−𝑇

𝜆𝑒−𝜆(𝑡−𝜉)𝑢(𝜉)𝑑𝜉 and∫𝑡−∞

𝜆𝑒−𝜆(𝑡−𝜉)𝑑𝜉 = 1.Meanwhile, lim sup𝑡→+∞{∫𝑡𝑡−𝑇 𝜆𝑒−𝜆(𝑡−𝜉)𝑢(𝜉)𝑑𝜉} ≤∫𝑡𝑡−𝑇

𝜆𝑒−𝜆(𝑡−𝜉)𝑑𝜉 × lim sup𝑡→+∞[𝑢(𝑡)] and, then,lim sup𝜏→+∞[𝑤(𝜏)] ≤ 𝑀𝑢(1 − 𝑒−𝜆𝑇) = 𝑘𝑀𝑢 = 𝑀𝑤.On the other hand,

�̇�/𝑢 = [1 − 𝑢(𝜏)] + (𝛽1V(𝜏) + 𝛼1)(1 − V(𝜏)/(𝑢(𝜏) +𝑑2))−(𝛾𝑤(𝜏)/(𝑤(𝜏)+𝑑1))(V(𝜏)/𝑢(𝜏)) and 1+ (𝛽1V(𝜏)+𝛼1)(1 − V(𝜏)/(𝑢(𝜏) + 𝑑2)) ≥ 𝑚(0)𝑢 = −(𝛽1/𝑑2)𝑀2V −(𝛼1/𝑑2)𝑀V + 1 + 𝛼1 + 𝛽1𝑚V. Then, 𝜏 > 𝑡2, and�̇� ≥ [𝑚(0)𝑢 − 𝛾𝑤(𝜏)V(𝜏)/(𝑢(𝜏)𝑤(𝜏) + 𝑚𝑑1) − 𝑢]𝑢(𝜏).So, 𝑑𝑢/𝑑𝑡 ≥ (𝐵2 − 𝐴2𝑢)𝑢 with 𝐴2 = 1 and 𝐵2 =𝑚(0)𝑢 −𝛾𝑀V𝑀𝑤/𝑚𝑑1.Therefore, applying property (7)of Lemma 3 one obtains lim inf 𝑡→+∞[𝑢(𝑡)] ≥ 𝑚(0)𝑢 −𝛾𝑀V𝑀𝑤/𝑚𝑑1.Consequently, 𝑙 ≥ 𝑚(0)𝑢 −𝛾𝑀V𝑀𝑤/(𝑙−𝜀)𝑑1 for𝑚 = 𝑙−𝜀.So, one obtains

𝑙2 − (𝑚(0)𝑢 + 𝜀) 𝑙 + 𝛾𝑀𝑢𝑀𝑤𝑑1 + 𝑚(0)𝑢 𝜀 ≥ 0. (12)If 𝑚(0)𝑢 ≥ 2√𝛾𝑘𝑀𝑢𝑀V/𝑑1 then [𝑚(0)𝑢 − 𝜀]2 −4𝛾𝑀𝑤𝑀V/𝑑1 ≥ 0.Thus one obtains lim inf 𝑡→+∞[𝑢(𝑡)] ≥ 𝑚𝜀𝑢 = (𝑚(0)𝑢 +√[𝑚(0)𝑢 − 𝜀]2 − 4𝛾𝑀𝑤𝑀V/𝑑1)/2with 0 ≤ 𝜀 ≤ 𝑚(0)𝑢 − 2√𝛾𝑘𝑀𝑢𝑀V/𝑑1 or 𝑚(0)𝑢 +2√𝛾𝑘𝑀𝑢𝑀V/𝑑1 ≤ 𝜀.

(4) It is known that lim inf 𝑡→+∞{∫𝑡𝑡−𝑇 𝜆𝑒−𝜆(𝑡−𝜉)𝑢(𝜉)𝑑𝜉} ≥∫𝑡𝑡−𝑇

𝜆𝑒−𝜆(𝑡−𝜉)𝑑𝜉 × lim inf 𝑡→+∞[𝑢(𝑡)]. So,lim inf 𝑡→+∞[𝑤(𝑡)] ≥ 𝑚𝑤 = 𝑘𝑚𝜀𝑢.

Remark 5. Let us consider the notations of Theorem 4.


(1) Let us consider the following assumptions:

𝛼2 > 𝛿,𝑚(0)𝑢 > 0,

𝑑1 [𝑚(0)𝑢 ]24𝛾𝑀𝑢𝑀V < 1

and 𝑇 < −1𝜆 ln[[1 − 𝑑1 [𝑚(0)𝑢 ]

2

4𝛾𝑀𝑢𝑀V ]].

(13)

𝛼2 > 𝛿,𝑚(0)𝑢 > 0

and 0 ≤ 𝛾 < 𝑑1 [𝑚(0)𝑢 ]2

4𝑘𝑀𝑢𝑀V .(14)

One remarks that each of assumptions (13) and (14)implies, respectively, condition (9).

(2) If𝑚(0)𝑢 ≥ 2√𝛾𝑘𝑀𝑢𝑀V/𝑑1, one has 0 < 𝑚(0)𝑢 /2 ≤ 𝑚𝜀𝑢 ≤lim inf 𝑡→+∞[𝑢(𝑡)]. So, we can adjust the minimumvalue of lim inf 𝑡→+∞[𝑢(𝑡)] by choosing the value of𝜀.

2.2. The Equilibria Points of Model (1.3)2.2.1. Points of Trivial Equilibria. If 0 < 𝛼2 ≤ 𝛿 thensystem (1.3) admits two points of trivial equilibria: 𝐸∗0 =(0; 0; 0), 𝐸∗1 = (1 + 𝛼1; 0; 𝑘(1 + 𝛼1)).

If 𝛼2 > 𝛿 ≥ 0 then system (1.3) admits three points oftrivial equilibria: 𝐸∗0 = (0; 0; 0), 𝐸∗1 = (1 + 𝛼1; 0; 𝑘(1 + 𝛼1)),and 𝐸∗2 = (0; (𝛼2 − 𝛿)𝑑2/𝛼2; 0).

2.2.2. Points of Interior Equilibria. Given 𝑝1(𝑥) = 𝑎14𝑥4 +𝑎13𝑥3 + 𝑎12𝑥2 + 𝑎11𝑥 + 𝑎10 such as𝑎14 = −𝑘𝛽22 ,𝑎13 = (𝑘 − 𝑑1) 𝛽22 − 2𝑘𝛽2𝛼2 + 𝑘𝛽2 (𝛽1𝛿 − 𝛾𝛽2) ,𝑎12 = 2𝛽2𝛼2 (𝑘 − 𝑑1) − 𝑘𝛼22 + 𝑑1𝛽22 + 𝑘𝛼1𝛿𝛽2

− 𝑘𝛾𝛽2𝛼2 + 𝛽1𝛽2𝛿𝑑1+ 𝑘 (𝛽2𝑑2 + 𝛼2 − 𝛿) (𝛽1𝛿 − 𝛾𝛽2) ,

𝑎11 = 𝛼1𝛿 (𝛽2𝑑1 + 𝑘𝛼2) + (𝑘 − 𝑑1) 𝛼22 + 2𝛽2𝛼2𝑑1+ 𝑘 (𝛽1𝛿 − 𝛾𝛽2) 𝑑2 (𝛼2 − 𝛿)+ (𝛽1𝛿𝑑1 − 𝑘𝛾𝛼2) (𝛽2𝑑2 + 𝛼2 − 𝛿) ,

𝑎10 = 𝑑1𝛼22 + 𝛼1𝛼2𝛿𝑑1 + (𝛽1𝛿𝑑1 − 𝑘𝛾𝛼2) 𝑑2 (𝛼2 − 𝛿) .

(15)

(1) System (1.3) does not admit any point of interiorequilibria if 𝛽2𝑀𝑢 + 𝛼2 − 𝛿 < 0.

(2) Any interior equilibrium point 𝐸∗3 = (𝑢∗; V∗, 𝑤∗) ofsystem (1.3) verifies the following relations:

𝑝1 (𝑢∗) = 0,V∗ = (𝛽2𝑢∗ + 𝛼2 − 𝛿)𝛽2𝑢∗ + 𝛼2 (𝑢∗ + 𝑑2)

with 𝛽2𝑢∗ + 𝛼2 − 𝛿 > 0,𝑤∗ = 𝑘𝑢∗.

(16)

Designating 𝑃𝑥𝑦 the projection over the plan 𝑥𝑦, we study theborder dynamics of the model.

3. Border Dynamics on the Plan 𝑢VOn 𝑢V plan, model (1.3) becomes

�̇� (𝜏) = [1 − 𝑢 (𝜏)] 𝑢 (𝜏) + (𝛽1V (𝜏) + 𝛼1) (1 − V (𝜏)𝑢 (𝜏) + 𝑑2)𝑢 (𝜏) ,V̇ (𝜏) = (𝛽2𝑢 (𝜏) + 𝛼2) (1 − V (𝜏)𝑢 (𝜏) + 𝑑2) V (𝜏) − 𝛿V,

𝑢, V, 𝑤 ∈ 𝐶1 ([0; +∞ [; R+) , 𝑢 (0) > 0, V (0) > 0, 𝑢0 = 𝜑 ∈ 𝐶 ([−𝑇; 0] ;R+) ,(3.1)

with (𝑑2, 𝛼1, 𝛼2, 𝛽1, 𝛽2, 𝛿, 𝑇, 𝜆) ∈ (R∗+)7.The equilibria points of model (3.1) are

𝑈(1)0 = 𝑃𝑢V (𝐸∗0 ) = (0, 0) ,𝑈(1)1 = 𝑃𝑢V (𝐸∗1 ) = (1 + 𝛼1; 0) ,

(17)

𝑈(1)2 = 𝑃𝑢V (𝐸∗2 ) = (0; (𝛼2 − 𝛿) 𝑑2𝛼2 ) , (18)𝑈(1)3 = (𝑢; (𝛽2𝑢 + 𝛼2 − 𝛿) (𝑢 + 𝑑2)𝛽2𝑢 + 𝛼2 )

if 𝑢 > 0 and (𝛽2𝑢 + 𝛼2 − 𝛿) > 0,(19)


where 𝑝0 (𝑢) = 𝑎03𝑢3 + 𝑎02𝑢2 + 𝑎01𝑢 + 𝑎00 = 0. (20)With 𝑎03 = −𝛽22 ,

𝑎02 = (𝛽2 − 2𝛼2 + 𝛿𝛽1) 𝛽2,𝑎01 = 𝛼2 (2𝛽2 − 𝛼2) + 𝛽1𝛿 (𝛽2𝑑2 + 𝛼2 − 𝛿)

+ 𝛿𝛽2𝛼1,𝑎00 = 𝛼22 + 𝛿𝛽1𝑑2 (𝛼2 − 𝛿) + 𝛼1𝛼2𝛿.

(21)

Let us give below the results on the permanence of themodel.Therefore, let us consider the notations ofTheorem 4. If 𝛼2 >𝛿 and𝑚(0)𝑢 > 0 thenmodel (3.1) is permanent. Among others,the set defined by

A(0)

= {(𝑢, V) ∈ R2+ : 𝑚(0)𝑢 ≤ 𝑢 ≤ 𝑀𝑢, and 𝑚V ≤ V ≤ 𝑀V}(22)

which is a bounded set, positively invariant for model(3.1).3.1. Local Stability of the Equilibria Points of Model (3.1).Performing the spectral study of Jacobian matrix of thesystem linearized around each of the points of equilibrium,one obtains, classically, the following conclusions:

(1) Stability of 𝑈(1)0 = (0; 0):(a) 𝑈(1)0 is an unstable node if 𝛼2 > 𝛿.(b) 𝑈(1)0 is a point unstable saddle repulsive follow-

ing direction 𝑢 and attractive following direc-tion V if 𝛼2 < 𝛿.

(2) Stability 𝑈(1)1 = (1 + 𝛼1; 0):(a) 𝑈(1)1 is stable if 𝛼2 + (1 + 𝛼1)𝛽2 < 𝛿.(b) 𝑈(1)1 is a point unstable saddle attractive follow-

ing direction𝑢 and repulsive following direction𝑤1 if 𝛼2 + (1 + 𝛼1)𝛽2 > 𝛿.(3) Stability of 𝑈(1)2 = (0; (𝛼2 − 𝛿)𝑑2/𝛼2) for 𝛼2 > 𝛿:

(a) 𝑈(1)2 is stable if 𝜓(𝛿) < 0.(b) 𝑈(1)2 is a point of unstable saddle if 𝜓(𝛿) > 0,

repulsive following 𝑤2, and attractive followingV. Consider

𝜓 (𝑥) = −𝑑2𝛽1𝑥2 + [𝑑2𝛽1𝛼2 + 𝛼1𝛼2] 𝑥 + 𝛼22 . (23)(4) Given𝑈(1)3 = (𝑢; V) interior equilibrium point of (3.1)

testifying the system (19)-(20) and J its associatedJacobian matrix.

(a) 𝑈(1)3 is stable (stable node or stable focus) ifdet(𝐽) > 0 and 𝑇𝑟(𝐽) < 0.

(b) 𝑈(1)3 is marginal or center if det(𝐽) ≥ 0 and𝑇𝑟(𝐽) = 0.(c) 𝑈(1)3 is unstable if det(𝐽) < 0 or (det(𝐽) > 0 and𝑇𝑟(𝐽) > 0), precisely:

(i) 𝑈(1)3 is a node or a focus if det(𝐽) > 0 and𝑇𝑟(𝐽) > 0,(ii) 𝑈(1)3 is a point unstable saddle if det(𝐽) < 0.

With 𝑇𝑟(𝐽) the trace and det(𝐽) determinant of J, vectors𝑤1 = ((1 + 𝛼1) [𝛽1 − 𝛼11 + 𝛼1 + 𝑑2 ] ; (1 + 𝛼1) (𝛽2 + 1)

+ 𝛼2 − 𝛿) .(24)

and𝑤2 = (𝜓(𝛿)/𝛼22 +𝛼2 −𝛿; [𝛽2𝑑2𝛿+𝛼2(𝛼2 −𝛿)](𝛼2 −𝛿)/𝛼22).3.2. Global Stability of 𝑈(1)3 = (𝑢; V). Now let us define theconditions for which the stability of the product and the stockof capital of the economy is global; i.e., it does not dependon the produced quantities and level of stock at the initialmoment. For this study, we define an appropriate Lyapunovfunction.

Theorem 6. Let us consider the following assumptions:

(3.1) 𝑎𝑑𝑚𝑖𝑡𝑠 𝑎 𝑢𝑛𝑖𝑞𝑢𝑒 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑝𝑜𝑖𝑛𝑡 𝑈(1)3= (𝑢; V) ,

(25)

0 < 𝛽1 < 4,𝛼2 > 𝛿,𝛽2𝑑2 < 2𝛼2,𝑑2 < 𝑑1,2𝛿𝛽1𝛼2 <

𝛼1𝑀𝑢 + 𝑑2 ,

(26)

𝛽1𝑀V𝑑2 +𝛼1𝑑2 <

𝛽1𝑑2 (2𝛼2 − 𝛿)𝛼2 (𝑀𝑢 + 𝑑2) +𝛿𝛽1 (1 + 𝑑2𝛽2/2𝛼2)𝛼2 , (27)

1 + 𝛼1𝑑2 − 1 +2𝛽1𝑑22𝛼21 < 0. (28)

where𝑀𝑢 and𝑀V are defined in Theorem 4.If assumptions (25), (26), (27), (28), (29), (30), (31), (35),

(36), and (37) are verified, then, the unique point of the interiorequilibrium ofmodel (3.1) is globally and asymptotically stable.


Proof. Let us consider system (3.1). Let us suppose thatassumption (35) is verified and, then, model (3.1) admits aunique interior equilibrium point 𝑈(1)3 = (𝑢; V) verifying

1 = 𝑢 − 𝛼1 (1 − V𝑢 + 𝑑2) ,𝛼2 − 𝛿 = −𝛽2𝑢 + (𝛽2𝑢 + 𝛼2) V𝑢 + 𝑑2 .

(29)

Let us note that 𝜆 = 𝛽1𝑑2/2𝛼2. Given 𝑉1 : R2 → R and𝑉2 : R2 → R one has𝑉1 (𝑢, V) = [𝑢 − 𝑢 − 𝑢 ln(𝑢𝑢)] = ∫

𝑢

𝑢[1 − 𝑢𝑥] 𝑑𝑥, (30)

𝑉2 (𝑢, V) = 𝜆 [V − V − V ln(VV)] = 𝜆∫V

V[1 − V𝑥] 𝑑𝑥. (31)

Considering the Lyapunov functional 𝑉 : R2 → R one has𝑉 (𝑢, V) = 𝑉1 (𝑢, V) + 𝑉2 (𝑢, V) . (32)

Let us pose

𝑔 (𝑢, V) = −1 + 𝛽1V2 + 𝛼1V(𝑢 + 𝑑2) (𝑢 + 𝑑2) ;ℎ (𝑢, V) = 12 [𝛽1 − 𝛽1 (V + V) + 𝛼1(𝑢 + 𝑑2) ]

+ 𝜆2 [𝛽2 − 𝛽2V𝑢 + 𝑑2 +(𝛽2𝑢 + 𝛼2) V(𝑢 + 𝑑2) (𝑢 + 𝑑2)] .

(33)

By using relation (29) and model (3.1), one obtains𝑑𝑉𝑑𝑡 ≤ [𝜙1 (𝑢, V)] (𝑢 − 𝑢)2 + [𝜙2 (𝑢, V)] (V − V)2 (34)

with 𝜙1 = 𝑔(𝑢, V) + |ℎ(𝑢, V)| and 𝜙2 = |ℎ(𝑢, V)| − 𝜆(𝛽2𝑢 +𝛼2)/(𝑢 + 𝑑2), ∀(𝑢, V) ∈ R2.Let us overestimate 𝑔(𝑢, V) and ℎ(𝑢, V) and, then, 𝜙1(𝑢, V)

and 𝜙2(𝑢, V).(1) One has 𝑔(𝑢, V) < 𝑅0 < 0, ∀(𝑢, V) ∈ R2, if 𝑅0 = −1 +(𝛽1𝑀2V + 𝛼1𝑀V)/𝑑22 < 0.(2) if 𝛽2 < 2𝛼2/𝑑2 then 𝜕ℎ/𝜕V < 0. So, ℎ(𝑀𝑢,𝑀V) ≤ℎ(𝑢, V) ≤ ℎ(0, 0). Therefore, ℎ(0, 0) < 0 if 𝛽2𝑑2 0. Then, (𝛽1/𝑑22)𝑀2V +(𝛼1/𝑑22)𝑀V < (1 + 𝛼1)/𝑑2. Moreover, 1/(𝑀𝑢 + 𝑑2) 𝛿,𝛽2 < 𝛼2𝑑2 ,

𝛿𝛽2𝛼2 <𝛼1𝑀𝑢 + 𝑑2 ,

(38)

−1 + 1 + 𝛼1𝑑2 +4𝛼2𝑑2𝛼21 < 0, (39)

𝛼12𝑑2 +𝛿𝛽2𝛼2 <

3 (𝛼2 − 𝛿)2 (𝑀𝑢 + 𝑑2) , (40)with constant𝑀𝑢 and𝑀V defined inTheorem 4.In fact, it is worth taking the Lyapunov functional 𝑉 :

R2 → R so that𝑉(𝑢, V) = 𝑉1(𝑢, V)+𝑉2(𝑢, V)with𝑉1 : R2 →R and 𝑉2 : R2 → R so that

𝑉1 (𝑢, V) = [𝑢 − 𝑢 − 𝑢 ln(𝑢𝑢)] = ∫𝑢

𝑢[1 − 𝑢𝑥] 𝑑𝑥, (41)

𝑉2 (𝑢, V) = [V − V − V ln(VV)] = ∫V

V[1 − V𝑥] 𝑑𝑥. (42)


Considering (38)–(40), one shows, similarly in 𝛽1 ̸= 0, that𝑑𝑉/𝑑𝑡 < 0, ∀(𝑢, V) ∈ R2.4. Border Dynamics of Plan 𝑢𝑤On the plan 𝑢𝑤model (1.3) becomes

�̇� (𝜏) = [1 + 𝛼1 − 𝑢 (𝜏)] 𝑢 (𝜏) ,�̇� = −𝜆 (𝑤 − 𝑢) − 𝜆𝑒−𝜆𝑇𝑢 (𝜏 − 𝑇) ,

𝑢, 𝑤 ∈ 𝐶1 ([0; +∞ [ ;R+) , 𝑢 (0) > 0, 𝑢0 = 𝜑 ∈ 𝐶 ([−𝑇; 0] ;R+)(4.1)

with (𝛼1, 𝑇, 𝜆) ∈ (R∗+)3.The equilibria of model (4.1) are𝑈(2)0 = 𝑃𝑢𝑤 (𝐸∗0 ) = (0, 0) ,𝑈(2)1 = 𝑃𝑢𝑤 (𝐸∗1 ) = (1 + 𝛼1; 𝑘 (1 + 𝛼1)) .

(43)

Let 𝑀(2)𝑢 = 1 + 𝛼1, 0 < 𝑚(2)𝑢 < 1 + 𝛼1, 𝑀(2)𝑤 = 𝑘(1 + 𝛼1), and𝑚(2)𝑤 = 𝑘𝑚(2)𝑢 .Themodel is permanent and the set defined byA(2) = {(𝑢, 𝑤) ∈ R2+ : 𝑚(2)𝑢 ≤ 𝑢 ≤ 𝑀(2)𝑢 , and 𝑚(2)𝑤 ≤ 𝑤≤ 𝑀(2)𝑤 }

(44)

is limited, positively invariant.Concerning the local stability, one obtains, for any delay𝑇, the following results:(1) 𝑈(2)0 = (0; 0) is an unstable saddle point, repulsive

along the direction 𝑢, and attractive along the direc-tion 𝑤.

(2) 𝑈(2)1 = (1 + 𝛼1; 𝑘(1 + 𝛼1)) is stable.The global stability of equilibrium 𝑈(2)1 is given in the

following theorem.

Theorem 8.

If 𝑇 < ln 2𝜆and 𝜆 < (1 + 𝛼1) (1 − ln 2)

(45)

and, then, the unique interior equilibrium point 𝑈(2)1 = (1 +𝛼1; 𝑘(1 + 𝛼1)) of model (4.1) is globally and asymptoticallystable.

Proof. Let us consider 𝑈(2)1 = (𝑢, 𝑤) with 𝑢 = 1 + 𝛼1 and𝑤 = 𝑘(1 + 𝛼1). Let us pose𝑈 = ln(𝑢/𝑢) and 𝑊 = ln(𝑤/𝑤). Then, one obtains thefollowing system:

�̇� (𝜏) = −𝑢 [𝑒𝑈(𝜏) − 1] ,�̇� (𝜏) = 𝜆𝑒−𝑤(𝜏)𝑘 [(𝑒𝑈(𝜏) − 1) − 𝑘 (𝑒𝑊(𝜏) − 1)

− 𝑒−𝜆𝑇 (𝑒𝑈(𝜏−𝑇) − 1)] .(46)

Let us note that 𝑉1(𝜏) = |𝑈(𝜏)|.

Then, the superior derivative𝐷+𝑉1(𝜏) of𝑉1(𝜏) in relationto time, alongside the solutions of (46), gives

𝐷+𝑉1 (𝜏) ≤ −𝑢 𝑒𝑈(𝜏) − 1 = − |𝑢 − 𝑢| . (47)Let us note that 𝑉21(𝜏) = |𝑊(𝜏)|. Considering (𝑢, 𝑤) ∈ A(2),one has 𝑚(2)𝑢 /𝑢 ≤ 𝑒𝑈(𝜏) ≤ 𝑀(2)𝑢 /𝑢 and 𝑚(2)𝑤 /𝑤 ≤ 𝑒𝑊(𝜏) ≤𝑀(2)𝑤 /𝑤. Hence, the superior derivative of 𝑉21(𝜏) in relationto time, alongside solutions (46), gives

𝐷+𝑉21 (𝜏) ≤ 𝜆 𝑒𝑈(𝜏) − 1 − 𝜆 𝑤∗

𝑚(2)𝑤𝑒𝑊(𝜏) − 1

+ 𝑀(2)𝑢 𝜆𝑒−𝜆𝑇𝑘 ∫𝜏

𝜏−𝑇

𝑒𝑈(𝑠) − 1 𝑑𝑠.(48)

Considering the functional: 𝑉22(𝜏) = (𝑀(2)𝑢 𝜆𝑒−𝜆𝑇/𝑘) ∫𝜏𝜏−𝑇

∫𝜏V |𝑒𝑈(𝑠) − 1|𝑑𝑠𝑑V then𝐷+𝑉22 (𝜏) ≤ 𝑇𝑀(2)𝑢 𝜆𝑒−𝜆𝑇𝑘 𝑒𝑈(𝜏) − 1

− 𝑀(2)𝑢 𝜆𝑒−𝜆𝑇𝑘 ∫𝜏

𝜏−𝑇

𝑒𝑈(𝑠) − 1 𝑑𝑠.(49)

Let us pose 𝑉2(𝜏) = 𝑉21(𝜏) + 𝑉22(𝜏).Thus,𝐷+𝑉2 (𝜏) ≤ 𝜆(1 + 𝑇𝑀(2)𝑢 𝑒−𝜆𝑇𝑘 ) 𝑒𝑈(𝜏) − 1

− 𝜆 𝑤∗𝑚(2)𝑤𝑒𝑊(𝜏) − 1 .

(50)

Then, 𝐷+𝑉2 (𝜏) ≤ 𝜆𝑢 (1 + 𝑇𝑀(2)𝑢 𝑒−𝜆𝑇𝑘 ) |𝑢 − 𝑢|

− 𝜆𝑚(2)𝑤 |𝑤 − 𝑤| .(51)

Let 𝑉(𝜏) = 𝑉1(𝜏) + 𝑉2(𝜏).Then,𝐷+𝑉 (𝜏) ≤ −[1 − 𝜆𝑢 (1 + 𝑇𝑀

(2)𝑢 𝑒−𝜆𝑇𝑘 )] |𝑢 − 𝑢|

− 𝜆𝑚(2)𝑤 |𝑤 − 𝑤| .(52)

So,𝐷+𝑉(𝜏) < 0 if (1 + (𝑇𝑒−𝜆𝑇/(1 − 𝑒−𝜆𝑇))𝑀(2)𝑢 < (1 + 𝛼1)/𝜆).Consequently, 𝐷+𝑉(𝜏) < 0 if (𝑇 < (ln 2)/𝜆 and 𝜆 < (1 +𝛼1)(1 − ln 2).5. Border Dynamics of the Plan V𝑤On the V𝑤 plan model (1.3) becomesV̇ (𝜏) = [𝛼2 − 𝛿 − 𝛼2V (𝜏)𝑑2 ] V (𝜏) ,

�̇� = −𝜆𝑤 − 𝜆𝑒−𝜆𝑇1[0;𝑇] (𝜏) 𝑢 (𝜏 − 𝑇) ,V, 𝑤 ∈ 𝐶1 ([0; +∞ [ ;R+) 𝑢 (0) > 0, 𝑢0 = 𝜑 ∈ 𝐶 ([−𝑇; 0] ;R+)

(5.1)


with (𝑑2, 𝛼2, 𝛿, 𝑇, 𝜆) ∈ (R∗+)4 and 1[0;𝑇](𝜏) = 1 if 0 ≤ 𝜏 ≤𝑇, 1[0;𝑇](𝜏) = 0 if 𝜏 > 𝑇.The equilibria of model (5.1) are as follows: for 𝜏 ≥ 𝑇,

𝑈(3)0 = 𝑃V𝑤 (𝑈0) = (0, 0) ,𝑈(3)2 = 𝑃V𝑤 (𝑈2) = ((𝛼2 − 𝛿) 𝑑2𝛼2 ; 0) .

(53)

Model (4.1) is nonpermanent. In fact, lim𝑡→+∞[𝑤(𝑡)] = 0.Let us now give the results upon stability of the equilibria

of model (5.1).(1) The equilibrium 𝑈(3)0 = (0; 0) is stable if 𝛼2 < 𝛿 and a

point unstable saddle if 𝛼2 > 𝛿.(2) The equilibrium 𝑈(3)2 = ((𝛼2 − 𝛿)𝑑2/𝛼2; 0) for 𝛼2 > 𝛿

is stable.Moreover, the equilibrium 𝑈(3)2 = ((𝛼2 − 𝛿)𝑑2/𝛼2; 0)is globally and asymptotically stable. In effect,lim𝑡→+∞[V(𝑡)] = (𝛼2−𝛿)𝑑2/𝛼2 and lim𝑡→+∞[𝑤(𝑡)] =0.

6. Local Stability of Model (1.3)Let us consider model (1.3).

Let us pose that 𝑈𝑖(𝜏) = 𝑢(𝜏) − 𝑢∗𝑖 , 𝑉𝑖(𝜏) = V(𝜏) − V∗𝑖 , and𝑊𝑖(𝜏) = 𝑤(𝜏) − 𝑤∗𝑖 for 𝑖 = 0, 1, 2, 3 with𝑢∗0 = 0,𝑢∗1 = 1 + 𝛼1,𝑢∗2 = 0,𝑢∗3 = 𝑢∗

(54)

V∗0 = 0,V∗1 = 0,V∗2 = (𝛼2 − 𝛿) 𝑑2𝛼2 ,V∗3 = V∗

(55)

𝑤∗0 = 0,𝑤∗1 = 𝑘 (1 + 𝛼1) ,𝑤∗2 = 0,𝑤∗3 = 𝑘𝑢∗.

(56)

The system linearized around the point of equilibrium 𝐸∗𝑖 =(𝑢∗𝑖 , V∗𝑖 , 𝑤∗𝑖 ), 𝑖 = 0, 1, 2, 3, is�̇�𝑖 (𝜏) = 𝐴(𝑖)11𝑈𝑖 (𝜏) + 𝐴(𝑖)12𝑉𝑖 (𝜏) + 𝐴(𝑖)13𝑊𝑖 (𝜏) ,�̇�𝑖 (𝜏) = 𝐴(𝑖)21𝑈𝑖 (𝜏) + 𝐴(𝑖)22𝑉𝑖 (𝜏) ,�̇�𝑖 (𝜏) = 𝜆𝑈𝑖 (𝜏) − 𝜆𝑊𝑖 (𝜏) − 𝜆𝑒−𝜆𝑇𝑈𝑖 (𝜏 − 𝑇) ,

(57)

with

𝐴(𝑖)11 = 1 − 2𝑢∗𝑖 + (𝛽1V∗𝑖 + 𝛼1) (1 − V∗𝑖𝑢∗𝑖 + 𝑑2)

+ (𝛽1V∗𝑖 + 𝛼1) 𝑢∗𝑖 V∗𝑖(𝑢∗𝑖 + 𝑑2)2 ,(58)

𝐴(𝑖)12 = [𝛽1 − 2𝛽1𝑢∗𝑖 + 𝑑2 V∗𝑖 − 𝛼1𝑢∗𝑖 + 𝑑2] 𝑢

∗𝑖 − 𝛾𝑤

∗𝑖𝑤∗𝑖 + 𝑑1 , (59)

𝐴(𝑖)13 = −𝛾𝑑1V∗𝑖(𝑤∗𝑖 + 𝑑1)2 ,

𝐴(𝑖)21 = [𝛽2 (1 − V∗𝑖𝑢∗𝑖 + 𝑑2) +

(𝛽2𝑢∗𝑖 + 𝛼2) V∗𝑖(𝑢∗𝑖 + 𝑑2)2 ] V∗𝑖 ,

(60)

𝐴(𝑖)22 = (𝛽2𝑢∗𝑖 + 𝛼2) (1 − 2V∗𝑖𝑢∗𝑖 + 𝑑2) − 𝛿,

𝐴(𝑖)23 = 0,(61)

𝐴(𝑖)31 = 𝜆,𝐴(𝑖)32 = 0,𝐴(𝑖)33 = −𝜆,𝐴(𝑖)34 = −𝜆𝑒−𝜆𝑇.

(62)

The characteristic equation of (57) is

Δ 𝑖 (𝑥, 𝑇) = 𝑃𝑖 (𝑥, 𝑇) + 𝑒−𝑥𝑇𝑄𝑖 (𝑥, 𝑇) = 0 (63)with 𝑃𝑖 (𝑥, 𝑇) = − [𝑥3 + 𝑎(𝑖)𝑥2 + 𝑏(𝑖)𝑥 + 𝑐(𝑖)] ,𝑄𝑖 (𝑥, 𝑇) = 𝑒−𝜆𝑇 [𝐷(𝑖)𝑥 + 𝐸(𝑖)] (64)where, 𝐷(𝑖) = −𝜆𝐴(𝑖)13,𝐸(𝑖) = 𝜆𝐴(𝑖)13𝐴(𝑖)22,𝑎(𝑖) = − (𝐴(𝑖)11 + 𝐴(𝑖)22 − 𝜆) ,

(65)

𝑏(𝑖) = − [𝜆 (𝐴(𝑖)11 + 𝐴(𝑖)22 + 𝐴(𝑖)13) + 𝐴(𝑖)21𝐴(𝑖)12 − 𝐴(𝑖)11𝐴(𝑖)22] , (66)𝑐(𝑖) = 𝜆 [𝐴(𝑖)11𝐴(𝑖)22 − 𝐴(𝑖)21𝐴(𝑖)12 + 𝐴(𝑖)13𝐴(𝑖)22] . (67)

6.1. Local Stability of 𝐸∗0 and 𝐸∗1 . We are in situations where𝑄𝑖(𝑥, 𝑇) = 0. Let us use the following criterion Routh-Hurwitz: the interior equilibrium 𝐸∗𝑖 is stable if 𝑎(𝑖) > 0,𝑏(𝑖) > 0, 𝑐(𝑖) > 0, and 𝑎(𝑖)𝑏(𝑖) − 𝑐(𝑖) > 0 and unstable if not.Theorem 9. Let us suppose that 𝛼2 > 𝛿. Then, the equilibria𝐸∗0 and 𝐸∗1 of model (1.3) are all unstable.Proof. Let us consider formula (63)–(67).


(1) The equation of 𝐸∗0 is Δ 0(𝑥, 𝑇) = 𝑥3 + 𝑎(0)𝑥2 + 𝑏(0)𝑥 +𝑐(0) = 0.One has the following: 𝑎(0) = 𝜆− (1+𝛼1 +𝛼2 −𝛿) and𝑏(0) = (1+𝛼1)(𝛼2−𝛿)−𝜆(1+𝛼1+𝛼2−𝛿). Let us supposethat (𝑎(0) > 0 and 𝑏(0) > 0). Then, one would have(1+𝛼1+𝛼2−𝛿) < 𝜆 < (1+𝛼1)(𝛼2−𝛿)/(1+𝛼1+𝛼2−𝛿).Consequently, (1+𝛼1)2+(𝛼2−𝛿)2+(1+𝛼1)(𝛼2−𝛿) < 0which is absurd. Hence, the equilibrium 𝐸∗0 of model(1.3) is unstable by application of the Routh-Hurwitzcriterion.

(2) One has Δ 1(𝑥, 𝑇) = 𝑥3 + 𝑎(1)𝑥2 + 𝑏(1)𝑥 + 𝑐(1) = 0.Meanwhile, 𝛼2 > 𝛿 and, then, 𝑐(1) = −𝜆(1+𝛼1)[𝛽2(1+𝛼1) +𝛼2 −𝛿] < 0 so the equilibrium 𝐸∗0 of model (1.3)is unstable by application of Routh-Hurwitz criterion.

6.2. Local Stability of𝐸∗2 and𝐸∗3 . Weare in the situationwhere𝑄𝑖(𝑥, 𝑇) ̸= 0. In order to assess the influence of delay 𝑇 overthe local stability of model (1.3), we use the results upon thelocal stability of model obtained taking 𝑇 = 0 into model(1.3) and the results of [5] upon the local stability of the delaysystems. Let us consider the characteristic equation (63).

(i) One has the following: 𝑄𝑖(𝑧, 𝑇) = 0 ⇒ 𝑧 =−𝐸(𝑖)/𝐷(𝑖) ∈ R. Then, 𝑃𝑖(𝑥, 𝑇) and 𝑄𝑖(𝑥, 𝑇) do nothave any common imaginary roots.

(ii) One has the following: 𝑃𝑖(𝑥, 𝑇) and 𝑄𝑖(𝑥, 𝑇) arepolynomials with real coefficients

so 𝑃𝑖(−𝑖𝜔, 𝑇) = 𝑃𝑖(𝑖𝜔, 𝑇) and 𝑄𝑖(−𝑖𝜔, 𝑇) = 𝑄𝑖(𝑖𝜔, 𝑇).(iii) If −(1/𝜆) ln(𝑐(𝑖)/𝐸(𝑖)) ̸= 𝑇 so 𝑃𝑖(0, 𝑇) + 𝑄𝑖(0, 𝑇) =𝐸(𝑖)𝑒−𝜆𝑇 − 𝑐(𝑖) ̸= 0.(iv) Let us pose 𝑥 = 𝜌𝑒𝑖𝜑.Then, |𝑃𝑖(𝑥, 𝑇)/𝑄𝑖(𝑥, 𝑇)| ∼ 1/𝜌4

when 𝜌 → +∞.Thus lim sup

|𝑥|→+∞

{𝑄𝑖 (𝑥, 𝑇)𝑃𝑖 (𝑥, 𝑇)

} = 0 < 1. (68)Let us consider the function defined on R by the following:𝐹𝑖(𝜔) = |𝑃𝑖(𝑖𝜔, 𝑇)|2 − |𝑄𝑖(𝑖𝜔, 𝑇)|2.

Then, 𝐹 (𝜔) = [𝑏(𝑖)𝜔 − 𝜔3]2 + [𝑐(𝑖) − 𝑎(𝑖)𝜔2]2− 𝑒−2𝜆𝑇 (𝐷(𝑖)𝜔2 + 𝐸(𝑖)) (69)

So 𝐹𝑖 (𝜔) = 𝜔6 + 𝑚2𝜔4 + 𝑚1𝜔2 + 𝑚0,where (70)𝑚2 = ([𝑎(𝑖)]2 − 2𝑏(𝑖)) ,𝑚1 = ([𝑏(𝑖)]2 − 2𝑎(𝑖)𝑐(𝑖) − 𝑒−2𝑇𝜆𝐷(𝑖)) ,𝑚0 = [𝑐(𝑖)]2 − 𝐸(𝑖)𝑒−2𝜆𝑇.

(71)

Let us determine 𝛿(𝑇∗𝑖 ) = sgn((𝑑Re(𝑥)/𝑑𝑇)|𝑥=𝑖𝜔(𝑇∗𝑖))

the sign of the real part of a solution 𝑥 of the characteristicequation Δ 𝑖(𝑥, 𝑇) = 0.

Lemma 10. Let us consider 𝑥, a solution of the characteristicequation Δ 𝑖(𝑥, 𝑇) = 𝑃𝑖(𝑥, 𝑇) + 𝑒−𝑥𝑇𝑄𝑖(𝑥, 𝑇) = 0, with

𝑃𝑖 (𝑥, 𝑇) = − [𝑥3 + 𝑎(𝑖)𝑥2 + 𝑏(𝑖)𝑥 + 𝑐(𝑖)]and 𝑄𝑖 (𝑥, 𝑇) = 𝑒−𝜆𝑇 {𝐷(𝑖)𝑥 + 𝐸(𝑖)} . (72)

Given 𝜔𝑖 = 𝜔(𝑇∗𝑖 ) positive root of 𝐹𝑖(𝜔) and 𝑇∗𝑖 the associateddelay certifying the following relation:

𝑇∗𝑖 = 1𝜔𝑖⋅ arctan{ (𝑎(𝑖)𝐷(𝑖) − 𝐸(𝑖)) 𝜔3 + (𝑏(𝑖)𝐸(𝑖) − 𝑐(𝑖)𝐷(𝑖)) 𝜔−𝐷(𝑖)𝜔4 + (−𝑎(𝑖)𝐸(𝑖) + 𝑏(𝑖)𝐷(𝑖)) 𝜔2 + 𝑐(𝑖)𝐸(𝑖)}+ 2𝑛𝜋𝜔𝑖 , 𝑛 ∈ N.

(73)

Let us pose 𝛿(𝑇∗𝑖 ) = sgn{(𝑑Re(𝑥)/𝑑𝑇)|𝑥=𝑖𝜔(𝑇∗𝑖)}.Then,

𝛿 (𝑇∗𝑖 ) = sgn {Λ 1𝑖𝐻3𝑖 + Λ 2𝑖𝐻1𝑖𝐻2𝑖} (74)with

𝐻1𝑖 = 𝜆2 + 𝜔2,𝐻2𝑖 = (𝜆𝐸(𝑖) − 𝐷(𝑖)𝜔2)2 + (𝐸(𝑖) + 𝜆𝐷(𝑖))2 𝜔2,𝐻3𝑖 = (𝑐(𝑖) − 𝑎(𝑖)𝜔2)2 + (𝑏(𝑖) − 𝜔)2 𝜔2,Λ 1𝑖 = (−𝑇∗𝑖 [𝐷(𝑖)]2 − 𝜆2 [𝐷(𝑖)]2)𝜔4 + {𝜆𝐷(𝑖)𝐸(𝑖)

− 𝑇∗𝑖 [(𝐸(𝑖) − 𝜆𝐷(𝑖))2 − 2𝜆𝐷(𝑖)𝐸(𝑖)] − 𝜆2 [𝐷(𝑖)]2}⋅ 𝜔2 + (𝜆3𝐷(𝑖)𝐸(𝑖) − 𝜆2 [𝐸(𝑖)]2 𝑇∗𝑖 ) ,

Λ 2𝑖 = −3𝑎(𝑖)𝜔4 + 2𝜔3 + (3𝑐(𝑖) + 𝑎(𝑖)𝑏(𝑖) − 2𝑏(𝑖)) 𝜔2− 𝑏(𝑖)𝑐(𝑖).

(75)

Proof. Given𝜔𝑖 = 𝜔(𝑇∗𝑖 ) positive root of 𝐹𝑖(𝜔). Let us assumethat there is a delay 𝑇∗𝑖 so that𝑇∗𝑖 = 1𝜔 (𝑇∗𝑖 ) arctan{−

Im [𝑃𝑖 (𝑖𝜔, 𝑇∗𝑖 ) /𝑄𝑖 (𝑖𝜔, 𝑇∗𝑖 )]Re [𝑃𝑖 (𝑖𝜔, 𝑇∗𝑖 ) /𝑄𝑖 (𝑖𝜔, 𝑇∗𝑖 )] }

+ 2𝑛𝜋𝜔 (𝑇∗𝑖 ) , 𝑛 ∈ N.(76)

Then, 𝑇∗𝑖 = 1𝜔𝑖⋅ arctan{ (𝑎(𝑖)𝐷(𝑖) − 𝐸(𝑖)) 𝜔3 + (𝑏(𝑖)𝐸(𝑖) − 𝑐(𝑖)𝐷(𝑖)) 𝜔−𝐷(𝑖)𝜔4 + (−𝑎(𝑖)𝐸(𝑖) + 𝑏(𝑖)𝐷(𝑖)) 𝜔2 + 𝑐(𝑖)𝐸(𝑖)}+ 2𝑛𝜋𝜔𝑖 , 𝑛 ∈ N.

(77)


Let us pose

𝛿 (𝑇∗𝑖 ) = sgn( 𝑑Re (𝑥)𝑑𝑇𝑥=𝑖𝜔(𝑇∗

𝑖)

)

= [sgn( 𝑑Re (𝑥)𝑑𝑇𝑥=𝑖𝜔(𝑇∗

𝑖)

)]−1 .(78)

One has

Δ 𝑖 (𝑥, 𝑇) = 𝑃𝑖 (𝑥, 𝑇) + 𝑒−𝑥𝑇𝑄𝑖 (𝑥, 𝑇) = 0 ⇒𝑑𝑃𝑖 (𝑥, 𝑇)𝑑𝑇 − [𝑇𝑑 (𝑥)𝑑𝑇 + 𝑥] 𝑒−𝑥𝑇𝑄𝑖 (𝑥, 𝑇)

+ 𝑒−𝑥𝑇𝑑𝑄𝑖 (𝑥, 𝑇)𝑑𝑇 = 0.(79)

Therefore,𝑃𝑖(𝑥, 𝑇) = −[𝑥3+𝑎(𝑖)𝑥2+𝑏(𝑖)𝑥+𝑐(𝑖)] and𝑄𝑖(𝑥, 𝑇) =𝑒−𝜆𝑇{𝐷(𝑖)𝑥 + 𝐸(𝑖)}. Thus, {𝑒−𝑥𝑇[𝐷(𝑖)𝑒−𝜆𝑇 − 𝑇𝑄𝑖(𝑥, 𝑇)] − (3𝑥2 +2𝑎(𝑖)𝑥 + 𝑏(𝑖))}(𝑑(𝑥)/𝑑𝑇) − 𝑒−𝑥𝑇{𝑥𝑄𝑖(𝑥, 𝑇) + 𝜆𝑒−𝜆𝑇(𝐷(𝑖)𝑥 +𝐸(𝑖))} = 0.Hence, [𝑑 (𝑥)𝑑𝑇 ]

−1

= {𝑒−𝑥𝑇 [𝐷(𝑖)𝑒−𝜆𝑇 − 𝑇𝑄𝑖 (𝑥, 𝑇)] − (3𝑥2 + 2𝑎(𝑖)𝑥 + 𝑏(𝑖))}𝑒−𝑥𝑇 {𝑥𝑄𝑖 (𝑥, 𝑇) + 𝜆𝑒−𝜆𝑇 (𝐷(𝑖)𝑥 + 𝐸(𝑖))} .Thus, [𝑑 (𝑥)𝑑𝑇 ]

−1

= − 𝑇𝑥 + 𝜆 + 𝐷(𝑖)

(𝑥 + 𝜆) (𝐷(𝑖)𝑥 + 𝐸(𝑖))− (3𝑥2 + 2𝑎(𝑖)𝑥 + 𝑏(𝑖))𝑥3 + 𝑎(𝑖)𝑥2 + 𝑏(𝑖)𝑥 + 𝑐(𝑖)

(80)

and, meanwhile, 𝛿(𝑇∗𝑖 ) = [sgn((𝑑Re(𝑥)/𝑑𝑇)|𝑥=𝑖𝜔(𝑇∗𝑖))]−1.

Then, 𝛿(𝑇∗𝑖 ) = sgn{Λ 1𝑖𝐻3𝑖 + Λ 2𝑖𝐻1𝑖𝐻2𝑖}with 𝐻1𝑖 = 𝜆2 + 𝜔2, 𝐻2𝑖 = (𝜆𝐸(𝑖) − 𝐷(𝑖)𝜔2)2

+ (𝐸(𝑖) + 𝜆𝐷(𝑖))2𝜔2, 𝐻3𝑖 = (𝑐(𝑖) − 𝑎(𝑖)𝜔2)2 + (𝑏(𝑖) −𝜔)2𝜔2, Λ 1𝑖 = (−𝑇∗𝑖 [𝐷(𝑖)]2 − 𝜆2[𝐷(𝑖)]2)𝜔4 + {𝜆𝐷(𝑖)𝐸(𝑖) −𝑇∗𝑖 [(𝐸(𝑖) − 𝜆𝐷(𝑖))2, −2𝜆𝐷(𝑖)𝐸(𝑖)] − 𝜆2[𝐷(𝑖)]2}𝜔2 + (𝜆3𝐷(𝑖)𝐸(𝑖) −𝜆2[𝐸(𝑖)]2𝑇∗𝑖 ), and Λ 2𝑖 = −3𝑎(𝑖)𝜔4 + 2𝜔3 + (3𝑐(𝑖) + 𝑎(𝑖)𝑏(𝑖) −2𝑏(𝑖))𝜔2 − 𝑏(𝑖)𝑐(𝑖).It is noticed that the analysis of the local stability of

the system with delay when 𝑄𝑖(𝑥, 𝑇) ̸= 0 depends on theexistence of positive root for function 𝐹𝑖(𝜔). The coefficients𝑚0 and 𝑚1, of 𝐹𝑖(𝜔), depend on 𝑇. Therefore, its positiveroot 𝜔 depends on 𝑇. Using the Cartan method and Viet

formulas, one proves on the one hand that 𝐹𝑖(𝜔) admits atleast a positive root if 𝑚0 < 0. On the other hand, if 𝑚0 ≥ 0,then, 𝐹𝑖(𝜔) admits either of two positive roots whereas 𝐹𝑖(𝜔)does not admit any positive root.

The condition 𝑚0 < 0 is equivalent to the followingassumption:

0 ≤ 𝑇 ≤ 𝑇𝑠 = − 12𝜆 ln[𝑐(𝑖)]2𝐸(𝑖)

with 𝐸(𝑖) > 0 and [𝑐(𝑖)]2 < 𝐸(𝑖).(81)

Let us denote det[𝐽(𝑖)] = 𝐴(𝑖)11𝐴(𝑖)22 −𝐴(𝑖)21𝐴(𝑖)12, 𝑝 = −𝑚1𝑚3 +𝑚22/3, and 𝑞 = −𝑚0𝑚23 − 2(𝑚2/3)3 + 𝑚1𝑚2𝑚3/3. Then, oneobtains the following proposition.

Proposition 11. Let us suppose that 𝐴(𝑖)13𝐴(𝑖)22 > 0.Let us pose 𝜆(𝑖)𝑚𝑎𝑥 = 𝐴(𝑖)13𝐴(𝑖)22/[det[𝐽(𝑖)] + 𝐴(𝑖)13𝐴(𝑖)22]2 and𝑇(𝑖)𝑠 = −(1/2𝜆) ln([𝑐(𝑖)]2/𝐸(𝑖)), 𝑖 = 2, 3.(1) 𝐹𝑖(𝜔) admits at least a positive root if

0 ≤ 𝜆 < 𝜆(𝑖)𝑚𝑎𝑥and 0 ≤ 𝑇 < 𝑇(𝑖)𝑠 .

(82)

(2) 𝐹𝑖(𝜔) does not admit any positive root if one of thefollowing conditions is verified:

𝑇(𝑖)𝑠 ≤ 𝑇and 27𝑞2 − 4𝑝3 > 0. (83)

𝑇(𝑖)𝑠 ≤ 𝑇,27𝑞2 − 4𝑝3 = 0

and [𝑚1𝑚2 − 𝑚0] > 0.(84)

𝑇(𝑖)𝑠 ≤ 𝑇,27𝑞2 − 4𝑝3 < 0,

𝑚2 < 0,𝑚1 < 0

and [𝑚1𝑚2 − 𝑚0] > 0.

(85)

Now, let us use the results ofTheorem 4 from [5] to studythe stability of these points of equilibriums 𝐸∗2 and 𝐸∗3 .Theorem 12. One poses

𝑥− = 𝛽1𝑑2 + 𝛼1 + 𝛾𝑑2/𝑑21 − √(𝛽1𝑑2 + 𝛼1 + 𝛾𝑑2/𝑑21)2 − 4𝛽1𝑑2 (𝛾𝑑2/𝑑21 − 1)2𝛽1𝑑2 . (86)


Let us consider one of these assumptions (82)–(85) and one ofthe following conditions:

𝛾𝑑2𝑑21 < 1, (87)𝛾𝑑2𝑑21 ≥ 1

and max (0; 𝑥−) ≤ 𝛿𝛼2 < 1,(88)

𝛾𝑑2𝑑21 ≥ 1,0 ≤ 𝛿𝛼2 < min (1; 𝑥−)

and 𝑇 ̸= − 1𝜆 ln(1 − 𝑑21𝜓 (𝛿)𝛾𝛼2 (𝛼2 − 𝛿)) ,

(89)

det (𝐽) = 𝐴(3)11𝐴(3)22 − 𝐴(3)21𝐴(3)12 > 0and 𝑇𝑟 (𝐽) = 𝐴(3)11 + 𝐴(3)22 < 0.

(90)

With 𝜓(𝛿) defined in formula (23), consider the following.(1) Let us suppose that one of assumptions (83)–(85) is

verified for i=2,3. Then, there is no change of stabilityfor equilibrium 𝐸∗𝑖 , 𝑖 = 2, 3.

(2) Let us suppose that assumptions (82) and one of theseassumptions (87)–(89) are verified for i=2.

(a) If 𝛿(𝑇∗2 ) > 0 then the equilibrium 𝐸∗2 is unstablefor any 𝑇 ∈ [0; 𝑇𝑠[.

(b) If 𝛿(𝑇∗2 ) < 0 then the equilibrium 𝐸∗2 is unstableif 0 ≤ 𝑇 < 𝑇∗2 and stable if 𝑇∗2 ≤ 𝑇.

(3) Let us suppose that assumption (82) is verified for i=3.

(a) In case 𝛿(𝑇∗3 ) > 0,(i) If assumption (90) is verified then the equi-

librium 𝐸∗3 is stable if 0 ≤ 𝑇 < 𝑇∗3 andunstable if 𝑇∗3 ≤ 𝑇.

(ii) If assumption (90) is not verified then theequilibrium 𝐸∗3 is unstable for any 𝑇 ∈[0; 𝑇𝑠[.

(b) Considering 𝛿(𝑇∗3 ) < 0,(i) If assumption (90) is verified then the equi-

librium 𝐸∗3 is stable for every 𝑇 ∈ [0; 𝑇𝑠[.(ii) If assumption (90) is not verified then the

equilibrium 𝐸∗3 is unstable if 0 ≤ 𝑇 < 𝑇∗3and stable if 𝑇∗3 ≤ 𝑇.

With 𝑇∗𝑖 and 𝛿(𝑇∗𝑖 ), 𝑖 = 2, 3 are the denotations defined inLemma 10.

Proof. Let us consider formula (63)–(67). Let us considerthe function 𝐹𝑖(𝜔) defined in relation to the formula (70)-(71) for any point of equilibrium 𝐸∗𝑖 , 𝑖 = 2, 3. Let us

consider assumptions (82). Then, 𝑇(𝑖)𝑠 exists so that, forany 𝑇 ≤ 𝑇(𝑖)𝑠 , 𝐹𝑖(𝜔) admits 𝜔𝑖 = 𝜔(𝑇∗𝑖 ) a positive root(see Proposition 11). Considering 𝑇∗𝑖 the associated delaydetermined from Lemma 10, consider the following:

(1) Let us suppose that one of assumptions (83)–(85) isverified in i=2,3. Then, ∀𝑇 > 𝑇(𝑖)𝑠 , 𝐹𝑖(𝜔) does notadmit any positive root. Thus, there is no change ofstability of the equilibrium 𝐸∗𝑖 , 𝑖 = 2, 3.

(2) Stability of𝐸∗2 : let us consider that one of assumptions(87)-(88) is verified.Then, 𝑐(2) < 0. If assumption (89)is verified, then, 𝑇 ̸= 𝑇𝑙 = −(1/𝜆) ln(𝑐(2)/𝐸). Hence,𝑃2(0, 𝑇)+𝑄2(0, 𝑇) = 𝐸(2)𝑒−𝜆𝑇−𝑐(2) ̸= 0 if assumptions(87)-(88) are verified. Then, let us apply Theorem 4from [5]. So, let us suppose that 0 ≤ 𝑇 < 𝑇(2)𝑠 . It isknown that 𝑇 = 0 and 𝐸∗2 = 𝑈(1)2 so 𝐸∗2 is unstablebecause 𝜓(𝛿) > 0 for 𝛿 < 𝛼2. Taking into account theconclusions of Theorem 4 from [5], one obtains thestability of 𝐸∗2 .

(3) Let us suppose that the assumptiondet(𝐽) = 𝐴(3)11𝐴(3)22−𝐴(3)21𝐴(3)12 > 0 of (90) is verified. Then, 𝑐 > 𝐵 so 𝑇 ≥0 > 𝑇𝑙 = −(1/𝜆) ln(𝑐/𝐵).Thus, let us applyTheorem 4from [5]. It is known that when 𝑇 = 0 and 𝐸(3)3 = 𝑈(1)3then 𝐸(3)3 is stable if assumption (90) is verified andunstable if not. Taking into account the conclusionsof Theorem 4 from [5], one gets the stability of 𝐸∗3 .

7. Global Stability of Model (1.3)Theorem13. Let us suppose thatmodel (1.3) is permanent andthat it admits a unique interior equilibrium point. If

𝛿 < 𝛼2,𝛽2𝑑2 − 𝛼2 < 0,𝑑2 < 𝑑1,𝛾 < 𝛼2,𝑇 < ln 2𝜆 ,12 < 𝜆,

(91)

1 + 𝛼1𝑑22 +𝛼2𝑀V𝑑22 +

4𝑑2𝛼21 (1 + 𝛼1𝑑1 + 𝜆) +

𝛽2𝛿𝛼2 + ln 2− 1 < 0,

(92)

𝛽1 𝛿𝛼2 +𝛾𝑀𝑢 + 𝑑1 <

𝛼2𝑀𝑢 + 𝑑2 , (93)then the unique interior equilibrium point 𝐸∗3 =(𝑢∗; V∗, 𝑘𝑢∗) of model (1.3) is globally and asymptotically

stable.


Proof. Let us consider model (1.3) and 𝐸∗3 = (𝑢∗; V∗, 𝑘𝑢∗) aunique interior equilibrium point. Let us pose 𝑈 = ln(𝑢/𝑢∗),𝑉 = ln(V/V∗), and𝑊 = ln(𝑤/𝑤∗).

Then, 𝑢 − 𝑢∗ = 𝑢∗[𝑒𝑈 − 1], V − V∗ = V∗[𝑒𝑉 − 1], and𝑤 − 𝑤∗ = 𝑤∗[𝑒𝑊 − 1].Thus,�̇� = 𝑢∗ [−1 + 𝛽1V2 + 𝛼1V(𝑢∗ + 𝑑2) (𝑢 + 𝑑2)

+ 𝛾𝑘V(𝑘𝑢∗ + 𝑑1) (𝑤 + 𝑑1) 𝑢] {𝑒𝑈 − 1} + V∗ [𝛽1− (𝛽1 (V + V∗) + 𝛼1)(𝑢∗ + 𝑑2) −

𝛾𝑘𝑘𝑢∗ + 𝑑1 ] {𝑒𝑉 − 1}+ 𝑤∗ [ −𝑑1(𝑘𝑢∗ + 𝑑

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