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Research article

A comprehensive study of a feedback control problem with a state-dependent implicit pantograph equation of Chandrasekhar type

  • Received: 18 October 2024 Revised: 31 December 2024 Accepted: 07 January 2025 Published: 16 January 2025
  • MSC : 34A12, 34C08, 34K43, 47H09, 47H10

  • In this research, we investigate the existence of at least one continuous solution of a problem with feedback control involving implicit pantograph equations of the Chandrasekhar type with state-dependent delay. In addition, we examine the possibility of the uniqueness of the solution under suitable assumptions. Furthermore, we analyze the problem's Hyers-Ulam stability and the continuous dependency of the unique solution on the original data and the parameter. Moreover, we look into this problem in the absence of feedback control. We provided a few instances to indicate our findings.

    Citation: Ahmed M. A. El-Sayed, Eman M. Al-Barg, Hanaa R. Ebead. A comprehensive study of a feedback control problem with a state-dependent implicit pantograph equation of Chandrasekhar type[J]. AIMS Mathematics, 2025, 10(1): 951-971. doi: 10.3934/math.2025045

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  • In this research, we investigate the existence of at least one continuous solution of a problem with feedback control involving implicit pantograph equations of the Chandrasekhar type with state-dependent delay. In addition, we examine the possibility of the uniqueness of the solution under suitable assumptions. Furthermore, we analyze the problem's Hyers-Ulam stability and the continuous dependency of the unique solution on the original data and the parameter. Moreover, we look into this problem in the absence of feedback control. We provided a few instances to indicate our findings.



    The Chandrasekhar integral equation arises in the context of radiative transfer theory; this type of equation is essential when examining the propagation of radiation through a medium that exhibits scattering [14]. The study of Chandrasekhar's integral equation is crucial in a variety of areas and encountered in several applications, including mathematical and computational methods, astrological applications, radiative transfer theory, and stellar atmospheres (see [5,11,12,48]). Several researchers have focused their attention on this type of equation as seen in [13,15,26,28,29] and the references therein.

    Feedback control system is a fundamental concept in engineering and automation that controls a system's behavior by modifying inputs based on output [1]. A controller is a biological system that regulates the operation of other biological processes. Integrative feedback control is essential for regulation, sensory adaptation, and long-term effects. Control variables are vital in dealing with unanticipated occurrences that disrupt real-world ecosystems, potentially altering biological features. These difficulties are converted into mathematical models [17,39,49]. Several authors investigated feedback problems. In [18], the authors established a necessary condition for a positive periodic solution of a feedback control model for chemostats. A positive periodic solution with feedback control involving a nonlinear neutral delay population problem has been analyzed in [36]. The author of [37] studied the asymptotic stability and solvability of nonlinear functional-integral equations with feedback control.

    Stability analysis is an extensive and varied field with deep theoretical roots and numerous applications in engineering, economics, biology, physics, and other disciplines. It is a prevalent topic in the mathematical sciences [6]. An equation or problem can be used to simulate a physical process if a minor change in it causes a corresponding small change in the outcome. Alsina and Ger [4] initially examined the Hyers-Ulam stability of differential equations. Several papers have been devoted to studying the Hyers-Ulam stability of differential and integral equations, for example, (see [2,3,42,43,44]). Another concept in stability theory is continuous dependency [40], which analyzes the behavior of mathematical solutions under different conditions. Hyers-Ulam stability assesses the problem's resilience to disruptions, while continuous dependency examines how even minor parameter changes affect the problem's unique solution. The study of the continuous dependence of the solution has been addressed through many research works [22,27].

    Pantograph differential equations are an effective tool for modeling systems with feedback interactions. From signal processing to finance, control systems, and neural networks, these equations aid in the capture of complicated dynamics in which the present state-depends on past states at scaled time intervals. Solving pantograph differential equations analytically is difficult because of their nonlocal nature; often, numerical methods such as finite difference schemes are used to obtain approximate solutions. Much research has been conducted on pantograph equations due to their significance in various research areas. For instance, in [38], Patade et al. investigated the analytical solution of the pantograph equation with two delays; they analyzed the existence, uniqueness, and stability of the solution. In [23], the authors studied the solvability and the Hyers-Ulam stability of non-local fractional orders of the pantograph equation with a feedback control.

    State-dependent (self-reference) differential and integral equations are a special new type of functional differential equations in which the deviation of the argument depends on the time and the state together. In most differential and integral equations with deviating arguments that are found in literature, the deviation of the argument involves only the time, although another important case in theory and practice involves deviating arguments that depend on both the state variable x and the time t (see for instance [9,10,20,34]). This kind of delay is widely utilized in nonlinear analysis and has a wide range of applications, including mechanical models [31], population models [7], infectious disease transmission [41], the two-body problem of classical electrodynamics [19], the dynamics of economical systems [8], and it has many applications in hereditary phenomena [35,45,46]. One of the first researches in self-reference differential equations was presented by Eder [20], he classifies the solutions to the differential equation x(t)=x(x(t)),tAR, and demonstrated the existence and uniqueness of the solution with the condition x(t0)=x0. Fe'ckan [25] introduced a generalization of Eder's results by studying the functional differential equation of the form x(t)=f(x(x(t))),tAR, where fC1(R). Buicá [10], examined the uniqueness of solution and data dependence of the problem x(t)=f(t,x(x(t))),t[a,b],x(t0)=x0, where t0,x0[a,b] and fC([a,b],[a,b]). Lin and Lu [34] studied the qualitative behavior of a state-dependent functional differential equation. Yang et al. [47] examined an ordinary differential equation with a state-dependent delay. EL-Sayed et al. [24] investigated a nonlocal boundary value problem of a state-dependent differential equation. In [22], the authors analyzed the state-dependent Chandrasekhar integral equation

    x(t)=b(t)+λx(t0tt+sa(s)x(s)ds,),t[0,1],

    they proved the existence of the solution for this equation and analyzed its continuous dependency on function a. In [30], Humphries et al. examined the state-dependent delay system

    x(t)=αx(t)+βx(taηx(tb)),

    where the delay a+ηx(tb) depends linearly on the state x with strength η. Lauran in [33], analyzed iterative and non iterative first order differential equations of the form

    dx(t)dt=f(t,x(t),x(λt))

    and

    dx(t)dt=f(t,x(t),x(x(t))),

    respectively, with the initial condition x(t0)=x0. A generalization of the result in [33] was introduced by Hashem et al. [27], they studied the system of the state dependent functional equations

    x(t)=f1(t,y(t),x(y(t))),t[0,b],
    y(t)=f2(t,x(t),y(x(t))),t[0,b],

    the authors proved the existence of a unique solution of the system and discussed the continuous dependence of the solution. In [21], the authors examined the existence of a unique solution of a feedback control problem with an implicit state-dependent pantograph equation; in addition, they investigated the Hyres-Ulam stability of the problem and the continuous dependence of the solution.

    Inspired by modern literature, we consider the state-dependent implicit pantograph equation of the Chandrasekhar type

    dxdt=b1(t)+λ1x(10tt+sg1(s,dyds)ds),x(0)=x0,a.e.t(0,1] (1.1)

    with the feedback control

    dydt=b2(t)+λ2y(10tt+sg2(s,dxds)ds),y(0)=y0,a.e.t(0,1]. (1.2)

    Where λi(0,1) and x0,y0R are the initial data.

    Our goal in this work is to investigate the existence and the uniqueness of the solution (x,y)X of the problems (1.1) and (1.2). We prove the continuous dependence of the solution on the initial data x0,y0 and the parameters λi. Furthermore, we establish the Hyres-Ulam stability of the problem. Next, as a particular case of our work, we discuss an issue of the same type without feedback control.

    We outline the main contributions of this paper as follows:

    ● We examine the feedback control problems (1.1) and (1.2) of the state-dependent pantograph equation of the Chandrasekhar type; we study the qualitative properties of the solution of (1.1) under the feedback control (1.2).

    ● We explore the problem (5.1) of the state-dependent pantograph equation of the Chandrasekhar type; we refer to the qualitative properties of the solution of (5.1) in the absence of the feedback control.

    This study enhances the qualitative analysis of a state-dependent pantograph differential equation with feedback control. The article is structured as follows: Section 2 presents the appropriate assumptions and proves the existence results for the solution of the implicit state-dependent problem (1.1) with the feedback control (1.2) using Schauder fixed point theorem. The suitable assumptions and proofs for the uniqueness of the solution will be provided in Section 3. Section 4 investigates the stability analysis of the problem due to the Hyers-Ulam stability. Additionally, we test the possibility of the solution resisting disturbances through the study of the continuous dependency on x0,y0 and the parameters λi. Furthermore, in Section 5, we examine a special case of our problem without the control variable; we present some results supporting the problem's existence and stability. In Section 6, we give some instances that illustrate the findings. Finally, Section 7 provides a conclusion.

    Let C(I) be the class of all continuous functions on I=[0,1] with the standard norm uC=suptI|u(t)|, and X=C(I)×C(I) be the Banach space with the norm (u,v)X=max{uC,vC}. Let L1(I) be the space of real functions defined and Lebesgue integrable on the interval I, which is equipped with the standard norm.

    Consider the problems (1.1) and (1.2) under the assumptions:

    (1) gi:I×RR satisfies Carathˊeodory condition [16], i.e., it is measurable in tI xR and continuous in xR, tI, and there exist functions ai:IR+=[0,),aiL1(I) and ai(t)Kitn,Ki(0,1) where nN, tI, such that

    |gi(t,x(t))|ai(t)|x(t)|,i=1,2.

    (2) bi:IR are continuous functions on I.

    (3) There exists a real positive root r of the algebraic equation

    Kλr2r+(B+λA)=0,

    where Kλr<1, K=max{Ki}, λ=max{λi}, B=max{bi}, A=max{x0,y0}.

    Let dxdt=u(t),dydt=v(t), then

    x(t)=x0+t0u(s)ds, (2.1)
    y(t)=y0+t0v(s)ds. (2.2)

    Then the problems (1.1) and (1.2) will be given by

    u(t)=b1(t)+λ1x(10tt+sg1(s,v(s))ds) (2.3)
    v(t)=b2(t)+λ2y(10tt+sg2(s,u(s))ds). (2.4)

    Where

    x(10tt+sg1(s,v(s))ds)=x0+10tt+sg1(s,v(s))ds0u(s)dsx0+u10tt+sg1(s,v(s))ds

    and

    y(10tt+sg2(s,u(s))ds)=y0+10tt+sg2(s,u(s))ds0v(s)dsy0+v10tt+sg2(s,u(s))ds.

    Define the operator F associated with (2.3) and (2.4) by

    F(u,v)=(F1u,F2v),

    where

    F1u(t)=b1(t)+λ1x(10tt+sg1(s,v(s))ds),
    F2v(t)=b2(t)+λ2y(10tt+sg2(s,u(s))ds).

    Theorem 2.1. Let the assumptions (1)–(3) be satisfied. Then the problems (2.3) and (2.4) has at least one solution.

    Proof. Define set QrX as

    Qr={(u,v)X:ur,vr}.

    Obviously, Qr is a closed convex bounded set. Now for (u,v)Qr,tI, we have

    |F1u(t)|=|b1(t)+λ1x(10tt+sg1(s,v(s))ds)||b1(t)|+λ1(|x0|+u10tt+s|g1(s,v(s))|ds)b1+λ1|x0|+λ1u10tt+s|g1(s,v(s))|dsb1+λ1|x0|+λ1uK110tt+ssn|v(s)|dsb1+λ1|x0|+λ1uK110|v(s)|dsb1+λ1|x0|+λ1uK1vb1+λ1|x0|+λ1K1r2.

    Similarly,

    |F2u(t)||b2(t)|+λ2(|y0|+v10tt+sg2(s,u(s))ds)b2+λ2|y0|+λ2v10tt+s|g2(s,u(s))|dsb2+λ2|y0|+λ2vK210tt+ssn|u(s)|dsb2+λ2|y0|+λ2vK210|u(s)|dsb2+λ2|y0|+λ2vK2ub2+λ2|y0|+λ2K2r2.

    Then

    F(u,v)X=(F1u,F2v)X=max{F1uC,F2vC}max{b1+λ1|x0|+λ1K1r2,b2+λ2|y0|+λ2K2r2}B+λA+λKr2=r.

    Then the class of functions {F(u,v)} is uniformly bounded on I. Let (u,v)Qr and t1,t2I with t1<t2 such that |t2t1|<δ, then

    |F1u(t2)F1u(t1)|=|b1(t2)+λ1x(10t2t2+sg1(s,v(s))ds)b1(t1)λ2x(10t1t1+sg1(s,v(s))ds)||b1(t2)b1(t1)|+λ1(x0+u10t2t2+sg1(s,v(s))ds)λ1(x0+u10t1t1+sg1(s,v(s))ds)|b1(t2)b1(t1)|+λ1u10t2t2+s|g1(s,v(s))|dsλ1u10t1t1+s|g1(s,v(s))|ds|b1(t2)b1(t1)|+λ1u10(t2t2+st1t1+s)|g1(s,v(s))|ds|b1(t2)b1(t1)|+λ1u10(t2t1)s(t2+s)(t1+s)|g1(s,v(s))|ds|b1(t2)b1(t1)|+λ1u|t2t1|101(t2+s)|g1(s,v(s))|ds|b1(t2)b1(t1)|+λ1u|t2t1|10K1snt2+s|v(s)|ds|b1(t2)b1(t1)|+λ1u|t2t1|K1v10st2+ssn1ds|b1(t2)b1(t1)|+λKr2|t2t1|.

    Then the class of functions {F1u} is equi-continuous.

    Similarly,

    |F2v(t2)F2v(t1)|=|b2(t2)+λ2y(10t2t2+sg2(s,u(s))ds)b2(t1)λ2y(10t1t1+sg2(s,u(s))ds)||b2(t2)b2(t1)|+λ2(y0+v10t2t2+s|g2(s,u(s))|ds)λ2(y0+v10t1t1+s|g2(s,u(s))|ds)|b2(t2)b2(t1)|+λ2v10t2t2+s|g2(s,u(s))|dsλ2v10t1t1+s|g2(s,u(s))|ds|b2(t2)b2(t1)|+λ2v10(t2t2+st1t1+s)|g2(s,u(s))|ds|b2(t2)b2(t1)|+λ2v10(t2t1)s(t2+s)(t1+s)|g2(s,u(s))|ds|b2(t2)b2(t1)|+λ2v|t2t1|101(t2+s)|g2(s,u(s))|ds|b2(t2)b2(t1)|+λ2v|t2t1|10K2snt2+s|u(s)|ds|b2(t2)b2(t1)|+λ2v|t2t1|K2u10st2+ssn1ds|b2(t2)b2(t1)|+λKr2|t2t1|.

    Then the class of functions {F2u} is equi-continuous. We deduce that F:QrQr and the class functions {F(u,v)} is equi-continuous. By Arzela-Theorem [32], {F(u,v)} is compact, then F is compact.

    Now, let {un},{vn}Qr such that un(t)u(t), vn(t)v(t), where n, then

    F1un(t)=b1(t)+λ1x(10tt+sg1(s,vn(s))ds),
    F2vn(t)=b2(t)+λ2y(10tt+sg2(s,un(s))ds)

    and

    limnF1un(t)=b1(t)+λ1limnx(10tt+sg1(s,vn(s))ds),
    limnF2vn(t)=b2(t)+λ2limny(10tt+sg2(s,un(s))ds).

    Then, from assumption (1), we have

    g1(s,vn(s))g1(s,v(s)),g2(s,un(s))g2(s,u(s))

    and

    |g1(s,v(s))|a1(t)|v(t)|L1[0,1],|g2(s,u(s))|a2(t)|u(t)|L1[0,1].

    Applying Lebesgue dominated convergence theorem [32], we have

    limnF1un(t)=b1(t)+λ1limnx(10tt+sg1(s,vn(s)))ds=b1(t)+λ1x(10tt+slimng1(s,vn(s)))ds)=b1(t)+λ1x(10tt+sg1(s,v(s)))ds)=F1u(t).

    Similarly,

    limnF2vn(t)=b2(t)+λ2limny(10tt+sg2(s,un(s)))ds=b2(t)+λ2y(10tt+slimng2(s,un(s)))ds)=b2(t)+λ2y(10tt+sg2(s,u(s)))ds)=F2v(t).

    Now,

    limnF(un,vn)=limn(F1un,F2vn)=(F1u,F2u)=F(u,v).

    Then F(u,v) is continuous. Now all conditions of Schauder's fixed point theorem [32] are satisfied, then the operator F has at least one fixed point (u,v)Qr, Consequently, there exist at least one solution of the problems (2.3) and (2.4).

    Corollary 2.1. Let the assumptions of Theorem 2.1 be satisfied, by using Eqs (2.1) and (2.2), we deduce that the problems (1.1) and (1.2) has at least one solution xC(I).

    Consider the following additional assumptions:

    (1) gi:I×RR satisfies the Lipschitz condition with positive Lipschitz constants Mi such that

    |gi(t,x)gi(t,y)|Mi|xy|,tI,x,yQrmax{Mi}=M,i=1,2.

    Theorem 3.1. Let the assumptions (1)–(3) and (1) be satisfied, If 0<λrM(1λrK)<1, then the solution of the problems (1.1) and (1.2) is unique.

    Proof. let (u1,v1),(u2,v2) be two solutions of (2.3) and (2.4), then

    |u1(t)u2(t)|=|λ1x(10tt+sg1(s,v1(s))ds)λ1x(10tt+sg1(s,v2(s))ds)||λ1u110tt+sg1(s,v1(s))dsλ1u210tt+sg1(s,v2(s))ds||λ1u110tt+sg1(s,v1(s))dsλ1u110tt+sg1(s,v2(s))ds+λ1u110tt+sg1(s,v2(s))dsλ1u210tt+sg1(s,v2(s))ds||λ1u110tt+sg1(s,v1(s))dsλ1u110tt+sg1(s,v2(s))ds|+|λ1u110tt+sg1(s,v2(s))dsλ1u210tt+sg1(s,v2(s))ds|λ1u110tt+s|g1(s,v1(s))g1(s,v2(s))|ds+λ1|u1u2|10tt+s|g1(s,v2(s))|dsλ1rM110tt+s|v1(s)v2(s)|ds+λ1u1u210tt+s|g1(s,v2(s))|dsλ1rM1v1v2+λ1u1u210tt+sa1(s)|v2(s)|dsλ1rM1v1v2+λ1rK1u1u210tt+ssndsλrMv1v2+λrKu1u2,

    and

    (1λrK)u1u2λrMv1v2,

    then

    u1u2λrM1λrKv1v2.

    Similarly,

    v1v2λrM1λrKu1u2.

    Then

    (u1,v1)(u2,v2)=(u1u2),(v1v2)=max{u1u2,v1v2}max{λrM1λrKv1v2,λrM1λrKu1u2}λrM1λrKmax{v1v2,u1u2}λrM1λrK(u1u2),(v1v2)λrM1λrK(u1,v1)(u2,v2).

    Then

    (1λrM1λrK)(u1,v1)(u2,v2)0.

    Since 0<λrM1λrK<1, then the solution of the problems (2.3) and (2.4) is unique.

    Corollary 3.1. Let the assumptions of Theorem 3.1 be satisfied; according to the Eqs (2.1) and (2.2), the solution (x,y)X of (1.1) and (1.2) is unique.

    Definition 4.1. Let the solution (x,y)X of (1.1) and (1.2) be exists, then the problems (1.1) and (1.2) is Hyers-Ulam stable if ϵ>0, there exists δ(ϵ)>0 such that for any δ-approximate solution (xs,ys)X of (1.1) and (1.2) satisfies

    max{|dxsdtb1(t)λ1xs(10tt+sg1(s,dysdt)ds))|,|dysdtb2(t)λ2ys(10tt+sg2(s,dxsdt)ds))|}<δ,

    implies

    (x,y)(xs,ys)Xϵ.

    Theorem 4.1. If the assumptions of Theorem 3.1 are met, then the problems (1.1) and (1.2) is Hyers-Ulam stable.

    Proof. Let

    max{|dxsdtb1(t)λ1xs(10tt+sg1(s,dysdt)ds))|,|dysdtb2(t)λ2ys(10tt+sg2(s,dysdt)ds))|}<δ,

    then

    |dxsdtb1(t)λ1xs(10tt+sg1(s,dysdt)ds))|<δ
    δ<dxsdtb1(t)λ1xs(10tt+sg1(s,dysdt)ds))<δ.

    Similarly,

    |dysdtb2(t)λ2ys(10tt+sg2(s,dxsdt)ds))|<δ
    δ<dysdtb2(t)λ2ys(10tt+sg2(s,dxsdt)ds))<δ.

    Let dxsdt=us and dysdt=vs, then

    xs(t)=x0+t0us(s)dsxs(10tt+sg1(s,vs(s))ds)x0+us10tt+sg1(s,vs(s))ds,
    ys(t)=y0+t0us(s)dsys(10tt+sg2(s,us(s))ds)y0+vs10tt+sg1(s,us(s))ds.

    Hence

    δ<us(t)b1(t)λ1xs(10tt+sg1(s,vs(s))ds)<δ,
    δ<vs(t)b2(t)λ2ys(10tt+sg2(s,us(s))ds)<δ.

    Then,

    |u(t)us(t)|=|b1(t)+λ1x(10tt+sg1(s,v(s))ds)us(t)|=|b1(t)+λ1x(10tt+sg1(s,v(s))ds)λ1xs(10tt+sg1(s,vs(s))ds)+λ1xs(10tt+sg1(s,vs(s))ds)us(t)||λ1x(10tt+sg1(s,v(s))ds)λ1xs(10tt+sg1(s,vs(s))ds)|+|b1(t)+λ1xs(10tt+sg1(s,vs(s))ds)us(t)||λ1(x0+u10tt+sg1(s,v(s))ds)λ1(x0+us10tt+sg1(s,vs(s))ds)|+δλ1|u10tt+sg1(s,v(s))dsus10tt+sg1(s,vs(s))ds|+δλ1|u10tt+sg1(s,v(s))dsu10tt+sg1(s,vs(s))ds+u10tt+sg1(s,vs(s))dsus10tt+sg1(s,vs(s))ds|+δλ1u10tt+s|g1(s,v(s))g1(s,vs(s))|ds+λ1|uus|10tt+s|g1(s,vs(s))|ds+δλ1uM110tt+s|v(s)vs(s)|ds+λ1uus10tt+sa1(s)|vs(s)|ds+δλ1rM1vvs+λ1rK1uus10tt+ssnds+δλ1rM1vvs+λ1rK1uus+δ,

    and

    (1λ1rK1)uusδ+λ1rM1vvs.

    Hence

    uusδ(1λ1rK1)+λ1rM1(1λ1rK1)vvs.

    Similarly,

    vvsδ(1λ2rK2)+λ2rM2(1λ2rK2)uus.

    Then

    (u,v)(us,vs)X=((uus),(vvs))X=max{(uus)C,(vvs)C}max{δ(1λ1rK1)+λ1rM1(1λ1rK1)vvs,δ(1λ2rK2)+λ2rM2(1λ2rK2)uus}δ(1λrk)+max{λ1rM1(1λ1rK1)vvs,λ2rM2(1λ2rK2)uus}δ(1λrK)+λrM(1λrK)max{vvs,uus}δ(1λrK)+λrM(1λrK)((uus),(vvs))Xδ(1λrK)+λrM(1λrK)(u,v)(us,vs)X,

    and

    (1λrM(1λrK))(u,v)(us,vs)δ(1λrK),

    then

    \|(u, v)-(u_s, v_s)\|\leq \frac{\delta}{1-(\lambda rM+\lambda rK)} = \epsilon.

    Now,

    \begin{eqnarray*} \|(x, y)-(x_s, y_s)\|_X& = &\|((x-x_s), (y-y_s))\|_X = \max\{\|(x-x_s)\|_C, \; \|(y-y_s)\|_C\\ &\leq &\max\{\|(u-u_s)\|_C, \|(v-v_s)\|_C\}\leq \|((u-u_s), (v-v_s))\|\\ &\leq& \|(u, v)-(u_s, v_s)\|\leq \epsilon. \end{eqnarray*}

    Then

    \|(x, y)-(x_s, y_s)\|_X \leq \epsilon.

    Then the problems (1.1) and (1.2) is Hyers-Ulam stable.

    Definition 4.2. The solution (u, v)\in Q_r of (2.3) and (2.4) depends continuously on x_0, \; y_0, \; \lambda if \forall\; \epsilon > 0 there exists \; \; \delta(\epsilon) > 0 such that

    \max \{|x_0-x_0^{*}|, \; |y_0-y_0^{*}|, \; |\lambda_i-\lambda_i^*|\} < \delta\; \Rightarrow\; \|(u, v)-(u^*, v^*)\|_X < \epsilon, \; i = 1, 2.

    where

    \begin{equation} u^*(t) = b_1(t)+\lambda_1^{*}x^*(\int_{0}^{1}\frac{t}{t+s}g_1(s, v^*(s))ds), \end{equation} (4.1)
    \begin{equation} v^*(t) = b_2(t)+\lambda_2^{*}y^*(\int_{0}^{1}\frac{t}{t+s}g_2(s, u^*(s))ds). \end{equation} (4.2)

    Theorem 4.2. Let the assumptions of Theorem 3.1 be satisfied, then (u, v) depends continuously on the initial data x_0, \; y_0 and the parameters \lambda_i, \; i = 1, 2.

    Proof. Let \delta(\epsilon) > 0 be given such that

    \max \{|x_0-x_0^{*}|, \; |y_0-y_0^{*}|, \; |\lambda_i-\lambda_i^{*}|\} < \delta, \; i = 1, 2.

    and let (u^*, v^*) be the solution of (4.1) and (4.2), then

    \begin{eqnarray*} & &|u(t)-u^*(t)|\\ & = & |\lambda_1 x(\int_{0}^{1}\frac{t}{t+s}g_1(s, v(s))ds)-\lambda_1^{*} x^*(\int_{0}^{1}\frac{t}{t+s}g_1(s, v^*(s))ds)| \\ &\leq& |\lambda_1(x_0+\|u\|\int_{0}^{1}\frac{t}{t+s}g_1(s, v(s))ds)-\lambda_1^{*}(x_0^{*}+\|u^*\|\int_{0}^{1}\frac{t}{t+s}g_1(s, v^*(s))ds)| \\ &\leq& |\lambda_1x_0-\lambda_1^{*}x_0^{*}|+\Big|\lambda_1\|u\|\int_{0}^{1}\frac{t}{t+s}g_1(s, v(s))ds-\lambda_1^{*}\|u^*\|\int_{0}^{1}\frac{t}{t+s}g_1(s, v^*(s))ds\Big| \\ &\leq&|\lambda_1 x_0-\lambda_1 x_0^{*}|+|\lambda_1 x_0^{*}-\lambda_1^{*} x_0^{*}|+\Big|\lambda_1\|u\|\int_{0}^{1}\frac{t}{t+s}g_1(s, v(s))ds-\lambda_1\|u\|\int_{0}^{1}\frac{t}{t+s}g_1(s, v^*(s))ds\\ &+&\lambda_1\|u\|\int_{0}^{1}\frac{t}{t+s}g_1(s, v^*(s))ds-\lambda_1^{*}\|u^*\|\int_{0}^{1}\frac{t}{t+s}g_1(s, v^*(s))ds\Big| \\ &\leq&\lambda_1 |x_0-x_0^{*}|+|x_0^{*}|\; |\lambda_1 -\lambda_1^{*}|+\lambda_1\|u\|\int_{0}^{1}\frac{t}{t+s}|g_1(s, v(s))-g_1(s, v^*(s))|ds\\ &+&\Big|\lambda_1\|u\|-\lambda_1^{*}\|u^*\|\Big|\int_{0}^{1}\frac{t}{t+s}|g_1(s, v^*(s))|ds \\ &\leq&\lambda_1 \delta+|x_0^{*}|\delta+\lambda_1\|u\|M_1\int_{0}^{1}\frac{t}{t+s}|v(s)-v^*(s)|ds\\ &+&\Big|\lambda_1\|u\|-\lambda_1\|u^*\|+\lambda_1\|u^*\|-\lambda_1^{*}\|u^*\|\Big|\int_{0}^{1}\frac{t}{t+s}|g_1(s, v^*(s))|ds \\ &\leq&\lambda_1 \delta+|x_0^{*}|\; \delta+\lambda_1\|u\|M_1\|v-v^*\|+\Big(\lambda_1\Big|\|u\|-\|u^*\|\Big|+\|u^*\|\Big|\lambda_1-\lambda_1^{*}\Big|\Big)\int_{0}^{1}\frac{t}{t+s}a_1(s)\; |v^*(s)|ds \\ &\leq&\lambda_1 \delta+|x_0^{*}|\; \delta+\lambda_1r\; M_1\|v-v^*\|+(\lambda_1\|u-u^*\|+\|u^*\|\; \delta)\int_{0}^{1}\frac{t}{t+s}a_1(s)\; |v^*(s)|ds \\ &\leq&\lambda_1 \delta+|x_0^{*}|\; \delta+\lambda_1r\; M_1\|v-v^*\|+(\lambda_1\|u-u^*\|+\|u^*\|\; \delta)r\; K_1\int_{0}^{1}\frac{t}{t+s}s^nds \\ &\leq&(\lambda_1+|x_0^{*}|+r^2K)\; \delta+\lambda_1 r\; M_1\|v-v^*\|+\lambda_1 r\; K_1\|u-u^*\|, \end{eqnarray*}

    and

    (1-\lambda_1 rK_1)\|u-u^*\|\leq (\lambda_1+|x_0^{*}|+r^2K_1)\; \delta+\lambda_1 r\; M_1\|v-v^*\|.

    Then

    \|u-u^*\|\leq \frac{\lambda_1+|x_0^{*}|+r^2K_1}{1-\lambda_1 rK_1}\delta+\frac{\lambda_1 r\; M_1}{1-\lambda_1 rK_1}\|v-v^*\|.

    Similarly,

    \|v-v^*\|\leq \frac{\lambda_2+|y_0^{*}|+r^2K_2}{1-\lambda_2 rK_2}\delta+\frac{\lambda_2 r\; M_2}{1-\lambda_2 rK_2}\|u-u^*\|.

    Then

    \begin{eqnarray*} \|(u, v)-(u^*, v^*)\|_X& = &\|((u-u^*), (v-v^*))\|_X = \max\{\|(u-u^*)\|_C, \; \|(v-v^*)\|_C\}\\ &\leq& \max\{\frac{\lambda_1+|x_0^{*}|+r^2K_1}{1-\lambda_1 r\; K_1}\delta+\frac{\lambda_1 r\; M_1}{1-\lambda_1 r\; K_1}\|v-v^*\|, \\ & & \frac{\lambda_2+|y_0^{*}|+r^2K_2}{1-\lambda_2 r\; K_2}\delta+\frac{\lambda_2 r\; M_2}{1-\lambda_2 r\; K_2}\|u-u^*\|\\ &\leq& \frac{\lambda+A+r^2K}{1-\lambda rK}\delta+\max\{\frac{\lambda_1 r\; M_1}{1-\lambda_1 r\; K_1}\|v-v^*\|, \frac{\lambda_2 r\; M_2}{1-\lambda_2 r\; K_2}\|u-u^*\|\}\\ &\leq& \frac{\lambda+A+r^2K}{1-\lambda rK}\delta+\frac{\lambda r\; M}{1-\lambda r\; K}\max\{\|v-v^*\|, \|u-u^*\|\}\\ & = & \frac{\lambda+A+r^2K}{1-\lambda rK}\delta+\frac{\lambda r\; M}{1-\lambda r\; K}\|((u-u^*), (v-v^*))\|_X\\ & = & \frac{\lambda+A+r^2K}{1-\lambda rK}\delta+\frac{\lambda r\; M}{1-\lambda r\; K}\|(u, v)-(u^*, v^*)\|_X, \end{eqnarray*}

    and

    (1-\frac{\lambda rM}{1-\lambda rK})\; \|(u, v)-(u^*, v^*)\|_X \leq \frac{\lambda+A+r^2K}{1-\lambda r\; K}\delta.

    Then

    \|(u, v)-(u^*, v^*)\|_X\leq \frac{\lambda+A+r^2K}{1-(\lambda r\; M+\lambda r\; K)}\delta = \epsilon.

    Thus

    \delta = \frac{1-(\lambda r\; M+\lambda r\; K)} {\lambda+A+r^2K}\epsilon.

    Which meas that \delta = \delta(\epsilon). Moreover, we have \lambda r\; M+\lambda r\; K < 1 and A, r, k, \lambda > 0, then \delta is positive. This means that the solution of (2.3) and (2.4) depends continuously on the initial data x_0, \; y_0 and the parameters \lambda_i, \; i = 1, 2.

    Definition 4.3. The solution (x, y) \in X of (1.1) and (1.2) depends continuously on u, \; v if \forall\; \epsilon > 0 there exists \delta(\epsilon) > 0 such that

    \max\{|u-u^{*}|, \; |v-v^*|\} < \delta(\epsilon)\; \Rightarrow\; \|(x, y)-(x^*, y^*)\| < \epsilon,

    where

    \begin{equation} x^*(t) = x_0+\int_{0}^{t}u^*(s)ds, \end{equation} (4.3)
    \begin{equation} y^*(t) = y_0+\int_{0}^{t}v^*(s)ds. \end{equation} (4.4)

    Theorem 4.3. Let the assumptions of Theorem 4.2 be satisfied, then the solution (x, y)\in X depends continuously on u, \; v .

    Proof. Let (x^*, y^*) be the solution of (4.3) and (4.4), then

    \begin{eqnarray*} |x(t)-x^*(t)| &\leq&|\int_{0}^{t}u(s)ds-\int_{0}^{t}u^*(s)ds|\\ &\leq&\int_{0}^{t}|u(s)-u^*(s)|ds \leq \|u-u^*\| \leq \delta = \epsilon, \end{eqnarray*}

    then

    \|x-x^*\|\leq \epsilon.

    Similarly,

    \begin{eqnarray*} |y(t)-y^*(t)| &\leq&|\int_{0}^{t}v(s)ds-\int_{0}^{t}v^*(s)ds|\\ &\leq&\int_{0}^{t}|v(s)-v^*(s)|ds \leq \|v-v^*\| \leq \delta = \epsilon, \end{eqnarray*}

    then

    \|y-y^*\|\leq \epsilon.

    Now,

    \begin{eqnarray*} \|(x, y)-(x^*, y^*)\|_X& = &\|((x-x^*), (y-y^*))\|_X = \max\{\|x-x^*\|_C, \; \|y-y^*\|_C\} \leq \epsilon. \end{eqnarray*}

    Then

    \|(x, y)-(x^*, y^*)\|_X \leq \epsilon.

    This means that the solution (x, y)\in X of (1.1) and (1.2) depends continuously on u, \; v .

    Corollary 4.1. Let the assumptions of Theorem 4.3 be satisfied, then the solution (x, y)\in X of (1.1) and (1.2) depends continuously on x_0, y_0 and the parameter \lambda_i .

    In the lack of feedback control; as a special case of our work, we can investigate the following problem of the state-dependent implicit pantograph differential equation of the chandrasekhar type.

    \begin{equation} \frac{dx}{dt} = b(t)+\lambda x(\int_{0}^{1}\frac{t}{t+s}\; g(s, \frac{dx}{dt})\; ds), \; \; x(0) = x_0, \; \; a.e. \; t\in(0, 1], \end{equation} (5.1)

    where \lambda\in(0, 1) , b : I\rightarrow R are continuous function and the function g : I\times R \rightarrow R satisfies Caratheodory condition.

    The existence of the unique solution of (5.1) will be studied. We will also prove the continuous dependence of the solution on the initial data x_0 and the parameter \lambda . Furthermore, we will establish the Hyres-Ulam stability. This problem can be addressed under the following assumptions:

    (i) g:I\times R \rightarrow R satisfies Carath \grave{e} odory condition[16]. (i.e., it is measurable in t \in I \forall x \in R and continuous in x \in R \forall t \in I ) and there exists a function a:I \rightarrow R^+ = [0, \infty), \; a \in L^1(I) and a(t)\leq Kt^n, \; K\in(0, 1), \; n \in N \forall \; t \in I , such that

    |g(t, x(t))|\leq a(t)\; |x(t)|.

    (ii) There exists a real positive root r of the algebraic equation

    K \lambda r^2-r+(\|b\|+\lambda|x_0|) = 0.

    such that K \lambda r < 1.

    We can formulate the problem as follows: Put \frac{dx}{dt} = y(t) , we get

    \begin{equation} y(t) = b(t)+\lambda x(\int_{0}^{1}\frac{t}{t+s}g(s, y(s))ds). \end{equation} (5.2)

    Using the same techniques, we can derive the following theorems:

    Theorem 5.1. Let the assumptions (i) (ii) be satisfied, then integral equation (5.2) has at least one solution y \in C(I) . Consequently, the problem (5.1) has at least one solution x \in C(I) .

    Theorem 5.2. Let the assumptions (i) (ii) of Theorem 5.1 be satisfied, if g satisfies lipschetz condition such that |g(t, x)-g(t, y)|\leq c|x-y| \forall t\in I and x, \; y \in Q_r where c is a positive constant. If \lambda r(c+K) < 1, then the solution of the problem (5.1) is unique.

    Theorem 5.3. Let the assumptions of Theorem (5.2) be satisfied, then the problem (5.1) is Hyers-Ulam stable.

    Theorem 5.4. Let the assumptions of Theorem 5.2 be satisfied, then y\in C(I) depends continuously on the initial data x_0 and the parameter \lambda .

    Theorem 5.5. Let the assumptions of Theorem 5.2 be satisfied, then the solution x\in C(I) depends continuously on y .

    Corollary 5.1. Let the assumptions of Theorems 5.4 and 5.5 be satisfied, then the solution x\in C(I) depends continuously on the initial data x_0 and the parameter \lambda .

    Example 1. Consider the problem

    \begin{equation} \frac{dx}{dt} = \frac{1}{3}(t^2+1)+\frac{1}{2}x(\int_{0}^{1}\frac{t}{t+s}\; \frac{\sin s^3}{5}\; \frac{dy}{dt}ds), \; \; x(0) = \frac{1}{7}, \; \; t\in(0, 1]. \end{equation} (6.1)
    \begin{equation} \frac{dy}{dt} = \frac{5}{2}\sin t+\frac{1}{3}x(\int_{0}^{1}\frac{t}{t+s} \frac{e^{-s}s^2}{7}\; \frac{dx}{dt}ds), \; \; y(0) = \frac{1}{10}, \; \; t\in(0, 1]. \end{equation} (6.2)

    Let \frac{dx}{dt} = u, \; \; \frac{dy}{dt} = v then

    \begin{equation*} u(t) = \frac{1}{3}(t^2+1)+\frac{1}{2}x(\int_{0}^{1}\frac{t}{t+s}\; \frac{\sin s^3}{5}\; v(s)ds), \end{equation*}
    \begin{equation*} v(t) = \frac{5}{2}\sin t+\frac{1}{3}x(\int_{0}^{1}\frac{t}{t+s} \frac{e^{-s}s^2}{7}\; u(s)ds). \end{equation*}

    Set

    b_1(t) = \frac{1}{3}(t^2+1), \; \; b_2(t) = \frac{5}{2}\sin t

    and

    g_1(t, v(t)) = \frac{\sin t^3}{5}\; v(t), \; \; g_2(t, u(t)) = \frac{e^{-t}t^2}{7}\; u(t),

    then

    |g_1(t, v(t))|\leq\frac{1}{5}t^3\; v(t), \; \; \; |g_2(t, u(t))|\leq\frac{1}{7}t^2\; u(t)

    and

    a_1(t)\leq \frac{1}{5}t^3, \; \; \; \; \; a_2(t)\leq \frac{1}{7}t^2.

    Where K = \max\{\frac{1}{5}, \frac{1}{7}\} = \frac{1}{5}, \lambda = \max\{\frac{1}{2}, \; \frac{1}{3}\} = \frac{1}{2} , A = \max\{\frac{1}{7}, \; \frac{1}{10}\} = \frac{1}{7} , B = \max\{0.6667, 0.043\} = 0.6667, \; M = \max\{\frac{1}{5}, \frac{1}{7}\} = \frac{1}{5} . Then we get r_1 = 0.8027, \; r_2 = 9.1973, such that \lambda r_1 K = 0.0803 < 1 and \lambda r_2 K = 0.9197 < 1. It is clear that all assumptions of Corollary 1 are satisfied. Hence there exist at least one solution (x, y) \in X of (6.1) and (6.2). Moreover, we have only r_1 satisfies \lambda r_1(M+K) = 0.1605 < 1. Thus all assumptions of Corollary 2 are satisfied, then the solution of problems (6.1) and (6.2) is unique.

    Example 2. Consider the problem

    \begin{equation} \frac{dx}{dt} = \frac{t^2}{2(4-t)}+\frac{1}{3}x(\int_{0}^{1}\frac{t}{t+s}(\frac{t^2}{4}\sin t\; \frac{dx}{dt})ds), \; \; x(0) = \frac{1}{2}, \; \; t\in(0, 1]. \end{equation} (6.3)

    Let \frac{dx}{dt} = y then

    \begin{equation*} y(t) = \frac{t^2}{2(4-t)}+\frac{1}{3}x(\int_{0}^{1}\frac{t}{t+s}(\frac{t^2}{4}\sin t\; y(s))ds), \; \; x(0) = \frac{1}{2}, \; \; t\in(0, 1]. \end{equation*}

    Set

    b(t) = \frac{t^2}{2(4-t)}

    and

    g(t, y(t)) = \frac{t^2}{4}\sin t\; y(t),

    thus

    |g(t, y(t))|\leq \frac{t^2}{4}\; y(t)

    and

    a(t)\leq \frac{1}{4}\; t^2.

    Where K = \frac{1}{4}, \lambda = \frac{1}{3} , x_0 = \frac{1}{2} , \|b\| = \frac{1}{6}, \; c = \frac{1}{4} . Then we get r_1 = 0.343, \; r_2 = 11.657 , such that \lambda r_1 K = 0.0286 < 1 and \lambda r_2 K = 0.9714 < 1.

    It is clear that all assumptions of Theorem 5.1 are satisfied. Hence there exist at least one solution x \in C[0, 1] of (6.3). Moreover, we have only r_1 satisfies \lambda r_1(c+K) = 0.0572 < 1.

    Thus all assumptions of Theorem 5.2 are satisfied, then the solution of Problem 6.3 is unique.

    Differential equations with control variables are frequently encountered within several domains such as control theory, optimization, dynamic systems, etc...The existence and uniqueness of solutions of this type of equations is crucial. Sometimes the equations may not have a unique solution, or solutions may need particular conditions to exist. Determining whether a solution is stable or not is often challenging. Stability analysis is critical for control systems to guarantee that small perturbations do not cause significant variations in system behavior. Research on feedback control problems with state-dependent delays has numerous applications in biology, ecology, physics, engineering, and other fields. This type of delay imparts memory effects into the system, which implies intricate and rich dynamics. In this paper, we investigate the existence of at least one continuous solution to a feedback control problem including implicit pantograph equations of the Chandrasekhar type. In addition, we analyzed the uniqueness of the solution in light of appropriate assumptions. Also, we investigated the problem's Hyers-Ulam stability and the continuous dependence of the solution on the original data and parameter. Furthermore, we introduce a brief study for the problem in the absence of feedback control. Finally, we presented few examples in both cases when the problem contains a control variable and in the absence of a control variable.

    The authors contributed equally to this paper.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors thank the referees for their useful suggestions and comments, which helped strengthen our paper.

    The authors declare no conflicts of interest.



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