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Research article

A comprehensive view of the solvability of non-local fractional orders pantograph equation with a fractal-fractional feedback control

  • Received: 09 April 2024 Revised: 28 May 2024 Accepted: 30 May 2024 Published: 11 June 2024
  • MSC : 74H10, 45G10

  • In this article, the solvability of the pantograph equation of fractional orders under a fractal-fractional feedback control was investigated. This investigation was located in the class of all continuous functions. The necessary conditions for the solvability of that problem and the continuous dependence of the solution on some parameters and the control variable were established with the help of some fixed point theorems. Additionally, the Hyers-Ulam stability of the issue was explored. Finally, some specific problems extended to the corresponding problem with integer orders were illustrated. The theoretical results were supported by numerical simulations and comparisons with existing results in the literature.

    Citation: A. M. A. El-Sayed, H. H. G. Hashem, Sh. M. Al-Issa. A comprehensive view of the solvability of non-local fractional orders pantograph equation with a fractal-fractional feedback control[J]. AIMS Mathematics, 2024, 9(7): 19276-19298. doi: 10.3934/math.2024939

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  • In this article, the solvability of the pantograph equation of fractional orders under a fractal-fractional feedback control was investigated. This investigation was located in the class of all continuous functions. The necessary conditions for the solvability of that problem and the continuous dependence of the solution on some parameters and the control variable were established with the help of some fixed point theorems. Additionally, the Hyers-Ulam stability of the issue was explored. Finally, some specific problems extended to the corresponding problem with integer orders were illustrated. The theoretical results were supported by numerical simulations and comparisons with existing results in the literature.



    Fractional-order differential and integral equations represent wide applications in physics, engineering, and biomedical engineering. Nonlocal situations often arise in mathematical and physical problems. In which the system's behavior is influenced by multiple elements or parameters. Nonlocal integral conditions are common in mathematical analysis when working with differential equations, sometimes involving constraints or objectives. Fixed point theorems are valuable tools for examining the solvability of differential equation problems extensively covered in various monographs and publications (see [1,2,3,4] and the references therein).

    Delay systems have been extensively used to describe the evolution of propagation and transportation or population movements [6,7]. In economic systems, decisions like investment strategies and the dynamics of commodity markets are spread out across time intervals, leading to the natural occurrence of delays. In the 19th century, Euler, Lagrange, and Laplace delved into the realm of delayed differential equations. Subsequently, in the late 1930s and early 1940s, Voltaire introduced various delayed differential equations while researching the predator-prey model, becoming the first to systematically investigate these equations.

    The pantograph differential equation is a type of delay differential equation initially developed through the study of an electric locomotive[8]. Mahler introduced pantograph equations in 1940 as part of his Number Theory [9]. The term "pantograph" originated from the work of Ockendon and Taylor, who examined the electric locomotive's catenary system. Their objective was to formulate an equation for analyzing the movement of a pantograph head on an electric locomotive that operates using a trolley overhead wire [8]. Kato and Mcleod [10] investigated the asymptotic properties and stability of solutions to the pantograph equation u(t)=au(kt)+bu(t).

    A pantograph (or "pan" or "panto") is a device installed on the roof of an electric train, tram, or electric bus to gather power by making contact with an overhead line [11]. An important application of the pantograph appears in engineering, particularly in designing machines and mechanisms that require precise scaling and copying of movements (see [12,13,14]). For instance, a pantograph can be utilized to move in a specific manner. Additionally, pantographs find relevance in electrodynamics [9] and number theory [15]. In 1971, Ockendon and Taylor [8] researched how electric current is collected by the pantograph of an electric locomotive using a delay equation, now known as the pantograph equation. Since then, numerous researchers have explored and applied it across various mathematical and scientific domains such as number theory, probability, electrodynamics, and medicine, as seen in [8,16,17] and the references therein.

    Much research has been conducted on fractional pantograph equations due to their significance in various research areas. For instance, in [18], Balachandran and Kiruthika examined the existence of solutions. Additionally, in [19], Jalilian and Ghasemi studied a fractional integro-differential equation of pantograph type along with suitable initial conditions.

    Let C(I)=C[0,1] be the class of continuous functions defined on I=[0,1], and the norm of xC(I) is defined by xC=suptI|x(t)|.

    Inspired by contemporary literature, we consider the nonlocal issue of the pantograph equation via Caputo fractional-order derivative Dς,ς{α,β,ρ},

    dxdt=f(t,ux(t),x(t),Dαx(γt)),t(0,1],γ(0,1), (1.1)

    satisfying

    x(τ)=x0+1τ0h(s,x(s),Dβx(γs))ds,τ(0,1] (1.2)

    equipped with the fractal-fractional feedback control

    dux(t)dtδ=λux(t)+g(t,x(t),Dρx(γt)),u0=ux(0),λ0,δ(0,1], (1.3)

    where α,β,ρ(0,1], τ is a fixed parameter and ddtδ denotes the fractal derivative of order δ (for more information on fractal derivatives, refer to [20,21]).

    In this study, we explore the presence of solutions xC(0,1] to the problem (1.1)–(1.3). The necessary conditions for the uniqueness of the solution will be provided. The continuous dependence of the unique solution xC(0,1] of (1.1) and (1.2) on y(t)=dx(t)dt, the function h, and the parameter x0 will be demonstrated. The Hyers-Ulam stability of the problem (1.1)–(1.3) will be examined. The feedback control problem of the linear version of the pantograph equation (Ambartsumian)

    dxdt=aux(t)+bx(t)+cDαx(γt),t(0,1], (1.4)
    x(τ)=x0+1τ0h(s,x(s),Dβx(γs))ds,τ(0,1] (1.5)

    equipped with the fractal-fractional feedback control (1.3) will be addressed. The continuation of α,β1 will be established. We outline the key contributions of this article as follows:

    ● We examine the feedback control problem of the fractional pantograph differential equation with arbitrary fractional orders (1.1) and (1.2) with the fractal-fractional feedback control (1.3).

    ● We explore the feedback control problem of the linear version of the pantograph (Ambartsumian) equations (1.4) and (1.5) with the fractal-fractional feedback control (1.3).

    This paper enhances the qualitative analysis of the feedback control fractional pantograph differential equation problem. The paper's structure is as follows: Section 2 covers key features, and demonstrates the existence-uniqueness of the solution, and Section 3 explores Ulam Hyers stability (UHRS) and the continuous dependence on some data. Furthermore, Section 4 presents a special case and an example. Finally, Section 5 provides a conclusion.

    The problem (1.1)–(1.3) will be investigated under the assumptions:

    (ⅰ) The function f:I×R3R is measurable in tI for all ui,vi,wiR,i=1,2, and satisfies the Lipschitz condition with a positive Lipschitz constant b

    |f(t,u1,v1,w1)f(t,u2,v2,w1)|b(|u1u2|+|v1v2|+|w1w2|).

    Observation 1. Based on assumption (i), we have

    |f(t,u1,v1,w1)||f(t,0,0,0)||f(t,u1,v1,w1)f(t,0,0,0)|b(|u1|+|v1|+|w1|).

    This implies that

    |f(t,u1,v1,w1)|a+b(|u1|+|v1|+|w1|),wherea=suptI|f(t,0,0,0)|.

    (ⅱ) h:I×R2R is measurable in tI for every u,vR and continuous in u,vR for every tI. There exist a function a2L1(I) and a positive constant b2 such that

    |h(t,u,v)|a2(t)+b2(|u|+|v|),suptIt0|a2(s)|dsM.

    (ⅲ) g:I×R2R, is continuous function for every u,vR, such that

    |g(t,u,v)|k[|u|+|v|]+m,wherem=suptI|g(t,0,0)|.

    (ⅳ) r1 and r2 represent positive solutions of the two simultaneous equations:

    |x0|+M+b2(r2+γr1)+2r1=r1,
    a+b1r2+b1(w+w1+w2r2+w3+w4)+2γr1b1=r2.

    Additionally, the condition b1(1+w2)<1 is satisfied.

    Let X be the Banach space consisting of all pairs (x,y) where x and y are functions in the space C(I). The norm of an element (x,y) in X is defined as the sum of the sup norms of x and y in C(I) respectively.

    The following lemma shows that the feedback control problem described in Eqs (1.1) and (1.2) equipped with a fractal feedback control (1.3) is equivalent to its respective integral equations.

    Lemma 2.1. If the solution of (1.1)–(1.3) exists, it can be expressed by the following coupled system

    x(t)=x0+1τ0h(s,x(s),γI1βy(γs))dsτ0y(s)ds+t0y(s)ds, (2.1)
    y(t)=f(t,ux(s),x(t),γI1αy(γt)), (2.2)

    along with the integral feedback equation

    ux(t)=u0eλtδ+δt0eλ(tδsδ)sδ1g(s,x(s),γI1ρy(γs))ds,u0=ux(0). (2.3)

    Proof. Let x represent the solution of the problem (1.1)–(1.3). Take y(t)=ddtx(t), then

    x(γt)=x(0)+γt0y(s)ds, (2.4)
    ddtx(γt)=γy(γt). (2.5)

    Applying the Riemann-Liouville fractional integral operators I1α, I1β and I1ρ to both sides of (2.5), and we get

    Dαx(γt)=I1αdx(γt)dt=γI1αy(γt), (2.6)
    Dβx(γt)=I1βdx(γt)dt=γI1βy(γt), (2.7)

    and

    Dρx(γt)=I1ρdx(γt)dt=γI1ρy(γt).

    Using the substitutions of Dαx(γt),Dβx(γt), and Dρx(γt) in the problem (1.1)-(1.2), we obtain the representation (2.1)-(2.2).

    On the contrary, assume x is a solution of (2.1) and differentiate both sides of (2.1). We get

    dxdt=y(t)=f(t,ux(t),x(t),Dαx(γt)),t(0,1]

    Put t=τ in Eq (2.1), then we can deduce

    x(τ)=x0+1τ0h(s,x(s),Dβx(γs))ds,τ(0,1].

    This demonstrates the equivalence between the problem (1.1)-(1.2) and the problem (2.1)-(2.2).

    Now, the fractal-fractional feedback control (1.3), can be expressed as

    dux(t)dtdtdtδ=λux(t)+g(t,x(t),Dρx(γt)),

    then

    1δtδ1dux(t)dt=λux(t)+g(t,x(t),Dρx(γt)),

    therefore,

    dux(t)dt=λδtδ1ux(t)+δtδ1g(t,x(t),Dρx(γt)).

    Multiply both terms by eλtδ

    eλtδdux(t)dt+eλtδλδtδ1ux(t)=eλtδδtδ1g(t,x(t),Dρx(γt)),

    and

    ddt(ux(t)eλtδ)=eλtδδtδ1g(t,x(t),Dρx(γt)).

    Integrate with respect to t, then

    ux(t)eλtδ=ux(0)+t0δsδ1eλsδg(s,x(s),Dρx(γs))ds.

    Hence

    ux(t)=u0eλtδ+t0δsδ1eλ(tδsδ)g(s,x(s),Dρx(γs))ds.

    Substitute for Dρx(γt), and we obtain

    ux(t)=u0eλtδ+t0δsδ1eλ(tδsδ)g(s,x(s),γI1ρy(γs))ds.

    For any real-valued function x, the solution of the fractal differential feedback control (1.3) denoted as ux(t) can be expressed as shown in (2.3).

    Lemma 2.2. The control variable ux(t) satisfies (1.3) and can be expressed by (2.3), then the solution ux(t) is bounded for u0>0.

    Proof.

    |ux(t)|u0|eλtδ|+δt0eλ(tδsδ)sδ1[k(|ux(s)|+|x(s)|+|γI1ρy(γs))|)+m]dsw+kt0eλ(tδsδ)sδ1|ux(s)|ds+kt0eλ(tδsδ)sδ1xds+kγt0eλ(tδsδ)sδ1I1ρyds+kt0eλ(tδsδ)sδ1mdsw+suptIt0eλ(tδsδ)sδ1k|ux(s)|ds+suptIt0eλ(tδsδ)sδ1kxds+kγsuptIt0eλ(tδsδ)sδ1I1ρyds+suptIt0eλ(tδsδ)sδ1mdsw+w1+w2r2+w3+w4,

    where

    suptIu0eλtδ=w,suptIt0eλ(tδsδ)sδ1k|ux(s)|ds=w1,suptIt0eλ(tδsδ)sδ1xds=w2,suptIt0eλ(tδsδ)sδ1mds=w3,suptIkγt0eλ(tδsδ)sδ1I1ρyds=w4,

    and

    |ux1(t)ux2(t)|δt0eλ(tδsδ)sδ1|g(s,ux1(s),x1(s),γI1ρy(γs))g(s,ux2(s),x2(s),γI1ρy(γs))|dskt0eλ(tδτδ)sδ1(|ux1(s)ux2(s)|+|x1(s)x2(s)|)dskeλδλ(ux1ux2+x1x2).

    Hence,

    ux1ux2Δx1x2,

    with keλδλ<1 and Δ=keλδλ(1keλδλ).

    Here, we demonstrate the existence of a continuous solution for the problem (1.1)-(1.2) equipped with a fractal differential constraint (1.3). To achieve this, we introduce the following theorem.

    Theorem 2.3. Assuming conditions (ⅰ)–(ⅳ) are met, then the constrained problem (1.1)–(1.3) has at least one solution xC(I).

    Proof. Define an operator F as F(x,y)=(F1y,F2x), where

    F1y(t)=x0+1τ0h(s,x(s),γI1βy(γs))dsτ0y(s)ds+t0y(s)ds,F2x(t)=f(t,ux(t),x(t),γI1αy(γt)).

    Define set UX as

    U={u=(x,y)X:xCr1,yCr2},

    then (x,y)X=xC+yCr where r=r1+r2, for positive values of r1 and r2 satisfying condition (iv).

    Obviously, U is a closed convex bounded set. For (x,y)U and tI, we have:

    |F1y(t)|=|x0+1τ0h(s,x(s),γI1βy(γs))dsτ0y(s)ds+t0y(s)ds||x0|+1τ0|h(s,x(s),γI1βy(γs)|ds+τ0|y(s)|ds+t0|y(s)|ds|x0|+1τ0(|a2(s)|+b2(|x(s)|+γI1β|y(s)|))ds+τ0|y(s)|ds+t0|y(s)|ds|x0|+1τ0(|a2(s)|+b2(r2+r1γΓ(2β)))ds+2r1|x0|+M+b2(r2+r1γΓ(2β))+2r1,

    and

    F1yC|x0|+M+b2(r2+γr1)+2r1=r1.

    In a similar manner,

    |F2x(t)|=|f(t,ux(t),x(t),γI1αy(γt))||a1(t)|+b1(|ux(t)|+|x(t)|+γI1α|y(γt)|)a+b1r2+b1(w+w1+w2r2+w3+w4)+γr1b1Γ(2α),

    and

    F2xa+b1r2+b1(w+w1+w2r2+w3+w4)+2γr1b1=r2.

    Therefore,

    FUX=F(x,y)X=(F1y,F2x)X=F1yC+F2xC=r1+r2=r.

    For each point u=(x,y)U, the function F(u) is also in U, showing that F maps the set U into itself. This implies that the set of functions {FU} is uniformly bounded on the interval I.

    Now, we show that the class F1y is equi-continuous. Let t1,t2I such that t2>t1 and t1t2∣≤δ, then

    |F1y(t2)F1y(t1)|=|x0+1τ0h(s,x(s),λI1βy(λs))dsτ0|y(s)|ds+t20y(s)dsx01τ0h(s,x(s),λI1βy(λs))ds+τ0|y(s)|dst10y(s)ds|t20|y(s)|dst10|y(s)|dst2t1|y(s)|ds.

    This demonstrates that the class of functions {F1y} is equi-continuous on the interval I in the space of continuous functions.

    Similarly,

    |F2x(t2)F2x(t1)|=|f(t2,ux(t2),x(t2),γt20(t2s)αΓ(1α)y(γs)ds)f(t1,ux(t1,x(t1),γt10(t1s)αΓ(1α)y(γs)ds)|=|f(t2,ux(t2),x(t2),γt20(t2s)αΓ(1α)y(γs)ds)f(t1,ux(t2),x(t2),γt20(t2s)αΓ(1α)y(γs)ds)+f(t1,ux(t2),x(t2),γt20(t2s)αΓ(1α)y(γs)ds)f(t1,ux(t2),x(t1),γt20(t2s)αΓ(1α)y(γs)ds)+f(t1,ux(t2),x(t1),γt20(t2s)αΓ(1α)y(γs)ds)f(t1,ux(t2),x(t1),γt10(t2s)αΓ(1α)y(γs)ds)+f(t1,ux(t2),x(t1),γt10(t2s)αΓ(1α)y(γs)ds)f(t1,ux(t2),x(t1),γt10(t1s)αΓ(1α)y(γs)ds)|+f(t1,ux(t2),x(t1),γt10(t1s)αΓ(1α)y(γs)ds)f(t1,ux(t1),x(t1),γt10(t1s)αΓ(1α)y(γs)ds)||f(t2,ux(t2),x(t2),γt20(t2s)αΓ(1α)y(γs)ds)f(t1,ux(t2),x(t2),γt20(t2s)αΓ(1α)y(γs)ds)|+|f(t1,ux(t2),x(t2),γt20(t2s)αΓ(1α)y(γs)ds)f(t1,ux(t2),x(t1),γt20(t2s)αΓ(1α)y(γs)ds)|+|f(t1,ux(t2),x(t1),γt20(t2s)αΓ(1α)y(γs)ds)f(t1,ux(t2),x(t1),γt10(t2s)αΓ(1α)y(γs)ds)|+|f(t1,ux(t2),x(t1),γt10(t2s)αΓ(1α)y(γs)ds)f(t1,ux(t2),x(t1),γt10(t1s)αΓ(1α)y(γs)ds)|+|f(t1,ux(t2),x(t1),γt10(t1s)αΓ(1α)y(γs)ds)f(t1,ux(t1),x(t1),γt10(t1s)αΓ(1α)y(γs)ds)|ϵ+b1|x(t2)x(t1)|+γb1t2t1(t2s)αΓ(1α)y(γt)ds+b1|ux(t2)ux(t1)|+γb1t10(t2s)αΓ(1α)(t1s)αΓ(1α)y(γt)dsϵ+b1|ux(t2)ux(t1)|+b1|x(t2)x(t1)|+γb1r2t2t11Γ(1α)(t2s)αds+γb1r2t10(t1s)α(t2s)αΓ(1α)(t2s)α(t1s)αds.

    This demonstrates that the class {F2x} is equi-continuous on the interval C(I).

    Fu(t2)Fu(t1)=F(x,y)(t2)F(x,y)(t1)=(F1y(t2),F2x(t2))(F1y(t1),F2x(t1))=(F1y(t2)F1y(t1),F2x(t2)F2x(t1)),

    which implies that

    Fu(t2)Fu(t1)F1y(t2)F1y(t1)+F2x(t2)F2x(t1).

    Then, the class of functions FU is equi-continuous on X.

    Thus, by the Arzela-Ascoli theorem [22], {FU} is relatively compact. Hence the operator F is compact.

    Now, we will show that the operator F is continuous.

    Let (xn,yn)U such that xnx,yny, then

    F1yn=x0+1τ0h(s,xn(s),λI1βyn(λs))dsτ0yn(s)ds+t0yn(s)ds,F2xn=f(t,uxn(t),xn(t),γI1αyn(γt)),limnF1yn=x0+limn1τ0h(s,xn(s),λI1βyn(λs))dslimnτ0yn(s)ds+limnt0yn(s)ds,

    and

    limnF2xn(t)=limnf(t,uxn(t),xn(t),γI1αyn(γt)).

    Since f,h are continuous in x,y, then

    f(t,uxn(t),xn(t),I1αyn(s))f(t,ux(t),x(t),γI1αy(γt)),h(s,xn(s),λI1βyn(λs))h(s,x(s),λI1βy(λs)).

    Since

    |h(t,x,y)|a2(t)+b2(|x|+|y|),

    and regarding to the Lebesgue dominated convergence theorem [22], we can derive

    limnF1yn(t)=x0+1τ0h(s,x(s),λI1βy(λs))dsτ0y(s)ds+t0y(s)ds=F1y(t).

    Then F1 is continuous. Also,

    limnF2xn(t)=f(t,ux(t),x(t),γI1αy(γt))=F2x(t),

    and F2 is continuous. Hence,

    limnF(xn,yn)=limn(F1y,F2x)=(F1y,F2x)=F(x,y).

    Therefore, the function F is continuous. As all the requirements of Schauder's fixed point theorem [22] are met, it follows that F must have at least one fixed point u=(x,y)U. Then the coupled system of integral equations

    x(t)=x0+1τ0h(s,x(s),λI1βy(λs))dsτ0y(s)ds+t0y(s)ds,y(t)=f(t,ux(t),x(t),γI1αy(γt)),

    has a solution u=(x,y)U. Consequently, the problem (1.1)-(1.2) with a fractal differential constraint (1.3) has a solution xC(I).

    Corollary 2.1. Let the assumptions of Theorem 2.3 be satisfied, then Dαx(γt), Dβx(γt)C(I).

    Proof. From Theorem 2.3, we have yC(I). By utilizing the definition and properties of the fractional operators [23], and making use of Eqs (2.6) and (2.7), we can conclude that

    Dαx(γt)=γI1αy(γt)C(I),Dβx(γt)=γI1βy(γt)C(I).

    For the uniqueness of the solution of the problem (1.1)-(1.2) with the fractal-fractional feedback control (1.3), we replace condition (ii) by the assumption:

    (ⅱ*) The function h:I×R2R is measurable for tI for all (xi,yi)R×R, where i=1,2, and h satisfies the Lipschitz condition

    |h(t,x1,y1)h(t,x2,y2)|b2(|x1x2|+|y1y2|),tI,xi,yiR

    with Lipschitz constant b2>0.

    Observation 2. From assumption (ii), we obtain:

    |h(t,x,y)||h(t,0,0)|+b2(|x|+|y|),

    and

    |h(t,x,y)|a2+b2(|x|+|y|),wherea2=suptI|h(t,0,0)|,tI.

    This demonstrates that assumption (ii) is satisfied.

    Theorem 3.1. If the conditions of Theorem 2.3 are met, assumption (ii) is replaced by (ii), and if the inequality

    b1(1+Δ)(b2γ+2)(12γb1)(1b2)<1

    holds, then the solution xC(I) of problem (1.1)-(1.2) (and, consequently, (1.4)-(1.5)) with the fractal-fractional feedback control (1.3) is unique.

    Proof. All conditions of Theorem 2.3 are met, implying that solutions to the problem (2.1)-(2.2) with feedback control (1.3) exists. Consider u1=(x1,y1) and u2=(x2,y2) as two solutions of the problem (2.1)-(2.2) with feedback control (1.3). Then, we can observe

    (x1,y1)(x2,y2)X=x1x2,y1y2X=x1x2C+y1y2C.

    Now,

    |x1x2|=|x0+1τ0h(s,x1(s),γI1βy1(γs))dsτ0y1(s)ds+t0y1(s)dsx01τ0h(s,x2(s),γI1βy2(s))ds+τ0y2(s)dst0y2(s)ds|1τ0|h(s,x1(s),γI1βy1(s))h(s,x2(s),γI1βy2(s))|ds+τ0|y1(s)y2(s)|ds+t0|y1(s)y2(s)|dsb21τ0[|x1(s)x2(s)|+γI1β|y1(γs)y2(γs)|]ds+τ0|y1(s)y2(s)|ds+t0|y1(s)y2(s)|dsb2x1x2C+b2γΓ(2β)y1y2C+2y1y2C,

    then

    x1x2C(b2γ+2)1b2y1y2C.

    Also

    |y1y2|=|f(t,ux1(t),x1(t),γI1αy1(γt))f(t,ux2(t),x2(t),γI1αy2(γt))||f(t,ux1(t),x1(t),γI1αy1(γt))f(t,ux2(t),x2(t),γI1αy1(γt))+f(t,ux1(t),x2(t),γI1αy1(γt))f(t,ux2(t),x2(t),γI1αy2(γt))|b1ux1ux2C+b1x1x2C+b1γI1α|y1(γt)y2(γt)|b1Δx1x2C+b1x1x2C+γb1Γ(2α)y1y2C,b1(1+Δ)x1x2C+2γb1y1y2C,

    then

    y1y2Cb1(1+Δ)x1x2C12γb1.

    Hence,

    x1x2Cb1(1+Δ)(b2γ+2)(12γb1)(1b2)x1x2C,(1b1(1+Δ)(b2γ+2)(12γb1)(1b2))x1x2C0,

    this gives x1=x2.

    In a similar manner,

    y1y2Cb1(1+Δ)12γb1x1x2Cb1(1+Δ)(b2γ+2)(12γb1)(1b2)y1y2C,y1y2C(1b1(1+Δ)(b2γ+2)(12γb1)(1b2))0,

    then y1=y2.

    Thus, the solution of the coupled system (2.1)-(2.2) with the fractal-fractional feedback control (1.3) is unique. Consequently, the solution xC(I) of the problem (1.1)-(1.2) with a feedback control (1.3) is also unique.

    Definition 3.1. Let the unique solution xC(I) of (1.1)-(1.2) with feedback control (1.3) exist. The problem (1.1)–(1.3) is Hyers-Ulam stable if ϵ>0, δ(ϵ)>0 such that for any approximate solution xsC(I) of (1.1)-(1.2) with a feedback control (1.3) satisfying

    |dxsdtf(t,uxs(t),xs(t),Dαxs(γt))|δ.

    Then

    xxsCϵ.

    Theorem 3.2. Assuming that the hypothesis of Theorem 3.1 is satisfied, the problem (1.1)-(1.2) with feedback control (1.3) is Hyers-Ulam stable. Consequently, (1.4)-(1.5) with feedback control (1.3) is Hyers-Ulam stable.

    Proof. Let us improve

    |dxs(t)dtf(t,uxs(t),xs(t),Dαxs(γt))|δ,

    then

    δdxs(t)dtf(t,uxs(t),xs(t),Dαxs(γt))δ,δ1ys(t)f(t,uxs(t),xs(t),γI1αys(γt))δ1,ys(t)=dxs(t)dt.

    Now,

    |y(t)ys(t)|=|f(t,,ux(t),x(t),γI1αy(γt))ys(t)f(t,uxs(t),xs(t),γI1αys(γt))+f(t,uxs(t),xs(t),γI1αys(γt))||f(t,ux(t),x(t),γI1αy(γt))f(t,uxs(t),xs(t),γI1αys(γt))|+|ys(t)f(t,uxs(t),xs(t),γI1αys(γt))|b1|ux(t)uxs(t)|+b1|x(s)xs(s)|+γb1I1α|y(γt)ys(γt)|+δb1ux1ux2C+b1xxsC+(2α)γb1Γ(3α)yysC+δb1Δx1x2C+b1xxsC+2γb1yysC+δ.

    Hence,

    \begin{eqnarray*} \|y-y_{s}\|_C(1- 2\gamma b_{1} ) \leq \delta+b_{1}(1+\Delta)\|x-x_{s}\|_C \end{eqnarray*}

    and

    \begin{eqnarray*} \|y-y_{s}\|_C\leq \frac{b_{1}(1+\Delta)\|x-x_{s}\|_C}{1-2\gamma b_{1} }+\frac{\delta}{1-2\gamma b_{1} }. \end{eqnarray*}

    Now,

    \begin{eqnarray*} |x(t) - x_{s}(t)|& = & \bigg|x_{0}+ \int_{0}^{1-\tau} h(s, x(s), \gamma I^{1-\beta}y(\gamma s))ds-\int_{0}^{\tau}y(s)ds+\int_{0}^{t}y(s)ds\\&-& x_{0}- \int_{0}^{1-\tau} h(s, x_s(s), \gamma I^{1-\beta}y_{s}(\gamma s))ds+\int_{0}^{\tau}y_{s}(s)ds-\int_{0}^{t}y_{s}(s)ds \bigg|\\ &\leq&\; b_{2}\int_{0}^{1-\tau}[ |x(s)-x_{s}(s)|+\gamma I^{1-\beta} |y(\gamma s)-y_{s}(\gamma s)|]ds\\ &+&\int_{0}^{\tau}|y( s)-y_{s}( s)|ds+\int_{0}^{t}|y(s)-y_{s}(s)|ds, \\ &\leq&\; b_{2} \|x-x_{s}\|_C+\frac{b_{2} \gamma }{\Gamma(2-\beta)}\|y-y_{s}\|_C+2\|y-y_{s}\|_C\\ &\leq&\; b_{2} \|x-x_{s}\|_C+\; b_{2} \gamma \|y-y_{s}\|_C+2\|y-y_{s}\|_C. \\ &\leq&\; b_{2} \|x-x_{s}\|_C+(2+b_{2}\gamma )\|y-y_{s}\|_C, \end{eqnarray*}

    and

    \begin{eqnarray*} \|x-x_{s}\|_C\leq \frac{(2+ b_{2})\|y-y_{s}\|_C}{1-\; b_{2} }. \end{eqnarray*}

    Substituting by \|y-y_{s}\|_C , we obtain

    \begin{eqnarray*} \|x-x_{s}\|_C&\leq& (2+ b_{2} \gamma )( \frac{b_{1}(1+ \Delta)\|x-x_{s}\|_C}{1-2\gamma b_{1} }+\frac{\delta}{1-2\gamma b_{1}} )\\ \\ &\leq& \frac{(2 +b_{2}) \delta}{1-2\gamma b_{1} }+\frac{(2 + b_{2})b_{1}(1+ \Delta) \|x-x_{s}\|_C}{1-2\gamma b_{1} }, \end{eqnarray*}
    \begin{eqnarray*} \bigg[1-\bigg(\frac{(2 +b_{2})b_{1}(1+ \Delta)}{1-2\gamma b_{1} }\bigg)\bigg] \; \|x-x_{s}\|_C\leq \frac{(2 +b_{2}) \delta}{1-2\gamma b_{1} }, \end{eqnarray*}

    and

    \begin{eqnarray*} \|x-x_{s}\|_C \leq \dfrac{\dfrac{(2 +b_{2}) \delta}{1-2\gamma b_{1} }}{1-\bigg(\dfrac{(2 +b_{2})b_{1}(1+ \Delta)}{1-2\gamma b_{1} }\bigg)}. \end{eqnarray*}

    Since

    \begin{eqnarray*} \dfrac{(2 +b_{2})b_{1}(1+ \Delta)}{1-2\gamma b_{1} } \leq 1 , \end{eqnarray*}

    then

    \begin{eqnarray*} \|x-x_{s}\|_C \leq \epsilon. \end{eqnarray*}

    Then, the problem (1.1)-(1.2) with feedback control (1.3) is Hyers-Ulam-stable.

    Definition 3.2. The unique solution x\in C(I) of (1.1)-(1.2) constrained with (1.3) depends continuously on y, \; h , and x_{0} , and if for all \epsilon > 0 , there exists \delta(\epsilon) > 0 such that

    \max\{|y -y^{*}|, \; |h -h^{*}|, \; |x_{0} -x_{0}^{*}|\leq \delta\}\; \; \Rightarrow \|x-x^{*}\|_C\leq \epsilon,

    where x^{*} and y^{*} are the solutions of the following respectively

    \begin{eqnarray} x^{*}(t) & = & x_{0}^{*}+\int_{0}^{1-\tau}h^{*}(s, x^{*}(s), \gamma I^{1-\beta}y^{*}(\gamma s))ds-\int_{0}^{\tau}y^{*}(s)ds + \int_{0}^{t} y^{*}(s)ds , \end{eqnarray} (3.1)
    \begin{eqnarray} y^{*}(t)& = & f\big(t, u_{x^{*}}(t), x^{*}(t), \gamma I^{1-\alpha}y^{*}(\gamma t)\big), \end{eqnarray} (3.2)

    and

    \begin{eqnarray} u_x^{*}(t)& = & u_0e^{-\lambda t^\delta}+\delta\int_0^t e^{-\lambda(t^\delta- s^\delta)}\; s^{\delta-1}\; g(s, x^{*}(s), \gamma I^{1-\rho} y^{*}(\gamma s))ds. \end{eqnarray} (3.3)

    Theorem 3.3. Suppose that the hypotheses of Theorem 3.1, are satisfied; then the solution x\in C(I) of (1.1)-(1.2) with feedback control (1.3) (consequently, (1.4)-(1.5) with feedback control (1.3)) depends continuously on y, \; h , and x_{0} ).

    Proof. Let x(t) and x^{*}(t) be the two solutions of (1.1) and (1.2), respectively, then

    \begin{eqnarray*} |x(t)-x^{*}(t)| & = &\bigg|x_{0}+\int_{0}^{1-\tau}h(s, x(s), \gamma I^{1-\beta}y(\gamma s))ds-\int_{0}^{\tau}y(s)ds +\int_{0}^{t}y(s)ds\\ &-&x_{0}^{*}-\int_{0}^{1-\tau}h^{*}(s, x^{*}(s), \gamma I^{1-\beta}y^{*}(\gamma s))ds+\int_{0}^{\tau}y^{*}(s)ds-\int_{0}^{t}y^{*}(s)ds\bigg|\\ &\leq&\; |x_{0}-x_{0}^{*}|+\bigg|\int_{0}^{1-\tau}h(s, x(s), \gamma I^{1-\beta}y(\gamma s))ds -h^{*}(s, x^{*}(s), \gamma I^{1-\beta}y^{*}(\gamma s))ds\\ &+&\int_{0}^{\tau}y(s)-y^{*}(s)ds+ \int_{0}^{t}y(s)-y^{*}(s)ds \bigg |\\ &\leq&\; |x_{0}-x_{0}^{*}|+\int_{0}^{1-\tau}|h(s, x(s), \gamma I^{1-\beta}y(\gamma s))-h(s, x(s), \gamma I^{1-\beta}y^{*}(\gamma s))\; ds\\ &+&\int_{0}^{1-\tau}|h(s, x(s), \gamma I^{1-\beta}y^{*}(\gamma s))-h(s, x^*(s), \gamma I^{1-\beta}y^{*}(\gamma s))|ds\\ &+&\int_{0}^{1-\tau} |h(s, x^*(s), \gamma I^{1-\beta}y^{*}(\gamma s))-h^{*}(s, , x^*(s), \gamma I^{1-\beta}y^{*}(\gamma s))|ds +2\|y-y^{*}\|_C\\ &\leq&\; |x_{0}-x_{0}^{*}|+b_{2}\int_{0}^{1-\tau}(|x( s)- x^{*}(s)|+\gamma I^{1-\beta}|y(\gamma s)- y^{*}(\gamma s)|)ds\\ &+& b_{2}\int_{0}^{1-\tau}\|h-h^{*}\|_Cds+2\|y-y^{*}\|_C\\ &\leq&\; \delta+b_{2}\bigg( \|x-x^{*}\|_C+\|y-y^{*}\|_C\frac{\gamma }{\Gamma(2-\beta)}+\|h-h^{*}\|_C\bigg) +2 \|y-y^{*}\|_C, \\ \|x-x^{*}\|_C &\leq&\; \bigg( \delta+b_{2}\bigg(\|y-y^{*}\|_C\frac{\gamma }{\Gamma(2-\beta)}+\|h-h^{*}\|_C\bigg) +2 \|y-y^{*}\|_C\bigg)(1-b_2 )^{-1}. \end{eqnarray*}

    Then

    \begin{eqnarray*} \|x-x^{*}\|_C &\leq&\; \bigg( \delta+b_{2}\delta (2+\frac{1}{\Gamma(2-\beta)})+b_{2} \delta \bigg)(1-b_2 )^{-1} = \epsilon, \end{eqnarray*}

    and, the solution x\in C(I) of (1.1)-(1.2) (consequently, (1.4)-(1.5)) with feedback control (1.3) depends continuously on y, \; h, \; and\; x_{0} .

    Definition 3.3. The unique solution x\in C(I) of (1.1)-(1.2) with feedback control (1.3) depends continuously on the control variable u_x , if for all \epsilon > 0 , there exists \delta(\epsilon) > 0 such that

    \max\{|u_x -u_{x^{*}}|\leq \delta\}\; \; \Rightarrow \|y-y^{*}\|_C\leq \epsilon,

    where x^* and y^* are the solutions of

    \begin{eqnarray} x^{*}(t) & = & x_{0}^{*}+\int_{0}^{1-\tau}h^{*}(s, x^{*}(s), \gamma I^{1-\beta}y^{*}(\gamma s))ds-\int_{0}^{\tau}y^{*}(s)ds + \int_{0}^{t} y^{*}(s)ds , \end{eqnarray} (3.4)
    \begin{eqnarray} y^{*}(t)& = & f\bigg(t, u_{x^{*}}(t), x^{*}(t), \gamma I^{1-\alpha}y^{*}(\gamma t)\bigg), \end{eqnarray} (3.5)

    constrained to (1.3), respectively.

    Proof. Let (x(t), y(t)) and (x^{*}(t), y^{*}(t)) be the two solutions of the problem (1.1)-(1.2) constrained to (1.3), respectively, then

    \begin{eqnarray*} |y-y^{*}|& = &|f(t, u_{x}(t), x(t), \gamma I^{1-\alpha}y(\gamma t))-f(t, u_{x^{*}}(t), x^{*}(t), \gamma I^{1-\alpha}y^{*}(\gamma t))|\\ &\leq &|f(t, u_{x}(t), x(t), \gamma I^{1-\alpha}y(\gamma t))-f(t, u_{x^{*}}(t), x^{*}(t), \gamma I^{1-\alpha}y(\gamma t))\\&+&f(t, u_{x}(t), x^{*}(t), \gamma I^{1-\alpha}y(\gamma t))-f(t, u_{x^{*}}(t), x^{*}(t), \gamma I^{1-\alpha}y^{*}(\gamma t))|\\ &\leq& b_{1} \|u_{x}-u_{x^{*}}\|_C+b_{1}\|x-x^{*}\|_C+b_{1} \gamma I^{1-\alpha} |y(\gamma t)-y^{*}(\gamma t)|\\ &\leq&b_{1}\; \delta +b_{1}\|x-x^{*}\|_C+\frac{\gamma b_{1}}{\Gamma(2-\alpha)}\|y-y^{*}\|_C, \\ &\leq& b_{1}\delta+b_{1}\|x-x^{*}\|_C+2 \gamma b_{1}\|y-y^{*}\|_C. \end{eqnarray*}
    \begin{eqnarray*} \|y-y^{*}\|_C\leq \frac{b_{1}\delta+b_{1}\|x-x^{*}\|_C}{1-2\gamma b_{1}}. \end{eqnarray*}

    Also,

    \begin{eqnarray*} |x-x^{*}| & = &|x_{0}+ \int_{0}^{1-\tau} h(s, x(s), \gamma I^{1-\beta}y(\gamma s))ds-\int_{0}^{\tau}y(s)ds+\int_{0}^{t}y_{1}(s)ds\\&-& x_{0}- \int_{0}^{1-\tau} h(s, x^{*}(s), \gamma I^{1-\beta}y^{*}(s))ds+\int_{0}^{\tau}y^{*}(s)ds-\int_{0}^{t}y^{*}(s)ds|\\ &\leq &\int_{0}^{1-\tau}| h(s, x(s), \gamma I^{1-\beta}y(s))- h(s, x^{*}(s), \gamma I^{1-\beta}y^{*}(s))|ds\\ &+&\int_{0}^{\tau}|y(s)-y^{*}(s)|ds+\int_{0}^{t}|y(s)-y^{*}(s)|ds\\ &\leq & b_{2}\int_{0}^{1-\tau}[|x(s)-x^{*}(s)|+\gamma I^{1-\beta} |y(\gamma s)-y^{*}(\gamma s)|]ds\\ &+&\int_{0}^{\tau}|y(s)-y^{*}(s)|ds + \int_{0}^{t}|y(s)-y^{*}(s)|ds\\ &\leq &b_{2} \|x-x^{*}\|_C+\frac{b_{2} \gamma }{\Gamma(2-\beta)}\|y-y^{*}\|_C+2 \|y-y^{*}\|_C, \end{eqnarray*}
    \begin{eqnarray*} \|x-x^{*} \|_C = \frac{(b_{2} \gamma +2)}{1-\; b_{2}}\|y-y^{*}\|_C. \end{eqnarray*}

    Therefore

    \begin{eqnarray*} \|y-y^{*}\|_C\leq \frac{b_{1}\delta}{1-2\gamma b_{1}}+ \frac{b_{1}(b_{2} \gamma +2)}{(1-2\gamma b_{1})(1-\; b_{2})}\|y-y^{*}\|_C. \end{eqnarray*}

    Then

    \begin{eqnarray*} \|y-y^{*}\|_C\leq \frac{b_{1}\delta}{1-2\gamma b_{1}} \bigg(1-\frac{b_{1}(b_{2} \gamma +2)}{(1-2\gamma b_{1})(1-\; b_{2})}\bigg)^{-1} = \epsilon. \end{eqnarray*}

    Thus, the solution x\in C(I) of the problem (1.1)-(1.2) constrained with (1.3) (consequently, (1.4)-(1.5) with feedback control (1.3)) depends continuously on feedback control u_x.

    In this section, we analyze cases in the presence and absence of a control variable, addressing the integer orders issue with an illustrative example.

    Here, we pinpoint specific instances that are valuable for qualitatively analyzing the nonlocal issue of the fractional pantograph differential equation with the fractal feedback control and are essential for various models and practical applications.

    ● For \; \lambda = 0, \; \delta\to 1 , then the fractional pantograph differential equation

    \begin{eqnarray} \frac{dx}{dt} = f(t, u_x(t), x(t), D^{\alpha}x(\gamma t)), \; \; \; t\in (0, 1], \; \; \; \gamma\in (0, 1), \end{eqnarray} (4.1)
    \begin{eqnarray} x(\tau) = x_{0}+ \int_{0}^{1-\tau} h(s, x(s), D^{\beta}x(\gamma s))ds, \; \; \; \tau\in (0, 1], \end{eqnarray} (4.2)

    equipped with the fractal feedback control

    \begin{equation} \frac{d u_x(t)}{dt} = g(t, x(t), D^{\rho}x(\gamma s)), \; \; u_0 = 0, \; \lambda \geq 0, \; \delta \in (0, 1], \end{equation} (4.3)

    which gives

    \begin{eqnarray} \frac{dx}{dt} = f\bigg(t, \int_0^t g(s, x(s), D^{\rho}x(\gamma s))\; ds, x(t), D^{\alpha}x(\gamma t)\bigg), \; \; \; t\in (0, 1], \; \; \; \gamma\in (0, 1), \end{eqnarray} (4.4)
    \begin{eqnarray} x(\tau) = x_{0}+ \int_{0}^{1-\tau} h(s, x(s), D^{\beta}x(\gamma s))ds, \; \; \; \tau\in (0, 1], \end{eqnarray} (4.5)

    under the assumptions of Theorem 3.3. The problem (4.4)-(4.5) depends continuously on the functions on y, \; h , and x_{0} . This case is the same result discussed in [24].

    ● If we put \tau = 1 in Eq (1.2), then the pantograph problem has the form

    \begin{eqnarray} \frac{dx}{dt} = f(t, u_x(t), x(t), D^{\alpha}x(\gamma t)), \; \; \; t\in (0, 1], \; \; \; \gamma\in (0, 1), \end{eqnarray} (4.6)

    with backward boundary condition

    \begin{eqnarray} x(1) = x_{0} \end{eqnarray} (4.7)

    equipped with the fractal-fractional feedback control

    \begin{equation} \frac{d u_x(t)}{dt^\delta} = -\lambda u_x(t)+ g(t, x(t), D^{\rho}x(\gamma t)), \; \; u_0 = u_x(0), \; \lambda \geq 0, \; \delta \in (0, 1], \end{equation} (4.8)

    under the assumptions of Theorem 3.3. The pantograph backward problem (4.6)-(4.7) equipped with the fractal-fractional feedback control (4.8) depends continuously on the functions y, \; h , and on the parameter x_{0} .

    We derive specific cases in the absence of the control variable, which are valuable for the qualitative analysis of certain functional integral equations and essential for various models and real problems.

    ● The pantograph problem becomes

    \begin{eqnarray} \frac{dx}{dt} = f(t, x(t), D^{\alpha}x(\gamma t)), \; \; \; t\in (0, 1], \; \; \; \gamma\in (0, 1), \end{eqnarray} (4.9)
    \begin{eqnarray} x(\tau) = x_{0}+ \int_{0}^{1-\tau} h(s, x(s), D^{\beta}x(\gamma s))ds, \; \; \; \; \; \tau\in (0, 1] , \end{eqnarray} (4.10)

    under the assumptions of Theorem 3.3. The problem (4.9)-(4.10) depends continuously on y, \; h , and x_{0} .

    Assuming the conditions of Theorem 2.3 are met, utilizing the characteristics of the fractional order derivative [23], we derive

    \begin{eqnarray*} \lim\limits_{\alpha \rightarrow1}\; \; \frac{dx}{dt} = \lim\limits_{\alpha \rightarrow 1} f(t, u_{x}(t), x(t), D^{\alpha}x(\gamma t)), \end{eqnarray*}

    yielding

    \begin{eqnarray*} \frac{dx}{dt}& = & f(t, u_{x}(t), x(t), \lim\limits_{\alpha\rightarrow 1} D^{\alpha}x(\gamma t)) \end{eqnarray*}

    and

    \begin{eqnarray} \frac{dx}{dt}& = & f(t, u_{x}(t), x(t), \frac{dx(\gamma t)}{dt}), \; \; \; \; \gamma \in (0, 1]. \end{eqnarray} (4.11)

    Moreover,

    \begin{eqnarray*} \lim\limits_{\beta \rightarrow1}\; \; x(\tau) = x_{0}+ \lim\limits_{\beta \rightarrow1}\int_{0}^{1-\tau} h(s, x(s), D^{\beta}x(\gamma s))ds. \end{eqnarray*}

    However,

    \begin{eqnarray*} |h(t, x(t), D^{\beta}x(\gamma t))| \leq |a_{2}(t)|+b_{2}(|x(t)|+|D^{\beta}x(\gamma t)|), \end{eqnarray*}

    thus

    \begin{eqnarray*} \lim\limits_{\beta \rightarrow1} h(t, x(t), D^{\beta}x(\gamma t)) = h(t, x(t), \lim\limits_{\beta \rightarrow1} D^{\beta}x(\gamma t)) = h(t, x(t), \frac{dx(\gamma t)}{dt}), \end{eqnarray*}

    and

    \begin{array}{l} x(\tau) = x_{0}+ \int_{0}^{1-\tau}\lim\limits_{\beta \rightarrow1}( h(s, x(s), D^{\beta}x(\gamma s)))ds.\\ \;\;\;\;\;x(\tau) = x_{0}+ \int_{0}^{1-\tau} h(s, x(s), \frac{dx(\gamma s)}{ds})ds. \end{array} (4.12)

    Therefore, we have established the subsequent corollary.

    Corollary 4.1. Assuming the conditions of Theorem 2.3 are met, then both of the two integer order problems:

    \begin{eqnarray*} \frac{dx}{dt}& = & f(t, u_{x}(t), x(t), \frac{d}{dt}x(\gamma t)), \; \; \; t\in (0, 1], \; \; \; \gamma\in (0, 1], \\ x(\tau)& = & x_{0}+ \int_{0}^{1-\tau} h(s, x(s), \frac{d}{ds}x(\gamma s))ds, \; \; \; \tau\in (0, 1], \end{eqnarray*}

    equipped with the fractal-fractional feedback control

    \begin{equation*} \frac{d u_x(t)}{dt^\delta} = -\lambda u_x(t)+ g(t, x(t), \frac{d}{dt}x(\gamma t))), \; \; u_0 = u_x(0), \; \lambda \geq 0, \; \delta \in (0, 1], \end{equation*}

    and

    \begin{eqnarray*} \frac{dx}{dt}& = &a\; u_{x}(t)+b\; x(t)+ c\; \frac{d}{dt}x(\gamma t), \; \; \; \; \; \gamma\in (0, 1], \\ x(\tau)& = & x_{0}+ \int_{0}^{1-\tau} h(s, x(s), \frac{d}{ds}x(\gamma s))ds, \; \; \; \tau\in (0, 1], \end{eqnarray*}

    equipped with the fractal-fractional feedback control

    \begin{equation*} \frac{d u_x(t)}{dt^\delta} = -\lambda u_x(t)+ g(t, x(t), \frac{d}{dt}x(\gamma t))), \; \; u_0 = u_x(0), \; \lambda \geq 0, \; \delta \in (0, 1], \end{equation*}

    are guaranteed to have at least one solution x \in C(I) .

    Corollary 4.2. Under the assumption that Theorem 2.3 is valid, the linear pantograph (Ambartsumian) problem (1.4) and (1.5) possesses at least one solution x \in C(I) .

    Proof. By considering

    \begin{eqnarray*} f(t, u_{x}(t), x(t), D^{\alpha}x(\gamma t)) = a\; u_{x}(t)+b\; x(t)+c\; D^{\alpha}x(\gamma t), \end{eqnarray*}

    the conclusions can be derived.

    Corollary 4.3. Let the hypothesis of Theorem 2.3 be valid; if we put \tau = 1 in (1.2), then the backward problem

    \begin{eqnarray} \frac{dx}{dt} = f(t, u_x(t), x(t), D^{\alpha}x(\gamma t)), \; \; \; t\in (0, 1], \end{eqnarray} (4.13)

    with

    \begin{eqnarray} x(1) = x_{0}, \; \; \; \end{eqnarray} (4.14)

    with a fractal feedback control (1.3) has a solution x \in C(I) . Consequently, if the hypotheses of Theorem 3.1 are valid, then it has a unique solution x \in C(I) .

    Corollary 4.4. Let the hypothesis of Corollary 4.2 be valid; if we put \tau = 1 in (1.5), then the backward problem

    \begin{eqnarray} \frac{dx}{dt}& = &a\; u_x(t)+b x(t)+ c D^{\alpha}x(\gamma t), \; \; \; \; \; \; \; t\in (0, 1], \end{eqnarray} (4.15)
    \begin{eqnarray} x(1)& = & x_{0}, \; \; \; \; \; \end{eqnarray} (4.16)

    with a fractal feedback control (1.3) has a solution x \in C(I) . Consequently, if the hypotheses of Corollary 4.2 are valid, it has a unique solution x \in C(I) .

    Example 1. Consider the problem

    \begin{eqnarray} \frac{dx}{dt} & = & (\frac{t}{2})^{2}+u_x(t)+\frac{1}{3} x(t)+ \frac{1}{3}D ^{\frac{1}{2}} x(\frac{t}{2}), \; \; \; \; \; \; t \in (0, \frac{1}{4}], \end{eqnarray} (4.17)
    \begin{eqnarray} x(\tau) & = & \frac{1}{4} + \int_{0}^{ 1 - \tau}\bigg(\frac{\sqrt{t}}{3}+x(s)+\frac{1}{3} D ^{\frac{1}{2}} x(\frac{s}{2})\bigg) ds, \end{eqnarray} (4.18)

    with fractal-fractional feedback control

    \begin{eqnarray} \frac{du(t)}{dt^{\frac{1}{2}}} = -0.4 u(t)+ e^{-\frac{7}{2}t}(\cos{t}+\frac{1}{3}t^3\; D ^{\frac{1}{2}} x(t)). \end{eqnarray} (4.19)

    Note that, this issue is a specific instance of a feedback control problem (1.1)–(1.3) as shown below

    \begin{eqnarray*} \alpha = \beta = \gamma = \rho = \frac{1}{2}, \; x_{0} = \frac{1}{4}. \end{eqnarray*}

    Set

    \begin{eqnarray*} f(t, u_x(t), x(t), D^{\alpha}x(\gamma t)) & = &(\frac{t}{2})^{2}+u_x(t)+\frac{1}{3} x(t)+ \frac{1}{3}D ^{\frac{1}{2}} x(\frac{t}{2}), \\ h(t, x(t), D^{\beta}x(\gamma t)) & = & \frac{\sqrt{t}}{3}+x(t)+\frac{1}{3} D ^{\frac{1}{2}}x(\frac{t}{2}). \end{eqnarray*}

    Thus, conditions (i), (ii) are satisfied with a^{*} = \frac{1}{4}, \; b_1\; = \; b_2\; = \frac{1}{3}, \; M = \frac{1}{6}.\; It is evident that all conditions of Theorem 2.1 are met as follows b_{1}(1+w_2) = 0.333 < 1. Therefore, there is at least one solution x \in C(I) for (4.17)–(4.19). Additionally, we have

    \frac{b_{1}(1+\Delta) \big(b_{2} \gamma +2) }{(1-2\gamma b_{1})(1-\; b_{2})}\thickapprox 0.1981 < 1.

    Thus, all assumptions of Theorem 3.1 are satisfied, then the solution of the problem (4.17)–(4.19) is unique.

    In some applicable differential equations, the state variable appears with a delayed argument. This kind of differential equation, which appears in many fields of science, is well-known as a delay differential equation. This equation involving fractional orders has been examined by several authors in [25,26]. In some studies, fractional-order delay differential equations have been found to exhibit interesting dynamical behaviors that differ from their integer-order counterparts. The presence of delays introduces memory effects into the system, leading to rich and complex dynamics. Researchers have explored various analytical and numerical techniques to study the stability, bifurcations, and oscillatory behavior of such systems. The investigation of fractional delay differential equations is an active area of research with applications in physics, engineering, biology, and other fields [27].

    In this research, the solvability of the fractional pantograph differential equation (1.1) with an integro-differential boundary condition (1.2) constrained to a fractal-fractional feedback control (1.3) was established. The existence of solutions to the problem (1.1)–(1.3) was proved, some sufficient conditions for the uniqueness of the solution were provided, and then the Hyers Ulam stability of the problem (1.1)–(1.3) was derived. Additionally, some continuous dependency results of the solution x on the fractional-order derivative y(t) , the parameter x_{0} , the function h , and on the control variable u_x were established. Finally, a few special cases and examples were presented.

    All authors contributed equally and significantly to writing this article. All authors read and approved the final manuscript.

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

    The authors express their thanks to the anonymous referees for their valuable remarks and comments that helped to improve our paper.

    The authors declare no conflicts of interest.



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