In this article, we analyze the existence and uniqueness of mild solution to the Stieltjes integral boundary value problem involving a nonlinear multi-term, Caputo-type sequential fractional integro-differential equation. Krasnoselskii's fixed-point theorem and the Banach contraction principle are utilized to obtain the existence and uniqueness of the mild solution of the aforementioned problem. Furthermore, the Hyers-Ulam stability is obtained with the help of established methods. Our proposed model contains both the integer order and fractional order derivatives. As a result, the exponential function appears in the solution of the model, which is a fundamental and naturally important function for integer order differential equations and its many properties. Finally, two examples are provided to illustrate the key findings.
Citation: Jiqiang Zhang, Siraj Ul Haq, Akbar Zada, Ioan-Lucian Popa. Stieltjes integral boundary value problem involving a nonlinear multi-term Caputo-type sequential fractional integro-differential equation[J]. AIMS Mathematics, 2023, 8(12): 28413-28434. doi: 10.3934/math.20231454
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In this article, we analyze the existence and uniqueness of mild solution to the Stieltjes integral boundary value problem involving a nonlinear multi-term, Caputo-type sequential fractional integro-differential equation. Krasnoselskii's fixed-point theorem and the Banach contraction principle are utilized to obtain the existence and uniqueness of the mild solution of the aforementioned problem. Furthermore, the Hyers-Ulam stability is obtained with the help of established methods. Our proposed model contains both the integer order and fractional order derivatives. As a result, the exponential function appears in the solution of the model, which is a fundamental and naturally important function for integer order differential equations and its many properties. Finally, two examples are provided to illustrate the key findings.
In the previous few decades, fractional differential equations have become of great interest for researchers due to its high accuracy and usability in numerous subjects of science and technology. A lot of physical and natural phenomena can be modeled through fractional differential equations, which provides better results compared to integer order differential equations. Due to this, fractional differential equations are counted as a special tool for modeling. Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, electrochemistry, aerodynamics, viscoelasticity, polymer rheology, economics, biology, electrodynamics of complex medium, etc. For details, see [2,6,7,8,9,12,14,15,16,17,19,20,22,24,28,29,31,37,39].
Additionally, fractional differential equations serve as an excellent tool for the description of the hereditary properties of various materials and processes. Consequently, the subject of the aforementioned equations is gaining great importance and attention from the researchers. Additionally, researchers are attracted to the enriched material on theoretical aspects and analytic/numerical methods for solving fractional order models. Furthermore, the mathematical models involving fractional order derivatives are more realistic and practical compared to the classical models. In the most recent years, many researchers have focused on the existence of solutions for fractional differential equations, for instance, see [3,4,10] and the references therein.
In 1940, while interacting with the mathematical community at the University of Wisconsin, Ulam expressed his concern regarding the stability of group homomorphisms [25]. The broader form of his views about the stability of functional equations is, "Impose constraints which converges the solutions of an inequality to the exact solutions of the corresponding equations". In 1941, by considering Banach spaces, Hyers gave half an answer to Ulam's question about the stability of functional equations [18]. Due to this contribution of Hyers, Ulam's problem was refereed to as Hyers-Ulam stability of functional equations. For the first time, Hyers-Ulam stability of linear differential equations were introduced by Obloza [21]. Along with generalization, the work of Obloza has been enhanced with different features by using new approaches as time progressed. For more details regarding Ulam's stability with different approaches, we recommend [27,33,34,35,36].
In [32], the authors studied a new class of impulsive implicit sequential fractional differential equations of the following form:
{cDβ(D+λ)u(x):=f(x,u(x),cDβu(x)),x∈(xk,wk], k=0,1,…,m, β∈(0,1],u(x):=Gk(x,u(x)),x∈(wk−1,xk], k=1,2,…,m,u(0):=0, u(wk):=0,k=0,1,2,…,m, | (1.1) |
where cDβ denotes the Caputo fractional derivative of order β, D denotes an ordinary derivative, with the lower limit 0, 0=x0<w0<x1<w1<⋯<xm<wm=T, and T is a pre-fixed number and β∈R+. The function f:[0,T]×R×R→R is continuous and Gk:[wk−1,xk]×R→R is also continuous for all k=1,2,…,m,
Binlin et al. [38] studied the existence and uniqueness (EU) of the solution, as well as the stability in the form of Ulam's problem, for the following FDEs with Stieltjes integral condition:
{CDα(D+λ)x(ξ)=ϕ(ξ),ξ∈[0,1]x(0)=0,CDβ00,ξx(1)=p∑i=1∫10Dβi0,ξξ(s)dxi(s). |
Bashir et al. [5] studied the EU of the solution of the nonelinear multi-term fractional integro-DE with anti-periodic conditions:
{(λC1Dvx(ξ)+λC2Dϱx(ξ))=f(ξ,x(ξ))+Iωg(ξ,x(ξ)),ξ∈[0,τ]x(0)=−x(τ),x′(0)=−x′(τ), |
where, λ1, λ2 ∈R, λ1≠0,v∈[1,2), ϱ∈(1,v), ω>0,f,g:[0,1]×R→R are appropriate functions.
In this article, we analyze the multi-term, nonlinear, sequential fractional differential equation with Stieltjes integral conditions of the following form:
{(λC1Dv+λC2Dϱ)(D+λ3)x(ξ)=f(ξ,x(ξ))+Iωg(ξ,x(ξ)),ξ∈[0,1]x(0)=0,x′(0)=0,CDβ00,ξx(1)=p∑i=1∫10Dβi0,ξx(s)dxi(s), | (1.2) |
where, v∈(1,2],ϱ∈(1,v),λ1,λ2,∈R,λ3∈R+,λ1≠0,p∈N,βi∈R for all i=0,1,⋯,p,0≤β1≤β2≤⋯≤βp<v, β0∈[1,v),f,g:[0,1]×R→R are appropriate functions, and the integrals presented in Boundary Condions (BCs) are Riemann-Stieltjes integrals with xi(i=1,2,⋯,p) functions of a bounded variation. In addition to general FDEs, the multi–point boundary conditions are more valuable than the classical initial/boundary conditions, because these conditions describe the characteristics of chemical, physical or others processes happening inside the domain.
Here, we present necessary preliminaries so that the paper will be self contained.
Let C=C([0,1],R) be the Banach space of all continuous functions endowed with the norm denoted by ‖.‖.
Definition 2.1. [1] The fractional integral of order v from 0 to ξ for the function x is defined by the following:
Iv0,ξx(ξ)=1Γ(v)∫ξ0(ξ−s)v−1x(s)ds,ξ>0, v>0 |
where Γ(.) is the Gamma function.
Definition 2.2. [1] The Caputo derivative of fractional order v from 0 to ξ for a function x can be defined as follows:
CDv0,ξx(ξ)=1Γ(n−v)∫ξ0(ξ−s)n−v−1xn(s)ds,wheren=⌊v⌋+1. |
Lemma 2.1. [1] The FDE CDv0,ξx(ξ)=0 with v>0, involving Caputo differential operator CDv0,ξ has a solution in the following form:
x(ξ)=c0+c1ξ+c2ξ2+⋯+cm−1ξm−1, |
where ck∈R,k=0, 1,⋯,m−1=:¯0,m−1 and m=⌊v⌋+1.
Lemma 2.2. [1] For each v>0, we have the following:
Iv0,ξ(CDv0,ξx(ξ))=c0+c1ξ+c2ξ2+⋯+cm−1ξm−1, |
where ck∈R, k=¯0,m−1 and m=⌊v⌋+1.
Lemma 2.3. [38] If ℜ(v)>0 and λ>0, then
CDv+eλξ=λveλξ and CDv−e−λξ=λve−λξ. |
Theorem 2.1. (Krasnoselskii's fixed point theorem [11]) If K is a closed, convex, non-empty subset of a Banach space M such that P and Q map K into M and,
(ⅰ) Px+Qy∈K(∀x,y∈K),
(ⅱ) P is compact and continuous,
(ⅲ) Q is a contraction mapping,
then ∃ y in K such that, Py+Qy=y.
Theorem 2.2. (Banach fixed point theorem) Every contraction mapping δ from B to B has a fixed point (unique), where B is a non-empty closed set in a Banach space X.
Definition 2.3. Consider a Cauchy problem ddtx(t)=f(t) with x(t0)=x0, then, a continuous function u is called its mild solution if
u(t)=x(t0)+∫tt0f(s)ds. |
Lemma 2.4. Suppose that f,g∈C([0,1]×R,R), then, the mild solution of (1.2) has the following form:
x(ξ)=−λ2λ1∫ξ0e−λ3(ξ−s)Iv−ϱ−1x(s)ds−λ1λ3λ1∫ξ0e−λ3(ξ−s)Iv−ϱx(s)ds+1λ1∫ξ0e−λ3(ξ−s)Ivf(s,x(s))ds+1λ1∫ξ0e−λ3(ξ−s)Iv+ωg(s,x(s))ds+1λ1λ23Δ[λ2λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−ϱ−βi−1x(s)dsdxi(s)+λ2λ3λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βi−ϱx(s)dsdxi(s)−1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βif(s,x(s))dsdxi(s)−1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βi+ωg(s,x(s))dsdxi(s)][λ3ξ−1+e−λ3ξ]+∇λ1λ3(1−e−λ3ξ)+1λ1λ23Δ[−λ2λ1∫10e−λ3(1−s)Iv−β0−ϱ−1x(s)ds−λ2λ3λ1∫10e−λ3(1−s)Iv−β0−ϱx(s)ds+1λ1∫10e−λ3(1−s)Iv−β0f(s,x(s))ds+1λ1∫10e−λ3(1−s)Iv−β0+ωg(s,x(s))ds][λ3ξ−1+e−λ3ξ]+∇λ21λ33Δ[(−λ3)β0e−λ3−p∑i=1∫10(−λ3)βie−λ3ξdxi(s)][λ3ξ−1+e−λ3ξ],where,∇=λ2Iv−ϱ−1x(0)+λ2Iv−ϱx(0)−Ivf(0,x(0))−Iv+ωg(0,x(0)).Δ=1λ1λ23[p∑i=1∫10(−λ3)βie−λ3ξ−(−λ3)β0e−λ3]≠0. | (2.1) |
Proof. Consider problem (1.2) and apply the fractional integral of order v to obtain the following:
(λ1+λ2Iv−ϱ)(D+λ3)x(ξ)=Ivf(ξ,x(ξ))+Iv+ωg(ξ,x(ξ))+C0+C1ξ. |
This implies the following:
λ1Dx(ξ)+λ1λ3x(ξ)+λ2Iv−ϱ−1x(ξ)+λ2λ3Iv−ϱx(ξ)=Ivf(ξ,x(ξ))+Iv+ωg(ξ,x(ξ))+C0+C1ξ. |
Equivalently, we obtain te following:
Dx(ξ)+λ3x(ξ)=−λ2λ1Iv−ϱ−1x(ξ)−λ2λ3λ1Iv−ϱx(ξ)+1λ1Ivf(ξ,x(ξ))+1λ1Iv+ωg(ξ,x(ξ))+1λ1C0+1λ1C1ξ. |
Now, multiplying by the integrating factor eλ3ξ and then integrating from 0 to ξ, we obtain the following:
x(ξ)=−λ2λ1∫ξ0e−λ3(ξ−s)Iv−ϱ−1x(s)ds−λ2λ3λ1∫ξ0e−λ3(ξ−s)Iv−ϱx(s)ds+1λ1∫ξ0e−λ3(ξ−s)Ivf(s,x(s))ds+1λ1∫ξ0e−λ3(ξ−s)Iv+ωg(s,x(s))ds+c0λ1∫ξ0e−λ3(ξ−s)ds+c1λ1∫ξ0se−λ3(ξ−s)ds+c2e−λ3ξ. | (2.2) |
Therefore,
∫ξ0e−λ3(ξ−s)ds=1λ3(1−e−λ3ξ) and,∫ξ0se−λ3(ξ−s)ds=1λ23[λ3ξ−1+e−λ3ξ]. |
Consequently, Eq (2.2) becomes the following:
x(ξ)=−λ2λ1∫ξ0e−λ3(ξ−s)Iv−ϱ−1x(s)ds−λ2λ3λ1∫ξ0e−λ3(ξ−s)Iv−ϱx(s)ds+1λ1∫ξ0e−λ3(ξ−s)Ivf(s,x(s))ds+1λ1∫ξ0e−λ3(ξ−s)Iv+ωg(s,x(s))ds+c0λ1λ3[1−e−λ3ξ]+c1λ1λ23[λ3ξ−1+e−λ3ξ]+c2e−λ3ξ. | (2.3) |
Now, the boundary condition x(0)=0 implies that c2=0. Thus, (2.3) implies the following:
x(ξ)=−λ2λ1∫ξ0e−λ3(ξ−s)Iv−ϱ−1x(s)ds−λ2λ3λ1∫ξ0e−λ3(ξ−s)Iv−ϱx(s)ds+1λ1∫ξ0e−λ3(ξ−s)Ivf(s,x(s))ds+1λ1∫ξ0e−λ3(ξ−s)Iv+ωg(s,x(s))ds+c0λ1λ3[1−e−λ3ξ]+c1λ1λ23[λ3ξ−1+e−λ3ξ]. | (2.4) |
Differentiating Eq (2.4) w.r.t ξ and then applying the boundary condition, we obtain the following:
c0=λ2Iv−ϱ−1x(0)+λ2λ3Iv−ϱx(0)−Ivf(0,x(0))−Iv+ωg(0,x(0)). |
Now, let β0∈[1,v), then, from [38], we obtain the following:
CDβ0x(1)=−λ2λ1∫10e−λ3(1−s)Iv−ϱ−β0−1x(s)ds−λ2λ3λ1∫10e−λ3(1−s)Iv−ϱ−β0x(s)ds+1λ1∫10e−λ3(1−s)Iv−β0f(s,x(s))ds+1λ1∫10e−λ3(1−s)Iv+ω−β0g(s,x(s))ds+c0λ1λ3(−λ3)β0e−λ3+c1λ1λ23(−λ3)β0e−λ3. | (2.5) |
Similarly, for 1≤β1≤β2≤⋯≤βp<v, we obtain the following:
CDβix(ξ)=−λ2λ1∫ξ0e−λ3(ξ−s)Iv−ϱ−βi−1x(s)ds−λ2λ3λ1∫ξ0e−λ3(ξ−s)Iv−ϱ−βix(s)ds+1λ1∫ξ0e−λ3(ξ−s)Iv−βif(s,x(s))ds+1λ1∫ξ0e−λ3(ξ−s)Iv+ω−βig(s,x(s))ds+c0λ1λ3(−λ3)βie−λ3ξ+c1λ1λ23(−λ3)βie−λ3ξ. | (2.6) |
Additionally,
p∑i=1∫10Dβix(ξ)dxi(s)=−λ2λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−ϱ−βi−1x(s)dsdxi(s)+c0λ1λ3p∑i=1∫10(−λ3)βie−λ3ξdxi(s)−λ2λ3λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−ϱ−βix(s)dsdxi(s)+1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βif(s,x(s))dsdxi(s)+1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv+ω−βig(s,x(s))dsdxi(s)+c1λ1λ23p∑i=1∫10(−λ3)βie−λ3ξdxi(s). | (2.7) |
From Eqs (2.5) and (2.7), we have the following:
c1=−λ2λ1Δ∫10e(1−s)Iv−ϱ−β0−1x(s)ds−λ2λ3λ1Δ∫10e(1−s)Iv−ϱ−β0x(s)ds+1λ1Δ∫10e(1−s)Iv−β0f(s,x(s))ds |
+1λ1Δ∫10e(1−s)Iv−β0+ωg(s,x(s))ds+∇λ1λ3Δ[(−λ3)β0e−λ3−p∑i=1∫10(−λ3)βie−λ3ξdxi(s)] |
+λ2λ1Δp∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−ϱ−βi−1x(s)dsdxi(s)+λ2λ3λ1Δp∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−ϱ−βix(s)dsdxi(s) |
−1λ1Δp∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−β0f(s,x(s))dsdxi(s)−1λ1Δp∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−β0+ωg(s,x(s))dsdxi(s). |
Now, Eq (2.4) becomes the following:
x(ξ)=−λ2λ1∫ξ0e−λ3(ξ−s)Iv−ϱ−1x(s)ds−λ1λ3λ1∫ξ0e−λ3(ξ−s)Iv−ϱx(s)ds+1λ1∫ξ0Ive−λ3(ξ−s)f(s,x(s))ds+1λ1∫ξ0e−λ3(ξ−s)Iv+ωg(s,x(s))ds+1λ1λ23Δ[λ2λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−ϱ−βi−1x(s)dsdxi(s)+λ2λ3λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βi−1x(s)dsdxi(s)−1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−β0f(s,x(s))dsdxi(s)−1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−β0+ωg(s,x(s))dsdxi(s)][λ3ξ−1+e−λ3ξ]+∇λ1λ3(1−e−λ3ξ)+1λ1λ23Δ[−λ2λ1∫10e−λ3(1−s)Iv−β0−ϱ−1x(s)ds−λ2λ3λ1∫10e−λ3(1−s)Iv−β0−ϱx(s)ds+1λ1∫10e−λ3(1−s)Iv−β0f(s,x(s))ds+1λ1∫10e−λ3(1−s)Iv−β0+ωg(s,x(s))ds][λ3ξ−1e−λ3ξ]+∇λ21λ33Δ[(−λ3)β0e−λ3−p∑i=1∫10(−λ3)βie−λ3ξDxi(s)][λ3ξ−1e−λ3ξ], |
which is the required proof.
Remark 2.1. According to Lemma 2, and the Counter-Example 1 in [13], Example 3.1 from [23] and Fact 2, from [23], the existence of continuous (even Holderian) solutions of the fractional-type integral forms is not sufficient to ensure the existence of solutions to the corresponding Caputo-type fractional differential problems. For this reason, we only obtain the mild solution.
Throughout the paper, in the case of the Stieltjes integral, we consider the same monotonic functions, which are functions of t and also functions of bounded variation.
The given problem can be converted into a fixed point problem as follows: δ(x)=x, where δ:C→C is defined by
δx(ξ)=−λ2λ1∫ξ0e−λ3(ξ−s)Iv−ϱ−1x(s)ds−λ1λ3λ1∫ξ0e−λ3(ξ−s)Iv−ϱx(s)ds+1λ1∫ξ0e−λ3(ξ−s)Ivf(s,x(s))ds+1λ1∫ξ0e−λ3(ξ−s)Iv+ωg(s,x(s))ds+1λ1λ23Δ[λ2λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−ϱ−βi−1x(s)dsdxi(s)+λ2λ3λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βi−1x(s)dsdxi(s)−1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βif(s,x(s))dsdxi(s)−1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βi+ωg(s,x(s))dsdxi(s)][λ3ξ−1+e−λ3ξ]+∇λ1λ3(1−e−λ3ξ)+1λ1λ23Δ[−λ2λ1∫10e−λ3(1−s)Iv−β0−ϱ−1x(s)ds−λ2λ3λ1∫10e−λ3(1−s)Iv−β0−ϱx(s)ds+1λ1∫10e−λ3(1−s)Iv−β0f(s,x(s))ds+1λ1∫10e−λ3(1−s)Iv−β0+ωg(s,x(s))ds][λ3ξ−1+e−λ3ξ]+∇λ21λ33Δ[(−λ3)β0e−λ3−p∑i=1∫10(−λ3)βie−λ3ξdxi(s)][λ3ξ−1+e−λ3ξ]. | (3.1) |
Theorem 3.1. H1: Suppose f,g∈C([0,1],R)→R and
|f(ξ,0)|≤N1<∞ and |g(ξ,0)|≤N2<∞, |
where N=max{N1,N2}, and
|f(ξ,x)−f(ξ,y)|≤L1|x−y|, |g(ξ,x)−g(ξ,y)|≤L2|x−y|, |
with L1,L2>0 ∀ ξ∈[0,1] and x, y∈R.
H2: L=max{L1,L2} and Ω1+L Ω2<1, where
Ω1=|λ2||λ1|λ3Γ(v−ϱ)(1−e−λ3)+|λ2λ3||λ1|λ3Γ(v−ϱ+1)(1−e−λ3)+1|Δλ1λ23|[|λ2||λ1|λ3p∑i=1(1−e−λ3ρ)ρv−ϱ−βi−1ℜiΓ(v−βi−ϱ)+ |
|λ2λ3||λ1|λ3p∑i=1(1−e−λ3ρ)ρv−ϱ−βiℜiΓ(v−βi−ϱ+1)][λ3−1+e−λ3]+1|Δλ1λ23|[|λ2||λ1|λ3Γ(v−ϱ−β0)(1−e−λ3)+ |
|λ2λ3||λ1|λ3Γ(v−ϱ−β0+1)(1−e−λ3)][λ3−1+e−λ3]. |
Ω2=(1−e−λ3)|λ1|λ3(1Γ(v+1)+1Γ(v+1+ω))+(1−e−λ3)|Δλ1λ23|λ3[1Γ(v−β0+1)+1Γ(v−β0+1+ω)][λ3−1+e−λ3]+ |
1|Δλ1λ23|[1|λ1|λ3p∑i=1(1−e−λ3ρ)ρv−βiℜiΓ(v−βi+1)+1|λ1|λ3p∑i=1(1−e−λ3ρ)ρv−βi+ωℜiΓ(v−βi+1+ω)][λ3−1+e−λ3]. |
Ω3=|∇|(1−e−λ3)|λ1λ3|+|∇||Δλ21λ33|[(λ3)β0e−λ3+p∑i=1(−λ3)βie−λ3ρℜi][λ3−1+e−λ3]. |
Then, Eq (1.2) has a unique mild solution in C.
Proof. Considering a closed ball Br={x∈C:‖x‖≤r}, we show that δBr⊂Br, where δ is defined by Eq (3.1) and r≥(NΩ2+Ω3)(1−Ω1−LΩ2)−1. For any x∈Br, it follows from condition (H1) that
|f(ξ,x)|≤L1‖x‖+N1≤L1r+N1 and g(ξ,x)≤L2r+N2. |
Now, as ‖δ(x)‖=supξ∈[0,1]|δx(ξ)|, the following holds true:
‖δ(x)‖=supξ∈[0,1]|−λ2λ1∫ξ0e−λ3(ξ−s)Iv−ϱ−1x(s)ds−λ1λ3λ1∫ξ0e−λ3(ξ−s)Iv−ϱx(s)ds+1λ1∫ξ0e−λ3(ξ−s)Ivf(s,x(s))ds+1λ1∫ξ0e−λ3(ξ−s)Iv+ωg(s,x(s))ds+1λ1λ23Δ[λ2λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−ϱ−βi−1x(s)dsdxi(s)+λ2λ3λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βi−ϱx(s)dsdxi(s)−1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βif(s,x(s))dsdxi(s)−1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βi+ωg(s,x(s))dsdxi(s)][λ3ξ−1+e−λ3ξ]+∇λ1λ3(1−e−λ3ξ)+1λ1λ23Δ[−λ2λ1∫10e−λ3(1−s)Iv−β0−ϱ−1x(s)ds−λ2λ3λ1∫10e−λ3(1−s)Iv−β0−ϱx(s)ds+1λ1∫10e−λ3(1−s)Iv−β0f(s,x(s))ds+1λ1∫10e−λ3(1−s)Iv−β0+ωg(s,x(s))ds][λ3ξ−1+e−λ3ξ]+∇λ21λ33Δ[(−λ3)β0e−λ3−p∑i=1∫10(−λ3)βie−λ3ξdxi(s)][λ3ξ−1+e−λ3ξ]|. |
Now, using the mean value Theorem [30] for Stieltjes integral with χ∈[0,1] and xi(1)=ℜi>0 and from (H1), we obtain the following:
‖δ(x)‖≤r|λ2||λ1|λ3Γ(v−ϱ)(1−e−λ3)+r|λ2λ3||λ1|λ3Γ(v−ϱ+1)(1−e−λ3)+(L1r+N1)(1−e−λ3)|λ1|λ3Γ(v+1)+(L2r+N2)(1−e−λ3)|λ1|λ3Γ(v+1+ω)+|∇||λ1λ2|(1−e−λ3)+1|Δλ1λ23|[r|λ2||λ1|p∑i=1(1−e−λ3χ)χv−ϱ−βi−1ℜiΓ(v−βi−ϱ)+r|λ2λ3||λ1|λ3p∑i=1(1−e−λ3χ)χv−ϱ−βiℜiΓ(v−βi−ϱ+1)+(L1r+N1)|λ1|λ3p∑i=1(1−e−λ3χ)χv−βiℜiΓ(v−βi+1)+(L2r+N2)|λ1|λ3p∑i=1(1−e−λ3χ)χv−βi+ωℜiΓ(v−βi+1+ω)][λ3−1+e−λ3]+1|Δλ1λ23|[r|λ2||λ1|λ3Γ(v−β0−ϱ)(1−e−λ3)+r|λ2λ3||λ1|λ3Γ(v−ϱ−β0+1)(1−e−λ3)+(L1r+N1)|λ1|λ3Γ(v−β0+1)(1−e−λ3)+(L2r+N2)|λ1|λ3Γ(v−β0+ω+1)(1−e−λ3)][λ3−1+e−λ3]+|∇||λ21λ33Δ||[(−λ3)β0e−λ3−p∑i=1(−λ3)βie−λ3χℜi]|[λ3−1+e−λ3] |
≤r[|λ2||λ1|λ3Γ(v−ϱ)(1−e−λ3)+|λ2λ3||λ1|λ3Γ(v−ϱ+1)(1−e−λ3)+1|Δλ1λ23|[|λ2||λ1|λ3Γ(v−β0−ϱ)(1−e−λ3) |
+|λ2λ3||λ1|λ3Γ(v−ϱ−β0+1)(1−e−λ3)][λ3−1+e−λ3]+1|Δλ1λ23|[|λ2||λ1|p∑i=1(1−e−λ3χ)χv−ϱ−βi−1ℜiΓ(v−βi−ϱ) |
+|λ2λ3||λ1|λ3p∑i=1(1−e−λ3χ)χv−ϱ−βiℜiΓ(v−βi−ϱ+1)][λ3−1+e−λ3]+(Lr+N)[(1−e−λ3)|λ1|λ3Γ(v+1)+(1−e−λ3)|λ1|λ3Γ(v+1+ω) |
+1|Δλ1λ23|[1|λ1|λ3p∑i=1(1−e−λ3χ)χv−βiℜiΓ(v−βi+1)+1|λ1|λ3p∑i=1(1−e−λ3χ)χv−βi+ωℜiΓ(v−βi+1+ω)][λ3−1+e−λ3]] |
+1|Δλ1λ23|[1|λ1|λ3Γ(v−β0+1)(1−e−λ3)+1|λ1|λ3Γ(v−β0+ω+1)(1−e−λ3)][λ3−1+e−λ3]] |
+|∇||λ21λ33Δ||[(−λ3)β0e−λ3−p∑i=1(−λ3)βie−λ3χℜi]|[λ3−1+e−λ3]+|∇||λ1λ2|(1−e−λ3). |
This implies the following:
‖δ(x)‖≤rΩ1+(Lr+N)Ω2+Ω3≤r. |
Next, to prove the contraction, consider the following:
‖δx−δy‖=supξ∈[0,1]|δx(ξ)−δy(ξ)|=supξ∈[0,1]|−λ2λ1∫ξ0e−λ3(ξ−s)Iv−ϱ−1(x(s)−y(s))ds−λ1λ3λ1∫ξ0e−λ3(ξ−s)Iv−ϱ(x(s)−y(s))ds+1λ1∫ξ0e−λ3(ξ−s)Iv(f(s,x(s))−f(s,y(s))ds+1λ1∫ξ0e−λ3(ξ−s)Iv+ω(g(s,x(s))−g(s,y(s))ds+1λ1λ23Δ[λ2λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−ϱ−βi−1(x(s)−y(s))dsdxi(s)+λ2λ3λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βi−ϱ(x(s)−y(s))dsdxi(s)−1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−β0(f(s,x(s))−f(s,y(s)))dsdxi(s)−1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−β0+ω(g(s,x(s))−g(s,y(s))dsdxi(s)][λ3ξ−1+e−λ3ξ]+1λ1λ23Δ[−λ2λ1∫10e−λ3(1−s)Iv−β0−ϱ−1(x(s)−y(s))ds−λ2λ3λ1∫10e−λ3(1−s)Iv−β0−ϱ(x(s)−y(s))ds+1λ1∫10e−λ3Iv−β0(f(s,x(s))−f(s,y(s)))ds+1λ1∫10e−λ3(1−s)Iv−β0+ω(g(s,x(s))−g(s,y(s))ds][λ3ξ−1+e−λ3ξ]|. |
Now, using the mean value theorem for the Stieltjes integral with χ∈[0,1] and xi(1)=ℜi>0 and from (H1), we obtain the following:
‖δx−δy‖≤|λ2||λ1|λ3Γ(v−ϱ)(1−e−λ3)‖x−y‖+|λ2λ3||λ1|λ3Γ(v−ϱ+1)(1−e−λ3)‖x−y‖+L1(1−e−λ3)|λ1|λ3Γ(v+1)‖x−y‖+L2(1−e−λ3)|λ1|λ3Γ(v+1+ω)‖x−y‖+‖x−y‖|Δλ1λ23|[|λ2||λ1|p∑i=1(1−e−λ3χ)χv−ϱ−βi−1ℜiΓ(v−βi−ϱ)+|λ2λ3||λ1|λ3p∑i=1(1−e−λ3χ)χv−ϱ−βiℜiΓ(v−βi−ϱ+1)+L1|λ1|λ3p∑i=1(1−e−λ3χ)χv−βiℜiΓ(v−βi+1)+L2|λ1|λ3p∑i=1(1−e−λ3χ)χv−βi+ωℜiΓ(v−βi+1+ω)][λ3−1+e−λ3]+‖x−y‖|Δλ1λ23|[|λ2||λ1|λ3Γ(v−β0−ϱ)(1−e−λ3)+|λ2λ3||λ1|λ3Γ(v−ϱ−β0+1)(1−e−λ3)+L1|λ1|λ3Γ(v−β0+1)(1−e−λ3)+L2|λ1|λ3Γ(v−β0+ω+1)(1−e−λ3)][λ3−1+e−λ3].≤‖x−y‖[|λ2||λ1|λ3Γ(v−ϱ)(1−e−λ3)+|λ2λ3||λ1|λ3Γ(v−ϱ+1)(1−e−λ3)+1|Δλ1λ23|[|λ2||λ1|p∑i=1(1−e−λ3χ)χv−ϱ−βi−1ℜiΓ(v−βi−ϱ)+|λ2λ3||λ1|λ3p∑i=1(1−e−λ3χ)χv−ϱ−βiℜiΓ(v−βi−ϱ+1)][λ3−1+e−λ3]+1|Δλ1λ23|[|λ2||λ1|λ3Γ(v−β0−ϱ)(1−e−λ3)+|λ2λ3||λ1|λ3Γ(v−ϱ−β0+1)(1−e−λ3)][λ3−1+e−λ3]]+L‖x−y‖[(1−e−λ3)|λ1|λ3Γ(v+1)+(1−e−λ3)|λ1|λ3Γ(v+1+ω)+1|Δλ1λ23|[1|λ1|λ3p∑i=1(1−e−λ3χ)χv−βiℜiΓ(v−βi+1)+1|λ1|λ3p∑i=1(1−e−λ3χ)χv−βi+ωℜiΓ(v−βi+1+ω)][λ3−1+e−λ3]+1|Δλ1λ23|[1|λ1|λ3Γ(v−β0+1)(1−e−λ3)+1|λ1|λ3Γ(v−β0+ω+1)(1−e−λ3)][λ3−1+e−λ3]]≤‖x−y‖[Ω1+LΩ2]. | (3.2) |
Thus,
‖δx−δy‖≤ [Ω1+LΩ2]|x−y‖. |
From (H2), we know that (Ω1+LΩ2) < 1, which demonstrates that the operator δ is contractive. Therefore, according to the Banach contraction principle, there exists a fixed point, which is the mild solution of the problem (1.2).
Theorem 3.2. Suppose that (H1) and the following condition are satisfied.
(H3): ∃ μ1, μ2 ∈ C([0,1],R+) such that |f(ξ,x)|≤μ1(ξ) and |g(ξ,x)|≤μ2, Ω1<1, then, ∃ at least one mild solution of problem (1.2) in C.
Proof. Suppose Kξ = {x∈C:‖x‖≤ξ} be a closed subset of Br and define δ1 and δ2 on Kξ→C as follows:
δ1x(ξ)=1λ1∫ξ0e−λ3(ξ−s)Ivf(s,x(s))ds+1λ1∫ξ0e−λ3(ξ−s)Iv+ωg(s,x(s))ds+∇λ1λ3(1−e−λ3ξ)−1λ1λ23Δ[1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βif(s,x(s))dsdxi(s)+1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βi+ωg(s,x(s))dsdxi(s)]×[λ3ξ−1+e−λ3ξ]+∇λ21λ33Δ[(−λ3)β0e−λ3−p∑i=1∫10(−λ3)βie−λ3ξdxi(s)][λ3ξ−1+e−λ3ξ]+1λ1λ23Δ[1λ1∫10e−λ3(1−s)Iv−β0f(s,x(s))ds+1λ1∫10e−λ3(1−s)Iv−β0+ωg(s,x(s))ds][λ3ξ−1+e−λ3ξ], | (3.3) |
and
δ2x(ξ)=−λ2λ1∫ξ0e−λ3(ξ−s)Iv−ϱ−1x(s)ds−λ1λ3λ1∫ξ0e−λ3(ξ−s)Iv−ϱx(s)ds+1λ1λ23Δ[λ2λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−ϱ−βi−1x(s)dsdxi(s)+λ2λ3λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−ϱ−βix(s)dsdxi(s)][λ3ξ−1+e−λ3ξ]+1λ1λ23Δ[−λ2λ1∫10e−λ3(1−s)Iv−β0−ϱ−1x(s)ds−λ2λ3λ1∫10e−λ3(1−s)Iv−β0−ϱx(s)ds][λ3ξ−1+e−λ3ξ]. | (3.4) |
Clearly, δ=δ1+δ2. Now, we verify the hypothesis of Krasnoselskii's fixed point theorem for x,y∈Kξ with ξ≥(μΩ2+Ω3)(1−Ω)−1. Presently,
‖δ1x(ξ)+δ2y(ξ)‖=supξ∈[0,1]|δ1x(ξ)+δ2y(ξ)|=supξ∈[0,1]|1λ1∫ξ0e−λ3(ξ−s)Ivf(s,x(s))ds+1λ1∫ξ0e−λ3(ξ−s)Iv+ωg(s,x(s))ds+∇λ1λ3(1−e−λ3ξ)−1λ1λ23Δ[1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βif(s,x(s))dsdxi(s)+1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βi+ωg(s,x(s))dsdxi(s)][λ3ξ−1+e−λ3ξ]+1λ1λ23Δ[1λ1∫10e−λ3(1−s)Iv−β0f(s,x(s))ds+1λ∫10e−λ3(1−s)Iv−β0+ωg(s,x(s))ds][λ3ξ−1+e−λ3ξ]+∇λ21λ33Δ[(−λ3)β0e−λ3−p∑i=1∫10(−λ3)βie−λ3ξdxi(s)][λ3ξ−1+e−λ3ξ]−λ2λ1∫ξ0e−λ3(ξ−s)Iv−ϱ−1x(s)ds−λ1λ3λ1∫ξ0e−λ3(ξ−s)Iv−ϱx(s)ds+1λ1λ23Δ[λ2λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−ϱ−βi−1x(s)dsdxi(s)+λ2λ3λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βi−1x(s)dsdxi(s)]×[λ3ξ−1+e−λ3ξ]+1λ1λ23Δ[−λ2λ1∫10e−λ3(1−s)Iv−β0−ϱ−1x(s)ds−λ2λ3λ1∫10e−λ3(1−s)Iv−β0−ϱx(s)ds][λ3ξ−1+e−λ3ξ]| |
≤(1−e−λ3)|λ1|λ3Γ(v+1)‖μ1‖+(1−e−λ3)|λ1|λ3Γ(v+1+ω)‖μ2‖+|∇||λ1λ2|(1−e−λ3)+1|Δλ1λ23|[‖μ1‖|λ1|λ3p∑i=1(1−e−λ3χ)χv−βiℜiΓ(v−βi+1)+‖μ2‖|λ1|λ3p∑i=1(1−e−λ3χ)χv−βi+ωℜiΓ(v−βi+1+ω)][λ3−1+e−λ3]+1|Δλ1λ23|[‖μ1‖|λ1|λ3Γ(v−β0+1)(1−e−λ3)+‖μ2‖|λ1|λ3Γ(v−β0+ω+1)(1−e−λ3)][λ3−1+e−λ3]+|∇||λ21λ33Δ|[(−λ3)β0e−λ3−p∑i=1(−λ3)βie−λ3χℜi][λ3−1+e−λ3]|λ2||λ1|λ3Γ(v−ϱ)(1−e−λ3)ξ+|λ2λ3||λ1|λ3Γ(v−ϱ+1)(1−e−λ3)ξ+1|Δλ1λ23|[ξ|λ2||λ1|λ3p∑i=1(1−e−λ3χ)χv−ϱ−βi−1ℜiΓ(v−βi−ϱ)+ξ|λ2λ3||λ1|λ3p∑i=1(1−e−λ3χ)χv−ϱ−βiℜiΓ(v−βi−ϱ+1)][λ3−1+e−λ3]+1|Δλ1λ23|[ξ|λ2||λ1|λ3Γ(v−β0−ϱ)(1−e−λ3)+ξ|λ2λ3||λ1|λ3Γ(v−ϱ−β0+1)(1−e−λ3)][λ3−1+e−λ3].≤μΩ2+Ω3+ξΩ1≤ξ. |
Thus, δ1x+δ2y∈Kξ.
(ⅱ) Next, we show that δ1 is continuous and compact. Since f and g are continuous thus δ1 is also continuous. Additionally, δ1 is uniformly bounded, i.e.,
‖δ1x‖≤‖μ‖Ω2+Ω3. |
For compactness, assume that
sup(ξ,x)∈[0,1]×Kξ|f(ξ,x)|=f1, sup(ξ,x)∈[0,1]×Kξ|g(ξ,x)|=g1 |
and
[λ3ξ−1+e−λ3ξ]=max{[λ3ξ1−1+e−λ3ξ1],[λ3ξ2−1+e−λ3ξ2]}. |
Then, for ξ1<ξ2, we have the following:
|δ1x(ξ2)−δ1x(ξ1)|=|[1λ1∫ξ20e−λ3(ξ2−s)Ivf(s,x(s))ds+1λ1∫ξ20e−λ3(ξ2−s)Iv+ωg(s,x(s))ds+∇λ1λ3(1−e−λ3ξ2)−1λ1λ23Δ[1λ1p∑i=1∫10∫ξ20e−λ3(ξ2−s)Iv−βif(s,x(s))dsdxi(s)+1λ1p∑i=1∫10∫ξ20e−λ3(ξ2−s)Iv−βi+ωg(s,x(s))dsdxi(s)]×[λ3ξ2−1+e−λ3ξ2]+1λ1λ23Δ[1λ1∫10e−λ3(1−s)Iv−β0f(s,x(s))ds+1λ∫10e−λ3(1−s)Iv−β0+ωg(s,x(s))ds]×[λ3ξ2−1+e−λ3ξ2]+∇λ21λ33Δ[(−λ3)β0e−λ3−p∑i=1∫10(−λ3)βie−λ3ξ2dxi(s)][λ3ξ2−1+e−λ3ξ2]]−[1λ1∫ξ10e−λ3(ξ1−s)Ivf(s,x(s))ds+1λ1∫ξ10e−λ3(ξ1−s)Iv+ωg(s,x(s))ds+∇λ1λ3(1−e−λ3ξ1)−1λ1λ23Δ[−1λ1p∑i=1∫10∫ξ10e−λ3(ξ1−s)Iv−βif(s,x(s))dsdxi(s)−1λ1p∑i=1∫10∫ξ10e−λ3(ξ1−s)Iv−βi+ωg(s,x(s))dsdxi(s)]×[λ3ξ1−1+e−λ3ξ1]+1λ1λ23Δ[1λ1∫10e−λ3(1−s)Iv−β0f(s,x(s))ds+1λ∫10e−λ3(1−s)Iv−β0+ωg(s,x(s))ds]×[λ3ξ1−1+e−λ3ξ1]+∇λ21λ33Δ[(−λ3)β0e−λ3−p∑i=1∫10(−λ3)βie−λ3ξ1dxi(s)][λ3ξ1−1+e−λ3ξ1]]|. |
≤|f1λ1Γ(v+1)(∫ξ10(ξv2e−λ3(ξ2−s)−ξv1e−λ3(ξ1−s))ds+∫ξ2ξ1ξv2e−λ3(ξ2−s)ds)+g1λ1Γ(v+1+ω)(∫ξ10(ξv+ω2e−λ3(ξ2−s)−ξv+ω1e−λ3(ξ1−s))ds+∫ξ2ξ1ξv+ω2e−λ3(ξ2−s)ds)−f1Δλ1λ23[1λ1p∑i=11Γ(v−βi+1)∫10(∫ξ10(ξv−βi2e−λ3(ξ2−s)−ξv−βi1e−λ3(ξ1−s))ds+∫ξ2ξ1ξv−βi2e−λ3(ξ2−s)ds)dxi(s)]×[λ3ξ−1+e−λ3ξ]−g1Δλ1λ23[1λ1p∑i=11Γ(v−βi+1+ω)∫10(∫ξ10(ξv−βi+ω2e−λ3(ξ2−s)−ξv−βi+ω1e−λ3(ξ1−s))ds+∫ξ2ξ1ξv−βi+ω2e−λ3(ξ2−s)ds)dxi(s)][λ3ξ−1+e−λ3ξ]+∇λ1λ3(e−λ3ξ1−e−λ3ξ2)+∇λ21λ33Δ[p∑i=1∫10(−λ3)βie−λ3ξ1dxi(s)−p∑i=1∫10(−λ3)βie−λ3ξ2dxi(s)][λ3ξ−1+e−λ3ξ]]|. |
≤|f1||λ1Γ(v+1)|λ3(ξv2(e−λ3(ξ2−ξ1)−e−λ3ξ2)−ξv1(1−e−λ3ξ1)+ξv2(1−e−λ3(ξ2−ξ1)))|g1||λ1Γ(v+1+ω)|λ3(ξv+ω2(e−λ3(ξ2−ξ1)−e−λ3(ξ2))−ξv+ω1(1−e−λ3ξ1)+ξv+ω2(1−e−λ3(ξ2−ξ1)))|f1||Δλ1λ23|[1|λ1|λ3p∑i=11Γ(v−βi+1)((χv−βi2(e−λ3(χ2−χ1)−e−λ3χ2)−χv−βi1(1−e−λ3χ1)+χv−βi2(1−e−λ3(χ2−χ1)))ℜi]×[λ3ξ−1+e−λ3ξ]+|g1||Δλ1λ23|[1|λ1|λ3p∑i=11Γ(v−βi+1+ω)((χv−βi+ω2(e−λ3(χ2−χ1)−e−λ3χ2)−χv−βi+ω1(1−e−λ3χ1)+χv−βi+ω2(1−e−λ3(χ2−χ1)))ℜi][λ3ξ−1+e−λ3ξ]+|∇||λ21λ33Δ|λ3[p∑i=1|(−λ3)βi(e−λ3χ1−e−λ3χ2)|ℜi][λ3ξ−1+e−λ3ξ].+|∇||λ1λ3|(e−λ3ξ1−e−λ3ξ2), |
where χ1,χ2∈[0,1]. If ξ1→ξ2, then χ1→χ2 and the right side approaches to 0 independent of x ∈ Kξ. Thus, δ1 is equicontinuous. Therefore, by the Arzela-Ascoli theorem, δ1 is relatively compact.
(ⅲ) As Ω1<1, therefore, δ2 is contraction. Thus, there is at least one mild solution to the problem in C.
Let ϵ>0, f,g∈([0,1]×R,R), and assume the following inequality:
|(λ1Dv+λ2Dϱ)(D+λ3)u(ξ)−f(ξ,u(ξ))−Iωg(ξ,u(ξ))|≤ϵ∀ξ∈[0,1]. | (4.1) |
Definition 4.1. (1.2) is said to be Hyers-Ulam stable if there is a constant k0∈R+ such that for each ϵ>0 and each red mild solution u∈C of (4.1), there is a mild solution x∈C of (1.2) with,
|u(ξ)−x(ξ)|≤k0ϵ∀ξ∈[0,1].
Remark 4.1. A function u∈C is a mild solution of (4.1) if there is ψ∈C depending on u, such that:
(ⅰ) (λ1Dv+λ2Dϱ)(D+λ3)u(ξ)=f(ξ,u(ξ))+Iωg(ξ,u(ξ))+ψ(ξ) ∀ ξ∈[0,1].
(ⅱ) |ψ(ξ)|≤ϵ∀ξ∈[0,1].
Lemma 4.1. If u∈C is a mild solution of the Inequality (4.1), then u is also a mild solution of the following inequality:
|u(ξ)−x(ξ)|≤k0ϵ.
where,
k0=[1|λ1|λ3Γ(v+1)(1−e−λ3ξ)+1|λ1λ23Δ|[1|λ1|λ3p∑i=1(1−e−λ3χ)χv−βiℜiΓ(v−βi+1)][λ3−1+e−λ3]+1|λ1λ23Δ|((1−e−λ3)|λ1|λ3Γ(v−β0+1))[λ3−1+e−λ3]]. |
Proof. If u is the mild solution of Inequality (4.1), then u will also be the mild solution of the following:
(λ1Dv+λ2Dϱ)(D+λ3)u(ξ)=f(ξ,u(ξ))+Iωg(t,u(ξ))+ψ(ξ),t∈[0,1]x(0)=0, x′(0)=0, CDβ00,ξu(1)=p∑i=1∫10Dβiu(ξ)dui(s). | (4.2) |
The mild solution of (4.2) is given by the following:
u(ξ)=−λ2λ1∫ξ0e−λ3(ξ−s)Iv−ϱ−1u(s)ds−λ1λ3λ1∫ξ0e−λ3(ξ−s)Iv−ϱu(s)ds+1λ1∫ξ0Ive−λ3(ξ−s)f(s,u(s))ds+1λ1∫ξ0e−λ3(ξ−s)Iv+ωg(s,u(s))ds+1λ1λ23Δ[λ2λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−ϱ−βi−1u(s)dsdui(s)+λ2λ3λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βi−1u(s)dsdui(s)−1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βif(s,u(s))dsdui(s)−1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βi+ωg(s,u(s))dsdui(s)][λ3−1+e−λ3ξ]+∇λ1λ3(1−e−λ3ξ)+1λ1λ23Δ[−λ2λ1∫10e−λ3(1−s)Iv−β0−ϱ−1u(s)ds−λ2λ3λ1∫10e−λ3(1−s)Iv−β0−ϱu(s)ds+1λ1∫10e−λ3(1−s)Iv−β0f(s,u(s))ds+1λ∫10e−λ3(1−s)Iv−β0+ωg(s,u(s))ds][λ3ξ−1+e−λ3ξ]+∇λ21λ33Δ[(−λ3)β0e−λ3−p∑i=1∫10(−λ3)βie−λ3ξdui(ξ)][λ3ξ−1+e−λ3ξ]+1λ1∫ξ0e−λ3(ξ−s)Ivψ(s))ds−1λ1λ23Δ[1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βiψ(s)dsdui(s)][λ3ξ−1+e−λ3ξ]+1λ1λ23Δ[1λ1∫10e−λ3(1−s)Iv−β0ψ(s)ds][λ3ξ−1+e−λ3ξ]. | (4.3) |
Let the terms free of ψ be denoted by x(ξ); then,
|u(ξ)−x(ξ)|≤1|λ1|∫ξ0e−λ3(ξ−s)Iv|ψ(s))|ds+1|λ1λ23Δ|[1λ1p∑i=1∫10∫ξ0e−λ3(ξ−s)Iv−βi|ψ(s)|dsdui(s)][λ3ξ−1e−λ3ξ]+1|λ1λ23Δ|[1λ1∫10e−λ3(1−s)Iv−β0|ψ(s)|ds][λ3ξ−1e−λ3ξ].≤ϵ[1|λ1|λ3Γ(v+1)(1−e−λ3)+1|λ1λ23Δ|[1|λ1|λ3p∑i=1(1−e−λ3χ)χv−βiℜiΓ(v−βi+1)][λ3−1+e−λ3]+1|λ1λ23Δ|((1−e−λ3)|λ1|λ3Γ(v−β0+1))[λ3−1+e−λ3]]. |
Assume that,
k0=[1|λ1|λ3Γ(v+1)(1−e−λ3)+1|λ1λ23Δ|[1|λ1|λ3p∑i=1(1−e−λ3χ)χv−βiℜiΓ(v−βi+1)][λ3−1+e−λ3]+1|λ1λ23Δ|((1−e−λ3)|λ1|λ3Γ(v−β0+1))[λ3−1+e−λ3]]. |
In view of (ⅱ) of Remark 3.2, we have the following:
|u(ξ)−x(ξ)|≤k0ϵ, |
which is the desired result.
In this section, we solve some examples using the obtained theorems.
Example 5.1.
{(6CD1.35+4CD1.1)(D+1)=e−2ξ|x(ξ)|(t+5)2(1+|x(ξ)|)+I2sinξ|x(ξ)|√(49+t),∀ξ∈[0.1],x(0)=0x′(0)=0CDβ00,ξx(1)=2∑i=1∫10Dβix(s)dxi(s), | (5.1) |
where,
p=2,v=1.35,ξ=1.1,ω=2,λ1=6,λ2=4,λ3=1,χ=0.1,β0=1.12,β1=1.2,β2=1.32,ℜ1=10,ℜ2=15. |
Since,
f(ξ,x(ξ))=e−2ξ|x(ξ)|(t+5)2(1+|x(ξ)|)andg(ξ,x(ξ))=sinξ|x(ξ)|√(49+t). |
From assumptions (H1) and (H2), we have the following:
|f(ξ,x(ξ))−f(ξ,y(ξ))|≤e−2ξ|x−y|(t+5)2≤125|x−y|,|g(ξ,x(ξ))−g(ξ,y(ξ))|≤sinξ|x−y|√(49+t)≤17|x−y|, |
where, L1=125, L2=17 and L=max{L1,L2}=17. Calculating Ω1 and Ω2 from the given data, we obtain Ω1=0.93461 and Ω2=0.136698. Additionally, Ω1+LΩ2≈0.95414<1. Hence, the given problem (5.1) has a unique mild solution in C for ξ∈[0,1].
Example 5.2.
{(5CD1.75−CD1.6)(D+2)x(ξ)=ξ2ξ3+17x(ξ)+I2cosξet+15x(ξ),∀ξ∈[0,1],x(0)=0x′(0)=0,CDβ00,ξx(1)=2∑i=1∫10Dβix(s)dxi(s), | (5.2) |
where,
λ1=5,λ2=−1,λ3=2,χ=0.5,p=2,ℜ1=30,ℜ2=15,v=1.75,ξ=1.6,β0=1.24,β1=1.28,β2=1.32. |
Therefore,
f(ξ,x(ξ))=ξ2ξ3+17x(ξ)and g(ξ,x(ξ))=cosξet+15x(ξ). |
From assumption (H1) and (H2), we have
|f(ξ,x(ξ))−f(ξ,y(ξ))|≤ξ2ξ3+17|x−y|≤117|x−y|,|g(ξ,x(ξ))−g(ξ,y(ξ))|≤cosξet+15|x−y|≤116|x−y|, |
where, L1=117, L2=116 and L=max{L1,L2}=116. Calculating Ω1 and Ω2 from the given data, we obtain Ω1=0.2188677, Ω2=0.324496. Additionally, Ω1+LΩ2≈0.239415<1. Hence, the given problem (5.2) has a unique mild solution in C for t∈[0,1].
The paper established sufficient conditions that showed the existence, uniqueness and Ulam's stability for the mild solutions of Problem (1.2). The conditions were obtained from the view of fixed point theorems. Furthermore, we demonstrated the obtained results using two examples. The multi-point boundary conditions can be used to describe the characteristics of chemical, physical or others processes occouring inside the domain. Thus, the obtained results can be fruitful in the mentioned processes. The Hyers-Ulam stability means that for any approximation in a specific region we will obtain an exact mild solution. Therefore, the obtained results of the Hyers-Ulam stability can be utilized in a numerical analysis and approximation theory of the related mentioned processes.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
This research is supported by the Anhui Provincial Natural Science Foundation Project (KJ2021A1175) and Funding project for cultivating top-notch talents in universities (gxgnfx2022096).
The authors declare no conflicts of interest.
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