Existence of Solutions of Impulsive Partial Hyperbolic Differential Inclusion of Fractional Order

Abstract

This paper is concerned with the existence of solutions of a class of Cauchy problems for hyperbolic partial fractional differential inclusions (HPFD) involving the Caputo fractional derivative with an impulse whose right hand side is convex and non-convex valued. Our results are achieved within the framework of the nonlinear alternative of Leray-Schauder type and contraction multivalued maps. A detailed example was provided to support the theorem.

1. Introduction

Fractional calculus has emerged as a powerful mathematical tool for modeling and analyzing complex systems that exhibit memory effects and long-range dependencies. It extends the classical calculus operators of differentiation and integration to non-integer orders, allowing for a more accurate representation of phenomena in various scientific and engineering domains. The topic is as old as differential calculus. Fractional calculus was credited to G.W. Leibniz (1697) and L. Euler (1730) (see [1]). Mathematicians, physicists and engineers have shown interest in the concept of fractional calculus and fractional order differential equations and inclusions in several applications involving rheology, control, porous media, viscoelasticity, electrochemistry, electromagnetism, etc.; many authors have found beneficial and valid results [2,3,4,5,6,7,8]. See the monographs referenced therein [6,7,9,10,11,12,13,14,15,16,17,18,19], for examples of recent developments in ordinary and partial fractional differential equations.

Motivated by the need to model real-world phenomena with more fidelity, researchers have explored the combination of hyperbolic systems and fractional calculus. Hyperbolic differential inclusions (HDI) of fractional order allow for the incorporation of memory effects and fractional derivatives, which can capture non-local or non-Markovian behaviour in the system dynamics. The fractional order derivatives in hyperbolic differential inclusions enable the inclusion of memory effects, allowing for a more accurate representation of real-world systems that exhibit long-range dependencies. HDI also consider a set-valued formulation that encompasses a range of possible solutions or uncertain parameters. This framework is particularly useful when dealing with uncertain or variable system parameters, providing a more robust characterization of the system behaviour.

Leibniz invented the notation $\frac{d^{n}y}{d^{x}n}$. Perhaps it was a naive play with symbols that prompted L’Hopital in 1695 to ask Leibniz “what if n be $\frac{1}{2}$?”

Differential inclusion is a differential equation with a discontinuous right-hand side. It is used to study ordinary differential equation with an inaccurately known right-hand side. The dynamics of evolving processes are often subjected to abrupt changes such as shocks, harvesting, and natural disasters. Often these short-term perturbations are treated as having acted instantaneously or in the form of ’impulses’. Impulsive differential equations such as

$$(\star )x^{\prime }=f(t,x),t\in [0,b]|(t_1,\cdots ,t_m)$$

subject to impulse effects

$$(\star \star )\Delta x=x\left(t_k^{+}\right)-x\left(t_k^{-}\right)=I_k\left(x\left(t_k^{-}\right)\right),k=1,2,\cdots ,m$$

with $f:\left([0,b]\right|\{t_1,\cdots ,t_m\})\times {\mathbb{R}}^n⟶{\mathbb{R}}^n$ and $I_k$ an impulse operator. Impulsive differential inclusion is the case when the right-hand side of $(\star )$ has discontinuities, differential inclusions such as $x^{\prime }\in F(t,x\left(t\right)),t\in [0,b]\setminus (t_1,\cdots ,t_m)$ subject to the impulse conditions $(\star \star )$, where $F:([0,b]\setminus \{t_1,\cdots ,t_m\})\times \mathbb{R}⟶2^{{\mathbb{R}}^n}$.

In this paper, the authors examine some simple cases of the Riemann-Liouville and Caputo fractional derivatives, and we also look into the existence of solutions to the system’s fractional order IVP.

$$\left({}^c{D}_{x_k}^{r}u\right)(x,y)\in F(x,y,u(x,y)),if(x,y)\in J_k,k=0,\cdots ,m$$

(1)

$$\begin{array}{cc}\hfill u(x_k,y)& =u(x_k^{-},y)+I_k(u(x_k^{-},y),\mathrm{if}y\in [0,b];k=1,\cdots ,m\hfill \\ \hfill u(x,0)& =\phi \left(x\right);x\in [0,a],u(0,y)=\psi \left(y\right);y\in [0,b].\hfill \end{array}$$

(2)

where $J_0=[0,x_1]\times [0,b],J_k=(x_k,x_{k+1}];k=1,\cdots ,m,a,b>0,0<x_1<\cdots <x_m<x_{m+1}=a,0<y_1<\cdots <y_m<y_{m+1}=b$. $F:J\times {\mathbb{R}}^n⟶P\left({\mathbb{R}}^n\right)$ is a compact valued multivalued map, $J=[0,a]\times [0,b],P\left({\mathbb{R}}^n\right)$ is a family of all subsets of ${\mathbb{R}}^n,I_k:{\mathbb{R}}^n⟶{\mathbb{R}}^n,k=1,\cdots ,m$ are given functions and $\phi :[0,a]⟶{\mathbb{R}}^n,\psi :[0,b]⟶{\mathbb{R}}^n$ are absolutely continuous functions with $\phi \left(0\right)=\psi \left(0\right)$.

We present two existence results for the problem (1) and (2), the first one is based on Banach’s contraction principle and the second one on the nonlinear alternative of Leray-Schauder type.

Definition 1

([3], Gamma Function). Let $\mu \in \mathbb{R}$, other than zero or a negative integer and $0<x<\infty$. We define the Gamma function, denoted by $\gamma (\mu ,x),$ and the complementary incomplete Gamma function, denoted by $\Gamma (\mu ,x),$ from the integrals, $\gamma (\mu ,x)={\int }_0^x{e}^{-ϵ}{ϵ}^{\mu -1}dϵ$ and $\Gamma (\mu ,x)={\int }_x^{\infty }{e}^{-ϵ}{ϵ}^{\mu -1}dϵ$, respectively.

The first substantive step towards the creation of the fractional calculus was taken in 1695 by Leibniz and Leo-Peter, with the introduction of the factorial function.

$$\begin{array}{cc}\hfill y& =x^n\hfill \\ \hfill \frac{dy}{dx}& =n{x}^{n-1}\hfill \\ \hfill \frac{d^{2}y}{d{x}^2}& =n(n-1)x^{n-2}=\frac{n(n-1)(n-2)!}{(n-2)!}{x}^{n-2}\hfill \\ \hfill & =\frac{n!}{(n-2)!}{x}^{n-2}\hfill \\ \hfill \frac{d^{3}y}{d{x}^3}& =n(n-1)(n-2)x^{n-3}=\frac{n(n-1)(n-2)(n-3)!}{(n-3)!}{x}^{n-3}\hfill \\ \hfill & =\frac{n!}{(n-3)!}{x}^{n-3}\hfill \\ \hfill & \cdots \hfill \end{array}$$

For $m<n$, we have

$$\frac{d^{m}y}{d{x}^m}=\frac{n!}{(n-m)!}{x}^{n-m}$$

(3)

Equation (3) is the power rule for derivative for classical calculus.

If $m\in \mathbb{N}⟶\alpha \in \mathbb{R}$, then

$$\frac{d^{\alpha }y}{d{x}^{\alpha }}=\frac{n!}{(n-\alpha )!}{x}^{n-\alpha },\forall \alpha \in \mathbb{R}.$$

(4)

In 1729, Euler extended the factorial function to Gamma function. He gave the integral representation of Gamma function and its properties.

If n is integer, then $\Gamma (n+1)=n!,\forall n\in \mathbb{N};\Gamma (x+1)=x\Gamma \left(x\right),\forall x\in \mathbb{R};\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }$.

Thus, Equation (4) becomes

$$D^{\alpha }y=\frac{\Gamma (n+1)}{\Gamma (n-\alpha +1)}{x}^{n-\alpha }$$

(5)

Equation (5) was first used for the power rule for fractional derivative.

Example 1.

Find the semi-derivative (half order) derivative of x.

Solution

$$D^{\frac{1}{2}}f\left(x\right)=\frac{n!}{\Gamma (n-\alpha +1)}{x}^{n-\alpha }$$

Here, $n=1,\alpha =\frac{1}{2}$

$$D^{\frac{1}{2}}\left(x\right)=\frac{1!}{\Gamma (1-\frac{1}{2}+1)}{x}^{1-\frac{1}{2}}$$

$$D^{\frac{1}{2}}\left(x\right)=\frac{1}{\Gamma \left(\frac{3}{2}\right)}{x}^{\frac{1}{2}}$$

Recall that $\Gamma \left(\frac{3}{2}\right)=\Gamma (1+\frac{1}{2})=\frac{1}{2}\Gamma \left(\frac{1}{2}\right)=\frac{1}{2}\sqrt{\pi }$

Thus

$$D^{\frac{1}{2}}\left(x\right)=\frac{1}{\frac{1}{2}\sqrt{\pi }}{x}^{\frac{1}{2}}=2\sqrt{\frac{x}{\pi }}$$

Example 2.

Find the semi-derivative (half order) derivative of a constant function, i.e, $f\left(x\right)=c$

Solution Here, $y=c{x}^0,n=0$ and $\alpha =\frac{1}{2}$

$$D^{\frac{1}{2}}\left[c{x}^0\right]=\frac{0!}{\Gamma (0-\frac{1}{2}+1)}{x}^{0-\frac{1}{2}}=\frac{1}{\Gamma \left(\frac{1}{2}\right)}{x}^{-\frac{1}{2}}=\frac{1}{\sqrt{x\pi }}$$

Which is not continuous at $x=0$. This technique violates the Polynomial function theorem which states that ’every polynomial function is continuous everywhere on $\mathbb{R}$’. Take for instance, if $y=x^2$, then the first derivative is $\dot{y}=2x$ and the second derivative is $\ddot{y}=2$. Each of these polynomials is continuous in classical derivative but the semi-derivative of function $y=c{x}^0$ is not continuous everywhere.

This result was not good enough. This lead to later discoveries.

2. Preliminaries

In this section, we introduce notations, definitions, and preliminary facts which are used throughout this paper. By C(J) we denote the Banach space of all continuous functions from J into ${\mathbb{R}}^n$ with the norm

$$\parallel u\parallel ={\displaystyle \underset{(x,y)\in J}{sup}}\parallel w(x,y)\parallel$$

where $|·|$ denotes a suitable complete norm on ${\mathbb{R}}^n$.

As usual, by $AC\left(J\right)$ we denote the space of absolutely continuous functions from J into ${\mathbb{R}}^n$ and $L^1\left(J\right)$ is the space of Lebegue-integrable functions.

Let $(X,\parallel .\parallel )$ be a Banach space. Denote $P\left(X\right)=\{Y\in P\left(X\right)\ne \varnothing \},P_{cl}\left(X\right)=\{Y\in P\left(X\right):Y$ is closed}, $P_b\left(X\right)=\{Y\in P\left(X\right):$ Y is bounded}, $P_{cp}\left(X\right)=\{Y\in P\left(X\right)$: Y is compact}, and $P_{cp,cv}\left(X\right)=\{Y\in P\left(X\right):Y$ is compact and convex}.

Definition 2

([2,17,20]). Riemann-Liouville partial fractional integration

Let $\alpha \in (0,\infty )$ and $u\in L^1(J;{\mathbb{R}}^n)$. The Riemann-Liouville partial fractional integral of order $\alpha$ of $u(x,y)$ with respect to x is defined by

$$I_{0,x}^{\alpha }u(x,y)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_0^x{(x-s)}^{\alpha -1}u(s,y)ds;$$

for almost all $x\in [0,a]$ and for almost all $y\in [0,b]$, where $\Gamma \left(\alpha \right)$ is the Euler Gamma function of $\alpha$ and provided that the integral exists.

Similarly, the Riemann-Liouville partial fractional integral of order $\alpha$ of $u(x,y)$ with respect to y is defined by

$$I_{0,y}^{\alpha }u(x,y)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_0^y{(y-s)}^{\alpha -1}u(x,s)ds.$$

for almost all $x\in [0,a]$ and for almost all $y\in [0,b]$.

Let $r=(r_1,r_2)\in (0,\infty )\times (0,\infty );$ $z_k=(x_k,0)$ and $u\in L^1(J,{\mathbb{R}}^n)$. The left-sided mixed Riemann-Liouville fractional integral of order r of $u(x,y)$ is defined by

$$\left(I_{z_k^{+}u}^r\right)(x,y)=\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}u(s,t)dtds,$$

where $\Gamma \left(r_1\right),\Gamma \left(r_2\right)$ are the Euler Gamma functions of $r_1,r_2$, respectively, provided that the integral exists.

Definition 3

([2,17]). Riemann-Liouville partial fractional integration

Let $\alpha \in (0,\infty )$ and $u\in L^1(J;{\mathbb{R}}^n)$. The Riemann-Liouville partial fractional derivative of order $\alpha$ of $u(x,y)$ with respect to x is defined by

$$\begin{array}{cc}\hfill D_{0,x}^{\alpha }u(x,y)& =\frac{\partial }{\partial x}\left(I_{0,x}^{n-\alpha }u(x,y)\right)\hfill \\ \hfill & =\frac{\partial }{\partial x^n}·\frac{1}{\Gamma (1-\alpha )}{\int }_0^x{(x-t)}^{n-\alpha -1}f\left(t\right)dt\hfill \end{array}$$

for almost all $x\in [0,a]$ and for almost all $y\in [0,b]$.

Let $r=(r_1,r_2)\in (0,\infty )\times (0,\infty )$, $z_k=(x_k,0)$ and $u\in L^1(J,{\mathbb{R}}^n)$. The left-sided mixed Riemann-Liouville fractional derivative of order r of $u(x,y)$ is defined by

$$\left(D_{x_k^{+}}^{r}u\right)(x,y)=\frac{{\partial }^2}{\partial x\partial y}\left(I_{z_k^{+}}^{1-r}u\right)(x,y),$$

and the right-sided mixed Riemann-Liouville fractional derivative of order r of u is defined by

$$\left(D_{x_k^{-}}^{r}u\right)(x,y)=\frac{{\partial }^2}{\partial x\partial y}\left(I_{z_k^{-}}^{1-r}\right)(x,y).$$

Definition 4

([2,17]). Caputo partial fractional derivative

Let $\alpha \in (0,\infty )$ and $u\in L^1(J;{\mathbb{R}}^n)$. The Caputo partial fractional derivative of order $\alpha ofu(x,y)$ with respect to x is defined by

$$\left({}^c{D}_{0,x}^{\alpha }u\right)(x,y)=I_{0,x}^{n-\alpha }\left(\frac{\partial }{\partial x}u(x,y)\right),$$

for almost all $x\in [0,a]$ and for almost all $y\in [0,b]$.

Let $r=(r_1,r_2)\in (0,\infty )\times (0,\infty )$,  $z_k=(x_k,0)$ and $u\in L^1(J,{\mathbb{R}}^n)$. The left-sided mixed Caputo fractional derivative of order r of u is defined by

$$\left({}^c{D}_{z_k^{+}}^{r}u\right)(x,y)=\left(I_{z_k^{+}}^{1-r}\frac{{\partial }^2}{\partial x\partial y}u\right)(x,y),$$

and the right-sided mixed Caputo fractional derivative of order r of u is defined by

$$\left({}^c{D}_{z_k^{-}}^{r}u\right)(x,y)=\left(I_{z_k^{-}}^{1-r}\frac{{\partial }^2}{\partial x\partial y}u\right)(x,y),$$

where $1-r$ means that $(1-r_1,1-r_2)\in (0,\infty )\times (0,\infty )$ and $D_{xy}^2=\frac{{\partial }^2}{\partial x\partial y}$ denote the mixed second order partial derivative.

Definition 5.

Grunwald-Letnikov left-sided derivative:

$${}^{GL}{D}_{a^{+}}^{\alpha }\left[f\left(x\right)\right]={\displaystyle \underset{h⟶0}{lim}}\frac{1}{h^{\alpha }}{\displaystyle \sum _{k=0}^n}{(-1)}^k\frac{\Gamma (\alpha +1)f(x-kh)}{\Gamma (k+1)(\alpha -k+1)},nh=x-a.$$

Definition 6. 

Weyl derivative:

$${}_x{D}_{\infty }^{\alpha }=D^{\alpha }\left[f\left(x\right)\right]={(-1)}^m{\left(\frac{d}{dt}\right)}^n{[}_x{W}_{\infty }^{\alpha }\left(f\left(x\right)\right)]$$

Weyl integral is defined as

$${}_x{W}_{\infty }^{\alpha }\left[f\left(x\right)\right]=\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^{\infty }{(t-x)}^{\alpha -1}f\left(t\right)dt$$

Definition 7.

Hadamard derivative:

$$D_{+}^{\alpha }\left[f\left(x\right)\right]=\frac{\alpha }{\Gamma (1-\alpha )}{\int }_0^x\frac{f\left(x\right)-f\left(t\right)}{{\left[ln\left(\frac{x}{t}\right)\right]}^{1+\alpha }}\frac{dt}{t}$$

Hadamard integral is defined by

$$I_{+}^{\alpha }\left[f\left(x\right)\right]=\frac{1}{\alpha }{\int }_0^x\frac{f\left(t\right)}{{\left[ln\left(\frac{t}{x}\right)\right]}^{1-\alpha }}·\frac{dt}{t},x>0,\alpha >0.$$

Definition 8.

He’s fractional derivative:

$$\frac{{\partial }^{\alpha u}}{\partial t^{\alpha }}=\frac{1}{\Gamma (n-\alpha )}\frac{d^n}{d{t}^n}{\int }_{t_0}^t{(s–t)}^{n-\alpha -1}[u_0\left(s\right)-u\left(s\right)]ds$$

Definition 9.

Atangana-Baleanu’s fractional derivative:

Let $f\in H^{\prime }(a,b),b>0,\alpha \in [0,1]$ then, the Atangana-Baleanu’s fractional derivative is given as:

$${}_b^{ABR}{D}_t^{\alpha }\left(f\left(t\right)\right)=\frac{B\left(\alpha \right)}{1-\alpha }\frac{d}{dt}{\int }_b^{t}f\left(x\right){\mathbb{E}}_{\alpha }\left[\frac{-\alpha {(t-x)}^{\alpha }}{1-\alpha }\right]dx$$

Example 3.

Find the Riemann-Liouville fractional derivative of a constant function.

Solution

i.e., $f\left(x\right)=c$

$$\begin{array}{cc}\hfill {}_a{D}_x^{\alpha }f\left(x\right)& ={\frac{d^n}{d{x}^n}}_a{I}_x^{n-\alpha }f\left(x\right){⟹}_a{D}_x^{\alpha }\left(c\right)={\frac{d^n}{d{x}^n}}_a{I}_x^{n-\alpha }\left(c\right)\hfill \\ \hfill {}_a{D}_x^{\alpha }\left(c\right)& =\frac{d^n}{d{x}^n}·\frac{1}{\Gamma (n-\alpha )}{\int }_a^x{(x-t)}^{n-\alpha -1}\left(c\right)dt{⟹}_a{D}_x^{\alpha }\left(c\right)=\frac{d^n}{d{x}^n}·\frac{c}{\Gamma (n-\alpha )}{\left[-\frac{{(x-t)}^{n-\alpha }}{n-\alpha }\right]}_a^x\hfill \\ \hfill {}_a{D}_x^{\alpha }\left(c\right)& =\frac{d^n}{d{x}^n}·\frac{c}{\Gamma (n-\alpha )}\frac{{(x-a)}^{n-\alpha }}{n-\alpha }{⟹}_a{D}_x^{\alpha }\left(c\right)=\frac{c}{(n-\alpha )\Gamma (n-\alpha )}\left[\frac{d^n}{d{x}^n}{(x-a)}^{n-\alpha }\right]\hfill \end{array}$$

Recall that $\frac{d^m}{d{x}^m}=\frac{n!}{(n-m)!}{x}^{n-m}$

Thus,

$$\begin{array}{cc}\hfill {}_a{D}_x^{\alpha }\left(c\right)& =\frac{c}{(n-\alpha )\Gamma (n-\alpha )}·\frac{(n-\alpha )!{(x-a)}^{n-\alpha -n}}{(n-\alpha -n)!}\hfill \\ \hfill {}_a{D}_x^{\alpha }\left(c\right)& =\frac{c}{(n-\alpha )\Gamma (n-\alpha )}·\frac{(n-\alpha )!{(x-a)}^{-\alpha }}{(-\alpha )!}\hfill \end{array}$$

Recall that $\Gamma (1+x)=x\Gamma \left(x\right)$ Hence, we have,

$$\begin{array}{cc}\hfill {}_a{D}_x^{\alpha }\left(c\right)& =\frac{c}{\Gamma (n-\alpha +1)}·\frac{\Gamma (n-\alpha +1){(x-a)}^{-\alpha }}{\Gamma (1-\alpha )}\hfill \\ \hfill {}_a{D}_x^{\alpha }\left(c\right)& =\frac{c}{\Gamma (1-\alpha )}{(x-a)}^{-\alpha }{⟹}_a{D}_x^{\alpha }\left(c\right)=\frac{c}{\Gamma (1-\alpha ){(x-a)}^{\alpha }}\ne 0\hfill \end{array}$$

Example 4.

Find the Caputo fractional derivative of a constant function.

Solution

i.e, $f\left(x\right)=c$

$$\begin{array}{cc}\hfill {}_a^c{D}_x^{\alpha }f\left(x\right)& {=}_a{I}_x^{n-\alpha }\left[\frac{d^n}{d{t}^n}f\left(t\right)\right]dt\hfill \\ \hfill {}_a^c{D}_x^{\alpha }f\left(x\right)& =\frac{1}{\Gamma (n-\alpha )}{\int }_a^x{(x-t)}^{n-\alpha -1}\left[\frac{d^n}{d{t}^n}\left(c\right)\right]dt\hfill \end{array}$$

However, $\frac{d^n}{d{t}^n}\left(c\right)=0$

$$\begin{array}{cc}\hfill {}_a^c{D}_x^{\alpha }f\left(x\right)& =\frac{1}{\Gamma (n-\alpha )}{\int }_a^x{(x-t)}^{n-\alpha -1}\left[0\right]dt\hfill \\ \hfill {}_a^c{D}_x^{\alpha }f\left(x\right)& =0\hfill \end{array}$$

Hence, the partial Caputo fractional derivative of a constant is zero and the partial Riemann-Liouville fractional derivative of a constant is not zero.

Example 5.

Find the Caputo fractional derivative of $x^{\beta }$ for $\beta >-1$.

Solution

$$\begin{array}{cc}\hfill {}_a^c{D}_x^{\alpha }f\left(x\right)& {=}_a{I}_x^{n-\alpha }\left[\frac{d^n}{d{t}^n}f\left(t\right)\right]dt\hfill \\ \hfill {}_a^c{D}_x^{\alpha }\left(x^{\beta }\right)& {=}_a{I}_x^{n-\alpha }\left[\frac{d^n}{d{t}^n}{t}^{\beta }\right]dt\hfill \\ \hfill {}_a^c{D}_x^{\alpha }\left(x^{\beta }\right)& =\frac{1}{\Gamma (n-\alpha )}{\int }_a^x{(x-t)}^{n-\alpha -1}·\left(\frac{d^n}{d{t}^n}{t}^{\beta }\right)dt\hfill \\ \hfill {}_a^c{D}_x^{\alpha }\left(x^{\beta }\right)& =\frac{1}{\Gamma (n-\alpha )}{\int }_a^x{(x-t)}^{n-\alpha -1}·\frac{\beta !}{(\beta -n)!}{t}^{\beta -n}dt\hfill \\ \hfill {}_a^c{D}_x^{\alpha }\left(x^{\beta }\right)& =\frac{\beta !}{(\beta -n)!\Gamma (n-\alpha )}{\int }_a^x{(x-t)}^{n-\alpha -1}{t}^{\beta -n}dt\hfill \\ \hfill {}_a^c{D}_x^{\alpha }\left(x^{\beta }\right)& =\frac{\Gamma (\beta +1)}{\Gamma (1+\beta -n)\Gamma (n-\alpha )}{\int }_a^x{x}^{n-\alpha -1}{\left(x-\frac{t}{x}\right)}^{n-\alpha -1}{t}^{\beta -n}dt\hfill \end{array}$$

Put $\frac{t}{x}=y⟹t=xy⟹dt=xdy$

$$t=a⟶x$$

$$y=\frac{a}{x}⟶1$$

$$\begin{array}{cc}\hfill {}_a^c{D}_x^{\alpha }\left(x^{\beta }\right)& =\frac{\Gamma (\beta +1)}{\Gamma (1+\beta -n)\Gamma (n-\alpha )}{\int }_{\frac{a}{x}}^1{x}^{n-\alpha -1}{(1-y)}^{n-\alpha -1}{\left(xy\right)}^{\beta -n}dy\hfill \\ \hfill {}_a^c{D}_x^{\alpha }\left(x^{\beta }\right)& =\frac{\Gamma (\beta +1)}{\Gamma (1+\beta -n)\Gamma (n-\alpha )}{x}^{n-\alpha -1+\beta -n+1}{\int }_{\frac{a}{x}}^1{(1-y)}^{n-\alpha -1}{y}^{\beta -n}dy\hfill \\ \hfill {}_a^c{D}_x^{\alpha }\left(x^{\beta }\right)& =\frac{\Gamma (\beta +1)}{\Gamma (1+\beta -n)\Gamma (n-\alpha )}{x}^{\beta -\alpha }{\int }_{\frac{a}{x}}^1{(1-y)}^{n-\alpha -1}{y}^{\beta -n}dy\hfill \end{array}$$

Putting $a=0$ in the last equation above.

Recall that Beta Function

$$B(n-\alpha ,\beta -n+1)={\int }_0^1{(1-y)}^{n-\alpha -1}{y}^{\beta -n}dy=\frac{\Gamma (n-\alpha )\Gamma (\beta -n+1)}{\Gamma (n-\alpha +\beta -n+1)}$$

$$\begin{array}{cc}\hfill {}_a^c{D}_x^{\alpha }\left(x^{\beta }\right)& =\frac{\Gamma (\beta +1)}{\Gamma (1+\beta -n)\Gamma (n-\alpha )}{x}^{\beta -\alpha }·\frac{\Gamma (n-\alpha )\Gamma (\beta -n+1)}{\Gamma (n-\alpha +\beta -n+1)}\hfill \\ \hfill & =\frac{\Gamma (\beta +1)}{\Gamma (1+\beta -n)\Gamma (n-\alpha )}{x}^{\beta -\alpha }·\frac{\Gamma (n-\alpha )\Gamma (\beta -n+1)}{\Gamma (\beta -\alpha +1)}\hfill \\ \hfill & =\frac{\Gamma (\beta +1)}{\Gamma (1+\beta -\alpha )}{x}^{\beta -\alpha }\hfill \end{array}$$

The partial Caputo fractional derivative and partial Riemann-Liouville fractional derivative of a function are not the same.

Definition 10

([1,21] Set-Valued Maps). Let X and Y be two sets. A set valued map T from X to Y is a map that associates with any $x\in X$ a subset $T\left(x\right)$ of Y. The image is not a point but a set. We will always assume that $T\left(x\right)$ is a non-empty. We will denote by $2^Y$ the set of all subsets of Y. In this case write $T:X⟶2^Y$. Some examples of set-valued maps arise under several instances, these include inverse images of non-bijective functions, solution sets of metric projections, sub-differential map of a convex function, normal and tangent cone maps of a convex set, etc.

Let X and Y be topological spaces.

i. A set valued map $F:X⟶2^Y$ is said to be Closed valued, open valued or compact valued if, for each $x\in X$. $F\left(x\right)$ is a closed, open or compact set, respectively.

ii. A set valued map $F:X⟶2^Y$ is said to be Closed, open or compact set-valued map if the Graph(F) is a closed, open or compact set with respect to the product topology of X and Y.

F is bounded on bounded sets if and only if there exist $c>0$ such that $\parallel Fx\parallel \le {c\left|x\right|}_x$, for $x\in D\left(T\right)$

F is said to be completely continuous if for every bounded subset $M\subset X$, the image $F\left(M\right)$ is relatively compact that is the closure of $F\left(M\right)$ of $F\left(M\right)$ is compact.

Let X and Y be normed linear space and $T:X⟶Y$ be multivalued map. A point $x\in X$ is said to be a fixed point at T if $x\in T\left(x\right)$. The fixed point set of the multivalued operator T will be denoted by FixT.

A multivalued map $G:X⟶P_{cl}\left({\mathbb{R}}^n\right)$ is said to be measurable if for every $v\in {\mathbb{R}}^n$, the function $x\mapsto d(v,G(x\left)\right)=inf\{\parallel v-z\parallel :z\in G(x\left)\right\}$ is measurable.

Let X be a topological space and $f:X⟶\mathbb{R}$, we say that f is lower semi-continuous, if the set $\{x\in X:f(x)>a\}$ is open for every $a\in \mathbb{R}$. We say that f is upper semi-continuous, if the set $\{x\in X:f(x)<a\}$ is open for every $a\in \mathbb{R}$

For set-valued maps, the definition of upper semicontinuity and lower semicontinuity reads as follows:

Let X,Y be topological spaces and $T:X⟶2^Y$ be a set-valued mapping. We say that T is upper semicontinuous (u.s.c.) on X if for each $x_0\in X$, the set $T\left(x_0\right)$ is a nonempty closed subset of X, and if for each open set N of X containing $T\left(x_0\right)$, there exists an open neighbourhood $N_0$ of $x_0$ such that $T\left(N_0\right)\subseteq N$.

We say that T is lower semi-continuous (l.s.c.) if the set $\{x\in X:T(x)\cap A\ne \varnothing \}$ is open for any open subset $A\subseteq X$.

Theorem 1

([4] Completely continuous). If $F:X⟶{\mathcal{P}}_{cl}\left(Y\right)$ is upper semi-continuous (u.s.c.), then Graph(G) is a closed subset of $X\times Y$, i.e., for every sequence ${\left\{x_n\right\}}_{n\in \mathbb{N}}\subset X$ and ${\left\{y_n\right\}}_{n\in \mathbb{N}}\subset Y$, if when $n⟶\infty ,x_n⟶x_{*},y_n⟶y_{*}$ and $y_n\in F\left(x_n\right)$, then $y_{*}\in G\left(x_{*}\right)$. Conversely, if F is completely continuous and has a closed graph, then it is upper semi-continuous.

Let G be a completely continuous multivalued map with nonempty compact values, then G is upper semicontinuous (u.s.c) if and only if G has a closed graph, i.e., $u_n⟶u,w_n⟶w,w_n\in G\left(u_n\right)$ implies $w\in G\left(u\right).$

Definition 11

([4] Caratheodory). A multivalued map $F:J⟶\mathcal{P}\left({\mathbb{R}}^n\right)$ is said to be Caratheodory if

1. 

$(x,y)\mapsto F(x,y,u)$ is measurable for each $u\in {\mathbb{R}}^n,$

2. 

$u\mapsto F(x,y,u)$ is upper semicontinuous for almost all $(x,y)\in J,$

F is said to be $L^1$-Caratheodory if (1), (2) and the following condition holds;

3. 

For each $c>0$, $\exists {\sigma }_c\in L_1(J,R_{+})$ such that

$${\parallel F(x,y,u)\parallel }_P=sup\{\parallel f\parallel :f\in F(x,y,u)\}$$

$$\le {\sigma }_c(x,y)forall\parallel u\parallel \le candfora.e.(x,y)\in J,$$

For each $u\in C(J,{\mathbb{R}}^n),$ define the set of selection of F by

$$S_{F,u}=\{w\in L^1(J,{\mathbb{R}}^n):w(x,y)\in F(x,y,u_{(x,y)})a.e.(x,y)\in J$$

Let $F:J\times {\mathbb{R}}^n⟶P({\mathbb{R}}^n($ be a multivalued map with nonempty compact values. Assign to F the multivalued operator $\mathcal{F}:C(J,{\mathbb{R}}^n)⟶L^1(J,{\mathbb{R}}^n)$ by letting $\mathcal{F}\left(u\right)=S_{F,u}$. We say F is l.s.c. if $\mathcal{F}$ is l.s.c. and has nonempty closed and decomposable values.

The above operator $\mathcal{F}$ is called the Niemytzki operator associated to F.

A multivalued operator $N:X⟶P_{cl}\left(X\right)$ is called

(a) 

$\gamma -$ Lipschitz if and only if there exists $\gamma >0$ such that

$$H_d(N\left(u\right),N\left(v\right))\le \gamma d(u,v)foreachu,v\in X,$$

(b) 

A contraction if and only if it is $\gamma -Lipschitzwith\gamma <1.$

Definition 12

(Convex Set). A set $A\subset {\mathbb{R}}^n$ is said to be convex if $\lambda x+(1-\lambda )y\in A$ whenever $x\in A,y\in A,$ and $\lambda \in [0,1]$. By definition it follows that an intersection of any number of convex sets is a convex set, and if $A\subset {\mathbb{R}}^n,B\subset {\mathbb{R}}^n$ are convex, $\alpha$ and $\beta$ are real numbers, then the set $\alpha A+\beta B$ is convex. If A is convex, then interior of A and closure of A are also convex sets.

Let X, Y be non-empty sets and $F:X⟶\mathcal{P}\left(Y\right)$ be a multivalued function. The single-valued operator $f:X⟶Y$ is called a selection of F if and only if $f\left(x\right)\in F\left(x\right)$, for each $x\in X$. The set of all selection functions for F is denoted by $S_F$.

Lemma 1

(Mazur). Let E be a normed space and ${\left\{x\right\}}_{k\in \mathbb{N}}\subset E$ be a sequence weakly converging to a limit $x\in E$. Then there exists a sequence of convex combinations $y_m={\sum }_{k=1}^m{\alpha }_{mk}{x}_k$ with ${\alpha }_{mk}>0$ for $k=1,2,\dots ,m$ and ${\sum }_{k=1}^m{\alpha }_{mk}=1$, which converges strongly to x.

Definition 13

([2] Multivalued Version of Nonlinear Alternative of Leray Schauder Fixed Point). let X be a Banach space and C a nonempty convex subset of X. Let U be a nonempty open subset of C with $0\in U$ and $T:\overline{U}⟶P\left(C\right)$ an upper semicontinuous and compact multivalued operator. Then either

(a) 

T has fixed points, or

(b) 

There exist $u\in \partial U$ and $\lambda \in (0,1)$ with $u\in \lambda T\left(u\right)$

Lemma 2

(Covitz–Nadler Fixed Point). Let $(X,d)$ be a complete metric space. If $N:X⟶P_{cl}\left(X\right)$ is a contraction, then N has fixed points.

Auxiliary Results To define the solutions of the problem (1) to (2), we shall consider the space

$$PC(J,{\mathbb{R}}^n)=\{u:J⟶{\mathbb{R}}^n|u\in C(J_k,\mathbb{R});k=1,\dots ,m,$$

and there exist

$$u(x_k^{-},y)$$

and $u(x_k^{+},y);k=1,\dots ,m,$ with $u(x_k^{-},y)=u(x_k,y)\},$ where $J_k=(x_k,x_{k+1}]\times (0,b]$. This set is a Banach space with the norm

$${\parallel u\parallel }_{PC}={\displaystyle \underset{(x,y)\in J}{sup}}\parallel u(x,y)\parallel$$

Set J′:= $J\left|\right\{(x_1,y),\dots ,(x_n,y),y\in [0,b]\}.$

Definition 14

(Covitz–Nadler Fixed Point). A function $u\in PC(J,{\mathbb{R}}^n)$ whose r-derivative exists on J′ is said to be a solution of (1) and (2) if there exists a function $f\in L^1(J,\mathbb{R})$ with $f(x,y)\in F(x,y,u(x,y\left)\right)$ such that $\left({}^c{D}_0^{r}u\right)=f(x,y)$ on J′ and u satisfies conditions (2).

Let $h\in C([x_k,x_{k+1}]\times [0,b],{\mathbb{R}}^{\mathbb{n}}),z_k=(x_k,0),$ and

$${\mu }_k(x,y)=u(x,0)+u(x_k^{+},y)-u(x_k^{+},0),k=0,\dots ,m.$$

For the existence of solutions for the problem (1) and (2), we need the following lemma:

Definition 15.

A function $u\in AC([x_k,x_{k+1}]\times [0,b],\mathbb{R});k=0,\dots ,m$ is a solution of the differential equation

$$\left({}^c{D}_{z_k}^{r}u\right)=h(x,y);(x,y)\in [x_k,x_{k+1}]\times [0,b],$$

if and only if $u(x,y)$ satisfies

$$u(x,y)={\mu }_{k(x,y)}+\left(I_{zk}^{r}h\right)(x,y);(x,y)\in [x_k,x_{k+1}]\times [0,b].$$

(6)

Proof. 

Let u(x,y) be a solution of

$$\left({}^c{D}_{z_k}^{r}u\right)=h(x,y);(x,y)\in [x_k,x_{k+1}]\times [0,b],$$

Then, taking into account the definition of Caputo fractional derivative, i.e.,

$$\left({}^c{D}_{z_k}^{r}u\right)=I_{{z_k}^{+}}^{1-r}\left(D_{xy}^{2}u\right)(x,y)$$

we have

$$I_{{z_k}^{+}}^{1-r}\left(D_{xy}^{2}u\right)(x,y)=h(x,y)$$

Taking the Caputo integral of both sides, we obtain

$$I_{{zk}^{+}}^r\left(I_{zk}^{1-r}{D}_{xy}^{2}u\right)(x,y)=\left(I_{{zk}^{+}}^{r}h\right)(x,y),$$

Then

$$I_{{zk}^{+}}^1\left(D_{xy}^{2}u\right)(x,y)=\left(I_{{zk}^{+}}^{r}h\right)(x,y),$$

Recall that

$$I_{{zk}^{+}}^1\left(D_{x,y}^{2}u\right)(x,y)=u(x,y)-u(x,0)-u(x_k^{+},y)+u(x_k^{+},0),$$

So, we have

$$\begin{array}{cc}\hfill u(x,y)& =u(x,0)+u(x_k^{+},y)-u(x_k^{+},0)+I_{{zk}^{+}}^1\left(D_{x,y}^{2}u\right)(x,y)\hfill \\ \hfill u(x,y)& =u(x,0)+u(x_k^{+},y)-u(x_k^{+},0)+\left(I_{{zk}^{+}}^{r}h\right)(x,y),\hfill \\ \hfill u(x,y)& ={\mu }_k(x,y)+\left(I_{{zk}^{+}}^{r}h\right)(x,y).\hfill \end{array}$$

Now $u(x,y)$ satisfies (6). It is clear that u(x,y) satisfy

$$\left({}^c{D}_{z_k}^{r}u\right)=h(x,y)on[x_k,x_{k+1}]\times [0,b],$$

In all what follows set

$${\mu }_0(x,y):=\mu (x,y);(x,y)\in J$$

 □

Lemma 3.

Let $0<r_1,r_2\le 1$ and let $h:J⟶{\mathbb{R}}^n$ be continuous. A function u is a solution of the fractional integral equation

$$u(x,y)=\left\{\begin{array}{cc}\mu (x,y)+\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_0^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds;\hfill & if(x,y)\in [0,x_1]\times [0,b],\hfill \\ \mu (x,y)+{\displaystyle \sum _{i=1}^k}\left(I_i\left(u(x_i^{-},y)\right)-(I_i\left(u(x_i^{-},0)\right)\right)\hfill \\ +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{i=1}^k}{\int }_{x_{i-1}}^{x_i}{\int }_0^y{(x_i-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds\hfill \\ +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds;\hfill & if(x,y)\in (x_k,x_{k+1}]\times [0,b],k=1,\dots ,m.\hfill \end{array}\right.$$

(7)

If and only if u is a solution of the fractional IVP

$${}^c{D}^{r}u(x,y)=h(x,y),if(x,y)\in J^{\prime }$$

(8)

$$u(x_k^{+},y)=u(x_k^{-},y)+I_k\left(u(x_k^{-},y)\right),k=1,\dots ,m.$$

(9)

Proof. 

Assuming u satisfies (10) and (11). If $(x,y)\in [0,x_1]\times [0,b]$, then

$${}^c{D}^{r}u(x,y)=h(x,y)$$

Lemma 2 implies

$$u(x,y)=\mu (x,y)+\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_0^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds$$

Similarly, if $(x,y)\in (x_1,x_2]\times [0,b]$, from Lemma 3, we have

$$\begin{array}{cc}\hfill u(x,y)& ={\mu }_1(x,y)+\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_1}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds\hfill \\ \hfill & =\phi \left(x\right)+u(x_1^{+},y)-u(x_1^{+},0)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_1}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds\hfill \\ \hfill & =\phi \left(x\right)+u(x_1^{-},y)-u(x_1^{-},0)+I_1\left(u(x_1^{-},y)\right)-I_1\left(u(x_1^{-},0)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_1}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds\hfill \\ \hfill & =\phi \left(x\right)+u(x_1,y)-u(x_1,0)+I_1\left(u(x_1^{-},y)\right)-I_1\left(u(x_1^{-},0)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_1}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds\hfill \\ \hfill & =\mu (x,y)+I_1\left(u(x_1^{-},y)\right)-I_1\left(u(x_1^{-},0)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_0^x{\int }_0^y{(x_1-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_1}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds\hfill \end{array}$$

In addition, if $(x,y)\in (x_2,x_3]\times [0,b]$, from Lemma 3, we obtain

$$\begin{array}{cc}\hfill u(x,y)& ={\mu }_2(x,y)+\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_2}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds\hfill \\ \hfill & =\phi \left(x\right)+u(x_2^{+},y)-u(x_2^{+},0)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_2}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds\hfill \\ \hfill & =\phi \left(x\right)+u(x_2^{-},y)-u(x_2^{-},0)+I_2\left(u(x_2^{-},y)\right)-I_2\left(u(x_2^{-},0)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_2}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds\hfill \\ \hfill & =\phi \left(x\right)+u(x_2,y)-u(x_2,0)+I_2\left(u(x_2^{-},y)\right)-I_2\left(u(x_2^{-},0)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_2}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds\hfill \\ \hfill & =\mu (x,y)+I_2\left(u(x_2^{-},y)\right)-I_2\left(u(x_2^{-},0)\right)+I_1\left(u(x_1^{-},y)\right)-I_1\left(u(x_1^{-},0)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_0^{x_1}{\int }_0^y{(x_1-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_1}^{x_2}{\int }_0^y{(x_2-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_2}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}h(s,t)dtds\hfill \end{array}$$

If $(x,y)\in (x_k,x_{k+1}]\times [0,b]$, then again from Lemma 2 we obtain Lemma 3. □

3. Results

In this section, we present the existence result to impulsive fractional order IVP for the system (1) and (2).

To define the solutions of (1) and (2), we shall consider the Banach space $PC(J,{\mathbb{R}}^n)=\{u:J⟶{\mathbb{R}}^n|$ there exist $0=x_0<x_1<x_2<\cdots <x_m<x_{m+1}=a$ such that $u(x_k^{-},y)$ and $u(x_k^{+},y)$ exist with $u(x_k^{-},y)=u(x_k,y);k=0,\cdots ,m$ and $u\in C(J_k,{\mathbb{R}}^n);k=0,\cdots ,m\}.$

Definition 16.

A function $u\in PC(J,{\mathbb{R}}^n)$ whose r-derivative exist on $J^{\prime }$ is a said to be a solution of (1) and (2) if there exists a function $f\in (J,{\mathbb{R}}^n)$ with $f(x,y)\in F(x,y,u(x,y\left)\right)$ such that $\left({}^c{D}_{x_k}^{r}u\right)(x,y)=f(x,y)$ on $J^{\prime }$ and u satisfies condition (2).

The Convex Case Here, we considered the existence of solutions for the IVP (1) and (2) when the right-hand side is compact and convex valued.

Theorem 2.

Assume the following hypotheses hold:

• H1: $F:J_k\times \mathbb{R}⟶P_{cp,cv}\left(\mathbb{R}\right),k=0,\cdots ,m$ is a Caratheodory multivalued map.
• H2: There exist $p\in L^{\infty }(J,{\mathbb{R}}_{+})$ and ${\psi }_{*}:[0,\infty )⟶(0,\infty )$ continuous and non decreasing such that

  $${\parallel F(x,y,u)\parallel }_p\le p(x,y){\psi }_{*}(\parallel u\parallel )$$

  for $(x,y)\in J,x\ne x_k,k=0,\cdots ,m$ and each $u\in {\mathbb{R}}^n$
• H3: There exist $l\in L^{\infty }(J,{\mathbb{R}}_{+})$ such that

  $$H_d(F(x,y,u),F(x,y,\overline{u}))\le l(x,y)\parallel u-\overline{u}\parallel a.e.u,\overline{u}\in {\mathbb{R}}^n$$

  and

  $$d(0,F(x,y,0\left)\right)\le l(x,y),a.e.(x,y)\in J$$
• H4: There exist constant $q_k,$ such that $\parallel I_k\left(u\right)\parallel \le q_k,k=1,\cdots ,m$ for each $u\in {\mathbb{R}}^n.$
• H5: There exists a constant $q_k^{*}$, such that

  $$\parallel I_k\left(u\right)-I_k\left(\overline{u}\right)\parallel \le q_k^{*}\parallel u-\overline{u}\parallel ,foreachu,\overline{u}\in {\mathbb{R}}^n,k=1,\cdots ,m.$$
• H6: There exists a number $M>0$ such that

  $$\frac{M}{{\parallel \mu \parallel }_{\infty }+2{\sum }_{k=1}^m{q}_k+\frac{2{\psi }_{*}\left(M\right)p^{*}{a}^{r_1}{b}^{r_2}}{\Gamma (r_1+1)\Gamma (r_2+1)}}>1,$$

  where $p^{*}={\parallel p\parallel }_{\infty }$. Then the IVP (1) and (2) have at least one solution on J.

Proof of Theorem 1.

We transform the problem (1) and (2) into a fixed point problem. Consider the setvalued operator $N:PC(J,{\mathbb{R}}^n)⟶P\left(PC(J,{\mathbb{R}}^n)\right)$ defined by $N\left(u\right)=\{h\in PC(J,{\mathbb{R}}^n)\}$ where f $\in S_{F,u}$.

$$\begin{array}{cc}\hfill h\left(u\right)(x,y)& =\mu (x,y)+{\displaystyle \sum _{0<x_k<x}}\left(I_k\left(u(x_k^{-},y)\right)-I_k\left(u(x_k^{-},0)\right)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}f(s,t)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}f(s,t)dtds.\hfill \end{array}$$

We shall show that operator N satisfies the assumptions of the nonlinear alternative of Leray–Schauder fixed point. The proof of this theorem will be given in several steps.

Step 1: We show that operator $N\left(u\right)$ is convex for each $u\in PC(J,{\mathbb{R}}^n)$.

Indeed, if $h_1,h_2$ belong to $N\left(u\right)$, then there exists $f_1,f_2\in S_{F,u}$ such that for each $(x,y)\in J$, we have:

$$\begin{array}{cc}\hfill h_i\left(u\right)(x,y)& =\mu (x,y)+{\displaystyle \sum _{0<x_k<x}}\left(I_k\left(u(x_k^{-},y)\right)-I_k\left(u(x_k^{-},0)\right)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}{f}_i(s,t)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}{f}_i(s,t)dtds.\hfill \end{array}$$

where $f_i\in S_{F,u},i=1,2$.

Let $0\le d\le 1$. Then for each $(x,y)\in J,$ we have:

$$\begin{array}{cc}\hfill (d{h}_1& +(1-d)h_2)(x,y)=\mu (x,y)+{\displaystyle \sum _{0<x_k<x}}\left(I_k\left(u(x_k^{-},y)\right)-I_k\left(u(x_k^{-},0)\right)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}\left(d{f}_1(s,t)+(1-d)f_2(s,t)\right)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}\left(d{f}_1(s,t)+(1-d)f_2(s,t)\right)dtds\hfill \end{array}$$

Since $S_{F,u}$ is convex (because F has convex values), we have

$$d{h}_1+(1-d)h_2\in N\left(u\right)$$

Step 2: We show that operator N maps bounded sets into bounded set in $PC(J,{\mathbb{R}}^n).$

Let $B_{{\eta }^{*}}=\{u\in PC(J,{\mathbb{R}}^n){:\parallel u\parallel }_{\infty }\le {\eta }^{*}\}$ be bounded set in $PC(J,{\mathbb{R}}^n)$ and $u\in B_{{\eta }^{*}}$. Then for each $h\in N\left(u\right)$, there exists $f\in S_{F,u}$ such that

$$\begin{array}{cc}\hfill h\left(u\right)(x,y)& =\mu (x,y)+{\displaystyle \sum _{0<x_k<x}}\left(I_k\left(u(x_k^{-},y)\right)-I_k\left(u(x_k^{-},0)\right)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}f(s,t)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}f(s,t)dtds.\hfill \end{array}$$

for $(x,y)\in J$. Now, we have

$$\begin{array}{cc}\hfill \parallel h(x,y)\parallel & =\parallel \mu (x,y)+{\displaystyle \sum _{0<x_k<x}}\left(I_k\left(u(x_k^{-},y)\right)-I_k\left(u(x_k^{-},0)\right)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}f(s,t)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}f(s,t)dtds\parallel \hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel h(x,y)\parallel & \le \parallel \mu (x,y)\parallel +{\displaystyle \sum _{0<x_k<x}}\left(\parallel I_k\left(u(x_k^{-},y)\right)\parallel +\parallel I_k\left(u(x_k^{-},0)\right)\parallel \right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}\parallel f(s,t)\parallel dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}\parallel f(s,t)\parallel dtds\hfill \end{array}$$

By applying hypotheses H2 and H4, we have

$$\begin{array}{cc}\hfill \parallel h(x,y)\parallel & \le \parallel \mu (x,y)\parallel +{\displaystyle \sum _{k=1}^m}\left(\parallel I_k\left(u(x_k^{-},y)\right)\parallel +\parallel I_k\left(u(x_k^{-},0)\right)\parallel \right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}p(s,t){\psi }_{*}(\parallel u\parallel )dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}{\psi }_{*}(\parallel u\parallel )dtds\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel h(x,y)\parallel & \le \parallel \mu (x,y)\parallel +{\displaystyle \sum _{k=1}^m}(q_k+q_k)\hfill \\ \hfill & +\frac{p^{*}{\psi }_{*}(\parallel u\parallel )}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}dtds\hfill \\ \hfill & +\frac{p^{*}{\psi }_{*}(\parallel u\parallel )}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}dtds\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel h(x,y)\parallel & \le \parallel \mu (x,y)\parallel +2{\displaystyle \sum _{k=1}^m}{q}_k+\frac{p^{*}{\psi }_{*}(\parallel u\parallel )}{r_1{r}_2\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{(x_k-s)}^{r_1}{{|}_0^x{(y-t)}^{r_2}|}_0^y\hfill \\ \hfill & +\frac{p^{*}{\psi }_{*}(\parallel u\parallel )}{r_1{r}_2\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{(x-s)}^{r_1}{{|}_0^x{(y-t)}^{r_2}|}_0^y\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel h(x,y)\parallel & \le \parallel \mu (x,y)\parallel +2{\displaystyle \sum _{k=1}^m}{q}_k+\frac{a^{r_1}{b}^{r_2}{p}^{*}{\psi }_{*}(\parallel u\parallel )}{\Gamma (r_1+1)\Gamma (r_2+1)}+\frac{a^{r_1}{b}^{r_2}{p}^{*}{\psi }_{*}(\parallel u\parallel )}{\Gamma (r_1+1)\Gamma (r_2+1)}\hfill \end{array}$$

Thus,

$${\parallel h\parallel }_{\infty }\le {\parallel \mu \parallel }_{\infty }+2{\displaystyle \sum _{k=1}^m}{q}_k+\frac{2p^{*}{\psi }_{*}(\parallel u\parallel )a^{r_1}{b}^{r_2}}{\Gamma (r_1+1)\Gamma (r_2+1)}:=\ell .$$

Step 3: We show that operator N maps bounded sets into equicontinuous sets of $PC(J,{\mathbb{R}}^n)$.

Let $({\tau }_1,y_1)\in J$ and $({\tau }_2,y_2)\in J$, ${\tau }_1<{\tau }_2$ and $y_1<y_2,B_{{\eta }^{*}}$ be a bounded set of $PC(J,{\mathbb{R}}^n)$ as in step 2 above. Let $u\in B_{{\eta }^{*}}andh\in N\left(u\right)$, then for each $(x,y)\in J$, we have

$$\begin{array}{cc}\hfill \parallel h\left(u\right)& ({\tau }_2,y_2)-h\left(u\right)({\tau }_1,y_1)\parallel =\parallel \mu ({\tau }_2,y_2)-\mu ({\tau }_1,y_1)+{\displaystyle \sum _{k=1}^m}\left(I_k(u(x_k^{-},y_1)-I_k(u(x_k^{-},y_2)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}\left[{(y_2-t)}^{r_2-1}-{(y_1-t)}^{r_2-1}\right]f(s,t,u(s,t))dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{\int }_{x_{k-1}}^{x_k}{\int }_{y_1}^{y_2}{(x_k-s)}^{r_1-1}{(y_2-t)}^{r_2-1}f(s,t,u(s,t))dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_0^{{\tau }_1}{\int }_0^{y_1}\left[{({\tau }_2-s)}^{r_1-1}{(y_2-t)}^{r_2-1}-{({\tau }_1-s)}^{r_1-1}{(y_1-t)}^{r_2-1}\right]f(s,t,u(s,t))dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{{\tau }_1}^{{\tau }_2}{\int }_{y_1}^{y_2}{({\tau }_2-s)}^{r_1-1}{(y_2-t)}^{r_2-1}f(s,t,u(s,t))dtds\parallel \hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel h& \left(u\right)({\tau }_2,y_2)-h\left(u\right)({\tau }_1,y_1)\parallel \le \parallel \mu ({\tau }_2,y_2)-\mu ({\tau }_1,y_1)\parallel +{\displaystyle \sum _{k=1}^m}\left[\parallel I_k(u(x_k^{-},y_1)-I_k(u(x_k^{-},y_2)\parallel \right]\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{\int }_{x_{k-1}}^{x_k}{\int }_0^{y_1}{(x_k-s)}^{r_1-1}\left[{(y_2-t)}^{r_2-1}-{(y_1-t)}^{r_2-1}\right]\parallel f(s,t,u(s,t))\parallel dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{\int }_{x_{k-1}}^{x_k}{\int }_{y_1}^{y_2}{(x_k-s)}^{r_1-1}{(y_2-t)}^{r_2-1}\parallel f(s,t,u(s,t))\parallel dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_0^{{\tau }_1}{\int }_0^{y_1}\left[{({\tau }_2-s)}^{r_1-1}{(y_2-t)}^{r_2-1}-{({\tau }_1-s)}^{r_1-1}{(y_1-t)}^{r_2-1}\right]\parallel f(s,t,u(s,t))\parallel dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{{\tau }_1}^{{\tau }_2}{\int }_{y_1}^{y_2}{({\tau }_2-s)}^{r_1-1}{(y_2-t)}^{r_2-1}\parallel f(s,t,u(s,t))\parallel dtds\hfill \end{array}$$

By hypotheses (H2) and (H4), we have

$$\begin{array}{cc}\hfill \parallel h& \left(u\right)({\tau }_2,y_2)-h\left(u\right)({\tau }_1,y_1)\parallel \le \parallel \mu ({\tau }_1,y_1)-\mu ({\tau }_2,y_2)\parallel +{\displaystyle \sum _{k=1}^m}\left[\parallel I_k(u(x_k^{-},y_1)-I_k(u(x_k^{-},y_2)\parallel \right]\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{\int }_{x_{k-1}}^{x_k}{\int }_{y_1}^{y_2}{(x_k-s)}^{r_1-1}{(y_2-t)}^{r_2-1}p(s,t){\psi }_{*}(\parallel u\parallel )dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{\int }_{x_{k-1}}^{x_k}{\int }_0^{y_1}{(x_k-s)}^{r_1-1}\left[{(y_2-t)}^{r_2-1}-{(y_1-t)}^{r_2-1}\right]p(s,t){\psi }_{*}(\parallel u\parallel )dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_0^{{\tau }_1}{\int }_0^{y_1}\left[{({\tau }_2-s)}^{r_1-1}{(y_2-t)}^{r_2-1}-{({\tau }_1-s)}^{r_1-1}{(y_1-t)}^{r_2-1}\right]p(s,t){\psi }_{*}(\parallel u\parallel )dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{{\tau }_1}^{{\tau }_2}{\int }_{y_1}^{y_2}{({\tau }_2-s)}^{r_1-1}{(y_2-t)}^{r_2-1}p(s,t){\psi }_{*}(\parallel u\parallel )dtds\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel h\left(u\right)& ({\tau }_2,y_2)-h\left(u\right)({\tau }_1,y_1)\parallel \le \parallel \mu ({\tau }_1,y_1)-\mu ({\tau }_2,y_2)\parallel +{\displaystyle \sum _{k=1}^m}\left[\parallel I_k(u(x_k^{-},y_1)-I_k(u(x_k^{-},y_2)\parallel \right]\hfill \\ \hfill & +\frac{p^{*}{\psi }_{*}(\parallel u\parallel )}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{\int }_{x_{k-1}}^{x_k}{\int }_0^{y_1}{(x_k-s)}^{r_1-1}\left[{(y_2-t)}^{r_2-1}-{(y_1-t)}^{r_2-1}\right]dtds\hfill \\ \hfill & +\frac{p^{*}{\psi }_{*}(\parallel u\parallel )}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{\int }_{x_{k-1}}^{x_k}{\int }_{y_1}^{y_2}{(x_k-s)}^{r_1-1}{(y_2-t)}^{r_2-1}dtds\hfill \\ \hfill & +\frac{p^{*}{\psi }_{*}(\parallel u\parallel )}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_0^{{\tau }_1}{\int }_0^{y_1}\left[{({\tau }_2-s)}^{r_1-1}{(y_2-t)}^{r_2-1}-{({\tau }_1-s)}^{r_1-1}{(y_1-t)}^{r_2-1}\right]dtds\hfill \\ \hfill & +\frac{p^{*}{\psi }_{*}(\parallel u\parallel )}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{{\tau }_1}^{{\tau }_2}{\int }_{y_1}^{y_2}{({\tau }_2-s)}^{r_1-1}{(y_2-t)}^{r_2-1}dtds\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel h\left(u\right)& ({\tau }_2,y_2)-h\left(u\right)({\tau }_1,y_1)\parallel \le \parallel \mu ({\tau }_1,y_1)-\mu ({\tau }_2,y_2)\parallel +{\displaystyle \sum _{k=1}^m}\left[\parallel I_k(u(x_k^{-},y_1)-I_k(u(x_k^{-},y_2)\parallel \right]\hfill \\ \hfill & +\frac{p^{*}{\psi }_{*}\left({\eta }^{*}\right)}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{\int }_{x_{k-1}}^{x_k}{\int }_0^{y_1}{(x_k-s)}^{r_1-1}\left[{(y_2-t)}^{r_2-1}-{(y_1-t)}^{r_2-1}\right]dtds\hfill \\ \hfill & +\frac{p^{*}{\psi }_{*}\left({\eta }^{*}\right)}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{\int }_{x_{k-1}}^{x_k}{\int }_{y_1}^{y_2}{(x_k-s)}^{r_1-1}\left[{(y_2-t)}^{r_2-1}-{(y_1-t)}^{r_2-1}\right]dtds\hfill \\ \hfill & +\frac{p^{*}{\psi }_{*}\left({\eta }^{*}\right)}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_0^{{\tau }_1}{\int }_0^{y_1}\left[{({\tau }_2-s)}^{r_1-1}{(y_2-t)}^{r_2-1}-{({\tau }_1-s)}^{r_1-1}{(y_1-t)}^{r_2-1}\right]dtds\hfill \\ \hfill & +\frac{p^{*}{\psi }_{*}\left({\eta }^{*}\right)}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{{\tau }_1}^{{\tau }_2}{\int }_{y_1}^{y_2}{({\tau }_2-s)}^{r_1-1}{(y_2-t)}^{r_2-1}dtds\hfill \end{array}$$

As ${\tau }_1⟶{\tau }_2$ and $y_1⟶y_2,$ the right hand side of the above inequality tends to zero, which yields equicontinuity of N. By invoking the Arzela-Ascoli theorem and as a consequence of steps 1–3, we conclude that operator N is completely continuous.

Step 4: We show that operator N has a closed graph.

Let $u_n⟶u_{*},h_n\in N\left(u_n\right)$ and $h_n⟶h_{*}$. We shall show that $h_{*}\in N\left(u_{*}\right)$. If $h_n\in N\left(u_n\right)$, it implies there must exists $f\in S_{F,u_n}$ such that for each $(x,y)\in J$

$$\begin{array}{cc}\hfill h_n(x,y)& =\mu (x,y)+{\displaystyle \sum _{0<x_k<x}}\left(I_k\left(u_n(x_k^{-},y)\right)-I_k\left(u_n(x_k^{-},0)\right)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}{f}_n(s,t)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}{f}_n(s,t)dtds.\hfill \end{array}$$

We need to show that there exists $f_{*}\in S_{F,u}$ such that, for each $(x,y)\in J.$

$$\begin{array}{cc}\hfill h_{*}(x,y)& =\mu (x,y)+{\displaystyle \sum _{0<x_k<x}}\left(I_k\left(u_{*}(x_k^{-},y)\right)-I_k\left(u_{*}(x_k^{-},0)\right)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}{f}_{*}(s,t)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}{f}_{*}(s,t)dtds.\hfill \end{array}$$

Since $F(x,y,·)$ is upper semicontinuous, for every $ϵ>0$, there exist $n_0\left(ϵ\right)\ge 0$ such that for every $n\ge n_0,$ we have:

$$f_n(x,y)\in F(x,y,u(x,y))\subset F(x,y,u_{*}(x,y))+ϵB(0,1)a.e.(x,y)\in J$$

Since $F(·,·,·)$ has compact values, then there must exist a subsequence $f_{n_m}$ such that $f_{n_m}(·,·)⟶f_{*}(x,y)$ as $m⟶\infty$ and

$$f_{*}(x,y)\in F(x,y,u_{*}(x,y)),\mathrm{a}.\mathrm{e}.(x,y)\in J\mathrm{for}\mathrm{every}w(x,y)\in F(x,y,u_{*}(x,y)).$$

We have

$$\begin{array}{cc}\hfill \parallel f_{n_m}(x,y)-u_{*}(x,y)\parallel & =\parallel f_{n_m}(x,y)-w(x,y)+w(x,y)-f_{*}(x,y)\parallel \hfill \\ \hfill & \le \parallel f_{n_m}(x,y)-w(x,y)\parallel +\parallel w(x,y)-f_{*}(x,y)\parallel \hfill \end{array}$$

Then,

$$\parallel f_{n_m}(x,y)-f_{*}(x,y)\parallel \le d\left(f_{n_m}(x,y),F(x,y,u_{*}(x,y))\right)$$

By a comparable relation, acquired by switching the positions of $f_{n_m}$ and $f_{*}$, it implies that

$$\begin{array}{cc}\hfill \parallel f_{n_m}(x,y)-u_{*}(x,y)\parallel & \le H_d\left(F(x,y,u_n(x,y)),F(x,y,u_{*}(x,y))\right)\hfill \\ \hfill & \le l(x,y)\parallel u_n-u_{*}{\parallel }_{\infty }.\hfill \end{array}$$

Let $l^{*}:={\parallel l\parallel }_{L^{\infty }}$, then by (H3) and (H5). We have for each $(x,y)\in J,$

$$\begin{array}{cc}\hfill \parallel h_{mn}(x,y)& -h_{*}(x,y)\parallel =\parallel \mu (x,y)-\mu (x,y)+{\displaystyle \sum _{0<x_k<x}}\left(I_k\left(u_n(x_k^{-},y)\right)-I_k\left(u_{*}(x_k^{-},0)\right)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}\left[f_{n_m}(s,t)-f_{*}(s,t)\right]dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}\left[f_{n_m}(s,t)-f_{*}(s,t)\right]dtds\hfill \end{array}$$

$$\begin{array}{cc}\hfill & \le {\displaystyle \sum _{k=1}^m}\parallel I_k\left(u_n(x_k^{-},y)\right)-I_k\left(u_{*}(x_k^{-},y)\right)\parallel +{\displaystyle \sum _{k=1}^m}\parallel I_k\left(u_n(x_k^{-},0)\right)-I_k\left(u_{*}(x_k^{-},0)\right)\parallel \hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}\parallel \left[f_{n_m}(s,t)-f_{*}(s,t)\right]\parallel dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}\parallel \left[f_{n_m}(s,t)-f_{*}(s,t)\right]\parallel dtds\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel h_{mn}(x,y)-h_{*}(x,y)\parallel & \le {\displaystyle \sum _{k=1}^m}\parallel u_{nm}-u_{*}{\parallel }_{\infty }{q}_k^{*}+{\displaystyle \sum _{k=1}^m}{\parallel u_{nm}-u_{*}\parallel }_{\infty }{q}_k^{*}\hfill \\ \hfill \parallel h_{mn}(x,y)-h_{*}(x,y)\parallel & +\frac{\parallel u_{nm}-u_{*}{\parallel }_{\infty }}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_0^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}l(s,t)dtds\hfill \\ \hfill \parallel h_{mn}(x,y)-h_{*}(x,y)\parallel & +\frac{\parallel u_{nm}-u_{*}{\parallel }_{\infty }}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_0^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}l(s,t)dtds\hfill \\ \hfill \parallel h_{mn}-h_{*}\parallel & \le 2{\displaystyle \sum _{k=1}^m}{\parallel u_{nm}-u_{*}\parallel }_{\infty }{q}_k^{*}\hfill \\ \hfill \parallel h_{mn}-h_{*}\parallel & +\frac{2\parallel u_{nm}-u_{*}{\parallel }_{\infty }}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_0^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}l(s,t)dtds\hfill \\ \hfill \parallel h_{mn}-h_{*}\parallel & \le 2\parallel u_{nm}-u_{*}{\parallel }_{\infty }{\displaystyle \sum _{k=1}^m}{q}_k^{*}+\frac{2a^{r_1}{b}^{r_2}{l}^{*}{\parallel u_{nm}-u_{*}\parallel }_{\infty }}{r_1{r}_2\Gamma \left(r_1\right)\Gamma \left(r_2\right)}\hfill \\ \hfill \parallel h_{mn}-h_{*}\parallel & \le 2\parallel u_{nm}-u_{*}{\parallel }_{\infty }{\displaystyle \sum _{k=1}^m}{q}_k^{*}+\frac{2a^{r_1}{b}^{r_2}{l}^{*}{\parallel u_{nm}-u_{*}\parallel }_{\infty }}{\Gamma (r_1+1)\Gamma (r_2+1)}\hfill \end{array}$$

$$\begin{array}{c}\hfill \parallel h_{mn}-h_{*}\parallel \le 2\parallel u_{nm}-u_{*}{\parallel }_{\infty }{\displaystyle \sum _{k=1}^m}{q}_k^{*}+\frac{2a^{r_1}{b}^{r_2}{l}^{*}{\parallel u_{nm}-u_{*}\parallel }_{\infty }}{r_1{r}_2\Gamma \left(r_1\right)\Gamma \left(r_2\right)}\end{array}$$

$$\begin{array}{cc}\hfill \parallel h_{mn}-h_{*}\parallel & \le 2\parallel u_{nm}-u_{*}{\parallel }_{\infty }{\displaystyle \sum _{k=1}^m}{q}_k^{*}+\frac{2a^{r_1}{b}^{r_2}{l}^{*}{\parallel u_{nm}-u_{*}\parallel }_{\infty }}{\Gamma (r_1+1)\Gamma (r_2+1)}\hfill \end{array}$$

Hence

$$\parallel h_{nm}-h_{*}{\parallel }_{\infty }\le \left[2{\displaystyle \sum _{k=1}^m}{q}_k^{*}+\frac{2l^{*}{a}^{r_1}{b}^{r_2}}{\Gamma (r_1+1)\Gamma (r_2+1)}\right]{\parallel u_{nm}-u_{*}\parallel }_{\infty }$$

$$⟶0asm⟶\infty .$$

Step 5: A priori bounds on solutions.

Let u be a possible solution to the problems (1) and (2). Then, there exists $f\in S_{F,u}$ such that, for each $(x,y)\in J$, we have

$$\begin{array}{cc}\hfill u(x,y)& =\mu (x,y)+{\displaystyle \sum _{x_1<x_k<x}}\left(I_k\left(u(x_k^{-},y)\right)-I_k(x_k^{-},0)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{x_1<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}f(s,t)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}f(s,t)dtds\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel u(x,y)\parallel & =\parallel \mu (x,y)+{\displaystyle \sum _{x_1<x_k<x}}\left(I_k\left(u(x_k^{-},y)\right)-I_k(x_k^{-},0)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{x_1<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}f(s,t)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}f(s,t)dtds\parallel \hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel u(x,y)\parallel & \le \parallel \mu (x,y)\parallel +{\displaystyle \sum _{x_1<x_k<x}}\left(\parallel I_k(u(x_k^{-},y)\parallel +\parallel I_k(x_k^{-},0)\parallel \right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{x_1<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}\parallel f(s,t)\parallel dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}\parallel f(s,t)\parallel dtds\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel u(x,y)\parallel & \le \parallel \mu (x,y)\parallel +{\displaystyle \sum _{x_1<x_k<x}}\left(\parallel I_k(u(x_k^{-},y)\parallel +\parallel I_k(x_k^{-},0)\parallel \right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{x_1<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}p(s,t){\psi }_{*}(\parallel u\parallel )dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}p(s,t){\psi }_{*}(\parallel u\parallel )dtds\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel u(x,y)\parallel & \le \parallel \mu (x,y)\parallel +{\displaystyle \sum _{k=1}^m}(q_k^{*}+q_k^{*})+\frac{p^{*}{\psi }_{*}(\parallel u\parallel )}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}dtds\hfill \\ \hfill & +\frac{p^{*}{\psi }_{*}(\parallel u\parallel )}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}dtds\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel u(x,y)\parallel & \le \parallel \mu (x,y)\parallel +2{\displaystyle \sum _{k=1}^m}{q}_k^{*}+\frac{p^{*}{a}^{r_1}{b}^{r_2{\psi }_{*}(\parallel u\parallel )}}{r_1{r}_2\Gamma \left(r_1\right)\Gamma \left(r_2\right)}+\frac{p^{*}{a}^{r_1}{b}^{r_2{\psi }_{*}(\parallel u\parallel )}}{r_1{r}_2\Gamma \left(r_1\right)\Gamma \left(r_2\right)}\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel u(x,y)\parallel & \le \parallel \mu (x,y)\parallel +2{\displaystyle \sum _{k=1}^m}{q}_k^{*}+\frac{2p^{*}{a}^{r_1}{b}^{r_2}{\psi }_{*}(\parallel u\parallel )}{\Gamma (r_1+1)\Gamma (r_2+1)}\hfill \\ \hfill {\parallel u\parallel }_{\infty }& \le {\parallel \mu \parallel }_{\infty }+2{\displaystyle \sum _{k=1}^m}{q}_k^{*}+\frac{2p^{*}{a}^{r_1}{b}^{r_2}{\psi }_{*}{(\parallel u\parallel }_{\infty })}{\Gamma (r_1+1)\Gamma (r_2+1)}\hfill \end{array}$$

Then,

$$\frac{{\parallel u\parallel }_{\infty }}{{\parallel \mu \parallel }_{\infty }+2{\sum }_{k=1}^m{q}_k^{*}+\frac{2p^{*}{a}^{r_1}{b}^{r_2}{\psi }_{*}{(\parallel u\parallel }_{\infty })}{\Gamma (r_1+1)\Gamma (r_2+1)}}\le 1.$$

Thus, there exists M such that ${\parallel u\parallel }_{\infty }\ne M$.

$$\mathrm{Let}U=\{u\in PC(J,{\mathbb{R}}^n):\parallel u{\parallel }_{\infty }<M\}$$

The operator $N:\overline{U}⟶P\left(PC(J,{\mathbb{R}}^n)\right)$ is upper semicontinuous and completely continuous. From the choice of U, there is no $u\in \partial U$ such that $u\in \lambda N\left(u\right)$ for some $\lambda \in (0,1)$. As consequence of the nonlinear alternative of the Lerey-Schauder fixed point theorem, we deduce that N has a fixed point $u\in \overline{U}$ which is a solution of the problem (1) and (2). □

The nonconvex case. Our results are based on the fixed-point theorem for contraction setvalued maps proved by Covitz and Nadler, and they are applicable to problems (1) and (2) with a nonconvex valued right hand side.

Theorem 3.

Assuming the below axiom and hypothesis H3 of Theorem 3 holds: $F:J\times {\mathbb{R}}^n⟶P_{cp}\left({\mathbb{R}}^n\right)$ has the property that $F(·,·,u):J⟶P_{cp}\left({\mathbb{R}}^n\right)$ is measurable.

If

$$2{\displaystyle \sum _{k=1}^m}{q}_k^{*}+\frac{2l^{*}{a}^{r_1}{b}^{r_2}}{\Gamma (r_1+1)\Gamma (r_2+1)}<1$$

(10)

then the IVP (1) and (2) have at least one solution on J.

Remarks: For each $u\in PC(J,{\mathbb{R}}^n)$, the set $S_{F,u}$ is nonempty since by (1), F has a measurable selection.

Proof of Theorem 2.

We will show that operator N, as described in Theorem 3, satisfies the assumptions of Lemma 2. The proof will be presented in two steps.

Step 1: Let $N\left(u\right)\in P_{cl}\left(PC(J,{\mathbb{R}}^n)\right)$ for each $u\in PC(J,{\mathbb{R}}^n)$. Indeed, let ${\left(u_n\right)}_n\ge 0\in N\left(u\right)$ such that $u_n⟶\overline{u}inPC(J,{\mathbb{R}}^n)$. Then $\overline{u}\in PC(J,{\mathbb{R}}^n)$ and there exists $f_n\in S_{F,u}$ such that for each $(x,y)\in J.$

$$u_n(x,y)=\mu (x,y)+{\displaystyle \sum _{0<x_k<x}}\left(I_k\left(u(x_k^{-},y)\right)-I_k\left(u(x_k^{-},0)\right)\right)$$

$$+\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}{f}_n(s,t)dtds$$

$$+\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}{f}_n(s,t)dtds$$

We can pass to a subsequence if necessary to obtain that $f_n(·,·)$ converges weakly to $f\in L_w^1(J,{\mathbb{R}}^n)$ [the space endowed with the weak topology] by exploiting the fact that F has compact values and from (H3). A common argument infers that $f_n(·,·)$ strongly converges to f and therefore $f\in S_{F,u}.$ Then, for each $(x,y)\in J,$

$$u_n(x,y)⟶\overline{u}(x,y),$$

where

$$\begin{array}{cc}\hfill \overline{u}(x,y)& =\mu (x,y)+{\displaystyle \sum _{0<x_k<x}}\left(I_k\left(u(x_k^{-},y)\right)-I_k\left(u(x_k^{-},0)\right)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}{f}_n(s,t)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}{f}_n(s,t)dtds\hfill \end{array}$$

So, $\overline{u}\in N\left(u\right).$

Step 2: There exist $\gamma <1$ such that

$$H_d(N\left(u\right),N\left(\overline{u}\right))\le \gamma {\parallel u-\overline{u}\parallel }_{\infty }\mathrm{for}\mathrm{each}u,\overline{u}\in PC(J,{\mathbb{R}}^n)$$

Let $u,\overline{u}\in PC(J,{\mathbb{R}}^n)$ and $h\in N\left(u\right)$. Then there exists $f(x,y)\in F(x,y,u(x,y\left)\right)$ such that for each $(x,y)\in J$

$$\begin{array}{cc}\hfill h(x,y)& =\mu (x,y)+{\displaystyle \sum _{0<x_k<x}}\left(I_k\left(u(x_k^{-},y)\right)-I_k\left(u(x_k^{-},0)\right)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}f(s,t)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}f(s,t)dtds\hfill \end{array}$$

From (H3), it follows that

$$H_d\left(F(x,y,u(x,y))\right),F(x,y,\overline{u}(x,y))\le l(x,y)\parallel u(x,y)-\overline{u}(x,y)\parallel$$

Hence, there exist $w(x,y)\in F(x,y,\overline{u}(x,y))$ such that

$$\parallel f(x,y)-w(x,y)\parallel \le l(x,y)\parallel u(x,existsverlineu(x,y)\parallel ,(x,y)\in J.$$

Consider $U:J⟶P\left({\mathbb{R}}^n\right)$ given by

$$U(x,y)=\{w\in PC(J,{\mathbb{R}}^n):\parallel f(x,y)-w(x,y)\parallel \le l(x,y)\parallel u(x,y)-\overline{u}(x,y)\parallel \}.$$

Since the multivalued operator $u(x,y)=U(x,y)\cap F(x,y,\overline{u}(x,y))$ is measurable, there exists a function $u_2(x,y)$ which is measurable selection for u. So $f(x,y)\in F(x,y,\overline{u}(x,y))$ and for each $(x,y)\in J.$

$$\parallel f(x,y)-\overline{f}(x,y)\parallel \le l(x,y)\parallel u(x,y)-\overline{u}(x,y)\parallel$$

Let us define for each $(x,y)\in J,$

$$\begin{array}{cc}\hfill \overline{h}(x,y)& =\mu (x,y)+{\displaystyle \sum _{0<x_k<x}}\left(I_k\left(\overline{u}(x_k^{-},y)\right)-I_k\left(\overline{u}(x_k^{-},0)\right)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}\overline{f}(s,t)dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}\overline{f}(s,t)dtds\hfill \end{array}$$

Then, we obtain

$$\begin{array}{cc}\hfill \parallel h(x,y)-\overline{h}(x,y)\parallel & =\parallel \mu (x,y)-\mu (x,y)+{\displaystyle \sum _{0<x_k<x}}\left(I_k\left(u(x_k^{-},y)\right)-I_k\left(\overline{u}(x_k^{-},y)\right)\right)\hfill \\ \hfill & +{\displaystyle \sum _{0<x_k<x}}\left(I_k\left(u(x_k^{-},0)\right)-I_k\left(\overline{u}(x_k^{-},0)\right)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{0<x_k<x}}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}\left[f(s,t)-\overline{f}(s,t)\right]dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}\left[f(s,t)-\overline{f}(s,t)\right]dtds\parallel \hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel h(x,y)-\overline{h}(x,y)\parallel & \le {\displaystyle \sum _{k=1}^m}\left(I_k\left(u(x_k^{-},y)\right)-I_k\left(\overline{u}(x_k^{-},y)\right)\right)+{\displaystyle \sum _{k=1}^m}\left(I_k\left(u(x_k^{-},0)\right)-I_k\left(\overline{u}(x_k^{-},0)\right)\right)\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\displaystyle \sum _{k=1}^m}{\int }_{x_{k-1}}^{x_k}{\int }_0^y{(x_k-s)}^{r_1-1}{(y-t)}^{r_2-1}\parallel f(s,t)-\overline{f}(s,t)\parallel dtds\hfill \\ \hfill & +\frac{1}{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_{x_k}^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}\parallel f(s,t)-\overline{f}(s,t)\parallel dtds\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel h(x,y)-\overline{h}(x,y)\parallel & \le {\displaystyle \sum _{k=1}^m}{q}_k^{*}\parallel u-\overline{u}\parallel +{\displaystyle \sum _{k=1}^m}{q}_k^{*}\parallel u-\overline{u}\parallel \hfill \\ \hfill & +\frac{\parallel u-\overline{u}\parallel }{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_0^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}l(s,t)dtds\hfill \\ \hfill & +\frac{\parallel u-\overline{u}\parallel }{\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{\int }_0^x{\int }_0^y{(x-s)}^{r_1-1}{(y-t)}^{r_2-1}l(s,t)dtds\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel h(x,y)-\overline{h}(x,y)\parallel & \le {\displaystyle \sum _{k=1}^m}2q_k^{*}\parallel u-\overline{u}+\frac{2\parallel u-\overline{u}\parallel }{r_1{r}_2\Gamma \left(r_1\right)\Gamma \left(r_2\right)}{a}^{r_1}{b}^{r_2}{l}^{*}\hfill \\ \hfill \parallel h(x,y)-\overline{h}(x,y)\parallel & \le {\displaystyle \sum _{k=1}^m}2q_k^{*}\parallel u-\overline{u}\parallel +\frac{2\parallel u-\overline{u}\parallel }{\Gamma (r_1+1)\Gamma (r_2+1)}{a}^{r_1}{b}^{r_2}{l}^{*}\hfill \\ \hfill \parallel h(x,y)-\overline{h}(x,y)\parallel & \le \left[2{\displaystyle \sum _{k=1}^m}{q}_k^{*}+\frac{2l^{*}{a}^{r_1}{b}^{r_2}}{\Gamma (r_1+1)\Gamma (r_2+1)}\right]\parallel u-\overline{u}\parallel \hfill \end{array}$$

Thus, for each $x,y\in J$

$$\parallel h-\overline{h}{\parallel }_{\infty }\le \left[2{\displaystyle \sum _{k=1}^m}{q}_k^{*}+\frac{2l^{*}{a}^{r_1}{b}^{r_2}}{\Gamma (r_1+1)\Gamma (r_2+1)}\right]{\parallel u-\overline{u}\parallel }_{\infty }$$

By a comparable relation, obtained by interchanging the positions of u and $\overline{u},$ it follows that

$$H_d(N\left(u\right),N\left(\overline{u}\right))\le \left[2{\displaystyle \sum _{k=1}^m}{q}_k^{*}+\frac{2l^{*}{a}^{r_1}{b}^{r_2}}{\Gamma (r_1+1)\Gamma (r_2+1)}\right]{\parallel u-\overline{u}\parallel }_{\infty }$$

So by (10), operator N is a contraction and thus by lemma (2.1), operator N has a fixed point u which is the solution to (1) and (2). □

Example We consider the existence of the following impulsive fractional differential inclusion as an application of the main results:

$$\begin{array}{cc}\hfill {}^c{D}_0^{r}u(x,y)& \in F(x,y,u(x,y));a.e(x,y)\in J,x\ne \frac{k}{k+1},k=1,2,\cdots ,m\hfill \\ \hfill u\left({\left(\frac{k}{k+1}\right)}^{+},y\right)& =u\left({\left(\frac{k}{k+1}\right)}^{-},y\right)+\frac{|u\left({\left(\frac{k}{k+1}\right)}^{-},y\right)|}{3k+u\left(\exists \frac{k}{k+1}{)}^{+},y\right)}\hfill \end{array}$$

For $y\in [0,1],k=1,2,\cdots ,m$, $u(x,0)=x$, $u(0,y)=y^2$; $x\in [0,1]$, $y\in [0,1]$, where $J=[0,1]\times [0,1]$, $r=(r_1,r_2)$ and $0<r_1,r_2\le 1.$

To show the existence of this IFDE above, set

$$I_k\left(u\right)=\frac{u}{3k+u};u\in {\mathbb{R}}_{+},k=1,2,\cdots ,m.$$

and

$$q_k\left(u\right)=\frac{u}{3k+u};u\in {\mathbb{R}}_{+}.$$

Then, for each $u\in {\mathbb{R}}_{+}$ and $k=1,2,\cdots ,m$, we have

$$|I_k\left(u\right)|\le q_k.$$

Hence condition (H4) of Theorem (3) is satisfied.

Let $u,\overline{u}\in {\mathbb{R}}_{+}$, then for each $(x,y)\in [0,1]\times [0,1]$, we have

$$\begin{array}{cc}\hfill |I_k\left(u\right)-I_k\left(\overline{u}\right)|& =|\frac{u}{3k+u}-\frac{\overline{u}}{3k+\overline{u}}|\hfill \\ \hfill & \le \frac{1}{3}|u-\overline{u}|\le \frac{1}{1+x^2+y^2}|u-\overline{u}|\hfill \end{array}$$

Hence condition (H5) of Theorem (3) is satisfied with

$$q_k^{*}=\frac{1}{1+x^2+y^2},(x,y)\in [0,1]\times [0,1].$$

Set

$$F(x,y,u(x,y))=\{u\in \mathbb{R}:f_1(x,y,u(x,y))\le u\le f_2(x,y,u(x,y))\}$$

where $f_1,f_2:[0,1]\times [0,1]\times \mathbb{R}⟶\mathbb{R}$. We assume that for each $(x,y)\in J$, $f_1(x,y,u(x,y))$ is lower semi-continuous (i.e., the set $\{z\in \mathbb{R}):f_1(x,y,z)>v\}$ is open for each $v\in \mathbb{R}$) and assume that for each $(x,y)\in J$, $f_2(x,y,u(x,y))$ is upper semi-continuous (i.e., the set $\{z\in \mathbb{R}):f_2(x,y,z)<v\}$ is open for each $v\in \mathbb{R}$). Assume that there are $p\in L^{\infty }(J,{\mathbb{R}}_{+})$ and $\psi :[0,\infty )⟶(0,\infty )$ continuous and non-decreasing such that

$$max\left(\right|f_1(x,y,z)|,|f_2(x,y,z)\left|\right)\le p(x,y)\psi \left(\right|z\left|\right),$$

for a.e $(x,y)\in J$ and $z\in \mathbb{R}.$

Thus, F is an compact, convex valued and upper semi-continuous function. Hence, problem (11) has at least one solution u on $[0,1]\times [0,1]$ since all hypotheses of Theorem (3) are satisfied.

4. Discussion

In this research, we have shown the necessary and sufficient conditions for the existence of solutions to the initial valued problem involving impulsive differential inclusion of fractional order using the nonlinear alternative of the Larey-Schauder fixed point theorem. Using (Theorem 3), one may rapidly establish if the initial value problem (HPFD) has at least one solution. The results (theorems) provided offer a straightforward way for determining if the initial valued problem for impulsive partial differential inclusion of fractional order has solutions or not.