Construction of a measure of noncompactness in Sobolev spaces with an application to functional integral-differential equations

Abstract

In this paper, first we introduce a measure of noncompactness in the Sobolev space $W^{k,1}(\Omega )$ and then, as an application, we study the existence of solutions for a class of the functional integral-differential equations using Darbo’s fixed point theorem associated with this new measure of noncompactness. Further, two examples are presented to verify the effectiveness and applicability of our main results.

Introduction

Sobolev spaces [11], i.e., the class of functions with derivatives in $L^p,$ play an outstanding role in the modern analysis. In the last decades, there has been increasing attempts to study these spaces. Their importance comes from the fact that solutions of partial differential equations are naturally found in Sobolev spaces. They also highlighted in approximation theory, calculus of variation, differential geometry, spectral theory etc.

On the other hand, integral-differential equations (IDE) have a great deal of applications in some branches of sciences. It arises especially in a variety of models from applied mathematics, biological science, physics and another phenomenon, such as the theory of electrodynamics, electromagnetic, fluid dynamics, heat and oscillating magnetic, etc. [9, 12, 18, 21, 24]. There have appeared recently a number of interesting papers [2, 6, 10, 19, 22, 23, 27] on the solvability of various integral equations with help of measures of noncompactness.

The first such measure was defined by Kuratowski [25]. Next, Banaś et al. [8] proposed a generalization of this notion which is more convenient in the applications. The technique of measures of noncompactness is frequently applicable in several branches of nonlinear analysis, in particular the technique turns out to be a very useful tool in the existence theory for several types of integral and integral-differential equations. Furthermore, they are often used in the functional equations, fractional partial differential equations, ordinary and partial differential equations, operator theory and optimal control theory [1, 3, 7, 13, 15,16,17, 26, 28, 29]. The most important application of measures of noncompactness in the fixed point theory is contained in the Darbo’s fixed point theorem [4, 5].

Now, in this paper, we introduce a new measure of noncompactness in the Sobolev space $W^{k,1}(\Omega )$ as a more effective approach. Then, we study the problem of existence of solutions of the functional integral-differential equation

$$ u(x)=\,p(x)+q(x)u(x)+\int _{\Omega }k\big (x,y\big )g\big (y,u(y),\frac{\partial u}{\partial x_1}(y),\ldots ,\frac{\partial u}{\partial x_n}(y),Tu(y)\big ) \mathrm{d}y. $$

We provide some notations, definitions and auxiliary facts which will be needed further on.

Throughout this paper, $\mathbb {R}_{+}$ indicates the interval $[0,+\infty )$ and for the Lebesgue measurable subset D of $\mathbb {R,}$ m(D) denotes the Lebesgue measure of D. Moreover, let $L^1(D)$ be the space of all Lebesgue integrable functions f on D equipped with the standard norm $\Vert f\Vert _{L^1(D)}=\int _D|f(x)| \mathrm{d}x$.

Let $(E,\Vert \cdot \Vert )$ be a real Banach space with zero element 0. The symbol $\overline{B}(x,r)$ stands for the closed ball centered at x with radius r and put $\overline{B}_{r}=\overline{B}(0,r)$. Denote by ${\mathfrak {M}}_E$ the family of nonempty and bounded subsets of E and by ${\mathfrak {N}}_E$ its subfamily consisting of all relatively compact sets of E. For a nonempty subset X of E, the symbols $\overline{X}$ and ConvX will denote the closure and the closed convex hull of X, respectively.

Definition 1.1

[8] A mapping $\mu :\mathfrak {M}_{E} \rightarrow \mathbb {R}_{+}$ is said to be a measure of noncompactness in E if it satisfies the following conditions:

• $1^{\circ }$ The family $\ker \mu =\{X\in \mathfrak {M}_{E}: \mu (X)=0\}$ is nonempty and $\ker \mu \subset \mathfrak {N}_{E}$.

• $2^{\circ }$ $X\subset Y\Rightarrow \mu (X)\le \mu (Y)$.

• $3^{\circ }$ $\mu (\overline{X})=\mu (X)$.

• $4^{\circ }$ $\mu (Conv X) = \mu (X)$.

• $5^{\circ }$ $\mu (\lambda X + (1 - \lambda )Y )\le \lambda \mu (X) + (1 -\lambda )\mu (Y )$ for $\lambda \in [0, 1]$.

• $6^{\circ }$ If $\{X_{n}\}$ is a sequence of closed sets from $\mathfrak {M}_{E}$ such that $X_{n+1} \subset X_{n}$ for $n=1,2,\ldots$ and if $\displaystyle {\lim \nolimits _{n\rightarrow \infty }} \mu (X_{n}) = 0$ then the set $X_{\infty }=\displaystyle {\bigcap \nolimits _{n=1}^{\infty }}X_n$ is nonempty. A measure of noncompactness $\mu$ is said to be regular if it additionally satisfies the following conditions:

• $7^{\circ }$ $\mu ( X\bigcup Y)=\max \{\mu (X),\mu (Y)\}.$

• $8^{\circ }$ $\mu ( X+ Y)\le \mu (X)+\mu (Y).$

• $9^{\circ }$ $\mu (\lambda X)=|\lambda |\mu (X)$ for $\lambda \in \mathbb {R}.$

• $10^{\circ }$ $\ker \mu =\mathfrak {N}_E.$

In what follows, we recall the well known Darbo’s fixed point theorem.

Theorem 1.2

[13] Let $\Omega$ be a nonempty, bounded, closed and convex subset of a Banach space E and let $F:\Omega \rightarrow \Omega$ be a continuous mapping such that there exists a constant $k\in [0,1)$ with the property

$$\begin{aligned} \mu (FX) \le k \mu (X), \end{aligned}$$

(1)

for any nonempty subset X of $\Omega$, where $\mu$ is a measure of noncompactness defined in E. Then, F has a fixed point in the set $\Omega$.

Construction of a measure of noncompactness in Sobolev spaces

In this section, we introduce a measure of noncompactness in the Sobolev space $W^{k,1}(\Omega )$.

Let $\Omega$ be a subset of $\mathbb {R}^n$ and $k\in \mathbb {N}$, we denote by $W^{k,1}(\Omega )$ the space of functions f which, together with all their distributional derivatives $D^\alpha f$ of order $|\alpha |\le k$, belong to $L^1(\Omega )$. Here $\alpha =(\alpha _1,\ldots ,\alpha _n)$ is a multi-index, i.e., each $\alpha _j$ is a nonnegative integer, $|\alpha |=\alpha _1+\cdots +\alpha _n$, and

$$\begin{aligned} D^{\alpha }=\partial ^{|\alpha |}/\partial x_1^{\alpha _1}\ldots \partial x_n^{\alpha _n}. \end{aligned}$$

Then, $W^{k,1}(\Omega )$ is equipped with the complete norm

$$\begin{aligned} \Vert f\Vert _{k,1}=\max _{0\le |\alpha |\le k}\Vert D^{\alpha }f\Vert _{L^1(\Omega )}. \end{aligned}$$

We present the following theorem which characterizes the compact subsets of the Sobolev spaces.

Theorem 2.1

[20] A subset $\mathcal {F}\subset W^{k,1}(\mathbb {R}^n)$ is totally bounded if, and only if, the following holds:

• (i) $\mathcal {F}$ is bounded, i.e., there is some M so that

  $$\begin{aligned} \int |D^{\alpha }f(x)| \mathrm{d}x<M,\ f\in \mathcal {F},\ |\alpha |\le k. \end{aligned}$$

• (ii) For every $\varepsilon >0$ there is some R so that

  $$\begin{aligned} \int _{\Vert x\Vert _{\mathbb {R}^n}>R}|D^{\alpha }f(x)| \mathrm{d}x<\varepsilon ,\ f\in \mathcal {F},\ |\alpha |\le k. \end{aligned}$$

• (iii) For every $\varepsilon >0$ there is some $\rho >0$ so that

  $$\begin{aligned} \int _{\mathbb {R}^n}|D^{\alpha }f(x+y)-D^{\alpha }f(x)| \mathrm{d}x<\varepsilon ,\ f\in \mathcal {F},\ |\alpha |\le k,\ \Vert y\Vert _{\mathbb {R}^n}< \rho . \end{aligned}$$

Now, we are going to describe a measure of noncompactness in $W^{k,1}(\Omega )$.

Theorem 2.2

Suppose $1\le k <\infty$ and U is a bounded subset of $W^{k,1}(\Omega )$. For $u \in U$, $\varepsilon >0$ and $0\le |\alpha |\le k$, let

$$\begin{aligned} \omega ^T(u,\varepsilon )= & \sup \{\Vert \mathcal {T}_h D^{\alpha }u-D^{\alpha }u\Vert _{L^1(B_T)}:h\in \Omega ,\ \Vert h\Vert _{\mathbb {R}^n} <\varepsilon ,0\le |\alpha |\le k\},\\ \omega ^T(U,\varepsilon )= & \sup \{\omega ^T(u,\varepsilon ):u\in U\},\\ \omega ^T(U)= & \displaystyle {\lim _{\varepsilon \rightarrow 0}}\omega ^T(U,\varepsilon ),\\ \omega (U)= & \displaystyle {\lim _{T\rightarrow \infty }}\omega ^T(U),\\ d(U)= & \displaystyle {\lim _{T\rightarrow \infty }}\sup \{\Vert D^{\alpha }u\Vert _{L^1(\Omega \backslash B_T)}:u\in U,0\le |\alpha |\le k\}, \end{aligned}$$

where $B_T=\{a\in \Omega :\Vert a\Vert _{\mathbb {R}^n} \le T\}$ and $\mathcal {T}_hu(t)=u(t+h)$.

Then $\omega _0:\mathfrak {M}_{W^{k,1}(\Omega )}\rightarrow \mathbb {R}$ given by

$$\begin{aligned} \omega _0(U)=\omega (U)+d(U) \end{aligned}$$

(2)

defines a measure of noncompactness in $W^{k,1}(\Omega )$.

Proof

Take $U\in \mathfrak {M}_{W^{k,1}(\Omega )}$ such that $\omega _0(U)=0$. Fix arbitrary $\alpha$ such that $0\le |\alpha |\le k$. Let $\eta >0$ be arbitrary, since $\omega _0(U)=0$,

$$\begin{aligned} \displaystyle {\lim _{T\rightarrow \infty }\lim _{\varepsilon \rightarrow 0}}\omega ^T(U, \varepsilon )=0. \end{aligned}$$

Thus, there exists small enough $\delta >0$ and large enough $T>0$ such that $\omega ^T(U, \delta )<\eta$. This implies that

$$\begin{aligned} \Vert \mathcal {T}_hD^{\alpha }u-D^{\alpha }u\Vert _{L^1(B_T)}<\eta \end{aligned}$$

for all $u\in U$ and $h\in \Omega$ such that $\Vert h\Vert _{\mathbb {R}^n}<\delta$. Since $\eta >0$ was arbitrary, we obtain

$$\begin{aligned} \displaystyle {\lim _{h\rightarrow 0}}\Vert \mathcal {T}_hD^{\alpha }u-D^{\alpha }u\Vert _{L^1(\Omega )}= \displaystyle {\lim _{h\rightarrow 0}}\displaystyle {\lim _{T\rightarrow \infty }}\Vert \mathcal {T}_hD^{\alpha }u-D^{\alpha }u\Vert _{L^1(B_T)}=0. \end{aligned}$$

Using again the fact that $\omega _0(U)=0$ we have

$$\begin{aligned} \displaystyle {\lim _{T\rightarrow \infty }}\sup \{\Vert D^{\alpha }u\Vert _{L^1(\Omega \backslash B_T)}:u\in U\}=0, \end{aligned}$$

and so for $\varepsilon >0$ there exists large enough $T>0$ such that

$$\begin{aligned} \Vert D^{\alpha }u\Vert _{L^1(\Omega \backslash B_T)}<\varepsilon \ \ \text{ for } \text{ all } \ \ u\in U. \end{aligned}$$

It follows then from Theorem 2.1 that U is totally bounded. Thus, $1^{\circ }$ holds.

$2^{\circ }$ is obvious by the definition of $\omega _0$.

Now, we check that condition $3^{\circ }$ holds. For this purpose, suppose that $U\in \mathfrak {M}_{W^{k,1}(\Omega )}$ and $\{u_n\}\subset U$ such that $u_n\rightarrow u\in \overline{U}$ in $W^{k,1}(\Omega )$. From the definition of $\omega ^T(U,\varepsilon ),$ we have

$$\begin{aligned} \Vert \mathcal {T}_hD^{\alpha }u_n-D^{\alpha }u_n\Vert _{L^1(B_T)}\le \omega ^T(U,\varepsilon ), \end{aligned}$$

for any $n\in \mathbb {N}$, $T>0$ and $h\in \Omega$ with $\Vert h\Vert _{\mathbb {R}^n}<\varepsilon$. Letting $n\rightarrow \infty ,$ we get

$$\begin{aligned} \Vert \mathcal {T}_hD^{\alpha }u-D^{\alpha }u\Vert _{L^1(B_T)}\le \omega ^T(U,\varepsilon ), \end{aligned}$$

for any $T>0$ and $h\in \Omega$ with $\Vert h\Vert _{\mathbb {R}^n}<\varepsilon$. Hence

$$\begin{aligned} \displaystyle {\lim _{T\rightarrow \infty }}\displaystyle {\lim _{\varepsilon \rightarrow 0}}\omega ^T(\overline{U},\varepsilon )\le \displaystyle {\lim _{T\rightarrow \infty }}\displaystyle {\lim _{\varepsilon \rightarrow 0}}\omega ^T(U,\varepsilon ). \end{aligned}$$

This concludes that $\omega (\overline{U})\le \omega (U)$. Similarly, we can show that

$$\begin{aligned} d(\overline{U})\le d(U), \end{aligned}$$

and thus

$$\begin{aligned} \omega _0(\overline{U})\le \omega _0(U). \end{aligned}$$

(3)

From (3) and $2^{\circ }$ we obtain $\omega _0(\overline{U})=\omega _0(U)$.

$4^{\circ }$ follows directly from $D^{\alpha }[Conv(U)]=Conv(D^{\alpha }U)$ and hence is omitted.

The proof of condition $5^{\circ }$ can be obtained by using the equality

$$\begin{aligned} D^{\alpha }(\lambda u_1+(1-\lambda )u_2)=\lambda D^{\alpha }u_1+(1-\lambda )D^{\alpha }u_2, \end{aligned}$$

for all $\lambda \in [0,1],$ $u_1\in X$ and $u_2\in Y$.

It remains only to verify $6^{\circ }$, suppose that $\{U_{n}\}$ is a sequence of closed and nonempty sets of $\mathfrak {M}_{W^{k,1}(\Omega )}$ such that $U_{n+1} \subset U_{n}$ for $n=1,2,\ldots$, and $\lim\nolimits _{n\rightarrow \infty } \omega _0(U_{n}) = 0$. Now, for any $n\in \mathbb {N}$, take $u_n\in U_n$ and set $\mathcal {G}=\overline{\{u_n\}}$.

Claim: $\mathcal {G}$ is a compact set in $W^{k,1}(\Omega )$.

Let $\varepsilon >0$ be fixed, since ${\lim\nolimits _{n\rightarrow \infty }} \omega _0(U_{n}) = 0$, there exists sufficiently large $m_1\in \mathbb {N}$ such that $\omega _0(U_{m_1}) <\varepsilon$. Hence, we can find small enough $\delta _1>0$ and large enough $T_1>0$ such that $\omega ^{T_1}(U_{m_1},\delta _1)<\varepsilon$ and $d(U_{m_1})<\varepsilon$. Therefore,

$$\begin{aligned} \Vert \mathcal {T}_hD^{\alpha }u_n-D^{\alpha }u_n\Vert _{L^1(B_{T_1})}<\varepsilon , \end{aligned}$$

and

$$\begin{aligned} \Vert D^{\alpha }u_n\Vert _{L^1(\Omega \backslash B_{T_1})}<\varepsilon , \end{aligned}$$

for all $n> m_1$, $0\le |\alpha |\le k$ and $h\in \Omega$ with $\Vert h\Vert _{\mathbb {R}^n}<\delta _1$. Thus we have

$\Vert \mathcal {T}_hD^{\alpha }u_n-D^{\alpha }u_n\Vert _{L^1(\Omega )}$

$$\begin{aligned}\le & {} \Vert \mathcal {T}_hD^{\alpha }u_n-D^{\alpha }u_n\Vert _{L^1(B_{T_1})} +\Vert \mathcal {T}_hD^{\alpha }u_n-D^{\alpha }u_n\Vert _{L^1(\Omega \backslash B_{T_1})}\\\le & {} \Vert \mathcal {T}_hD^{\alpha }u_n-D^{\alpha }u_n\Vert _{L^1(B_{T_1})}+\Vert \mathcal {T}_hD^{\alpha }u_n\Vert _{L^1(\Omega \backslash B_{T_1})}+\Vert D^{\alpha }u_n\Vert _{L^1(\Omega \backslash B_{T_1})}\\< & {} 3\varepsilon . \end{aligned}$$

On the other hand, we know that the set $\{u_1,u_2,\ldots ,u_{m_1}\}$ is compact, hence there exist $\delta _2>0$ and $T_2>0$ such that

$$\begin{aligned} \Vert \mathcal {T}_hD^{\alpha }u_n-D^{\alpha }u_n\Vert _{L^1(B_{T_2})}<\varepsilon , \end{aligned}$$

(4)

for all $n=1,2,\ldots ,m_1$, $0\le |\alpha |\le k$ and $h\in \Omega$ with $\Vert h\Vert _{\mathbb {R}^n}<\delta _2$.

Furthermore,

$$\begin{aligned} \Vert D^{\alpha }u_n\Vert _{L^1(\Omega \backslash B_{T_2})}<\varepsilon , \end{aligned}$$

(5)

which implies that

$$\begin{aligned} \Vert \mathcal {T}_hD^{\alpha }u_n-D^{\alpha }u_n\Vert _{L^1(\Omega )}<3\varepsilon , \end{aligned}$$

for all $n=1,2,\ldots ,m_1$.

Thus,

$$\begin{aligned} \Vert \mathcal {T}_hD^{\alpha }u_n-D^{\alpha }u_n\Vert _{L^1(\Omega )}<3\varepsilon , \end{aligned}$$

and

$$\begin{aligned} \Vert D^{\alpha }u_n\Vert _{L^1(\Omega \backslash B_{T})}<\varepsilon <3\varepsilon , \end{aligned}$$

(6)

for all $n\in \mathbb {N}$, $\Vert h\Vert _{\mathbb {R}^n}<\min \{\delta _1,\delta _2\}$ and $T=\max \{T_1,T_2\}.$ Therefore, all the hypotheses of Theorem 2.1 are satisfied, that completes the proof of the claim.

Using the above claim, there exists a subsequence $\{u_{n_j}\}$ and $u_0\in W^{k,1}(\Omega )$ such that $u_{n_j}\rightarrow u_0$. Since $u_n\in U _n$, $U_{n+1} \subset U_{n}$ and $U_n$ is closed for all $n\in \mathbb {N}$, we yield

$$\begin{aligned} u_0\in \bigcap _{n=1}^{\infty } U_n=U_{\infty }, \end{aligned}$$

that finishes the proof of $6^{\circ }$. $\square$

We now investigate the regularity of $\omega _0$.

Theorem 2.3

The measure of noncompactness $\omega _0$ defined in (2) is regular.

Proof

Suppose that $X, Y\in \mathfrak {M}_{W^{k,1}(\Omega )}$. First, notice that for all $\varepsilon >0$, $\lambda \in \mathbb {R}$ and $T>0$ we have

$$\begin{aligned} \omega ^T(X\cup Y,\varepsilon )= & {} \max \{\omega ^T(X,\varepsilon ),\omega ^T( Y,\varepsilon )\},\\ \omega ^T(X+ Y,\varepsilon )\le & {} \omega ^T(X,\varepsilon )+\omega ^T( Y,\varepsilon ),\\ \omega ^T(\lambda X,\varepsilon )= & {} |\lambda |\omega ^T(X,\varepsilon ),\\ \displaystyle {\sup _{u\in X\cup Y}\Vert D^{\alpha }u\Vert _{L^1(\Omega \backslash B_{T})}}= & {} \max \{\displaystyle {\sup _{u\in X}\Vert D^{\alpha }u\Vert _{L^1(\Omega \backslash B_{T})}},\displaystyle {\sup _{u\in Y}\Vert D^{\alpha }u\Vert _{L^1(\Omega \backslash B_{T})}}\},\\ \displaystyle {\sup _{u\in X+ Y}\Vert D^{\alpha }u\Vert _{L^1(\Omega \backslash B_{T})}}\le & {} \displaystyle {\sup _{u\in X}\Vert D^{\alpha }u\Vert _{L^1(\Omega \backslash B_{T})}}+\displaystyle {\sup _{u\in Y}\Vert D^{\alpha }u\Vert _{L^1(\Omega \backslash B_{T})}},\\ \displaystyle {\sup _{u\in \lambda X}\Vert D^{\alpha }u\Vert _{L^1(\Omega \backslash B_{T})}}= & {} |\lambda | \displaystyle {\sup _{u\in X}\Vert D^{\alpha }u\Vert _{L^1(\Omega \backslash B_{T})}}. \end{aligned}$$

Then, the hypotheses $7^{\circ }$–$9^{\circ }$ hold. Next, we show that $10^{\circ }$ holds. Take $U\in \mathfrak {N}_{W^{k,1}(\Omega )}$. Thus, the closure of U in $W^{k,1}(\Omega )$ is compact. By Theorem 2.1, for all $|\alpha |\le k$ and for all $\varepsilon >0$, there exists $T>0$ such that $\Vert D^{\alpha }u\Vert _{L^1(\Omega \backslash B_{T})}<\varepsilon$ for all $u\in U,$ and there exists $\delta >0$ such that $\Vert \mathcal {T}_hD^{\alpha }u-D^{\alpha }u\Vert _{L^1(B_T)}<\varepsilon$ for all $h\in \Omega$ with $\Vert h\Vert _{\mathbb {R}^n}<\delta$. Then, for all $u\in U$ we have

$$\begin{aligned} \omega ^T(u,\delta )=\sup \{\Vert \mathcal {T}_hD^{\alpha }u-D^{\alpha }u\Vert _{L^1( B_T)}:h\in \Omega ,\ \Vert h\Vert _{\mathbb {R}^n}<\delta \}\le \varepsilon . \end{aligned}$$

Therefore,

$$\begin{aligned} \omega ^T(U,\delta )=\sup \{\Vert \omega ^T(u,\delta )\Vert :u\in U\}=0. \end{aligned}$$

It yields that

$$\begin{aligned} \omega (U)=\displaystyle {\lim _{T\rightarrow \infty }}\displaystyle {\lim _{\delta \rightarrow 0}}\omega ^T(U,\delta )=0. \end{aligned}$$

Furthermore,

$$\begin{aligned} d(U)=\displaystyle {\lim _{T\rightarrow \infty }}\sup \{\Vert D^\alpha u\Vert _{L^1(\Omega \backslash B_T)}:u\in U\}=0. \end{aligned}$$

Then $\omega _0(U)=0$ and $\ker (\omega _0)= \mathfrak {N}_{W^{k,1}(\Omega )}$. $\square$

Theorem 2.4

Let $Q=\{u\in W^{k,1}(\mathbb {R}^n):\Vert u\Vert _{1,1}\le 1\}.$ Then $\omega _0(Q)=3$.

Proof

Applying the same strategy as ([4], Theorem [14]), we observe that $\omega _0(Q)\le 3.$ It remains to verify $\omega _0(Q)\ge 3.$ For any $k\in \mathbb {N}$, there exists $E_k\subset \mathbb {R}^n$ such that $m(E_k)=\frac{1}{10k}$, $diam(E_k)\le \frac{1}{k}$, $E_k\cap B_k=\emptyset$ and $E_k\subset B_{2k}.$ Define $f_k:\mathbb {R}^n\rightarrow \mathbb {R}$ by

$$\begin{aligned} f_k(x)= {\left\{ \begin{array}{ll} 10k, &{} {x\in E_k,}\\ 0, &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

In addition, observe that $\Vert f_k\Vert _{1,1}=1$, $\Vert D^\alpha \mathcal {T}_{\beta _k}f_k-D^\alpha f_k\Vert _{L^1(B_{2k})} =2$ and

$\Vert D^\alpha f_k\Vert _{L^1(\mathbb {R}^n\backslash B_k)}=1$ for all $k\in \mathbb {N}$, where $\beta _k=(\frac{1}{k},\ldots ,\frac{1}{k})\in \mathbb {R}^n$. Thus, we conclude that $\omega _0(Q)\ge \omega _0(\{f_k\})=3.$ $\square$

Application

In this section, we study the existence of solutions for some functional integral-differential equations. We also provide some illustrative examples to verify effectiveness and applicability of our results.

We start with some preliminaries which we need in subsequence.

Lemma 3.1

[14] Let $\Omega$ be a Lebesgue measurable subset of $\mathbb {R}^n$ and $1\le p\le \infty .$ If $\{f_n\}$ is convergent to f in the $L^p$-norm, then there is a subsequence $\{f_{n_k}\}$ which converges to f a.e., and there is $g\in L^p(\Omega )$, $g\ge 0,$ such that

$$\begin{aligned} |f_{n_k}(x)|\le g(x)\ \ \ {\text{for}}\ \ a.e. \ \ x\in \Omega . \end{aligned}$$

Definition 3.2

[4] We say that a function $f:\mathbb {R}^n\times \mathbb {R}^m\rightarrow \mathbb {R}$ satisfies the Carath$\acute{e}$odory conditions if the function f(., u) is measurable for any $u\in \mathbb {R}^m$ and the function f(x, .) is continuous for almost all $x\in \mathbb {R}^n.$

Let $\Omega$ be a subset of $\mathbb {R}^n$ and $k\in \mathbb {N}$, we denote by $BC^k(\Omega )$ the space of functions f which are bounded and k-times continuously differentiable on $\Omega$ with the standard norm

$$\begin{aligned} \Vert f\Vert _{BC^k(\Omega )}=\max _{0\le |\alpha |\le k}\Vert D^\alpha f\Vert _u, \end{aligned}$$

where $\Vert D^\alpha f\Vert _u=\sup \{|D^\alpha f(x)|:x\in \Omega \}$.

Theorem 3.3

Let $\Omega$ be a subset of $\mathbb {R}^n$ with $m(\Omega )<\infty$. Assume that the following conditions are satisfied:

• (i) $p\in W^{1,1}(\Omega ),$ $q\in BC^1(\Omega )$ and

  $$\begin{aligned} \lambda :=\sup \{\Vert q\Vert _u+\Vert \frac{\partial q}{\partial x_i}\Vert _u:i=1,\ldots ,n\}<1. \end{aligned}$$

  (7)

• (ii) $g:\Omega \times \mathbb {R}^{n+2} \rightarrow \mathbb {R}$ satisfies the Carath $\acute{e}$ odory conditions and there exist a bounded continuous function $a:\Omega \rightarrow \mathbb {R}_+$ with $|a(x)|\le M$ for all $x\in \Omega$ and some $M>0$ and a concave, lower semi-continuous and nondecreasing function $\zeta :\mathbb {R_+}\rightarrow \mathbb {R_+}$ such that

  $$\begin{aligned} |g(x,u_0,u_1,\ldots ,u_{n+1})|\le a(x)\zeta (\max _{0\le i\le n+1}|u_i|). \end{aligned}$$

  (8)

• (iii) $k:\Omega \times \Omega \rightarrow \mathbb {R}$ satisfies the Carath$\acute{e}$odory conditions and has a derivative of order 1 with respect to the first argument. Moreover, there exist $g_1,g_3\in W^{1,1}(\Omega )$ and $g_2\in L^\infty (\Omega )$ such that

  $$\begin{aligned} |k(x,y)|\le g_1(x)g_2(y),\ |k(x_1,y)-k(x_2,y)|\le g_2(y)|g_3(x_1)-g_3(x_2)|, \end{aligned}$$

  and

  $$\begin{aligned} \left|\frac{\partial k}{\partial x_i}(x,y)\right|\le g_1(x)g_2(y),\ \left|\frac{\partial k}{\partial x_i}(x_1,y)-\frac{\partial k}{\partial x_i}(x_2,y)\right|\le g_2(y)|g_3(x_1)-g_3(x_2)|, \end{aligned}$$

  for almost $x,y,x_1,x_2\in \Omega$ and $1\le i\le n$.

• (iv) There exists a positive solution $r_0$ of the inequality

  $$\begin{aligned} \Vert p\Vert _{1,1}+\lambda r+Mm(\Omega )\Vert g_1\Vert _{L^1(\Omega )}\Vert g_2\Vert _{L^\infty }\zeta \left(\frac{1}{m(\Omega )}\Vert u\Vert _{1,1}\right)\le r. \end{aligned}$$

  (9)

• (v) $T:W^{1,1}(\Omega ) \rightarrow L^1(\Omega )$ is a continuous operator such that for any $x\in W^{1,1}(\Omega )$ we have

  $$\begin{aligned} \Vert T(x)\Vert _{L^1(\Omega )}\le \Vert x\Vert _{1,1}. \end{aligned}$$

Then, the functional integral-differential equation

$$ u(x)=p(x)+q(x)u(x)+\int _{\Omega }k\big (x,y\big )g\big (y,u(y),\frac{\partial u}{\partial x_1}(y),\ldots ,\frac{\partial u}{\partial x_n}(y),Tu(y)\big ) \mathrm{d}y $$

(10)

has at least one solution in the space $W^{1,1}(\Omega )$.

Proof

We define the operator $F:W^{1,1}(\Omega )\rightarrow W^{1,1}(\Omega )$ by

$$\begin{aligned} Fu(x)=p(x)+q(x)u(x)+\int _{\Omega }k\big (x,y\big )g\big (y,u(y),\frac{\partial u}{\partial x_1}(y),\ldots ,\frac{\partial u}{\partial x_n}(y),Tu(y)\big ) \mathrm{d}y. \end{aligned}$$

Obviously, Fu is measurable for any $u\in W^{1,1}(\Omega )$. Also, for any $x\in \Omega$ we have

$$\begin{aligned} \frac{\partial (Fu)}{\partial x_i}(x)= & {} \frac{\partial p}{\partial x_i}(x)+ \frac{\partial q}{\partial x_i}(x)u(x)+q(x)\frac{\partial u}{\partial x_i}(x)\\ & \quad+ \int _{\Omega }\frac{\partial k}{\partial x_i}\big (x,y\big )g\big (y,u(y), \frac{\partial u}{\partial x_1}(y), \ldots , \frac{\partial u}{\partial x_n}(y),Tu(y)\big ) \mathrm{d}y, \end{aligned}$$

and Fu has measurable derivatives. We show that, $Fu\in W^{1,1}(\Omega )$. Using our assumptions, for arbitrarily fixed $x\in \Omega$, we have

$$\begin{aligned} |Fu(x)|\le & {} |p(x)|+|q(x)||u(x)|\\ & \quad + {} \Big |\int _{\Omega }k\big (x,y\big )g\big (y,u(y),\frac{\partial u}{\partial x_1}(y),\ldots ,\frac{\partial u}{\partial x_n}(y),Tu(y)\big ) \mathrm{d}y\Big |. \end{aligned}$$

According to the Jensen’s inequality, we deduce

$$\begin{aligned} \Vert Fu\Vert _{L^1(\Omega )}\le \Vert p\Vert _{L^1(\Omega )}+\Vert q\Vert _u\Vert u\Vert _{L^1(\Omega )}+Mm(\Omega )\Vert g_1\Vert _{L^1(\Omega )}\Vert g_2\Vert _{L^\infty }\zeta \left( \frac{1}{m(\Omega )}\Vert u\Vert _{1,1}\right) . \end{aligned}$$

By the same argument as above,

$$\begin{aligned} |\frac{\partial (Fu)}{\partial x_i}(x)|\le & {} |\frac{\partial p}{\partial x_i}(x)|+|\frac{\partial q}{\partial x_i}(x)| |u(x)|+|q(x)||\frac{\partial u}{\partial x_i}(x)|\\ & \quad + {} \Big |\int _{\Omega }\frac{\partial k}{\partial x_i} \big (x,y\big )g\big (y,u(y),\frac{\partial u}{\partial x_1}(y),\ldots ,\frac{\partial u}{\partial x_n}(y),Tu(y)\big ) \mathrm{d}y\Big | \end{aligned}$$

and

$$\begin{aligned} \Vert \frac{\partial (Fu)}{\partial x_i}\Vert _{L^1(\Omega )}\le & {} \Vert \frac{\partial p}{\partial x_i}(x)\Vert _{L^1(\Omega )}+\Vert \frac{\partial q}{\partial x_i}\Vert _{u}\Vert u\Vert _{L^1(\Omega )}+\Vert q\Vert _u\Vert \frac{\partial u}{\partial x_i}\Vert _{L^1(\Omega )}\\ & \quad + Mm(\Omega )\Vert g_1\Vert _{L^1(\Omega )}\Vert g_2\Vert _{L^\infty }\zeta \left( \frac{1}{m(\Omega )}\Vert u\Vert _{1,1}\right) .\end{aligned}$$

Thus, we obtain

$$\begin{aligned} \Vert Fu\Vert _{1,1}\le \Vert p\Vert _{1,1}+\lambda \Vert u\Vert _{1,1}+M m(\Omega )\Vert g_1\Vert _{L^1(\Omega )}\Vert g_2\Vert _{L^\infty }\zeta \left( \frac{1}{m(\Omega )}\Vert u\Vert _{1,1}\right) . \end{aligned}$$

(11)

Due to (11) and using condition (iv), we derive that F is a mapping from $\bar{B}_{r_0}$ into $\bar{B}_{r_0}$. Now, we show that the map F is continuous. Let $\{u_m\}$ be an arbitrary sequence in $W^{1,1}(\Omega )$ which converges to $u\in W^{1,1}(\Omega )$. By Lemma 3.1 there is a subsequence $\{u_{m_k}\}$ which converges to u a.e., $\{\frac{\partial u_{m_k}}{\partial x_i}\}$ converges to $\{\frac{\partial u}{\partial x_i}\}$ a.e., $\{Tu_{m_k}\}$ converges to Tu a.e. and there is $h\in L^1(\Omega )$, $h\ge 0,$ such that

$$\begin{aligned} \max \{|u_{m_k}(y)|,|\frac{\partial u_{m_k}}{\partial x_1}(y)|,|\frac{\partial u_{m_k}}{\partial x_2}(y)|,\ldots ,|Tu_{m_k}(y)|\}\le h(y) \ \ \text {for}\ \ \text{ a.e. }\ \ y\in \Omega . \end{aligned}$$

Since $u_{m_k}\rightarrow u$ almost everywhere and g satisfies the Carath$\acute{e}$odory conditions, it follows that

$$\begin{aligned} g(y,u_{m_k}(y),\frac{\partial u_{m_k}}{\partial x_1}(y),\ldots ,Tu_{m_k}(y))\rightarrow g(y,u(y),\frac{\partial u}{\partial x_1}(y),\ldots ,Tu(y)), \end{aligned}$$

for almost all $y\in \Omega$.

From condition (ii) we have

$$\begin{aligned} g(y,u_{m_k}(y),\frac{\partial u_{m_k}}{\partial x_1}(y),\ldots ,Tu_{m_k}(y))\le a(y)\zeta (h(y))\ \ \text {for}\ \ \text{ a.e. }\ \ y\in \Omega . \end{aligned}$$

As a consequence of the Lebesgue’s Dominated Convergence Theorem, it yields that

$$\begin{aligned} \int g(y,u_{m_k}(y),\frac{\partial u_{m_k}}{\partial x_1}(y),\ldots ,Tu_{m_k}(y)) \mathrm{d}y\rightarrow \int g(y,u(y),\frac{\partial u}{\partial x_1}(y),\ldots ,Tu(y)) \mathrm{d}y, \end{aligned}$$

(12)

for almost all $y\in \Omega .$ Inequality (12) and condition (iii) imply that

$$\begin{aligned} \Vert Fu_{m_k}-Fu\Vert _{1,1}\rightarrow 0\ \ \text {and}\ \ \left\Vert \frac{\partial Fu_{m_k}}{\partial x_i}- \frac{\partial Fu}{\partial x_i}\right\Vert _{1,1}\rightarrow 0\ \ \text {as}\ \ k\rightarrow \infty \ \ (1\le i\le n). \end{aligned}$$

Therefore, $F:W^{1,1}(\Omega ) \longrightarrow W^{1,1}(\Omega )$ is continuous.

To finish, the proof we have to verify that condition (1) is satisfied. We fix arbitrary $T>0$ and $\varepsilon > 0$. Let U be a nonempty and bounded subset of $\bar{B}_{r_0}$. Choose $u\in U$ and $x,h\in B_T$ with $\Vert h\Vert _{\mathbb {R}^n}\le \varepsilon$, then we have

$$\begin{aligned} & \int _{B_T}|Fu(x)-Fu(x+h)| \mathrm{d}x\\ \quad & \quad {}\le \int _{B_T}|p(x)-p(x+h)| \mathrm{d}x+\int _{B_T}|q(x)-q(x+h)||u(x)| \mathrm{d}x\\ & \qquad {}+\int _{B_T}|q(x+h)||u(x)-u(x+h)| \mathrm{d}x\\ & \qquad +\int _{B_T}\int _{\Omega }|k(x,y)-k(x+h,y)|\big |g\big (y,u(y),\frac{\partial u}{\partial x_1}(y),\ldots ,\frac{\partial u}{\partial x_n}(y),Tu(y)\big )\big | \mathrm{d}y \mathrm{d}x\\ & \quad {}\le \omega ^T(p,\varepsilon )+\int _{B_T}|q(x)-q(x+h)||u(x)| \mathrm{d}x+\lambda \omega ^T(U,\varepsilon )\\ & \qquad {}+Mm(\Omega )\Vert g_2\Vert _{L^\infty }\zeta \left( \frac{1}{m(\Omega )}\Vert u\Vert _{1,1}\right) \omega ^T(g_3,\varepsilon ). \end{aligned}$$

(13)

Obviously, ${\omega }^T(p, \varepsilon )\rightarrow 0$, ${\omega }^T (g_3,\varepsilon )\rightarrow 0$ and by continuity of q,

$$\begin{aligned} \int _{B_T}|q(x)-q(x+h)||u(x)| \mathrm{d}x\rightarrow 0, \end{aligned}$$

as $\varepsilon \rightarrow 0$. Then the right hand side of (13) tends to $\lambda \omega ^T(U)$ as $\varepsilon \rightarrow 0$.

By a similar argument and using condition (i), for each $i=1,\ldots ,n$, we get

$$\begin{aligned} & \int _{B_T}\left|\frac{\partial (Fu)}{\partial x_i}(x)-\frac{\partial (Fu)}{\partial x_i}(x+h)\right| \mathrm{d}x\\ & \quad {}\le \int _{B_T}\left|\frac{\partial p}{\partial x_i}(x)-\frac{\partial p}{\partial x_i}(x+h)\right| \mathrm{d}x +\int _{B_T}\left|\frac{\partial q}{\partial x_i}(x)-\frac{\partial q}{\partial x_i}(x+h)||u(x)\right| \mathrm{d}x\\ & \qquad {}+\int _{B_T}\left|\frac{\partial q}{\partial x_i}(x+h)||u(x)-u(x+h)\right| \mathrm{d}x+\int _{B_T}\left|q(x)||\frac{\partial u}{\partial x_i}(x)-\frac{\partial u}{\partial x_i}(x+h)\right| \mathrm{d}x\\ & \qquad {}+\int _{B_T}\left|\frac{\partial u}{\partial x_i}(x+h)||q(x)-q(x+h)\right| \mathrm{d}x+\int _{B_T}\int _{\Omega }\left|\frac{\partial k}{\partial x_i}(x,y)-\frac{\partial k}{\partial x_i}(x+h,y)\right|\\ & \qquad {}\times \left|g\big (y,u(y),\frac{\partial u}{\partial x}(y),\ldots ,\frac{\partial u}{\partial x_n}(y),Tu(y)\big ) \right| \mathrm{d}y \mathrm{d}x\\ & \quad {}\le {\omega }^T(p, \varepsilon )+\int _{B_T}\left|\frac{\partial q}{\partial x_i}(x)-\frac{\partial q}{\partial x_i}(x+h)||u(x)\right| \mathrm{d}x+\lambda \omega ^T(U,\varepsilon )\\ & \qquad {}+\int _{B_T}\left|\frac{\partial u}{\partial x_i}(x+h)||q(x)-q(x+h)\right| \mathrm{d}x\\ & \qquad {}+Mm(\Omega )\Vert g_2\Vert _{L^\infty }\zeta \left(\frac{1}{m(\Omega )}\Vert u\Vert _{1,1}\right)\omega ^T(g_3,\varepsilon ). \end{aligned}$$

(14)

Applying the same reasoning as above, the right hand side of (14) tends to $\lambda \omega ^T(U)$ as $\varepsilon \rightarrow 0,$ too. Regarding to (13) and (14), and since u is an arbitrary element of U, then $\omega ^T(FU)\le \lambda \omega ^T(U)$. Letting $T\rightarrow \infty$, we deduce

$$\begin{aligned} \omega (FU)\le \lambda \omega (U). \end{aligned}$$

(15)

Next, let us fix an arbitrary number $T>0$. Then, taking into account our hypotheses, for an arbitrary function $u\in U$ we derive

$$\begin{aligned} \Vert Fu\Vert _{L^1(\Omega \backslash B_T)} &\le \Vert p\Vert _{L^1(\Omega \backslash B_T)}+\Vert q\Vert _u\Vert u\Vert _{L^1(\Omega \backslash B_T)}\\&\quad + Mm(\Omega )\Vert g_1\Vert _{L^1(\Omega \backslash B_T)}\Vert g_2\Vert _{L^\infty (\Omega \backslash B_T))}\zeta \left( \frac{1}{m(\Omega )}\Vert u\Vert _{1,1}\right) . \end{aligned}$$

Now, since

$$\begin{aligned} \Vert p\Vert _{L^1(\Omega \backslash B_T)}\rightarrow 0, \ \ \Vert g_1\Vert _{L^1(\Omega \backslash B_T)}\rightarrow \ 0\ \text{ as } \ \ T\rightarrow \infty , \end{aligned}$$

then

$$\begin{aligned} \lim _{T\rightarrow \infty }\Vert Fu\Vert _{L^1(\Omega \backslash B_T)}\le \lambda d(U). \end{aligned}$$

Similarly,

$$\begin{aligned} \lim _{T\rightarrow \infty } \left\{ \left\Vert \frac{\partial (Fu)}{\partial x_i}\right\Vert _{L^1(\Omega \backslash B_T)}:i=1,\ldots ,n\right\}\le \lambda d(U). \end{aligned}$$

These relations imply that

$$\begin{aligned} d(FU)\le \lambda d(U). \end{aligned}$$

(16)

Finally, from (15) and (16) we conclude that $\omega _0(FU)\le \lambda \omega _0(U)$.

According to Theorem 1.2, we obtain that the operator F has a fixed point x in $\bar{B}_{r_0}$, and thus functional integral-differential equation (10) has at least one solution in the space $W^{1,1}(\Omega )$. $\square$

Now, we present two examples which verify the effectiveness and applicability of Theorem 3.3.

Example 3.4

Consider the following functional integral-differential equation

$$\begin{aligned} u(x_1,x_2,x_3)= & {} \root 4 \of {x_1^5}+ e^{-(x_1+x_2+x_3+1)}u(x_1,x_2,x_3)\nonumber \\ \quad+ & \int _0^1\int _0^1\int _0^1\frac{e^{-(x_1+x_2+x_3)}}{(y_1+1)^3(y_2+2)^2(y_3+5)}\nonumber \\ \quad\times & {} \cos \Big (y_1u(x_1,x_2,x_3)\frac{\partial u}{\partial x_1}(y_1,y_2,y_3)+y_2\frac{\partial u}{\partial x_2}(y_1,y_2,y_3)\nonumber \\ \quad+ & {} y_3\frac{\partial u}{\partial x_3}(y_1,y_2,y_3)+\frac{1}{2}u(y_1,y_2,y_3)\Big ) \mathrm{d}y_1 \mathrm{d}y_2 \mathrm{d}y_3. \end{aligned}$$

(17)

Eq. (17) is a special case of Eq. (10) with

$\Omega = [0, 1]\times [0, 1]\times [0, 1],p(x_1,x_2,x_3)=\root 4 \of {x_1^5},q(x_1,x_2,x_3)=e^{-(x_1+x_2+x_3+1 )},$ $g(y_1,y_2,y_3, u_0, u_1, u_2,u_3,u_4)=\cos (y_1 u_0 u_1+y_2u_2+y_3u_3+u_4)$, $Tu=\frac{1}{2}u$,

$$\begin{aligned} k(x_1,x_2,x_3,y_1,y_2,y_3)=\frac{e^{-(x_2+y_2+y_3+1)}}{(y_1+1)^3(y_2+2)^2(y_3+5)}, \end{aligned}$$

$$\begin{aligned} g_1(x_1,x_2,x_3)=g_3(x_1,x_2,x_3)=e^{-(x_1+x_2+x_3+1)}, \end{aligned}$$

and

$$\begin{aligned} g_2(x_1,x_2,x_3)=\frac{1}{(y_1+1)^2(y_2+2)(y_3+5)}. \end{aligned}$$

It is easy to see that $p\in W^{1,1}(\Omega )$, $q\in BC^1(\Omega )$ and $\lambda =2e^{-1}$. Also, g satisfies Carath$\acute{e}$odory conditions and if we define $a(x_1,x_2,x_3)=\zeta (x)=1$, then condition (ii) of Theorem 3.3 holds. We observe that $g_1,g_3\in L^1(\Omega ), g_2\in L^\infty (\Omega )$ and k satisfies condition (iii). Moreover, it can be easily shown that each number $r \ge 4$ satisfies the inequality in condition (iv), i.e.,

$$\begin{aligned} \Vert p\Vert _{1,1}+\lambda r+M\Vert g_1\Vert _{L^1(\Omega )}\Vert g_2\Vert _{L^\infty }\zeta (r)\le 1+2e^{-1}r+\dfrac{(1-e^{-1})^3}{10}\le r. \end{aligned}$$

Thus, as the number $r_0$ we can take $r_0= 4$. Consequently, all the conditions of Theorem 3.3 are satisfied. Hence the functional integral-differential equation (17) has at least one solution in the space $\displaystyle W^{1,1}(\Omega )$.

Example 3.5

Consider the following functional integral-differential equation

$$\begin{aligned} u(x)=\frac{u(x)}{x+2}+\int _0^1\frac{\root 5 \of {y^3u(y)+3 u^{(2)}(y)+u^{(3)}(y)}}{1+(u')^2(y)e^{\sin (u^{(2)}(y)+1)}} \mathrm{d}y. \end{aligned}$$

(18)

Eq. (18) is a special case of Eq. (10) with

$$\begin{aligned}&p(x)=0, q(x)=\frac{1}{x+2},k(x,y)=e^{x-y}, T(u)=0, \Omega =[0, 1]\\ \text {and}\\&g(y, u_0, u_1, u_2, u_3,u_4)=\frac{\root 5 \of {y^3 u_0+3 u_2+u_4}}{1+u_1^2 e^{\sin ( u_2+1)}}.T(u)=0 \end{aligned}$$

It is easy to see that $q\in BC^1(\Omega )$ and $\lambda =\dfrac{3}{4}$. Also, g satisfies Carath$\acute{e}$odory conditions and if we define $a(x)=\root 5 \of {5}$ and $\zeta (x)=\root 5 \of {x}$, then condition (ii) of Theorem 3.3 holds. Moreover, k is continuous and has a continuous derivative of order 1 with respect to the first argument. On the other hand, $g_1(x)=g_3(x)=e^x$ and $g_2(x)=e^{-x}$. It can be easily shown that each number $r \ge 10$ satisfies the inequality in condition (iv), i.e.,

$$\begin{aligned} \Vert p\Vert _{1,1}+\lambda r+M\Vert g_1\Vert _{L^1(\Omega )}\Vert g_2\Vert _{L^{\infty }}\zeta (r)\le \frac{3}{4}r+\root 5 \of {5} (e-1)(1-e^{-1})\root 5 \of {r}\le r. \end{aligned}$$

Hence, as the number $r_0$ we can take $r_0=10$. Consequently, all the conditions of Theorem 3.3 are satisfied. It implies that the functional integral-differential equation (18) has at least one solution in the space $\displaystyle W^{1,1}(\Omega )$.