Study on the existence of solutions for a generalized functional integral equation in ${L}^{1}$ spaces
 Lijuan Yang^{1},
 Jing Wang^{1} and
 Ganshan Yang^{2, 3}Email author
https://doi.org/10.1186/1029242X2013235
© Yang et al.; licensee Springer. 2013
Received: 4 November 2012
Accepted: 24 April 2013
Published: 8 May 2013
Abstract
Using a nonlinear alternative theorem of Krasnosel’skii type proved recently by Smaïl Djebali and Zahira Sahnoun, we investigate, in this paper, the existence of solutions for a generalized mixedtype functional integral equation in ${L}^{1}$ space. We also present some examples of the integral equation to confirm the efficiency of our results.
MSC:47H10, 45A05.
Keywords
1 Introduction
The functional integral equations describe many physical phenomena in various areas of natural science, mathematical physics, mechanics and population dynamics [1–4]. The theory of integral equations is developing rapidly with the help of tools in functional analysis, topology and fixedpoint theory (see, for instance, [5–8]) and it serves as a useful tool, in turn, for other branches of mathematics, for example, for differential equations (see [9, 10]). A fixed point theorem, frequently used to solve integral equations, is a theorem proved by Krasnosel’skii in 1958 (see, for instance, [11, 12]). The Krasnosel’skii theorem asserts that $A+B$ has a fixed point in a closed, convex nonempty subset M of X if A, B satisfy the following conditions:

A is compact and continuous;

B is a strict contraction;

$AM+BM\subseteq M$.
However, the Krasnosel’skii fixed point theorem sometimes turns out to be restrictive for some equations due to the weak topology of the problem. In order to use this result and its variant, one has to find a selfmapped closed convex set M so that $A+B$ maps M into itself or the weaker one: $x=Ax+By\phantom{\rule{0.25em}{0ex}}(y\in M)\Rightarrow x\in M$. From the application point of view, this condition is also generally strict and is hard to achieve. To relax these conditions, a new effort is made in [13] by establishing a new variant of nonlinear Krasnosel’skii type fixed point theorem for nonself maps.
Let us first recall the nonlinear alternative Krasnosel’skii fixed point theorem established in [13], which plays a central role in our discussion.
Theorem 1 Let $S\ni 0$ be an open subset of a Banach space X and let $\overline{S}$ be the closure of S. Let $A:S\to X$ and $B:X\to X$ be two mappings satisfying:

A is continuous, $A(\overline{S})$is relatively weakly compact, and A verifies the condition H1.

B is a contraction and verifies the condition H2.
Then either the equation $Ax+Bx=x$ admits a solution in $\overline{S}$, or there exists an element $x\in \partial S$ (∂S denotes the boundary of S) such that $x=\lambda Ax+\lambda B(\frac{x}{\lambda})$ for some $\lambda \in (0,1)$, where conditions H1 and H2 are given in Section 2.
The advantage of Theorem 1 lies in that in applying Theorem 1, one does not need to verify that the involved operator maps a closed convex subset onto itself.
The outline of this paper is as follows. In Section 2, we introduce some basic facts and use them to obtain our aims in Section 3. In the last section, we present some examples that verify the application of this kind of nonlinear integral equation.
2 Preliminaries
2.1 The weak MNC
for each $M\in B(X)$.
The following Lemma 1 comes from [14].
 (i)
${M}_{1}\subseteq {M}_{2}$ implies $\omega ({M}_{1})\le \omega ({M}_{2})$.
 (ii)
$\omega ({M}_{1})=0$ if and only if ${M}_{1}$ is relatively weakly compact.
 (iii)
$\omega (\overline{{M}_{1}^{w}})=\omega ({M}_{1})$ (where $\overline{{M}_{1}^{w}}$ is the weak closure of ${M}_{1}$).
 (iv)
$\omega ({M}_{1}\cup {M}_{2})=max\{\omega ({M}_{1}),\omega ({M}_{2})\}$.
 (v)
$\omega (\lambda {M}_{1})=\lambda \omega ({M}_{1})$ for all $\lambda \in \mathcal{R}$.
 (vi)
$\omega (co({M}_{1}))=\omega ({M}_{1})$ ($co({M}_{1})$ refers to the convex hull of ${M}_{1}$).
 (vii)
$\omega ({M}_{1}+{M}_{2})\le \omega ({M}_{1})+\omega ({M}_{2})$.
The map $\omega (\cdot )$ is called the De Blasi measure of weak noncompactness.
for all bounded $M\subset {L}^{1}(\mathrm{\Omega},X)$, where $meas(\cdot )$ represents the Lebesgue measure, X is a finitedimensional Banach space.
Let J be a nonlinear operator from X into itself. In what follows, we need the following two conditions:
H1. If ${({x}_{n})}_{n\in N}$ is a weakly convergent sequence in X, then ${(J{x}_{n})}_{n\in N}$ has a strongly convergent subsequence in X;
H2. If ${({x}_{n})}_{n\in N}$ is a weakly convergent sequence in X, then ${(J{x}_{n})}_{n\in N}$ has a weakly convergent subsequence in X.
2.2 The superposition operator
 (i)
for any $x\in X$, the map $t\to f(t,x)$ is measurable from Ω to Y;
 (ii)
for almost all $t\in \mathrm{\Omega}$, the map $x\to f(t,x)$ is continuous from X to Y.
Definition 1 (Nemytskii’s operator)
Let $f:\mathrm{\Omega}\times X\to Y$ be a Carathéodory function, Nemytskii’s operator associated with f, ${N}_{f}:m(\mathrm{\Omega},X)\to m(\mathrm{\Omega},Y)$ is defined by ${N}_{f}x(t)=f(t,x(t))$, $\mathrm{\forall}t\in \mathrm{\Omega}$.
The superposition operator enjoys several nice properties. Specifically, we have the following results.
Lemma 2 [16]
where ${L}_{+}^{1}(\mathrm{\Omega})$ stands for the positive cone of the space ${L}^{1}(\mathrm{\Omega})$.
Lemma 3 [17]
Let X, Y be two finitedimensional Banach spaces and let Ω be a bounded domain of ${\mathcal{R}}^{n}$. If $f:\mathrm{\Omega}\times X\to Y$ is a Carathéodory function and ${N}_{f}$ maps ${L}^{1}(\mathrm{\Omega},X)$ into ${L}^{1}(\mathrm{\Omega},Y)$, then ${N}_{f}$ satisfies the condition H2.
We give a fixed point lemma for bilinear forms.
Lemma 4 Let X be a Banach space and let $B:X\times X\to X$ be a bilinear map. Let ${\parallel \cdot \parallel}_{X}$ denote the norm in X. If for all ${x}_{1},{x}_{2}\in X$, ${\parallel B({x}_{1},{x}_{2})\parallel}_{X}\le \eta {\parallel {x}_{1}\parallel}_{X}{\parallel {x}_{2}\parallel}_{X}$. Then for all $y\in X$ satisfying $4\eta {\parallel y\parallel}_{X}<1$, the equation $x=y+B(x,x)$ has a solution $x\in X$ satisfying and uniquely defined by the condition ${\parallel x\parallel}_{X}\le {\parallel y\parallel}_{X}$.
Remark 1 The proof of this lemma also shows that $x={lim}_{k\to \mathrm{\infty}}{x}_{k}$, where the approximate solutions ${x}_{k}$ are defined by ${x}_{0}=y$ and ${x}_{k}=y+B({x}_{k1},{x}_{k1})$. Moreover, ${\parallel {x}_{k}\parallel}_{X}\le 2{\parallel y\parallel}_{X}$ for all k.
3 Main results
In this section, we investigate the solvability of the nonlinear functional integral Eq. (1) in the space ${L}^{1}(\mathrm{\Omega},X)$ by applying Theorem 1.
 (a)
The function $g:\mathrm{\Omega}\times X\to X$ is a measurable function, $g(\cdot ,0)\in {L}^{1}(\mathrm{\Omega},X)$ and g is a contraction with respect to the second variable, i.e., there exists an $L\in [0,1)$ such that $\parallel g(t,x)g(t,y)\parallel \le L\parallel xy\parallel $ for almost all $t\in \mathrm{\Omega}$ and all $x,y\in X$.
 (b)
${f}_{i}:\mathrm{\Omega}\times X\to X$, $i=1,2$ satisfy Carathéodory conditions and ${N}_{{f}_{i}}$, $i=1,2$, act from ${L}^{1}(\mathrm{\Omega},X)$ into itself continuously.
 (c)
The operators T and A are linear and bounded on ${L}^{1}(\mathrm{\Omega},X)$.
 (d)
The Urysohn operator U defined as before maps continuously ${L}^{1}(\mathrm{\Omega},X)$ into ${L}^{1}(\mathrm{\Omega},X)$.
 (e)$\parallel u(t,s,x)\parallel \le \kappa (t,s)\{\xi (s)+\mu \parallel x\parallel \}$ for $(t,s)\in {\mathrm{\Omega}}^{2}$ and $x\in X$, where ξ belongs to ${L}_{+}^{1}(\mathrm{\Omega})$, μ is a nonnegative constant and $\kappa :\mathrm{\Omega}\times \mathrm{\Omega}\to {\mathcal{R}}^{+}$ is a measurable function such that its associated integral operator K defined by$(K\rho )(t)={\int}_{\mathrm{\Omega}}\kappa (t,s)\rho (s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\rho \in {L}^{1}(\mathrm{\Omega}),t\in \mathrm{\Omega},$(4)
 (f)There exists a constant $N>0$ independent of ${\lambda}^{\ast}\in (0,1)$ such that any solution of the integral equation$y(t)={\lambda}^{\ast}g(t,\frac{1}{{\lambda}^{\ast}}y(t))+{\lambda}^{\ast}T{N}_{{f}_{1}}U{N}_{{f}_{2}}Ay(t),\phantom{\rule{1em}{0ex}}t\in \mathrm{\Omega},$
satisfies ${\parallel y\parallel}_{{L}^{1}(\mathrm{\Omega},X)}\ne N$.
Remark 2 It is deserved to mention that though the Urysohn operator U maps ${L}^{1}(\mathrm{\Omega},X)$ into itself, it does not have to be continuous. Sufficient conditions showing that U maps ${L}^{1}(\mathrm{\Omega},X)$ into itself and is continuous can be found in [18].
Before going on, we give crucial Lemma 5.
Lemma 5 Let X be a finitedimensional Banach space and let Ω be a compact subset of ${\mathcal{R}}^{n}$. If the conditions (b)(e) are satisfied, then the operator ${N}_{{f}_{1}}U{N}_{{f}_{2}}A$ satisfies the condition H2.
for any bounded subset S of ${L}^{1}(\mathrm{\Omega},X)$.
Next, let ${({y}_{n})}_{n\in N}$ be a weakly convergent sequence of ${L}^{1}(\mathrm{\Omega},X)$. Owing to (6), we infer that $\omega \{{N}_{{f}_{1}}U{N}_{{f}_{2}}A({y}_{n}):n\in N\}=0$. This shows that the set $\{{N}_{{f}_{1}}U{N}_{{f}_{2}}A({y}_{n}):n\in N\}$ is relatively weakly compact in ${L}^{1}(\mathrm{\Omega},X)$. This completes the proof. □
where $l(t)=\parallel g(t,0)\parallel \in {L}_{+}^{1}(\mathrm{\Omega})$.
This shows that the operator ℬ is continuous and maps a bounded set of ${L}^{1}(\mathrm{\Omega},X)$ into a bounded set of ${L}^{1}(\mathrm{\Omega},X)$. According to Lemma 3, we obtain ℬ satisfies the condition H2.
Now we are in a position to state our main result.
Theorem 2 Let X be a finitedimensional Banach space and let Ω be a bounded domain of ${\mathcal{R}}^{n}$. Assume that the conditions (a)(f) hold true. Then Eq. (1) admits at least one solution in ${L}^{1}(\mathrm{\Omega},X)$.
So, ℬ is a strict contraction mapping on ${L}^{1}(\mathrm{\Omega},X)$, and from Remark 3, ℬ satisfies the condition H2.
Claim 2. Clearly, by the assumptions (b)(d), $\mathcal{A}=T{N}_{{f}_{1}}U{N}_{{f}_{2}}A$ is continuous on ${L}^{1}(\mathrm{\Omega},X)$. Now we check that $\mathcal{A}$ satisfies the condition H1. To do this, let ${({y}_{n})}_{n\in N}$ be a weakly convergent sequence of ${L}^{1}(\mathrm{\Omega},X)$. By Lemma 5, ${({N}_{{f}_{1}}U{N}_{{f}_{2}}A({y}_{n}))}_{n\in N}$ has a weakly convergent subsequence, say ${({N}_{{f}_{1}}U{N}_{{f}_{2}}A({y}_{{n}_{k}}))}_{k\in N}$. Furthermore, the continuity of the linear operator T implies its weak continuity on ${L}^{1}(\mathrm{\Omega},X)$ for almost all $t\in \mathrm{\Omega}$. Thus, the sequence ${(T{N}_{{f}_{1}}U{N}_{{f}_{2}}A({y}_{{n}_{k}}))}_{k\in N}$, i.e., ${(\mathcal{A}({y}_{{n}_{k}}))}_{k\in N}$ converges pointwisely for almost all $t\in \mathrm{\Omega}$. By the Vitali convergence theorem [[19], p.150], we conclude that ${(\mathcal{A}({y}_{{n}_{k}}))}_{k\in N}$ converges strongly in ${L}^{1}(\mathrm{\Omega},X)$. Hence, $\mathcal{A}$ satisfies the condition H1.
Owing to (5), we deduce that $\omega (\mathcal{A}(\overline{S}))=0$, and hence $\mathcal{A}(\overline{S})$ is relatively weakly compact.
Finally, thanks to the assumption (f), the second situation of Theorem 1 does not occur. Now, applying Theorem 1, we get that $\mathcal{A}+\mathcal{B}$ has a fixed point in $\overline{S}$, that is to say, Eq. (1) has a solution in $\overline{S}$. This completes the proof. □
Remark 4 The requirement that X should be a finitedimensional Banach space comes from the usage of the relation (3) proved in [15] for bounded subsets in the space of Lebesgue integrable functions with values in a finitedimensional Banach space.
By Theorem 2, we can get a special existence criterion for Eq. (1).
 (i)There exists a continuous function $h:[0,\mathrm{\infty})\to [0,\mathrm{\infty})$ such that $h(u)>0$ whenever $u>0$ and${\int}_{\mathrm{\Omega}}{\parallel {N}_{{f}_{1}}U{N}_{{f}_{2}}Ay(t)\parallel}_{X}\phantom{\rule{0.2em}{0ex}}dt\le h\left({\parallel y\parallel}_{{L}^{1}(\mathrm{\Omega},X)}\right)\phantom{\rule{1em}{0ex}}\mathit{\text{for every}}y\in {L}^{1}(\mathrm{\Omega},X).$
 (ii)$\underset{\theta \in [0,\mathrm{\infty})}{sup}\left(\frac{(1L)\theta}{{\parallel l\parallel}_{{L}_{+}^{1}}+\parallel T\parallel h(\theta )}\right)>1,$
where $l(t):=\parallel g(t,0)\parallel $ and L is the constant in the assumption (a). Then Eq. (1) has a solution in ${L}^{1}(\mathrm{\Omega},X)$.
Assuming that ${\parallel y\parallel}_{{L}^{1}(\mathrm{\Omega},X)}=N$. Equation (9) implies $\frac{(1L)N}{{\parallel l\parallel}_{{L}_{+}^{1}}+\parallel T\parallel h(N)}\le 1$ contradicting (7). So, each solution of (8) satisfies ${\parallel y\parallel}_{{L}_{+}^{1}}\ne N$. Accordingly, by Theorem 2, Eq. (1) has a solution $y\in {L}^{1}(\mathrm{\Omega},X)$. This completes the proof. □
 (iii)
$L+\mu {b}_{1}{b}_{2}\parallel K\parallel \parallel A\parallel <1$, where the constants ${b}_{1}$, ${b}_{2}$, μ are these in Lemma 2 and the hypothesis (f), $\parallel K\parallel $ denotes the norm of the operator K defined in (4).
Then Eq. (1) has a solution in ${L}^{1}(\mathrm{\Omega},X)$.
which is a contradiction. So, the hypothesis (g) is satisfied and the result then follows from Theorem 2. This completes the proof. □
4 Examples
In this section, we provide some examples of a classical integral and functional equation considered in nonlinear analysis which are a particular case of Eq. (1).
is also a special case of Eq. (1) with $T=B$ and ${f}_{2}(t,y)=y$.
with $0\le s\le t\le 1$.
We can suppose T to be an arbitrary linear bounded operator on $L[0,1]$. It is easy to see that the function g satisfies the assumption (a) with $L=\frac{1}{6}$, the function $u(t,s,x)\le \frac{exp(2s)}{exp(t)}(exp(s)+x)$ with $k(t,s)=\frac{exp(2s)}{exp(t)}$ and $\mu =1$.
Declarations
Acknowledgements
The work is supported by the National Natural Science Foundation of China (No. 10861014, No. 11161057).
Authors’ Affiliations
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