Abstract
We investigate the existence of solutions for a nonlinear integral inclusion of Urysohn–Stieltjes type. As applications, we give a Chandrasekhar quadratic integral equation and a nonlinear Chandrasekhar integral inclusion.
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1 Introduction
The integral equations of Urysohn–Stieltjes (U-S) type have been studied by some authors; see, for example, [3, 5, 11–15], and [16–22], and reference therein.
The quadratic Chandrasekhar integral equation
has been studied in some papers; see, for example, [1, 4, 7–10], and [24] and references therein.
Our aim is to study the existence of solutions \(x\in C[0,1]\) of the U-S nonlinear functional integral inclusion
As applications, we will prove the existence of solutions \(x\in C[0,1]\) of the nonlinear Chandrasekhar functional integral inclusion
and the Chandrasekhar quadratic integral equation
The paper is organized as follows. In Sect. 2, we establish the existence and uniqueness results for single-valued nonlinear U-S equations. We also prove the continuous dependence of the unique solution on the \(g_{i} \) (\(i=1,2\)). As an application, we discuss some particular cases by presenting the existence of solutions of nonlinear Chandrasekhar quadratic functional integral equations. In Sect. 3, we add conditions to our problem in order to obtain a new existence result with an application. Our results are generalized in Sect. 4, where we discuss the existence of solutions for set-valued equation (1.1) with continuous dependence on the set \(S_{F}\) and demonstrate a particular case of inclusion by presenting the existence of solutions for set-valued Chandrasekhar nonlinear functional integral equations.
2 Single-valued problem
Here we consider the nonlinear single-valued functional integral equation of U-S type
2.1 Existence of solutions I
Consider the U-S functional integral equation (2.1) under the following assumptions:
-
(i)
\(a: [0,1]\rightarrow [0,1] \) is a continuous function, with \(a = \sup_{t\in [0,1]} \vert a(t) \vert \).
-
(ii)
-
a)
\(f:[0,1]\times [0,1]\times R \times R \rightarrow R \) is a continuous function, and there exist two continuous functions \(m_{1},k_{1}:[0,1]\times [0,1]\rightarrow R\) such that
$$ \bigl\vert f(t,s,x,y) \bigr\vert \leq m_{1}(t,s)+k_{1}(t,s) \bigl( \vert x \vert + \vert y \vert \bigr). $$ -
b)
\(h:[0,1]\times [0,1]\times R \rightarrow R \) is a continuous function, and there exist two continuous functions \(m_{2},k_{2}:[0,1]\times [0,1]\rightarrow R\) such that
$$ \bigl\vert h(t,s,x) \bigr\vert \leq m_{2}(t,s)+k_{2}(t,s) \vert x \vert . $$ -
c)
\(k=\sup \{ k_{i}(t,s):t, s\in [0,1]\}\), and \(m=\sup \{ m_{i}(t,s):t, s\in [0,1], i=1,2\}\).
-
a)
-
(iii)
\(g_{i}:[0,1]\times R \rightarrow R\), \(i=1,2\), are continuous functions with
$$ \mu =\max \bigl\{ \sup \bigl\vert g_{i}(t,1) \bigr\vert +\sup \bigl\vert g_{i}(t,0) \bigr\vert , \text{on } [0,1] \bigr\} . $$ -
(iv)
For all \(t_{1},t_{2}\in I\), \(t_{1}< t_{2}\), the functions \(s \rightarrow g_{i}(t_{2},s)-g_{i}(t_{1},s)\) are nondecreasing on \([0,1]\).
-
(v)
\(g_{i}(0,s)= 0\) for \(s \in [0,1]\).
-
(vi)
\(k\mu +k^{2}\mu ^{2}<1\).
Let E be a Banach space with the norm \(\Vert \cdot \Vert _{E}\), and let \(I = [0, 1]\). Denote by \(C = C (I, E)\) the space of all continuous functions on I taking values in the space E. This space becomes a Banach space with supnorm
Remark 2.1
(see [11])
Note that the function \(s\rightarrow g(t,s)\) is nondecreasing on the interval \([0,1]\). Indeed, for \(s_{1}, s_{2}\in [0,1]\) with \(s_{1}< s_{2}\), from assumptions (iv) and (v) we obtain
Lemma 2.2
([11])
Assume that a function g satisfies assumption (v). Then for arbitrary \(s_{1}, s_{2}\in I\) with \(s_{1}< s_{2}\), the function \(t\rightarrow g(t,s_{2})-g(t,s_{1})\) is nondecreasing on I.
Indeed, take \(t_{1},t_{2}\in [0,1]\) such that \(t_{1}< t_{2}\). Then by assumption (vi) we get
For the existence of at least one solution of the U-S nonlinear functional integral equation (2.1), we have the following theorem.
Theorem 2.3
Let the assumptions (i)–(vi) be satisfied. Then the functional integral equation (2.1) has at least one solution \(x\in C[0,1]\).
Proof
Define the operator A by
and define let the set
where
It is clear that \(Q_{r}\) is a nonempty, bounded, closed, and convex set.
Let \(x\in Q_{r} \). Then
This proves that the operator \(A: Q_{r} \rightarrow Q_{r}\) and the class \(\{A x\}\) is uniformly bounded on \(Q_{r}\).
Then, for \(x\in Q_{r} \) and \(y(s)=\int _{0}^{1}h(s,\theta ,x(\theta )) \,d_{\theta }g_{2}(s,\theta )\), define the set
Then from the uniform continuity of the function \(f: [0,1]\times [0,1]\times Q_{r}\times Q_{r} \rightarrow R \) and assumption (ii) we deduce that \(\theta (\delta ) \rightarrow 0 \) as \(\delta \rightarrow 0\), independently of \(x \in Q_{r}\).
Now let \(t_{2}, t_{1} \in [0,1]\), \(\vert t_{2}-t_{1} \vert <\delta\). Then we have
This inequality means that the class of functions \(\{A{x}\} \) is equicontinuous.
Therefore by the Arzelà–Ascoli theorem [25] A is compact.
Let \(\{x_{n}\}\subset Q_{r}\), \(x_{n}\rightarrow x\). Then
and from assumption (ii) (see [23]) we get
This proves that \(Ax_{n}(t)\rightarrow A x(t)\) and A is continuous.
Now (see [23]) A has at least one fixed point \(x \in Q_{r}\), and (2.1) has at least one solution \(x \in Q_{r}\subset C[0,1]\). □
2.2 Uniqueness of the solution
To prove the existence of a unique solution of U-S functional integral equation (2.1), let us replace condition (ii) by
- \((\mathrm{ii})^{*}\):
-
-
a)
the function \(f:I\times I\times R \times R \rightarrow R \) is continuous and satisfies the Lipschitz condition
$$ \bigl\vert f(t,s,x_{1},y_{1})-f(t,s,x_{2},y_{2}) \bigr\vert \leq k_{1} \bigl( \vert x_{1}-x_{2} \vert + \vert y_{1}-y_{2} \vert \bigr). $$ -
b)
\(h:I\times I\times R \rightarrow R \) is continuous and satisfies the Lipschitz condition
$$ \bigl\vert h(t,s,x)-h(t,s,y) \bigr\vert \leq k_{2} \vert x-y \vert . $$
-
a)
By condition \((\mathrm{ii})^{*}\) we have
Then
and
where \(m_{1}=\sup_{t\times s\in I\times I} \vert f(t,s,0,0) \vert \), and
Then
and
where \(m_{2}=\sup_{t\times s\in I\times I} \vert h(t,s,0) \vert \), \(m=\max \{m_{1},m_{2}\}\), and \(k=\max \{k_{1},k_{2}\}\).
Theorem 2.4
Let conditions (i), \((\mathit{ii})^{*}\), (iii), and (iv)–(v) be satisfied with \(\mu k+k^{2}\mu ^{2}\leq 1\). Then the functional integral equation (2.1) has unique solution \(x \in C[0,1] \).
Proof
Let \(x_{1}\), \(x_{2}\) be solutions of the integral equation (2.1). Then
Hence we have
and
which implies
□
2.2.1 Continuous dependence of solution on functions \(g_{i}(t,s)\)
Here we show that the solution of U-S functional integral equation (2.1) continuously depends on the functions \(g_{i}\).
Definition 2.5
The solutions of functional integral equation (2.1) continuously depends on the functions \(g_{i}(t,s)\), \(i=1,2\), if for every \(\epsilon >0\), there exists \(\delta >0 \) such that
Theorem 2.6
Let the assumptions of Theorem 2.4be satisfied. Then the solution of (2.1) depends continuously on functions \(g_{i}(t,s)\), \(i=1,2\).
Proof
Let \(\delta >0\) be such that \(\vert g_{i}(t,s)-g_{i}^{*}(t,s) \vert \leq \delta \) for all \(t \geq 0 \). Then
Taking the supremum over \(t\in I\), we get
Then
Now we get that the solution of (2.1) continuously depends on the functions \(g_{i}\), \(i=1,2\). □
3 Existence of solutions II
Now we replace assumptions (ii) a), (vi) by
- (\(\mathrm{ii} ^{*}\)):
-
- \(\mathrm{a}^{*}\)):
-
\(f:[0,1]\times [0,1]\times R \times R \rightarrow R \) is a function, and there exist two continuous functions \(m_{1},k_{1}:[0,1]\times [0,1]\rightarrow R \) such that
$$ \bigl\vert f(t,s,x,y) \bigr\vert \leq m_{1}(t,s)+k_{1}(t,s) \vert x \vert \cdot \vert y \vert . $$
- (\(\mathrm{vi} ^{*}\)):
-
There exists a positive root l of the algebraic equation
$$ \mu ^{2}k^{2} l^{2}+ \bigl(k\mu ^{2}m-1 \bigr) l+(a+m \mu )=0. $$
Theorem 3.1
Let the assumptions of Theorem 2.3be satisfied with (ii) a) and (vi) replaced by (\(ii^{*}\)) \(a^{*})\) and (\(vi^{*}\)), respectively. Then equation (2.1) has at least one solution \(x\in C[0,1]\).
Proof
Define the operator \(A^{*}\) by
and define the set
where l is a positive root of the algebraic equation
It is clear that \(Q_{l}\) is a nonempty, bounded, closed, and convex set.
Now let \(x\in Q_{l} \). Then
This proves that \(A^{*}: Q_{l} \rightarrow Q_{l} \) and the class \(\{A^{*} x\}\) is uniformly bounded on \(Q_{l}\).
Now for \(x\in Q_{r} \) and \(y(s)=\int _{0}^{1}h(s,\theta ,x(\theta )) \,d_{\theta}g_{2}(s,\theta )\), define the set
Then from the uniform continuity of the function \(f: [0,1]\times [0,1]\times Q_{l}\times Q_{l} \rightarrow R \) and assumption (\(\mathrm{ii} ^{*}\)) we deduce that \(\theta (\delta ) \rightarrow 0 \) as \(\delta \rightarrow 0\), independently of \(x \in Q_{l}\).
Now let \(t_{2}, t_{1} \in [0,1] \) be such that \(\vert t_{2}-t_{1} \vert <\delta \). Then we have
This inequality means that the class of functions \(\{A^{*}{x}\} \) is equicontinuous. Therefore \(A^{*} \) is compact by the Arzelà–Ascoli theorem [25].
Let \(\{x_{n}\}\subset Q_{l}\), \(x_{n}\rightarrow x\). Then
and by assumption (\(\mathrm{ii} ^{*}\)) (see [23]) we get
This proves that \(A^{*}x_{n}(t)\rightarrow A^{*} x(t)\) and \(A^{*}\) is continuous. So (see [23]) \(A^{*}\) has at least one fixed point \(x \in Q_{r}\), and (2.1) has at least one solution \(x \in Q_{l}\subset C([0,1])\). □
3.1 Application
Let in equation (2.1), \(h(t,s,x(s))=b_{2}(t)x(s)\),
and
Then \(g_{1}\), \(g_{2}\) satisfy our assumptions (iii)–(v), and we obtain the nonlinear Chandrasekhar functional integral equation
Let, in equation (3.1), \(f(t,s,x(s),y(s))=b_{1}(s)x(s)\cdot y(s)\), where
Then we obtain the Chandrasekhar quadratic functional integral equation of the form
Now, under the assumptions of Theorem 3.1, the Chandrasekhar quadratic functional integral equation (3.2) has at least one solution \(x\in C[0,1]\).
3.2 Example
Consider the following Chandrasekhar quadratic functional integral equation:
First, note that equation (3.3) is a particular case of equation (3.2) if we put
\(b_{1}(s)=\frac{2\cos (s) }{7e^{2s} (1+\cos ^{2} (s))}\), \(b_{2}(s)=\frac{\sin (s)}{4(1+\sin ^{2}(s))}\), with \(k_{1}=\frac{2}{7}\) and \(k_{2}=\frac{1}{4}\).
Thus conditions (i), (\(\mathrm{ii} ^{*}\)) and (iii) are satisfied with \(a=\frac{1}{10}\), \(k=\frac{1}{4}\), and \(m=0\). By all facts established above, we deduce that condition (\(\mathrm{vi} ^{*}\)) of the form
has a positive solution l. For example, if \(l \thickapprox 0.1\) or \(l \thickapprox 33\), then assumption (\(\mathrm{vi} ^{*}\)) will be satisfied if we choose one of this values.
As all the conditions of Theorem 3.1 are satisfied, equation (3.3) has at least one solution \(x\in C[0,1]\).
4 Set-valued problem
Consider the U-S nonlinear functional integral inclusion (1.1),
under the following assumptions:
- (i):
-
\(a: [0,1]\rightarrow [0,1] \) is a continuous function.
- \((\mathrm{ii})^{***}\):
-
-
(a)
\(F : [0,1] \times [0,1] \times R \times R \rightarrow P(R) \), is a Lipschitzian set-valued map with a nonempty compact convex subset of \(2^{R}\), with a Lipschitz constant \(k_{1} > 0\):
$$ \bigl\Vert F(t, s,x_{1},y_{1}) - F(t,s,x_{2},y_{2}) \bigr\Vert \leq k_{1} \bigl( \vert x_{1}-x_{2} \vert + \vert y_{1}-y_{2} \vert \bigr). $$Remark. From this assumption and Theorem 1 from [2, Sect. 9, Chap. 1] on the existence of Lipschitzian selection we deduce that the set of Lipschitz selections of F is not empty and there exists \(f\in F\) such that
$$ \bigl\vert f(t,s,x_{1},y_{1})-f(t,s,x_{2},y_{2}) \bigr\vert \leq k_{1} \bigl( \vert x_{1}-x_{2} \vert + \vert y_{1}-y_{2} \vert \bigr). $$ -
(b)
\(h:[0,1]\times [0,1]\times R \rightarrow R \) is a continuous function such that
$$ \bigl\vert h(t,s,x) \bigr\vert \leq m_{2}(t,s)+k_{2}(t,s) \vert x \vert . $$ -
(c)
\(k=\sup_{(t, s)\in [0,1]\times [0,1]} k_{i}(t,s)\) and \(m=\sup_{(t, s)\in [0,1]\times [0,1]} m_{i}(t,s)\).
-
(a)
- (iii):
-
\(g_{i}:[0,1]\times R \rightarrow R\), \(i=1,2\), are continuous with
$$ \mu =\max \bigl\{ \sup \bigl\vert g_{i} \bigl(t,\varphi (t) \bigr) \bigr\vert +\sup \bigl\vert g_{i}(t,0) \bigr\vert \text{ on } [0,1] \bigr\} . $$ - (iv):
-
For all \(t_{1},t_{2}\in [0,1]\), \(t_{1}< t_{2}\), the functions \(s\rightarrow g_{i}(t_{2},s)-g_{i}(t_{1},s)\) are nondecreasing on \([0,1]\).
- (v):
-
\(g_{i}(0,s)= 0\) for any \(s \in [0,1]\).
- (vi):
-
\(k\mu +k^{2}\mu ^{2}<1\).
4.1 Existence of solution
Theorem 4.1
Let assumptions \((i)\)–\((ii)^{***}\), and \((iv)\)–\((vi)\) be satisfied. Then (1.1) has at least one solution \(x\in C[0,1]\).
Proof
By assumption \((\mathrm{ii})^{***}\)-(a) it is clear that the set of Lipschitz selection of F is nonempty. So, the solution of the single-valued (2.1) where \(f \in S_{F}\) is a solution to (1.1).
Note that the Lipschitz selection \(f: [0,1]\times [0,1]\times R \times R \rightarrow R\) satisfies
From this condition with \(m_{1}=\sup_{(t,s) \in I \times I} \vert f(t,s,0,0) \vert \) we have
Then
and
that is, assumption (ii) of Theorem 2.3 is satisfied. So, all conditions of Theorem 2.3 hold.
Note that if \(x \in C(I,R)\) is a solution of (2.1), then x is a solution to (1.1). □
4.1.1 Continuous dependence on the set of selection \(S_{F}\)
Here we study the continuous dependence on the set \(S_{F}\) of all selections of the set-valued function F.
Definition 4.2
The solution of (1.1) continuously depends on the set \(S_{F}\) if for all \(\epsilon >0\), there exists \(\delta >0 \) such that if
then \(\Vert x - x^{*} \Vert < \epsilon \).
Now we have the following theorem.
Theorem 4.3
Let the assumptions of Theorem 4.1be satisfied with
Then the solution of (1.1) continuously depends on the set \(S_{F}\) of all Lipschitzian selections of F.
Proof
For two solutions \(x(t)\) and \(x^{*}(t)\) of (1.1) corresponding to two selections \(f, f^{*} \in S_{F}\), we have
Now, taking the supremum over \(t\in I\), we get
Hence
Thus from last inequality we get
This proves the continuous dependence of the solution on the set \(S_{F}\). □
4.2 Set-valued Chandrasekhar nonlinear quadratic functional integral inclusion
Now, as an application of the nonlinear set-valued functional integral equations of U-S type (1.1), we have the following. Let the functions \(g_{i} \) be defined by
and
Let, in (1.1), \(h(t,s,x(s))=b_{2}(s) x(s)\) and \(F (t,s,x(s),y(s) )=F (b_{1}(s) x(s),y(s) )\), where
Further, since the functions \(g_{i}\) satisfy assumptions (iii)–(v) (see [6]), we obtain the nonlinear Chandrasekhar functional integral inclusion
Now we can state the following existence result for (4.1).
Theorem 4.4
Under the assumptions of Theorem 4.1, inclusion (4.1) has at least one continuous solution \(x \in C[0,1]\).
4.3 Example
Consider the following nonlinear Chandrasekhar functional integral inclusion:
Note that this inclusion is a particular case of inclusion (4.1) if we choose \(F:[0,1]\times \mathbb{R}\to 2^{\mathbb{R}^{+}}\) in (4.2) as follows:
Further, note that now the terms involved in (4.1) have the form
with \(b_{1}(s)=\frac{1}{{s^{2}+1}}\) and \(b_{2}(s)=\frac{\sqrt{s}}{e^{s+1}}\).
Let \(f:[0,1]\times {R}\to {R}\) be a continuous map. Note that if \(f \in S_{F}\), then we have
and
Thus conditions (i) and \((\mathrm{ii})^{*}\) are satisfied with \(a=e\), \(k_{1}=\frac{\sqrt{\pi }}{e^{2} (\pi +1 )}\), and \(k_{2}=\frac{1}{e^{2}}\).
Moreover, we have
This shows that assumption (vii) is satisfied. So, as all the conditions of Theorem 4.4 are satisfied, inclusion (4.2) has at least one solution \(x \in C[0,1]\).
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El-Sayed, A., Al-Issa, S. & Omar, Y. On Chandrasekhar functional integral inclusion and Chandrasekhar quadratic integral equation via a nonlinear Urysohn–Stieltjes functional integral inclusion. Adv Differ Equ 2021, 137 (2021). https://doi.org/10.1186/s13662-021-03298-9
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DOI: https://doi.org/10.1186/s13662-021-03298-9