On Chandrasekhar functional integral inclusion and Chandrasekhar quadratic integral equation via a nonlinear Urysohn–Stieltjes functional integral inclusion

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.

Our aim is to study the existence of solutions x ∈ C[0, 1] of the U-S nonlinear functional integral inclusion x(t)a(t) ∈ 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.

Single-valued problem
Here we consider the nonlinear single-valued functional integral equation of U-S type

Existence of solutions I
Consider the U-S functional integral equation ( (iv) For all t 1 , t 2 ∈ I, t 1 < t 2 , the functions s → g i (t 2 , s)g i (t 1 , s) are nondecreasing on [0, 1].

Lemma 2.2 ([11])
Assume that a function g satisfies assumption (v). Then for arbitrary s 1 , s 2 ∈ I with s 1 < s 2 , the function t → g(t, s 2 )g(t, s 1 ) is nondecreasing on I.
Indeed, take t 1 , t 2 ∈ [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. Proof Define the operator A by and define let the set It is clear that Q r is a nonempty, bounded, closed, and convex set. Let x ∈ Q r . Then This proves that the operator A : Q r → Q r and the class {Ax} is uniformly bounded on Q r . Then, for x ∈ Q r and y(s) Then from the uniform continuity of the function f : s, x(s), y(s) d s g 1 (t 2 , s) ≤ a(t 2 )a(t 1 ) + 1 0 θ (δ) d s g 1 (t 2 , s) This inequality means that the class of functions {Ax} is equicontinuous.
Therefore by the Arzelà-Ascoli theorem [25] A is compact.
and from assumption (ii) (see [23]) we get This proves that Ax n (t) → Ax(t) and A is continuous. Now (see [23]) A has at least one fixed point x ∈ Q r , and (2.1) has at least one solution x ∈ Q r ⊂ C[0, 1].

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 (ii) * a) the function f : I × I × R × R → R is continuous and satisfies the Lipschitz condition Then Proof Let x 1 , x 2 be solutions of the integral equation (2.1). Then Hence we have which implies

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 > 0, there exists δ > 0 such that Proof g 1 (t, s)) Taking the supremum over t ∈ I, we get
Now we get that the solution of (2.1) continuously depends on the functions g i , i = 1, 2.

Existence of solutions II
Now we replace assumptions (ii) a), (vi) by 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 ∈ Q l . Then
This proves that A * : Q l → Q l and the class {A * x} is uniformly bounded on Q l . Now for x ∈ Q r and y(s) = 1 0 h(s, θ , x(θ )) d θ g 2 (s, θ ), define the set Then from the uniform continuity of the function f : [0, 1] × [0, 1] × Q l × Q l → R and assumption (ii * ) we deduce that θ (δ) → 0 as δ → 0, independently of x ∈ Q l . Now let t 2 , t 1 ∈ [0, 1] be such that |t 2t 1 | < δ. 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 } ⊂ Q l , x n → x. Then and by assumption (ii * ) (see [23]) we get This proves that A * x n (t) → A * x(t) and A * is continuous. So (see [23]) A * has at least one fixed point x ∈ Q r , and (2.1) has at least one solution x ∈ Q l ⊂ C([0, 1]). Then g 1 , g 2 satisfy our assumptions (iii)-(v), and we obtain the nonlinear Chandrasekhar functional integral equation

Application
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 ∈ C[0, 1].

Set-valued problem
Consider the U-S nonlinear functional integral inclusion (1.1), under the following assumptions: , is a Lipschitzian set-valued map with a nonempty compact convex subset of 2 R , with a Lipschitz constant k 1 > 0: 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 ∈ F such that f (t, s, x 1 , y 1 )f (t, s, x 2 , y 2 ) ≤ k 1 |x 1x 2 | + |y 1y 2 | .  Note that the Lipschitz selection f :

Existence of solution
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 * ∈ S F , we have Now, taking the supremum over t ∈ I, we get
Thus from last inequality we get xx * ≤ .
This proves the continuous dependence of the solution on the set S F .

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 Further, since the functions g i satisfy assumptions (iii)-(v) (see [6]), we obtain the nonlinear Chandrasekhar functional integral inclusion x(t) ∈ a(t) +
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 ∈ C[0, 1].