Solvability of a Quadratic Integral

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Mathematics

and Applications

JMA No 43, pp 47-66 (2020)

COPYRIGHT c by Publishing House of Rzesz´ow University of Technology P.O. Box 85, 35-959 Rzesz´ow, Poland

Solvability of a Quadratic Integral

Equation of Fredholm Type Via a Modified Argument

˙Ilyas Dal and ¨ Omer Faruk Temizer

Abstract: This article concerns with the existence of solutions of the a quadratic integral equation of Fredholm type with a modified argument,

x(t) = p(t) + (F x) (t) Z 1

0

k(t, τ )x(q (τ ))dτ,

where p, k are functions and F is an operator satisfying the given conditions. Using the properties of the H¨older spaces and the classical Schauder fixed point theorem, we obtain the existence of solutions of the equation under certain assumptions. Also, we present two concrete examples in which our result can be applied.

AMS Subject Classification: 45G10, 45M99, 47H10.

Keywords and Phrases: Fredholm equation; H¨older condition; Schauder fixed point theorem.

1. Introduction

Integral equations arise from naturally in many applications in describing numer- ous real world problems (see, for instance, the books [2, 3] and references therein).

Quadratic integral equations arise naturally in applications of real world problems.

For example, problems in the theory of radiative transfer in the theory of neutron transport and in the kinetic theory of gases lead to the quadratic equation [12, 20].

There are many interesting existence results for all kinds of quadratic integral equations, one can refer to [6, 1].

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The study of differential equations with a modified arguments arise in a wide variety of scientific and technical application, including the modelling of problems from the natural and social sciences such as physics, biological and economics sciences. A special class of these differential equations have linear modifications of their arguments, and have been studied by several authors, [7] - [23].

Recently, Bana´s and Nalepa [7] have studied the space of real functions defined on a given bounded metric space and having the growths tempered by a given modulus of continuity, and derive the existence theorem in the space of functions satisfying the H¨older condition for some quadratic integral equations of Fredholm type

x(t) = p(t) + x(t) Z b

a

k(t, τ ) x(τ )dτ. (1.1)

Further, Caballero et al. [9] have studied the solvability of the following quadratic integral equation of Fredholm type

x(t) = p(t) + x(t) Z 1

0

k(t, τ ) x(q (τ ))dτ (1.2)

in H¨older spaces. The purpose of this paper is to investigate the existence of solutions of the following integral equation of Fredholm type with a modified argument in H¨older spaces

x(t) = p(t) + (F x) (t) Z 1

0

k(t, τ ) x(q (τ ))dτ, t ∈ I = [0, 1] (1.3) where p, k, q and F are functions satisfying the given conditions. To do this, we will use a recent result about the relative compactness in H¨older spaces and the classical Schauder fixed point theorem.

Notice that equation (1.1) in [9] is a particular case of (1.3), for (F x)(τ ) = x(τ ).

The obtained result in this paper is more general than the result in [9].

2. Preliminaries

Let we introduce notations, definitions and theorems which are used throughout this paper.

By C[a, b], we denote the space of continuous functions on [a, b] equipped with usually the supremum norm

kxk_∞= sup{|x(t)| : t ∈ [a, b]}

for x ∈ C[a, b]. For a fixed α with 0 < α 6 1, we write Hα[a, b] to denote the set of all the real valued functions x defined on [a, b] and satisfying the H¨older condition with α, that is, there exists a constant H such that the inequality

|x(t) − x(s)| 6 H|t − s|^α (2.1)

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holds for all t, s ∈ [a, b]. One can easily seen that H_α[a, b] is a linear subspaces of C[a, b]. In the sequel, for x ∈ H_α[a, b], by H_x^α we will denote the least possible constant for which inequality (2.1) is satisfied. Rather, we put

H_x^α= sup |x(t) − x(s)|

|t − s|^α : t, s ∈ [a, b], t 6= s

. (2.2)

The space H_α[a, b] with 0 < α 6 1 can be equipped with the norm:

kxkα= |x(a)| + sup |x(t) − x(s)|

|t − s|^α : t, s ∈ [a, b], t 6= s

(2.3) for x ∈ H_α[a, b]. In [7], the authors proved that (H_α[a, b], k · k_α) with 0 < α 6 1 is a Banach space. The following lemmas in [7] present some results related to the H¨older spaces and norm.

Lemma 2.1. For 0 < α 6 1 and x ∈ H^α[a, b], the following inequality is satisfied kxk∞6 max {1, (b − a)^α} kxkα.

In particular, the inequality kxk∞6 kxk^α holds, for a = 0 and b = 1.

Lemma 2.2. For 0 < α < γ 6 1, we have

Hγ[a, b] ⊂ Hα[a, b] ⊂ C[a, b].

Moreover, for x ∈ Hγ[a, b] the following inequality holds kxkα6 max1, (b − a)^γ−α kxkγ.

In particular, the inequality kxk_∞6 kxk^α6 kxk^γ is satisfied for a = 0 and b = 1.

Now we present the important theorem which is the sufficient condition for relative compactness in the spaces H_α[a, b] with 0 < α 6 1.

Theorem 2.3. [9] Suppose that 0 < α < β 6 1 and that A is a bounded subset of H_β[a, b] (this means that kxk_β6 M for certain constant M > 0, for all x ∈ A) then A is a relatively compact subset of H_α[a, b].

Lemma 2.4. [9] Suppose that 0 < α < β 6 1 and by Br^β we denote the closed ball centered at θ with radius r in the space H_β[a, b], i.e., B_r^β= {x ∈ H_β[a, b] : kxk_β6 r}.

Then B_r^β is a closed subset of H_α[a, b].

Corollary 2.5. Suppose that 0 < α < β 6 1 then B^βr is a compact subset of the space H_α[a, b], [9].

Theorem 2.6 (Schauder’s fixed point theorem). Let L be a nonempty, convex, and compact subset of a Banach space (X, k·k) and let T : L → L be a continuity mapping.

Then T has at least one fixed point in L, [24].

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3. Main Result

In this section, we will study the solvability of the equation (1.3) in the space H_α[0, 1]

(0 < α 6 1). We will use the following assumptions:

(i) p ∈ Hβ[0, 1], 0 < β 6 1.

(ii) k : [0, 1] × [0, 1] → R is a continuous function such that it satisfies the H¨older condition with exponent β with respect to the first variable, that is, there exists a constant kβ > 0 such that:

|k(t, τ ) − k(s, τ )| 6 kβ|t − s|^β,

for any t, s, τ ∈ [0, 1].

(iii) q : [0, 1] → [0, 1] is a measurable function.

(iv) The operator F : Hβ[0, 1] → Hβ[0, 1] is continuous with respect to the norm k · kα for 0 < α < β 6 1 and there exists a function f : R⁺ → R⁺= [0, ∞) which is non-decreasing such that it holds the inequality

kF xkβ6 f (kxk^β), for any x ∈ H_β[0, 1].

(v) There exists a positive solution r0 of the inequality kpkβ+ (2K + k_β)rf (r) 6 r, where K is a constant is satisfying the following inequality,

K = sup

Z 1 0

|k(t, τ )|dτ : t ∈ [0, 1]

.

Theorem 3.1. Under the assumptions (i)-(v), Equation (1.3) has at least one solution belonging to the space H_α[0, 1].

Proof. Consider the operator T below that defined on the space Hβ[0, 1] by

(T x)(t) = p(t) + (F x) (t) Z 1

0

k(t, τ )x(q (τ ))dτ, t ∈ [0, 1].

We will firstly prove that T transforms the space H_β[0, 1] into itself. For arbitrarily fixed x ∈ H_β[0, 1] and t, s ∈ [0, 1] with (t 6= s), taking into account assumptions

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(i), (ii) and (iii), we obtain

|(T x)(t) − (T x)(s)|

|t − s|^β

=

p(t) + (F x) (t)R1

0 k(t, τ )x (q(τ )) dτ − p(s) − (F x) (s)R1

0 k(s, τ )x (q(τ )) dτ

|t − s|^β

6 1

|t − s|^β

"

|p(t) − p(s)| +

(F x) (t) Z 1

0

k(t, τ )x (q(τ )) dτ

− (F x) (s) Z 1

0

k(s, τ )x (q(τ )) dτ

#

6 |p(t) − p(s)|

|t − s|^β

+ 1

|t − s|^β

(F x) (t) Z 1

0

k(t, τ )x (q(τ )) dτ − (F x) (s) Z 1

0

+ 1

|t − s|^β

(F x) (s) Z 1

0

k(t, τ )x (q(τ )) dτ − (F x) (s) Z 1

0

k(s, τ )x (q(τ )) dτ 6 |p(t) − p(s)|

|t − s|^β +|(F x) (t) − (F x) (s)|

|t − s|^β

Z 1 0

|k(t, τ )| |x (q(τ ))| dτ

+|(F x) (s)|R1

0 |k(t, τ ) − k(s, τ )| |x (q(τ ))| dτ

|t − s|^β 6 |p(t) − p(s)|

|t − s|^β +|(F x) (t) − (F x) (s)|

|t − s|^β kxk_∞ Z 1

0

|k(t, τ )| dτ

+kF xk_∞kxk_∞R1

0 |k(t, τ ) − k(s, τ )| dτ

|t − s|^β

6 |p(t) − p(s)|

|t − s|^β +|(F x) (t) − (F x) (s)|

|t − s|^β kxk_∞K +kF xk_∞kxk_∞R1

0 kβ|t − s|^βdτ

|t − s|^β 6 Hp^β+ H_{F x}^β kxk_∞K + kF xk_∞kxk_∞kβ.

By using the facts that kxk_∞6 kxkβ and H_x^β6 kxkβ concluded Lemma 2.1 and the definition kxk_β, respectively we infer that

|(T x)(t) − (T x)(s)|

|t − s|^β 6 Hp^β+ (K + kβ)kxkβkF xk_β.

From this inequality, we have T x ∈ H_β[0, 1] . This proves that the operator T

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maps the space H_β[0, 1] into itself. On the other hand we can write

kT xk_β= |(T x) (0)| + sup |(T x)(t) − (T x)(s)|

|t − s|^β : t, s ∈ [0, 1] , t 6= s

6 |(T x) (0)| + Hp^β+ (K + kβ)kxkβkF xk_β

6 |p(0)| + |(F x) (0)|

Z 1 0

|k(0, τ )| |x (q(τ ))| dτ + H_p^β+ (K + k_β)kxk_βkF xk_β

6 kpkβ+ kF xk_∞kxk_∞ Z 1

0

|k(0, τ )| dτ + (K + kβ)kxkβkF xk_β 6 kpkβ+ K kF xk_βkxk_β+ (K + kβ)kxkβkF xk_β

= kpk_β+ (2K + kβ)kxkβkF xk_β 6 kpkβ+ (2K + k_β)kxk_βf

kxk_β

, (3.1)

for any x ∈ H_β[0, 1]. So, if we take x in B_r^β

0then by assumption (v) we get T x ∈ B_r^β

0. As a result, it follows that T transforms the ball

B_r^β

0 = {x ∈ H_β[0, 1] : kxk_β 6 r0} into itself. That is,

T : B^β_r₀ → B_r^β₀.

Next, we will prove that the operator T is continuous on B^β_r₀, according to the induced norm by k · kα, where 0 < α < β 6 1. To do this, let us take any fixed y ∈ Br^β₀ and arbitrary ε > 0. Since the operator F : Hβ[0, 1] → Hβ[0, 1] is continuous on Hβ[0, 1]

with respect to the norm k · k_α, there exists δ > 0 such that the inequality

kF x − F yk_α< ε 4 (K + kβ) r0

is satisfied for all x ∈ B_r^β

0, such that kx − yk_α6 δ and 0 < δ < ε

2 (2K + kβ) f (r0).

Then, for any x ∈ B_r^β

0 and t, s ∈ [0, 1] with t 6= s and 0 < α 6 1 we have

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|[(T x)(t) − (T y)(t)] − [(T x)(s) − (T y)(s)]|

|t − s|^α

= h

(F x) (t)R1

0 k(t, τ )x (q(τ )) dτ − (F y) (t)R1

0 k(t, τ )y(q(τ ))dτi

|t − s|^α

− h

(F x) (s)R1

0 k(s, τ )x (q(τ )) dτ − (F y) (s)R1

0 k(s, τ )y(q(τ ))dτi

|t − s|^α

= 1

|t − s|^α

(F x) (t) Z 1

0

k(t, τ )x (q(τ )) dτ − (F y) (t) Z 1

0

+

(F y) (t) Z 1

0

k(t, τ )x (q(τ )) dτ − (F y) (t) Z 1

0

k(t, τ )y(q(τ ))dτ

−

(F x) (s) Z 1

0

k(s, τ )x (q(τ )) dτ − (F y) (s) Z 1

0

−

(F y) (s) Z 1

0

k(s, τ )x (q(τ )) dτ − (F y) (s) Z 1

0

k(s, τ )y(q(τ ))dτ

= 1

|t − s|^α

[(F x) (t) − (F y) (t)]

Z 1 0

+

(F y) (t) Z 1

0

k(t, τ ) [x (q(τ )) − y (q(τ ))] dτ

−

[(F x) (s) − (F y) (s)]

Z 1 0

−

(F y) (s) Z 1

0

k(s, τ ) [x (q(τ )) − y (q(τ ))] dτ

= 1

|t − s|^α

{[(F x) (t) − (F y) (t)] − [(F x) (s) − (F y) (s)]}

Z 1 0

+

[(F x) (s) − (F y) (s)]

Z 1 0

−

[(F x) (s) − (F y) (s)]

Z 1 0

+

(F y) (t) Z 1

0

k(t, τ ) [x (q(τ )) − y (q(τ ))] dτ

−

(F y) (s) Z 1

0

k(s, τ ) [x (q(τ )) − y (q(τ ))] dτ

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= 1

|t − s|^α

{[(F x) (t) − (F y) (t)] − [(F x) (s) − (F y) (s)]}

Z 1 0

+

[(F x) (s) − (F y) (s)]

Z 1 0

(k(t, τ ) − k(s, τ )) x (q(τ )) dτ

+

(F y) (t) Z 1

0

k(t, τ ) [x (q(τ )) − y (q(τ ))] dτ

−

(F y) (s) Z 1

0

k(s, τ ) [x (q(τ )) − y (q(τ ))] dτ

. From the last inequality it follows that

|[(T x)(t) − (T y)(t)] − [(T x)(s) − (T y)(s)]|

|t − s|^α

6 1

|t − s|^α|[(F x) (t) − (F y) (t)] − [(F x) (s) − (F y) (s)]|

Z 1 0

+ 1

|t − s|^α|(F x) (s) − (F y) (s)|

Z 1 0

(k(t, τ ) − k(s, τ )) x (q(τ )) dτ

+ 1

|t − s|^α

(F y) (t) Z 1

0

k(t, τ ) [x (q(τ )) − y (q(τ ))] dτ

− (F y) (s) Z 1

0

k(s, τ ) [x (q(τ )) − y (q(τ ))] dτ 6 |[(F x) (t) − (F y) (t)] − [(F x) (s) − (F y) (s)]|

|t − s|^α kxk_∞

Z 1 0

|k(t, τ )| dτ

+ |[(F x) (s) − (F y) (s)] − [(F x) (0) − (F y) (0)]| kxk_∞ Z 1

0

|k(t, τ ) − k(s, τ )|

|t − s|^α dτ + |(F x) (0) − (F y) (0)| kxk_∞

Z 1 0

|k(t, τ ) − k(s, τ )|

|t − s|^α dτ

+ 1

|t − s|^α

(F y) (t) Z 1

0

k(t, τ ) [x (q(τ )) − y (q(τ ))] dτ

− (F y) (s) Z 1

0

k(t, τ ) [x (q(τ )) − y (q(τ ))] dτ

+ 1

|t − s|^α

(F y) (s) Z 1

0

k(t, τ ) [x (q(τ )) − y (q(τ ))] dτ

− (F y) (s) Z 1

0

k(s, τ ) [x (q(τ )) − y (q(τ ))] dτ

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6 HF x−F y^α kxk_∞K + sup

u,v∈[0,1]

|[(F x) (u) − (F y) (u)] − [(F x) (v) − (F y) (v)]| kxk_∞ Z 1

0

|k(t, τ ) − k(s, τ )|

|t − s|^α dτ

+ |(F x) (0) − (F y) (0)| kxk_∞ Z 1

0

|k(t, τ ) − k(s, τ )|

|t − s|^α dτ +|(F y) (t) − (F y) (s)|

|t − s|^α

Z 1 0

|k(t, τ )| |x (q(τ )) − y (q(τ ))| dτ

+ |(F y) (s)|

Z 1 0

|k(t, τ ) − k(s, τ )|

|t − s|^α |x (q(τ )) − y (q(τ ))| dτ 6 K kxk∞kF x − F ykα

+ sup

u,v∈[0,1]

|[(F x) (u) − (F y) (u)] − [(F x) (v) − (F y) (v)]| kxk_∞ Z 1

0

kβ|t − s|^β

|t − s|^α dτ

+ |(F x) (0) − (F y) (0)| kxk_∞ Z 1

0

kβ|t − s|^β

|t − s|^α dτ +|(F y) (t) − (F y) (s)|

|t − s|^α

Z 1 0

|k(t, τ )| |x (q(τ )) − y (q(τ ))| dτ

+ |(F y) (s)|

Z 1 0

kβ|t − s|^β

|t − s|^α |x (q(τ )) − y (q(τ ))| dτ.

In view of the inequalities kxk_∞6 kxkα, H_x^β6 kxkα,we derive the following inequli- ties

|[(T x)(t) − (T y)(t)] − [(T x)(s) − (T y)(s)]|

|t − s|^α

6 K kxk∞kF x − F ykα+ kβkxk_∞|t − s|^β−α· sup

u,v∈[0,1],u6=v

|[(F x) (u) − (F y) (u)] − [(F x) (v) − (F y) (v)]|

|u − v|^α |u − v|^α

+ kβkxk_∞|t − s|^β−α|(F x) (0) − (F y) (0)| + KH_{F y}^α kx − yk_∞ + kβkF yk_∞kx − yk_∞|t − s|^β−α

6 K kxkβkF x − F yk_α+ 2kβkxk_βkF x − F yk_α + K kF yk_αkx − yk_α+ kβkF yk_αkx − yk_α

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= (K + 2k_β) kxk_βkF x − F yk_α

+ (K + kβ) kF yk_αkx − yk_α. (3.2)

Since kyk_α6 kykβ6 r0(see Lemma 2.2 ) and from the assumption (iv) and (3.2) we deduce the following inequality

|[(T x)(t) − (T y)(t)] − [(T x)(s) − (T y)(s)]|

|t − s|^α

6 (K + 2kβ) kxk_βkF x − F yk_α+ (K + k_β) kF yk_βkx − yk_α 6 (K + 2kβ) kxk_βkF x − F yk_α+ (K + k_β) f

kyk_β

kx − yk_α

6 (K + 2kβ) r₀kF x − F yk_α+ (K + k_β) f (r₀) δ. (3.3) On the other hand,

|(T x) (0) − (T y) (0)| =

(F x) (0) Z 1

0

k(0, τ )x (q(τ )) dτ − (F y) (0) Z 1

0

k(0, τ )y (q(τ )) dτ 6

(F x) (0) Z 1

0

k(0, τ )x (q(τ )) dτ − (F x) (0) Z 1

0

k(0, τ )y (q(τ )) dτ +

(F x) (0) Z 1

0

k(0, τ )y (q(τ )) dτ − (F y) (0) Z 1

0

k(0, τ )y (q(τ )) dτ 6 |(F x) (0)|

Z 1 0

|k(0, τ )| |x (q(τ )) − y (q(τ ))| dτ

+ |(F x) (0) − (F y) (0)|

Z 1 0

|k(0, τ )| |y (q(τ ))| dτ From the last inequality it follows that

|(T x) (0) − (T y) (0)| 6 K kF xk_∞kx − yk_∞+ K kyk_∞kF x − F yk_∞ 6 K kF xkβkx − yk_α+ K kyk_βkF x − F yk_α 6 Kf

kxk_β

kx − yk_α+ K kyk_βkF x − F yk_α

6 Kf (r⁰) δ + Kr0kF x − F yk_α. (3.4) From (3.3) and (3.4), it follows that

kT x − T yk_α

= |(T x) (0) − (T y) (0)| + H_{T x−T y}^α

= |(T x) (0) − (T y) (0)| + sup

t,s∈[0,1],t6=s

|[(T x) (t) − (T y) (t)] − [(T x) (s) − (T y) (s)]|

|t − s|^α

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6 Kf (r⁰) δ + Kr₀kF x − F yk_α+ (K + 2k_β) r₀kF x − F yk_α+ (K + k_β) f (r₀) δ

= 2 (K + k_β) r₀kF x − F yk_α+ (2K + k_β) f (r₀) δ

<ε 2 +ε

2 = ε.

This show that the operator T is continuous at the point y ∈ B^β_r

0. We conclude that T is continuous on B^β_r

0with respect to the norm k·k_α. In addition the set B_r^β

0 is compact subset of the space H_α[0, 1] from [9] (see [9; the appendix at the p. 9]). Therefore, applying the classical Schauder fixed point theorem, we complete the proof.

4. Examples

In this section, we provide an example illustrating the main results in the above.

Example 1. Let us consider the quadratic integral equation:

x(t) = ln

√4

n sin t + ˆn + 1

+ x²(t) Z 1

0

p3

mt³+ τ x

1

τ + 1

dτ (4.1)

where t ∈ [0, 1] and n, ˆn, m are the suitable non-negative constants.

Observe that (4.1) is a particular case of (1.3) if we put p(t) = ln √⁴

n sin t + ˆn + 1, k(t, τ ) =√³

mt³+ τ and q (τ ) = _{τ +1}¹ . The operator F defined by (F x) (t) = x²(t) for all t ∈ [0, 1].

Since functions s, h : R⁺ → R⁺ defined by s (t) = ln (t + 1), h (t) = √⁴ t are concav and s (0) = 0, h (0) = 0, then from Lemma 4.4 in [9] these functions are subadditive. If we consider the result of subadditivity and the inequalities ln x < x for x > 0 and |sin x − sin y| 6 |x − y| for x, y ∈ R, we can write

|p(t) − p(s)| = ln√₄

n sin t + ˆn + 1

− ln√₄

n sin s + ˆn + 1 6 ln

√4

n sin t + ˆn −√⁴

n sin s + ˆn

<

√4

n sin t + ˆn −√⁴

n sin s + ˆn 6

p4

n |sin t − sin s|

6√⁴

n |t − s|¹⁴. It means that p ∈ H1

4[0, 1] and, moreover, H

1

p4 = √⁴

n. We can take the constants α and β as 0 < α < ¹₄ and β = ¹₄. Therefore, assumption (i) of Theorem (3.1) is

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satisfied. Note that kpk¹

4 = |p(0)| + sup |p(t) − p(s)|

|t − s|¹⁴ : t, s ∈ [0, 1], t 6= s

= |p(0)| + H

1

p4 = ln√₄ ˆ n + 1

+√⁴ n.

Further, we have

|k(t, τ ) − k(s, τ )| =

p3

mt³+ τ −p³

ms³+ τ 6p|mt³ ³− ms³|

=√³

mp|t³ ³− s³|

=√³

mp|t − s|³ p|t³ ²+ ts + s²| 6√³

3m|t − s|¹³

=√³

3m|t − s|¹⁴|t − s|¹²¹ 6√³

3m|t − s|¹⁴

for all t, s ∈ [0, 1]. Assumption (ii) of Theorem (3.1) is satisfied with k_β= k1 4 =√³

3m.

It is clear that q (τ ) =_{τ +1}¹ satisfies assumption (iii). The constant K is given by

K = sup

Z 1 0

|k(t, τ )|dτ : t ∈ [0, 1]

= sup

Z 1 0

p3

mt³+ τ

dτ : t ∈ [0, 1]

= Z 1

0

√3

m + τ dτ

= 3 4

p3

(m + 1)⁴−√³ m⁴

. For all x ∈ Hβ[0, 1] ,

kF xk_β= |(F x) (0)| + sup

(|(F x) (t) − (F x) (s)|

|t − s|^β : t, s ∈ [0, 1] , t 6= s )

= x²(0)

+ sup

( x²(t) − x²(s)

|t − s|^β : t, s ∈ [0, 1] , t 6= s )

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= x²(0)

+ sup

(|x (t) − x (s)| |x (t) + x (s)|

|t − s|^β : t, s ∈ [0, 1] , t 6= s )

6 x²(0)

+ 2 kxk_∞sup

(|x (t) − x (s)|

|t − s|^β : t, s ∈ [0, 1] , t 6= s )

6 x²(0)

+ 2 kxk_βsup

(|x (t) − x (s)|

|t − s|^β : t, s ∈ [0, 1] , t 6= s )

6 kxk²β+ 2 kxk²_β= 3 kxk²_β.

Therefore, F is an operator from H_β[0, 1] into H_β[0, 1] and we can chose the function f : R⁺ → R⁺ as f (x) = 3x². This function is non-decreasing and satisfies the inequality in assumption (iv).

Now, we will show that the operator F is continuous on the Hβ[0, 1] with respect to the norm k.k_α. To this end, fix arbitrarily y ∈ Hβ[0, 1] and ε > 0. Assume that x ∈ Hβ[0, 1] is an arbitrary function and kx − yk_α< δ, where δ is a positive number such that 0 < δ <

q

kyk²_α+₃^ε− kyk_α. Then, for arbitrary t, s ∈ [0, 1] we obtain

(F x − F y) (t) − (F x − F y) (s)

= x²(t) − y²(t) − x²(s) − y²(s)

= (x (t) − y (t)) (x (t) + y (t)) − (x (s) − y (s)) (x (s) + y (s))

= [x (t) − y (t) − (x (s) − y (s))] (x (t) + y (t)) + (x (s) − y (s)) (x (t) + y (t))

− (x (s) − y (s)) (x (s) + y (s))

= [x (t) − y (t) − (x (s) − y (s))] (x (t) + y (t))

+ (x (s) − y (s)) [x (t) + y (t) − (x (s) + y (s))] . (4.2) By (4.2), we have

|(F x − F y) (t) − (F x − F y) (s)|

6 kx + yk∞|x (t) − y (t) − (x (s) − y (s))| + kx − yk_∞|x (t) + y (t) − (x (s) + y (s))|

6 kx + ykα|x (t) − y (t) − (x (s) − y (s))| + kx − yk_α|x (t) + y (t) − (x (s) + y (s))| . (4.3) By (4.3), we observe

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sup |(F x − F y) (t) − (F x − F y) (s)|

|t − s|^α : t, s ∈ [0, 1] , t 6= s

6 kx + ykα sup

t,s∈[0,1], t6=s

|x (t) − y (t) − (x (s) − y (s))|

|t − s|^α + kx − yk_α sup

t,s∈[0,1], t6=s

|x (t) + y (t) − (x (s) + y (s))|

|t − s|^α 6 kx + ykαkx − yk_α+ kx − yk_αkx + yk_α

= 2 kx + yk_αkx − yk_α. (4.4)

From (4.4), it follows

kF x − F yk_α= |(F x − F y) (0)| + sup

t6=s

|(F x − F y) (t) − (F x − F y) (s)|

|t − s|^α : t, s ∈ [0, 1]

6

x²(0) − y²(0)

+ 2 kx + yk_αkx − yk_α

= |x (0) − y (0)| |x (0) + y (0)| + 2 kx + yk_αkx − yk_α 6 kx − yk∞kx + yk_∞+ 2 kx + yk_αkx − yk_α 6 3 kx + ykαkx − yk_α

6 3 kx − ykα(kx − yk_α+ 2 kyk_α) 6 3δ (δ + 2 kykα)

< ε. (4.5)

So that, the inequality

kF x − F yk_α6 3δ (δ + 2 kykα) < ε

is satisfied for all x ∈ Hβ[0, 1], where 0 < δ <

q

kyk²_α+ ε − kyk_α. Therefore, we can chose the positive number δ as δ = ¹₂

q

kyk²_α+ ε − kyk_α. This shows that the operator F is continuous at the point y ∈ H_β[0, 1] . Since y is chosen arbitrarily, we deduce that F is continuous on Hβ[0, 1] with respect to the norm k.k_α.

In this case, the inequality appearing in assumption (v) of Theorem (3.1) takes the following form

kpk¹

4 + (2K + k¹

4)rf (r) 6 r which is equivalent to

ln√₄ ˆ n + 1

+√⁴ n + 3

2

p3

(m + 1)⁴−√³ m⁴

+√³ 3m

3r³6 r. (4.6)

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Obviously, there exists a positive number r₀ satisfying (4.6) provided that the constants n, ˆn and m can chosen as suitable. For example, if one chose n = ₂¹₁₆, ˆn = 0 and m = 1, r₀= ¹₄, then the inequality

kpk¹

4 + (2K + k¹

4)rf (r)

= ln√₄ ˆ n + 1

+√⁴ n + 3

2

p3

(m + 1)⁴−√³ m⁴

+√³ 3m

3r³

≈ 0, 23696 < 1 4.

Finally, applying Theorem (3.1) we conclude that equation (4.1) has at least one solution in the space Hα[0, 1] with 0 < α < ¹₄.

Example 2. Let us consider the quadratic integral equation

x(t) = ln t 7+ 1

+ (ax (t) + b) Z 1

0

pmt²+ τ x (e^τ) dτ, t ∈ [0, 1]. (4.7)

Set p(t) = ln ₇^t+ 1 , k(t, τ ) =√

mt²+ τ , q (τ ) = e^τ for t, τ ∈ [0, 1] and m are non-negative constant. The operator F defined by (F x) (t) = ax (t) + b, where a and b are any real number.

In what follows, we will prove that assumption (i)-(v) of Threom (3.1) are satisfied. Since function p : R⁺ → R⁺ defined by p(t) = ln ₇^t+ 1, is concav and p (0) = 0, then from Lemma 4.4 in [9] these functions are subadditive. By the result of subadditive

|p(t) − p(s)| =

ln t 7 + 1

− lns 7 + 1

6 ln

t − s 7

< |t − s|

7 6 1

7|t − s|¹²

where we have used that ln x < x for x > 0 . This says that p ∈ H1

2[0, 1] (i.e. β = ¹₂) and, moreover, H

1

p2 =¹₇. Therefore, assumption (i) of Theorem (3.1) is satisfied. Note that

kpk1

2 = |p(0)| + sup |p(t) − p(s)|

|t − s|¹² : t, s ∈ [0, 1], t 6= s

= |p(0)| + Hp¹² = Hp¹² = 1 7.

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Further, we have

|k(t, τ ) − k(s, τ )| =

pmt²+ τ −p

ms²+ τ 6p|mt²− ms²|

=√

mp|t²− s²|

=√

mp|t − s|p|t + s|

6√ m√

2|t − s|¹² 6√

2m|t − s|¹²

for all t, s ∈ [0, 1]. Assumption (ii) of Theorem (3.1) is satisfied with kβ= k¹

2 =√ 2m.

It is clear that q (τ ) = e^τ satisfies assumption (iii). In our case, the constant K is given by

K = sup

Z 1 0

|k(t, τ )|dτ : t ∈ [0, 1]

= sup

Z 1 0

pmt²+ τ

dτ : t ∈ [0, 1]

= Z 1

0

√m + τ dτ

=2 3

p(m + 1)³−√ m³

. For all x ∈ H_β[0, 1]

kF xk_β= |(F x) (0)| + sup

(|(F x) (t) − (F x) (s)|

|t − s|^β : t, s ∈ [0, 1] , t 6= s )

= |ax (0) + b| + sup

(|ax (t) + b − ax (s) − b|

|t − s|^β : t, s ∈ [0, 1] , t 6= s )

= |a| |x (0)| + |b| + sup

(|x (t) − x (s)| |a|

|t − s|^β : t, s ∈ [0, 1] , t 6= s )

6 |a| |x (0)| + |b| + |a| sup

(|x (t) − x (s)|

|t − s|^β : t, s ∈ [0, 1] , t 6= s )

6 |a| |x (0)| + sup

(|x (t) − x (s)|

|t − s|^β : t, s ∈ [0, 1] , t 6= s )!

+ |b|

6 |a| kxkβ+ |b| .

Therefore, F is an operator from Hβ[0, 1] into Hβ[0, 1] and we can chose the function f : R⁺→ R⁺ as f (x) = |a| x + |b| . This function is non-decreasing and satisfies the inequality in Assumption (iv).

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Now, we will show that the operator F is continuous on the H_β[0, 1] with respect to the norm k.k_α. To this end, fix arbitrarily y ∈ H_β[0, 1] and ε > 0. Assume that x ∈ H_β[0, 1] is an arbitrary function and kx − yk_α< δ, where δ is a positive number such that 0 < δ < _|a|^ε (in this place a 6= 0. It is obvious that if a is zero, the operator F is continuous).

Then, for arbitrary t, s ∈ [0, 1] we obtain kF x − F yk_α= |(F x − F y) (0)| + sup

t6=s

|(F x − F y) (t) − (F x − F y) (s)|

|t − s|^α

= |ax (0) − ay (0)| + sup

t6=s

|(ax (t) − ay (t)) − (ax (s) − ay (s))|

|t − s|^α

= |a| |x (0) − y (0)| + |a| sup

t6=s

|(x (t) − y (t)) − (x (s) − y (s))|

|t − s|^α

= |a| |x (0) − y (0)| + sup

t6=s

|(x (t) − y (t)) − (x (s) − y (s))|

|t − s|^α

!

= |a| kx − yk_α 6 |a| δ

< ε.

This shows that the operator F is continuous at the point y ∈ H_β[0, 1] . Since y was chosen arbitrarily, we deduce that F is continuous on Hβ[0, 1] with respect to the norm k.k_α.

In this case, the inequality appearing in assumption (v) of Theorem (3.1) takes the following form

kpk¹

2 + (2K + k1

2)rf (r) 6 r which is equivalent to

1 7 + 4

3

p(m + 1)³−√ m³

+√ 2m

r (|a| r + |b|) 6 r. (4.8)

Obviously, there exists a number positive r0 satisfying (4.8) provided that the constants a, b and m can chosen as suitable. For example, if one chose a = ₁₀¹, b = ₆₀¹ and m =¹₂, r0= ¹₆, then the inequality

kpk1

2 + (2K + k1

2)r0f (r0)

= 1 7+ 4

3

p(m + 1)³−√ m³

+√ 2m

r0(|a| r0+ |b|)

≈ 0, 15939 < 1 6.

Therefore, using Theorem (3.1), we conclude that equation (4.7) has at least one solution in the space H_α[0, 1] with 0 < α < ¹₂ = β.

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DOI: 10.7862/rf.2020.3

˙Ilyas Dal

email: ilyasdal@live.com ORCID: 0000-0002-1417-5341

Department of Mathematical Education In¨on¨u University

44280-Malatya TURKEY

Omer Faruk Temizer¨

email: omer.temizer@inonu.edu.tr ORCID: 0000-0002-3843-5945 E˘gitim Fak¨ultesi, A-Blok

˙Inönü Üniversitesi

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44280-Malatya TURKEY

Received 19.08.2019 Accepted 03.03.2020