On a Cubic Integral Equation of Urysohn Type with Linear Perturbation

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Mathematics

and Applications

JMA No 41, pp 29-38 (2018)

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

On a Cubic Integral Equation of Urysohn Type with Linear Perturbation

of Second Kind

Hamed Kamal Awad, Mohamed Abdalla Darwish and Mohamed M.A. Metwali

Abstract: In this paper, we concern by a very general cubic integral equation and we prove that this equation has a solution in C[0, 1]. We apply the measure of noncompactness introduced by Bana´s and Olszowy and Darbo’s fixed point theorem to establish the proof of our main result.

AMS Subject Classification: 45G10, 45M99, 47H09.

Keywords and Phrases: Cubic integral equation; Darbo’s fixed point theorem; Mono- tonicity measure of noncompactness.

1. Introduction

Cubic integral equations have several useful applications in modeling numerous prob- lems and events of the real world (cf. [3, 8, 9, 12, 13, 18, 19]).

In this paper we consider the cubic Urysohn integral equation with linear perturbation of second kind

x(τ ) = φ(τ ) + ϕ(τ, x(τ )) + x²(τ ) Z 1

0

u(τ, s, (Λx)(s)) ds, τ ∈ I = [0, 1]. (1.1) In the above equation, we consider φ : I → R, ϕ : I × R → R, u : I × I × R → R are given functions and Λ : C(I) → C(I) is an operator verifies special assumption which will state in Section 3.

Eq.(1.1) is of interest since it contains many includes several integral equations studied earlier as special cases, see [1, 2, 6, 7, 10, 11, 14, 15, 16, 20, 21, 22] and references therein. By using the measure of noncompactness related to monotonicity associated with fixed point theorem due to Darbo, we show that Eq.(1.1) has at least one solution in C(I) which is nondecreasing on the interval I.

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2. Auxiliary Facts and Results

In this section, we present some definitions and results which we will use further on.

Let (E, k · k) be a real Banach space with zero element 0. Let B(x, r) be the closed ball centered at x with radius r. We denote by B_r the closed ball B(0, r). Next, let X be a subset of E, we denote by X and ConvX the closure and convex closure of X, respectively. We use the symbols λX and X + Y for the usual algebraic operations on the sets. Moreover, the symbol ME stands for the family of all nonempty and bounded subsets of E and the symbol NE stands for its subfamily consisting of all relatively compact subsets.

Now, we state the definition of a measure of noncompactness [4]:

Definition 2.1. A function µ : M_E→ R+ is called a measure of noncompactness in E if it verifies the following assumptions:

(1) The family kerµ 6= ∅ and kerµ ⊂ N_E, where kerµ = {X ∈ M_E : µ(X) = 0}.

(2) µ(X) ≤ µ(Y ), if X ⊂ Y .

(3) µ(X) = µ(X) and µ(ConvX) = µ(X).

(4) µ(λX + (1 − λ)Y ) ≤ λµ(X) + (1 − λ)µ(Y ), 0 ≤ λ ≤ 1.

(5) If Xn ∈ ME, Xn = Xn, Xn+1 ⊂ Xn for n = 1, 2, 3, ... and lim

n→∞µ(Xn) = 0, then ∩^∞_n=1X_n6= ∅.

Notice that kerµ is said to be the kernel of the measure of noncompactness µ.

In the following, we will work in the Banach space C(I) of all real functions defined and continuous on I = [0, 1] equipped with the standard norm kxk = max{|x(τ )| : τ ∈ I}. We recall the measure of noncompactness in C(I) which we will need in the next section (see [5]).

Let ∅ 6= X ⊂ C(I). For x ∈ X and ε ≥ 0 we denote by ω(x, ε) the modulus of continuity of the function x as follows

ω(x, ε) = sup{|x(τ ) − x(t)| : τ, t ∈ I, |τ − t| ≤ ε}.

Next, we put ω(X, ε) = sup{ω(x, ε) : x ∈ X} and ω0(X) = lim

ε→0ω(X, ε). Moreover, we define

d(x) = sup{|x(τ ) − x(t)| − [x(τ ) − x(t)] : τ, t ∈ I, τ ≥ t}

and

d(X) = sup{d(x) : x ∈ X}.

Notice that d(X) = 0 if and only if all functions belonging to X are nondecreasing on I.

Finally, we define the function µ on the family M_C(I)as follows µ(X) = ω₀(X) + d(X).

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Notice that the function µ is a measure of noncompactness in C(I) [5].

We present a fixed point theorem due to Darbo [17] which we will need in the proof of our main result. First, we make use of the following definition.

Definition 2.2. Let ∅ 6= M be a subset of a Banach space E and let P : M → E be a continuous mapping which maps bounded sets onto bounded sets. The operator Psatisfies the Darbo condition (with a constant κ ≥ 0) with respect to a measure of noncompactness µ if for any bounded subset X of M we have

µ(PX) ≤ κµ(X).

If P verifies the Darbo condition with κ < 1 then it is a contraction operator with respect to µ.

Theorem 2.3. Let ∅ 6= Ω be a closed, bounded and convex subset of the space E and let P : Ω → Ω be a contraction mapping with respect to the measure of noncompactness µ.

Then P has a fixed point in the set Ω.

Notice that the assumptions of the above theorem gives us that the set FixP of all fixed points of P belongs to Ω is an element of kerµ [4].

3. The Main Result

We consider Eq.(1.1) and assume that the following assumptions are verified:

(a1) The function φ : I → R is continuous, nonnegative and nondecreasing on I.

(a2) The function ϕ : I × R → R is continuous, ϕ : I × R⁺→ R+ and

∃ c ≥ 0 : |ϕ(τ, x1) − ϕ(τ, x2)| ≤ c|x1− x2| ∀ (x1, x2) ∈ R²& τ ∈ I.

(a3) The superposition operator Φ generated by the function ϕ satisfies for any nonnegative function x the condition d(Φx) ≤ cd(x), where c is the same c appears in assumption (a2).

(a₄) The function u : I × I × R → R is a continuous, u : I × I × R+ → R+ and for arbitrary fixed t ∈ I and x ∈ R the function τ → u(τ, t, x) is nondecreasing on I.

Moreover,

∃ Ψ : R+→ R+(nondecreasing) : |u(τ, t, x)| ≤ Ψ(|x|) ∀ (τ, t) ∈ I²& x ∈ R.

(a₅) The operator Λ : C(I) → C(I) is continuous and

∃ ψ : R+→ R+(nondecreasing) : |(Λx)(τ )| ≤ ψ(kxk) for any τ ∈ I, x ∈ C(I).

Moreover, for every nonnegative function x ∈ C(I), the function Λx is nonnegative and nondecreasing on I.

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(a₆) The inequality

kφk + cr + ϕ^∗+ r²Ψ(ψ(r)) ≤ r (3.1) has a positive solution r0such that c+2r0Ψ(ψ(r0)) < 1, where ϕ^∗= max

0≤τ ≤1ϕ(τ, 0).

Under the above assumptions, we state our main result as follows.

Theorem 3.1. Let the assumptions (a1) − (a6) be verified, then the cubic Urysohn integral equation (1.1) has at least one solution x ∈ C(I) which is nondecreasing on I.

Proof. Let F be an operator defined on C(I) by

(Fx)(τ ) = φ(τ ) + ϕ(τ, x(τ )) + x²(τ )(U x)(t), (3.2) where U is the Urysohn integral operator

(U x)(τ ) = Z 1

0

u(τ, t, (Λx)(t)) dt. (3.3)

For better readability, we will write the proof in seven steps.

Step 1: F maps the space C(I) into itself.

Notice that for a given x ∈ C(I), according to assumptions (a₁) − (a₅), we have Fx ∈ C(I). Therefore, the operator F maps C(I) into itself.

Step 2: F maps the ball Br₀ into itself.

For all τ ∈ I, we have

|(Fx)(τ )| ≤

φ(τ ) + ϕ(τ, x(τ )) + x²(τ ) Z 1

0

u(τ, t, (Λx)(t)) dt

≤ |φ(τ )| + |ϕ(τ, x(τ )) − ϕ(τ, 0)| + |ϕ(τ, 0)|

+|x²(τ )|

Z 1 0

|u(τ, t, (Λx)(t))| dt

≤ kφk + ckxk + ϕ^∗+ kxk²Ψ(ψ(kxk)) Z 1

0

ds

= kφk + ckxk + ϕ^∗+ kxk²Ψ(ψ(kxk)).

From the above estimate, we get

kFxk ≤ kφk + ckxk + ϕ^∗+ kxk²Ψ(ψ(kxk)).

Therefore, if we have kxk ≤ r0, we obtain

kFxk ≤ kφk + cr0+ ϕ^∗+ r²₀Ψ(ψ(r0)) ≤ r0,

in view of the assumption (a₆). Consequently, the operator F maps the ball B_r₀ into itself.

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Further, let B_r⁺

0 be the subset of B_r₀ given by B_r⁺

0= {x ∈ B_r₀ : x(τ ) ≥ 0, for τ ∈ I}.

Notice that, the set ∅ 6= B_r⁺₀ is closed, bounded and convex.

Step 3: F maps continuously the ball B⁺_r₀ into itself.

In view of the above facts about B_r⁺₀ and assumptions (a1) − (a4), we infer that F maps the set B_r⁺

0 into itself.

Step 4: The operator F is continuous on B_r⁺

0.

To establish this, let us fix arbitrarily ε > 0 and y ∈ B_r⁺₀. By assumption (a4), we can find δ > 0 such that for arbitrary x ∈ B_r⁺₀ with kx − yk ≤ δ we have that kΛx − Λyk ≤ ε. Indeed, for each τ ∈ I we have

|(Fx)(τ ) − (Fy)(τ )|

≤ |ϕ(τ, x(τ )) − ϕ(τ, y(τ ))|

+

x²(τ ) Z 1

0

u(τ, t, (Λx)(t)) dt − y²(τ ) Z 1

0

u(τ, t, (Λy)(t)) dt

≤ c|x(τ ) − y(τ )| +

x²(τ ) Z 1

0

u(τ, t, (Λx)(t)) dt − y²(τ ) Z 1

0

u(τ, t, (Λx)(t)) dt +

y²(τ ) Z 1

0

u(τ, t, (Λx)(t)) dt − y²(τ ) Z 1

0

u(τ, t, (Λy)(t)) dt

≤ c|x(τ ) − y(τ )| + |x²(τ ) − y²(τ )|

Z 1 0

|u(τ, t, (Λx)(t))| dt

+|y²(τ )|

Z 1 0

|u(τ, t, (Λx)(t)) − u(τ, t, (Λy)(t))| dt.

Therefore, we have

kFx − Fyk ≤ ckx − yk + 2r0Ψ(ψ(r0))kx − yk + r₀²ω^∗(u, ε), (3.4) where we denoted

ω^∗(u, ε) = sup{|u(τ, t, x) − u(τ, t, y)| : τ, t ∈ I, x, y ∈ [0, ψ(r₀)], |x − y| ≤ ε}.

From assumption (a4) we infer that ω^∗(u, ε) → 0 as ε → 0 and therefore, the operator Fis continuous in B_r⁺₀.

Step 5: An estimate of F with respect to the term related to continuity ω0. Let ∅ 6= X ⊂ B_r⁺₀, fix an arbitrarily number ε > 0 and choose x ∈ X and τ1, τ2∈ I such that |τ2− τ1| ≤ ε. Without restriction of the generality, we may assume that τ₁≤ τ2. In the view of our assumptions, we have

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|(Fx)(τ2) − (Fx)(τ1)|

≤ |φ(τ₂) − φ(τ₁)| + |ϕ(τ₂, x(τ₂)) − ϕ(τ₁, x(τ₁))|

+

x²(τ2) (U x)(τ2) − x²(τ2) (U x)(τ1) +

x²(τ2) (U x)(τ1) − x²(τ1) (U x)(τ1)

≤ ω(φ, ε) + |ϕ(τ₂, x(τ₂)) − ϕ(τ₁, x(τ₂))| + |ϕ(τ₁, x(τ₂)) − ϕ(τ₁, x(τ₁))|

+ x²(τ2)

|(U x)(τ2) − (U x)(τ1)| +

x²(τ2) − x²(τ1)

|(U x)(τ1)|

≤ ω(φ, ε) + γr₀(ϕ, ε) + c ω(x, ε) + |x(τ2)|² |(U x)(τ2) − (U x)(τ1)|

+ |x(τ2) − x(τ1)| |x(τ2) + x(τ1)| |(U x)(τ1)|

≤ ω(φ, ε) + γr₀(ϕ, ε) + c ω(x, ε) +kxk²

Z 1 0

|u(τ2, t, (Λx)(t)) − u(τ1, t, (Λx)(t))| dt + 2kxkω(x, ε)Ψ(ψ(kxk))

≤ ω(φ, ε) + γr0(ϕ, ε) + c ω(x, ε) + kxk²ωψ(kxk)(u, ε) + 2kxkω(x, ε)Ψ(ψ(kxk)), where we denoted

γ_r₀(ϕ, ε) = sup {|ϕ(τ₂, x) − ϕ(τ₁, x)| : τ₁, τ₂∈ I, x ∈ [0, r0], |τ₂− τ1| ≤ ε}

and

ω_b(u, ε) = sup {|u(τ₂, t, y) − u(τ₁, t, y)| : t, τ₁, τ₂∈ I, y ∈ [0, b], |τ₂− τ₁| ≤ ε} . Hence,

ω(Fx, ε) ≤ ω(φ, ε) + γr₀(ϕ, ε) + c ω(x, ε) + r²₀ω_ψ(r₀₎(u, ε) + 2r0ω(x, ε)Ψ(ψ(r0)).

Consequently,

ω(FX, ε) ≤ ω(φ, ε) + γr₀(ϕ, ε) + (c + 2r0Ψ(ψ(r0))) ω(X, ε) + r₀²ω_ψ(r₀₎(u, ε).

Since the function φ is continuous on I, the function ϕ is uniformly continuous on I × [0, r0] and the function u is uniformly continuous the set I × I × [0, ψ(r0)], then we obtain

ω0(FX) ≤ (c + 2r0Ψ(ψ(r0))) ω0(X). (3.5) Step 6: An estimate of F with respect to the term related to monotonicity d.

Fix an arbitrary x ∈ X and τ₁, τ₂ ∈ I with τ₂ > τ₁. Then, taking into account our assumption, we get

|(Fx)(τ2) − (Fx)(τ1)| − ((Fx)(τ2) − (Fx)(τ1))

=

φ(τ2) + ϕ(τ2, x(τ2)) + x²(τ2) Z 1

0

u(τ2, t, (Λx)(t)) dt

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−φ(τ1) − ϕ(τ1, x(τ1)) − x²(τ1) Z 1

0

−

φ(τ₂) + ϕ(τ₂, x(τ₂)) + x²(τ₂) Z 1

0

u(τ₂, t, (Λx)(t)) dt

−φ(τ1) − ϕ(τ1, x(τ1)) − x²(τ1) Z 1

0

≤ [|φ(τ2) − φ(τ₁)| − (φ(τ₂) − φ(τ₁))]

+ [|ϕ(τ2, x(τ2)) − ϕ(τ1, x(τ1))| − (ϕ(τ2, x(τ2)) − ϕ(τ1, x(τ1)))]

+

x²(τ2) Z 1

0

u(τ2, t, (Λx)(t)) dt − x²(τ1) Z 1

0

u(τ2, t, (Λx)(t)) dt +

x²(τ1) Z 1

0

u(τ2, t, (Λx)(t)) dt − x²(τ1) Z 1

0

−

x²(τ2)

Z 1 0

u(τ2, t, (Λx)(t)) dt − x²(τ1) Z 1

0

−

x²(τ1)

Z 1 0

u(τ2, t, (Λx)(t)) dt − x²(τ1) Z 1

0

≤ |ϕ(τ2, x(τ2)) − ϕ(τ1, x(τ1))| − (ϕ(τ2, x(τ2)) − ϕ(τ1, x(τ1))) +|x²(τ2) − x²(τ1)| − (x²(τ2) − x²(τ1))

Z 1 0

+x²(τ1)

Z 1 0

u(τ2, t, (Λx)(t)) dt − Z 1

0

−

Z 1 0

u(τ2, t, (Λx)(t)) dt − Z 1

0

≤ d(Φx) + 2kxkΨ(ψ(kxk))d(x).

The above estimate gives us that

d(Fx) ≤ cd(x) + 2r0Ψ(ψ(r0))d(x), and consequently,

d(FX) ≤ (c + 2r0Ψ(ψ(r0)))d(X). (3.6)

Step 7: F is a contraction with respect to the measure of noncompactness µ.

By adding (3.5) and (3.6), we get

ω₀(FX) + d(FX) ≤ (c + 2r₀Ψ(ψ(r₀)))ω₀(X) + (c + 2r₀Ψ(ψ(r₀)))d(X) or

µ(FX) ≤ (c + 2r₀Ψ(ψ(r₀)))µ(X).

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Since c + 2r₀Ψ(ψ(r₀)) < 1, then the operator F is contraction with respect to the measure of noncompactness µ.

Finally, Theorem 2.3 guarantees that Eq.(1.1) has at least one solution x ∈ C(I) which is nondecreasing on I. This completes the proof.

4. Example

Let us consider the cubic Urysohn integral equation

x(τ ) =

√τ

8 + τ x(τ )

1 + τ² +x²(τ ) 4

Z 1 0

arctan τRt

0sx²(s) ds 1 + t²

!

dt. (4.1)

Here, φ(τ ) =

√τ

8 and this function verifies assumption (a₁) and kφk = 1/8. Also, ϕ(τ, x) =_1+τ^{τ x}2 and this function verifies assumption (a2) with

|ϕ(τ, x) − ϕ(τ, y)| ≤ 1

2|x − y| ∀ t ∈ I & (x, y) ∈ R².

Moreover, the function ϕ verifies assumption (a₃). Indeed, for arbitrary nonnegative function x ∈ C(I) and τ₁, τ₂∈ I with τ1≤ τ2, we have

d(Φx) = |(Φx)(τ₂) − (Φx)(τ₁)| − ((Φx)(τ₂) − (Φx)(τ₁))

= |ϕ(τ2, x(τ2)) − ϕ(τ1, x(τ1))| − (ϕ(τ2, x(τ2)) − ϕ(τ1, x(τ1)))

=

τ2

1 + τ₂²x(τ2) − τ1

1 + τ₁²x(τ1)

−

τ2

1 + τ₂²x(τ2) − τ1

1 + τ₁²x(τ1)

≤ τ2

1 + τ₂²|x(τ2) − x(τ1)| +

τ2

1 + τ₂² − τ1

1 + τ₁²

x(τ1)

− τ2

1 + τ₂²(x(τ2) − x(τ1)) −

τ2

1 + τ₂²− τ1

1 + τ₁²

x(τ1)

= τ₂

1 + τ₂²[|x(τ₂) − x(τ₁)| − (x(τ₂) − x(τ₁))]

= τ₂

1 + τ₂²d(x) ≤ 1 2d(x).

The function u(τ, t, x) = arctan_1+t^{τ x}₂ satisfies assumption (a₄), we have |u(τ, t, x)| ≤ |x|

which means Ψ(r) = r. Moreover, the operator (Λx)(τ ) = Rτ

0 tx²(t) dt verifies assumption (a5) with ψ(r) = r².

Therefore, the inequality (3.1) has the form ¹₈ +^r₂+ r⁴ ≤ r or ¹₄+ r + 2r⁴≤ 2r.

This inequality admits r0= 1/2 as a positive solution. Moreover, c + 2r0Ψ(ψ(r0)) = 1

2 +1 4 = 3

4 < 1.

Consequently, Theorem 3.1 guarantees that equation (4.1) has a continuous nondecreasing solution.

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DOI: 10.7862/rf.2018.3 Hamed Kamal Awad

email: hamedk66@sci.dmu.edu.eg ORCID: 0000-0003-4866-3164

Department of Mathematics, Faculty of Science Damanhour University

Damanhour EGYPT

Mohamed Abdalla Darwish email: dr.madarwish@gmail.com ORCID: 0000-0002-4245-4364

Damanhour EGYPT

Mohamed M.A. Metwali email: m.metwali@yahoo.com ORCID: 0000-0003-1091-8619

Damanhour EGYPT

Received 09.11.2018 Accepted 11.12.2018