The Existence and Attractivity of Solutions of an Urysohn Integral Equation on an Unbounded Interval

Definition 1. A mapping μ : 𝔐_E → ℝ₊ is said to be a measure of noncompactness in E if it satisfies the following conditions.

• (1^°)

  The family ker  μ = {X ∈ 𝔐_E : μ(X) = 0} is nonempty and ker  μ ⊂ 𝔑_E.

• (2^°)

  X ⊂ Y⇒μ(X) ≤ μ(Y).

• (3^°)

  $$.

• (4^°)

  μ(λX + (1 − λ)Y) ≤ λμ(X)+(1 − λ)μ(Y) for λ ∈ [0,1].

• (5^°)

  If (X_n) is a sequence of closed sets from 𝔐_E such that X_n+1 ⊂ X_n for n = 1,2, … and if lim _n→∞ μ(X_n) = 0, then the intersection set $$ is nonempty.

The family ker  μ appearing in 1° is called the kernel of the measure of noncompactness  μ.

Observe that the set X_∞ from the axiom 5^° is a member of the family ker  μ. Indeed, since μ(X_∞) ≤ μ(X_n) for any natural number n, we infer that μ(X_∞) = 0. Consequently, X_∞ ∈ ker  μ. This simple observation will be essential in our further investigations.

Now, we formulate a fixed point theorem of Darbo type which will be used further on [21].

Theorem 2. Let Ω be a nonempty, bounded, closed, and convex subset of the Banach space E, and let F : Ω → Ω be a continuous mapping. Assume that there exists a constant k ∈ [0,1) such that μ(FX) ≤ kμ(X) for any nonempty subset X of Ω. Then, F has a fixed point in the set Ω.

Remark 3. Denote by Fix F the set of all fixed points of the operator F belonging to Ω. It can be easily seen [21] that the set Fix F belongs to the family ker  μ.

In what follows, we will work in the Banach space BC(ℝ₊) consisting of all real functions x = x(t) defined, continuous, and bounded on ℝ₊. This space will be endowed with the standard supremum norm

Now, we recall the construction of a measure of noncompactness in the space BC(ℝ₊) which was introduced in [21]. To this end, fix a nonempty and bounded subset X of the space BC(ℝ₊) and positive numbers ε > 0, T > 0. For x ∈ X, denote by ω^T(x, ε)  the modulus of continuity of the function x on the interval [0, T]; that is,

Next, we put Further, for a fixed number t ∈ ℝ₊ let us put Finally, let us consider the function μ defined on the family $$ by the formula It can be shown [21] that the function μ is a measure of noncompactness in the space BC(ℝ₊). Moreover, the kernel ker  μ  of this measure consists of all nonempty and bounded subsets X of BC(ℝ₊) such that functions from X are locally equicontinuous on ℝ₊, and the thickness of the bundle formed by functions from X tends to zero at infinity. This property in combination with Remark 3 permits us to characterize solutions of the integral equation considered in the sequel.

For further purposes, we introduce now the concept of attractivity (stability) of solutions of operator equations in the space BC(ℝ₊). To this end, assume that Ω is a nonempty subset of the space BC(ℝ₊). Moreover, let F be an operator defined on Ω with values in BC(ℝ₊). Let us consider the operator equation of the form

Definition 4. We say that solutions of (6) are attractive (or locally attractive) if there exists a ball B(x₀, r) in the space BC(ℝ₊) such that B(x₀, r)∩Ω ≠ ∅, and for arbitrary solutions x = x(t), y = y(t) of (6) belonging to the set B(x₀, r)∩Ω we have that

In the case when limit (7) is uniform with respect to the set B(x₀, r)∩Ω, that is, when for each ε > 0 there exists T > 0 such that for all x, y ∈ B(x₀, r)∩Ω being solutions of (6) and for any t ≥ T, we will say that solutions of (6) are uniformly attractive (or asymptotically stable).

Notice that the previous definition comes from [16, 19].

Remark 5. It is worthwhile mentioning that in the theory of improper Riemann integral with a parameter there has been considered the concept of the uniform convergence of the improper integral with respect to that parameter (cf. [22]). In order to recall this concept, suppose that there is a given function z(t, s) = z :   ℝ₊ × ℝ₊ → ℝ  such that the improper integral

does exist for every fixed t ∈ ℝ₊.

We say that the integral (15) is uniformly convergent with respect to t ∈ ℝ₊ if

uniformly with respect to t ∈ ℝ₊.

Equivalently (cf. [10]), the integral (15) is uniformly convergent with respect to t ∈ ℝ₊ if

Let us observe that if integrals appearing in assumption (v), that is, the integrals

are uniformly convergent with respect to t ∈ ℝ₊, then the inequalities from assumption (vi) are satisfied (cf. [10]). Indeed, this conclusion follows easily from the inequalities which are valid for any T > 0.

It may be also shown that the converse implications are, in general, not valid [10].

Remark 6. It can be also shown [10] that the requirements concerning the function g(t, s) imposed in assumption (v) are independent, that is, there exist functions g_i : ℝ₊ × ℝ₊ → ℝ₊  (i = 1,2) such that for each t ∈ ℝ₊ there exist the integrals $$ and such that the integral $$ is uniformly convergent but $$ while $$, but the integral $$ is not uniformly convergent.

Now, we formulate our last assumption.

• (vii)

  The inequality

  $$ (20)

has a positive solution r₀ such that

Remark 7. Assume that r₀ > 0 satisfies the first inequality from assumption (vii); that is,

Then, we obtain

Thus, the second inequality from assumption (vii) is satisfied provided at least one of the quantities ∥a∥, $$, and $$ does not vanish.

Now, we are prepared to formulate our main result.

Theorem 8. Under assumptions (i)-(vii), (9) has at least one solution x = x(t) in the space BC(ℝ₊). Moreover, all solutions of (9) are uniformly attractive.

Proof. Consider the operator U defined on the space BC(ℝ₊) by the formula

Notice that in view of assumptions (i), (ii), (iv), and (v), the function t → (Ux)(t) is well defined on the interval ℝ₊. We show that this function is continuous on ℝ₊. To this end, fix arbitrarily T > 0 and ε > 0. Next, take arbitrary numbers t, s ∈ [0, T] with |t − s | ≤ ε. Then, in view of imposed assumptions we obtain where we denoted Obviously, in the previous performed calculations we should put ∥x∥ in place of d.

Further, let us notice that in view of assumptions (ii) and (iv), the function f is uniformly continuous on the set [0, T]×[−∥x∥, ∥x∥], while the function u is uniformly continuous on the set [0, T]×[0, T]×[−∥x∥, ∥x∥]. Hence, we derive that $$ and $$ as ε → 0. Next, let us observe that based on assumption (vi) we can choose a number T so large that the last terms in estimate (25), that is, the integrals

are sufficiently small. The same is also true with regard to the integral Thus, taking into account the all facts established above and estimate (25), we infer that the function Ux is continuous on the whole interval [0, T] for each T > 0 being big enough. This implies that Ux is continuous on the whole interval ℝ₊.

In what follows, we show that the function Ux is bounded on ℝ₊. Indeed, keeping in mind our assumptions, for an arbitrary fixed t ∈ ℝ₊, we get the following estimates:

Hence, we obtain the following evaluation which implies that the function Ux is bounded on ℝ₊. Combining this fact with the continuity of the function Ux on ℝ₊, we conclude that the operator U transforms the space BC(ℝ₊) into itself.

Further, observe that from (30) we get

Linking the previous inequality with assumption (vii), we deduce that the operator U maps the ball $$ into itself, where r₀ > 0 is a number indicated in assumption (vii).

Now, let us take a nonempty subset X of the ball $$.

Fix numbers ε > 0 and T > 0 and choose an arbitrary function x ∈ X. Then, in virtue of estimate (25), for an arbitrary t ∈ [0, T] we obtain

Hence, we derive the following estimate Further, keeping in mind the properties of the terms of the right hand side of the perviously obtained inequality, which were mentioned earlier (cf. assumptions (ii) and (iv)), we deduce that the following estimate holds Consequently, in view of assumption (vi) we have

In what follows assume, as previously, that X is a fixed nonempty subset of the ball $$. Next, take arbitrary elements x, y ∈ X. Then, for an arbitrarily fixed t ∈ ℝ₊, in virtue of imposed assumptions, we obtain:

Hence, we get Taking into account assumption (v), from the previous inequality we derive the following one: Obviously, the previous inequality implies the following estimate: Finally, let us observe that by combining (35) and (39) we have where μ is the measure of noncompactness defined by formula (5).

In the last step of our proof, we show that the operator U is continuous on the ball $$. To this end, fix an arbitrary number ε > 0 and take $$ such that ∥x − y∥ ≤ ε. Then, from estimate (36) we obtain

The previous inequality in conjunction with assumption (iv) implies that the operator U is a continuous self-mapping of the ball $$.

Finally, using the above established facts and (40) and taking into account assumption (vii) and Theorem 2, we infer that the operator U has at least one fixed point x in the ball $$. Obviously, every function x = x(t) being a fixed point of the operator U, is a solution of (9). Moreover, keeping in mind Remark 3, we conclude that the set FixU of all fixed points of the operator U belonging to the ball $$ (equivalently: the set FixU of all solutions of (9) belonging to the ball $$) is a member of ker  μ. Hence, in view of the description of the kernel ker  μ given in Section 2 we infer that all solutions of (9) belonging to the ball $$ are uniformly attractive (asymptotically stable).

The proof is complete.

Example 9. Let us consider the following quadratic Urysohn integral equation:

where t ∈ ℝ₊ and α, β are positive constants.

Observe that this equation is a special case of (9) if we put

It is easily to check that for the previous functions there are satisfied assumptions of Theorem 8. Indeed, the function a = a(t) is an element of the space BC(ℝ₊) and | | a|| = α. This means that there is verified assumption (i). Next, notice that f is continuous on ℝ₊ ×   ℝ and f(t, 0) = βt²/(1 + t²). Thus, the function t → f(t, 0) belongs to BC(ℝ₊), and there is satisfied assumption (ii). Moreover, we have that $$. Apart from this, for x, y ∈ ℝ and for t ∈ ℝ₊ we obtain This implies that the function f(t, x) satisfies assumption (iii) with k = β.

Further, let us note that the function u(t, s, x) is continuous on the set ℝ₊ × ℝ₊ × ℝ. For arbitrarily fixed t, s ∈ ℝ₊, and x, y ∈ ℝ we obtain

Hence, using the inequality (cf. [9]) which is satisfied for all x, y ∈ ℝ and for any fixed a, a ≥ 0, we obtain This implies that the function u(t, s, x) satisfies assumption (iv) with h(r) = r^2/3 and g(t, s) = 1/(1 + t + s²). Obviously, the function h : ℝ₊ → ℝ₊ and is continuous and increasing on ℝ₊, while the function g(t, s) transforms ℝ₊ × ℝ₊ into ℝ₊ and is continuous on ℝ₊ × ℝ₊.

Moreover, we get

Thus, the function g(t, s) is integrable over ℝ₊. Next, we obtain Apart from this, in view of (48) we can easily to obtain that Further, let us fix arbitrarily a number T > 0. Then, similarly as above, we get This allows us to deduce the following equality: Finally, let us take into account the function Calculating the indefinite integral of the function u(t, s, 0), we obtain Hence, we get Hence, taking into account that we derive the following equality:

Now, fix arbitrarily a number T > 0. Then, from (54) we get

This allows us to derive the following estimate: Consequently, this yields that the following equality holds:

Next, observe that taking into account (48), (52), (55), and (60), we conclude that the functions g(t, s) and u(t, s, 0) satisfy assumptions (v) and (vi).

Now, we are coming to the last assumption of Theorem 8, that is, assumption (vii). Notice, that in the case of our Equation (42), in view of estimates (44), (50), and (55), the first inequality from assumption (vii) has the form

Hence, after some simplification, we obtain Thus, if for fixed positive α and β there exists a number r₀ > 0 satisfying inequality (62), then the second inequality from assumption (vii) is automatically satisfied. Indeed, it is an immediate consequence of Remark 7. In such a case (42) has solutions belonging to the ball $$ which are uniformly attractive.

For example, it is easy to check that if we take α = β = 1/8 then the number r₀ = 4/5 satisfies inequality (62). This means that (42) with α = β = 1/8 has solutions belonging to the ball B_4/5 being uniformly attractive.