Asymptotic Behavior of Solutions to a Vector Integral Equation with Deviating Arguments

Theorem 1. Under hypotheses (H0), (H1), and (H2), the integral equation (E) has a solution y(t) asymptotically equal to ω(t) as t → ∞.

Remark 2. This theorem represents a generalization of the one presented in the work [7] in two aspects. First, we have made the jump to deal with integral equations in the setting of infinite dimensional spaces. And second, we have included deviating arguments in the equation.

Proof of Theorem 1. First observe that it suffices to find a solution to the integral equation (E~) in $$, and this will be achieved by proving the existence of a fixed point in $$ of the operator:

We proceed to check that the conditions of the Schauder fixed point Theorem are fulfilled. First observe that $$ is a nonempty, bounded, closed, and convex subset of the Banach space 𝒞_b(ℝ⁺, X). Also, by (H1), ∥Tx(t)∥_X ≤ g(t) for all t ∈ ℝ⁺ and all $$. So $$ provided Tx ∈ 𝒞(ℝ⁺, X) for all $$.

In fact, we will prove four assertions. (a) $$ is uniformly equicontinuous on ℝ⁺. This will give the desired continuity of Tx for each $$. (b) T is uniformly continuous on $$, hence it will be continuous on $$. (c) $$ is relatively compact in X. Thus, (a) and (c) will imply, by the Arzelà-Ascoli theorem, that each sequence in $$ has a subsequence which converges uniformly on each compact subset of ℝ⁺ to a given function in 𝒞(ℝ⁺, X) (actually in $$). Finally, in order to obtain that $$ is relatively compact in 𝒞_b(ℝ⁺, X), we will use the “funnel” structure of $$ to prove (d), that any sequence in $$ which converges uniformly on each compact subset of ℝ⁺ to a given function in $$ must indeed converge uniformly to that function in all of ℝ⁺. With all these assertions, the Schauder fixed point theorem can be applied to conclude the existence of a fixed point of T, as we want.

Start fixing an arbitrary ɛ > 0 once for all. In what follows, we will build up different objects indexed by this ɛ, (t_ɛ, τ_ɛ, D_ɛ, B_ɛ), knowing that even if for each assertion we have to start taking an arbitrary ɛ > 0, the objects will vary accordingly but not the way to obtain them.

Since g(t) → 0 as t → ∞, there exists t_ɛ > 0 such that

We start proving (d), as it is quite independent of the rest of assertions. Assume that a sequence $$ converges uniformly on each compact subset of ℝ⁺ to a function $$, and let us show that indeed converges uniformly to u in all of ℝ⁺. For the ɛ > 0 above (so for any ɛ > 0), find the corresponding t_ɛ > 0 to satisfy (8). Since u_n, n ∈ ℕ, and u all belong to $$, then

In particular,

Now, since {u_n} converges uniformly to u in [0, t_ɛ], there exists n_ɛ ∈ ℕ such that

This tells us that if n ≥ n_ɛ, then $$, proving that the convergence of {u_n} to u is uniform on ℝ⁺, and thus (d) is proven.

Next, continue building up other objects associated with the arbitrary ɛ fixed above. Observe that, by (8) and (H1),

Also, since h(τ) → 0 as τ → ∞, there exists τ_ɛ > 0 such that

The continuity of ω and the γ_j′s, j = 1, …, N, and the uniform bound for functions in $$ (given by the bound of g) imply that there exists a bounded set D_ɛ⊆X^N, depending on τ_ɛ, ω, the γ_j′s, and g, but not on $$, such that

Now, observe that B_ɛ : = [0, t_ɛ + 1]×[0, τ_ɛ] × D_ɛ is bounded in ℝ⁺ × ℝ⁺ × X^N, so by (H0),

By the uniform continuity of f on B_ɛ, there exists δ_ɛ ∈ (0,1) such that, whenever (t₁, s₁, y_1,1, …, y_N,1) ∈ B_ɛ and (t₂, s₂, y_1,2, …, y_N,2) ∈ B_ɛ with |t₁ − t₂ | < δ_ɛ, |s₁ − s₂ | < δ_ɛ, and $$, j = 1, …, N, then

Notice that if t₁, t₂ ∈ ℝ⁺ with |t₁ − t₂ | < δ_ɛ and, without loss of generality, t₁ ≤ t₂, then several cases are possible. If t₂ ≥ t_ɛ + 1, then, as δ_ɛ < 1, t₂ ≥ t₁ > t₂ − δ_ɛ > t_ɛ + 1 − 1 = t_ɛ, and so, by (12),

If, on the other hand, t₂ ≤ t_ɛ + 1, then, by (15), Γ(ω + x)([0, τ_ɛ])⊆D_ɛ for all $$, and, by (17) and (14), we have, for any $$,

Now notice that if $$ with $$ and t ∈ ℝ⁺, then, again, two cases are possible. If t ≥ t_ɛ, then, by (12),

For the compactness of $$ in X, it suffices to show that $$ is totally bounded [13, page 298]; that is, for the given ɛ > 0 (so for any ɛ > 0) there exists a finite covering of $$ with balls of radii not bigger than ɛ. Observe first that, by (12), ∥Tx(t)∥_X < ɛ/2 for all $$ and all t ≥ t_ɛ, that is,

Now, in order to control the elements of $$, observe that each of these can be decomposed as the sum of a “head” and a “tail,”

The “head” can be approximated by Riemann sums, which, in turn, are nothing else but τ_ɛ times a convex linear combination of elements of −f(B_ɛ), that is, the “head” is an element of $$. By (16), f(B_ɛ) has compact closure in X, so by Mazur′s theorem [14], $$ is compact, and therefore it can be covered with a finite number of balls, say B₁, …, B_ℓ, of radii not bigger than ɛ/(2τ_ɛ). This yields a finite covering of $$ with balls of radii not bigger than ɛ/2, precisely the collection $$. On the other hand, by (14), the “tail” of each of the above integrals is bounded by h(τ_ɛ) < ɛ/4, so they are elements of B_X(0, ɛ/4). All this can be summarized as follows:

At the end, by (22) and (24), we have

Theorem 3. Let ω, γ_j, j = 1, …, N, Γ be as for Theorem 1, and let q : ℝ⁺ × ℝ⁺ → ℝ  be continuous. Under hypotheses (H0^′), (H1), and (H2)  (where f(t, s, y₁, …, y_N) = q(t, s)  F(s, y₁, …, y_N)), the integral equation (E₁) has a solution y(t) asymptotically equal to ω(t) as t → ∞.

Remark 4. Observe, as in [7], that if the kernel q is the one we started with, q(t, s) = (s − t) ₊, then hypothesis (H2) is redundant, because in that case, if s ≥ τ, we always have, for τ ≤ t, (s − t) ₊ ≤ (s − τ) ₊ ≤ 2(s−τ/2)₊, and for τ > t,

In general, hypothesis (H2) is superfluous whenever there exist constants, a ∈ (0,1) and A > 1, such that

Proof of Theorem 3. We proceed as before. Consider initially the same nonempty, bounded, closed, and convex subset $$ as before, as well as the same operator T. The idea is to find a fixed point for T using the Schauder fixed point theorem. For that, we start repeating the same scheme of four assertions established in the previous theorem. The two assertions that do not depend on the uniform continuity of f on bounded sets are done the same way as before; hence we omit their proofs. These are (c), that $$ is compact in X, and (d), that uniform convergence on compact subsets of ℝ⁺ of a sequence in $$ turns into actual uniform convergence on ℝ⁺.

Let us now prove (a), that $$ is uniformly equicontinuous on ℝ⁺. This will finish showing that $$, and, by (c), (d), and the Arzelà-Ascoli theorem, that $$ has compact closure in 𝒞_b(R⁺, X).

Let ɛ > 0 be fixed. Find, as before, t_ɛ > 0, τ_ɛ > 0, and D_ɛ bounded subset of X^N, so as to satisfy (8), (12), (13), (14), and (15). Now, observe that [0, τ_ɛ] × D_ɛ is bounded in ℝ⁺ × X^N, so by (H0′), F([0, τ_ɛ] × D_ɛ) is relatively compact in X; hence there exists M_ɛ > 0 such that

Notice now that q is uniformly continuous on R_ɛ = [0, t_ɛ + 1]×[0, τ_ɛ]. So there is δ_ɛ ∈ (0,1) such that, whenever (t₁, s₁) and (t₂, s₂) are in R_ɛ with |t₁ − t₂ | < δ and |s₁ − s₂ | < δ, then

Now take t₁, t₂ ∈ ℝ⁺ with |t₁ − t₂ | < δ_ɛ and, without loss of generality, assume that t₁ ≤ t₂. Then again, several cases are possible. If t₂ ≥ t_ɛ + 1, then, as δ_ɛ < 1, t₂ ≥ t₁ > t₂ − δ_ɛ > t_ɛ + 1 − 1 = t_ɛ, and so, by (12),

If, on the other hand, t₂ ≤ t_ɛ + 1, then, by (31), (30), and (14), we have, for any $$,

To finish the proof, we have to prove (b), that T is uniformly continuous. It is here that we restrict the domain of definition of T. Observe that $$ is nonempty, closed and convex. Also, $$ because $$ and $$ is closed and convex. More is true, since $$ has compact closure, then, by Mazur′s Theorem, $$ is compact too. One more thing, T leaves invariant $$:

With all of this, it suffices to prove that T is continuous on this new T-invariant set. Let us prove that, indeed, T is uniformly continuous on $$. Let ɛ > 0 be given. Find that t_ɛ > 0 and τ_ɛ > 0 as before, so as to satisfy (8), (12), (13), and (14). The continuity of ω and the γ_j′s, j = 1, …, N, and the compactness of $$ tells us that $$ is a compact subset of X^N, giving us the opportunity to say that F is uniformly continuous on [0, τ_ɛ] × D_ɛ and, consequently, that f(t, s, y₁, …, y_N) = q(t, s)  F(s, y₁, …, y_N) is uniformly continuous on B_ɛ : = [0, t_ɛ + 1]×[0, τ_ɛ] × D_ɛ. So there exists δ_ɛ ∈ (0,1) such that, for (t₁, s₁, y_1,1, …, y_N,1) ∈ B_ɛ and (t₂, s₂, y_1,2, …, y_N,2) ∈ B_ɛ with |t₁ − t₂ | < δ_ɛ, |s₁ − s₂ | < δ_ɛ, and $$, j = 1, …, N, then

This concludes the proof of the theorem.

Remark 5. Instead of working with functions defined on ℝ⁺, we could have worked with functions defined on any interval of the type [t₀, ∞), obtaining a result completely similar. Also, many times, one is just interested in giving partial solutions; that is, solutions defined not on the whole interval [t₀, ∞), but on some other interval [t₁, ∞) with t₁ ≥ t₀.

Theorem 6. Let F : ℝ⁺ × X^N → X be a continuous function mapping bounded sets into relatively compact ones, such that,

Then, for any ω ∈ 𝒞²(ℝ⁺, X), (1) has a solution y ∈ 𝒞²(ℝ⁺, X) with y(t) − ω(t) → 0 as t → ∞.

Proof. Take ω ∈ 𝒞²(ℝ⁺, X), consider the corresponding integral equation (2), and define

Observe that, by (37), g(t) → 0 as t → ∞. This allows us to consider the set

With this, Theorem 3 applies (hypothesis (H2) need not be verified by Remark 4) and then there exists y ∈ 𝒞(ℝ⁺, X), solution of (2) with y(t) − ω(t) → 0 as t → ∞. Finally, it is just an exercise to check that y is twice continuously differentiable and that satisfies the differential equation (1).