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

Authors: Cristóbal González; Antonio Jiménez-Melado
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

In this paper, we propose the study of an integral equation, with deviating arguments, of the typey(t)=ω(t)-∫0∞‍f(t,s,y(γ1(s)),…,y(γN(s)))ds,t≥0, in the context of Banach spaces, with the intention of giving sufficient conditions that ensure the existence of solutions with the same asymptotic behavior at ∞ as ω(t). A similar equation, but requiring a little less restrictive hypotheses, is y(t)=ω(t)-∫0∞‍q(t,s)F(s,y(γ1(s)),…,y(γN(s)))ds,t≥0. In the case of q(t,s)=(t-s)+, its solutions with asymptotic behavior given by ω(t) yield solutions of the second order nonlinear abstract differential equation y′′(t)-ω′′(t)+F(t,y(γ1(t)),…,y(γN(t)))=0, with the same asymptotic behavior at ∞ as ω(t).


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## Body

## 1. Introduction

From the pioneering work of Atkinson [1], and subsequent works found in the literature (see, e.g., [2–10] for recent papers on the subject), we consider the following differential problem, with deviating arguments:
(1)y′′(t)-ω′′(t)+F(t,y(γ1(t)),…,y(γN(t)))=0,t≥0,
with the task of finding solutions y with the same behavior at ∞ as ω. Solutions with this prescription are given by the solutions of the following integral equation:
(2)y(t)=ω(t)-∫t∞‍(s-t)F(s,y(γ1(s)),…,y(γN(s)))ds,t≥0,
which, by writing f(t,s,y(γ1(s)),…,y(γN(s))) = (s-t)+F(s,y(γ1(s)),…,y(γN(s))), is of the type
(3)y(t)=ω(t)-∫0∞‍f(t,s,y(γ1(s)),…,y(γN(s)))ds,t≥0.The purpose of this note is to provide conditions that ensure the existence of solutions to the above integral equation, whose asymptotic behavior at∞ is the same as that of ω, thus giving a procedure to show existence of solutions with prescribed asymptotic behavior of differential equation of the type (1). Our wish is to also work out this integral equation in the setting of Banach spaces.Denote byℝ+ the set [0,∞) of nonnegative real numbers. Assume that {γj}j=1N is a finite set of continuous mappings from ℝ+ to ℝ+, that (X,∥·∥X) is a Banach space (with norm ∥·∥X), and also that ω is a continuous mapping from ℝ+ to X. Finally, assume that f:ℝ+×ℝ+×XN→X is a given continuous mapping with certain regularity and integrability conditions, to be specified later. In order to give a better aspect to our equation, define, for each continuous y:ℝ+→X, the mapping Γ(y):ℝ+→XN given by
(4)Γ(y)(t)=(y(γ1(t)),…,y(γN(t))),t≥0.Then, our equation becomes(E)y(t)=ω(t)-∫0∞‍f(t,s,Γ(y)(s))ds,t≥0,
which, by writing x(t)=y(t)-ω(t), t∈ℝ+, is transformed into
(E~)x(t)=-∫0∞‍f(t,s,Γ(ω+x)(s))ds,t≥0.A bit more of notation and preliminary results are needed. As customary,BX(x,r) denotes the open ball in X centered at x with radius r. The closure in X of any set A⊆X is written A¯, and its closed convex hull, co¯(A). The space of continuous X-valued functions defined on ℝ+ is denoted by 𝒞(ℝ+,X), while the space of bounded ones is 𝒞b(ℝ+,X). The latter is a Banach space when endowed with the sup norm ∥·∥∞, (i.e., for x∈𝒞b(ℝ+,X), ∥x∥∞=supt∈ℝ+∥x(t)∥X).The Schauder fixed point theorem states that any continuous operatorT defined on a nonempty, bounded, closed and convex subset C of a Banach space has necessarily a fixed point, provided that T(C) is a relatively compact subset of C. We will also be needing a well-known version of the Arzelà-Ascoli theorem which, in the case that occupies us, is as follows: if a family ℱ⊆𝒞(ℝ+,X) is equicontinuous at each t∈ℝ+, and each section ℱ(t):={u(t):u∈ℱ} is relatively compact in X, then each sequence {un}⊆ℱ contains a subsequence that converges uniformly on compact subsets of ℝ+ to a given X-valued function u.Let us also say a word about vector integrals. For a brief introduction see, for example, [11] or [12]. If f:[a,b]→X is a bounded vector function, the Riemann sum of f associated to a finite partition of [a,b], ℘={a=t0<t1<⋯<tn=b} (with norm ∥℘∥=maxj|tj-tj-1|), and to a selection {tj*}j=1n, tj*∈[tj-1,tj], is ∑j=1n‍f(tj*)(tj-tj-1). Without specifying the type of limit alluded, we say that f is Riemann integrable on [a,b] if there exists the limit of its Riemann sums as the norm of the partitions of [a,b] tends to 0, in which case, this limit (which is unique) is called the integral of f over [a,b] and is denoted by ∫ab‍f(t)dt. That is, being a little bit sloppy with the notation, we have
(5)∫abf(t)dt=lim∥℘∥→0∑j‍f(tj*)(tj-tj-1).Observe that wheneverf is integrable on [a,b], then each Riemann sum associated to f is (b-a) times a linear convex combination of elements of f([a,b]). Therefore, the integral of f over [a,b], being a limit of Riemann sums, is a multiple of an element of the closed convex hull of f([a,b]), that is,
(6)∫ab‍f(t)dt=lim∥℘∥→0(b-a)×∑j‍f(tj*)tj-tj-1b-a∈(b-a)co¯f([a,b]).


## 2. Existence of Solutions

We begin this section enumerating the conditions that will be basic for our results on existence of solutions of(E). Take f, γj, and ω as above and, in analogy with what would happen with continuous mappings on finite dimensional spaces, assume that
(H0)fisuniformlycontinuousonboundedsetsofℝ+×ℝ+×XN,andmapsboundedsetsintorelativelycompactsetsofX.The following conditions have already been motivated in previous work [7]. Recall that ω:ℝ+→X is continuous, but not necessarily bounded. 						(H1)Thereexistsg:ℝ+⟶ℝ+bounded,withg(t)→t→∞0,suchthatforC={y∈𝒞(ℝ+,X):∥y(t)-ω(t)∥X≤g(t),t∈ℝ+},∫0∞‍∥f(t,s,Γ(y)(s))∥Xds≤g(t),∀t∈ℝ+,∀y∈C.(H2)Thereexistsh:ℝ+⟶ℝ+withh(τ)→τ⟶∞0,suchthat∫τ∞‍∥f(t,s,Γ(y)(s))∥Xds≤h(τ),∀t,τ∈ℝ+∀y∈C.The result on existence of solutions to the integral equation is the following.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 Theorem1.
First observe that it suffices to find a solution to the integral equation(E~) in C~={x∈𝒞b(ℝ+,X):∥x(t)∥X≤g(t),t∈ℝ+}, and this will be achieved by proving the existence of a fixed point in C~ of the operator:
(7)T:x∈C~↦Tx,Tx(t)=-∫0∞‍f(t,s,Γ(ω+x)(s))ds,t≥0.
We proceed to check that the conditions of the Schauder fixed point Theorem are fulfilled. First observe thatC~ is a nonempty, bounded, closed, and convex subset of the Banach space 𝒞b(ℝ+,X). Also, by (H1), ∥Tx(t)∥X≤g(t) for all t∈ℝ+ and all x∈C~. So TC~⊆C~ provided Tx∈𝒞(ℝ+,X) for all x∈C~.
In fact, we will prove four assertions. (a)TC~ is uniformly equicontinuous on ℝ+. This will give the desired continuity of Tx for each x∈C~. (b) T is uniformly continuous on C~, hence it will be continuous on C~. (c) TC~(ℝ+) is relatively compact in X. Thus, (a) and (c) will imply, by the Arzelà-Ascoli theorem, that each sequence in TC~ has a subsequence which converges uniformly on each compact subset of ℝ+ to a given function in 𝒞(ℝ+,X) (actually in C~). Finally, in order to obtain that TC~ is relatively compact in 𝒞b(ℝ+,X), we will use the “funnel” structure of C~ to prove (d), that any sequence in C~ which converges uniformly on each compact subset of ℝ+ to a given function in C~ 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 arbitraryɛ>0 once for all. In what follows, we will build up different objects indexed by this ɛ, (tɛ, τɛ, Dɛ, Bɛ), 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.
Sinceg(t)→0 as t→∞, there exists tɛ>0 such that
(8)g(t)<ɛ2,∀t≥tɛ.
We start proving (d), as it is quite independent of the rest of assertions. Assume that a sequence{un}⊆C~ converges uniformly on each compact subset of ℝ+ to a function u∈C~, 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 tɛ>0 to satisfy (8). Since un, n∈ℕ, and u all belong to C~, then
(9)∥un(t)-u(t)∥X≤2g(t),∀n∈ℕ,∀t∈ℝ+.
In particular,(10)∥un(t)-u(t)∥X≤2g(t)<ɛ,∀n∈ℕ,∀t≥tɛ.
Now, since{un} converges uniformly to u in [0,tɛ], there exists nɛ∈ℕ such that
(11)∥un(t)-u(t)∥X<ɛ,∀n≥nɛ,∀t∈[0,tɛ].
This tells us that ifn≥nɛ, then ∥un-u∥∞≤ɛ, proving that the convergence of {un} to u is uniform on ℝ+, and thus (d) is proven.
Next, continue building up other objects associated with the arbitraryɛ fixed above. Observe that, by (8) and (H1),
(12)∥Tx(t)∥X≤∫0∞‍∥f(t,s,Γ(ω+x)(s))∥Xds≤g(t)<ɛ2,∀t≥tɛ∀x∈C~.
Also, sinceh(τ)→0 as τ→∞, there exists τɛ>0 such that
(13)h(τ)<ɛ4,∀τ≥τɛ,
so, by (H2),
(14)∫τɛ∞‍∥f(t,s,Γ(ω+x)(s))∥Xds≤h(τɛ)<ɛ4,∀t∈ℝ+,∀x∈C~.
The continuity ofω and the γj's, j=1,…,N, and the uniform bound for functions in C~ (given by the bound of g) imply that there exists a bounded set Dɛ⊆XN, depending on τɛ, ω, the γj's, and g, but not on x∈C~, such that
(15)Γ(ω+x)(s)=(ω(γ1(s))+x(γ1(s)),…,ω(γN(s))+x(γN(s)))∈Dɛ,∀x∈C~,∀s∈[0,τɛ].
Now, observe thatBɛ:=[0,tɛ+1]×[0,τɛ]×Dɛ is bounded in ℝ+×ℝ+×XN, so by (H0),
(16)fisuniformlycontinuousonBɛ,andf(Bɛ)isrelativelycompactinX.
By the uniform continuity off on Bɛ, there exists δɛ∈(0,1) such that, whenever (t1,s1,y1,1,…,yN,1)∈Bɛ and (t2,s2,y1,2,…,yN,2)∈Bɛ with |t1-t2|<δɛ, |s1-s2|<δɛ, and ∥yj,1-yj,2∥X<δɛ, j=1,…,N, then
(17)∥f(t1,s1,y1,1,…,yN,1)-f(t2,s2,y1,2,…,yN,2)∥X<ɛ4τɛ.
Notice that ift1,t2∈ℝ+ with |t1-t2|<δɛ and, without loss of generality, t1≤t2, then several cases are possible. If t2≥tɛ+1, then, as δɛ<1,  t2≥t1>t2-δɛ>tɛ+1-1=tɛ, and so, by (12),
(18)∥Tx(t1)-Tx(t2)∥X≤∥Tx(t1)∥X+∥Tx(t2)∥X<ɛ,∀x∈C~.
If, on the other hand,t2≤tɛ+1, then, by (15), Γ(ω+x)([0,τɛ])⊆Dɛ for all x∈C~, and, by (17) and (14), we have, for any x∈C~,
(19)∥Tx(t1)-Tx(t2)∥X≤∫0τɛ‍∥f(t2,s,Γ(ω+x)(s))-f(t1,s,Γ(ω+x)(s))∥Xds+∫τɛ∞‍∥f(t1,s,Γ(ω+x)(s))∥Xds+∫τɛ∞‍∥f(t2,s,Γ(ω+x)(s))∥Xds<τɛɛ4τɛ+2ɛ4<ɛ.
This proves (a), the uniform equicontinuity of TC~ over ℝ+.
Now notice that ifx1,x2∈C~ with ∥x1-x2∥∞<δɛ and t∈ℝ+, then, again, two cases are possible. If t≥tɛ, then, by (12),
(20)∥Tx1(t)-Tx2(t)∥X≤∥Tx1(t)∥X+∥Tx2(t)∥X≤2g(t)<ɛ,
while for t≤tɛ, using (15), (17), (14), and (13),
(21)∥Tx1(t)-Tx2(t)∥X≤∫0τɛ‍∥f(t,s,Γ(ω+x2)(s))-f(t,s,Γ(ω+x1)(s))∥Xds+∫τɛ∞‍∥f(t,s,Γ(ω+x1)(s))∥Xds+∫τɛ∞‍∥f(t,s,Γ(ω+x2)(s))∥Xds<τɛɛ4τɛ+2h(τɛ)<ɛ,
This proves that ∥Tx1-Tx2∥∞<ɛ, showing (b), that T is uniformly continuous on C~.
For the compactness ofTC~(ℝ+)¯ in X, it suffices to show that TC~(ℝ+) is totally bounded [13, page 298]; that is, for the given ɛ>0 (so for any ɛ>0) there exists a finite covering of TC~(ℝ+) with balls of radii not bigger than ɛ. Observe first that, by (12), ∥Tx(t)∥X<ɛ/2 for all x∈C~ and all t≥tɛ, that is,
(22)TC~([tɛ,∞))⊆BX(0,ɛ2).
Now, in order to control the elements ofTC~([0,tɛ]), observe that each of these can be decomposed as the sum of a “head” and a “tail,”
(23)Tx(t)=-∫0τɛ‍f(t,s,Γ(ω+x)(s))ds-∫τɛ∞‍f(t,s,Γ(ω+x)(s))ds,∀x∈C~andallt∈[0,tɛ].
The “head” can be approximated by Riemann sums, which, in turn, are nothing else butτɛ times a convex linear combination of elements of -f(Bɛ), that is, the “head” is an element of τɛco¯(-f(Bɛ)). By (16), f(Bɛ) has compact closure in X, so by Mazur's theorem [14], co¯(-f(Bɛ)) is compact, and therefore it can be covered with a finite number of balls, say B1,…,Bℓ, of radii not bigger than ɛ/(2τɛ). This yields a finite covering of τɛco¯(-f(Bɛ)) with balls of radii not bigger than ɛ/2, precisely the collection {τɛBj}j=1ℓ. On the other hand, by (14), the “tail” of each of the above integrals is bounded by h(τɛ)<ɛ/4, so they are elements of BX(0,ɛ/4). All this can be summarized as follows:
(24)TC~([0,tɛ])⊆τɛco¯(-f(Bɛ))+BX(0,ɛ4)⊆(⋃j=1ℓ‍τɛBj)+BX(0,ɛ4)=⋃j=1ℓ‍(τɛBj+BX(0,ɛ4));
that is, TC~([0,tɛ]) can be covered with a finite collection of balls of radii smaller than ɛ, because each τɛBj+BX(0,ɛ/4) is readily seen to be a ball of radius not bigger than 3ɛ/4.
At the end, by (22) and (24), we have
(25)TC~(ℝ+)⊆TC~([0,tɛ])∪TC~([tɛ,∞))⊆⋃j=1ℓ‍(τɛBj+BX(0,ɛ4))∪BX(0,ɛ2);
that is, we have given a finite covering of TC~(ℝ+) with balls of radii not bigger than ɛ. With this we conclude (c) and, with all four assertions proved, the theorem too.Coming back to the integral equation (2) underlying the differential equation (1), we just need to adapt the hypotheses presented above to the function f(t,s,y1,…,yN)=(s-t)+F(s,y1,…,yN), to obtain a corresponding result on existence of solutions to (2) with asymptotic behavior given by ω(t). Notice that these hypotheses, (H0), (H1), and (H2), are natural generalizations from the 1-dimensional case. However, the proof in the abstract setting has shown many more properties for the operator T that needed in order to show the existence of a fixed point. This, somehow, is telling us that the hypotheses could be weakened. For the moment, we content ourselves noticing that, in (2), the corresponding hypothesis of uniform continuity on bounded sets is “not needed,” because it will be “consequence” of the other hypotheses, and a little trick of changing the domain of definition of the operator. Thus, for our next result, we will be using the following hypothesis:
(H0′)F:ℝ+×XN⟶Xiscontinuousandmapsboundedsetsintorelativelycompactones.
Also, instead of considering just the integral equation (2), we generalize a little bit to a convolution type integral equation,
(E1)y(t)=ω(t)-∫0∞‍q(t,s)F(s,Γ(y)(s))ds,t≥0,
with a kernel q(t,s) satisfying the right properties for us. Observe that, when q(t,s)=(s-t)+, and 0≤t1≤t2,
(26)q(t1,s)-q(t2,s)=(s-t1)+-(s-t2)+={0if0≤s≤t1s-t1ift1≤s≤t2t2-t1ift2≤s≤t2-t1.That is,q(t,s) satisfies a Lipschitz condition on the first variable, independent of the second. Actually, much less is needed, just continuity of q(t,s) suffices.Theorem 3.
Letω, γj, 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,…,yN)=q(t,s)F(s,y1,…,yN)), 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,
(27)(s-t)+=s-t≤s≤2(s-τ2)=2(s-τ2)+.
Consequently, from hypothesis (H1), taking h(τ)=3g(τ/2), we have h(τ)→0 as τ→∞, and
(28)∫τ∞‍(s-t)+∥F(s,Γ(y)(s))∥Xds≤3g(τ2)=h(τ),∀τ,t∈ℝ+,∀y∈C.
In general, hypothesis(H2) is superfluous whenever there exist constants, a∈(0,1) and A>1, such that
(29)|q(t,s)|≤A|q(aτ,s)|,∀t,s,τ∈ℝ+,withs≥τ.Proof of Theorem3.
We proceed as before. Consider initially the same nonempty, bounded, closed, and convex subsetC~ 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 TC~(ℝ+) is compact in X, and (d), that uniform convergence on compact subsets of ℝ+ of a sequence in C~ turns into actual uniform convergence on ℝ+.
Let us now prove (a), thatTC~ is uniformly equicontinuous on ℝ+. This will finish showing that TC~⊆C~, and, by (c), (d), and the Arzelà-Ascoli theorem, that TC~ has compact closure in 𝒞b(R+,X).
Letɛ>0 be fixed. Find, as before, tɛ>0, τɛ>0, and Dɛ bounded subset of XN, so as to satisfy (8), (12), (13), (14), and (15). Now, observe that [0,τɛ]×Dɛ is bounded in ℝ+×XN, so by (H0′), F([0,τɛ]×Dɛ) is relatively compact in X; hence there exists Mɛ>0 such that
(30)∥F(s,Γ(ω+x)(s))∥X≤Mɛ,∀s∈[0,τɛ],∀x∈C~.
Notice now thatq is uniformly continuous on Rɛ=[0,tɛ+1]×[0,τɛ]. So there is δɛ∈(0,1) such that, whenever (t1,s1) and (t2,s2) are in Rɛ with |t1-t2|<δ and |s1-s2|<δ, then
(31)|q(t1,s1)-q(t2,s2)|<ɛ4τɛMɛ.
Now taket1,t2∈ℝ+ with |t1-t2|<δɛ and, without loss of generality, assume that t1≤t2. Then again, several cases are possible. If t2≥tɛ+1, then, as δɛ<1, t2≥t1>t2-δɛ>tɛ+1-1=tɛ, and so, by (12),
(32)∥Tx(t1)-Tx(t2)∥X≤∥Tx(t1)∥X+∥Tx(t2)∥X<ɛ,∀x∈C~.
If, on the other hand,t2≤tɛ+1, then, by (31), (30), and (14), we have, for any x∈C~,
(33)∥Tx(t1)-Tx(t2)∥X≤∫0τɛ‍|q(t2,s)-q(t1,s)|×∥F(s,Γ(ω+x)(s))∥Xds+∫τɛ∞|q(t1,s)|∥F(s,Γ(ω+x)(s))∥Xds‍+∫τɛ∞|q(t2,s)|∥F(s,Γ(ω+x)(s))∥Xds‍<ɛ4τɛMɛMɛτɛ+2ɛ4<ɛ.
This proves the uniform equicontinuity of TC~ over ℝ+.
To finish the proof, we have to prove (b), thatT is uniformly continuous. It is here that we restrict the domain of definition of T. Observe that co¯(TC~) is nonempty, closed and convex. Also, co¯(TC~)⊆C~ because TC~⊆C~ and C~ is closed and convex. More is true, since TC~ has compact closure, then, by Mazur's Theorem, co¯(TC~) is compact too. One more thing, T leaves invariant co¯(TC~):
(34)co¯(TC~)⊆C~⟹T(co¯(TC~))⊆TC~⊆co¯(TC~).
With all of this, it suffices to prove thatT is continuous on this new T-invariant set. Let us prove that, indeed, T is uniformly continuous on co¯(TC~). Let ɛ>0 be given. Find that tɛ>0 and τɛ>0 as before, so as to satisfy (8), (12), (13), and (14). The continuity of ω and the γj's, j=1,…,N, and the compactness of co¯(TC~) tells us that Dɛ={Γ(ω+x)(s):s∈[0,τɛ],x∈co¯(TC~)} is a compact subset of XN, giving us the opportunity to say that F is uniformly continuous on [0,τɛ]×Dɛ and, consequently, that f(t,s,y1,…,yN)=q(t,s)F(s,y1,…,yN) is uniformly continuous on Bɛ:=[0,tɛ+1]×[0,τɛ]×Dɛ. So there exists δɛ∈(0,1) such that, for (t1,s1,y1,1,…,yN,1)∈Bɛ and (t2,s2,y1,2,…,yN,2)∈Bɛ with |t1-t2|<δɛ, |s1-s2|<δɛ, and ∥yj,1-yj,2∥X<δɛ, j=1,…,N, then
(35)∥f(t1,s1,y1,1,…,yN,1)-f(t2,s2,y1,2,…,yN,2)∥X<ɛ4τɛ.
Now, it is a matter of repeating the same steps as was done in Theorem 1 to prove the uniform continuity of T.
This concludes the proof of the theorem.Remark 5.
Instead of working with functions defined onℝ+, 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.Next, we just mention an easy result on the existence of solutions of the underlying differential equation, just to illustrate the type of functionsF that could generate a condition like (H1). This result is inspired from [7, Thm. 2] and [8, Thm. 1].Theorem 6.
LetF:ℝ+×XN→X be a continuous function mapping bounded sets into relatively compact ones, such that,
(36)∥F(t,y1,…,yN)∥X≤ψ(t),∀t∈ℝ+,∀(y1,…,yN)∈XN,
where
(37)ψ:ℝ+⟶ℝ+,satisfies∫0∞‍sψ(s)ds<∞.
Let also {γj}j=1N be a set of continuous functions from ℝ+ to ℝ+.
Then, for anyω∈𝒞2(ℝ+,X), (1) has a solution y∈𝒞2(ℝ+,X) with y(t)-ω(t)→0 as t→∞.Proof.
Takeω∈𝒞2(ℝ+,X), consider the corresponding integral equation (2), and define
(38)g(t)=∫t∞sψ(s)ds,t≥0.
Observe that, by (37), g(t)→0 as t→∞. This allows us to consider the set
(39)C={y∈𝒞(ℝ+,X):∥y(t)-ω(t)∥X≤g(t),t≥0},
which, adopting the notation used throughout the paper, gives by (36), for y∈C and t≥0,
(40)∫t∞‍∥(s-t)F(s,Γ(y)(s))∥Xds≤∫t∞‍(s-t)ψ(s)ds≤g(t).
With this, Theorem3 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).



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*Source: 957696-2013-12-19.xml*