# Existence and Uniqueness of Homoclinic Solution for a Class of Nonlinear Second-Order Differential Equations

Authors: Lijuan Chen; Shiping Lu
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

The authors study the existence and uniqueness of a set with2kT-periodic solutions for a class of second-order differential equations by using Mawhin's continuation theorem and some analysis methods, and then a unique homoclinic orbit is obtained as a limit point of the above set of 2kT-periodic solutions.


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## Body

## 1.   Introduction

In this paper, we study the existence and uniqueness of homoclinic solutions for the following nonlinear second-order differential equations:(1.1)u′′(t)+g(u′(t))+h(u(t))=f(t),
where u(t)∈R,  g,  h and f are all in C(R,R).As usual we say that a nonzero solutionu(t) of (1.1) is homoclinic (to 0) if u(t)→0 and u′(t)→0 as |t|→+∞.Equation (1.1) is important in the applied sciences such as nonlinear vibration of masses, see [1–3] and the references therein. But most of the authors in those papers are interested in the study of problems of periodic solutions. Recently, the existence of homoclinic solutions for some second-order ordinary differential equation (system) has been extensively studied by using critical point theory, see [4–13] and the references therein. For example, in [9], by using the Mountain Pass theorem, Lv et al. discussed the existence of homoclinic solutions for the following second-order Hamiltonian systems:
(1.2)u′′(t)-L(t)u(t)+∇w(t,u(t))=0,
and in [13], the authors by means of variational method studied the problem of homoclinic solutions for the forced pendulum equation without the first derivative term. But, as far as we know, there were few papers studying the existence of homoclinic solution for the equation such as (1.1). This is due to the fact that (1.1) contains the first derivative term g(u′(t)). This implies that the differential equation is not the Euler Lagrange equation associated with some functional I:W2kT1,p→R. So the method of critical point theory (or variational method) in [4–13] cannot be applied directly. Although paper [13] discussed the existence of homoclinic solutions for the following equation containing the first derivative term:
(1.3)x′′(t)+f(t)x′(t)+β(t)x(t)+g(t,x(t))=0,
the term containing the first derivative is only linear with respect to x′(t).In order to investigate the homoclinic solutions to (1.1), firstly, we study the existence of 2kT-periodic solutions to the following equation for each k∈N:
(1.4)u′′(t)+g(u′(t))+h(u(t))=fk(t),
where fk:R→R is a 2kT-periodic function such that
(1.5)fk(t)={f(t),t∈[-kT,kT-ε0]f(kT-ε0)+f(-kT)-f(kT-ε0)ε0(t-kT+ε0),t∈[kT-ε0,kT],T>0 is a given constant, and ε0∈(0,T) is a constant independent of k. Then a homoclinic solution to (1.1) is obtained as a limit point of the set {uk(t)}, where uk(t) is an arbitrary 2kT-periodic solution to (1.4) for each k∈N.The significance of present paper is that we not only investigate the existence of homoclinic solution to (1.1), but also study the uniqueness of the homoclinic solution and, the existence of 2kT-periodic solutions to (1.4) is obtained by using Mawhin’s continuation theorem [14], not by using the methods of critical point theory, which is quite different from the approaches of [4–13, 15]. Furthermore, the method to obtain the homoclinic solution to (1.1) is also different from the corresponding ones of [15].


## 2.  Main Lemmas

For eachk∈N, let C2kT={x∣x∈C(R,R),x(t+2kT)≡x(t)}, C2kT1={x∣x∈C1(R,R),x(t+2kT)≡x(t)}, the norms of C2kT and C2kT1 are defined by ∥·∥∞=maxt∈[-kT,kT]|x(t)| and ∥x∥C2kT1=max{∥x∥∞,∥x′∥∞}, respectively, then C2kT and C2kT1 are all Banach spaces. Furthermore for x∈C2kT, ∥x∥r=(∫-kTkT‍|x(t)|rdt)1/r, where r∈(1,+∞).Lemma 2.1 (see [12]).
Leta>0 and q∈W1,p(R,R), then for every t∈R, the following inequality holds:
(2.1)|q(t)|≤(2a)-1/μ(∫t-at+a‍|q(s)|μds)1/μ+a(2a)-1/p(∫t-at+a‍|q′(s)|pds)1/p,
where μ,p∈(1,+∞) are constants.Lemma 2.2 (see [12]).
Letq∈W2kT1,p(R,Rn), then the following inequality holds:
(2.2)∥q∥∞≤T-1/ν(∫-kTkT‍|q(s)|νds)1/ν+T(p-1)/p(∫-kTkT‍|q′(s)|pds)1/p,
where ν and p are constants with ν>1 and p>1.In order to use Mawhin’s continuation theorem for investigating the existence of2kT-periodic solutions to (1.4), we give some definitions associated with Mawhin’s continuation theorem.Definition 2.3 (see [14]).
LetX and Y be two Banach spaces with norms ∥x∥X and ∥x∥Y, respectively. A linear operator
(2.3)L:D(L)⊂X→Y
is said to be a Fredholm opeartor with index zero provided that (1)
ImL is a closed subset of Y;(2)
dimKerL=codimImL<∞.IfL:D(L)⊂X→Y is a Fredholm operator with index zero, then X=kerL⊕X1 and Y=ImL⊕Y1. Let P:X→kerL and Q:Y→Y1 be the continuous projectors. Clearly, kerL∩(D(L)∩X1)={0}, thus the restriction LP:=L|D(L)∩X1 is invertible. Denote by KP the inverse of LP.Definition 2.4 (see [14]).
LetX and Y be two Banach spaces with norms ∥x∥X and ∥x∥Y, respectively, and the operator
(2.4)L:D(L)⊂X→Y
is a Fredholm operator with index zero, Ω⊂X is an open bounded set with D(L)∩Ω≠ϕ. A continuous operator N:Ω¯⊂X→Y is said to be L-compact in Ω¯, provided that (1)
KP(I-Q)N(Ω¯) is a relative compact set of X;(2)
QN(Ω¯) is a bounded set of Y.Lemma 2.5 (see [14]).
Suppose thatX and Y are two Banach spaces, and L:D(L)⊂X→Y is a Fredholm operator with index zero. Furthermore, Ω⊂X is an open bounded subset and N:Ω¯→Y is L-compact on Ω¯. If all the following conditions hold:(1)
Lx≠λNx, for all x∈∂Ω∩D(L),λ∈(0,1);(2)
Nx∉ImL, for all x∈∂Ω∩KerL;(3)
deg{JQN,Ω∩KerL,0}≠0,whereJ:ImQ→KerL is an isomorphism. Then equation Lx=Nx has a solution on Ω¯∩D(L).Lemma 2.6.
Assume that there are positive constantsm,  m1,  n,  l0, and l1 with l0≥l1, such that the following conditions hold.(A1)
supt∈R|f(t)|<+∞, ∫R‍|f(t)|(l0+1)/l0dt<+∞,∫R‍|f(t)|(l1+1)/l1dt<+∞ and ∫R‍|f(t)|2dt<+∞.(A2)
(2.5)-m1|x|l0+1≤xg(x)≤-m|x|l0+1,∀x∈R,xh(x)≤-n|x|l1+1,∀x∈R.(A3)
h∈C1(R,R) with h′(x)≤0 for all x∈R.Then for everyk∈N, (1.4) possesses a 2kT-periodic solution.Remark 2.7.
From (1.5), we see
(2.6)∥fk∥∞≤supt∈R|f(t)|<+∞,∀k∈N,∥fk∥(l0+1)/l0=(∫-kTkT‍|fk(s)|(l0+1)/l0ds)l0/(l0+1)≤(∫-kTkT-ε0‍|fk(s)|(l0+1)/l0ds)l0/(l0+1)+(∫kT-ε0kT‍|fk(s)|(l0+1)/l0ds)l0/(l0+1)≤(∫R‍|f(s)|(l0+1)/l0ds)l0/(l0+1)+ε0l0/(l0+1)supt∈R|f(t)|,∀k∈N,
which together with assumption (A1) yields that ∥fk∥∞ and ∥fk∥(l0+1)/l0 are two constants independent of k∈N.Similarly, we have that∥fk∥(l1+1)/l1<+∞ and ∥fk∥2<+∞ are two constants independent of k∈N.Proof.
SetX=C2kT1,  Y=C2kT,  L:D(L)⊂X→Y,  Lu=u′′,   where D(L)={u∣u∈C2kT2}, and
(2.7)N:C2kT1→C2kT,[Nu](t)=-g(u′(t))-h(u(t))+fk(t).
Clearly, KerL=R, ImL={y∈Y:∫-kTkT‍y(s)ds=0}, which implies that ImL is a closed subset of X, and dimKerL=codimImL=1<+∞. So L is a Fredholm operator with index zero. Let
(2.8)P:X→KerL,Q:Y→Y/ImL
be defined respectively by Px=(1/2kT)∫02kT‍x(s)ds,Qx=(1/2kT)∫02kT‍x(s)ds and let
(2.9)LP=L|X∩KerP:X∩KerP→ImL.
Then LP has a unique continuous pseudo-inverse LP-1 on ImL defined by (LP-1y)(t)=[Fy](t), where
(2.10)[Fy](t)=∫02kT‍G(t,s)y(s)ds,G(t,s)={s(2kT-t)2kT,0≤s<t,t(2kT-s)2kT,t≤s≤2kT.
For each open bounded set Ω⊂C2kT, from the above formula, it is easy to see that the mapper N is L-compact on Ω¯.
Step  1. For each k∈N, let Ω1={x∈C2kT1:Lx=λNx,λ∈(0,1)}, that is,
(2.11)Ω1={x∈C2kT1:x′′(t)+λg(x′(t))+λh(x(t))=λfk(t),λ∈(0,1)}.
We will show that Ω1 is bounded in C2kT1. Suppose that u∈Ω1, then
(2.12)u′′(t)+λg(u′(t))+λh(u(t))=λfk(t),λ∈(0,1).
Multiplying both sides of (2.12) by u′(t) and integrating on the interval [-kT,kT], we have from assumption (A2) that
(2.13)m∫-kTkT‍|u′(t)|l0+1dt≤-∫-kTkT‍u′(t)g(u′(t))dt=-∫-kTkT‍fk(t)u′(t)dt.
By using hólder inequality, we get
(2.14)∥u′∥l0+1l0+1≤1m∥fk∥(l0+1)/l0·∥u′∥l0+1,
which together with the conclusion of Remark 2.7 shows
(2.15)∥u′∥l0+1≤(1m∥fk∥(l0+1)/l0)1/l0∶=β1.
Clearly, β1 is a constant independent of k and λ.
Multiplying both sides of (2.12) by u′′(t) and integrating on the interval [-kT,kT], we have
(2.16)∥u′′∥22-λ∫-kTkT‍(u′(t))2h′(u(t))dt=λ∫-kTkT‍fk(t)u′′(t)dt.
It follows from (2.16) and assumption (A3) that
(2.17)∥u′′∥22≤∫-kTkT‍|fk(t)u′′(t)|dt≤∥fk∥2·∥u′′∥2,
which implies
(2.18)∥u′′∥2≤∥fk∥2∶=β2.
By using Lemma 2.2, we have
(2.19)∥u′∥∞≤T-1/(l0+1)∥u′∥l0+1+T1/2∥u′′∥2≤T-1/(l0+1)β1+T1/2β2∶=β.
Clearly, β is a constant independent of k and λ.
On the other hand, multiplying both sides of (2.12) by u(t) and integrating on the interval [-kT,kT], we have
(2.20)-∫-kTkT‍|u′(t)|2dt+λ∫-kTkT‍h(u(t))u(t)dt=-λ∫-kTkT‍g(u′(t))u(t)dt+λ∫-kTkT‍fk(t)u(t)dt.
It follows from assumption (A2) that
(2.21)λn∫-kTkT‍(u(t))l1+1dt+∫-kTkT‍(u′(t))2dt≤λm1∫-kTkT‍|u′(t)|l0|u(t)|dt+λ∫-kTkT‍|fk(t)||u(t)|dt,
which together with (2.19) and l0≥l1 results in
(2.22)n∫-kTkT‍|u(t)|l1+1dt≤m1∫-kTkT‍|u′(t)|l0·|u(t)|dt+∫-kTkT‍|fk(t)u(t)|dt≤m1(∫-kTkT‍|u′(t)|l0(l1+1)/l1dt)l1/(l1+1)·(∫-kTkT‍|u(t)|l1+1dt)1/(l1+1)+(∫-kTkT‍|fk(t)|(l1+1)/l1dt)l1/(l1+1)·(∫-kTkT‍|u(t)|l1+1dt)1/(l1+1)=m1(∫-kTkT‍|u′(t)|l0+1|u′(t)|(l0-l1)/l1dt)l1/(l1+1)·∥u∥l1+1+∥fk∥(l1+1)/l1∥u∥l1+1≤m1β(l0-1)/(l1+1)(∫-kTkT‍|u′(t)|l0+1dt)l1/(l1+1)∥u∥l1+1+∥fk∥(l1+1)/l1∥u∥l1+1=m1β(l0-l1)/(l1+1)(∥u′∥l0+1)l1(l0+1)/(l1+1)∥u∥l1+1+∥fk∥(l1+1)/l1∥u∥l1+1≤m1β(l0-1)/(l1+1)β1l1(l0+1)/(l1+1)∥u∥l1+1+∥fk∥(l1+1)/l1∥u∥l1+1
that is,
(2.23)∥u∥l1+1l1+1≤1n[m1β(l0-l1)/(l1+1)β1l1(l0+1)/(l1+1)+∥fk∥(l1+1)/l1]∥u∥l1+1.
Therefore
(2.24)∥u∥l1+1≤(1n)1/l1[m1β(l0-l1)/(l1+1)β1l1(l0+1)/(l1+1)+∥fk∥(l1+1)/l1]1/l1∶=α1,
where α1 is a constant independent of k and λ. By using Lemma 2.2 again, we get
(2.25)∥u∥∞≤T-1/(l1+1)∥u∥l1+1+Tl0/(l0+1)∥u′∥l0+1≤T-1/(l1+1)α1+Tl0/(l0+1)β1∶=α.
Obviously, α is a constant independent of k and λ. Therefore, if u∈Ω1, then by (2.19) we see that
(2.26)∥u∥C2kT1=max{∥u∥∞,∥u′∥∞}≤max{α,β}+supt∈R|f(t)|∶=M~.
Clearly, M~>0 is a constant independent of k and λ; that is, Ω1 is uniformly bounded for all k∈N and λ∈(0,1).
Step  2. From assumptions (A2) and (A3), we see that there must be a constant M1>0 such that -h(M1)+fk¯>0 and -h(-M1)+fk¯<0. Set Ω2={u(t)∈C2kT1:∥u∥C2kT1<M}, where M=max{M1,M~}. We will show that Nu∉ImL, for all u∈∂Ω2∩KerL.
In fact, by assumption (A2), we see thatg(0)=0, and if u∈∂Ω2∩KerL, then u(t)≡M or u≡-M. So
(2.27)QN(u)=-12kT∫-kTkT‍[h(M)-fk(t)]dt=-h(M)+fk¯≥-h(M1)+fk¯>0,∀u(t)≡M,QN(u)=-12kT∫-kTkT‍[h(-M)-fk(t)]dt=-h(-M)+fk¯≤-h(-M1)+fk¯<0,∀u(t)≡-M,
where fk¯=(1/2kT)∫-kTkT‍fk(t)dt. This implies that Nu∉ImL, for all u∈∂Ω2∩KerL.
Step  3. Set J:ImQ→KerL, Jx=x, we will show deg{JQN,Ω2∩KerL,0}≠0.
LetH(x,μ)=μx+(1-μ)JQNx, for all x∈Ω2∩KerL, when x∈∂(Ω2∩KerL), we have x=±M and
(2.28)H(M,μ)=μM+(1-μ)JQN(M)=μM+(1-μ)QN(M)>0,H(-M,μ)=-μM+(1-μ)JQN(-M)=-μM+(1-μ)QN(-M)<0.
So for all μ∈[0,1],  H(∂(Ω2∩KerL),μ)≠0, and then
(2.29)deg{JQN,Ω2∩KerL,0}=deg{H(·,0),Ω2∩KerL,0}=deg{H(·,1),Ω2∩KerL,0}=deg{I,Ω2∩KerL,0}=1.
Therefore, by Lemma 2.5, (1.4) has a 2kT-periodic solution uk∈Ω2¯.Remark 2.8.
Suppose that all the conditions in Lemma2.6 hold. We see that for each k∈N, (1.4) has a 2kT-periodic solution uk∈Ω2. This implies that
(2.30)∥uk∥∞≤M,∥uk′∥∞≤M.
Furthermore, as same as the proof of step  1 in Lemma 2.6 with replacing u(t) by uk(t), we have
(2.31)∥uk∥l1+1≤α1,∥uk′∥l0+1≤β1,
where α1 and β1 are two positive constants independent of k∈N.Lemma 2.9 (see [12]).
Letuk∈C2kT1 be the 2kT-periodic solution for (1.4) and satisfies (2.30) and (2.31) for all k∈N. Then there exists a function u0∈C1(R,R) such that for each interval [c,d]⊂R, there is a subsequence {ukj} of {uk}k∈N with ukj′(t)→u0′(t) uniformly on [c,d].


## 3.  Main Results

Theorem 3.1.
Suppose that assumptions (A1), (A2), and (A3) in Lemma2.6 hold. Then (1.1) has a unique homoclinic solution.Proof.
Since assumptions (A1), (A2), onsisting of Kuratowski operations we used following principles and (A3) in Lemma2.6 hold, by using Lemma 2.6, we see that (1.4) has a 2kT-periodic solution uk(t) satisfying (2.30) and (2.31) for each k∈N. It follows from Lemma 2.9 that there exists a u0∈C1(R,R) such that for each interval [c,d]⊂R, there is a subsequence {ukj} of {uk}k∈N satisfying ukj′(t)→u0′(t) uniformly on [c,d]. Below, we will show that u0(t) is just a unique homoclinic solution to (1.1).
Step  1. We show that u0 is a solution of (1.1).
In view ofukj(t) being a 2kjT-periodic solution to (1.4), we have
(3.1)ukj′′(t)+g(ukj′(t))+h(ukj(t))=fkj(t),fort∈[-kjT,kjT],j∈N.
Take a,b∈R such that a<b, there exists j0∈N such that for all j>j0(3.2)ukj′′(t)+g(ukj′(t))+h(ukj(t))=f(t),fort∈[a,b].
Integrating (3.2) from a to t∈[a,b], we have
(3.3)ukj′(t)-ukj′(a)=∫at‍[-g(ukj′(s))-h(ukj(s))+f(s)]ds,fort∈[a,b].
Since Lemma 2.9 shows that ukj→u0 uniformly on [a,b] and ukj′→u0′ uniformly on [a,b] as j→∞, let j→∞ in (3.3), we get
(3.4)u0′(t)-u0′(a)=∫at‍[-g(u0′(s))-h(u0(s))+f(s)]ds,fort∈[a,b].
In view of a and b are arbitrary, (3.4) shows that u0(t) is a solution of (1.1).
Step  2. We prove that u0(t)→0, as t→±∞.
Obviously, for everyi∈N, there exists ji∈N such that for all j>ji, we have
(3.5)∫-iTiT‍[|ukj(t)|l1+1+|ukj′(t)|l0+1]dt≤∫-kjTkjT‍[|ukj(t)|l1+1+|ukj′(t)|l0+1]dt≤α1l1+1+β1l0+1∶=M2.
It follows that
(3.6)∫-∞+∞‍[|u0(t)|l1+1+|u0′(t)|l0+1]dt=limi→+∞∫-iTiT‍[|u0(t)|l1+1+|u0′(t)|l0+1]dt=limi→+∞limj→+∞∫-iTiT‍[|ukj(t)|l1+1+|ukj′(t)|l0+1]dt≤M2,
and then
(3.7)∫|t|≥r‍[|u0(t)|l1+1+|u0′(t)|l0+1]dt→0,asr→+∞,
which yields
(3.8)∫|t|≥r‍|u0(t)|l1+1dt→0,∫|t|≥r‍|u0′(t)|l0+1dt→0,asr→+∞.
By using Lemma 2.1, as t→±∞,
(3.9)|u0(t)|≤(2a)-1/(l1+1)(∫t-at+a‍|u0(s)|l1+1ds)1/(l1+1)+a·(2a)-1/(l0+1)(∫t-at+a‍|u0′(s)|l0+1ds)1/(l0+1)→0.
So we have u0(t)→0, as t→±∞.
Step  3. We will show that
(3.10)u0′(t)→0,ast→±∞.
From the Remark 2.8 and Lemma 2.9, we have
(3.11)|u0(t)|≤M,|u0′(t)|≤M,forallt∈R,
which together with (1.1) implies that
(3.12)∥u0′′∥∞≤gM+hM+supt∈R|f(t)|,
where gM=max|x|≤M|g(x)| and hM=max|x|≤M|h(x)|. If u0′(t)↛0, as t→±∞, then there exist a ε0∈(0,1/2) and a sequence {tk} such that
(3.13)|t1|<|t2|<|t3|<⋯,|tk|+1<|tk+1|,k∈N,|u0′(tk)|≥2ε0,k∈N.
From this, we have for t∈[tk,tk+ε0/(1+M1)](3.14)|u0′(t)|=|u0′(tk)+∫tkt‍u0′′(s)ds|≥|u0′(tk)|-∫tkt‍|u0′′(s)|ds≥ε0.
It follows that
(3.15)∫-∞+∞‍|u0′(t)|l0+1dt≥∑k=1∞‍∫tktk+ε0/(1+M1)‍|u0′(t)|l0+1dt=∞,
which contradicts (3.6), and so (3.10) holds.
Step  4. Finally, we will prove that (1.1) possesses a unique homoclinic solution. In order to do it, let u(t)=u1(t)-u2(t), where u1(t) and u2(t) are two arbitrary homoclinic solutions of (1.1). Then
(3.16)u(t)→0ast→±∞.
We will show that
(3.17)u(t)≡0.
If (3.17) does not hold, then there must be a t*∈R such that
(3.18)u(t*)>0
or
(3.19)u(t*)<0.
Ifu(t*)>0, then from (3.16), we see that there is a constant X>0 such that t*∈(-X,X) and u(t)<u(t*)/2 for t∈(-∞,X)∪(X,+∞). Let t**∈[-X,X] such that u(t**)=maxt∈[-X,X]u(t), then
(3.20)u(t**)≥u(t*)>0,(3.21)u(t**)≥u(t*)>supt∈(-∞,X)∪(X,+∞)u(t),
that is,
(3.22)u(t**)=maxt∈Ru(t).
So u′(t**)=0 and u′′(t**)≤0, and then from (1.1), we see
(3.23)-[h(u1(t**))-h(u2(t**))]=u1′′(t**)-u2′′(t**)=u′′(t**)≤0.
By using the condition (A3), we have that
(3.24)u(t**)=u1(t**)-u2(t**)≤0,
which contradicts to (3.20). This contradiction implies that (3.18) does not hold. Similarly, we can prove that (3.19) does not hold, either. So u(t)≡0.
As an application, we consider the following example:(3.25)u′′(t)-m(u′(t))3-n(u(t))=et/2e-t+et,
where m,n>0 are constants and, f(t)=et/2/(e-t+et). Corresponding to (1.1), we can chose l0=3 and l1=1 such that assumptions (A2) and (A3) hold. Furthermore, by the direct calculation, we can easily obtain that
(3.26)∫R‍|f(t)|(l1+1)/l1dt=∫-∞+∞‍|f(t)|2dt=π4<∞,∫R‍|f(t)|(l0+1)/l0dt=∫-∞+∞‍|f(t)|4/3dt=32<∞.
This implies that assumption (A1) also holds. So by applying Theorem 3.1, we know that (3.25) possesses a unique homoclinic solution.



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*Source: 615303-2012-12-25.xml*