# Periodic Solutions of Some Impulsive Hamiltonian Systems with Convexity Potentials

Authors: Dezhu Chen; Binxiang Dai
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We study the existence of periodic solutions of some second-order Hamiltonian systems with impulses. We obtain some new existence theorems by variational methods.


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## Body

## 1. Introduction

Consider the following systems:(1.1)u¨(t)=f(t,u(t)),a.e.t∈[0,T],Δu˙(tk)=gk(u(tk-)),k=1,2,…,m,u(T)-u(0)=u˙(T)-u˙(0)=0,
where k∈ℤ,u∈ℝn,Δu˙(tk)=u˙(tk+)-u˙(tk-) with u˙(tk±)=limt→tk±u˙(t),gk(u)=graduGk(u),Gk∈C1(ℝn,ℝn) for each k∈ℤ, there exists an m∈ℤ such that 0=t0<t1<⋯<tm<tm+1=T, and we suppose that f(t,u)=graduF(t,u) satisfies the following assumption. (A)F(t,x) is measurable in t for x∈ℝn and continuously differentiable in x for a.e. t∈[0,T], and there exist a∈C(ℝ+,ℝ+),b∈L1(0,T;ℝ+) such that
(1.2)|F(t,x)|+|f(t,x)|≤a(|x|)b(t),for allx∈ℝn and t∈[0,T].Many solvability conditions for problem (1.1) without impulsive effect are obtained, such as, the coercivity condition, the convexity conditions (see [1–4] and their references), the sublinear nonlinearity conditions, and the superlinear potential conditions. Recently, by using variational methods, many authors studied the existence of solutions of some second-order differential equations with impulses. More precisely, Nieto in [5, 6] considers linear conditions, [7–10] the sublinear conditions, and [11–16] the sublinear conditions and the other conditions. But to the best of our knowledge, except [7] there is no result about convexity conditions with impulsive effects. By using different techniques, we obtain different results from [7].We recall some basic facts which will be used in the proofs of our main results. Let(1.3)HT1={u:[0,T]→ℝnabsolutelycontinuous;u(0)=u(T),u˙(t)∈L2(0,T;ℝn)},
with the inner product
(1.4)〈u,v〉=∫0T‍(u(t),v(t))dt+∫0T‍(u˙(t),v˙(t))dt,∀u,v∈HT1,
where (·,·) denotes the inner product in ℝn. The corresponding norm is defined by
(1.5)∥u∥=(∫0T‍(u(t),u(t))dt+∫0T‍(u˙(t),u˙(t))dt)1/2,∀u∈HT1.The spaceHT1 has some important properties. For u∈HT1, let u-=(1/2T)∫0T‍u(t)dt, and u~=u(t)-u-. Then one has Sobolev's inequality (see Proposition 1.3 in [1]):
(1.6)∥u~∥∞2≤T12∫0T‍|u˙(t)|2dt.Consider the corresponding functionalφ on HT1 given by
(1.7)φ(u)=12∫0T‍|u˙(t)|2dt+∫0T‍F(t,u(t))dt+∑k=1m‍Gk(u(tk)).It follows from assumption(A) and the continuity of gk one has that φ is continuously differentiable and weakly lower semicontinuous on HT1. Moreover, we have
(1.8)〈φ′(u),v〉=∫0T‍(u˙(t),v˙(t))dt+∫0T‍(f(t,u(t)),v(t))dt+∑k=1m‍(gk(u(tk)),v(tk)),
for u,v∈HT1 and φ′ is weakly continuous and the weak solutions of problem (1.1) correspond to the critical points of φ (see [8]).Theorem 1.1 ([2, Theorem  1.1]).
Suppose thatV and W are reflexive Banach spaces, φ∈C1(V×W,R), φ(v,·) is weakly upper semi-continuous for all v∈V, and φ(·,w):V→R is convex for all w∈W and φ′ is weakly continuous. Assume that
(1.9)φ(0,w)→-∞
as ∥w∥→∞ and for every M>0,
(1.10)φ(v,w)→+∞,
as ∥v∥→∞ uniformly for ∥w∥≤M. Then φ has at least one critical point.


## 2. Main Results

Theorem 2.1.
Assume that assumption(A) holds. If further(H1)
F(t,·) is convex for a.e. t∈[0,T], and(H2)
there existη, θ>0 such that Gk(x)≥η|x|+θ, for all x∈ℝn, then (1.1) possesses at least one solution in HT1.Remark 2.2.
(
H
1
) implies there exists a point x- for which
(2.1)∫0T‍∇F(t,x-)dt=0.Proof of Theorem2.1.
It follows Remark2.2, (1.6), and (H2) that
(2.2)φ(u)=12∫0T‍|u˙(t)|2dt+∫0T‍(F(t,u(t))-F(t,x-))dt+∫0T‍F(t,x-)dt+∑k=1m‍Gk(u(tk))=12∫0T‍|u˙(t)|2dt+∫0T‍F(t,x-)dt+∫0T‍(f(t,x-),u(t)-x-)dt+∑k=1m‍Gk(u(tk))≥12∫0T‍|u˙(t)|2dt+∫0T‍F(t,x-)dt+∫0T‍(f(t,x-),u~)dt+∑k=1m‍η|u~+u-|+mθ≥12∫0T‍|u˙(t)|2dt-(∫0T‍|f(t,x-)|dt)∥u~∥∞+mη|u-|-mη∥u~∥∞+mθ≥12∫0T‍|u˙(t)|2dt-C0(∫0T‍|u˙(t)|2dt)1/2+mη|u-|+mθ,
for all u∈HT1 and some positive constant C0. As ∥u∥→∞ if and only if (|u|2+∥u˙∥22)1/2→∞, we have φ(u)→+∞ as ∥u∥→∞. By Theorem 1.1 and Corollary 1.1 in [1], φ has a minimum point in HT1, which is a critical point of φ. Hence, problem (1.1) has at least one weak solution.Theorem 2.3.
Assume that assumption(A) and (H1) hold. If further(H3)
there existη, θ>0 and α∈(0,2) such that Gk(x)≤η|x|α+θ for all x∈ℝn and(H4)
there exist someβ>α and γ>0 such that
(2.3)|x|-β∫0T‍F(t,x)dt≤-γ,for |x|≥M and t∈[0,T], where M is a constant, then (1.1) possesses at least one solution in HT1.Remark 2.4.
We can find that our condition(H4) is very different from condition (vii) in [7] since we prove this by the saddle point theorem substituted for the least action principle.Proof of Theorem2.3.
We proveφ satisfies the (PS) condition at first. Suppose {un} is such an sequence that {φ(un)} is bounded and limn→∞φ′(un)=0. We will prove it has a convergent subsequence. By (H3) and (1.6), we have
(2.4)∑k=1m‍Gk(u(tk))≤∑k=1m‍η|u~(tk)+u-|α+mθ≤4mη(|u~(tk)|α+|u-|α)+mθ≤C1∥u˙∥2α+C2|u-|α+C3,
for some positive constants C1, C2, C3. By Remark 2.2, (1.6), and (2.4), we have
(2.5)φ(un)=12∫0T‍|u˙n(t)|2dt+∫0T‍(F(t,un(t))-F(t,x-))dt+∫0T‍F(t,x-)dt+∑k=1m‍Gk(un(tk))=12∫0T‍|u˙n(t)|2dt+∫0T‍F(t,x-)dt+∫0T‍(f(t,x-),un(t)-x-)dt+∑k=1m‍Gk(un(tk))=12∫0T‍|u˙n(t)|2dt+∫0T‍F(t,x-)dt+∫0T‍(f(t,x-),u~n)dt+∑k=1m‍Gk(un(tk))≥12∫0T‍|u˙n(t)|2dt-(∫0T‍|f(t,x-)|dt)∥u~n∥∞-C1∥u˙n∥2α-C2|u-n|α-C4≥12∫0T‍|u˙n(t)|2dt-C5(∫0T‍|u˙n(t)|2dt)1/2-C1∥u˙n∥2α-C2|u-n|α-C4,
for some positive constants C4, C5, which implies that
(2.6)C|u-n|α/2≥(∫0T‍|u˙n(t)|2dt)1/2-C6,
for some positive constants C, C6. By (1.6), the above inequality implies that
(2.7)∥u~n∥∞≤C7(|u-n|α/2+1),
for the positive constant C7. The one has
(2.8)|un(t)|≥|u-n|-|u~n|≥|u-n|-∥u~n∥∞≥|u-n|-C7(|u-n|α/2+1),∀t∈[0,T].
If {|u-n|} is unbounded, we may assume that, going to a subsequence if necessary,
(2.9)|u-n|→∞asn→∞.
By (2.8) and (2.9), we have
(2.10)|un(t)|≥12|u-n|,
for all large n and every t∈[0,T]. By (2.10) and (H4), one has |un(t)|≥M for all large n. It follows from (H4), (2.4), (2.6), (2.7), and above inequality that
(2.11)φ(un)≤(C|u-n|α/2+C6)2-∫0T‍γ|un(t)|βdt+C2∥u~∥∞α+C2|u-|α+C3≤(C|u-n|α/2+C6)2-2-β|u-n|βTγ+C8(|u-n|α/2+1)α+C2|u-|α+C3,
for large n and the positive constant C8, which contradicts the boundedness of φ(un) since β>α. Hence (|u-n|) is bounded. Furthermore, (un) is bounded by (2.6). A similar calculation to Lemma 3.1 in [9] shows that φ satisfies the (PS) condition. We now prove that φ satisfies the other conditions of the saddle point theorem. Assume that H~T1={u∈HT1:u-=0}, then HT1=H~T1⊕ℝn. From above calculation, one has
(2.12)φ(u)≥12∫0T‍|u˙(t)|2dt-C5(∫0T‍|u˙(t)|2dt)1/2-C1∥u˙∥2α-C4,
for all u∈H~T1, which implies that
(2.13)φ(u)→+∞,
as ∥u∥→∞ in H~T1. Moreover, by (H3) and (H4) we have
(2.14)φ(x)=∫0T‍F(t,x)dt+∑k=1m‍Gk(x)≤-Tγ|x|β+mη|x|α+mθ,
for |x|>M, which implies that
(2.15)φ(x)→-∞,
as |x|→∞ in ℝn since β>α. Now Theorem 2.3 is proved by (2.13), (2.15), and the saddle point theorem.Theorem 2.5.
Assume that assumption(A) holds. Suppose that F(t,·),Gk(x) are concave and satisfy (H5)
Gk(x)≤-η|x|+θ for some positive constant η,θ>0, then (1.1) possesses at least one solution in HT1.Proof of Theorem2.5.
Consider the corresponding functionalφ on ℝn×H~T1 given by
(2.16)φ(u)=-12∫0T‍|u˙(t)|2dt-∫0T‍F(t,u(t))dt-∑k=1m‍Gk(u(tk)),
which is continuously differentiable, bounded, and weakly upper semi-continuous on HT1. Similar to the proof of Lemma 3.1 in [2], one has that φ(x+w) is convex in x∈ℝn for every w∈H~T1. By the condition, we have -Gk(x+w)≥-2Gk((1/2)x)+Gk(-w). Similar to the proof of Theorem 3.1, we have
(2.17)φ(x+w)=-12∫0T‍|w˙|2dt-∫0T‍F(t,x+w)dt-∑k=1m‍Gk(x+w)≥-12∫0T‍|w˙|2dt-(∫0T‍|f(t,x-)|dt)∥w∥∞-∑k=1m‍Gk(x+w)+C9≥-12∫0T‍|w˙|2dt-C0(∫0T‍|w˙|2dt)1/2-2Gk(12x)+Gk(-w)+C9,
which means φ(x+w)→+∞ as |x|→∞, uniformly for w∈H~T1 with ∥w∥≤M by (H5) and (1.6). On the other hand,
(2.18)φ(w)=-12∫0T‍|w˙|2dt-∫0T‍F(t,w)dt-∑k=1m‍Gk(w)≤-12∫0T‍|w˙|2dt+C0(∫0T‍|w˙|2dt)1/2+mη∥w∥∞+C9,
which implies that φ(w)→-∞ as ∥w∥→∞∈H~T1 by (H5) and (1.6). We complete our proof by Theorem 1.1.



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