New periodic solutions with a prescribed energy for a class of Hamiltonian systems

Abstract

We consider a class of second order Hamiltonian systems with a $C^{2}$ potential function. The existence of new periodic solutions with a prescribed energy is established by the use of constrained variational methods.

1 Introduction

In this paper, we examine the existence of periodic solutions for second order Hamiltonian systems

$$\begin{aligned}& \ddot{q}+V^{\prime}(q)=0, \end{aligned}$$

(1.1)

$$\begin{aligned}& \frac{1}{2}\vert \dot{q}\vert ^{2}+V(q)=h, \end{aligned}$$

(1.2)

with a fixed energy. The first major result in this direction we would like to highlight can be derived from the work of Benci [1], Gluck-Ziller [2], and Hayashi [3], which is based on the earlier work of Seifert [4] in 1948 and following the highly influential papers of Rabinowitz [5, 6] in 1978 and 1979. Utilizing the Jacobi metric and a very involved interplay between geodesic methods and algebraic topology, the following general theorem is established.

Theorem 1.1

Suppose $V\in C^{1}(\mathbb{R}^{n},\mathbb{R})$. If the potential well

$$ \bigl\{ x\in\mathbb{R}^{n}:V(x)\leq h\bigr\} $$

is bounded and non-empty, then the system (1.1)-(1.2) has a periodic solution with energy h. Furthermore, if

$$ V^{\prime}(x)\neq0, \quad \forall x\in\bigl\{ x\in\mathbb{R}^{n} :V(x)=h \bigr\} , $$

then the system (1.1)-(1.2) has a non-constant periodic solution with energy h.

For the existence of multiple periodic solutions for (1.1)-(1.2) with compact energy surfaces, we can refer the reader to Groessen [7] and Long [8] and the references therein.

For the weakly attractive potential V defined on an open subset Ω of $\mathbb{R}^{n}$, Ambrosetti and Coti Zelati [9] (Theorem 16.7) proved the following.

Theorem 1.2

Suppose $V\in C^{2}(\Omega,\mathbb{R})$ satisfies

$(V10)$ :

$3\langle V^{\prime}(x),x\rangle+\langle V''(x)x,x\rangle\neq0$, $\forall x\in\Omega$;

$(V11)$ :

$\langle V^{\prime}(x),x\rangle>0$, $\forall x\in \Omega$;

$(V12)$ :

$\exists\alpha\in(0,2)$, such that $\langle V^{\prime}(x), x\rangle\geq-\alpha V(x)$, $\forall x\in\Omega$;

$(V13)$ :

$\exists\beta\in(0,2)$ and $r>0$ such that $\langle V^{\prime}(x), x\rangle\leq-\beta V(x)$, $\forall0<\vert x\vert <r$;

$(V14)$ :

$G_{\infty}\geq0$; where $G_{\infty}=\lim_{\vert x\vert \rightarrow\infty }\inf G(x)$, $G(x)=[V(x)+\frac{1}{2}\langle V^{\prime}(x),x\rangle]$.

Then $\forall h<0$, the system (1.1)-(1.2) (referred to as $(P_{h})$) has at least one non-constant weak periodic solution with the given energy h.

Using a simpler constrained variational minimizing method, we obtain the following result.

Theorem 1.3

Suppose $V\in C^{2}(\mathbb{R}^{n},\mathbb{R})$ and $h \in\mathbb{R}$ satisfy

$(V_{1})$ :

$V(-q)=V(q)$;

$(V_{2})$ :

$\langle V^{\prime}(q),q\rangle>0$, $\forall q\neq0$;

$(V_{3})$ :

$3\langle V^{\prime}(q),q\rangle+\langle V''(q)q,q\rangle>0$, $\forall q\neq0$;

$(V_{4})$ :

$\exists\mu_{1}>0$, $\mu_{2}\geq0$, such that $\langle V^{\prime}(q), q\rangle\geq\mu_{1} V(q)-\mu_{2}$;

$(V_{5})$ :

$\lim_{\vert q\vert \rightarrow\infty }\sup[V(q)+\frac{1}{2}\langle V^{\prime}(q),q\rangle]\leq A$;

$(V_{6})$ :

$\frac{\mu_{2}}{\mu_{1}}< h< A$.

Then the system (1.1)-(1.2) has at least one non-constant periodic solution with the given energy h.

Remark 1.4

Comparing Theorem 16.7 of Ambrosetti and Coti Zelati [9] with our Theorem 1.3, we notice that our condition $(V_{2})$ corresponds to their $(V11)$, our condition $(V_{3})$ corresponds to their $(V10)$, our condition $(V_{4})$ corresponds to their $(V12)$ and $(V13)$, our conditions $(V_{5})$ and $(V_{6})$ correspond to their $(V14)$. Since the potential in Theorem 16.7 of Ambrosetti and Coti Zelati has a singularity, but the potential in Theorem 1.3 has no singularity, the two theorems are essentially different.

Remark 1.5

Take for $V(x)$ the following $C^{\infty}$ function:

$$\begin{aligned}& V(x)=e^{\frac{-1}{\vert x\vert }}, \quad \forall x\neq0; \\& V(0)=0. \end{aligned}$$

Then $V(x)$ satisfies ($V_{1}$)-($V_{5}$) in Theorem 1.3 if we take $\mu_{1}=\mu_{2}>0$ and $A=1$, but $(V_{6})$ does not hold.

Proof of Theorem 1.3

We verify ($V_{1}$)-($V_{5}$) by calculation:

(1) It is obvious for $(V_{1})$.

(2) For $(V_{2})$ and $(V_{3})$, we notice that

$$\begin{aligned}& \bigl\langle V^{\prime}(x),x\bigr\rangle =\frac{1}{\vert x\vert }e^{\frac{-1}{\vert x\vert }}>0,\quad \forall x\neq0, \\& \bigl\langle V''(x)x,x\bigr\rangle =e^{\frac{-1}{\vert x\vert }} \biggl(\frac{-2}{\vert x\vert }+\frac{1}{\vert x\vert ^{2}}\biggr), \\& 3\bigl\langle V^{\prime}(x),x\bigr\rangle +\bigl\langle V''(x)x,x \bigr\rangle =e^{\frac {-1}{\vert x\vert }}\biggl(\frac{1}{\vert x\vert }+\frac{1}{\vert x\vert ^{2}}\biggr) >0,\quad \forall x\neq0. \end{aligned}$$

(3) For $(V_{4})$, we set

$$ w(x)=\biggl(\frac{1}{\vert x\vert }-\mu_{1}\biggr)e^{\frac {-1}{\vert x\vert }}; \quad x\neq0, w(0)=0. $$

We will prove $w(x)>-\mu_{1}$; in fact,

$$ w^{\prime}(x)=\biggl[\frac{1}{\vert x\vert }-(1+\mu_{1})\biggr] \frac {x}{\vert x\vert ^{3}}e^{\frac{-1}{\vert x\vert }}; \quad x\neq0, w^{\prime}(0)=0. $$

From $w^{\prime}(x)=0$, we have $x=-\frac{1}{1+\mu_{1}}$ or 0 or $\frac{1}{1+\mu_{1}}$.

It is easy to see that $w(x)$ is strictly increasing on $(-\infty ,-\frac{1}{1+\mu_{1}}]$ and $[0,\frac{1}{1+\mu_{1}}]$, but strictly decreasing on $[\frac {-1}{1+\mu_{1}},0]$ and $[\frac{1}{1+\mu_{1}},+\infty)$. We notice that

$$\lim_{\vert x\vert \rightarrow+\infty}w(x)=-\mu_{1} $$

and

$$w(0)=0. $$

So

$$w(x)>-\mu_{1}. $$

When we take $\mu_{2}=\mu_{1}>0$, $(V_{4})$ holds.

(4) For $(V_{5})$, we have

$$\begin{aligned}& V(x)+\frac{1}{2}\bigl\langle V^{\prime}(x),x\bigr\rangle =e^{\frac{-1}{\vert x\vert }}\biggl(1+\frac{1}{2}\frac {1}{\vert x\vert }\biggr) < 1,\quad \forall x\neq0; \\& V(0)+\frac{1}{2}\bigl\langle V^{\prime}(0),0\bigr\rangle =0. \end{aligned}$$

 □

Corollary 1.6

Given $a>0$, $n\in\mathbb{N}$, define $V(x)=a\vert x\vert ^{2n}+e^{\frac{-1}{\vert x\vert }}$, $x\neq0$; $V(0)=0$. Then, for $h>1$, the system (1.1)-(1.2) has at least one non-constant periodic solution with the given energy h.

Remark 1.7

The potential $V(x)=e^{\frac{-1}{\vert x\vert }}$, $\forall x\neq0$; $V(0)=0$ in Remark 1.5 is noteworthy since the potential function is non-convex and bounded which satisfies neither of the conditions of Theorems 1.1, Offin’s geometrical conditions [10], nor Berg-Pasquotto-Vandervorst’s complex topological assumptions [11]. For this potential, the potential well $\{x\in\mathbb{R}^{n}:V(x)\leq h\}$ is a bounded set if $h<1$, but for $h\geq1$ it is $\mathbb{R}^{n}$ - an unbounded set. We also notice that the symmetrical condition on the potential simplified our Theorem 1.2 and its proof. It would be interesting to obtain non-constant periodic solutions when the symmetrical condition is deleted.

2 A few lemmas

Let

$$ H^{1}=W^{1,2}\bigl(\mathbb{R}_{\mathrm{per}}, \mathbb{R}^{n}\bigr)=\bigl\{ u:\mathbb{R}\rightarrow \mathbb{R}^{n},u(t+1)=u(t), u\in L^{2}[0,1],\dot{u}\in L^{2}[0,1]\bigr\} $$

denotes the periodic functional space of period 1. Then the standard $H^{1}$ norm is

$$ \Vert u\Vert =\Vert u\Vert _{H^{1}}= \biggl( \int^{1}_{0}\vert \dot{u}\vert ^{2} \,dt \biggr)^{1/2}+ \biggl( \int^{1}_{0}\vert u\vert ^{2}\,dt \biggr)^{1/2}. $$

Lemma 2.1

[12]

For $u\in H^{1}$, define

$$\begin{aligned}& g(u)= \int_{0}^{1}\biggl[V(u)+\frac{1}{2}\bigl\langle V^{\prime}(u),u\bigr\rangle \biggr]\,dt, \\& M=\bigl\{ u\in H^{1}:g(u)=h\bigr\} . \end{aligned}$$

For $u,v\in H^{1}$ and $s \in\mathbb{R}$, let

$$\phi(s)=g(u+sv). $$

Then

$$\phi^{\prime}(0)=\bigl\langle g'(u),v\bigr\rangle = \frac{1}{2} \int_{0}^{1}\bigl\{ 3 \bigl\langle V^{\prime}(u),v\bigr\rangle +\bigl\langle V''(u)v,u \bigr\rangle \bigr\} \,dt $$

and

$$\bigl\langle g'(u),u\bigr\rangle =\frac{1}{2} \int_{0}^{1}\bigl\{ 3 \bigl\langle V^{\prime }(u),u\bigr\rangle +\bigl\langle V''(u)u,u \bigr\rangle \bigr\} \,dt; $$

therefore, if $(V_{3})$ holds, then on M, $g'(u)\neq0$, which implies M is a $C^{1}$ manifold with codimension 1 in $H^{1}$.

Let

$$ f(u)=\frac{1}{4} \int^{1}_{0}\vert \dot{u}\vert ^{2} \,dt \int^{1}_{0}\bigl\langle V^{\prime}(u),u\bigr\rangle \,dt $$

(2.1)

and $\widetilde{u}\in M$ such that $f^{\prime}(\widetilde{u})=0$ and $f(\widetilde{u})>0$. Set

$$ \frac{1}{T^{2}}=\frac{\int^{1}_{0}\langle V^{\prime}(\widetilde {u}),\widetilde{u}\rangle\,dt}{\int^{1}_{0}\vert \dot{\widetilde {u}}\vert ^{2}\,dt}. $$

If $(V_{2})$ holds, then $\widetilde{q}(t)=\widetilde{u}(t/T)$ is a non-constant T-periodic solution for (1.1)-(1.2).

When the potential is even, then by Palais’ symmetrical principle [13] and Lemma 2.1 we have the following.

Lemma 2.2

[12]

Let

$$ F=\biggl\{ u\in M:u\biggl(t+\frac{1}{2}\biggr)=-u(t)\biggr\} $$

(2.2)

and suppose ($V_{1}$)-($V_{3}$) hold. If $\widetilde{u}\in F$ is such that $f^{\prime}(\widetilde{u})=0$ and $f(\widetilde{u})>0$, then $\widetilde{q}(t)=\widetilde{u}(\frac{t}{T})$ is a non-constant T-periodic solution for (1.1)-(1.2); in addition, we have

$$ \forall u\in F,\quad \int_{0}^{1}u(t)\,dt=0. $$

Wirtinger’s inequality [14] implies

$$\int^{1}_{0}\vert \dot{u}\vert ^{2} \,dt\geq(2\pi)^{2} \int^{1}_{0}\vert u\vert ^{2}, $$

from which it follows that $(\int^{1}_{0}\vert \dot{u}\vert ^{2}\,dt )^{1/2}$ is an equivalent norm for the space $H^{1}$.

Lemma 2.3

Let X be a Banach space and $F\subset X$ a weakly closed subset. Suppose Φ defined on F is Gateaux-differentiable, weakly lower semi-continuous and bounded from below on F. Suppose further that Φ satisfies the following $(\mathit{WPS})_{\inf\Phi ,F}$ condition:

• If $\{x_{n}\}\subset F$ such that $\Phi(x_{n}) \rightarrow c$ and $\Vert \Phi'(x_{n})\Vert \rightarrow0$, then $\{x_{n}\}$ has a weakly convergent subsequence.

Then Φ attains its infimum on F.

Proof

By Ekeland’s variational principle [15, 16], we know that there is a sequence $\{x_{n}\}\subset F$ satisfying

$$\Phi(x_{n})\rightarrow\inf\Phi\quad \hbox{and}\quad \bigl\Vert \Phi'(x_{n})\bigr\Vert \rightarrow0. $$

Since Φ satisfies the $(\mathit{WPS})_{\inf\Phi,F}$ condition, $\{x_{n}\}$ has a weakly convergent subsequence which as a weak limit x. Because $F\subset X$ is a weakly closed subset, we have $x\in F$. Finally, by the weakly lower semi-continuous assumption on Φ, we conclude that Φ attains its infimum on F. □

3 The proof of Theorem 1.3

We prove Theorem 1.3 by the following sequence of lemmas. In the following, f and F are defined as in (2.1) and (2.2), respectively.

Lemma 3.1

If ($V_{1}$)-($V_{6}$) hold, then, for any given $c>0$, f satisfies the $(\mathit{PS})_{c,F}$ condition; that is, if $\{u_{n}\}\subset F$ satisfies

$$\begin{aligned} f(u_{n})\rightarrow c>0 \quad \textit{and}\quad f|_{F}^{\prime}(u_{n}) \rightarrow0, \end{aligned}$$

(3.1)

then $\{u_{n}\}$ has a strongly convergent subsequence.

Proof

We first prove that under our assumptions the constrained set $F\neq\emptyset$. For any given $u\in H^{1}$ satisfying $u(t)\neq0$, $\forall t\in[0,1]$ and for $a>0$, let

$$\begin{aligned} g_{u}(a)=g(au)= \int^{1}_{0}\biggl[V(au)+\frac{1}{2}\bigl\langle V^{\prime }(au),au\bigr\rangle \biggr] \,dt. \end{aligned}$$

(3.2)

By the assumption $(V_{3})$, we have

$$\begin{aligned} \frac{d}{da}g_{u}(a)>0 \end{aligned}$$

(3.3)

and so $g_{u}$ is strictly increasing. Since $V\in C^{2}$, we know that, for any given $a>0$,

$$\biggl[V\bigl(au(t)\bigr)+\frac{1}{2}\bigl\langle V^{\prime} \bigl(au(t)\bigr),au(t)\bigr\rangle \biggr] $$

is uniformly continuous on $[0,1]$.

Hence by $(V_{5})$, we have

$$\begin{aligned} \lim_{a\rightarrow+\infty}g_{u}(a)\leq \int^{1}_{0}\lim_{a\rightarrow +\infty}\sup \biggl[V(au)+\frac{1}{2}\bigl\langle V^{\prime}(au),au\bigr\rangle \biggr] \,dt\leq A. \end{aligned}$$

(3.4)

By $(V_{4})$, we notice that

$$\begin{aligned} g_{u}(0)=V(0)\leq\frac{\mu_{2}}{\mu_{1}}. \end{aligned}$$

(3.5)

Since $\frac{\mu_{2}}{\mu_{1}}< h< A$, we see that the equation $g_{u}(a)=h$ has a unique solution $a(u)$ with ${a(u)u\in M}$.

By $f(u_{n})\rightarrow c$, we have

$$\begin{aligned} \frac{1}{4} \int^{1}_{0}\bigl\vert \dot{u_{n}}(t) \bigr\vert ^{2}\,dt\cdot \int^{1}_{0}\bigl\langle V^{\prime}(u_{n}),u_{n} \bigr\rangle \,dt\rightarrow c, \end{aligned}$$

(3.6)

and by $(V_{4})$ we see that

$$\begin{aligned} h= \int^{1}_{0}\biggl[V(u_{n})+ \frac{1}{2}\bigl\langle V^{\prime}(u_{n}),u_{n} \bigr\rangle \biggr]\,dt\leq\biggl(\frac{1}{\mu_{1}}+\frac{1}{2}\biggr) \int_{0}^{1}\bigl\langle V^{\prime }(u_{n}),u_{n} \bigr\rangle \,dt+\frac{\mu_{2}}{\mu_{1}}. \end{aligned}$$

(3.7)

By (3.6) and (3.7), we have

$$\begin{aligned} \int_{0}^{1}\bigl\langle V^{\prime}(u_{n}),u_{n} \bigr\rangle \,dt\geq\frac{h-\frac{\mu_{2}}{\mu_{1}}}{\frac{1}{2}+\frac {1}{\mu_{1}}}. \end{aligned}$$

(3.8)

Condition $(V_{6})$ provides $h>\frac{\mu_{2}}{\mu_{1}}$. Then (3.6) and (3.8) imply $\int^{1}_{0}\vert \dot{u_{n}}(t)\vert ^{2}\,dt$ is bounded and $\Vert u_{n}\Vert =\Vert \dot{u}_{n}\Vert _{L^{2}}$ is bounded.

We know that $H^{1}$ is a reflexive Banach space, so $\{u_{n}\}$ has a weakly convergent subsequence; furthermore, by the embedding theorem the weakly convergent subsequence also uniformly converges to some $u\in H^{1}$. The standard argument can show that $\{u_{n}\}$ has a subsequence which converges under the $H^{1}$ norm. We omit the details of this standard demonstration. □

Lemma 3.2

$f(u)$ is weakly lower semi-continuous on F.

Proof

For any $u_{n}\subset F$ with $u_{n}\rightharpoonup u$, by Sobolev’s embedding theorem we have the uniform convergence

$$ \bigl\vert u_{n}(t)-u(t)\bigr\vert _{\infty}\rightarrow0. $$

Since $V\in C^{1}(\mathbb{R}^{n},\mathbb{R})$, we have

$$ \bigl\vert V\bigl(u_{n}(t)\bigr)-V\bigl(u(t)\bigr)\bigr\vert _{\infty}\rightarrow0. $$

By the weakly lower semi-continuity of the norm, we see that

$$ \liminf\biggl[ \int^{1}_{0}\vert \dot{u}_{n}\vert ^{2}\,dt\biggr]^{\frac {1}{2}}\geq\biggl( \int^{1}_{0}\vert \dot{u}\vert ^{2} \,dt\biggr)^{\frac{1}{2}}, $$

and so

$$ \liminf\biggl( \int^{1}_{0}\vert \dot{u}_{n}\vert ^{2}\,dt\biggr)\geq \int^{1}_{0}\vert \dot{u}\vert ^{2} \,dt. $$

Then

$$\begin{aligned} \liminf f(u_{n})&=\liminf\frac{1}{4} \int^{1}_{0}\vert \dot{u_{n}}\vert ^{2}\,dt \int^{1}_{0}\bigl\langle V^{\prime}(u_{n}),u_{n} \bigr\rangle \,dt \\ &\geq\frac{1}{4} \int^{1}_{0}\vert \dot{u}\vert ^{2} \,dt \int^{1}_{0}\bigl\langle V^{\prime}(u),u\bigr\rangle \,dt=f(u). \end{aligned}$$

 □

Lemma 3.3

F is a weakly closed subset in $H^{1}$.

Proof

This follows easily from Sobolev’s embedding theorem and $V\in C^{1}(\mathbb{R}^{n},\mathbb{R})$. □

Lemma 3.4

The functional $f(u)$ has a positive lower bound on F.

Proof

By the definitions of $f(u)$, F, and the assumption $(V_{2})$, we have

$$ f(u)=\frac{1}{4} \int^{1}_{0}\vert \dot{u}\vert ^{2} \,dt \int^{1}_{0}\bigl\langle V^{\prime}(u),u\bigr\rangle \,dt\geq0,\quad \forall u\in F. $$

We claim further that

$$\inf f(u)>0 ; $$

otherwise, $(V_{2})$ implies $u(t)=\mathit{const}$, and by the symmetrical property $u(t+1/2)=-u(t)$ we have $u(t)=0$, $\forall t\in\mathbb{R}$. But assumptions $(V_{4})$ and $(V_{6})$ imply

$$ V(0)\leq\frac{\mu_{2}}{\mu_{1}}< h, $$

which contradicts the definition of F since $V(0)=h$ if we have $0\in F$. Now by Lemmas 3.1-3.4 and Lemma 2.3, we see that $f(u)$ attains the infimum on F and we know that the minimizer is non-constant. □