Now, we are going to seek a function u in E such that
We consider the associated functional of (3.1),
\begin{array}{rl}I(u)& =\frac{1}{2}{\int}_{\mathrm{\Omega}}[{{u}_{t}}^{2}+{{u}_{x}}^{2}]\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt{\int}_{\mathrm{\Omega}}G(x,t,u)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ =\frac{1}{2}({\parallel {P}_{+}u\parallel}^{2}{\parallel {P}_{}u\parallel}^{2}){\int}_{\mathrm{\Omega}}G(x,t,u)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt,\end{array}
(3.2)
where
G(x,t,u)={\int}_{0}^{u}g(x,t,s)\phantom{\rule{0.2em}{0ex}}ds.
By (g1), I is well defined.
We recall the critical point theorem for the indefinite functional (cf. [6]).
Let
{B}_{r}=\{u\in H\parallel u\parallel \le r\},\phantom{\rule{2em}{0ex}}\partial {B}_{r}=\{u\in H\parallel u\parallel =r\}.
Theorem 3.1 (Critical point theorem for the indefinite functional)
Let E be a real Hilbert space with E={E}_{1}\oplus {E}_{2} and {E}_{2}={E}_{1}^{\perp}. Suppose that I\in {C}^{1}(E,R) satisfies (PS), and that

(I1)
I(u)=\frac{1}{2}(Lu,u)+bu, where Lu={L}_{1}{P}_{1}u+{L}_{2}{P}_{2}u and {L}_{i}:{E}_{i}\to {E}_{i} is bounded and selfadjoint, i=1,2,

(I2)
{b}^{\prime} is compact, and

(I3)
there exist a subspace \tilde{E}\subset E and sets S\subset E, Q\subset \tilde{E} and constants \alpha >\omega such that

(i)
S\subset {E}_{1} and I{}_{S}\ge \alpha,

(ii)
Q is bounded and I{}_{\partial Q}\le \omega,

(iii)
S and ∂Q link.
Then I possesses a critical value c\ge \alpha.
The eigenvalues of {A}_{1} are {\lambda}_{jk}={j}^{2}{(\frac{2\pi k}{T})}^{2}, where j is odd and k is even, j=1,2,\dots and k=0,1,2,\dots . Thus {A}_{1}^{1} is a compact operator.
It is convenient for the following to rearrange the eigenvalues {\lambda}_{jk}, where j is odd and k is even, by increasing magnitude: from now on we denote by {({\eta}_{i}^{})}_{i\ge 1}={\lambda}_{jk}<0 the sequence of negative eigenvalues, by {({\eta}_{i}^{+})}_{i\ge 1}={\lambda}_{jk}>0 the sequence of positive ones, so that
\cdots \le {\eta}_{i}^{}\le \cdots \le {\eta}_{2}^{}\le {\eta}_{1}^{}<{\eta}_{1}^{+}\le {\eta}_{2}^{+}\le \cdots \le {\eta}_{i}^{+}\le \cdots .
(3.3)
We note that each eigenvalue has a finite multiplicity and that {\eta}_{i}^{}\to \mathrm{\infty}, {\eta}_{i}^{+}\to +\mathrm{\infty} as i\to \mathrm{\infty}.
We shall show that the functional I satisfies the geometrical assumptions of the critical point theorem for the indefinite functional.
By the following lemma, I(u)\in {C}^{1}(E,R) and the solutions of (3.1) coincide with the nonzero critical points of I(u).
Lemma 3.1 Assume that g satisfies (g1)(g3). Then I(u) is continuous and Fréchet differentiable in E with the Fréchet derivative
\begin{array}{rl}{I}^{\prime}(u)h& ={\int}_{\mathrm{\Omega}}[{u}_{t}\cdot {h}_{t}+{u}_{x}\cdot {h}_{x}g(x,t,u)h]\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ =({P}_{+}u,{P}_{+}h)({P}_{}u,{P}_{}h){\int}_{\mathrm{\Omega}}g(x,t,u)h\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\end{array}
for all h\in E. Moreover, if we set
F(u)={\int}_{\mathrm{\Omega}}G(x,t,u)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt,
then {F}^{\prime}(u) is continuous with respect to weak convergence, {F}^{\prime}(u) is compact, and
{F}^{\prime}(u)h={\int}_{\mathrm{\Omega}}g(x,t,u)h\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.25em}{0ex}}h\in E.
This implies that I\in {C}^{1}(E,R) and F(u) is weakly continuous.
For the proof of Lemma 3.1, refer to [6].
Lemma 3.2 Assume that g satisfies (g1)(g3). The problem
\begin{array}{r}{u}_{tt}{u}_{xx}\phantom{\rule{1em}{0ex}}\mathit{\text{in}}\phantom{\rule{0.25em}{0ex}}E,\\ u(0,t)=u(\pi ,t)=0,\\ u(x,t+T)=u(x,t),\\ {u}_{t}(x,0)={u}_{t}(x,T)\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in [0,\pi ]\end{array}
(3.4)
has only a trivial solution.
Proof Let Au={u}_{tt}{u}_{xx}. Since H\cap N(A)=\{0\}, E\cap N(A)=\{0\}. Thus (3.4) has only a trivial solution. □
Lemma 3.3 Assume that g satisfies (g1)(g3). Then I(u) satisfies the PalaisSmale condition: If for a sequence ({u}_{m}), I({u}_{m}) is bounded from above and {I}^{\prime}({u}_{m})\to 0 as m\to \mathrm{\infty}, then ({u}_{m}) has a convergent subsequence.
Proof Let ({u}_{m}) be a sequence with I({u}_{m})\le M and {I}^{\prime}({u}_{m})\to 0 as m\to \mathrm{\infty}. It suffices to show that ({u}_{m}) is bounded. By contradiction, we suppose that \parallel {u}_{m}\parallel \to \mathrm{\infty} as m\to \mathrm{\infty}. Let {w}_{m}=\frac{{u}_{m}}{\parallel {u}_{m}\parallel}. Then \parallel {w}_{m}=1\parallel and {w}_{m} converges weakly to an element, say w. Then we have
\frac{M}{{\parallel {u}_{m}\parallel}^{2}}\le \frac{I({u}_{m})}{{\parallel {u}_{m}\parallel}^{2}}=\frac{1}{2}{\int}_{\mathrm{\Omega}}[{\left{({w}_{m})}_{t}\right}^{2}+{\left{({w}_{m})}_{xx}\right}^{2}]{\int}_{\mathrm{\Omega}}\frac{G(x,t,{u}_{m})}{{\parallel {u}_{m}\parallel}^{2}}.
(3.5)
Since G(x,t,u) is bounded, letting m∞ in (3.5), we have
0\le lim{P}_{+}{w}_{m}lim{P}_{}{w}_{m}={\int}_{\mathrm{\Omega}}[{{w}_{t}}^{2}+{{w}_{xx}}^{2}].
(3.6)
On the other hand, we have
0\u27f5\frac{{I}^{\prime}({u}_{m})}{{\parallel {u}_{m}\parallel}^{2}}={\int}_{\mathrm{\Omega}}[{\left{({w}_{m})}_{t}\right}^{2}+{\left{({w}_{m})}_{xx}\right}^{2}]{\int}_{\mathrm{\Omega}}\frac{g(x,t,{u}_{m})}{{\parallel {u}_{m}\parallel}^{2}}.
(3.7)
Letting m\to \mathrm{\infty} in (3.7), we have
0=\underset{m\to \mathrm{\infty}}{lim}{\parallel {P}_{+}{w}_{m}\parallel}^{2}\underset{m\to \mathrm{\infty}}{lim}{\parallel {P}_{}{w}_{m}\parallel}^{2}={\int}_{\mathrm{\Omega}}[{{w}_{t}}^{2}+{{w}_{xx}}^{2}]={\parallel {P}_{+}w\parallel}^{2}{\parallel {P}_{}w\parallel}^{2}.
(3.8)
By (3.6) and (3.8), {lim}_{m\to \mathrm{\infty}}{\parallel {P}_{+}{w}_{m}\parallel}^{2}={\parallel {P}_{+}w\parallel}^{2} and {lim}_{m\to \mathrm{\infty}}{\parallel {P}_{}{w}_{m}\parallel}^{2}={\parallel {P}_{}w\parallel}^{2}. Thus {lim}_{m\to \mathrm{\infty}}\parallel {w}_{m}\parallel =\parallel w\parallel, so {w}_{m} converges strongly to w and by (3.8), w is a weak solution of the problem
{w}_{tt}{w}_{xx}=0\phantom{\rule{1em}{0ex}}\text{in}E.
By Lemma 3.2, w=0, which is absurd for the fact that \parallel w\parallel =1. Thus ({u}_{m}) is bounded. □
Let
Q=({\overline{B}}_{R}\cap {E}_{})\oplus \{ree\in \partial {B}_{1}\cap {E}_{+}\phantom{\rule{0.25em}{0ex}}0<r<R\}.
Lemma 3.4 Assume that g satisfies the conditions (g1)(g3). There exist a subspace \tilde{E}\subset E and sets \partial {B}_{\rho}\subset E, Q\subset \tilde{E} and constants \rho >0 and \alpha >0 such that

(i)
\partial {B}_{\rho}\subset {E}_{+} and I{}_{\partial {B}_{\rho}}\ge \alpha,

(ii)
there are e\in \partial {B}_{1}\cap {E}_{+} and R>\rho such that if Q=({\overline{B}}_{R}\cap {E}_{})\oplus \{re0<r<R\}, then I{}_{\partial Q}\le 0,

(iii)
there exists {u}_{0}\in E\mathrm{\setminus}Q such that \parallel {u}_{0}\parallel >R and I({u}_{0})\le 0,

(iv)
\partial {B}_{\rho} and ∂Q link.
Proof (i) Let u\in {E}_{+}. Since G(x,t,u) is bounded, there exists a constant C>0 such that
\begin{array}{rcl}I(u)& =& \frac{1}{2}{\parallel {P}_{+}u\parallel}^{2}\frac{1}{2}{\parallel {P}_{}u\parallel}^{2}{\int}_{\mathrm{\Omega}}G(x,t,u)\\ \ge & \frac{1}{2}{\parallel {P}_{+}u\parallel}^{2}C\end{array}
for C>0. Then there exist a constant \rho >0 and a constant \alpha >0 such that if u\in \partial {B}_{\rho}\cap {E}_{+}, then I(u)\ge \alpha.
(ii) Let us choose an element e\in {B}_{1}\cap {E}_{+}. Let u\in ({\overline{B}}_{r}\cap {E}_{})\oplus \{re0<r\}. Then u=v+w, v\in {B}_{r}\cap {E}_{}, w=re. We note that
\text{if}v\in {B}_{r}\cap {E}_{},\text{then}{\int}_{\mathrm{\Omega}}[{{v}_{t}}^{2}+{{v}_{x}}^{2}]\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\le {\eta}_{i}^{}{\parallel v\parallel}_{{L}^{2}(\mathrm{\Omega})}^{2}0.
Thus we have
\begin{array}{rcl}I(u)& =& \frac{1}{2}{r}^{2}\frac{1}{2}{\parallel v\parallel}^{2}{\int}_{\mathrm{\Omega}}G(x,t,v+re)\\ \le & \frac{1}{2}{r}^{2}+\frac{1}{2}{\eta}_{1}^{}{\parallel v\parallel}_{{L}^{2}(\mathrm{\Omega})}^{2}+{C}_{1}\end{array}
for {C}_{1}>0. Then there exists R>\rho >0 such that if u\in Q=({\overline{B}}_{R}\cap {E}_{})\oplus \{re0<r<R\}, then I(u){}_{\partial Q}\le 0.
(iii) Let us choose {u}_{0}\in E\mathrm{\setminus}Q. By (ii), I({u}_{0})\le 0.
(iv) S and Q link at the set \{e\}. □
Proof of Theorem 1.1 By Lemma 3.1, I(u) is continuous and Fréchet differentiable in E. By Lemma 3.3, the functional I satisfies the (PS) condition. We note that I(0)=0. By Lemma 3.4, there are constants \rho >0, \alpha >0 and a bounded neighborhood {B}_{\rho} of 0 such that I{}_{\partial {B}_{\rho}\cap {E}_{+}}\ge \alpha, and there are e\in \partial {B}_{1}\cap {E}_{+} and R>\rho such that if Q=({\overline{B}}_{R}\cap {E}_{})\oplus \{re0<r<R\}. Let us set
\mathrm{\Gamma}=\{\gamma \in C(\overline{Q},H)\gamma =\mathrm{id}\text{on}\partial Q\}.
By Theorem 3.1, I possesses a critical value c\ge \alpha. Moreover, c can be characterized as
c=\underset{\gamma \in \mathrm{\Gamma}}{inf}\underset{u\in Q}{max}I(\gamma (u)).
Thus we prove that I has at least one nontrivial critical point, we denote by \tilde{u} a critical point of I such that I(\tilde{u})=c. We claim that c is bounded. In fact, we have
c\le \underset{0\le \tau \le 1}{max}I(\tau {u}_{0}),
and by (g3),
\begin{array}{rcl}I(\tau {u}_{0})& =& {\tau}^{2}(\frac{1}{2}{\parallel {P}_{+}{u}_{0}\parallel}^{2}\frac{1}{2}{\parallel {P}_{}{u}_{0}\parallel}^{2}){\int}_{\mathrm{\Omega}}G(x,t,\tau {u}_{0})\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ \le & {\tau}^{2}{\parallel {u}_{0}\parallel}^{2}{\int}_{\mathrm{\Omega}}G(x,t,\tau {u}_{0})\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ \le & {C}_{1}{\tau}^{2}{\parallel {u}_{0}\parallel}^{2}+{C}_{1}\end{array}
for some constant {C}_{1}>0. Since 0\le \tau \le 1, c is bounded:
We claim that \tilde{u} is bounded. In fact, by contradiction, {\tilde{u}}_{tt}+{\tilde{u}}_{xx}=g(x,t,\tilde{u}) and, for any K>0, {max}_{\mathrm{\Omega}}\tilde{u}(x,t)>K imply that
c=I(\tilde{u})=\frac{1}{2}({\parallel {P}_{+}\tilde{u}\parallel}^{2}{\parallel {P}_{}\tilde{u}\parallel}^{2}){\int}_{\mathrm{\Omega}}G(x,t,\tilde{u})\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt
is not bounded, which is absurd to the fact that c=I(\tilde{u}) is bounded. Thus \tilde{u} is bounded, so (1.1) has at least one bounded weak solution, and hence we prove Theorem 1.1. □