Consider the two-point boundary value problem

\{\begin{array}{l}{x}^{\u2033}(t)=f(t,x(t)),\phantom{\rule{1em}{0ex}}t\in (0,1),\\ x(0)=x(1)=0.\end{array}

(3.1)

Here *f* is a continuous function from ℝ into ℝ.

Let H={H}_{0}^{1}(0,1) with the inner product and norm

(u,v)={\int}_{0}^{1}{u}^{\prime}{v}^{\prime}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{2em}{0ex}}\parallel u\parallel ={\left({\int}_{0}^{1}{\left({u}^{\prime}\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}},

and P=\{u\in H\mid u(t)\ge 0\}, then *H* is a lattice under the partial ordering induced by *P*.

Let X={L}^{2}(0,1) with the cone

{P}_{1}=\{u\in X\mid u(t)\ge 0\}.

For any x\in H, it is evident that

{\left(|x(t)|\right)}^{\prime}=\{\begin{array}{ll}{x}^{\prime}(t)& \text{if}x(t)0,\\ 0& \text{if}x(t)=0,\\ -{x}^{\prime}(t)& \text{if}x(t)0,\end{array}

and hence {\parallel x\parallel}_{1}=\parallel x\parallel and so condition (iv) of Theorem 2.1 holds with {M}_{1}={M}_{2}=1.

Let

(Bx)(t)={\int}_{0}^{1}G(t,s)x(s)\phantom{\rule{0.2em}{0ex}}ds

(3.2)

with

G(t,s)=\{\begin{array}{ll}t(1-s),& 0\le t\le s\le 1,\\ s(1-t),& 0\le s\le t\le 1.\end{array}

It is easy to see that B:{P}_{1}\to P is a positive bounded linear operator satisfying {(Bx)}^{\u2033}(t)=-x. For any x\in {P}_{1}, y\in P, we have

\begin{array}{rcl}(Bx,y)& =& {\int}_{0}^{1}{(Bx)}^{\prime}(t){y}^{\prime}(t)\phantom{\rule{0.2em}{0ex}}dt\\ =& -{\int}_{0}^{1}{(Bx)}^{\u2033}(t)y(t)\phantom{\rule{0.2em}{0ex}}dt\\ =& {\int}_{0}^{1}x(t)y(t)\phantom{\rule{0.2em}{0ex}}dt\ge 0\end{array}

and so condition (1) of Theorem 2.1 holds.

As is well known, *B* has an unbounded sequence of eigenvalues:

{\lambda}_{n}={n}^{2}\pi ,\phantom{\rule{1em}{0ex}}n=1,2,\dots .

The algebraic multiplicities of every eigenvalue are simple, and the spectral radius r(B)={\lambda}_{1}^{-1}=\frac{1}{{\pi}^{2}} and {\lambda}_{1}B(sin\pi t)=sin\pi t.

**Theorem 3.1** *Let* f(t,u):[0,1]\times \mathbb{R}\to \mathbb{R} *be continuous*. *Suppose that there exists* \epsilon >0 *such that*

\underset{u\to -\mathrm{\infty}}{lim\hspace{0.17em}sup}\frac{f(t,u)}{u}\le {\lambda}_{1}-\epsilon ,\phantom{\rule{1em}{0ex}}\mathit{\text{uniformly on}}t\in [0,1],

(3.3)

{\lambda}_{1}+\epsilon \le \underset{u\to +\mathrm{\infty}}{lim\hspace{0.17em}inf}\frac{f(t,u)}{u}\le \underset{u\to +\mathrm{\infty}}{lim\hspace{0.17em}sup}\frac{f(t,u)}{u}<+\mathrm{\infty},\phantom{\rule{1em}{0ex}}\mathit{\text{uniformly on}}t\in [0,1],

(3.4)

\underset{u\to 0}{lim\hspace{0.17em}sup}\left|\frac{f(t,u)}{u}\right|\le {\lambda}_{1}-\epsilon ,\phantom{\rule{1em}{0ex}}\mathit{\text{uniformly on}}t\in [0,1].

(3.5)

*Then* (3.1) *has at least one nontrivial solution*.

*Proof* We define the functional \mathrm{\Phi}:H\to \mathbb{R} as

\mathrm{\Phi}(u)={\int}_{0}^{1}(\frac{1}{2}{|{u}^{\prime}(t)|}^{2}-{\int}_{0}^{u(t)}f(t,\tau )\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}dt.

(3.6)

One has that {\mathrm{\Phi}}^{\prime}(x)y={\int}_{0}^{1}{x}^{\prime}{y}^{\prime}\phantom{\rule{0.2em}{0ex}}dt-{\int}_{0}^{1}(Fx)(t)y\phantom{\rule{0.2em}{0ex}}dt={\int}_{0}^{1}({x}^{\prime}-{(BFx)}^{\prime}(t)){y}^{\prime}\phantom{\rule{0.2em}{0ex}}dt, which implies that {\mathrm{\Phi}}^{\prime}(x)=x-BFx in *H*, where *B* is given by (3.2) and (Fx)(t)=f(t,x(t)). Furthermore, from the properties of *H*, the regularity properties of Φ, it is easy to assert that the solutions to (3.1) are precisely the critical points of Φ.

By means of (3.3), (3.4) and (3.5), we know that there exist {R}_{0}>r>0 such that

\begin{array}{c}f(u)\ge ({\lambda}_{1}-\frac{\epsilon}{2})u,\phantom{\rule{1em}{0ex}}u\le -{R}_{0},\hfill \\ f(u)\ge ({\lambda}_{1}+\frac{\epsilon}{2})u,\phantom{\rule{1em}{0ex}}u\ge {R}_{0},\hfill \\ |f(u)|\le ({\lambda}_{1}-\frac{\epsilon}{2})|u|,\phantom{\rule{1em}{0ex}}|u|\le r,\hfill \end{array}

and therefore there exists a constant M>0 such that

\begin{array}{c}Fx\ge ({\lambda}_{1}-\frac{\epsilon}{2})x-M,\phantom{\rule{1em}{0ex}}-x\in P,\hfill \\ Fx\ge ({\lambda}_{1}+\frac{\epsilon}{2})x-M,\phantom{\rule{1em}{0ex}}x\in P,\hfill \\ |Fx|\le ({\lambda}_{1}-\frac{\epsilon}{2})|x|,\phantom{\rule{1em}{0ex}}\parallel x\parallel \le r,\hfill \end{array}

and so conditions (ii), (iii) and (iv) of Theorem 2.1 hold.

From the above discussion, in order to apply Theorem 2.1, we only need to verify that Φ satisfies the PS condition. Suppose that \{{x}_{n}\}\subset H satisfies {\mathrm{\Phi}}^{\prime}({x}_{n})\to 0 as n\to \mathrm{\infty} and |\mathrm{\Phi}({x}_{n})|\le C. Taking the inner product of {\mathrm{\Phi}}^{\prime}({x}_{n}) and {x}_{n}^{-}, we have

\circ (1)\parallel {x}_{n}^{-}\parallel ={\parallel {x}_{n}^{-}\parallel}^{2}-{\int}_{0}^{1}f(t,{x}_{n}){x}_{n}^{-}\phantom{\rule{0.2em}{0ex}}dt.

We have

\begin{array}{rcl}-{\int}_{0}^{1}f(t,{x}_{n}){x}_{n}^{-}\phantom{\rule{0.2em}{0ex}}dt& \ge & -{\int}_{x(t)\le {R}_{0}}f(t,{x}_{n}){x}_{n}^{-}\phantom{\rule{0.2em}{0ex}}dt\\ \ge & -{\int}_{x(t)\le -{R}_{0}}f(t,{x}_{n}){x}_{n}^{-}\phantom{\rule{0.2em}{0ex}}dt-{C}_{1}\parallel {x}_{n}^{-}\parallel ,\end{array}

which implies

-{\int}_{0}^{1}f(t,{x}_{n}){x}_{n}^{-}\phantom{\rule{0.2em}{0ex}}dt\ge -({\lambda}_{1}-\frac{1}{2}\epsilon ){\int}_{0}^{1}{\left|{x}_{n}^{-}\right|}^{2}\phantom{\rule{0.2em}{0ex}}dt-{C}_{1}\parallel {x}_{n}^{-}\parallel .

Then we have

\circ (1)\parallel {x}_{n}^{-}\parallel \ge {\parallel {x}_{n}^{-}\parallel}^{2}-({\lambda}_{1}-\frac{1}{2}\epsilon ){\int}_{0}^{1}{\left|{x}_{n}^{-}\right|}^{2}\phantom{\rule{0.2em}{0ex}}dt-{C}_{1}\parallel {x}_{n}^{-}\parallel \ge \frac{\epsilon}{2{\lambda}_{1}}{\parallel {x}_{n}^{-}\parallel}^{2}-{C}_{1}\parallel {x}_{n}^{-}\parallel ,

which gives a bound for \{{x}_{n}^{-}\}. Then both \mathrm{\Phi}({x}_{n}^{-}) and \mathrm{\Phi}({x}_{n}^{+}) are bounded. In order to find a bound for \{{x}_{n}^{+}\}, we use a contradiction argument and assume that \parallel {x}_{n}^{+}\parallel \to \mathrm{\infty} as n\to \mathrm{\infty}. Defining {v}_{n}=\frac{{x}_{n}^{+}}{\parallel {x}_{n}^{+}\parallel} and selecting a subsequence if necessary, we have {v}_{n}\to v weakly in *H* and strongly in *X* as n\to \mathrm{\infty} for some v\in H, v\ge 0. Since {\int}_{0}^{u}f(t,\tau )\phantom{\rule{0.2em}{0ex}}d\tau \le {C}_{2}{|u|}^{2} for u\ge 0 and \mathrm{\Phi}({x}_{n}^{+}) is bounded, it follows from (3.6) that

\frac{1}{2}=\frac{\mathrm{\Phi}({x}_{n}^{+})}{{\parallel {x}_{n}^{+}\parallel}^{2}}+{\int}_{0}^{1}\frac{{\int}_{0}^{{x}_{n}^{+}}f(t,\tau )\phantom{\rule{0.2em}{0ex}}d\tau}{{\parallel {x}_{n}^{+}\parallel}^{2}}\phantom{\rule{0.2em}{0ex}}dt\le \circ (1)+{C}_{2}{\int}_{0}^{1}{v}_{n}^{2}\phantom{\rule{0.2em}{0ex}}dt,

which implies that v\ne 0. The embedding theorem and the boundedness of \{{x}_{n}^{-}\} in *H* guarantee that there exists {M}_{1}>0 such that {x}_{n}^{-}\ge -{M}_{1} for all *n*. So f(t,u) is bounded for -{M}_{1}\le u\le {R}_{0}. Taking the inner product of \frac{{\mathrm{\Phi}}^{\prime}({x}_{n})}{\parallel {x}_{n}^{+}\parallel} and sin\pi t, we see that

\begin{array}{rcl}({v}_{n},sin\pi t)& =& (\frac{{\mathrm{\Phi}}^{\prime}({x}_{n})}{\parallel {x}_{n}^{+}\parallel},sin\pi t)-(\frac{{x}_{n}^{-}}{\parallel {x}_{n}^{+}\parallel},sin\pi t)+{\int}_{0}^{1}\frac{f(t,{x}_{n})}{\parallel {x}_{n}^{+}\parallel}sin\pi t\phantom{\rule{0.2em}{0ex}}dt\\ \ge & \circ (1)+{\int}_{0}^{1}\frac{({\lambda}_{1}+\frac{\epsilon}{2}){x}_{n}^{+}-M}{\parallel {x}_{n}^{+}\parallel}sin\pi t\phantom{\rule{0.2em}{0ex}}dt\\ \ge & \circ (1)+({\lambda}_{1}+\frac{\epsilon}{2}){\int}_{0}^{1}{v}_{n}sin\pi t\phantom{\rule{0.2em}{0ex}}dt.\end{array}

Letting n\to \mathrm{\infty}, we obtain

{\lambda}_{1}{\int}_{0}^{1}{v}_{n}sin\pi t\phantom{\rule{0.2em}{0ex}}dt\ge ({\lambda}_{1}+\frac{\epsilon}{2}){\int}_{0}^{1}{v}_{n}sin\pi t\phantom{\rule{0.2em}{0ex}}dt,

which is a contradiction. Thus, \{{x}_{n}\} is bounded in *H* and it has a convergent subsequence, as a standard consequence. □