## Boundary Value Problems

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# Multiple positive doubly periodic solutions for a singular semipositone telegraph equation with a parameter

Boundary Value Problems20132013:7

DOI: 10.1186/1687-2770-2013-7

Accepted: 29 December 2012

Published: 16 January 2013

## Abstract

In this paper, we study the multiplicity of positive doubly periodic solutions for a singular semipositone telegraph equation. The proof is based on a well-known fixed point theorem in a cone.

MSC:34B15, 34B18.

### Keywords

semipositone telegraph equation doubly periodic solution singular cone fixed point theorem

## 1 Introduction

Recently, the existence and multiplicity of positive periodic solutions for a scalar singular equation or singular systems have been studied by using some fixed point theorems; see [19]. In [10], the authors show that the method of lower and upper solutions is also one of common techniques to study the singular problem. In addition, the authors [11] use the continuation type existence principle to investigate the following singular periodic problem:
${\left({|{u}^{\prime }|}^{p-2}{u}^{\prime }\right)}^{\prime }+h\left(u\right){u}^{\prime }=g\left(u\right)+c\left(t\right).$
More recently, using a weak force condition, Wang [12] has built some existence results for the following periodic boundary value problem:
$\left\{\begin{array}{c}{u}_{tt}-{u}_{xx}+{c}_{1}{u}_{t}+{a}_{11}\left(t,x\right)u+{a}_{12}\left(t,x\right)v={f}_{1}\left(t,x,u,v\right)+{\chi }_{1}\left(t,x\right),\hfill \\ {v}_{tt}-{v}_{xx}+{c}_{2}{v}_{t}+{a}_{21}\left(t,x\right)u+{a}_{22}\left(t,x\right)v={f}_{2}\left(t,x,u,v\right)+{\chi }_{2}\left(t,x\right).\hfill \end{array}$

The proof is based on Schauder’s fixed point theorem. For other results concerning the existence and multiplicity of positive doubly periodic solutions for a single regular telegraph equation or regular telegraph system, see, for example, the papers [1317] and the references therein. In these references, the nonlinearities are nonnegative.

On the other hand, the authors [18] study the semipositone telegraph system
$\left\{\begin{array}{c}{u}_{tt}-{u}_{xx}+{c}_{1}{u}_{t}+{a}_{1}\left(t,x\right)u={b}_{1}\left(t,x\right)f\left(t,x,u,v\right),\hfill \\ {v}_{tt}-{v}_{xx}+{c}_{2}{v}_{t}+{a}_{2}\left(t,x\right)v={b}_{2}\left(t,x\right)g\left(t,x,u,v\right),\hfill \end{array}$

where the nonlinearities f, g may change sign. In addition, there are many authors who have studied the semipositone equations; see [19, 20].

Inspired by the above references, we are concerned with the multiplicity of positive doubly periodic solutions for a general singular semipositone telegraph equation
$\left\{\begin{array}{c}{u}_{tt}-{u}_{xx}+c{u}_{t}+a\left(t,x\right)u=\lambda f\left(t,x,u\right),\hfill \\ u\left(t+2\pi ,x\right)=u\left(t,x+2\pi \right)=u\left(t,x\right),\hfill \end{array}$
(1)
where $c>0$ is a constant, $\lambda >0$ is a positive parameter, $a\left(t,x\right)\in C\left(R×R,R\right)$, $f\left(t,x,u\right)$ may change sign and is singular at $u=0$, namely,
$\underset{u\to {0}^{+}}{lim}f\left(t,x,u\right)=+\mathrm{\infty }.$

The main method used here is the following fixed-point theorem of a cone mapping.

Lemma 1.1 [21]

Let E be a Banach space, and $K\subset E$ be a cone in E. Assume ${\mathrm{\Omega }}_{1}$, ${\mathrm{\Omega }}_{2}$ are open subsets of E with $0\in {\mathrm{\Omega }}_{1}$, ${\overline{\mathrm{\Omega }}}_{1}\subset {\mathrm{\Omega }}_{2}$, and let $T:K\cap \left({\overline{\mathrm{\Omega }}}_{2}\setminus {\mathrm{\Omega }}_{1}\right)\to K$ be a completely continuous operator such that either
1. (i)

$\parallel Tu\parallel \le \parallel u\parallel$, $u\in K\cap \partial {\mathrm{\Omega }}_{1}$ and $\parallel Tu\parallel \ge \parallel u\parallel$, $u\in K\cap \partial {\mathrm{\Omega }}_{2}$; or

2. (ii)

$\parallel Tu\parallel \ge \parallel u\parallel$, $u\in K\cap \partial {\mathrm{\Omega }}_{1}$ and $\parallel Tu\parallel \le \parallel u\parallel$, $u\in K\cap \partial {\mathrm{\Omega }}_{2}$.

Then T has a fixed point in $K\cap \left({\overline{\mathrm{\Omega }}}_{2}\setminus {\mathrm{\Omega }}_{1}\right)$.

The paper is organized as follows. In Section 2, some preliminaries are given. In Section 3, we give the main result.

## 2 Preliminaries

Let ${\mathrm{\top }}^{2}$ be the torus defined as
${\mathrm{\top }}^{2}=\left(R/2\pi Z\right)×\left(R/2\pi Z\right).$
Doubly 2π-periodic functions will be identified to be functions defined on ${\mathrm{\top }}^{2}$. We use the notations
${L}^{p}\left({\mathrm{\top }}^{2}\right),\phantom{\rule{2em}{0ex}}C\left({\mathrm{\top }}^{2}\right),\phantom{\rule{2em}{0ex}}{C}^{\alpha }\left({\mathrm{\top }}^{2}\right),\phantom{\rule{2em}{0ex}}D\left({\mathrm{\top }}^{2}\right)={C}^{\mathrm{\infty }}\left({\mathrm{\top }}^{2}\right),\dots$

to denote the spaces of doubly periodic functions with the indicated degree of regularity. The space ${D}^{\prime }\left({\mathrm{\top }}^{2}\right)$ denotes the space of distributions on ${\mathrm{\top }}^{2}$.

By a doubly periodic solution of Eq. (1) we mean that a $u\in {L}^{1}\left({\mathrm{\top }}^{2}\right)$ satisfies Eq. (1) in the distribution sense, i.e.,
${\int }_{{\mathrm{\top }}_{2}}u\left({\phi }_{tt}-{\phi }_{xx}-c{\phi }_{t}+a\left(t,x\right)\phi \right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx=\lambda {\int }_{{\mathrm{\top }}^{2}}f\left(t,x,u\right)\phi \phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx.$
First, we consider the linear equation
(2)

where $c>0$, $\mu \in R$, and $h\left(t,x\right)\in {L}^{1}\left({\mathrm{\top }}^{2}\right)$.

Let ${\text{£}}_{\xi }$ be the differential operator
${\text{£}}_{\xi }u={u}_{tt}-{u}_{xx}+c{u}_{t}-\xi u,$
acting on functions on ${\mathrm{\top }}^{2}$. Following the discussion in [14], we know that if $\xi <0$, ${\text{£}}_{\xi }$ has the resolvent ${R}_{\xi }$,
${R}_{\xi }:{L}^{1}\left({\mathrm{\top }}^{2}\right)\to C\left({\mathrm{\top }}^{2}\right),\phantom{\rule{2em}{0ex}}{h}_{i}\left(t,x\right)↦{u}_{i}\left(t,x\right),$

where $u\left(t,x\right)$ is the unique solution of Eq. (2), and the restriction of ${R}_{\xi }$ on ${L}^{p}\left({\mathrm{\top }}^{2}\right)$ ($1) or $C\left({\mathrm{\top }}^{2}\right)$ is compact. In particular, ${R}_{\xi }:C\left({\mathrm{\top }}^{2}\right)\to C\left({\mathrm{\top }}^{2}\right)$ is a completely continuous operator.

For $\xi =-{c}^{2}/4$, the Green function $G\left(t,x\right)$ of the differential operator ${\text{£}}_{\xi }$ is explicitly expressed; see Lemma 5.2 in [14]. From the definition of $G\left(t,x\right)$, we have

For convenience, we assume the following condition holds throughout this paper:

(H1) $a\left(t,x\right)\in C\left({\mathrm{\top }}^{2},R\right)$, $0\le a\left(t,x\right)\le \frac{{c}^{2}}{4}$ on ${\mathrm{\top }}^{2}$, and ${\int }_{{\mathrm{\top }}^{2}}a\left(t,x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx>0$.

Finally, if −ξ is replaced by $a\left(t,x\right)$ in Eq. (2), the author [13] has proved the following unique existence and positive estimate result.

Lemma 2.1 Let $h\left(t,x\right)\in {L}^{1}\left({\mathrm{\top }}^{2}\right)$. Then Eq. (2) has a unique solution $u\left(t,x\right)=P\left[h\left(t,x\right)\right]$, $P:{L}^{1}\left({\mathrm{\top }}^{2}\right)\to C\left({\mathrm{\top }}^{2}\right)$ is a linear bounded operator with the following properties:
1. (i)

$P:C\left({\mathrm{\top }}^{2}\right)\to C\left({\mathrm{\top }}^{2}\right)$ is a completely continuous operator;

2. (ii)
If $h\left(t,x\right)>0$, a.e $\left(t,x\right)\in {\mathrm{\top }}^{2}$, $P\left[h\left(t,x\right)\right]$ has the positive estimate
$\underline{G}{\parallel h\parallel }_{{L}^{1}}\le P\left[h\left(t,x\right)\right]\le \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{{L}^{1}}}{\parallel h\parallel }_{{L}^{1}}.$
(3)

## 3 Main result

Theorem 3.1 Assume (H1) holds. In addition, if $f\left(t,x,u\right)$ satisfies

(H2) ${lim}_{u\to {0}^{+}}f\left(t,x,u\right)=+\mathrm{\infty }$, uniformly $\left(t,x\right)\in {\mathrm{\top }}^{2}$,

(H3) $f:{\mathrm{\top }}^{2}×\left(0,+\mathrm{\infty }\right)\to \left(-\mathrm{\infty },+\mathrm{\infty }\right)$ is continuous,

(H4) there exists a nonnegative function $h\left(t,x\right)\in C\left({\mathrm{\top }}^{2}\right)$ such that
$f\left(t,x,u\right)+h\left(t,x\right)\ge 0,\phantom{\rule{1em}{0ex}}\left(t,x\right)\in {\mathrm{\top }}^{2},u>0,$

(H5) ${\int }_{{\mathrm{\top }}^{2}}{F}_{\mathrm{\infty }}\left(t,x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx=+\mathrm{\infty }$, where the limit function ${F}_{\mathrm{\infty }}\left(t,x\right)={lim inf}_{u\to +\mathrm{\infty }}\frac{f\left(t,x,u\right)}{u}$,

then Eq. (1) has at least two positive doubly periodic solutions for sufficiently small λ.

$C\left({\mathrm{\top }}^{2}\right)$ is a Banach space with the norm $\parallel u\parallel ={max}_{\left(t,x\right)\in {\mathrm{\top }}^{2}}|u\left(t,x\right)|$. Define a cone $K\subset C\left({\mathrm{\top }}^{2}\right)$ by
$K=\left\{u\in C\left({\mathrm{\top }}^{2}\right):u\ge 0,u\left(t,x\right)\ge \delta \parallel u\parallel \right\},$

where $\delta =\frac{{\underline{G}}^{2}{\parallel a\parallel }_{{L}^{1}}}{\overline{G}}\in \left(0,1\right)$. Let $\partial {K}_{r}=\left\{u\in K:\parallel u\parallel =r\right\}$, ${\left[u\right]}^{+}=max\left\{u,0\right\}$. By Lemma 2.1, it is easy to obtain the following lemmas.

Lemma 3.2 If $h\left(t,x\right)\in C\left({\mathrm{\top }}^{2}\right)$ is a nonnegative function, the linear boundary value problem
$\left\{\begin{array}{c}{u}_{tt}-{u}_{xx}+c{u}_{t}+a\left(t,x\right)u=\lambda h\left(t,x\right),\hfill \\ u\left(t+2\pi ,x\right)=u\left(t,x+2\pi \right)=u\left(t,x\right)\hfill \end{array}$
has a unique solution $\omega \left(t,x\right)$. The function $\omega \left(t,x\right)$ satisfies the estimates
$\lambda \underline{G}{\parallel h\parallel }_{{L}^{1}}\le \omega \left(t,x\right)=\lambda P\left(h\left(t,x\right)\right)\le \lambda \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{{L}^{1}}}{\parallel h\parallel }_{{L}^{1}}.$
Lemma 3.3 If the boundary value problem
$\left\{\begin{array}{c}{u}_{tt}-{u}_{xx}+c{u}_{t}+a\left(t,x\right)u=\lambda \left[f\left(t,x,{\left[u\left(t,x\right)-\omega \left(t,x\right)\right]}^{+}\right)+h\left(t,x\right)\right],\hfill \\ u\left(t+2\pi ,x\right)=u\left(t,x+2\pi \right)=u\left(t,x\right)\hfill \end{array}$

has a solution $\stackrel{˜}{u}\left(t,x\right)$ with $\parallel \stackrel{˜}{u}\parallel >\lambda \frac{{\overline{G}}^{2}}{{\underline{G}}^{3}{\parallel a\parallel }_{{L}^{1}}^{2}}{\parallel h\parallel }_{{L}^{1}}$, then ${u}^{\ast }\left(t,x\right)=\stackrel{˜}{u}\left(t,x\right)-\omega \left(t,x\right)$ is a positive doubly periodic solution of Eq. (1).

Proof of Theorem 3.1 Step 1. Define the operator T as follows:
$\left(Tu\right)\left(t,x\right)=\lambda P\left[f\left(t,x,{\left[u\left(t,x\right)-\omega \left(t,x\right)\right]}^{+}\right)+h\left(t,x\right)\right].$

We obtain the conclusion that $T\left(K\mathrm{\setminus }\left\{u\in K:{\left[u\left(t,x\right)-\omega \left(t,x\right)\right]}^{+}=0\right\}\right)\subseteq K$, and $T:K\mathrm{\setminus }\left\{u\in K:{\left[u\left(t,x\right)-\omega \left(t,x\right)\right]}^{+}=0\right\}\to K$ is completely continuous.

For any $u\in K\mathrm{\setminus }\left\{u\in K:{\left[u\left(t,x\right)-\omega \left(t,x\right)\right]}^{+}=0\right\}$, then ${\left[u\left(t,x\right)-\omega \left(t,x\right)\right]}^{+}>0$, and T is defined. On the other hand, for $u\in K\mathrm{\setminus }\left\{u\in K:{\left[u\left(t,x\right)-\omega \left(t,x\right)\right]}^{+}=0\right\}$, the complete continuity is obvious by Lemma 2.1. And we can have
$\begin{array}{rcl}\left(Tu\right)\left(t,x\right)& =& \lambda P\left[f\left(t,x,{\left[u\left(t,x\right)-\omega \left(t,x\right)\right]}^{+}\right)+h\left(t,x\right)\right]\\ \ge & \lambda \underline{G}{\parallel f\left(t,x,{\left[u\left(t,x\right)-\omega \left(t,x\right)\right]}^{+}\right)+h\left(t,x\right)\parallel }_{{L}^{1}}\\ \ge & \underline{G}\frac{\underline{G}{\parallel a\parallel }_{{L}^{1}}}{\overline{G}}\parallel T\left(u\right)\parallel \\ \ge & \delta \parallel Tu\parallel .\end{array}$

Thus, $T\left(K\mathrm{\setminus }\left\{u\in K:u\left(t,x\right)\le \omega \left(t,x\right)\right\}\right)\subseteq K$.

Now we prove that the operator T has one fixed point $\stackrel{˜}{u}\in K$ and $\parallel \stackrel{˜}{u}\parallel >\lambda \frac{{\overline{G}}^{2}}{{\underline{G}}^{3}{\parallel a\parallel }_{{L}^{1}}^{2}}{\parallel h\parallel }_{{L}^{1}}$ for all sufficiently small λ.

Since ${\int }_{{\mathrm{\top }}^{2}}{F}_{\mathrm{\infty }}\left(t,x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx=+\mathrm{\infty }$, there exists ${r}_{1}\ge 2$ such that
${\int }_{{\mathrm{\top }}^{2}}\frac{f\left(t,x,u\right)}{u}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\ge \frac{1}{\delta },\phantom{\rule{1em}{0ex}}u\ge \delta {r}_{1}.$
Furthermore, we have ${\int }_{{\mathrm{\top }}^{2}}f\left(t,x,\delta {r}_{1}\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\ge {r}_{1}\ge 2$. It follows that
Let $\mathrm{\Phi }\left(t,x\right)=max\left\{f\left(t,x,u\right):\frac{\delta }{2}{r}_{1}\le u\le {r}_{1}\right\}+h\left(t,x\right)$. Then $\mathrm{\Phi }\in {L}^{1}\left({\mathrm{\top }}^{2}\right)$ and ${\int }_{{\mathrm{\top }}^{2}}\mathrm{\Phi }\left(t,x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx>0$. Set
${\lambda }^{\ast }=min\left\{\frac{{\delta }^{2}}{2\underline{G}{\parallel h\parallel }_{{L}^{1}}},\frac{2\underline{G}{\parallel a\parallel }_{{L}^{1}}}{\overline{G}{\parallel \mathrm{\Phi }\parallel }_{{L}^{1}}}\right\}.$
For any $u\in \partial {K}_{{r}_{1}}$ and $0<\lambda <{\lambda }^{\ast }$, we can verify that
$\begin{array}{rcl}u\left(t,x\right)-\omega \left(t,x\right)& \ge & \delta \parallel u\parallel -\omega \left(t,x\right)\\ =& \delta {r}_{1}-\omega \left(t,x\right)\\ \ge & \delta {r}_{1}-\lambda \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{{L}^{1}}}{\parallel h\parallel }_{{L}^{1}}\\ \ge & \delta {r}_{1}-\frac{\delta {r}_{1}}{2}\\ =& \frac{\delta {r}_{1}}{2}.\end{array}$
Then we have
$\begin{array}{rcl}\parallel Tu\parallel & =& \lambda \parallel P\left[f\left(t,x,{\left[u\left(t,x\right)-\omega \left(t,x\right)\right]}^{+}\right)+h\left(t,x\right)\right]\parallel \\ \le & \lambda \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{{L}^{1}}}{\parallel f\left(t,x,{\left[u\left(t,x\right)-\omega \left(t,x\right)\right]}^{+}\right)+h\left(t,x\right)\parallel }_{{L}^{1}}\\ \le & \lambda \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{{L}^{1}}}{\parallel \mathrm{\Phi }\left(t,x\right)\parallel }_{{L}^{1}}\\ <& 2\le {r}_{1}=\parallel u\parallel .\end{array}$
On the other hand,
$\underset{u\to +\mathrm{\infty }}{lim inf}\frac{f\left(t,x,u-\omega \left(t,x\right)\right)}{u}=\underset{u\to +\mathrm{\infty }}{lim inf}\frac{f\left(t,x,u\right)}{u}={F}_{\mathrm{\infty }}\left(t,x\right).$
By the Fatou lemma, one has
Hence, there exists a positive number ${r}_{2}>\delta {r}_{2}>{r}_{1}$ such that
${\int }_{{\mathrm{\top }}^{2}}\frac{f\left(t,x,u-\omega \left(t,x\right)\right)+h\left(t,x\right)}{u}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\ge {\lambda }^{-1}{\delta }^{-1}{\underline{G}}^{-1}{\left(4{\pi }^{2}\right)}^{-1},\phantom{\rule{1em}{0ex}}u\ge \delta {r}_{2}.$
Hence, we have
${\int }_{{\mathrm{\top }}^{2}}f\left(t,x,u-\omega \left(t,x\right)\right)+h\left(t,x\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}dx\ge {\lambda }^{-1}{\underline{G}}^{-1}{\left(4{\pi }^{2}\right)}^{-1}{r}_{2},\phantom{\rule{1em}{0ex}}u\ge \delta {r}_{2}.$
For any $u\in \partial {K}_{{r}_{2}}$, we have $\delta {r}_{2}=\delta \parallel u\parallel \le u\left(t,x\right)\le \parallel u\parallel ={r}_{2}$. On the other hand, since $0<\lambda <{\lambda }^{\ast }$, we can get
$\begin{array}{rcl}u\left(t,x\right)-\omega \left(t,x\right)& \ge & \delta {r}_{2}-\omega \left(t,x\right)\\ \ge & \delta \frac{{r}_{2}}{\delta }-\lambda \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{{L}^{1}}}\\ \ge & \delta {r}_{2}-\delta \\ >& 0.\end{array}$
From above, we can have
$\begin{array}{rcl}\parallel Tu\parallel & \ge & \lambda P\left[f\left(t,x,{\left[u\left(t,x\right)-\omega \left(t,x\right)\right]}^{+}\right)+h\left(t,x\right)\right]\\ \ge & \lambda \underline{G}{\parallel f\left(t,x,{\left[u\left(t,x\right)-\omega \left(t,x\right)\right]}^{+}\right)+h\left(t,x\right)\parallel }_{{L}^{1}}\\ \ge & \lambda \underline{G}4{\pi }^{2}{\lambda }^{-1}{\underline{G}}^{-1}{\left(4{\pi }^{2}\right)}^{-1}{r}_{2}\\ =& {r}_{2}.\end{array}$
Therefore, by Lemma 1.1, the operator T has a fixed point $\stackrel{˜}{u}\left(t,x\right)\in K$ and

So, Eq. (1) has a positive solution $\stackrel{ˆ}{u}\left(t,x\right)=\stackrel{˜}{u}\left(t,x\right)-\omega \left(t,x\right)\ge \delta$.

Step 2. By conditions (H2) and (H3), it is clear to obtain that
${u}_{0}=inf\left\{u\in K:f\left(t,x,u\right)\le 0,\left(t,x\right)\in {\mathrm{\top }}^{2}\right\}>0.$
Let ${r}_{4}=min\left\{\frac{\delta }{2},\frac{\delta \parallel {u}_{0}\parallel }{2}\right\}$. For any $u\in \left(0,{r}_{4}\right]$, we have $f\left(t,x,u\right)>0$. Then define the operator A as follows:
$\left(Au\right)\left(t,x\right)=\lambda \stackrel{ˆ}{P}\left[f\left(t,x,u\left(t,x\right)\right)\right].$

It is easy to prove that $A\left(K\cap \left\{u\in C\left({\mathrm{\top }}^{2}\right):0<\parallel u\parallel <{r}_{4}\right\}\right)\subseteq K$, and $A:K\cap \left\{u\in C\left({\mathrm{\top }}^{2}\right):0<\parallel u\parallel <{r}_{4}\right\}\to K$ is completely continuous.

And for any $\rho >0$, define
$M\left(\rho \right)=max\left\{f\left(t,x,u\right):u\in {R}^{+},\delta \rho \le u\le \rho ,\left(t,x\right)\in {\mathrm{\top }}^{2}\right\}>0.$
Furthermore, for any $u\in \partial {K}_{{r}_{4}}$, we have
$\begin{array}{rcl}\parallel Au\parallel & =& \lambda \parallel \stackrel{ˆ}{P}\left[f\left(t,x,u\left(t,x\right)\right)\right]\parallel \\ \le & \lambda \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{{L}^{1}}}{\parallel f\left(t,x,u\left(t,x\right)\right)\parallel }_{{L}^{1}}\\ \le & \lambda \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{{L}^{1}}}M\left({r}_{4}\right)4{\pi }^{2}.\end{array}$
Thus, from the above inequality, there exists $\overline{\lambda }$ such that
Since ${lim}_{u\to {0}^{+}}f\left(t,x,u\right)=\mathrm{\infty }$, then there is $0<{r}_{3}<\frac{{r}_{4}}{2}$ such that
where μ satisfies $\lambda \underline{G}\mu \delta >1$. For any $u\in \partial {K}_{{r}_{3}}$, then we have
By Lemma 2.1, it is clear to obtain that
$\begin{array}{rcl}\parallel Au\parallel & =& \lambda \parallel \stackrel{ˆ}{P}\left[f\left(t,x,u\left(t,x\right)\right)\right]\parallel \\ \ge & \lambda \underline{G}{\parallel f\left(t,x,u\left(t,x\right)\right)\parallel }_{{L}^{1}}\\ \ge & \lambda \underline{G}\mu \delta {r}_{3}\\ >& {r}_{3}=\parallel u\parallel .\end{array}$

Therefore, by Lemma 1.1, A has a fixed point in $\overline{u}\left(t,x\right)\in K$ and $\parallel \overline{u}\parallel \le {r}_{4}\le \frac{\delta }{2}$, which is another positive periodic solution of Eq. (1).

Finally, from Step 1 and Step 2, Eq. (1) has two positive doubly periodic solutions $\stackrel{ˆ}{u}\left(t,x\right)$ and $\overline{u}\left(t,x\right)$ for sufficiently small λ. □

Example

Consider the following problem:
$\left\{\begin{array}{c}{u}_{tt}-{u}_{xx}+2{u}_{t}+{sin}^{2}\left(t+x\right)u=\lambda \left[\frac{1}{u}+min\left\{{u}^{2},\frac{u}{|1-\frac{t}{\pi }||1-\frac{x}{\pi }|}\right\}-10\right],\hfill \\ u\left(t+2\pi ,x\right)=u\left(t,x+2\pi \right)=u\left(t,x\right).\hfill \end{array}$

It is clear that $f\left(t,x,u\right)$ satisfies the conditions (H1)-(H5).

## Declarations

### Acknowledgements

The authors would like to thank the referees for valuable comments and suggestions for improving this paper.

## Authors’ Affiliations

(1)
College of Science, Hohai University
(2)
Department of Mathematics, Nanjing University of Aeronautics and Astronautics

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