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

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.

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 [1–9]. 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:

$${\left({\left|u^{\prime }\right|}^{p-2}{u}^{\prime }\right)}^{\prime }+h(u)u^{\prime }=g(u)+c(t).$$

More recently, using a weak force condition, Wang [12] has built some existence results for the following periodic boundary value problem:

$$\{\begin{array}{c}{u}_{tt}-u_{xx}+c_1{u}_t+a_{11}(t,x)u+a_{12}(t,x)v=f_1(t,x,u,v)+{\chi }_1(t,x),\hfill \\ v_{tt}-v_{xx}+c_2{v}_t+a_{21}(t,x)u+a_{22}(t,x)v=f_2(t,x,u,v)+{\chi }_2(t,x).\hfill \end{array}$$

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 [13–17] and the references therein. In these references, the nonlinearities are nonnegative.

On the other hand, the authors [18] study the semipositone telegraph system

$$\{\begin{array}{c}{u}_{tt}-u_{xx}+c_1{u}_t+a_1(t,x)u=b_1(t,x)f(t,x,u,v),\hfill \\ v_{tt}-v_{xx}+c_2{v}_t+a_2(t,x)v=b_2(t,x)g(t,x,u,v),\hfill \end{array}$$

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

$$\{\begin{array}{c}{u}_{tt}-u_{xx}+c{u}_t+a(t,x)u=\lambda f(t,x,u),\hfill \\ u(t+2\pi ,x)=u(t,x+2\pi )=u(t,x),\hfill \end{array}$$

(1)

where $c>0$ is a constant, $\lambda >0$ is a positive parameter, $a(t,x)\in C(R\times R,R)$, $f(t,x,u)$ may change sign and is singular at $u=0$, namely,

$$\underset{u\to 0^{+}}{lim}f(t,x,u)=+\mathrm{\infty }.$$

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 $T:K\cap ({\overline{\mathrm{\Omega }}}_2\setminus {\mathrm{\Omega }}_1)\to K$ be a completely continuous operator such that either

• (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

• (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 $K\cap ({\overline{\mathrm{\Omega }}}_2\setminus {\mathrm{\Omega }}_1)$.

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

$${\mathrm{\top }}^2=(R/2\pi Z)\times (R/2\pi Z).$$

Doubly 2π-periodic functions will be identified to be functions defined on ${\mathrm{\top }}^2$. We use the notations

$$L^p\left({\mathrm{\top }}^2\right),C\left({\mathrm{\top }}^2\right),C^{\alpha }\left({\mathrm{\top }}^2\right),D\left({\mathrm{\top }}^2\right)=C^{\mathrm{\infty }}\left({\mathrm{\top }}^2\right),\dots$$

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

By a doubly periodic solution of Eq. (1) we mean that a $u\in L^1({\mathrm{\top }}^2)$ satisfies Eq. (1) in the distribution sense, i.e.,

$${\int }_{{\mathrm{\top }}_2}u({\phi }_{tt}-{\phi }_{xx}-c{\phi }_t+a(t,x)\phi )dtdx=\lambda {\int }_{{\mathrm{\top }}^2}f(t,x,u)\phi dtdx.$$

First, we consider the linear equation

$$u_{tt}-u_{xx}+c{u}_t-\xi u=h(t,x),\text{in }{D}^{\prime }\left({\mathrm{\top }}^2\right),$$

(2)

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

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 $R_{\xi }$,

$$R_{\xi }:L^1\left({\mathrm{\top }}^2\right)\to C\left({\mathrm{\top }}^2\right),h_i(t,x)\mapsto u_i(t,x),$$

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

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

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

(H1) $a(t,x)\in C({\mathrm{\top }}^2,R)$, $0\le a(t,x)\le \frac{c^2}{4}$ on ${\mathrm{\top }}^2$, and ${\int }_{{\mathrm{\top }}^2}a(t,x)dtdx>0$.

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

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

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

• (ii) If $h(t,x)>0$, a.e $(t,x)\in {\mathrm{\top }}^2$, $P[h(t,x)]$ has the positive estimate

  $$\underline{G}{\parallel h\parallel }_{L^1}\le P[h(t,x)]\le \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{L^1}}{\parallel h\parallel }_{L^1}.$$

  (3)

3 Main result

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

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

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

(H4) there exists a nonnegative function $h(t,x)\in C({\mathrm{\top }}^2)$ such that

$$f(t,x,u)+h(t,x)\ge 0,(t,x)\in {\mathrm{\top }}^2,u>0,$$

(H5) ${\int }_{{\mathrm{\top }}^2}{F}_{\mathrm{\infty }}(t,x)dtdx=+\mathrm{\infty }$, where the limit function $F_{\mathrm{\infty }}(t,x)={lim\hspace{0.17em}inf}_{u\to +\mathrm{\infty }}\frac{f(t,x,u)}{u}$,

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

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

$$K=\{u\in C\left({\mathrm{\top }}^2\right):u\ge 0,u(t,x)\ge \delta \parallel u\parallel \},$$

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

Lemma 3.2 If $h(t,x)\in C({\mathrm{\top }}^2)$ is a nonnegative function, the linear boundary value problem

$$\{\begin{array}{c}{u}_{tt}-u_{xx}+c{u}_t+a(t,x)u=\lambda h(t,x),\hfill \\ u(t+2\pi ,x)=u(t,x+2\pi )=u(t,x)\hfill \end{array}$$

has a unique solution $\omega (t,x)$. The function $\omega (t,x)$ satisfies the estimates

$$\lambda \underline{G}{\parallel h\parallel }_{L^1}\le \omega (t,x)=\lambda P(h(t,x))\le \lambda \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{L^1}}{\parallel h\parallel }_{L^1}.$$

Lemma 3.3 If the boundary value problem

$$\{\begin{array}{c}{u}_{tt}-u_{xx}+c{u}_t+a(t,x)u=\lambda [f(t,x,{[u(t,x)-\omega (t,x)]}^{+})+h(t,x)],\hfill \\ u(t+2\pi ,x)=u(t,x+2\pi )=u(t,x)\hfill \end{array}$$

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

Proof of Theorem 3.1 Step 1. Define the operator T as follows:

$$(Tu)(t,x)=\lambda P[f(t,x,{[u(t,x)-\omega (t,x)]}^{+})+h(t,x)].$$

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

For any $u\in K\mathrm{\setminus }\{u\in K:{[u(t,x)-\omega (t,x)]}^{+}=0\}$, then ${[u(t,x)-\omega (t,x)]}^{+}>0$, and T is defined. On the other hand, for $u\in K\mathrm{\setminus }\{u\in K:{[u(t,x)-\omega (t,x)]}^{+}=0\}$, the complete continuity is obvious by Lemma 2.1. And we can have

$$\begin{array}{rcl}(Tu)(t,x)& =& \lambda P[f(t,x,{[u(t,x)-\omega (t,x)]}^{+})+h(t,x)]\\ \ge & \lambda \underline{G}{\parallel f(t,x,{[u(t,x)-\omega (t,x)]}^{+})+h(t,x)\parallel }_{L^1}\\ \ge & \underline{G}\frac{\underline{G}{\parallel a\parallel }_{L^1}}{\overline{G}}\parallel T(u)\parallel \\ \ge & \delta \parallel Tu\parallel .\end{array}$$

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

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

Since ${\int }_{{\mathrm{\top }}^2}{F}_{\mathrm{\infty }}(t,x)dtdx=+\mathrm{\infty }$, there exists $r_1\ge 2$ such that

$${\int }_{{\mathrm{\top }}^2}\frac{f(t,x,u)}{u}dtdx\ge \frac{1}{\delta },u\ge \delta r_1.$$

Furthermore, we have ${\int }_{{\mathrm{\top }}^2}f(t,x,\delta r_1)dtdx\ge r_1\ge 2$. It follows that

Let $\mathrm{\Phi }(t,x)=max\{f(t,x,u):\frac{\delta }{2}{r}_1\le u\le r_1\}+h(t,x)$. Then $\mathrm{\Phi }\in L^1({\mathrm{\top }}^2)$ and ${\int }_{{\mathrm{\top }}^2}\mathrm{\Phi }(t,x)dtdx>0$. Set

$${\lambda }^{\ast }=min\{\frac{{\delta }^2}{2\underline{G}{\parallel h\parallel }_{L^1}},\frac{2\underline{G}{\parallel a\parallel }_{L^1}}{\overline{G}{\parallel \mathrm{\Phi }\parallel }_{L^1}}\}.$$

For any $u\in \partial K_{r_1}$ and $0<\lambda <{\lambda }^{\ast }$, we can verify that

$$\begin{array}{rcl}u(t,x)-\omega (t,x)& \ge & \delta \parallel u\parallel -\omega (t,x)\\ =& \delta r_1-\omega (t,x)\\ \ge & \delta r_1-\lambda \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{L^1}}{\parallel h\parallel }_{L^1}\\ \ge & \delta r_1-\frac{\delta r_1}{2}\\ =& \frac{\delta r_1}{2}.\end{array}$$

Then we have

$$\begin{array}{rcl}\parallel Tu\parallel & =& \lambda \parallel P[f(t,x,{[u(t,x)-\omega (t,x)]}^{+})+h(t,x)]\parallel \\ \le & \lambda \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{L^1}}{\parallel f(t,x,{[u(t,x)-\omega (t,x)]}^{+})+h(t,x)\parallel }_{L^1}\\ \le & \lambda \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{L^1}}{\parallel \mathrm{\Phi }(t,x)\parallel }_{L^1}\\ <& 2\le r_1=\parallel u\parallel .\end{array}$$

On the other hand,

$$\underset{u\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{f(t,x,u-\omega (t,x))}{u}=\underset{u\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{f(t,x,u)}{u}=F_{\mathrm{\infty }}(t,x).$$

By the Fatou lemma, one has

Hence, there exists a positive number $r_2>\delta r_2>r_1$ such that

$${\int }_{{\mathrm{\top }}^2}\frac{f(t,x,u-\omega (t,x))+h(t,x)}{u}dtdx\ge {\lambda }^{-1}{\delta }^{-1}{\underline{G}}^{-1}{\left(4{\pi }^2\right)}^{-1},u\ge \delta r_2.$$

Hence, we have

$${\int }_{{\mathrm{\top }}^2}f(t,x,u-\omega (t,x))+h(t,x)dtdx\ge {\lambda }^{-1}{\underline{G}}^{-1}{\left(4{\pi }^2\right)}^{-1}{r}_2,u\ge \delta r_2.$$

For any $u\in \partial K_{r_2}$, we have $\delta r_2=\delta \parallel u\parallel \le u(t,x)\le \parallel u\parallel =r_2$. On the other hand, since $0<\lambda <{\lambda }^{\ast }$, we can get

$$\begin{array}{rcl}u(t,x)-\omega (t,x)& \ge & \delta r_2-\omega (t,x)\\ \ge & \delta \frac{r_2}{\delta }-\lambda \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{L^1}}\\ \ge & \delta r_2-\delta \\ >& 0.\end{array}$$

From above, we can have

$$\begin{array}{rcl}\parallel Tu\parallel & \ge & \lambda P[f(t,x,{[u(t,x)-\omega (t,x)]}^{+})+h(t,x)]\\ \ge & \lambda \underline{G}{\parallel f(t,x,{[u(t,x)-\omega (t,x)]}^{+})+h(t,x)\parallel }_{L^1}\\ \ge & \lambda \underline{G}4{\pi }^2{\lambda }^{-1}{\underline{G}}^{-1}{\left(4{\pi }^2\right)}^{-1}{r}_2\\ =& r_2.\end{array}$$

Therefore, by Lemma 1.1, the operator T has a fixed point $\tilde{u}(t,x)\in K$ and

So, Eq. (1) has a positive solution $\hat{u}(t,x)=\tilde{u}(t,x)-\omega (t,x)\ge \delta$.

Step 2. By conditions (H2) and (H3), it is clear to obtain that

$$u_0=inf\{u\in K:f(t,x,u)\le 0,(t,x)\in {\mathrm{\top }}^2\}>0.$$

Let $r_4=min\{\frac{\delta }{2},\frac{\delta \parallel u_0\parallel }{2}\}$. For any $u\in (0,r_4]$, we have $f(t,x,u)>0$. Then define the operator A as follows:

$$(Au)(t,x)=\lambda \hat{P}\left[f(t,x,u(t,x))\right].$$

It is easy to prove that $A(K\cap \{u\in C({\mathrm{\top }}^2):0<\parallel u\parallel <r_4\})\subseteq K$, and $A:K\cap \{u\in C({\mathrm{\top }}^2):0<\parallel u\parallel <r_4\}\to K$ is completely continuous.

And for any $\rho >0$, define

$$M(\rho )=max\{f(t,x,u):u\in R^{+},\delta \rho \le u\le \rho ,(t,x)\in {\mathrm{\top }}^2\}>0.$$

Furthermore, for any $u\in \partial K_{r_4}$, we have

$$\begin{array}{rcl}\parallel Au\parallel & =& \lambda \parallel \hat{P}\left[f(t,x,u(t,x))\right]\parallel \\ \le & \lambda \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{L^1}}{\parallel f(t,x,u(t,x))\parallel }_{L^1}\\ \le & \lambda \frac{\overline{G}}{\underline{G}{\parallel a\parallel }_{L^1}}M(r_4)4{\pi }^2.\end{array}$$

Thus, from the above inequality, there exists $\overline{\lambda }$ such that

$$\parallel Au\parallel <\parallel u\parallel ,\text{for }u\in \partial K_{r_4},0<\lambda <\overline{\lambda }.$$

Since ${lim}_{u\to 0^{+}}f(t,x,u)=\mathrm{\infty }$, then there is $0<r_3<\frac{r_4}{2}$ such that

$$f(t,x,u)\ge \mu u,\text{for }u\in R^{+}\text{ with }0<u\le r_3,$$

where μ satisfies $\lambda \underline{G}\mu \delta >1$. For any $u\in \partial K_{r_3}$, then we have

$$f(t,x,u)\ge \mu u(t,x),\text{for }(t,x)\in {\mathrm{\top }}^2.$$

By Lemma 2.1, it is clear to obtain that

$$\begin{array}{rcl}\parallel Au\parallel & =& \lambda \parallel \hat{P}\left[f(t,x,u(t,x))\right]\parallel \\ \ge & \lambda \underline{G}{\parallel f(t,x,u(t,x))\parallel }_{L^1}\\ \ge & \lambda \underline{G}\mu \delta r_3\\ >& r_3=\parallel u\parallel .\end{array}$$

Therefore, by Lemma 1.1, A has a fixed point in $\overline{u}(t,x)\in K$ 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 $\hat{u}(t,x)$ and $\overline{u}(t,x)$ for sufficiently small λ. □

Example

Consider the following problem:

$$\{\begin{array}{c}{u}_{tt}-u_{xx}+2u_t+{sin}^2(t+x)u=\lambda [\frac{1}{u}+min\{u^2,\frac{u}{|1-\frac{t}{\pi }||1-\frac{x}{\pi }|}\}-10],\hfill \\ u(t+2\pi ,x)=u(t,x+2\pi )=u(t,x).\hfill \end{array}$$

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