Consider the following linear PIBVP:

$\begin{array}{r}{(p(t){x}^{\prime})}^{\prime}+q(t)x=0,\\ x(0)=x(T),\phantom{\rule{2em}{0ex}}{\int}_{0}^{T}x(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s=0,\end{array}$

(2.1)

here $p(t)\in {C}^{1}(R,R)$ is a given *T*-periodic function in $t\in R$, and $p(t)>0$; $q(t)\in C(R,R)$ is a *T*-periodic function. Assume that

(L1) there exist two constants

*m* and

*M* such that

$m\le \frac{q(t)}{p(t)}\le M$

for all $t\in R$. Moreover, *m* and *M* suit (A2).

**Theorem 2** *Assume that* (*L* 1) *and* (*A* 2) *are satisfied*, *then PIBVP* (2.1) *has only a trivial solution*.

In order to prove Theorem 2, let us give some following concepts.

First, for any interval

$[\alpha ,\beta ]\subset [0,T]$, define

$\begin{array}{rcl}{\mathcal{O}}_{\alpha ,\beta}& =& \{u(t)\in {L}^{2}(0,T):{u}^{\prime}(t)\text{is absolutely continuous on}[\alpha ,\beta ],\\ \text{and}u(t)=0\text{for any}t\in [0,\alpha ]\cup [\beta ,T]\}.\end{array}$

It is clear that

${\mathcal{O}}_{\alpha ,\beta}$ is a linear space with the norm as follows:

$\parallel u\parallel =\underset{t\in [0,T]}{max}|u(t)|+\underset{t\in [0,T]}{max}|{u}^{\prime}(t)|.$

Define a bilinear form on

${\mathcal{O}}_{\alpha ,\beta}$ as follows:

${H}_{\alpha ,\beta}(u,v)={\int}_{0}^{T}[p(t){u}^{\prime}(t){v}^{\prime}(t)-q(t)u(t)v(t)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t$

for any

$u(t)\in {\mathcal{O}}_{\alpha ,\beta}$ and

$v(t)\in {\mathcal{O}}_{\alpha ,\beta}$. Let

where *N* suits assumption (L1), and ${a}_{m}$, ${b}_{m}$, ${c}_{0}$, ${c}_{k}$ and ${d}_{k}$ are some constants. Then ${\mathcal{O}}_{\alpha ,\beta}={\mathcal{X}}_{\alpha ,\beta}\oplus {\mathcal{Y}}_{\alpha ,\beta}$.

From

$p(t)\in {C}^{1}(R,R)$ and

$p(t)>0$, we can obtain that there exist two constants

${M}_{1}$ and

${M}_{2}$ such that

$0\le {M}_{1}\le p(t)\le {M}_{2}$

for all

$t\in R$. Then from assumptions (L1) and (A2), we have

$\begin{array}{rcl}{H}_{\alpha ,\beta}(x,x)& =& {\int}_{0}^{T}p(t)({x}^{\mathrm{\prime}2}(t)-\frac{q(t)}{p(t)}{x}^{2}(t))\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\\ \ge & {\int}_{0}^{T}p(t)({x}^{\mathrm{\prime}2}(t)-M{x}^{2}(t))\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\\ \ge & \frac{2{\pi}^{2}{M}_{1}}{T}\sum _{i=N+1}^{\mathrm{\infty}}({i}^{2}-{(N+1)}^{2})({a}_{i}^{2}+{b}_{i}^{2})\\ \ge & 0\end{array}$

for all

$x\in {\mathcal{X}}_{\alpha ,\beta}$, and

$\begin{array}{rcl}{H}_{\alpha ,\beta}(y,y)& =& {\int}_{0}^{T}p(t)({y}^{\mathrm{\prime}2}(t)-\frac{q(t)}{p(t)}{y}^{2}(t))\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\\ \le & {\int}_{0}^{T}p(t)({y}^{\mathrm{\prime}2}(t)-m{y}^{2}(t))\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\\ \le & \frac{2{\pi}^{2}{M}_{2}}{T}\sum _{k=1}^{N}({k}^{2}-{N}^{2})({c}_{k}^{2}+{d}_{k}^{2})-m{M}_{2}T{c}_{0}^{2}\\ \le & 0\end{array}$

for all $y\in {\mathcal{Y}}_{\alpha ,\beta}$. Thus, ${H}_{\alpha ,\beta}$ is positive definite on ${\mathcal{X}}_{\alpha ,\beta}$ and negative definite on ${\mathcal{Y}}_{\alpha ,\beta}$. By the lemma in [1], we assert that if ${H}_{\alpha ,\beta}(u,v)=0$ for all $u\in {\mathcal{O}}_{\alpha ,\beta}$, then $v\equiv 0$.

For every

*x* on

$[\alpha ,\beta ]$ with

$x(\alpha )=x(\beta )=0$, we introduce an auxiliary function

${x}^{\alpha ,\beta}(t)=\{\begin{array}{cc}x(t),\hfill & t\in [\alpha ,\beta ],\hfill \\ 0,\hfill & t\in [0,\alpha ]\cup [\beta ,T].\hfill \end{array}$

The following lemma is very useful in our proofs.

**Lemma 1** *If* $p(t)$,

$q(t)$ *are continuous and satisfy* (

*L* 1)

*and* (

*A* 2),

*then the following two points boundary value problem* $\begin{array}{r}{(p(t){x}^{\prime})}^{\prime}+q(t)x=0,\\ x(\alpha )=x(\beta )=0\end{array}$

(2.2)

*has only a trivial solution*.

*Proof* It is clear that 0 is a solution of two points boundary value problem (2.2). If

$v(t)$ is a solution of problem (2.2), then

${v}^{\alpha ,\beta}(t)\in {\mathcal{O}}_{\alpha ,\beta}$. For any

$u\in {\mathcal{O}}_{\alpha ,\beta}$, we have

${\int}_{\alpha}^{\beta}[{(p(t){v}^{\prime}(t))}^{\prime}u(t)+q(t)u(t)v(t)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t=0$

by using (2.2). Integrating the first terms by parts, we derive

$-{H}_{\alpha ,\beta}(u,{v}^{\alpha ,\beta})=-{\int}_{0}^{T}[p(t){u}^{\prime}(t){{v}^{\alpha ,\beta}}^{\prime}(t)-q(t)u(t){v}^{\alpha ,\beta}(t)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t=0.$

By assumption (L1), ${H}_{\alpha ,\beta}$ is positive definite on ${\mathcal{X}}_{\alpha ,\beta}$ and negative definite on ${\mathcal{Y}}_{\alpha ,\beta}$. These show ${v}^{\alpha ,\beta}(t)\equiv 0$ for $t\in [0,T]$, *that is*, $v(t)\equiv 0$ for $t\in [\alpha ,\beta ]$. The proof of Lemma 1 is ended. □

*Proof of Theorem 2* It is clear that PIBVP (2.1) has at least one solution, for example, ${x}_{\ast}\equiv 0$. Assume that PIBVP (2.1) possesses a nontrivial solution ${x}^{\ast}\ne 0$. The proof is divided into three parts.

Case 1: ${x}^{\ast}(0)={x}^{\ast}(T)=0$. By Lemma 1 ($\alpha =0$ and $\beta =T$), PIBVP (2.1) has only a trivial solution. This contradicts ${x}^{\ast}\ne 0$.

Case 2:

${x}^{\ast}(0)={x}^{\ast}(T)=\eta >0$. Denote

$S=\{t\in [0,T]:{x}^{\ast}(t)=0\}.$

Take

$a=\underset{t\in S}{inf}t\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}b=\underset{t\in S}{sup}t.$

From

${\int}_{0}^{2\pi}{x}^{\ast}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s=0$, there are at least two points in the set

*S*, which implies that

$0<a<b<2\pi $ and

${x}^{\ast}(a)={x}^{\ast}(b)=0$. By Lemma 1 (

$\alpha =a$ and

$\beta =b$) the two points boundary value problem

$\begin{array}{r}{(p(t){x}^{\prime})}^{\prime}+q(t)x=0,\\ x(a)=x(b)=0\end{array}$

(2.3)

only has a trivial solution. Hence we obtain

${x}^{\ast}(t)\equiv 0$,

$t\in [a,b]$. By the definitions of

*a* and

*b*, one has

${x}^{\ast}(t)>0\phantom{\rule{1em}{0ex}}\text{for all}t\in [0,a)\cup (b,T].$

From

${\int}_{0}^{T}{x}^{\ast}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s=0$, we get

${\int}_{0}^{a}{x}^{\ast}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s+{\int}_{b}^{T}{x}^{\ast}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s=0.$

This contradicts ${\int}_{0}^{a}{x}^{\ast}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s>0$ and ${\int}_{b}^{T}{x}^{\ast}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s>0$.

Case 3: ${x}^{\ast}(0)={x}^{\ast}(T)=\eta <0$. This case is similar to Case 2.

Thus, we complete the proof of Theorem 2. □

**Theorem 3** *If* $p(t)$,

$q(t)$ *are continuous and satisfy* (

*L* 1)

*and* (

*A* 2),

*then the following PIBVP* $\begin{array}{r}{(p(t){x}^{\prime})}^{\prime}+q(t)x=h(t),\\ x(0)=x(T),\phantom{\rule{2em}{0ex}}{\int}_{0}^{T}x(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s=0\end{array}$

(2.4)

*has a unique solution*.

*Proof* Let

${x}_{\ast 1}(t)$ and

${x}_{\ast 2}(t)$ be two linear independent solutions of the following linear equation:

${(p(t){x}^{\prime})}^{\prime}+q(t)x=0.$

Assume that

${x}_{\ast}(t)={c}_{1}{x}_{\ast 1}(t)+{c}_{2}{x}_{\ast 2}(t)$ is a solution of PIBVP (2.1), where

${c}_{1}$ and

${c}_{2}$ are constants. Then by the boundary value conditions of (2.1),

$\{\begin{array}{c}({x}_{\ast 1}(0)-{x}_{\ast 1}(T)){c}_{1}+({x}_{\ast 2}(0)-{x}_{\ast 2}(T)){c}_{2}=0,\hfill \\ {\int}_{0}^{T}{x}_{\ast 1}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s{c}_{1}+{\int}_{0}^{T}{x}_{\ast 2}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s{c}_{2}=0.\hfill \end{array}$

By Theorem 3, PIBVP (2.1) has only a trivial solution, which shows

$\left|\begin{array}{cc}{x}_{\ast 1}(0)-{x}_{\ast 1}(T)& {x}_{\ast 2}(0)-{x}_{\ast 2}(T)\\ {\int}_{0}^{T}{x}_{\ast 1}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s& {\int}_{0}^{T}{x}_{\ast 2}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\end{array}\right|\ne 0.$

(2.5)

Let

${x}_{\star}(t)={c}_{3}{x}_{\ast 1}(t)+{c}_{4}{x}_{\ast 2}(t)+{x}_{\ast 0}(t)$ be a solution of PIBVP (2.4), where

${x}_{\ast 0}(t)$ is a solution of the equation

${(p(t){x}^{\prime})}^{\prime}+q(t)x=h(t).$

From the boundary value conditions, we have

$\{\begin{array}{c}({x}_{\ast 1}(0)-{x}_{\ast 1}(T)){c}_{3}+({x}_{\ast 2}(0)-{x}_{\ast 2}(T)){c}_{4}={x}_{\ast 0}(0)-{x}_{\ast 0}(T),\hfill \\ {\int}_{0}^{T}{x}_{\ast 1}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s{c}_{3}+{\int}_{0}^{T}{x}_{\ast 2}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s{c}_{4}={\int}_{0}^{T}{x}_{\ast 0}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s.\hfill \end{array}$

From (2.5) constants ${c}_{3}$, ${c}_{4}$ are unique. Thus, PIBVP (2.4) has only one solution. □