If a function $x\in {C}^{2}(J,R)$ satisfies equations (1.1) and (1.2), we call *x* a solution of (1.1) and (1.2). Let ${C}^{1}(J,R)$ be a Banach space with the norm $\parallel x\parallel =max\{{|x|}_{0},{|{x}^{\prime}|}_{0}\}$, where ${|x|}_{0}={max}_{t\in J}|x(t)|$, ${|{x}^{\prime}|}_{0}={max}_{t\in J}|{x}^{\prime}(t)|$.

Let

$\lambda >0$,

$\delta (t)\in C(J,R)$ and consider the anti-periodic boundary value problem

$\{\begin{array}{c}{x}^{\u2033}(t)-{\lambda}^{2}x(t)=\delta (t),\phantom{\rule{1em}{0ex}}t\in J,\hfill \\ x(0)+x(T)=0,\phantom{\rule{2em}{0ex}}{x}^{\prime}(0)+{x}^{\prime}(T)=0.\hfill \end{array}$

(2.1)

**Lemma 2.1** *x* *is a solution of* (2.1)

*if and only if* *x* *satisfies* $x(t)={\int}_{0}^{T}G(t,s)\delta (s)\phantom{\rule{0.2em}{0ex}}ds,$

(2.2)

*where*
$G(t,s)=\{\begin{array}{cc}\frac{{e}^{\lambda (t-s)}-{e}^{\lambda (T-t+s)}}{2\lambda (1+{e}^{\lambda T})},\hfill & 0\le s<t\le T,\hfill \\ \frac{{e}^{\lambda (s-t)}-{e}^{\lambda (T+t-s)}}{2\lambda (1+{e}^{\lambda T})},\hfill & 0\le t\le s\le T.\hfill \end{array}$

*Proof* Suppose

$x(t)$ is a solution of (2.1) and denote

$D=\frac{d}{dt}$, then the first equation of (2.1) can be rewritten as

$(D-\lambda )(D+\lambda )x(t)=\delta (t).$

(2.3)

Let

$y(t)=(D+\lambda )x(t),$

(2.4)

then from (2.3), we have

$(D-\lambda )y(t)=\delta (t).$

Multiplying both sides of the above equation by

${e}^{-\lambda t}$ and integrating from 0 to

*t* yields

where $y(0)={x}^{\prime}(0)+\lambda x(0)$.

Similarly, multiplying the two sides of (2.4) by

${e}^{\lambda t}$ and integrating from 0 to

*t* yields

$x(t)={e}^{-\lambda t}[x(0)+{\int}_{0}^{t}{e}^{\lambda s}y(s)\phantom{\rule{0.2em}{0ex}}ds].$

(2.5)

By direct computation, we get

Substituting (2.6) into (2.5),

$\begin{array}{rcl}x(t)& =& \frac{1}{2\lambda}\{({x}^{\prime}(0)+\lambda x(0)){e}^{\lambda t}+(\lambda x(0)-{x}^{\prime}(0)){e}^{-\lambda t}\\ -{\int}_{0}^{t}({e}^{-\lambda (t-s)}-{e}^{\lambda (t-s)})\delta (s)\phantom{\rule{0.2em}{0ex}}ds\}.\end{array}$

(2.7)

Further from (2.7),

$\begin{array}{rcl}{x}^{\prime}(t)& =& \frac{1}{2\lambda}\{({x}^{\prime}(0)+\lambda x(0))\lambda {e}^{\lambda t}+(\lambda x(0)-{x}^{\prime}(0))(-\lambda ){e}^{-\lambda t}\\ -{\int}_{0}^{t}(-\lambda {e}^{-\lambda (t-s)}-\lambda {e}^{\lambda (t-s)})\delta (s)\phantom{\rule{0.2em}{0ex}}ds\},\end{array}$

Taking into account

$x(0)+x(T)=0$,

${x}^{\prime}(0)+{x}^{\prime}(T)=0$, we obtain

${x}^{\prime}(0)+\lambda x(0)=\frac{-1}{{e}^{\lambda T}+1}{\int}_{0}^{T}{e}^{\lambda (T-s)}\delta (s)\phantom{\rule{0.2em}{0ex}}ds,$

(2.8)

and

$\lambda x(0)-{x}^{\prime}(0)=\frac{1}{{e}^{-\lambda T}+1}{\int}_{0}^{T}{e}^{-\lambda (T-s)}\delta (s)\phantom{\rule{0.2em}{0ex}}ds.$

(2.9)

Substituting (2.8) and (2.9) into (2.7), we get

$\begin{array}{rcl}x(t)& =& \frac{1}{2\lambda}\{\frac{-1}{{e}^{\lambda T}+1}{\int}_{0}^{T}{e}^{\lambda (T-s)}\delta (s)\phantom{\rule{0.2em}{0ex}}ds\cdot {e}^{\lambda t}+\frac{1}{{e}^{-\lambda T}+1}{\int}_{0}^{T}{e}^{-\lambda (T-s)}\delta (s)\phantom{\rule{0.2em}{0ex}}ds\cdot {e}^{-\lambda t}\\ -{\int}_{0}^{t}[{e}^{-\lambda (t-s)}-{e}^{\lambda (t-s)}]\delta (s)\phantom{\rule{0.2em}{0ex}}ds\}\\ =& \frac{1}{2\lambda}\{\frac{1}{1+{e}^{\lambda T}}[-{\int}_{0}^{T}{e}^{\lambda (T+t-s)}\delta (s)\phantom{\rule{0.2em}{0ex}}ds+(1+{e}^{\lambda T}){\int}_{0}^{t}{e}^{\lambda (t-s)}\delta (s)\phantom{\rule{0.2em}{0ex}}ds]\\ +\frac{1}{1+{e}^{-\lambda T}}[{\int}_{0}^{T}{e}^{-\lambda (T+t-s)}\delta (s)\phantom{\rule{0.2em}{0ex}}ds-(1+{e}^{-\lambda T}){\int}_{0}^{t}{e}^{-\lambda (t-s)}\delta (s)\phantom{\rule{0.2em}{0ex}}ds]\}\\ =& \frac{1}{2\lambda}\{\frac{1}{1+{e}^{\lambda T}}[{\int}_{0}^{t}{e}^{\lambda (t-s)}\delta (s)\phantom{\rule{0.2em}{0ex}}ds-{\int}_{t}^{T}{e}^{\lambda (T+t-s)}\delta (s)\phantom{\rule{0.2em}{0ex}}ds]\\ +\frac{1}{1+{e}^{-\lambda T}}[-{\int}_{0}^{t}{e}^{-\lambda (t-s)}\delta (s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{t}^{T}{e}^{-\lambda (T+t-s)}\delta (s)\phantom{\rule{0.2em}{0ex}}ds]\}\\ =& \frac{1}{2\lambda}[{\int}_{0}^{t}\frac{{e}^{\lambda (t-s)}}{1+{e}^{\lambda T}}\cdot \delta (s)\phantom{\rule{0.2em}{0ex}}ds-{\int}_{0}^{t}\frac{{e}^{-\lambda (t-s)}}{1+{e}^{-\lambda T}}\cdot \delta (s)\phantom{\rule{0.2em}{0ex}}ds\\ +{\int}_{t}^{T}\frac{{e}^{-\lambda (T+t-s)}}{1+{e}^{-\lambda T}}\cdot \delta (s)\phantom{\rule{0.2em}{0ex}}ds-{\int}_{t}^{T}\frac{{e}^{\lambda (T+t-s)}}{1+{e}^{\lambda T}}\cdot \delta (s)\phantom{\rule{0.2em}{0ex}}ds]\\ =& \frac{1}{2\lambda (1+{e}^{\lambda T})}\{{\int}_{0}^{t}[{e}^{\lambda (t-s)}-{e}^{\lambda (T-t+s)}]\delta (s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{t}^{T}[{e}^{\lambda (s-t)}-{e}^{\lambda (T+t-s)}]\delta (s)\phantom{\rule{0.2em}{0ex}}ds\}\\ =& {\int}_{0}^{T}G(t,s)\delta (s)\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

That is, $x(t)$ is a solution of (2.2).

On the other hand, assume

$x(t)$ is a solution of (2.2). Then

${x}^{\prime}(t)={\int}_{0}^{T}\frac{\partial G(t,s)}{\partial t}\delta (s)\phantom{\rule{0.2em}{0ex}}ds={\int}_{0}^{T}{G}^{\ast}(t,s)\delta (s)\phantom{\rule{0.2em}{0ex}}ds,$

where

${G}^{\ast}(t,s)=\frac{1}{2({e}^{\lambda T}+1)}\{\begin{array}{cc}{e}^{\lambda (t-s)}+{e}^{\lambda (T-t+s)},\hfill & 0\le s<t\le T,\hfill \\ -{e}^{\lambda (s-t)}-{e}^{\lambda (T+t-s)},\hfill & 0\le t\le s\le T.\hfill \end{array}$

And

$\begin{array}{rcl}{x}^{\u2033}(t)& =& {\int}_{0}^{t}\frac{[\lambda {e}^{\lambda (t-s)}-\lambda {e}^{\lambda (T-t+s)}]}{2({e}^{\lambda T}+1)}\delta (s)\phantom{\rule{0.2em}{0ex}}ds+\frac{[1+{e}^{\lambda T}]}{2({e}^{\lambda T}+1)}\delta (t)\\ +{\int}_{t}^{T}\frac{[\lambda {e}^{\lambda (s-t)}-\lambda {e}^{\lambda (T+t-s)}]}{2({e}^{\lambda T}+1)}\delta (s)\phantom{\rule{0.2em}{0ex}}ds-\frac{[-1-{e}^{\lambda T}]}{2({e}^{\lambda T}+1)}\delta (t)\\ =& {\lambda}^{2}{\int}_{0}^{T}G(t,s)\delta (s)\phantom{\rule{0.2em}{0ex}}ds+\delta (t)\\ =& {\lambda}^{2}x(t)+\delta (t).\end{array}$

Direct computation yields

$x(0)+x(T)=0,\phantom{\rule{2em}{0ex}}{x}^{\prime}(0)+{x}^{\prime}(T)=0.$

Hence, $x(t)$ is a solution of (2.1). This proof is complete. □

For later use, we present the following estimations:

$\underset{(t,s)\in J\times J}{max}|G(t,s)|=\frac{{e}^{\lambda T}-1}{2\lambda (1+{e}^{\lambda T})},\phantom{\rule{2em}{0ex}}\underset{(t,s)\in J\times J}{max}|{G}^{\ast}(t,s)|=\frac{1}{2}.$

(2.10)

**Remark 2.1** The integral equation (2.2) we obtained is much simpler than that in [3] which needs a double integral.

Combining Lemma 2.1 and equation (1.1), we can easily get

**Theorem 2.1** *The anti*-

*periodic boundary value problem* (1.1)

*and* (1.2)

*is equivalent to the following integral equation*:

$x(t)={\int}_{0}^{T}G(t,s)[f(s,x(s),{x}^{\prime}(s))-{\lambda}^{2}x(s)]\phantom{\rule{0.2em}{0ex}}ds,$

(2.11)

*where* $\lambda >0$ *and* $G(t,s)$ *is defined in Lemma * 2.1.

Define an operator

$T:{C}^{1}(J,R)\to {C}^{1}(J,R)$ by

$Tx(t):={\int}_{0}^{T}G(t,s)[f(s,x(s),{x}^{\prime}(s))-{\lambda}^{2}x(s)]\phantom{\rule{0.2em}{0ex}}ds.$

(2.12)

**Lemma 2.2** $T:{C}^{1}(J,R)\to {C}^{1}(J,R)$ *is completely continuous*.

*Proof* Noting the continuity of *f*, this follows in a standard step-by-step process and so is omitted. □

In view of Theorem 2.1, we obtain

**Theorem 2.2** $x\in {C}^{2}(J,R)$ *is a solution of the anti*-*periodic boundary value problem* (1.1) *and* (1.2) *if and only if* $x\in {C}^{1}(J,R)$ *is the fixed point of the operator* *T*.