In this section, some existence theorems are presented.

**Theorem 4.1** *Let* {\alpha}_{1}, {\beta}_{1} *and* {K}_{1} *be nonnegative constants and* \lambda >0. *If* *f* *is continuous and satisfies*

|f(t,p,q)-{\lambda}^{2}p|\le {\alpha}_{1}|p|+{\beta}_{1}|q|+{K}_{1},\phantom{\rule{1em}{0ex}}(t,p,q)\in J\times R\times R,

*with*

\lambda ({e}^{\lambda T}+1)-{\beta}_{1}({e}^{\lambda T}-1)>0,

*and*

\frac{{e}^{\lambda T}+1}{{e}^{\lambda T}-1}{\lambda}^{2}-\lambda {\beta}_{1}-{\alpha}_{1}>0,

*then the anti*-*periodic boundary problem* (1.1) *and* (1.2) *has at least one solution*.

*Proof* Consider the family (3.1). We want to show the conditions of Theorem 3.2 hold for some positive constants *M* and *N*.

Let x(t) be a solution to (3.1) and consider the equivalent equation (3.2), that is,

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

(4.1)

For each t\in [0,T], we have

\begin{array}{rcl}|x(t)|& =& |\mu {\int}_{0}^{T}G(t,s)[f(s,x(s),{x}^{\prime}(s))-{\lambda}^{2}x(s)]\phantom{\rule{0.2em}{0ex}}ds|\\ \le & {\int}_{0}^{T}|G(t,s)|\cdot |f(s,x(s),{x}^{\prime}(s))-{\lambda}^{2}x(s)|\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\int}_{0}^{T}|G(t,s)|[{\alpha}_{1}|x(s)|+{\beta}_{1}|{x}^{\prime}(s)|+{K}_{1}]\phantom{\rule{0.2em}{0ex}}ds.\end{array}

Since

\begin{array}{rcl}{\int}_{0}^{T}|G(t,s)|\phantom{\rule{0.2em}{0ex}}ds& =& {\int}_{0}^{t}|G(t,s)|\phantom{\rule{0.2em}{0ex}}ds+{\int}_{t}^{T}|G(t,s)|\phantom{\rule{0.2em}{0ex}}ds\\ =& {\int}_{0}^{t}\frac{|{e}^{\lambda (t-s)}-{e}^{\lambda (T-t+s)}|}{2\lambda (1+{e}^{\lambda T})}\phantom{\rule{0.2em}{0ex}}ds+{\int}_{t}^{T}\frac{|{e}^{\lambda (s-t)}-{e}^{\lambda (T+t-s)}|}{2\lambda (1+{e}^{\lambda T})}\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\int}_{0}^{t}\frac{{e}^{\lambda (t-s)}+{e}^{\lambda (T-t+s)}}{2\lambda (1+{e}^{\lambda T})}\phantom{\rule{0.2em}{0ex}}ds+{\int}_{t}^{T}\frac{{e}^{\lambda (s-t)}+{e}^{\lambda (T+t-s)}}{2\lambda (1+{e}^{\lambda T})}\phantom{\rule{0.2em}{0ex}}ds\\ =& \frac{-{e}^{\lambda (t-s)}+{e}^{\lambda (T-t+s)}}{2{\lambda}^{2}(1+{e}^{\lambda T})}{|}_{0}^{t}+\frac{{e}^{\lambda (s-t)}-{e}^{\lambda (T+t-s)}}{2{\lambda}^{2}(1+{e}^{\lambda T})}{|}_{t}^{T}\\ =& \frac{{e}^{\lambda T}-1}{{\lambda}^{2}({e}^{\lambda T}+1)},\end{array}

it follows that

{|x|}_{0}\le \frac{{e}^{\lambda T}-1}{{\lambda}^{2}({e}^{\lambda T}+1)}[{\alpha}_{1}{|x|}_{0}+{\beta}_{1}{\left|{x}^{\prime}\right|}_{0}+{K}_{1}].

(4.2)

Differentiating both sides of (4.1), we get

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

then

|{x}^{\prime}(t)|\le {\int}_{0}^{T}|{G}^{\ast}(t,s)|[{\alpha}_{1}{|x|}_{0}+{\beta}_{1}{\left|{x}^{\prime}\right|}_{0}+{K}_{1}]\phantom{\rule{0.2em}{0ex}}ds,

and because of

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

Therefore,

{\left|{x}^{\prime}\right|}_{0}\le \frac{{e}^{\lambda T}-1}{\lambda ({e}^{\lambda T}+1)}[{\alpha}_{1}{|x|}_{0}+{\beta}_{1}{\left|{x}^{\prime}\right|}_{0}+{K}_{1}].

The rearrangement yields

{\left|{x}^{\prime}\right|}_{0}\le \frac{({e}^{\lambda T}-1)({\alpha}_{1}{|x|}_{0}+{K}_{1})}{\lambda ({e}^{\lambda T}+1)-{\beta}_{1}({e}^{\lambda T}-1)}.

(4.3)

By substituting (4.3) into (4.2) and rearranging, we obtain

{|x|}_{0}\le \frac{{K}_{1}}{\frac{{e}^{\lambda T}+1}{{e}^{\lambda T}-1}{\lambda}^{2}-\lambda {\beta}_{1}-{\alpha}_{1}}:={M}_{1}.

So,

{\left|{x}^{\prime}\right|}_{0}\le \frac{({e}^{\lambda T}-1)({\alpha}_{1}{M}_{1}+{K}_{1})}{\lambda ({e}^{\lambda T}+1)-{\beta}_{1}({e}^{\lambda T}-1)}:={N}_{1}.

Hence, Theorem 3.2 holds for positive constants M={M}_{1}+1 and N={N}_{1}+1. The solvability of (1.1) and (1.2) now follows. □

**Theorem 4.2** *Assume there exist nonnegative constants* {\alpha}_{2}, {K}_{2} *and* \lambda >0 *such that*

|f(t,p,q)-{\lambda}^{2}p|<{\alpha}_{2}[pf(t,p,q)+{q}^{2}]+{K}_{2},\phantom{\rule{1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.1em}{0ex}}(t,p,q)\in J\times R\times R,

*then the anti*-*periodic boundary value problem* (1.1) *and* (1.2) *has at least one solution*.

*Proof* Suppose x(t) is a solution of (3.1), and in view of (2.10), we have

\begin{array}{rcl}{|x|}_{0}& =& \underset{t\in J}{max}|\mu {\int}_{0}^{T}G(t,s)[f(s,x(s),{x}^{\prime}(s))-{\lambda}^{2}x(s)]\phantom{\rule{0.2em}{0ex}}ds|\\ \le & \mu \underset{t\in J}{max}{\int}_{0}^{T}|G(t,s)|\cdot \left|[f(s,x(s),{x}^{\prime}(s))-{\lambda}^{2}x(s)]\right|\phantom{\rule{0.2em}{0ex}}ds\\ \le & \underset{t\in J}{max}{\int}_{0}^{T}|G(t,s)|[\mu {\alpha}_{2}(x(s)f(s,x(s),{x}^{\prime}(s))+{|{x}^{\prime}(s)|}^{2})+\mu {K}_{2}]\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\int}_{0}^{T}\underset{t\in J}{max}|G(t,s)|\left[{\alpha}_{2}\right(x(s)(\mu f(s,x(s),{x}^{\prime}(s))\\ +{\lambda}^{2}(1-\mu )x(s))+{|{x}^{\prime}(s)|}^{2})+{K}_{2}]\phantom{\rule{0.2em}{0ex}}ds\\ \le & \frac{{e}^{\lambda T}-1}{2\lambda (1+{e}^{\lambda T})}{\int}_{0}^{T}[{\alpha}_{2}(x(s){x}^{\u2033}(s)+{|{x}^{\prime}(s)|}^{2})+{K}_{2}]\phantom{\rule{0.2em}{0ex}}ds\\ =& \frac{{e}^{\lambda T}-1}{2\lambda (1+{e}^{\lambda T})}{\int}_{0}^{T}[{\alpha}_{2}\frac{d}{ds}({x}^{\prime}(s)x(s))+{K}_{2}]\phantom{\rule{0.2em}{0ex}}ds\\ =& \frac{{e}^{\lambda T}-1}{2\lambda (1+{e}^{\lambda T})}[{\alpha}_{2}(x(T){x}^{\prime}(T)-x(0){x}^{\prime}(0))+{K}_{2}T]\\ \le & \frac{{e}^{\lambda T}-1}{2\lambda (1+{e}^{\lambda T})}{K}_{2}T:={M}_{2}.\end{array}

Similarly,

\begin{array}{rcl}{\left|{x}^{\prime}\right|}_{0}& =& |\mu \underset{t\in J}{max}{\int}_{0}^{T}{G}^{\ast}(t,s)[f(s,x(s),{x}^{\prime}(s))-{\lambda}^{2}x(s)]\phantom{\rule{0.2em}{0ex}}ds|\\ \le & \underset{t\in J}{max}{\int}_{0}^{T}|{G}^{\ast}(t,s)|[{\alpha}_{2}\mu (x(s)f(s,x(s),{x}^{\prime}(s))+{|{x}^{\prime}(s)|}^{2})+{K}_{2}]\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\int}_{0}^{T}\underset{t\in J}{max}|{G}^{\ast}(t,s)|\left[{\alpha}_{2}\right(x(s)(\mu f(s,x(s),{x}^{\prime}(s))\\ +{\lambda}^{2}(1-\mu )x(s))+{|{x}^{\prime}(s)|}^{2})+{K}_{2}]\phantom{\rule{0.2em}{0ex}}ds\\ \le & \frac{1}{2}{\int}_{0}^{T}[{\alpha}_{2}(x(s){x}^{\u2033}(s)+{|{x}^{\prime}(s)|}^{2})+{K}_{2}]\phantom{\rule{0.2em}{0ex}}ds\\ \le & \frac{1}{2}{\int}_{0}^{T}[{\alpha}_{2}(x(s){x}^{\u2033}(s)+{|{x}^{\prime}(s)|}^{2})+{K}_{2}]\phantom{\rule{0.2em}{0ex}}ds\\ =& \frac{1}{2}{\alpha}_{2}[x(T){x}^{\prime}(T)-x(0){x}^{\prime}(0)]+\frac{1}{2}{K}_{2}T.\\ =& \frac{1}{2}{K}_{2}T:={N}_{2}.\end{array}

Therefore, Theorem 3.2 holds for positive constants M={M}_{2}+1 and N={N}_{2}+1. The solvability of (1.1) and (1.2) now follows. □

**Example 4.1** Consider the anti-periodic boundary value problem

\{\begin{array}{c}{x}^{\u2033}(t)=x(t)+x(t){({x}^{\prime}(t))}^{2}+sint,\phantom{\rule{1em}{0ex}}t\in [0,1],\hfill \\ x(0)=-x(1),\phantom{\rule{2em}{0ex}}{x}^{\prime}(0)=-{x}^{\prime}(1).\hfill \end{array}

(4.4)

We claim (4.4) has at least one solution.

*Proof* Let T=1, and f(t,p,q)=p+p{q}^{2}+sint in Theorem 4.2. Choose \lambda =1, we get for (t,p,q)\in [0,1]\times {R}^{2} that

|f(t,p,q)-p|=|p{q}^{2}+sint|\le |p|{q}^{2}+sint,

and

pf(t,p,q)+{q}^{2}={p}^{2}+{p}^{2}{q}^{2}+psint+{q}^{2}\ge {p}^{2}+{p}^{2}{q}^{2}+{q}^{2}-|p|.

Note {min}_{v\ge 0}\{{v}^{2}-v\}>-1, we have {p}^{2}{q}^{2}+{q}^{2}-|p|{q}^{2}={q}^{2}({p}^{2}-|p|+1)>0. Thus, for {\alpha}_{2}=1, {K}_{2}=2

\begin{array}{rcl}|f(t,p,q)-p|& \le & |p|{q}^{2}+sint\le |p|{q}^{2}+1\\ \le & {q}^{2}+{p}^{2}{q}^{2}+1\\ \le & {q}^{2}+{p}^{2}{q}^{2}+{p}^{2}-|p|+2\\ \le & {\alpha}_{2}[pf(t,p,q)+{q}^{2}]+{K}_{2}.\end{array}

Then the conclusion follows from Theorem 4.2. □

Now, we reconsider the problem (1.2) and (1.3). The following result is obtained.

**Theorem 4.3** *Suppose* f:[0,T]\times R\to R *is continuous*. *If there exist nonnegative constants* {\alpha}_{3}, {K}_{3} *and* \lambda >0 *such that*

|f(t,p)-{\lambda}^{2}p|<{\alpha}_{3}pf(t,p)+{K}_{3},\phantom{\rule{1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.1em}{0ex}}(t,p)\in [0,T]\times R,

*then* (1.2) *and* (1.3) *has at least one solution*.

*Proof* The proof is similar to Theorem 4.2 and here we omit it. □

An example to highlight the Theorem 4.3 is presented.

**Example 4.2** Consider the anti-periodic boundary value problem given by

\{\begin{array}{c}{x}^{\u2033}(t)={x}^{3}+x+t,\phantom{\rule{1em}{0ex}}t\in [0,10],\hfill \\ x(0)+x(10)=0,\phantom{\rule{2em}{0ex}}{x}^{\prime}(0)+{x}^{\prime}(10)=0.\hfill \end{array}

(4.5)

We claim (4.5) has at least one solution.

*Proof* Let f(t,p)={p}^{3}+p+t and see that |f(t,p)-p|\le {|p|}^{3}+10 for (t,p)\in [0,10]\times R. For {\alpha}_{3}, {K}_{3} and *λ* to be chosen below, see that

Thus, the conditions of Theorem 4.3 hold and the solvability follows. □

**Remark 4.1** The results of [3] do not apply to the above example since |f(t,p)| grows more than linearly in |p|. Therefore, we improve the previous results.

Finally, in order to illustrate our main results, we use the ‘bvp4c’ package in MATLAB to simulate. As shown in Figure 1(a) and (b), numerical simulations also suggest that Examples 4.1 and 4.2 with the given coefficients admit at least one solution.