### 3.1 First-order periodic boundary value problems

We consider the following first-order periodic boundary value problem:

$\{\begin{array}{c}{x}^{\prime}(t)=f(t,x(t)),\phantom{\rule{1em}{0ex}}t\in (0,1),\hfill \\ x(0)=x(1),\hfill \end{array}$

(3.1)

where $f:[0,1]\times [0,+\mathrm{\infty})\to \mathbb{R}$ is continuous and $f(0,x)=f(1,x)$ for all $x\in \mathbb{R}$.

Consider the Banach spaces

$X=Y=C[0,1]$ endowed with the norm

$\parallel x\parallel ={max}_{t\in [0,1]}|x(t)|$. Define the cone

*K* in X by

$K=\{x\in X:x(t)\ge 0,t\in [0,1]\}.$

Let

*L* be the linear operator from

$domL\subset X$ to

*Y* with

$domL=\{x\in X:{x}^{\prime}\in C[0,1],x(0)=x(1)\},$

and

$Lx(t)={x}^{\prime}(t),\phantom{\rule{1em}{0ex}}x\in domL,t\in [0,1].$

Let us define

$N:X\to Y$ by

$Nx(t)=f(t,x(t)),\phantom{\rule{1em}{0ex}}t\in [0,1].$

Then (3.1) is equivalent to the equation

It is obvious that

*L* is a Fredholm operator of index zero with

Next we define the projections

$P:X\to X$,

$Q:Y\to Y$ by

and the isomorphism

$J:ImQ\to ImP$ as

$Jy=y$. Note that for

$y\in ImL$, the inverse operator

${L}_{1}^{-1}:ImL\to domL\cap KerP$

of

$L{|}_{domL\cap KerP}:domL\cap KerP\to ImL$

is given by

$\left({L}_{1}^{-1}y\right)(t)={\int}_{0}^{1}K(t,s)y(s)\phantom{\rule{0.2em}{0ex}}ds,$

where

$K(t,s)=\{\begin{array}{cc}s+1,\hfill & 0\le s<t\le 1,\hfill \\ s,\hfill & 0\le t\le s\le 1.\hfill \end{array}$

Set

$G(t,s)=1+K(t,s)-{\int}_{0}^{1}K(t,s)\phantom{\rule{0.2em}{0ex}}ds.$

We can verify that

$G(t,s)=\{\begin{array}{cc}\frac{3}{2}-(t-s),\hfill & 0\le s<t\le 1,\hfill \\ \frac{1}{2}+(s-t),\hfill & 0\le t\le s\le 1,\hfill \end{array}$

and

$\frac{1}{2}\le G(t,s)\le \frac{3}{2},\phantom{\rule{1em}{0ex}}t,s\in [0,1].$

To state the existence result, we introduce two conditions:

(H_{1}) $f(t,b)<0$ for all $t\in [0,1]$,

(H_{2}) $f(t,x)>0$ for all $(t,x)\in [0,1]\times [0,a]$.

**Theorem 3.1** *Assume that there exist two positive numbers* $0<a<b$ *such that* (H_{1}), (H_{2}) *and*

(H_{3}) $f(t,x)\ge -\frac{2}{3}x$ *for all* $(t,x)\in [0,1]\times [0,b]$

*hold*. *Then* (3.1) *has at least one positive periodic solution* ${x}^{\ast}\in K$ *with* $a\le \parallel {x}^{\ast}\parallel \le b$.

*Proof* First, we note that *L*, as defined, is Fredholm of index zero, ${L}_{1}^{-1}$ is compact by the Arzela-Ascoli theorem and thus $L-\lambda N$ is A-proper for $\lambda \in [0,1]$ by [[20], Lemma 2(a)].

For each

$x\in K$, then by condition (H

_{3}),

Thus $(P+JQN+{L}_{1}^{-1}(I-Q)N)(K)\subset K$.

Let

${\mathrm{\Omega}}_{1}=\{x\in X:\parallel x\parallel <a\},\phantom{\rule{2em}{0ex}}{\mathrm{\Omega}}_{2}=\{x\in X:\parallel x\parallel <b\}.$

Clearly,

${\mathrm{\Omega}}_{1}$ and

${\mathrm{\Omega}}_{2}$ are bounded open sets and

$\theta \in {\mathrm{\Omega}}_{1}\subset {\overline{\mathrm{\Omega}}}_{1}\subset {\mathrm{\Omega}}_{2}.$

We now show that

$(P+JQN)x+{L}_{1}^{-1}(I-Q)Nx\ngeqq x\phantom{\rule{1em}{0ex}}\text{for any}x\in \partial {\mathrm{\Omega}}_{2}\cap K.$

(3.2)

In fact, if there exists

${x}_{3}\in \partial {\mathrm{\Omega}}_{2}\cap K$ such that

$(P+JQN){x}_{3}+{L}_{1}^{-1}(I-Q)N{x}_{3}\ge {x}_{3}.$

Then

${x}_{3}^{\prime}(t)\le f(t,{x}_{3}(t)),\phantom{\rule{1em}{0ex}}t\in [0,1].$

Let

${t}_{1}\in [0,1]$ be such that

${x}_{3}({t}_{1})=b$. Clearly, the function

${x}_{3}^{2}$ attains a maximum on

$[0,1]$ at

$t={t}_{1}$. Therefore

$2{x}_{3}({t}_{1}){x}_{3}^{\prime}({t}_{1})=0$. As a consequence,

$0=2b{x}_{3}^{\prime}({t}_{1})\le 2bf({t}_{1},{x}_{3}({t}_{1}))=2bf({t}_{1},b),$

which is a contradiction to (H_{1}). Therefore (3.2) holds.

On the other hand, we claim that

$(P+JQN)x+{L}_{1}^{-1}(I-Q)Nx\nleqq x\phantom{\rule{1em}{0ex}}\text{for any}x\in \partial {\mathrm{\Omega}}_{1}\cap K.$

(3.3)

In fact, if not, there exists

${x}_{4}\in \partial {\mathrm{\Omega}}_{1}\cap K$ such that

$(P+JQN){x}_{4}+{L}_{1}^{-1}(I-Q)N{x}_{4}\le {x}_{4}.$

For any

${x}_{4}\in \partial {\mathrm{\Omega}}_{1}\cap K$, we have

$\parallel {x}_{4}\parallel =a$, then

$0\le {x}_{4}(t)\le a$ for

$t\in [0,1]$. By condition (H

_{2}), we have

$\begin{array}{rcl}{x}_{4}(t)& \ge & (P+JQN){x}_{4}(t)+{L}_{1}^{-1}(I-Q)N{x}_{4}(t)\\ =& {\int}_{0}^{1}{x}_{4}(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{1}G(t,s)f(s,{x}_{4}(s))\phantom{\rule{0.2em}{0ex}}ds\\ >& {\int}_{0}^{1}{x}_{4}(s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\text{for any}t\in [0,1],\end{array}$

which is a contradiction. As a result, (3.3) is verified.

It follows from (3.2), (3.3) and Theorem 2.1 that there exists ${x}^{\ast}\in K\cap ({\overline{\mathrm{\Omega}}}_{2}\mathrm{\setminus}{\mathrm{\Omega}}_{1})$ such that $L{x}^{\ast}=N{x}^{\ast}$ with $a\le \parallel {x}^{\ast}\parallel \le b$. □

**Remark 3.1** In [18], the following condition is required instead of (H_{2}):

(H

^{∗}) there exist

$a\in (0,b)$,

${t}_{0}\in [0,1]$,

$r\in (0,1]$, and continuous functions

$g:[0,1]\to [0,\mathrm{\infty})$,

$h:(0,a]\to [0,\mathrm{\infty})$ such that

$f(t,x)\ge g(t)h(x)$ for all

$t\in [0,1]$ and

$x\in (0,a]$,

$h(x)/{x}^{r}$ is nonincreasing on

$(0,a]$ with

$\frac{h(a)}{{2}^{r-1}}{\int}_{0}^{1}G({t}_{0},s)g(s)\phantom{\rule{0.2em}{0ex}}ds\ge a.$

Obviously, our condition (H_{2}) is much weaker and less strict compared with (H^{∗}). Moreover, (H_{2}) is easier to check than (H^{∗}). So, our result generalizes and improves [[18], Theorem 5].

**Remark 3.2** From the proof of Theorem 3.1, we can see that condition (H_{2}) can be replaced by one of the following two relatively weaker conditions:

(${\mathrm{H}}_{2}^{\ast}$) $f(t,x)\ge 0$ for all $(t,x)\in [0,1]\times [0,a]$ and $f(t,\cdot )$ is positive for almost everywhere on $[0,a]$.

(${\mathrm{H}}_{2}^{\ast \ast}$) ${lim}_{x\to {0}^{+}}{min}_{t\in [0,1]}f(t,x)>0$.

**Remark 3.3** Finally in this section, we note that conditions (H_{1}) and (H_{2}) can be replaced by the following asymptotic conditions:

(${\mathrm{H}}_{1}^{\prime}$) ${lim}_{x\to +\mathrm{\infty}}\frac{f(t,x)}{x}<0$ uniformly for *t*;

(${\mathrm{H}}_{2}^{\prime}$) ${lim}_{x\to {0}^{+}}\frac{f(t,x)}{x}>0$ uniformly for *t*.

**Example 3.1** Let the nonlinearity in (3.1) be

$f(t,x)=c(t){x}^{\alpha}+\mu d(t){x}^{\beta}-kx,$

where $0<\alpha <1<\beta $, $c(t),d(t)\in C[0,1]$ are positive 1-periodic functions, $k\in (0,2/3)$ and $\mu >0$ is a positive parameter. Then (3.1) has at least one positive 1-periodic solution for each $0<\mu <{\mu}^{\ast}$, here ${\mu}^{\ast}$ is some positive constant.

*Proof* We will apply Theorem 3.1 with

$f(t,x)=c(t){x}^{\alpha}+\mu d(t){x}^{\beta}-kx$. Since

$k\in (0,2/3)$, it is easy to see that (H

_{3}) holds. Set

$T(x)=\frac{kx-{c}^{\ast}{x}^{\alpha}}{{d}^{\ast}{x}^{\beta}},$

where

${c}^{\ast}=\underset{t}{max}c(t),\phantom{\rule{2em}{0ex}}{d}^{\ast}=\underset{t}{max}d(t).$

Since

$0<\alpha <1<\beta $, we have

$T\left({0}^{+}\right)=-\mathrm{\infty},\phantom{\rule{2em}{0ex}}T(+\mathrm{\infty})=0.$

One may easily see that there exists

$b>0$ such that

$T(b)=\frac{kb-{c}^{\ast}{b}^{\alpha}}{{d}^{\ast}{b}^{\beta}}=\underset{x>0}{sup}T(x)>0.$

Let

${\mu}^{\ast}=\frac{kb-{c}^{\ast}{b}^{\alpha}}{{d}^{\ast}{b}^{\beta}}.$

Then, for each

$\mu \in (0,{\mu}^{\ast})$, we have

$\begin{array}{rcl}f(t,b)& =& c(t){b}^{\alpha}+\mu d(t){b}^{\beta}-kb\\ <& {c}^{\ast}{b}^{\alpha}+{\mu}^{\ast}{d}^{\ast}{b}^{\beta}-kb\\ =& 0,\end{array}$

which implies that (H_{1}) holds.

On the other hand, we have

which implies that (${\mathrm{H}}_{2}^{\prime}$) holds. Now we have the desired result. □

### 3.2 Second-order periodic boundary value problems

Let

$f:[0,1]\times [0,+\mathrm{\infty})\to \mathbb{R}$ be continuous and

$f(0,x)=f(1,x)$ for all

$x\in \mathbb{R}$. We will discuss the existence of positive solutions of the second-order periodic boundary value problem

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

(3.4)

Since some parts of the proof are in the same line as that of Theorem 3.1, we will outline the proof with the emphasis on the difference.

Let

*X*,

*Y* be Banach spaces and the cone

*K* be as in Section 3.1. In this case, we may define

$domL=\{x\in X:{x}^{\u2033}\in C[0,1],x(0)=x(1),{x}^{\prime}(0)={x}^{\prime}(1)\},$

and let the linear operator

$L:domL\to Y$ be defined by

$Lx=-{x}^{\u2033},\phantom{\rule{1em}{0ex}}\text{for}x\in domL.$

Then

*L* is Fredholm of index zero,

$KerL=\{x\in domL:x(t)\equiv \mathrm{constants}\},$

and

$ImL=\{y\in Y:{\int}_{0}^{1}y(s)\phantom{\rule{0.2em}{0ex}}ds=0\}.$

Thus it is clear that (3.4) is equivalent to

We use the same projections

*P*,

*Q* as in Section 3.1 and define the isomorphism

$J:ImQ\to ImP$ as

where

$\beta =\frac{1}{6}$. It is easy to verify that the inverse operator

${L}_{1}^{-1}:ImL\to domL\cap KerP$ of

$L{|}_{domL\cap KerP}:domL\cap KerP\to ImL$ is

$\left({L}_{1}^{-1}y\right)(t)={\int}_{0}^{1}\mathrm{\Lambda}(t,s)y(s)\phantom{\rule{0.2em}{0ex}}ds,$

where

$\mathrm{\Lambda}(t,s)=\{\begin{array}{cc}\frac{s}{2}(1-2t+s),\hfill & 0\le s<t\le 1,\hfill \\ \frac{1}{2}(1-s)(2t-s),\hfill & 0\le t\le s\le 1.\hfill \end{array}$

Set

$H(t,s)=\frac{1}{6}+\mathrm{\Lambda}(t,s)-{\int}_{0}^{1}\mathrm{\Lambda}(t,s)\phantom{\rule{0.2em}{0ex}}ds.$

We can verify that

$H(t,s)=\{\begin{array}{cc}\frac{1}{4}+\frac{s}{2}(1-2t+s)+\frac{{t}^{2}}{2}-\frac{t}{2},\hfill & 0\le s<t\le 1,\hfill \\ \frac{1}{4}+\frac{1}{2}(1-s)(2t-s)+\frac{{t}^{2}}{2}+\frac{t}{2},\hfill & 0\le t\le s\le 1,\hfill \end{array}$

and

$\frac{1}{8}\le H(t,s)\le \frac{1}{4},\phantom{\rule{1em}{0ex}}t,s\in [0,1].$

**Theorem 3.2** *Assume that there exist two positive numbers* $0<a<b$ *such that* (H_{1}), (H_{2}) *and*

(H_{4}) $f(t,x)\ge -4x$ *for all* $(t,x)\in [0,1]\times [0,b]$

*hold*. *Then* (3.4) *has at least one positive periodic solution* ${x}^{\ast}\in K$ *with* $a\le \parallel {x}^{\ast}\parallel \le b$.

*Proof* It is again easy to show that $L-\lambda N$ is A-proper for $\lambda \in [0,1]$ by [[20], Lemma 2(a)].

For each

$x\in K$, then by condition (H

_{4}),

Thus $(P+JQN+{L}_{1}^{-1}(I-Q)N)(K)\subset K$.

Let

${\mathrm{\Omega}}_{3}=\{x\in X:\parallel x\parallel <a\},\phantom{\rule{2em}{0ex}}{\mathrm{\Omega}}_{4}=\{x\in X:\parallel x\parallel <b\}.$

Clearly,

${\mathrm{\Omega}}_{3}$ and

${\mathrm{\Omega}}_{4}$ are bounded and open sets and

$\theta \in {\mathrm{\Omega}}_{3}\subset {\overline{\mathrm{\Omega}}}_{3}\subset {\mathrm{\Omega}}_{4}.$

Next, we show that

$(P+JQN)x+{L}_{1}^{-1}(I-Q)Nx\ngeqq x,\phantom{\rule{1em}{0ex}}\text{for any}x\in \partial {\mathrm{\Omega}}_{4}\cap K.$

(3.5)

On the contrary, suppose that there exists

${x}_{5}\in \partial {\mathrm{\Omega}}_{4}\cap K$ such that

$(P+JQN){x}_{5}+{L}_{1}^{-1}(I-Q)N{x}_{5}\ge {x}_{5}.$

Then

$-{x}_{5}^{\u2033}(t)\le f(t,{x}_{5}(t)),\phantom{\rule{1em}{0ex}}t\in [0,1].$

Let

${t}_{2}\in [0,1]$ such that

${x}_{5}({t}_{2})={max}_{t\in [0,1]}{x}_{5}(t)=b$. Using the boundary conditions, we have

${t}_{2}\in (0,1)$. In this case,

${x}_{5}^{\prime}({t}_{2})=0$,

${x}_{5}^{\u2033}({t}_{2})\le 0$. This gives

$0\le -{x}_{5}^{\u2033}({t}_{2})\le f({t}_{2},{x}_{5}({t}_{2}))=f({t}_{2},b),$

which is a contradiction to condition (H_{1}). Therefore (3.5) holds.

Finally, similar to the proof of (3.3), it follows from condition (H

_{2}) that

$(P+JQN)x+{L}_{1}^{-1}(I-Q)Nx\nleqq x,\phantom{\rule{1em}{0ex}}\text{for any}x\in \partial {\mathrm{\Omega}}_{3}\cap K.$

Consequently all conditions of Theorem 2.1 are satisfied. Therefore, there exists ${x}^{\ast}\in K\cap ({\overline{\mathrm{\Omega}}}_{4}\mathrm{\setminus}{\mathrm{\Omega}}_{3})$ such that $L{x}^{\ast}=N{x}^{\ast}$ with ${x}^{\ast}\in K$ and $a\le \parallel {x}^{\ast}\parallel \le b$ and the assertion follows. □