In this section, we state and prove the new existence results for (1.1). The proof is based on the following well-known fixed point theorem on compression and expansion of cones, which we state here for the convenience of the reader, after introducing the definition of a cone.

**Definition 3.1** Let

*X* be a Banach space and let

*K* be a closed, nonempty subset of

*X*.

*K* is a cone if

- (i)
$\alpha u+\beta v\in K$ for all $u,v\in K$ and all $\alpha ,\beta >0$,

- (ii)
$u,-u\in K$ implies $u=0$.

We also recall that a compact operator means an operator which transforms every bounded set into a relatively compact set. Let us define the function $\omega (t)=\lambda {\int}_{0}^{T}G(t,s)\phantom{\rule{0.2em}{0ex}}ds$ and use ${\parallel \cdot \parallel}_{1}$ to denote the usual ${L}^{1}$-norm over $(0,T)$, by $\parallel \cdot \parallel $ we denote the supremum norm of $\mathbb{C}[0,T]$.

**Lemma 3.2** [22]

*Let* *X* *be a Banach space and* *K* (⊂

*X*)

*be a cone*.

*Assume that* ${\mathrm{\Omega}}_{1}$,

${\mathrm{\Omega}}_{2}$ *are open subsets of* *X* *with* $0\in {\mathrm{\Omega}}_{1}$,

${\overline{\mathrm{\Omega}}}_{1}\subset {\mathrm{\Omega}}_{2}$,

*and let* $\mathcal{A}:K\cap ({\overline{\mathrm{\Omega}}}_{2}\setminus {\mathrm{\Omega}}_{1})\to K$

*be a completely continuous operator such that either*
- (i)
$\parallel \mathcal{A}u\parallel \ge \parallel u\parallel $, $u\in K\cap \partial {\mathrm{\Omega}}_{1}$ *and* $\parallel \mathcal{A}u\parallel \le \parallel u\parallel $, $u\in K\cap \partial {\mathrm{\Omega}}_{2}$; *or*

- (ii)
$\parallel \mathcal{A}u\parallel \le \parallel u\parallel $, $u\in K\cap \partial {\mathrm{\Omega}}_{1}$ *and* $\parallel \mathcal{A}u\parallel \ge \parallel u\parallel $, $u\in K\cap \partial {\mathrm{\Omega}}_{2}$.

*Then* $\mathcal{A}$ *has a fixed point in* $K\cap ({\overline{\mathrm{\Omega}}}_{2}\setminus {\mathrm{\Omega}}_{1})$.

*Let*
$X=\mathbb{C}[0,T]$
*and define*
$K=\{x\in X:x(t)\ge 0\mathit{\text{and}}\underset{0\le t\le T}{min}x(t)\ge \sigma \parallel x\parallel \},$

*where* *σ* *is as in* (2.14).

*One may readily verify that* *K* *is a cone in* *X*.

*Now*,

*suppose that* $F:[0,T]\times \mathbb{R}\times \mathbb{R}\to [0,\mathrm{\infty})$ *is a continuous function*.

*Define an operator* $(\mathcal{A}x)(t)={\int}_{0}^{T}G(t,s)F(s,x(s),{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds$

*for* $x\in X$ *and* $t\in [0,T]$.

**Lemma 3.3** $\mathcal{A}:X\to K$ *is well defined*.

*Proof* Let

$x\in X$, then we have

$\begin{array}{rcl}\underset{0\le t\le T}{min}(\mathcal{A}x)(t)& =& \underset{0\le t\le T}{min}{\int}_{0}^{T}G(t,s)F(s,x(s),{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & A{\int}_{0}^{T}F(s,x(s),{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\\ =& \sigma B{\int}_{0}^{T}F(s,x(s),{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \sigma \underset{0\le t\le T}{max}{\int}_{0}^{T}G(t,s)F(s,x(s),{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\\ =& \sigma \parallel \mathcal{A}x\parallel .\end{array}$

This implies that $\mathcal{A}(X)\subset K$ and the proof is completed. □

It is easy to prove.

**Lemma 3.4** $\mathcal{A}$ *is continuous and completely continuous*.

Now we present our main result.

**Theorem 3.5** *Suppose that* (1.1) *satisfies* (H). *Furthermore*, *assume that*

(H

_{1})

$f:[0,T]\times {\mathbb{R}}^{+}\times \mathbb{R}\to \mathbb{R}$ *is continuous and there exists a constant* $M>0$ *such that* $F(t,x,y)=f(t,x,y)+M\ge 0\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}(t,x,y)\in [0,T]\times {\mathbb{R}}^{+}\times \mathbb{R}.$

(H_{2}) ${lim}_{x\to {0}^{+}}f(t,x,y)=+\mathrm{\infty}$ *and* ${lim}_{x\to +\mathrm{\infty}}f(t,x,y)/x=+\mathrm{\infty}$ *uniformly* $(t,y)\in {\mathbb{R}}^{2}$.

*Then* (1.1) *has at least two positive* *T*-*periodic solutions for sufficiently small* *λ*.

*Proof* To show that (1.1) has a positive solution, we should only show that

$-{x}^{\u2033}+a(t){x}^{\prime}+b(t)x=\lambda F(t,x(t)-M\omega (t),{x}^{\prime}(t)-M{\omega}^{\prime}(t))$

(3.1)

has a positive solution

*x* satisfying (1.3) and

$x(t)>M\omega (t)$ for

$t\in [0,T]$. If it is right, then

$\varphi (t)=x(t)-M\omega (t)$ is a solution of (1.1) since

$\begin{array}{r}-{\varphi}^{\u2033}(t)+a(t){\varphi}^{\prime}(t)+b(t)\varphi (t)\\ \phantom{\rule{1em}{0ex}}=-{(x(t)-M\omega (t))}^{\u2033}+a(t){(x(t)-M\omega (t))}^{\prime}+b(t)(x(t)-M\omega (t))\\ \phantom{\rule{1em}{0ex}}=\lambda F(t,x(t)-M\omega (t),{x}^{\prime}(t)-M{\omega}^{\prime}(t))-\lambda M\\ \phantom{\rule{1em}{0ex}}=\lambda f(t,x(t)-M\omega (t),{x}^{\prime}(t)-M{\omega}^{\prime}(t))\\ \phantom{\rule{1em}{0ex}}=\lambda f(t,\varphi (t),{\varphi}^{\prime}(t)),\end{array}$

where $-{\omega}^{\u2033}(t)+a(t){\omega}^{\prime}(t)+b(t)\omega (t)=\lambda $ is used.

Problem (3.1)-(1.3) is equivalent to the following fixed point of the operator equation:

$x(t)=(\mathcal{A}x)(t),$

where

$\mathcal{A}$ is a completely continuous operator defined by

$(\mathcal{A}x)(t)=\lambda {\int}_{0}^{T}G(t,s)F(s,x(s)-M\omega (s),{x}^{\prime}(s)-M{\omega}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds.$

Since

${lim}_{x\to +\mathrm{\infty}}\frac{f(t,x,y)}{x}=+\mathrm{\infty}$, there exists

${r}_{1}\ge MT$ such that

$x\ge \sigma {r}_{1}\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\frac{f(t,x,y)}{x}\ge \frac{1}{\sigma}.$

For $r>0$, let ${\mathrm{\Omega}}_{r}=\{x\in K:\parallel x\parallel <r\}$ and note that $\partial {\mathrm{\Omega}}_{r}=\{x\in K:\parallel x\parallel =r\}$.

First we show

$\parallel \mathcal{A}x\parallel \le \parallel x\parallel \phantom{\rule{1em}{0ex}}\text{for}x\in K\cap \partial {\mathrm{\Omega}}_{{r}_{1}}.$

Let

$h(t)=max\{f(t,x,{x}^{\prime}):\frac{\sigma}{2}{r}_{1}\le x\le {r}_{1}\}$ and

${\lambda}^{\ast}=min\{{\sigma}^{2}/2A,MT/B{\parallel h\parallel}_{1}\}$. For any

$x\in \partial \mathrm{\Omega}{r}_{1}$ and

$0<\lambda <{\lambda}^{\ast}$, we can verify that

$\begin{array}{rcl}x(t)-M\omega (t)& \ge & \sigma \parallel x\parallel -M\omega (t)=\sigma {r}_{1}-M\omega (t)\\ \ge & \sigma {r}_{1}-\lambda MBT\ge \sigma {r}_{1}-\frac{\sigma {r}_{1}}{2}=\frac{\sigma {r}_{1}}{2}.\end{array}$

Then we have

$\begin{array}{rcl}(\mathcal{A}x)(t)& =& \lambda {\int}_{0}^{T}G(t,s)F(s,x(s)-M\omega (s),{x}^{\prime}(s)-M{\omega}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\\ \le & \lambda B{\int}_{0}^{T}F(s,x(s)-M\omega (s),{x}^{\prime}(s)-M{\omega}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\\ \le & \lambda B{\parallel h\parallel}_{1}\\ \le & {r}_{1}=\parallel x\parallel .\end{array}$

This implies $\parallel \mathcal{A}x\parallel \le \parallel x\parallel $.

In view of the assumption

$\underset{x\to +\mathrm{\infty}}{lim}\frac{f(t,x,{x}^{\prime})}{x}=\underset{x\to +\mathrm{\infty}}{lim}\frac{F(t,x-M\omega ,{x}^{\prime}-M{\omega}^{\prime})}{x-M{\omega}^{\prime}}=+\mathrm{\infty},$

then there is

${r}_{2}>\sigma {r}_{2}>{r}_{1}$ such that

$\frac{F(t,x-M\omega ,{x}^{\prime}-M{\omega}^{\prime})}{x-M\omega}\ge {\lambda}^{-1}{\sigma}^{-1}{A}^{-1}{T}^{-1},\phantom{\rule{1em}{0ex}}x\ge \sigma {r}_{2}.$

Hence, we have

$F(t,x-M\omega ,{x}^{\prime}-M{\omega}^{\prime})\ge {\lambda}^{-1}{A}^{-1}{T}^{-1}{r}_{2},\phantom{\rule{1em}{0ex}}x\ge \sigma {r}_{2}.$

Next, we show that

$\parallel \mathcal{A}x\parallel \ge \parallel x\parallel \phantom{\rule{1em}{0ex}}\text{for}x\in K\cap \partial {\mathrm{\Omega}}_{{r}_{2}}.$

To see this, let

$x\in K\cap \partial \mathrm{\Omega}{r}_{2}$, then

$\begin{array}{rcl}\parallel \mathcal{A}x\parallel & =& \underset{0\le t\le T}{max}|\lambda {\int}_{0}^{T}G(t,s)F(s,x(s)-M\omega (s),{x}^{\prime}(s)-M{\omega}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds|\\ \ge & \lambda A{\int}_{0}^{T}F(s,x(s)-M\omega (s),{x}^{\prime}(s)-M{\omega}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \lambda A{\int}_{0}^{T}{\lambda}^{-1}{A}^{-1}{T}^{-1}{r}_{2}\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {r}_{2}=\parallel x\parallel .\end{array}$

It follows from Lemma 3.2 that

$\mathcal{A}$ has a fixed point

${\tilde{x}}_{1}(t)$ such that

${\tilde{x}}_{1}(t)\in {\overline{\mathrm{\Omega}}}_{{r}_{2}}\setminus {\mathrm{\Omega}}_{{r}_{1}}$, which is a positive periodic solution of (3.1) for

$\lambda <{\lambda}^{\ast}$ satisfying

${r}_{1}<\parallel {\tilde{x}}_{1}\parallel <{r}_{2}.$

So, equation (1.1) has a positive solution ${x}_{1}(t)={\tilde{x}}_{1}(t)-M\omega (t)\ge \sigma {r}_{1}-\frac{\sigma MT}{2}\ge \frac{\sigma MT}{2}$.

On the other hand, since

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

hence, there exists a positive number

$0<{r}_{3}<{r}_{1}$ such that

$f(t,x,y)>0,\phantom{\rule{1em}{0ex}}x\in {\mathbb{R}}^{+}\text{with}0x\le {r}_{3}\frac{\sigma MT}{2},$

problem (1.1)-(1.3) is equivalent to the following fixed point of the operator equation:

$x(t)=\left({\mathcal{A}}^{\prime}x\right)(t),$

where

${\mathcal{A}}^{\prime}$ is a continuous and completely continuous operator defined by

$\left({A}^{\prime}x\right)(t)=\lambda {\int}_{0}^{T}G(t,s)f(s,x(s),{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds.$

And for any

$\rho >0$, define

$\mathrm{\Delta}(\rho )=max\{f(t,x,y):x\in {\mathbb{R}}^{+},\sigma \rho \le x\le \rho ,(t,y)\in [0,T]\times \mathbb{R}\}.$

Furthermore, for any

$x\in K\cap \partial \mathrm{\Omega}{r}_{3}$, we have

$\begin{array}{rcl}\left({A}^{\prime}x\right)(t)& =& \lambda {\int}_{0}^{T}G(t,s)f(s,x(s),{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\\ \le & \lambda B{\int}_{0}^{T}f(s,x(s),{x}^{\prime}(s)))\phantom{\rule{0.2em}{0ex}}ds\\ \le & \lambda B\mathrm{\Delta}({r}_{3})T.\end{array}$

Thus, from the above inequality, there exists

${\lambda}^{\ast \ast}$ such that

$\parallel {\mathcal{A}}^{\prime}x\parallel <\parallel x\parallel \phantom{\rule{1em}{0ex}}\text{for}x\in \partial {\mathrm{\Omega}}_{{r}_{3}},0\lambda {\lambda}^{\ast \ast}.$

Since

${lim}_{x\to {0}^{+}}f(t,x,{x}^{\prime})=+\mathrm{\infty}$, then there is a positive number

$0<{r}_{4}<\sigma {r}_{3}<{r}_{3}$ such that

$f(t,x,{x}^{\prime})>\gamma x,\phantom{\rule{1em}{0ex}}x\in {\mathbb{R}}^{+}\text{with}0x\le {r}_{4},$

where *γ* satisfies $\lambda \gamma \sigma AT>1$.

If

$x\in K\cap \partial \mathrm{\Omega}{r}_{4}$, then

$\begin{array}{rcl}\parallel {\mathcal{A}}^{\prime}x\parallel & =& \underset{0\le t\le T}{max}|\lambda {\int}_{0}^{T}G(t,s)f(s,x(s),{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds|\\ \ge & \lambda A{\int}_{0}^{T}f(s,x(s),{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \lambda A{\int}_{0}^{T}\gamma x\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \lambda A{\int}_{0}^{T}\gamma \sigma \parallel x\parallel \phantom{\rule{0.2em}{0ex}}ds\\ \ge & \parallel x\parallel .\end{array}$

It follows from Lemma 3.2 that

${\mathcal{A}}^{\prime}$ has a fixed point

${x}_{2}(t)$ such that

${x}_{2}(t)\in {\overline{\mathrm{\Omega}}}_{{r}_{3}}\setminus {\mathrm{\Omega}}_{{r}_{4}}$, which is a positive periodic solution of (1.1) for

$\lambda <{\lambda}^{\ast \ast}$ satisfying

${r}_{4}<\parallel {x}_{2}\parallel <{r}_{3}.$

Noting that

${r}_{4}<\parallel {x}_{2}\parallel <{r}_{3}<\frac{\sigma {r}_{1}}{2}<\parallel {x}_{1}\parallel <{r}_{2},$

we can conclude that ${x}_{1}$ and ${x}_{2}$ are the desired distinct positive periodic solutions of (1.1) for $\lambda <min\{{\lambda}^{\ast},{\lambda}^{\ast \ast}\}$. □

**Example** Let the nonlinearity in (1.1) be

$f(t,x,y)=(1+{|y|}^{\gamma})(c(t){x}^{-\alpha}+d(t){x}^{\beta}+e(t)),\phantom{\rule{1em}{0ex}}0\le t\le T,$

where $\alpha >0$, $\beta >1$, $\gamma \ge 0$, $c(t),d(t),e(x)\in \mathbb{C}[0,T]$. It is clear that $f(t,x,y)$ satisfies conditions (H_{1}), (H_{2}). Then (1.1) has at least two positive *T*-periodic solutions for sufficiently small *λ*.