In this section, we state and prove the new existence results for (1.1). The proof is based on the following wellknown 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 Tperiodic 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:
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,xM\omega ,{x}^{\prime}M{\omega}^{\prime})}{xM{\omega}^{\prime}}=+\mathrm{\infty},
then there is {r}_{2}>\sigma {r}_{2}>{r}_{1} such that
\frac{F(t,xM\omega ,{x}^{\prime}M{\omega}^{\prime})}{xM\omega}\ge {\lambda}^{1}{\sigma}^{1}{A}^{1}{T}^{1},\phantom{\rule{1em}{0ex}}x\ge \sigma {r}_{2}.
Hence, we have
F(t,xM\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 Tperiodic solutions for sufficiently small λ.