For convenience, we introduce the following notations:

\begin{array}{c}{f}^{0}=\underset{y\to 0}{lim\hspace{0.17em}sup}\underset{t\in J}{max}\frac{f(t,y)}{y},\phantom{\rule{2em}{0ex}}{f}_{0}=\underset{y\to 0}{lim\hspace{0.17em}inf}\underset{t\in J}{min}\frac{f(t,y)}{y},\hfill \\ {f}^{\mathrm{\infty}}=\underset{y\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\underset{t\in J}{max}\frac{f(t,y)}{y},\phantom{\rule{2em}{0ex}}{f}_{\mathrm{\infty}}=\underset{y\to \mathrm{\infty}}{lim\hspace{0.17em}inf}\underset{t\in J}{min}\frac{f(t,y)}{y}.\hfill \end{array}

We also define as [50]{i}_{0} = number of zeros in the set \{{f}^{0},{f}^{\mathrm{\infty}}\} and {i}_{\mathrm{\infty}} = number of infinities in the set \{{f}_{0},{f}_{\mathrm{\infty}}\}. Sun and Li [51] pointed out that {i}_{0},{i}_{\mathrm{\infty}}=0,1\text{or}2, and there are six possible cases: (i) {i}_{0}=0 and {i}_{\mathrm{\infty}}=0; (ii) {i}_{0}=0 and {i}_{\mathrm{\infty}}=1; (iii) {i}_{0}=0 and {i}_{\mathrm{\infty}}=2; (iv) {i}_{0}=1 and {i}_{\mathrm{\infty}}=0; (v) {i}_{0}=1 and {i}_{\mathrm{\infty}}=1; and (vi) {i}_{0}=2 and {i}_{\mathrm{\infty}}=0. By using Krasnosel’skii’s fixed point theorem in a cone, some results are obtained for the existence of at least one or two positive solutions of problem (1.1) for \alpha (t)\ge t on *J* under the above six possible cases.

### 3.1 For the case \alpha (t)\ge ton *J* under {i}_{0}=1and {i}_{\mathrm{\infty}}=1

In this subsection, we discuss the existence of a single positive solution for problem (1.1) for \alpha (t)\ge t on *J* under {i}_{0}=1 and {i}_{\mathrm{\infty}}=1.

For convenience, we introduce the following notations:

\gamma ={\int}_{0}^{1}\omega (s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{2em}{0ex}}{\gamma}_{1}={\int}_{\xi}^{1}\omega (s)\phantom{\rule{0.2em}{0ex}}ds.

**Theorem 3.1** *Assume that* (H_{1})-(H_{4}) *hold*. *If* {i}_{0}=1 *and* {i}_{\mathrm{\infty}}=1, *then problem* (1.1) *has at least one positive solution*.

*Proof* First, we consider the case {f}^{0}=0 and {f}_{\mathrm{\infty}}=\mathrm{\infty}. Since {f}^{0}=0, then there exists {r}_{1}>0 such that

f(t,y)\le \frac{{c}_{m}}{\beta \gamma {c}_{M}}y\phantom{\rule{1em}{0ex}}(\mathrm{\forall}t\in J,0\le y\le {r}_{1}).

Since 0\le t\le \alpha (t)\le 1 on *J*, it follows from 0\le y(t)\le {r}_{1} on *J* that

0\le y(\alpha (t))\le {r}_{1}\phantom{\rule{1em}{0ex}}\text{for}t\in J.

Let r=min\{{r}_{1},\frac{1}{{c}_{M}}{r}_{1}\}. Then, for y\in K\cap \partial {\mathrm{\Omega}}_{r}, we have 0\le y(t)\le r\le {r}_{1} for t\in J, and then

c(\alpha (t))y(\alpha (t))\le {c}_{M}\parallel y\parallel ={c}_{M}r\le {r}_{1},\phantom{\rule{1em}{0ex}}t\in J.

Consequently, for any t\in J and y\in K\cap \partial {\mathrm{\Omega}}_{r}, (2.9) and (2.12) imply

\begin{array}{rl}(Ty)(t)& ={\int}_{0}^{1}H(t,s)\omega (s){c}^{-1}(s)f(s,c(\alpha (s))y(\alpha (s)))\phantom{\rule{0.2em}{0ex}}ds\\ \le \beta {c}_{m}^{-1}{\int}_{0}^{1}\omega (s)\frac{{c}_{m}}{\gamma \beta {c}_{M}}c(\alpha (s))y(\alpha (s))\phantom{\rule{0.2em}{0ex}}ds\\ \le \beta {c}_{m}^{-1}\frac{{c}_{m}}{\gamma \beta {c}_{M}}{c}_{M}{\int}_{0}^{1}\omega (s)y(\alpha (s))\phantom{\rule{0.2em}{0ex}}ds\\ \le \frac{1}{\gamma}{\int}_{0}^{1}\omega (s)\phantom{\rule{0.2em}{0ex}}ds\parallel y\parallel \\ =\parallel y\parallel ,\end{array}

which implies

\parallel Ty\parallel \le \parallel y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in K\cap \partial {\mathrm{\Omega}}_{r}.

(3.1)

Next turning to {f}_{\mathrm{\infty}}=\mathrm{\infty}, there exists \stackrel{\u02c6}{r} satisfying 0<{r}_{1}<\stackrel{\u02c6}{r} such that

f(t,y)\ge \frac{{c}_{M}}{{c}_{m}\delta {\beta}^{\ast}{\gamma}_{1}}y,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in [\xi ,1],y\ge \stackrel{\u02c6}{r}.

Since \xi \le t\le \alpha (t)\le 1 on *J*, it follows from y(t)\ge \stackrel{\u02c6}{r} on [\xi ,1] that

y(\alpha (t))\ge \stackrel{\u02c6}{r}\phantom{\rule{1em}{0ex}}\text{for}t\in [\xi ,1].

Let R>max\{\stackrel{\u02c6}{r},\frac{\stackrel{\u02c6}{r}}{\delta {c}_{m}}\}. Then, for y\in K\cap \partial {\mathrm{\Omega}}_{R}, we have

c(\alpha (t))y(\alpha (t))\ge {c}_{m}y(\alpha (t))\ge {c}_{m}\delta \parallel y\parallel \ge \stackrel{\u02c6}{r},\phantom{\rule{1em}{0ex}}t\in [\xi ,1].

Hence, for y\in K\cap \partial {\mathrm{\Omega}}_{R}, it follows from (2.9) and (2.12) that

\begin{array}{rl}(Ty)(t)& ={\int}_{0}^{1}H(t,s)\omega (s){c}^{-1}(s)f(s,c(\alpha (s))y(\alpha (s)))\phantom{\rule{0.2em}{0ex}}ds\\ \ge {\beta}^{\ast}{\int}_{0}^{1}\omega (s){c}^{-1}(s)f(s,c(\alpha (s))y(\alpha (s)))\phantom{\rule{0.2em}{0ex}}ds\\ \ge {\beta}^{\ast}{c}_{M}^{-1}{\int}_{\xi}^{1}\omega (s)\frac{{c}_{M}}{{c}_{m}\delta {\beta}^{\ast}{\gamma}_{1}}c(\alpha (s))y(\alpha (s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge {\beta}^{\ast}{c}_{M}^{-1}\frac{{c}_{M}}{{c}_{m}\delta {\beta}^{\ast}{\gamma}_{1}}{c}_{m}{\int}_{\xi}^{1}\omega (s)y(\alpha (s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge \frac{1}{\delta {\gamma}_{1}}{\int}_{\xi}^{1}\omega (s)\phantom{\rule{0.2em}{0ex}}ds\delta \parallel y\parallel \\ =\parallel y\parallel ,\end{array}

which implies

\parallel Ty\parallel \ge \parallel y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in K\cap \partial {\mathrm{\Omega}}_{R}.

(3.2)

Thus by (i) of Lemma 2.6, it follows that *T* has a fixed point *y* in K\cap ({\overline{\mathrm{\Omega}}}_{R}\mathrm{\setminus}{\mathrm{\Omega}}_{r}) with

r\le \parallel y\parallel \le R.

Lemma 2.5 implies that problem (1.1) has at least one positive solution *x* with

{c}_{m}r\le \parallel x\parallel \le {c}_{M}R.

This gives the proof of Theorem 3.1. □

**Remark 3.1** For {i}_{0}=1 and {i}_{\mathrm{\infty}}=1, there is another case {f}^{\mathrm{\infty}}=0 and {f}_{0}=\mathrm{\infty}. However, at the moment, we give no information on the existence of a positive solution for problem (1.1) if we change {f}^{0}=0 and {f}_{\mathrm{\infty}}=\mathrm{\infty} into {f}^{\mathrm{\infty}}=0 and {f}_{0}=\mathrm{\infty} in Theorem 3.1.

### 3.2 For the case \alpha (t)\ge ton *J* under {i}_{0}=0and {i}_{\mathrm{\infty}}=0

In this subsection, we discuss the existence for the positive solutions of problem (1.1) under {i}_{0}=0 and {i}_{\mathrm{\infty}}=0. For convenience, we introduce the following notations:

{f}_{0}^{\rho}=max\{\underset{t\in J}{max}\frac{f(t,y)}{\rho}:y\in [0,\rho ]\}

and

l=\frac{{c}_{m}}{\beta \gamma},\phantom{\rule{2em}{0ex}}L=\frac{{c}_{M}}{{c}_{m}{\beta}^{\ast}\delta {\gamma}_{1}}.

Now, we shall state and prove the following main result.

**Theorem 3.2** *Suppose that* (H_{1})-(H_{4}) *hold and* \alpha (t)\ge t *on* *J*. *In addition*, *let the following two conditions hold*:

(H_{5}) *There exist* l>0 *and* {\rho}_{1}>0 *such that* {f}_{0}^{{\rho}_{1}}\le l;

(H_{6}) *There exist* \eta >0 *and* {\rho}_{2}>0 *such that* f(t,y)\ge \eta *for* t\in J, y\ge {\rho}_{2}; *furthermore*, {\rho}_{1}\ne {\rho}_{2}.

*Then problem* (1.1) *has at least one positive solution*.

*Proof* Without loss of generality, we may assume that {\rho}_{1}<{\rho}_{2}. Considering {f}_{0}^{{\rho}_{1}}\le l, we have f(t,y)\le l{\rho}_{1} for 0\le y\le {\rho}_{1}, t\in J.

Since 0\le t\le \alpha (t)\le 1 on *J*, it follows from 0\le y(t)\le {\rho}_{1} on *J* that

0\le y(\alpha (t))\le {\rho}_{1}.

Let \rho =min\{{\rho}_{1},\frac{1}{{c}_{M}}{\rho}_{1}\}. Then, for y\in K\cap \partial {\mathrm{\Omega}}_{\rho}, we have 0\le y(t)\le \rho \le {\rho}_{1} for t\in J, and then

c(\alpha (t))y(\alpha (t))\le {c}_{M}\parallel y\parallel ={c}_{M}\rho \le {\rho}_{1},\phantom{\rule{1em}{0ex}}t\in J.

Consequently, for any t\in J and y\in K\cap \partial {\mathrm{\Omega}}_{\rho}, (2.9) and (2.12) imply

\begin{array}{rl}(Ty)(t)& ={\int}_{0}^{1}H(t,s)\omega (s){c}^{-1}(s)f(s,c(\alpha (s))y(\alpha (s)))\phantom{\rule{0.2em}{0ex}}ds\\ \le \beta {c}_{m}^{-1}l{\rho}_{1}{\int}_{0}^{1}\omega (s)\phantom{\rule{0.2em}{0ex}}ds\\ =\beta {c}_{m}^{-1}l{\rho}_{1}\gamma \\ ={\rho}_{1},\end{array}

which implies

\parallel Ty\parallel \le \parallel y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in K\cap \partial {\mathrm{\Omega}}_{\rho}.

(3.3)

On the other hand, from (H_{6}), when {\rho}_{2} is fixed, there exists \eta >0 such that

f(t,y)\ge \eta \ge max\{{\rho}_{2},\frac{{\rho}_{2}}{\delta {c}_{m}}\}\times \frac{{c}_{M}}{{\beta}^{\ast}{\gamma}_{1}}

for t\in J and y\ge {\rho}_{2}. Since 0\le t\le \alpha (t)\le 1 on *J*, it follows from y(t)\ge {\rho}_{2} on *J* that

y(\alpha (t))\ge {\rho}_{2}.

Let \overline{\rho}=max\{{\rho}_{2},\frac{{\rho}_{2}}{\delta {c}_{m}}\}. Then, for y\in K\cap \partial {\mathrm{\Omega}}_{\overline{\rho}}, we have

c(t)y(t)\ge {c}_{m}y(t)\ge {c}_{m}\delta \parallel y\parallel \ge {\rho}_{2},\phantom{\rule{1em}{0ex}}t\in [\xi ,1].

Hence, for y\in K\cap \partial {\mathrm{\Omega}}_{\overline{\rho}}, it follows from (2.9) and (2.12) that

\begin{array}{rl}(Ty)(t)& ={\int}_{0}^{1}H(t,s)\omega (s){c}^{-1}(s)f(s,c(\alpha (s))y(\alpha (s)))\phantom{\rule{0.2em}{0ex}}ds\\ \ge {\beta}^{\ast}{\int}_{0}^{1}\omega (s){c}^{-1}(s)f(s,c(\alpha (s))y(\alpha (s)))\phantom{\rule{0.2em}{0ex}}ds\\ \ge {\beta}^{\ast}{c}_{M}^{-1}{\int}_{\xi}^{1}\omega (s)\eta \phantom{\rule{0.2em}{0ex}}ds\\ \ge {\beta}^{\ast}{c}_{M}^{-1}\eta {\int}_{\xi}^{1}\omega (s)\phantom{\rule{0.2em}{0ex}}ds\\ ={\beta}^{\ast}{c}_{M}^{-1}\eta {\gamma}_{1}\\ \ge {\beta}^{\ast}{c}_{M}^{-1}{\gamma}_{1}\overline{\rho}\frac{{c}_{M}}{{\beta}^{\ast}{\gamma}_{1}}\\ =\overline{\rho},\end{array}

which implies

\parallel Ty\parallel \ge \parallel y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in K\cap \partial {\mathrm{\Omega}}_{\overline{\rho}}.

(3.4)

Thus by (i) of Lemma 2.6, it follows that *T* has a fixed point *y* in K\cap ({\overline{\mathrm{\Omega}}}_{{\rho}_{2}}\mathrm{\setminus}{\mathrm{\Omega}}_{{\rho}_{1}}) with

\rho \le \parallel y\parallel \le \overline{\rho}.

Thus, it follows from Lemma 2.5 that problem (1.1) has at least one positive solution *x* with

{c}_{m}\rho \le \parallel x\parallel \le {c}_{M}\overline{\rho}.

This finishes the proof of Theorem 3.2. □

We remark that condition (H_{5}) in Theorem 3.2 can be replaced by the following condition:

{({\mathrm{H}}_{5})}^{\prime} {f}^{0}\le l,

which is a special case of (H_{5}).

**Corollary 3.1** *Suppose that* (H_{1})-(H_{4}), {({\mathrm{H}}_{5})}^{\prime}, (H_{6}) *hold and* \alpha (t)\ge t *on* *J*. *Then problem* (1.1) *has at least one positive solution*.

*Proof* We show that {({\mathrm{H}}_{5})}^{\prime} implies (H_{5}). Suppose that {({\mathrm{H}}_{5})}^{\prime} holds. Then there exists a positive number {\rho}_{1}\ne {\rho}_{2} such that

\frac{f(t,y)}{y}\le l,\phantom{\rule{1em}{0ex}}t\in J,0<y\le {\rho}_{1}.

Hence, we obtain

f(t,y)\le ly\le l{\rho}_{1},\phantom{\rule{1em}{0ex}}t\in J,0<y\le {\rho}_{1}.

Therefore, (H_{5}) holds. Hence, by Theorem 3.2, problem (1.1) has at least one positive solution. □

**Theorem 3.3** *Suppose that* (H_{1})-(H_{5}) *hold and* \alpha (t)\ge t *on* *J*. *In addition*, *let the following condition hold*:

(H_{7}) {f}_{\mathrm{\infty}}\ge L.

*Then problem* (1.1) *has at least one positive solution*.

*Proof* The proof is similar to those of (3.2) and (3.3), respectively. □

**Corollary 3.2** *Suppose that* (H_{1})-(H_{4}), {({\mathrm{H}}_{5})}^{\prime}, (H_{7}) *hold and* \alpha (t)\ge t *on* *J*. *Then problem* (1.1) *has at least one positive solution*.

### 3.3 For the case \alpha (t)\ge ton *J* under {i}_{0}=1and {i}_{\mathrm{\infty}}=0or {i}_{0}=0and {i}_{\mathrm{\infty}}=1

In this subsection, we discuss the existence for the positive solutions of problem (1.1) for the case \alpha (t)\ge t on *J* under {i}_{0}=1 and {i}_{\mathrm{\infty}}=0 or {i}_{0}=0 and {i}_{\mathrm{\infty}}=1.

**Theorem 3.4** *Suppose that* (H_{1})-(H_{4}) *hold*, \alpha (t)\ge t *on* *J* *and* {f}^{0}\in [0,l) *and* {f}_{\mathrm{\infty}}\in (L,\mathrm{\infty}). *Then problem* (1.1) *has at least one positive solution*.

*Proof* The proof is similar to that of Theorem 3.2. □

**Theorem 3.5** *Suppose that* (H_{1})-(H_{4}) *hold*, \alpha (t)\ge t *on* *J* *and* {f}_{0}\in (L,\mathrm{\infty}) *and* {f}^{\mathrm{\infty}}\in [0,l). *Then problem* (1.1) *has at least one positive solution*.

*Proof* Consider {f}_{0}\in (L,\mathrm{\infty}), then there exists {\rho}_{1}>0 such that f(t,y)>Ly for 0\le y\le {\rho}_{1}, t\in J.

Since 0\le t\le \alpha (t)\le 1 on *J*, it follows from 0\le y(t)\le {\rho}_{1} on *J* that

0\le y(\alpha (t))\le {\rho}_{1}.

Let \rho =min\{{\rho}_{1},\frac{1}{{c}_{M}}{\rho}_{1}\}. Then, for y\in K\cap \partial {\mathrm{\Omega}}_{\rho}, we have 0\le y(t)\le \rho \le {\rho}_{1} for t\in J, and then

c(\alpha (t))y(\alpha (t))\le {c}_{M}\parallel y\parallel ={c}_{M}\rho \le {\rho}_{1},\phantom{\rule{1em}{0ex}}t\in J.

Consequently, for y\in K\cap \partial {\mathrm{\Omega}}_{\rho}, it follows from (2.9) and (2.12) that

\begin{array}{rl}(Ty)(t)& ={\int}_{0}^{1}H(t,s)\omega (s){c}^{-1}(s)f(s,c(\alpha (s))y(\alpha (s)))\phantom{\rule{0.2em}{0ex}}ds\\ \ge {\beta}^{\ast}{\int}_{0}^{1}\omega (s){c}^{-1}(s)f(s,c(\alpha (s))y(\alpha (s)))\phantom{\rule{0.2em}{0ex}}ds\\ \ge {\beta}^{\ast}{c}_{M}^{-1}{\int}_{\xi}^{1}\omega (s)Lc(\alpha (s))y(\alpha (s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge {\beta}^{\ast}{c}_{M}^{-1}L{c}_{m}\delta \parallel y\parallel {\int}_{\xi}^{1}\omega (s)\phantom{\rule{0.2em}{0ex}}ds\\ ={\beta}^{\ast}{c}_{M}^{-1}L{c}_{m}\delta \parallel y\parallel {\gamma}_{1}\\ \ge \parallel y\parallel ,\end{array}

which implies

\parallel Ty\parallel \ge \parallel y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in K\cap \partial {\mathrm{\Omega}}_{\rho}.

(3.5)

Next, turn to {f}^{\mathrm{\infty}}\in [0,l). In fact, we can show that {f}^{\mathrm{\infty}}\in [0,l) implies (H_{5}).

Let \tau \in ({f}^{\mathrm{\infty}},l). Then there exists r>\tau such that {max}_{t\in J}f(t,y)\le \tau y for y\in [r,\mathrm{\infty}). Let

\beta =max\{\underset{t\in J}{max}f(t,y):0\le y\le r\}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\rho}_{1}^{\ast}>max\{\frac{\beta}{l-\tau},\rho ,{c}_{M}\rho \}.

Then we have

\underset{0\le t\le 1}{max}f(t,y)\le \tau y+\beta \le \tau {\rho}_{1}^{\ast}+\beta <l{\rho}_{1}^{\ast},\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in [0,{\rho}_{1}^{\ast}].

This implies that {f}_{0}^{{\rho}_{1}^{\ast}}\le l. Hence, {f}^{\mathrm{\infty}}\in [0,l) implies that (H_{5}).

Similarly to the proof of (3.3), we have

\parallel Ty\parallel \le \parallel y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in K\cap \partial {\mathrm{\Omega}}_{{\rho}^{\ast}},

(3.6)

where {\rho}^{\ast}=min\{{\rho}_{1}^{\ast},\frac{1}{{c}_{M}}{\rho}_{1}^{\ast}\}.

Thus by (ii) of Lemma 2.6, it follows that *T* has a fixed point *y* in K\cap ({\overline{\mathrm{\Omega}}}_{{\rho}^{\ast}}\mathrm{\setminus}{\mathrm{\Omega}}_{\rho}) with

\rho \le \parallel y\parallel \le {\rho}^{\ast}.

This finishes the proof of Theorem 3.5. □

From Theorems 3.4 and 3.5, we have the following result.

**Corollary 3.3** *Suppose that* {f}^{0}=0 *and condition* (H_{6}) *in Theorem * 3.2 *hold*. *Then problem* (1.1) *has at least one positive solution*.

**Theorem 3.6** *Suppose that* (H_{1})-(H_{4}), \alpha (t)\ge t *on* *J*, {f}^{0}\in (0,l) *and* {f}_{\mathrm{\infty}}=\mathrm{\infty}. *Then problem* (1.1) *has at least one positive solution*.

*Proof* The proof is similar to that of Theorem 3.2. □

**Theorem 3.7** *Suppose* (H_{1})-(H_{4}), \alpha (t)\ge t *on* *J*, {f}_{0}=\mathrm{\infty} *and* {f}^{\mathrm{\infty}}\in (0,l). *Then problem* (1.1) *has at least one positive solution*.

*Proof* The proof is similar to that of Theorem 3.2. □

From Theorems 3.6 and 3.7, the following corollaries are easily obtained.

**Corollary 3.4** *Suppose that* {f}^{0}=\mathrm{\infty} *and condition* (H_{5}) *in Theorem * 3.2 *hold*. *Then problem* (1.1) *has at least one positive solution*.

**Corollary 3.5** *Suppose that* {f}_{\mathrm{\infty}}=\mathrm{\infty} *and condition* (H_{5}) *in Theorem * 3.2 *hold*. *Then problem* (1.1) *has at least one positive solution*.

### 3.4 For the case \alpha (t)\ge ton *J* under {i}_{0}=0and {i}_{\mathrm{\infty}}=2or {i}_{0}=2and {i}_{\mathrm{\infty}}=0

In this subsection we study the existence of multiple positive solutions for problem (1.1) for the case \alpha (t)\ge t on *J* under {i}_{0}=0 and {i}_{\mathrm{\infty}}=2 or {i}_{0}=2 and {i}_{\mathrm{\infty}}=0.

Combining the proofs of Theorems 3.1 and 3.2, the following theorem is easily proved.

**Theorem 3.8** *Suppose that* (H_{1})-(H_{4}), \alpha (t)\ge t *on* *J*, {i}_{0}=0 *and* {i}_{\mathrm{\infty}}=2 *and condition* (H_{5}) *of Theorem * 3.2 *hold*. *Then problem* (1.1) *has at least two positive solutions*.

**Corollary 3.6** *Suppose that* (H_{1})-(H_{4}), \alpha (t)\ge t *on* *J*, {i}_{0}=0 *and* {i}_{\mathrm{\infty}}=2 *and condition* {({\mathrm{H}}_{5})}^{\prime} *of Corollary * 3.1 *hold*. *Then problem* (1.1) *has at least two positive solutions*.

**Remark 3.2** Noticing Remark 3.1, at the moment we give no information on the existence of a positive solution for problem (1.1) under {i}_{0}=2 and {i}_{\mathrm{\infty}}=0.