**Theorem 3.1** *Let a cone* *P* *be normal and condition* ({H}_{1}) *be satisfied*. *If* ({H}_{2}) *and* ({H}_{3}) *or* ({H}_{4}) *and* ({H}_{5}) *are satisfied*, *then BVP* (1.1) *has at least one positive solution*.

*Proof* Set

{Q}_{1}=\{u\in Q:u(t)\ge t(1-t)u(s),\mathrm{\forall}t,s\in [0,1]\}.

It is clear that {Q}_{1} is a cone of the Banach space C[I,E] and {Q}_{1}\subset Q. For any u\in {Q}_{1}, by (2.1), we can obtain A(Q)\subset {Q}_{1}, then

A({Q}_{1})\subset {Q}_{1}.

We first assume that ({H}_{2}) and ({H}_{3}) are satisfied. Let

W=\{u\in {Q}_{1}|Au\ge u\}.

In the following, we prove that *W* is bounded.

For any u\in W, we have \theta \le u\le Au, that is, \theta \le u(t)\le Au(t), t\in I. And so \parallel u(t)\parallel \le N\parallel Au(t)\parallel, set v(t)=\parallel u(t)\parallel, by ({H}_{2})

\begin{array}{rcl}v(t)& \le & N\parallel Au(t)\parallel \\ \le & N{\int}_{0}^{1}G(t,s)\parallel f(s,{\int}_{0}^{1}G(s,\tau )u(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,u(s))\parallel \phantom{\rule{0.2em}{0ex}}ds\\ \le & {\int}_{0}^{1}G(t,s)({a}_{1}\parallel {\int}_{0}^{1}G(s,\tau )u(\tau )\phantom{\rule{0.2em}{0ex}}d\tau \parallel +{b}_{1}\parallel u(s)\parallel +{c}_{1})\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\int}_{0}^{1}G(t,s)({a}_{1}{\int}_{0}^{1}G(s,\tau )\parallel u(\tau )\parallel \phantom{\rule{0.2em}{0ex}}d\tau +{b}_{1}\parallel u(s)\parallel )\phantom{\rule{0.2em}{0ex}}ds+{c}_{1}.\end{array}

(3.1)

For \psi \in C[I,\mathbb{R}], let {L}_{1}\psi ={a}_{1}{T}^{2}\psi +{b}_{1}T\psi, then {L}_{1}:C[I,\mathbb{R}]\to C[I,\mathbb{R}] is a bounded linear operator. From (3.1), one deduces that

((I-{L}_{1})v)(t)\le {c}_{1}.

Since {\pi}^{2} is the first eigenvalue of *T*, by ({H}_{2}), the first eigenvalue of {L}_{1}, r({L}_{1})=\frac{{a}_{1}}{{\pi}^{4}}+\frac{{b}_{1}}{{\pi}^{2}}<1. Therefore, by [14], the inverse operator {(I-{L}_{1})}^{-1} exists and

{(I-{L}_{1})}^{-1}=I+{L}_{1}+{L}_{1}^{2}+\cdots +{L}_{1}^{n}+\cdots .

It follows from {L}_{1}(K)\subset K that {(I-{L}_{1})}^{-1}(K)\subset K. So, we know that v(t)\le {(I-{L}_{1})}^{-1}{c}_{1}, t\in [0,1] and *W* is bounded.

Taking {R}_{2}>max\{{r}_{1},supW\}, we have

Au\ngeqq u,\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in \partial {B}_{{R}_{2}}\cap {Q}_{1}.

(3.2)

Next, we are going to verify that for any {r}_{3}\in (0,{r}_{1}),

If this is false, then there exists {u}_{1}\in \partial {B}_{{r}_{3}}\cap {Q}_{1} such that A{u}_{1}\le {u}_{1}. This together with ({H}_{3}) yields

\begin{array}{rcl}\phi ({u}_{1}(t))& \ge & \phi ((A{u}_{1})(t))=\phi ({\int}_{0}^{1}G(t,s)f(s,(S{u}_{1})(s),{u}_{1}(s))\phantom{\rule{0.2em}{0ex}}ds)\\ =& {\int}_{0}^{1}G(t,s)\phi \left(f(s,(S{u}_{1})(s),{u}_{1}(s))\right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {\int}_{0}^{1}G(t,s)({a}_{2}\phi ((S{u}_{1})(s))+{b}_{2}\phi ({u}_{1}(s)))\phantom{\rule{0.2em}{0ex}}ds\\ =& {\int}_{0}^{1}G(t,s)({a}_{2}{\int}_{0}^{1}G(s,\tau )\phi ({u}_{1}(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau +{b}_{2}\phi ({u}_{1}(s)))\phantom{\rule{0.2em}{0ex}}ds\\ =& ({a}_{2}{T}^{2}+{b}_{2}T)\phi ({u}_{1}(t)),\phantom{\rule{1em}{0ex}}t\in I.\end{array}

For \psi \in C[I,\mathbb{R}], let {L}_{2}\psi ={a}_{2}{T}^{2}\psi +{b}_{2}T\psi, then the above inequality can be written in the form

\phi ({u}_{1}(t))\ge {L}_{2}\left(\phi ({u}_{1}(t))\right).

(3.4)

It is easy to see that

\phi ({u}_{1}(t))\ne 0,\phantom{\rule{1em}{0ex}}t\in I.

In fact, \phi ({u}_{1}(t))=0 implies {u}_{1}(t)=\theta for t\in I, and consequently, {\parallel {u}_{1}\parallel}_{C}=0 in contradiction to {\parallel {u}_{1}\parallel}_{C}={r}_{3}. Now, notice that {L}_{2} is a {u}_{0}-positive operator with {u}_{0}(t)=sin\pi t, then by Lemma 2.2, we have \phi ({u}_{1}(t))={\mu}_{0}sin\pi t for some {\mu}_{0}>0. This together with (\frac{{a}_{2}}{{\pi}^{4}}+\frac{{b}_{2}}{{\pi}^{2}})sin\pi t={L}_{2}(sin\pi t) and (3.4) implies that

{\mu}_{0}sin\pi t=\phi ({u}_{1}(t))\ge {L}_{2}\left(\phi ({u}_{1}(t))\right)={L}_{2}({\mu}_{0}sin\pi t)={\mu}_{0}(\frac{{a}_{2}}{{\pi}^{4}}+\frac{{b}_{2}}{{\pi}^{2}})sin\pi t,

which is a contradiction to \frac{{a}_{2}}{{\pi}^{4}}+\frac{{b}_{2}}{{\pi}^{2}}>1. So, (3.3) holds.

By Lemma 2.4, *A* is a strict set contraction on {Q}_{{r}_{3},{R}_{2}}=\{u\in {Q}_{1}:{r}_{3}\le {\parallel u\parallel}_{C}\le {R}_{2}\}. Observing (3.2) and (3.3) and using Theorem 2.1, we see that *A* has a fixed point on {Q}_{{r}_{3},{R}_{2}}.

Next, in the case that ({H}_{4}) and ({H}_{5}) are satisfied, by the method as in establishing (3.3), we can assert from ({H}_{5}) that for any {r}_{4}\in (0,{r}_{2}),

Au\ngeqq u,\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in \partial {B}_{{r}_{4}}\cap {Q}_{1}.

(3.5)

Let

({L}_{3}v)(t)={\int}_{0}^{1}G(t,s)({a}_{3}{\int}_{0}^{1}G(s,\tau )v(\tau )\phantom{\rule{0.2em}{0ex}}d\tau +{b}_{3}v(s))\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}v\in C[I,\mathbb{R}].

It is clear that {L}_{3}:C[I,\mathbb{R}]\to C[I,\mathbb{R}] is a completely continuous linear {u}_{0}-operator with {u}_{0}(t)=sin\pi t and {L}_{3}:{K}_{1}\to {K}_{1} in which {K}_{1}=\{v\in K\subset C[I,\mathbb{R}]:v(t)\ge t(1-t)v(s),\mathrm{\forall}t,s\in I\}. In addition, the spectral radius r({L}_{3})=\frac{{a}_{3}}{{\pi}^{4}}+\frac{{b}_{3}}{{\pi}^{2}} and sin\pi t is the positive eigenfunction of {L}_{3} corresponding to its first eigenvalue {\lambda}_{1}={(r({L}_{3}))}^{-1}.

Let

({L}_{\delta}v)(t)={\int}_{\delta}^{1-\delta}G(t,s)({a}_{3}{\int}_{0}^{1}G(s,\tau )v(\tau )\phantom{\rule{0.2em}{0ex}}d\tau +{b}_{3}v(s))\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}v\in C[I,\mathbb{R}],

where \delta \in (0,\frac{1}{2}). It is clear that {L}_{\delta}:C[I,\mathbb{R}]\to C[I,\mathbb{R}] is a completely continuous linear {u}_{0}-operator with {u}_{0}(t)=t(1-t) and {L}_{\delta}({K}_{1})\subset {K}_{1}. Thus, the spectral radius r({L}_{\delta})\ne 0 and {L}_{\delta} has a positive eigenfunction corresponding to its first eigenvalue {\lambda}_{\delta}={(r({L}_{\delta}))}^{-1}.

Take {\delta}_{n}\in (0,1/2) (n=1,2,\dots) satisfying {\delta}_{1}\ge {\delta}_{2}\ge \cdots \ge {\delta}_{n}\ge \cdots and {\delta}_{n}\to 0 (n\to \mathrm{\infty}). For m>n, v\in {K}_{1}, we have

({L}_{{\delta}_{n}}v)(t)\le ({L}_{{\delta}_{m}}v)(t)\le ({L}_{3}v)(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in I.

By [12], we have r({L}_{{\delta}_{n}})\le r({L}_{{\delta}_{m}})\le r({L}_{3}). Let {\lambda}_{{\delta}_{n}}={(r({T}_{{\delta}_{n}}))}^{-1}, by Gelfand’s formula, we have {\lambda}_{{\delta}_{n}}\ge {\lambda}_{{\delta}_{m}}\ge {\lambda}_{1}. Let {\lambda}_{{\delta}_{n}}\to {\tilde{\lambda}}_{1} as n\to \mathrm{\infty}.

In the following, we prove that {\tilde{\lambda}}_{1}={\lambda}_{1}.

Let {v}_{{\delta}_{n}} be the positive eigenfunction of {T}_{{\delta}_{n}} corresponding to {\lambda}_{{\delta}_{n}}, *i.e.*,

\begin{array}{rcl}{v}_{{\delta}_{n}}(t)& =& {\lambda}_{{\delta}_{n}}({L}_{\delta}{v}_{{\delta}_{n}})(t)\\ =& {\lambda}_{{\delta}_{n}}{\int}_{{\delta}_{n}}^{1-{\delta}_{n}}G(t,s)({a}_{3}{\int}_{0}^{1}G(s,\tau ){v}_{{\delta}_{n}}(\tau )\phantom{\rule{0.2em}{0ex}}d\tau +{b}_{3}{v}_{{\delta}_{n}}(s))\phantom{\rule{0.2em}{0ex}}ds,\end{array}

(3.6)

satisfying \parallel {v}_{{\delta}_{n}}\parallel =1. Without loss of generality, by standard argument, we may suppose by the Arzela-Ascoli theorem and {\lambda}_{{\delta}_{n}}\to {\tilde{\lambda}}_{1} that {v}_{{\delta}_{n}}\to {\tilde{v}}_{0} as n\to \mathrm{\infty}. Thus, \parallel {\tilde{v}}_{0}\parallel =1 and by (3.6), we have

{\tilde{v}}_{0}(t)={\tilde{\lambda}}_{1}{\int}_{0}^{1}G(t,s)({a}_{3}{\int}_{0}^{1}G(s,\tau ){\tilde{v}}_{0}(\tau )\phantom{\rule{0.2em}{0ex}}d\tau +{b}_{3}{\tilde{v}}_{0}(s))\phantom{\rule{0.2em}{0ex}}ds,

that is, {\tilde{v}}_{0}(t)={\tilde{\lambda}}_{1}({L}_{3}{\tilde{v}}_{0})(t). This together with Lemma 2.2 guarantees that {\tilde{\lambda}}_{1}={\lambda}_{1}.

By the above argument, it is easy to see that there exists a \delta \in (0,\frac{1}{2}) such that

1<r({L}_{\delta})<\frac{{a}_{3}}{{\pi}^{4}}+\frac{{b}_{3}}{{\pi}^{2}}.

Choose

{R}_{3}>max\{{R}_{1},{r}_{2},\frac{30N{R}_{1}}{\delta (1-\delta )}\}.

(3.7)

Now, we assert that

If this is not true, then there exists {u}_{2}\in {Q}_{1} with {\parallel u\parallel}_{C}={R}_{3} such that A{u}_{2}\le {u}_{2}, then A{u}_{2}(t)\le {u}_{2}(t). Moreover, by the definition of {Q}_{1}, we know

\begin{array}{c}{u}_{2}(t)\ge t(1-t){u}_{2}(s),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t,s\in I,\hfill \\ (S{u}_{2})(t)={\int}_{0}^{1}G(t,s){u}_{2}(s)\phantom{\rule{0.2em}{0ex}}ds\ge t(1-t){\int}_{0}^{1}{s}^{2}{(1-s)}^{2}\phantom{\rule{0.2em}{0ex}}ds\cdot {u}_{2}(\tau ),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t,\tau \in I.\hfill \end{array}

Thus, N\parallel {u}_{2}(t)\parallel \ge t(1-t){\parallel {u}_{2}\parallel}_{C}, N\parallel (S{u}_{2})(t)\parallel \ge \frac{t(1-t)}{30}{\parallel {u}_{2}\parallel}_{C}, which implies by (3.7) we have

\underset{t\in [\delta ,1-\delta ]}{min}\parallel u(t)\parallel \ge \frac{\delta (1-\delta )}{N}{\parallel u\parallel}_{C}\ge {R}_{1},

and

\underset{t\in [\delta ,1-\delta ]}{min}\parallel (Su)(t)\parallel \ge \frac{\delta (1-\delta )}{30N}{\parallel u\parallel}_{C}\ge {R}_{1}.

So, by ({H}_{4}), we get

\begin{array}{rcl}\phi ({u}_{2}(t))& \ge & \phi ((A{u}_{2})(t))\\ =& \phi ({\int}_{0}^{1}G(t,s)f(s,(S{u}_{2})(s),{u}_{2}(s))\phantom{\rule{0.2em}{0ex}}ds)\\ \ge & {\int}_{\delta}^{1-\delta}G(t,s)\phi \left(f(s,(S{u}_{2})(s),{u}_{2}(s))\right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {\int}_{\delta}^{1-\delta}G(t,s)({a}_{3}\phi ((S{u}_{2})(s))+{b}_{3}\phi ({u}_{2}(s)))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {\int}_{\delta}^{1-\delta}G(t,s)({a}_{3}{\int}_{0}^{1}G(s,\tau )\phi ({u}_{2}(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau +{b}_{3}\phi ({u}_{2}(s)))\phantom{\rule{0.2em}{0ex}}ds\\ =& {L}_{\delta}\left(\phi ({u}_{2}(t))\right).\end{array}

It is easy to see that

\phi ({u}_{2}(t))\ne 0,\phantom{\rule{1em}{0ex}}t\in I.

In fact, \phi ({u}_{2}(t))=0 implies {u}_{2}(t)=\theta for t\in I, and consequently, {\parallel {u}_{2}\parallel}_{C}=0 in contradiction to {\parallel {u}_{2}\parallel}_{C}={R}_{3}. Now, notice that {L}_{\delta} is a {u}_{0}-positive operator with {u}_{0}(t)=t(1-t). Then by Lemma 2.2, we have \phi ({u}_{2}(t))={\mu}_{0}{v}_{\delta}(t) for some {\mu}_{0}>0, where {v}_{\delta} is the positive eigenfunction of {L}_{\delta} corresponding to {\lambda}_{\delta}. This together with r({L}_{\delta}){v}_{\delta}(t)={L}_{\delta}({v}_{\delta}(t)) implies that

{\mu}_{0}{v}_{\delta}(t)=\phi ({u}_{2}(t))\ge {L}_{\delta}\left(\phi ({u}_{2}(t))\right)={L}_{3}({\mu}_{0}{v}_{\delta}(t))={\mu}_{0}r({L}_{\delta}){v}_{\delta}(t),

which is a contradiction to r({L}_{\delta})>1. So, (3.8) holds.

By Lemma 2.4, *A* is a strict set contraction on {Q}_{{r}_{4},{R}_{3}}=\{u\in {Q}_{1}:{r}_{4}\le {\parallel u\parallel}_{C}\le {R}_{3}\}. Observing (3.5) and (3.8) and using Theorem 2.1, we see that *A* has a fixed point on {Q}_{{r}_{4},{R}_{3}}. This together with Lemma 2.3 implies that BVP (1.1) has at least one positive solution. □

**Theorem 3.2** *Let a cone* *P* *be normal*. *Suppose that conditions* ({H}_{1}), ({H}_{3}), ({H}_{4}) *and* ({H}_{6}) *are satisfied*. *Then BVP* (1.1) *has at least two positive solutions*.

*Proof* We can take the same {Q}_{1}\subset C[I,E] as in Theorem 3.1. As in the proof of Theorem 3.1, we can also obtain that A({Q}_{1})\subset {Q}_{1}. And we choose {r}_{5}, {R}_{4} with {R}_{4}>\eta >{r}_{5}>0 such that

On the other hand, it is easy to see that

Au\ngeqq u,\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in \partial {B}_{\eta}\cap {Q}_{1}.

(3.11)

In fact, if there exists {u}_{3}\in {Q}_{1} with {\parallel {u}_{3}\parallel}_{C}=\eta such that A{u}_{3}\ge {u}_{3}, then observing {max}_{t,s\in I}G(t,s)=\frac{1}{4} and {\parallel S{u}_{2}\parallel}_{C}\le \eta, we get

\begin{array}{rcl}\theta & \le & {u}_{3}(t)\le {\int}_{0}^{1}G(t,s)f(s,(S{u}_{3})(s),{u}_{3}(s))\phantom{\rule{0.2em}{0ex}}ds\\ \le & \frac{1}{4}{\int}_{0}^{1}f(s,(S{u}_{3})(s),{u}_{3}(s))\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in I,\end{array}

and so

\parallel {u}_{3}(t)\parallel \le \frac{N}{4}{\int}_{0}^{1}\parallel f(s,(S{u}_{3})(s),{u}_{3}(s))\parallel \phantom{\rule{0.2em}{0ex}}ds\le \frac{1}{4}NM,\phantom{\rule{1em}{0ex}}t\in I,

(3.12)

where, by virtue of ({H}_{5}),

M=\underset{t\in I,x,y\in P\cap {\mathrm{\Omega}}_{\eta}}{sup}\parallel f(t,x,y)\parallel <\frac{4\eta}{N}.

(3.13)

It follows from (3.12) and (3.13) that

\eta ={\parallel {u}_{3}\parallel}_{C}\le \frac{4\eta}{N}M<\eta ,

a contradiction. Thus (3.11) is true.

By Lemma 2.4, *A* is a strict set contraction on {Q}_{{r}_{5},\eta}=\{u\in {Q}_{1}|{r}_{5}\le {\parallel u\parallel}_{C}\le \eta \}, and also on {Q}_{\eta ,{R}_{4}}=\{u\in {Q}_{1}|\eta \le {\parallel u\parallel}_{C}\le {R}_{4}\}. Observing (3.9), (3.10), (3.11) and applying, respectively, Theorem 2.1 to *A*, {Q}_{{r}_{5},\eta} and {Q}_{\eta ,{R}_{4}}, we assert that there exist {u}_{1}\in {Q}_{{r}_{5},\eta} and {u}_{2}\in {Q}_{\eta ,{R}_{4}} such that A{u}_{1}={u}_{1} and A{u}_{2}={u}_{2} and, by Lemma 2.3 and (3.11), S{u}_{1}, S{u}_{2} are positive solutions of BVP (1.1). □

**Theorem 3.3** *Let a cone* *P* *be normal*. *Suppose that conditions* ({H}_{1}), ({H}_{2}) *and* ({H}_{5}) *and* ({H}_{7}) *are satisfied*. *Then BVP* (1.1) *has at least two positive solutions*.

*Proof* We can take the same {Q}_{1}\subset C[I,E] as in Theorem 3.1. As in the proof of Theorem 3.1, we can also obtain that A({Q}_{1})\subset {Q}_{1}. And we choose {r}_{6}, {R}_{5} with {R}_{5}>\eta >{r}_{6}>0 such that

On the other hand, it is easy to see that

In fact, if there exists {u}_{4}\in {Q}_{1} with {\parallel {u}_{4}\parallel}_{C}=\eta such that A{u}_{4}\ge {u}_{4}, then

\theta \le (A{u}_{4})(t)\le {u}_{4}(t),\phantom{\rule{1em}{0ex}}t\in I.

Observing {u}_{4}(t)\in {Q}_{1} and \frac{t(1-t)}{30}{u}_{4}(s)\le (S{u}_{4})(t)\le {u}_{4}(s), we get

\begin{array}{rcl}\eta & =& \parallel \phi \parallel {\parallel {u}_{4}\parallel}_{C}\ge \phi ({u}_{4}(t))\ge \phi ((A{u}_{4})(t))\\ =& {\int}_{0}^{1}G(t,s)\phi \left(f(s,(S{u}_{4})(s),{u}_{4}(s))\right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {\int}_{\tau}^{1-\tau}G(t,s)\phi \left(f(s,(S{u}_{4})(s),{u}_{4}(s))\right)\phantom{\rule{0.2em}{0ex}}ds\\ >& \alpha \eta \tau (1-\tau ){\int}_{\tau}^{1-\tau}s(1-s)\phantom{\rule{0.2em}{0ex}}ds=\eta ,\end{array}

which is a contradiction. Hence, (3.16) holds.

By Lemma 2.4, *A* is a strict set contraction on {Q}_{{r}_{6},\eta}=\{u\in {Q}_{1}|{r}_{6}\le {\parallel u\parallel}_{C}\le \eta \} and also on {Q}_{\eta ,{R}_{5}}=\{u\in {Q}_{1}|\eta \le {\parallel u\parallel}_{C}\le {R}_{5}\}. Observing (3.14), (3.15), (3.16) and applying, respectively, Theorem 2.1 to *A*, {Q}_{{r}_{6},\eta} and {Q}_{\eta ,{R}_{5}}, we assert that there exist {u}_{1}\in {Q}_{{r}_{6},\eta} and {u}_{2}\in {Q}_{\eta ,{R}_{5}} such that A{u}_{1}={u}_{1} and A{u}_{2}={u}_{2} and, by Lemma 2.3 and (3.16), S{u}_{1}, S{u}_{2} are positive solutions of BVP (1.1). □