**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)

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). □