In this section we will consider the existence of positive solutions of equation (1.1) for
$a\ge {\lambda}_{1}({\mathrm{\Omega}}_{0})$ and *c* large.
Our first result provides a necessary condition for existence of positive solutions for *c* large.

#### Theorem 3.1.

*Assume that $a\mathrm{\ge}{\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}\mathrm{\Omega}\mathrm{)}$ and $h\mathrm{\not\equiv}\mathrm{0}$ in $\mathrm{\Omega}\mathrm{\setminus}{\overline{\mathrm{\Omega}}}_{\mathrm{0}}$.
Then there exists $\mathrm{0}\mathrm{<}{\overline{c}}_{a}\mathrm{<}\mathrm{+}\mathrm{\infty}$ such that equation (1.1) does not have a nonnegative solution for $c\mathrm{>}{\overline{c}}_{a}$.*

#### Proof.

Let *u* be a nonnegative solution of (1.1).
Using the comparison Lemma 2.2, we have $u\le {\underset{\xaf}{U}}_{a}$ in $\mathrm{\Omega}\setminus {\overline{\mathrm{\Omega}}}_{0}$, where ${\underset{\xaf}{U}}_{a}$ is the minimal boundary blow-up solution given in Theorem 2.3.
Let ${\mathrm{\Omega}}_{n}=\{x\in \mathrm{\Omega}:d(x,{\mathrm{\Omega}}_{0})>\frac{1}{n}\}$.
Clearly, $h\not\equiv 0$ in ${\mathrm{\Omega}}_{n}$ for all *n* large.
Fix such an *n* and let ${\phi}_{n,1}>0$ be the first eigenfunction of $-\mathrm{\Delta}$ in ${\mathrm{\Omega}}_{n}$ with Dirichlet boundary condition. Multiplying equation (1.1) with ${\phi}_{n,1}$ and then integrating over ${\mathrm{\Omega}}_{n}$, we obtain

${\int}_{{\mathrm{\Omega}}_{n}}-\mathrm{\Delta}u{\phi}_{n,1}=a{\int}_{{\mathrm{\Omega}}_{n}}u{\phi}_{n,1}-{\int}_{{\mathrm{\Omega}}_{n}}b(x){u}^{2}{\phi}_{n,1}-c{\int}_{{\mathrm{\Omega}}_{n}}h(x){\phi}_{n,1},$

which implies

$c{\int}_{{\mathrm{\Omega}}_{n}}h(x){\phi}_{n,1}\le (a-{\lambda}_{1}({\mathrm{\Omega}}_{n})){\int}_{{\mathrm{\Omega}}_{n}}u{\phi}_{n,1}-{\int}_{\partial {\mathrm{\Omega}}_{n}}u{\partial}_{\nu}{\phi}_{n,1},$

where ν is the outer normal vector of $\partial {\mathrm{\Omega}}_{n}$.
Therefore, we have

$c{\int}_{{\mathrm{\Omega}}_{n}}h(x){\phi}_{n,1}\le (a-{\lambda}_{1}({\mathrm{\Omega}}_{n})){\int}_{{\mathrm{\Omega}}_{n}}{\underset{\xaf}{U}}_{a}{\phi}_{n,1}-{\int}_{\partial {\mathrm{\Omega}}_{n}}{\underset{\xaf}{U}}_{a}{\partial}_{\nu}{\phi}_{n,1}$

if $a>{\lambda}_{1}({\mathrm{\Omega}}_{n})$, and

$c{\int}_{{\mathrm{\Omega}}_{n}}h(x){\phi}_{n,1}\le -{\int}_{\partial {\mathrm{\Omega}}_{n}}{\underset{\xaf}{U}}_{a}{\partial}_{\nu}{\phi}_{n,1}$

if $a\le {\lambda}_{1}({\mathrm{\Omega}}_{n})$.
Therefore, *c* has a bound independent of *u*.
∎

In the rest of this section we will assume that $h\equiv 0$ in $\overline{\mathrm{\Omega}}\setminus {\mathrm{\Omega}}_{0}$. In order to prove the existence of positive solutions of (1.1) for $a\ge {\lambda}_{1}({\mathrm{\Omega}}_{0})$ and *c* large, we first show that (1.1) has an unstable solution for large *c* as *a* approaches ${\lambda}_{1}({\mathrm{\Omega}}_{0})$ from bellow.
Using this and the fact that unstable solutions can not blow up as $a\nearrow {\lambda}_{1}({\mathrm{\Omega}}_{0})$, we prove the existence of a positive solution for $a={\lambda}_{1}({\mathrm{\Omega}}_{0})$ by considering the limit of these unstable solutions.
Next, using the degree theory on cones we are able to establish the same result for *a* in a right neighborhood of ${\lambda}_{1}({\mathrm{\Omega}}_{0})$.
We start this procedure by first considering the behavior of ${\widehat{c}}_{a}$ as $a\nearrow {\lambda}_{1}({\mathrm{\Omega}}_{0})$.

#### Theorem 3.2.

*Assume that $h\mathrm{\equiv}\mathrm{0}$ in $\mathrm{\Omega}\mathrm{\setminus}{\overline{\mathrm{\Omega}}}_{\mathrm{0}}$. Then ${\widehat{c}}_{a}\mathrm{\to}\mathrm{+}\mathrm{\infty}$ as $a\mathrm{\nearrow}{\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}$.*

#### Proof.

Fix $c>0$ arbitrary large.
We will show that (1.1) has a positive solution for *a* sufficiently close ${\lambda}_{1}({\mathrm{\Omega}}_{0})$ by constructing an ordered pair of sub and super solutions.
Clearly, ${u}_{a}$ is a super solution of (1.1).
To construct a sub-solution, we consider

$\underset{\xaf}{u}=\{\begin{array}{cc}{w}_{n}(x),\hfill & x\in \mathrm{\Omega}\setminus {\overline{\mathrm{\Omega}}}_{0},\hfill \\ n+m{\phi}_{1}^{*}(x),\hfill & x\in {\overline{\mathrm{\Omega}}}_{0},\hfill \end{array}$

where *m* and *n* are large positive constants (given below), ${\phi}_{1}^{*}(x)$ denotes the first eigenfunction of $-\mathrm{\Delta}$ in ${\mathrm{\Omega}}_{0}$ with Dirichlet boundary condition, and ${w}_{n}$ is the positive solution of the equation

$-\mathrm{\Delta}w=aw-b(x){w}^{2}\mathit{\hspace{1em}}\text{in}\mathrm{\Omega}\setminus {\overline{\mathrm{\Omega}}}_{0},w=0\mathit{\hspace{1em}}\text{on}\partial \mathrm{\Omega},w=n\mathit{\hspace{1em}}\text{on}\partial {\mathrm{\Omega}}_{0},$

whose existence is shown in [8, Lemma 2.3].
We claim that $\underset{\xaf}{u}$ is a weak sub-solution of (1.1). First note that $\underset{\xaf}{u}$ is continuous. Next let $0\le \varphi \in {C}_{c}^{\mathrm{\infty}}(\mathrm{\Omega})$. We have

${\int}_{\mathrm{\Omega}}}\nabla \underset{\xaf}{u}\nabla \varphi ={\displaystyle {\int}_{\mathrm{\Omega}\setminus {\overline{\mathrm{\Omega}}}_{0}}}\nabla {w}_{n}\nabla \varphi +m{\displaystyle {\int}_{{\mathrm{\Omega}}_{0}}}\nabla {\phi}_{1}^{*}\nabla \varphi $$={\displaystyle {\int}_{\mathrm{\Omega}\setminus {\overline{\mathrm{\Omega}}}_{0}}}(-\mathrm{\Delta}{w}_{n})\varphi +m{\displaystyle {\int}_{{\mathrm{\Omega}}_{0}}}(-\mathrm{\Delta}{\phi}_{1}^{*})\varphi +{\displaystyle {\int}_{\partial {\mathrm{\Omega}}_{0}}}(m{\partial}_{\nu}{\phi}_{1}^{*}-{\partial}_{\nu}{w}_{n})\varphi $$=a{\displaystyle {\int}_{{\mathrm{\Omega}}_{0}\setminus {\overline{\mathrm{\Omega}}}_{0}}}({w}_{n}-b(x){w}_{n}^{2})\varphi +{\lambda}_{1}({\mathrm{\Omega}}_{0})m{\displaystyle {\int}_{{\mathrm{\Omega}}_{0}}}{\phi}_{1}^{*}\varphi +{\displaystyle {\int}_{\partial {\mathrm{\Omega}}_{0}}}(m{\partial}_{\nu}{\phi}_{1}^{*}-{\partial}_{\nu}{w}_{n})\varphi .$

Therefore,

${\int}_{\mathrm{\Omega}}}\nabla \underset{\xaf}{u}\nabla \varphi -{\displaystyle {\int}_{\mathrm{\Omega}}}(a\underset{\xaf}{u}+b(x){\underset{\xaf}{u}}^{2}-ch(x))\varphi \le {\displaystyle {\int}_{{\mathrm{\Omega}}_{0}}}\left(m({\lambda}_{1}({\mathrm{\Omega}}_{0})-a){\parallel {\phi}_{1}^{*}\parallel}_{\mathrm{\infty}}+c{\parallel h(x)\parallel}_{\mathrm{\infty}}-an\right)\varphi +{\displaystyle {\int}_{\partial {\mathrm{\Omega}}_{0}}}(m{\partial}_{\nu}{\phi}_{1}^{*}-{\partial}_{\nu}{w}_{n})\varphi .$

Now we assume that $a\in [{\lambda}_{1}({\mathrm{\Omega}}_{0})-{\u03f5}_{0},{\lambda}_{1}({\mathrm{\Omega}}_{0}))$ for some ${\u03f5}_{0}$ small.
First we take *n* large enough such that $({\lambda}_{1}({\mathrm{\Omega}}_{0})-{\u03f5}_{0})n-c{\parallel h\parallel}_{\mathrm{\infty}}\ge 1$.
Next, using the fact that ${\mathrm{max}}_{\partial {\mathrm{\Omega}}_{0}}{\partial}_{\nu}{\phi}_{1}^{*}<0$, we choose *m* sufficiently large so that $m{\partial}_{\nu}{\phi}_{1}^{*}-{\partial}_{\nu}{w}_{n}<0$ on $\partial {\mathrm{\Omega}}_{0}$.
Finally, we choose *a* close enough to ${\lambda}_{1}({\mathrm{\Omega}}_{0})$ such that ${u}_{a}>n+m{\phi}_{1}^{*}$ in ${\overline{\mathrm{\Omega}}}_{0}$ and $m({\lambda}_{1}({\mathrm{\Omega}}_{0})-a){\parallel {\phi}_{1}^{*}\parallel}_{\mathrm{\infty}}<1$.
Then $\underset{\xaf}{u}$ is a weak sub-solution and, in addition, $\underset{\xaf}{u}\le {u}_{a}$ (note that $h\equiv 0$ on $\mathrm{\Omega}\setminus {\overline{\mathrm{\Omega}}}_{0}$).
So, equation (1.1) has a positive solution between $\underset{\xaf}{u}$ and ${u}_{a}$. This completes the proof.
∎

Before stating our next result, we recall that by the anti-maximum principle (see [3]), there exists ${\delta}_{h}>0$ such that for ${\lambda}_{1}({\mathrm{\Omega}}_{0})<a<{\lambda}_{1}({\mathrm{\Omega}}_{0})+{\delta}_{h}$, the following equation has a unique positive solution ${\psi}_{a}$:

$-\mathrm{\Delta}\psi =a\psi -h(x),\text{in}{\mathrm{\Omega}}_{0},\psi =0\mathit{\hspace{1em}}\text{on}\partial {\mathrm{\Omega}}_{0}.$(3.1)

#### Lemma 3.3.

*There exists $\mathrm{0}\mathrm{<}\delta \mathrm{<}{\delta}_{h}$ such that if ${u}_{n}$ is a sequence of nonnegative solutions of equation (1.1), with $a\mathrm{=}{a}_{n}$, $c\mathrm{=}{c}_{n}$ such that ${\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}\mathrm{\Omega}\mathrm{)}\mathrm{<}{a}_{n}\mathrm{<}{\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}\mathrm{+}\delta $, ${c}_{n}\mathrm{\ge}\mathrm{0}$ and ${\mathrm{\parallel}{u}_{n}\mathrm{\parallel}}_{\mathrm{\infty}}\mathrm{\to}\mathrm{+}\mathrm{\infty}$.
Then ${u}_{n}\mathrm{\to}\mathrm{+}\mathrm{\infty}$ uniformly on ${\overline{\mathrm{\Omega}}}_{\mathrm{0}}$.*

We note that this lemma is in fact valid for arbitrary nonnegative $h(x)$.

#### Proof.

Up to a subsequence, ${a}_{n}\to {a}^{*}$ with ${\lambda}_{1}({\mathrm{\Omega}}_{0})\le {a}^{*}\le {\lambda}_{1}({\mathrm{\Omega}}_{0})+\delta $. We divide the proof into several steps.
*Step 1: ${u}_{n}\mathrm{\to}\mathrm{+}\mathrm{\infty}$ uniformly in every compact subsets of ${\mathrm{\Omega}}_{\mathrm{0}}$.*
Note that ${c}_{n}\lesssim {\parallel {u}_{n}\parallel}_{\mathrm{\infty}}$.
Thus, up to a subsequence, ${c}_{n}/{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}\to \alpha <+\mathrm{\infty}$. Let ${v}_{n}={u}_{n}/{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}$. From equation (1.1) we have

$-\mathrm{\Delta}{v}_{n}={a}_{n}{v}_{n}-b(x){\parallel {u}_{n}\parallel}_{\mathrm{\infty}}{v}_{n}^{2}-\frac{{c}_{n}}{{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}}h(x).$(3.2)

Since $-\mathrm{\Delta}{v}_{n}\le {a}_{n}{v}_{n}$ and ${\parallel {v}_{n}\parallel}_{\mathrm{\infty}}=1$, there exists ${v}_{0}\in {H}_{0}^{1}(\mathrm{\Omega})$ such that ${v}_{n}\to {v}_{0}$ weakly in ${H}_{0}^{1}(\mathrm{\Omega})$ and strongly in ${L}^{p}(\mathrm{\Omega})$ for $p>1$.
Multiplying (3.2) by ${v}_{n}/{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}$ and then integrating over Ω, we get

$\frac{1}{{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}}{\int}_{\mathrm{\Omega}}{|\nabla {v}_{n}|}^{2}=\frac{{a}_{n}}{{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}}{\int}_{\mathrm{\Omega}}{v}_{n}^{2}-{\int}_{\mathrm{\Omega}}b(x){v}_{n}^{3}-\frac{{c}_{n}}{{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}}{\int}_{\mathrm{\Omega}}h(x){v}_{n}.$

Thus, ${\int}_{\mathrm{\Omega}}b(x){v}_{0}^{3}=0$, and so ${v}_{0}\equiv 0$ in $\overline{\mathrm{\Omega}}\setminus {\mathrm{\Omega}}_{0}$.

Next, multiplying (3.2) by $\varphi \in {C}_{c}^{\mathrm{\infty}}({\mathrm{\Omega}}_{0})$ and integrating over ${\mathrm{\Omega}}_{0}$, we have

${\int}_{{\mathrm{\Omega}}_{0}}\nabla {v}_{n}\nabla \varphi ={a}_{n}{\int}_{{\mathrm{\Omega}}_{0}}{v}_{n}{\varphi}_{n}-\frac{{c}_{n}}{{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}}{\int}_{{\mathrm{\Omega}}_{0}}h(x)\varphi .$

Passing *n* to infinity, we obtain

${\int}_{{\mathrm{\Omega}}_{0}}\nabla {v}_{0}\nabla \varphi ={a}^{*}{\int}_{{\mathrm{\Omega}}_{0}}{v}_{0}\varphi -\alpha {\int}_{{\mathrm{\Omega}}_{0}}h(x)\varphi .$

Therefore, ${v}_{0}$ is a nonnegative weak solution of the equation

$-\mathrm{\Delta}{v}_{0}={a}^{*}{v}_{0}-\alpha h(x),\text{in}{\mathrm{\Omega}}_{0},{v}_{0}=0\mathit{\hspace{1em}}\text{on}\partial {\mathrm{\Omega}}_{0}.$

If ${a}^{*}={\lambda}_{1}({\mathrm{\Omega}}_{0})$, then, thanks to the Fredholm alternative, we have $\alpha =0$, since $h(x)\ge 0$ and $h(x)\not\equiv 0$. Therefore, ${v}_{0}=0$ or ${v}_{0}={\phi}_{1}^{*}>0$ in ${\mathrm{\Omega}}_{0}$ (recall that ${\phi}_{1}^{*}$ is the first eigenfunction of $-\mathrm{\Delta}$ with Dirichlet boundary condition in ${\mathrm{\Omega}}_{0}$).

We claim that ${v}_{0}\ne 0$.
In fact, if ${v}_{0}=0$, then, by equation (3.2), $-\mathrm{\Delta}{v}_{n}\le {a}_{n}{v}_{n}$, so that

$0\le {v}_{n}\le ({a}_{n}+1){(-\mathrm{\Delta}+1)}^{-1}{v}_{n}\to 0$

uniformly in Ω, as ${v}_{n}\to 0$ in ${L}^{p}(\mathrm{\Omega})$ for $p>1$.
This contradicts ${\parallel {v}_{n}\parallel}_{\mathrm{\infty}}=1$.

Next if ${\lambda}_{1}({\mathrm{\Omega}}_{0})<{a}^{*}\le {\lambda}_{1}({\mathrm{\Omega}}_{0})+\delta $, then assuming that $\alpha =0$, we have ${v}_{0}=0$, as ${a}^{*}\ne {\lambda}_{i}({\mathrm{\Omega}}_{0})$.
But this will again yield a contradiction as above.
Thus, $\alpha >0$ and ${v}_{0}=\alpha {\psi}_{{a}^{*}}>0$, the unique positive solution of equation (3.1).

Now standard elliptic estimates (see [11]) imply that ${u}_{n}\to +\mathrm{\infty}$ uniformly in every compact subset of ${\mathrm{\Omega}}_{0}$, completing the proof of step 1.

Next we let ${u}_{n}({z}_{n})={\mathrm{min}}_{x\in {\overline{\mathrm{\Omega}}}_{0}}{u}_{n}(x)$ and show that ${u}_{n}({z}_{n})\to \mathrm{\infty}$.
This will be done in the next two steps.
First note that since by assumption $\partial {\mathrm{\Omega}}_{0}$ is ${C}^{2,\gamma}$, it satisfies a uniform interior ball assumption, i.e., there exists $R>0$ such that for every $x\in \partial {\mathrm{\Omega}}_{0}$, there exists a ball ${B}_{x}={B}_{R}(y)$ with radius *R* and center *y* such that ${B}_{x}\subset {\overline{\mathrm{\Omega}}}_{0}$ and ${B}_{x}\cap \partial {\mathrm{\Omega}}_{0}=\{x\}$. Let *K* be a compact subset of ${\mathrm{\Omega}}_{0}$ such that ${B}_{R/2}(y)\subset K$ for all $x\in \partial {\mathrm{\Omega}}_{0}$.

*Step 2: Let ${u}_{n}\mathit{}\mathrm{(}{z}_{n}\mathrm{)}\mathrm{=}{\mathrm{min}}_{x\mathrm{\in}{\overline{\mathrm{\Omega}}}_{\mathrm{0}}}\mathit{}{u}_{n}\mathit{}\mathrm{(}x\mathrm{)}$. If $\mathrm{\{}{u}_{n}\mathit{}\mathrm{(}{z}_{n}\mathrm{)}\mathrm{\}}$ is bounded, then ${z}_{n}\mathrm{\in}\mathrm{\partial}\mathit{}{\mathrm{\Omega}}_{\mathrm{0}}$ for all **n* large.
Using some of the ideas of the proof of [8, Lemma 3.3], we argue by contradiction by assuming that for a subsequence, still denoted by $({z}_{n})$,
we have ${z}_{n}\in {\mathrm{\Omega}}_{0}$.
Then step 1 implies ${z}_{n}\to \partial {\mathrm{\Omega}}_{0}$. Hence, there exists ${x}_{n}\in \partial {\mathrm{\Omega}}_{0}$ such that ${z}_{n}\in {B}_{{x}_{n}}\setminus {B}_{R/2}({y}_{n})$.
Next we consider the auxiliary functions

${\eta}_{n}(x)={e}^{-\sigma {|x-{y}_{n}|}^{2}}-{e}^{-\sigma {R}^{2}},$

and show that there exists a sequence ${\beta}_{n}\to \mathrm{\infty}$ such that

${u}_{n}(x)\ge {u}_{n}({z}_{n})+{\beta}_{n}{\eta}_{n}(x)\mathit{\hspace{1em}}\text{in}{B}_{{x}_{n}}\setminus {B}_{R/2}({y}_{n}).$(3.3)

To start with, we fix a suitably chosen large $\sigma >0$, so that ${\eta}_{n}$ satisfy the following properties on ${B}_{{x}_{n}}$:

${\eta}_{n}(x)=0,$$x\in \partial {B}_{{x}_{n}},$${\eta}_{n}(x)>0,$$x\in {B}_{{x}_{n}}\setminus {B}_{R/2}({y}_{n}),$(3.4)${\eta}_{n}(x)={e}^{-\sigma {R}^{2}/4}-{e}^{-\sigma {R}^{2}}<{e}^{-\sigma {R}^{2}/4},$$x\in \partial {B}_{\frac{R}{2}}({y}_{n}),$${\partial}_{{\nu}_{n}}{\eta}_{n}({x}_{n})=2\sigma {R}^{2}{e}^{-\sigma {R}^{2}}>0,$$\text{where}{\nu}_{n}={\displaystyle \frac{{y}_{n}-{x}_{n}}{|{y}_{n}-{x}_{n}|}},$(3.5)

and

$\mathrm{\Delta}{\eta}_{n}+{a}_{n}{\eta}_{n}-{e}^{-\sigma {R}^{2}}h(x)=(4{\sigma}^{2}{|x-{y}_{n}|}^{2}-2\sigma N+{a}_{n}){e}^{-\sigma {|x-{y}_{n}|}^{2}}-({a}_{n}+h(x)){e}^{-\sigma {R}^{2}}$$>(4{\sigma}^{2}{|x-{y}_{n}|}^{2}-2\sigma N-h(x)){e}^{-\sigma {R}^{2}}>0,x\in {B}_{{x}_{n}}\setminus {B}_{R/2}({y}_{n}).$

Note that the choice of σ depends only on *R* and the function $h(x)$.
Let ${\alpha}_{n}={\mathrm{min}}_{K}{u}_{n}(x)$.
The rest of the proof proceeds by considering the two cases ${a}^{*}={\lambda}_{1}({\mathrm{\Omega}}_{0})$ and ${a}^{*}>{\lambda}_{1}({\mathrm{\Omega}}_{0})$ separately.

If ${a}^{*}={\lambda}_{1}({\mathrm{\Omega}}_{0})$, since ${u}_{n}/{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}\to {\phi}_{1}^{*}>0$ in ${\mathrm{\Omega}}_{0}$ and ${c}_{n}/{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}\to 0$, then ${\alpha}_{n}/{c}_{n}\to +\mathrm{\infty}$.
Hence, we can choose ${\beta}_{n}\to \mathrm{\infty}$ so that ${c}_{n}{e}^{\sigma {R}^{2}}\le {\beta}_{n}\le \frac{1}{2}{\alpha}_{n}{e}^{\sigma {R}^{2}/4}$.
Therefore, for $x\in {B}_{{x}_{n}}\setminus {B}_{R/2}({y}_{n})$, we have

$(-\mathrm{\Delta}-{a}_{n})\left({u}_{n}(x)-({u}_{n}({z}_{n})+{\beta}_{n}{\eta}_{n}(x))\right)\ge ({\beta}_{n}{e}^{-\sigma {R}^{2}}-{c}_{n})h(x)\ge 0$

and, in addition,

${u}_{n}(x)\ge {u}_{n}({z}_{n})+{\beta}_{n}{\eta}_{n}(x),x\in \partial ({B}_{{x}_{n}}\setminus {B}_{R/2}({y}_{n})).$

Since ${a}_{n}\to {\lambda}_{1}({\mathrm{\Omega}}_{0})<{\lambda}_{1}({B}_{{x}_{n}})$, the maximum principle finally yields (3.3).
Now taking $x={z}_{n}$ in (3.3), we have ${\beta}_{n}{\eta}_{n}({z}_{n})\le 0$, which, since ${z}_{n}\in {B}_{{x}_{n}}\setminus {B}_{R/2}({y}_{n})$, contradicts (3.4).

In the second case, that is, ${a}^{*}>{\lambda}_{1}({\mathrm{\Omega}}_{0})$, by step 1 we have
$0<\alpha ={lim}_{n\to +\mathrm{\infty}}{c}_{n}/{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}$ and ${v}_{n}\to {v}_{0}=\alpha {\psi}_{{a}^{*}}$.
Thus,

$\underset{n\to +\mathrm{\infty}}{lim}\frac{{\alpha}_{n}}{{c}_{n}}=\underset{n\to +\mathrm{\infty}}{lim}\frac{\frac{{\alpha}_{n}}{{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}}}{\frac{{c}_{n}}{{\parallel {u}_{n}\parallel}_{\mathrm{\infty}}}}\ge \frac{\alpha {\mathrm{min}}_{x\in K}{\psi}_{{a}^{*}}(x)}{\alpha}\ge \underset{x\in K}{\mathrm{min}}{\psi}_{{a}^{*}}(x).$

From (3.1) and Remark 2.9, we have

${\psi}_{{a}^{*}}=\frac{{h}_{1}}{{a}^{*}-{\lambda}_{1}({\mathrm{\Omega}}_{0})}{\phi}_{1}^{*}+{(\mathrm{\Delta}+{a}^{*})}^{-1}(\stackrel{~}{h}(x)).$

Hence, by further decreasing δ, we can guarantee that

$\underset{x\in K}{\mathrm{min}}{\psi}_{{a}^{*}}(x)=\underset{x\in K}{\mathrm{min}}\left(\frac{{h}_{1}}{{a}^{*}-{\lambda}_{1}({\mathrm{\Omega}}_{0})}{\phi}_{1}^{*}+{(\mathrm{\Delta}+{a}^{*})}^{-1}(\stackrel{~}{h}(x))\right)>{e}^{3\sigma {R}^{2}/4}.$

Now we may proceed as in the previous case, obtaining (3.3) and a contradiction as before.
The proof of step 2 is now complete.

*Step 3: Let ${u}_{n}\mathit{}\mathrm{(}{z}_{n}\mathrm{)}\mathrm{=}{\mathrm{min}}_{x\mathrm{\in}{\overline{\mathrm{\Omega}}}_{\mathrm{0}}}\mathit{}{u}_{n}\mathit{}\mathrm{(}x\mathrm{)}$. Then
${u}_{n}\mathit{}\mathrm{(}{z}_{n}\mathrm{)}\mathrm{\to}\mathrm{\infty}$ .*
To argue by contradiction, we assume that ${u}_{n}({z}_{n})$ is bounded, i.e., ${u}_{n}({z}_{n})\le M$ for some $M>0$ independent of *n*.
Then, by step 2, we have ${z}_{n}\in \partial {\mathrm{\Omega}}_{0}$. We follow the argument in step 2 where now ${z}_{n}={x}_{n}$, and conclude

${u}_{n}(x)\ge {u}_{n}({x}_{n})+{\beta}_{n}{\eta}_{n}(x)\mathit{\hspace{1em}}\text{in}{B}_{{x}_{n}}\setminus {B}_{R/2}({y}_{n})$(3.6)

for some sequence ${\beta}_{n}\to \mathrm{\infty}$.
Lemma 2.3 in [8] implies that the equation

$-\mathrm{\Delta}{w}_{n}={a}_{n}{w}_{n}-b(x){w}_{n}^{2},\text{in}\mathrm{\Omega}\setminus {\overline{\mathrm{\Omega}}}_{0},{w}_{n}=0\mathit{\hspace{1em}}\text{on}\partial \mathrm{\Omega},{w}_{n}={u}_{n}({x}_{n})\mathit{\hspace{1em}}\text{on}\partial {\mathrm{\Omega}}_{0},$

has a unique positive solution and by the comparison lemma, ${u}_{n}(x)\ge {w}_{n}(x)$ in $\overline{\mathrm{\Omega}}\setminus {\mathrm{\Omega}}_{0}$.
Similarly, *w*, the unique positive solution of

$-\mathrm{\Delta}w={a}^{*}w-b(x){w}^{2},\text{in}\mathrm{\Omega}\setminus {\overline{\mathrm{\Omega}}}_{0},w=0\mathit{\hspace{1em}}\text{on}\partial \mathrm{\Omega},w=M\mathit{\hspace{1em}}\text{on}\partial {\mathrm{\Omega}}_{0},$

satisfies ${w}_{n}(x)\le w(x)$ in $\overline{\mathrm{\Omega}}\setminus {\mathrm{\Omega}}_{0}$.
Thus, ${\parallel {w}_{n}\parallel}_{{L}^{\mathrm{\infty}}(\mathrm{\Omega}\setminus {\overline{\mathrm{\Omega}}}_{0})}$ is bounded, and therefore standard elliptic estimates imply that $\{{w}_{n}\}$ is bounded in ${C}^{1}(\overline{\mathrm{\Omega}}\setminus {\mathrm{\Omega}}_{0})$, and so, in particular, $|\nabla {w}_{n}({x}_{n})|$ remains uniformly bounded.
Since ${u}_{n}({x}_{n})={w}_{n}({x}_{n})$ and ${u}_{n}(x)\ge {w}_{n}(x)$ in $\overline{\mathrm{\Omega}}\setminus {\mathrm{\Omega}}_{0}$, we have

${\partial}_{{\nu}_{n}}({x}_{n})\le {\partial}_{{\nu}_{n}}{w}_{n}({x}_{n})\le {M}_{0}$(3.7)

for some ${M}_{0}>0$ independent of *n*.
On the other hand, using (3.6) and taking into account (3.5), we obtain

${\partial}_{{\nu}_{n}}{u}_{n}({x}_{n})\ge {\beta}_{n}{\partial}_{{\nu}_{n}}{\eta}_{n}({x}_{n})\to +\mathrm{\infty},$

contradicting (3.7). This completes the proof of step 3, and therefore the proof of the lemma.
∎

At this point for ${\lambda}_{1}(\mathrm{\Omega})<a<{\lambda}_{1}({\mathrm{\Omega}}_{0})$, we define

${\underset{\xaf}{c}}_{a}=sup\{c\ge 0:\text{equation (1.1) has a nonnegative solution on}\partial P\},$

where $\partial P$ is the boundary of the cone

$P=\{u\in {C}_{0}^{1}(\overline{\mathrm{\Omega}}):u(x)>0\text{for}x\in \mathrm{\Omega}\text{and}\frac{\partial u}{\partial \nu}(x)0\text{for}x\in \partial \mathrm{\Omega}\}$

and ${C}_{0}^{1}(\overline{\mathrm{\Omega}})$ denotes the subspace of functions in ${C}^{1}(\overline{\mathrm{\Omega}})$ which are zero on the boundary of Ω.
Note that $u=0$ is a nonnegative solution of (1.1) with $c=0$, so ${\underset{\xaf}{c}}_{a}$ is well defined and ${\underset{\xaf}{c}}_{a}\le {\widehat{c}}_{a}$.
In the following lemma we show that ${\underset{\xaf}{c}}_{a}$ does not go to infinity as $a\nearrow {\lambda}_{1}({\mathrm{\Omega}}_{0})$.

#### Lemma 3.4.

${lim\; sup}_{a\nearrow {\lambda}_{1}({\mathrm{\Omega}}_{0})}{\underset{\xaf}{c}}_{a}$
* is bounded.*

#### Proof.

Assume on the contrary that there exists a sequence ${a}_{n}$ such that ${a}_{n}\nearrow {\lambda}_{1}({\mathrm{\Omega}}_{0})$ and ${\underset{\xaf}{c}}_{{a}_{n}}\to +\mathrm{\infty}$.
Let ${u}_{n}\in \partial P$ be a sequence of nonnegative solutions of equation (1.1) with $a={a}_{n}$ and ${\underset{\xaf}{c}}_{{a}_{n}}-\u03f5\le {c}_{n}\le {\underset{\xaf}{c}}_{{a}_{n}}$ for some $\u03f5>0$.
Since ${c}_{n}\lesssim {\parallel {u}_{n}\parallel}_{\mathrm{\infty}}$, by Lemma 3.3, we have ${u}_{n}\to +\mathrm{\infty}$ uniformly in ${\overline{\mathrm{\Omega}}}_{0}$.
Thus, for large *n*, we have ${u}_{n}>0$ on $\partial {\mathrm{\Omega}}_{0}$. Now clearly we can choose ${M}_{n}>0$ large so that

$(-\mathrm{\Delta}+{M}_{n}){u}_{n}=({a}_{n}+{M}_{n}){u}_{n}-b(x){u}_{n}^{2}\ge 0\mathit{\hspace{1em}}\text{in}\mathrm{\Omega}\setminus {\overline{\mathrm{\Omega}}}_{0}.$

Hence, by the maximum principle and the Hopf boundary lemma ${u}_{n}>0$ in Ω and $\partial u/\partial \nu <0$ on $\partial \mathrm{\Omega}$, contradicting ${u}_{n}\in \partial P$.
∎

Our next result shows that for ${\underset{\xaf}{c}}_{a}<c<{\widehat{c}}_{a}$ equation (1.1) has an unstable positive solution.

#### Lemma 3.5.

*There exists $\delta \mathrm{>}\mathrm{0}$ such that if $a\mathrm{\in}\mathrm{(}{\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}\mathrm{-}\delta \mathrm{,}{\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}\mathrm{)}$ and ${\underset{\mathrm{\xaf}}{c}}_{a}\mathrm{<}c\mathrm{<}{\widehat{c}}_{a}$, then (1.1) has an unstable positive solution.*

#### Proof.

Since ${\widehat{c}}_{a}\to +\mathrm{\infty}$ as *a* approaches ${\lambda}_{1}({\mathrm{\Omega}}_{0})$, Lemma 3.4 implies the existence of $\delta >0$ such that ${\underset{\xaf}{c}}_{a}<{\widehat{c}}_{a}$ if $a\in ({\lambda}_{1}({\mathrm{\Omega}}_{0})-\delta ,{\lambda}_{1}({\mathrm{\Omega}}_{0}))$.
Next, given such an *a*, we fix ${\underset{\xaf}{c}}_{a}<{c}_{0}<{\widehat{c}}_{a}$ and $M>{\widehat{c}}_{a}$.
For the positive constants σ, ${K}_{\sigma}$ and $c\in [{c}_{0},M]$, we define

${T}_{\sigma}=\{u\in {C}_{0}^{1}(\overline{\mathrm{\Omega}}):\sigma {\phi}_{1}\le u\le {u}_{a}\text{in}\mathrm{\Omega}\text{and}{\displaystyle \frac{\partial {u}_{a}}{\partial \nu}}\le {\displaystyle \frac{\partial u}{\partial \nu}}\le \sigma {\displaystyle \frac{\partial {\phi}_{1}}{\partial \nu}}\text{on}\partial \mathrm{\Omega}\},$

where ${\phi}_{1}$ is the first eigenfunction of $-\mathrm{\Delta}$ in Ω with Dirichlet boundary condition and ${A}_{\sigma ,c}:{T}_{\sigma}\to {C}_{0}^{1}(\overline{\mathrm{\Omega}})$ is such that

${A}_{\sigma ,c}={(-\mathrm{\Delta}+{K}_{\sigma})}^{-1}\left((a+{K}_{\sigma})u-b(x){u}^{2}-ch(x)\right).$

Firstly, taking into account the definition of ${\underset{\xaf}{c}}_{a}$, a simple limiting argument implies that for σ sufficiently small, equation (1.1) has no solution on $\partial {T}_{\sigma}$ for $c\in [{c}_{0},M]$ (where $\partial {T}_{\sigma}$ denotes the relative boundary of ${T}_{\sigma}$ in *P*).
Fixing such a σ, we then take ${K}_{\sigma}$ sufficiently large so that ${A}_{\sigma ,c}$ maps ${T}_{\sigma ,c}$ into *P*.
Now for $c\in [{c}_{0},M]$, we have

$\mathrm{deg}(I-{A}_{\sigma ,c},{T}_{\sigma},0)=\mathrm{deg}(I-{A}_{\sigma ,M},{T}_{\sigma},0)=0.$(3.8)

On the other hand, by Theorem 1.3, for any $c<{\widehat{c}}_{a}$, (1.1) has a unique positive stable solution ${\overline{u}}_{a,c}$.
Moreover, using the fixed point index calculation of Dancer in [5], it is easily seen that

$\mathrm{ind}({A}_{\sigma ,c},{\overline{u}}_{a,c})=1.$(3.9)

Thus, (3.8) and (3.9) imply that for $c\in [{c}_{0},{\widehat{c}}_{a})$, equation (1.1) has another solution ${\underset{\xaf}{u}}_{a,c}$ in ${T}_{\sigma}$ which, by the uniqueness of stable solutions, satisfies ${\mu}_{1}({\underset{\xaf}{u}}_{a,c})\le 0$.
We claim that in fact ${\mu}_{1}({\underset{\xaf}{u}}_{a,c})<0$.
Indeed, if ${\mu}_{1}({\underset{\xaf}{u}}_{a,c})=0$, then applying the saddle-node bifurcation theorem, there exist $\u03f5>0$ and an open neighborhood $O\subset C(\overline{\mathrm{\Omega}})$ of ${\underset{\xaf}{u}}_{a,c}$, such that the solution set of (1.1) in $I=(c-\u03f5,c+\u03f5)\times O$ is an $\supset $-shaped curve.
Moreover, the upper part of the curve consists of stable solutions.
However, by the uniqueness of the curve of stable solutions, this can only happen if $c={\widehat{c}}_{a}$.
This completes the proof.
∎

#### Theorem 3.6.

*Assume that $a\mathrm{=}{\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}$.
There exists ${c}_{\mathrm{0}}\mathrm{>}{\mathrm{lim\; sup}}_{a\mathrm{\nearrow}{\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}}\mathit{}{\underset{\mathrm{\xaf}}{c}}_{a}$ such that for all $c\mathrm{>}{c}_{\mathrm{0}}$, equation (1.1) has a positive solution.*

#### Proof.

First fix $c>{lim\; sup}_{a\nearrow {\lambda}_{1}({\mathrm{\Omega}}_{0})}{\underset{\xaf}{c}}_{a}$.
Lemma 3.5 implies the existence of a sequence of unstable positive solutions ${u}_{n}$ of equation (1.1) with $a={a}_{n}<{\lambda}_{1}({\mathrm{\Omega}}_{0})$ and ${a}_{n}\nearrow {\lambda}_{1}({\mathrm{\Omega}}_{0})$.
We claim that ${\parallel {u}_{n}\parallel}_{\mathrm{\infty}}$ is bounded.
Otherwise, ${\parallel {u}_{n}\parallel}_{\mathrm{\infty}}\to +\mathrm{\infty}$, and therefore by Lemma 3.3, ${u}_{n}\to +\mathrm{\infty}$ uniformly in ${\overline{\mathrm{\Omega}}}_{0}$.
Hence, there exist ${n}_{1}$ and ${n}_{2}$ such that ${u}_{{n}_{2}}>{u}_{{n}_{1}}$ in ${\overline{\mathrm{\Omega}}}_{0}$ and ${a}_{{n}_{2}}>{a}_{{n}_{1}}$.
Now an application of the comparison lemma in $\mathrm{\Omega}\setminus {\overline{\mathrm{\Omega}}}_{0}$ easily implies ${u}_{{n}_{2}}\ge {u}_{{n}_{1}}$ in all of Ω (note that ${u}_{{n}_{2}}$ is a super solution and ${u}_{{n}_{1}}$ a solution of the logistic equation $-\mathrm{\Delta}u={a}_{{n}_{1}}u-b(x){u}^{2}$ on $\mathrm{\Omega}\setminus {\overline{\mathrm{\Omega}}}_{0}$).
Also, as ${a}_{{n}_{1}}<{a}_{{n}_{2}}$, we have that ${u}_{{n}_{1}}$ is a sub-solution of (1.1) with $a={a}_{{n}_{2}}$.
Therefore, equation (1.1) for $a={a}_{{n}_{2}}$ has a minimal solution ${u}_{0}$ between ${u}_{{n}_{1}}$ and ${u}_{{n}_{2}}$ with ${\mu}_{1}({u}_{0})\ge 0$ (see [17, Theorem 4.1]).
Hence, ${\mu}_{1}({u}_{{n}_{2}})\ge {\mu}_{1}({u}_{0})\ge 0$, contradicting the fact that ${u}_{{n}_{2}}$ is unstable.

Therefore, ${\parallel {u}_{n}\parallel}_{\mathrm{\infty}}$ is bounded and ${u}_{n}\to {u}_{c}^{*}$ weakly in ${H}_{0}^{1}(\mathrm{\Omega})$ and strongly in ${L}^{p}(\mathrm{\Omega})$ for some ${u}_{c}^{*}\in {H}_{0}^{1}(\mathrm{\Omega})$.
Obviously, ${u}_{c}^{*}$ is a nonnegative solution of (1.1) with $a={\lambda}_{1}({\mathrm{\Omega}}_{0})$.

Therefore, for all $c>{lim\; sup}_{a\nearrow {\lambda}_{1}({\mathrm{\Omega}}_{0})}{\underset{\xaf}{c}}_{a}$, equation (1.1) has a nonnegative solution ${u}_{c}^{*}$.
Since $c\lesssim {\parallel {u}_{c}^{*}\parallel}_{\mathrm{\infty}}$, by Lemma 3.3, we get ${u}_{c}^{*}\to +\mathrm{\infty}$ uniformly in ${\overline{\mathrm{\Omega}}}_{0}$ as $c\to +\mathrm{\infty}$.
Therefore, there exists ${c}_{0}$ such that ${u}_{c}^{*}>0$ in ${\overline{\mathrm{\Omega}}}_{0}$ for $c>{c}_{0}$, and a further application of the maximum principle in $\mathrm{\Omega}\setminus {\overline{\mathrm{\Omega}}}_{0}$ (see the proof of Lemma 3.4) yields ${u}_{c}^{*}>0$ in Ω.
∎

#### Lemma 3.7.

*Let ${\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}\mathrm{\le}a\mathrm{\le}{\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}\mathrm{+}\delta $ with δ as in Lemma 3.3. There exists ${c}_{\mathrm{1}}\mathrm{=}{c}_{\mathrm{1}}\mathit{}\mathrm{(}a\mathrm{)}$ such that every positive solution of (1.1) with $c\mathrm{\ge}{c}_{\mathrm{1}}$ is nondegenerate.*

#### Proof.

Fix an *a*. Assume on the contrary that ${u}_{n}$ is a sequence of solutions of (1.1) corresponding to ${c}_{n}\to +\mathrm{\infty}$ and ${\mu}_{{i}_{n}}({u}_{n})=0$ for some ${i}_{n}\in \mathbb{N}$.
From Lemma 3.3, we have ${u}_{n}\to +\mathrm{\infty}$ uniformly in ${\overline{\mathrm{\Omega}}}_{0}$.
Thus, for a subsequence ${u}_{{n}_{1}}<{u}_{{n}_{2}}<{u}_{{n}_{3}}<\mathrm{\cdots}$ in ${\overline{\mathrm{\Omega}}}_{0}$ and therefore (by an application of the comparison lemma, as in the proof of Theorem 3.6) in all of Ω.
Since ${\mu}_{i}(u)={\lambda}_{i}(-a+2b(x)u)$, we have ${\mu}_{i}({u}_{{n}_{j}})>0$ for all $j>1$ and $i\ge {i}_{{n}_{1}}$.
On the other hand, since ${\mu}_{{i}_{{n}_{j}}}({u}_{{n}_{j}})=0$, we have ${i}_{{n}_{j}}<{i}_{{n}_{1}}$. Therefore, there exists a fixed $k\le {i}_{{n}_{1}}$ such that ${\mu}_{k}({u}_{{n}_{j}})=0$ for all $j\in \mathbb{N}$.
This contradicts the fact that the sequence ${u}_{{n}_{j}}$ is strictly increasing, and therefore so is ${\mu}_{k}({u}_{{n}_{j}})$.
∎

#### Lemma 3.8.

*Let ${\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}\mathrm{\le}a\mathrm{\le}{\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}\mathrm{+}\delta $ with δ as in Lemma 3.3.
There exists ${c}_{\mathrm{2}}\mathrm{=}{c}_{\mathrm{2}}\mathit{}\mathrm{(}a\mathrm{)}$ such that for $c\mathrm{\ge}{c}_{\mathrm{2}}$, equation (1.1) has a at most one positive solution.*

#### Proof.

Fix an *a*.
Assume on the contrary that ${u}_{n}$ and ${\overline{u}}_{n}$ are two sequences of positive solutions of equation (1.1) corresponding to ${c}_{n}\to +\mathrm{\infty}$.
If we subtract the equations for ${u}_{n}$ and ${\overline{u}}_{n}$, then we obtain that ${\lambda}_{{i}_{n}}(-a+b(x)({u}_{n}+{\overline{u}}_{n}))=0$ for some ${i}_{n}\in \mathbb{N}$.
By Lemma 3.3, we have ${u}_{n}+{\overline{u}}_{n}\to +\mathrm{\infty}$ uniformly in ${\overline{\mathrm{\Omega}}}_{0}$.
Thus, there exists a subsequence ${u}_{{n}_{j}}+{\overline{u}}_{{n}_{j}}$ such that ${u}_{{n}_{1}}+{\overline{u}}_{{n}_{1}}<{u}_{{n}_{2}}+{\overline{u}}_{{n}_{2}}<{u}_{{n}_{3}}+{\overline{u}}_{{n}_{3}}<\mathrm{\cdots}$ in ${\overline{\mathrm{\Omega}}}_{0}$, and therefore by the comparison lemma in all of Ω.
Therefore, ${\lambda}_{i}(-a+b(x)({u}_{{n}_{j}}+{\overline{u}}_{{n}_{j}}))>0$ for all $j>1$ and $i\ge {i}_{{n}_{1}}$.
On the other hand, since ${\lambda}_{{i}_{{n}_{j}}}(-a+b(x)({u}_{{n}_{j}}+{\overline{u}}_{{n}_{j}}))=0$, we have ${i}_{{n}_{j}}<{i}_{{n}_{1}}$.
We can now continue as in the proof of Lemma 3.7 and reach a contradiction as before.
∎

The next two results prepare the ground for the application of degree theory arguments in order to prove our main existence result on positive solutions of (1.1) for all ${\lambda}_{1}({\mathrm{\Omega}}_{0})\le a\le {\lambda}_{1}({\mathrm{\Omega}}_{0})+\delta $ and *c* large.
In what follows, we let

${c}^{*}=\mathrm{max}\{{c}_{0},{c}_{1}({\lambda}_{1}({\mathrm{\Omega}}_{0})),{c}_{2}({\lambda}_{1}({\mathrm{\Omega}}_{0}))\}.$

Note that the above results imply that equation (1.1) for $a={\lambda}_{1}({\mathrm{\Omega}}_{0})$ and $c\ge {c}^{*}$ has a unique positive solution which, in addition, is nondegenerate.

#### Lemma 3.9.

*Let $\delta \mathrm{>}\mathrm{0}$ be defined as in Lemma 3.3 and let $d\mathrm{\ge}{c}^{\mathrm{*}}$ be a given constant.
There exists $K\mathrm{>}\mathrm{0}$, depending on δ and **d*, such that if *u* is a nonnegative solution of (1.1) with ${\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}\mathrm{\le}a\mathrm{\le}{\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}\mathrm{+}\delta $ and ${c}^{\mathrm{*}}\mathrm{\le}c\mathrm{\le}d$, then ${\mathrm{\parallel}u\mathrm{\parallel}}_{\mathrm{\infty}}\mathrm{<}K$.

#### Proof.

Assume on the contrary that there exists a sequence ${u}_{n}$ of nonnegative solutions of (1.1) with $a={a}_{n}$ and ${c}^{*}\le c={c}_{n}\le d$, where ${\lambda}_{1}({\mathrm{\Omega}}_{0})\le {a}_{n}\le {\lambda}_{1}({\mathrm{\Omega}}_{0})+\delta $, and ${\parallel {u}_{n}\parallel}_{\mathrm{\infty}}\to +\mathrm{\infty}$.
Now, by Lemma 3.3, we have ${u}_{n}\to +\mathrm{\infty}$ in ${\overline{\mathrm{\Omega}}}_{0}$.
Denoting the unique positive solution of (1.1) for $a={\lambda}_{1}({\mathrm{\Omega}}_{0})$ and $c=d$ by ${u}^{*}$, we have for ${n}_{0}$ large, ${u}_{{n}_{0}}>{u}^{*}$ in ${\overline{\mathrm{\Omega}}}_{0}$, and then by the comparison lemma in all of Ω (note that ${u}_{{n}_{0}}$ is a super solution and ${u}^{*}$ a solution of the logistic equation $-\mathrm{\Delta}u={\lambda}_{1}({\mathrm{\Omega}}_{0})u-b(x){u}^{2}$ on $\mathrm{\Omega}\setminus {\overline{\mathrm{\Omega}}}_{0}$).
Hence, ${u}^{*},{u}_{{n}_{0}}$ is an ordered pair of sub-super solution of (1.1) with $a={\lambda}_{1}({\mathrm{\Omega}}_{0})$ and $c=d$, and therefore (1.1) has a solution ${u}_{0}$ (achieved with the iteration process starting at ${u}_{{n}_{0}}$) between ${u}_{c}^{*}$ and ${u}_{{n}_{0}}$ with ${\mu}_{1}({u}_{0})\ge 0$. This contradicts Theorem 2.6.
∎

#### Lemma 3.10.

*There exists ${c}_{\mathrm{3}}\mathrm{>}\mathrm{0}$ such that (1.1) has no solution on $\mathrm{\partial}\mathit{}P$, provided that ${\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}\mathrm{\le}a\mathrm{\le}{\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}\mathrm{+}\delta $ with δ as in Lemma 3.3 and $c\mathrm{\ge}{c}_{\mathrm{3}}$.
*

#### Proof.

Assume on the contrary that ${u}_{n}\in \partial P$ is a sequence of solutions of (1.1) with $a={a}_{n}$ and $c={c}_{n}$ with ${\lambda}_{1}({\mathrm{\Omega}}_{0})\le a\le {\lambda}_{1}({\mathrm{\Omega}}_{0})+\delta $ and ${c}_{n}\to +\mathrm{\infty}$.
By Lemma 3.3, ${u}_{n}\to +\mathrm{\infty}$ uniformly in ${\overline{\mathrm{\Omega}}}_{0}$, and therefore, by the maximum principle (as in the proof of Lemma 3.4), ${u}_{n}>0$ in Ω for large *n*, contradicting ${u}_{n}\in \partial P$.
∎

We are now ready to state the main result of this paper.

#### Theorem 3.11.

*Let ${\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}\mathrm{\le}a\mathrm{\le}{\lambda}_{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{\Omega}}_{\mathrm{0}}\mathrm{)}\mathrm{+}\delta $ with δ as in Lemma 3.3.
Then, for $c\mathrm{\ge}\mathrm{max}\mathit{}\mathrm{\{}{c}^{\mathrm{*}}\mathrm{,}{c}_{\mathrm{3}}\mathrm{\}}$ equation (1.1) has an unstable positive solution.
Furthermore, for each **a*, there exists $c\mathit{}\mathrm{(}a\mathrm{)}$, such that for $c\mathrm{\ge}c\mathit{}\mathrm{(}a\mathrm{)}$, the positive solution is unique and nondegenerate.

#### Proof.

The only statement that requires a proof is the existence of a positive solution for $c\ge \mathrm{max}\{{c}^{*},{c}_{3}\}$. The proof uses similar arguments as in the proof of Theorem 3.6, so we shall brief them here.
Fixing a $c\ge \mathrm{max}\{{c}^{*},{c}_{3}\}$, we define

${T}_{e,K}=\{u\in {C}_{0}^{1}(\overline{\mathrm{\Omega}}):e{\phi}_{1}\le u\le K\text{in}\mathrm{\Omega}\text{and}\frac{\partial}{\partial \nu}(u-e{\phi}_{1})\le 0\text{on}\partial \mathrm{\Omega}\},$

where *e* and *K* are positive constants.
First notice that by Lemmas 3.9 and 3.10, we can choose *e* sufficiently small and *K* sufficiently large so that (1.1) has no solution on $\partial {T}_{e,K}$ (the relative boundary of ${T}_{e,K}$ in *P*) for ${\lambda}_{1}({\mathrm{\Omega}}_{0})\le a\le {\lambda}_{1}({\mathrm{\Omega}}_{0})+\delta $.
Next we define ${A}_{a}:{C}_{0}^{1}(\overline{\mathrm{\Omega}})\to {C}_{0}^{1}(\overline{\mathrm{\Omega}})$ by

${A}_{a}(u)={(-\mathrm{\Delta}+L)}^{-1}\left((a+L)u-b(x){u}^{2}-ch(x)\right).$

By taking $L>0$ sufficiently large, we may assume that $(a+L)u-b(x){u}^{2}-ch(x)$ is increasing in $[e{\phi}_{1}(x),K]$ for all $x\in \mathrm{\Omega}$ and that ${A}_{a,c}$ maps ${T}_{e,K}$ into *P*.
Indeed, if $u\in {T}_{e,K}$ and ${u}_{0}={A}_{a}(u)$, then we have

${u}_{0}={(-\mathrm{\Delta}+L)}^{-1}\left((a+L)u-b(x){u}^{2}-ch(x)\right)$$\ge {(-\mathrm{\Delta}+L)}^{-1}\left((a+L)e{\phi}_{1}-b(x){e}^{2}{\phi}_{1}^{2}-ch(x)\right)\ge 0,$

as $(a+L)e{\phi}_{1}-b(x){e}^{2}{\phi}_{1}^{2}-ch(x)\ge 0$ for $L>0$ large. Hence, for *e* small and *L* large, $\mathrm{deg}(I-{A}_{a},{T}_{e,K},0)$ is admissible for ${\lambda}_{1}({\mathrm{\Omega}}_{0})\le a\le {\lambda}_{1}({\mathrm{\Omega}}_{0})+\delta $.
Now, by Lemmas 3.7 and 3.8, we have

$\mathrm{deg}(I-{A}_{{\lambda}_{1}({\mathrm{\Omega}}_{0})},{T}_{e,K},0)\ne 0.$

Hence,

$\mathrm{deg}(I-{A}_{a},{T}_{e,K},0)=\mathrm{deg}({A}_{{\lambda}_{1}({\mathrm{\Omega}}_{0})},{T}_{e,K},0)\ne 0$

for all $a\in [{\lambda}_{1}({\mathrm{\Omega}}_{0}),{\lambda}_{1}({\mathrm{\Omega}}_{0})+\delta ]$. The proof is now complete.
∎

Finally, it is an interesting open problem to study existence of positive solutions of (1.1) for all *c* large when *a* is large and away from ${\lambda}_{1}({\mathrm{\Omega}}_{0})$.
In particular, for any $a>{\lambda}_{1}({\mathrm{\Omega}}_{0})$, it is easily seen (through a familiar limiting argument) that the existence of a positive solution to the equation

$-\mathrm{\Delta}u=au-h(x)\mathit{\hspace{1em}}\text{in}{\mathrm{\Omega}}_{0},u=0\mathit{\hspace{1em}}\text{on}\partial {\mathrm{\Omega}}_{0},$

is a necessary condition for existence of a positive solutions to (1.1) as $c\nearrow \mathrm{\infty}$.
Although the techniques used in this paper seem inadequate to deal with the question of sufficiency of this necessary condition, we do believe that some of the ideas used here should be of value in dealing with this problem.

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