Throughout this subsection, we assume that $p<{2}^{*}$, so that weak solutions of ((${P_{\lambda}}$)) are critical points of the ${C}^{1}$ functional ${I}_{\lambda}$, defined on *X* by

${I}_{\lambda}(u):={\displaystyle \frac{1}{2}}E(u)-{\displaystyle \frac{1}{p}}A(u)-{\displaystyle \frac{\lambda}{q}}B(u),$

where

$E(u)={\displaystyle {\int}_{\mathrm{\Omega}}}{|\nabla u|}^{2},A(u)={\displaystyle {\int}_{\mathrm{\Omega}}}{|u|}^{p},B(u)={\displaystyle {\int}_{\partial \mathrm{\Omega}}}b(x){|u|}^{q}.$

Let us recall that $X={H}^{1}(\mathrm{\Omega})$ is equipped with the usual norm

$\parallel u\parallel ={\left[{\displaystyle {\int}_{\mathrm{\Omega}}}({|\nabla u|}^{2}+{u}^{2})\right]}^{\frac{1}{2}}.$

We shall study the geometry of ${I}_{\lambda}$ to obtain non-negative weak solutions of ((${P_{\lambda}}$)) for a small $\lambda >0$.

The next result will be used repeatedly in this section.

#### Lemma 1.4.

*The following statements hold:*

(i)

*If *
$({u}_{n})$
* is a sequence such that *
${u}_{n}\rightharpoonup {u}_{0}$
* in *
*X*
* and *
$limE({u}_{n})=0$
*, then *
${u}_{0}$
* is a constant and *
${u}_{n}\to {u}_{0}$
* in *
*X*.

(ii)

*Assume *
${\int}_{\partial \mathrm{\Omega}}b<0$
*. If *
$v\not\equiv 0$
* and *
$B(v)\ge 0$
*, then *
*v*
* is not a constant.*

#### Proof.

(i) Since ${u}_{n}\rightharpoonup {u}_{0}$ in *X* and *E* is weakly lower semicontinuous, we have $0\le E({u}_{0})\le limE({u}_{n})=0$.
Hence, $E({u}_{0})=0$, which implies that ${u}_{0}$ is a constant. In addition, if ${u}_{n}\nrightarrow {u}_{0}$ in *X*, then $E({u}_{0})<limE({u}_{n})=0$, which is a contradiction. Therefore, ${u}_{n}\to {u}_{0}$ in *X*.

(ii) If *v* is a non-zero constant and $B(v)\ge 0$, then
$B(v)={|v|}^{p}{\int}_{\partial \mathrm{\Omega}}b<0$, which yields a contradiction.
∎

Let us introduce some useful subsets of *X*:

${E}^{+}=\{u\in X:E(u)>0\},$${A}^{+}=\{u\in X:A(u)>0\},$${B}^{\pm}=\{u\in X:B(u)\gtrless 0\},{B}_{0}=\{u\in X:B(u)=0\},{B}_{0}^{\pm}={B}^{\pm}\cup {B}_{0}.$

The Nehari manifold associated to ${I}_{\lambda}$ is given by

${N}_{\lambda}:=\{u\in X\setminus \{0\}:\u3008{I}_{\lambda}^{\prime}(u),u\u3009=0\}=\{u\in X\setminus \{0\}:E(u)=A(u)+\lambda B(u)\}.$

We shall use the splitting

${N}_{\lambda}={N}_{\lambda}^{+}\cup {N}_{\lambda}^{-}\cup {N}_{\lambda}^{0},$

where

${N}_{\lambda}^{\pm}:=\{u\in {N}_{\lambda}:\u3008{J}_{\lambda}^{\prime}(u),u\u3009\gtrless 0\}=\{u\in {N}_{\lambda}:E(u)\lessgtr \lambda {\displaystyle \frac{p-q}{p-2}}B(u)\}$$=\{u\in {N}_{\lambda}:E(u)\gtrless {\displaystyle \frac{p-q}{2-q}}A(u)\}$

and

${N}_{\lambda}^{0}=\{u\in {N}_{\lambda}:\u3008{J}_{\lambda}^{\prime}(u),u\u3009=0\}.$

Here, ${J}_{\lambda}(u)=\u3008{I}_{\lambda}^{\prime}(u),u\u3009=E(u)-A(u)-\lambda B(u)$.

Note that any nontrivial weak solution of ((${P_{\lambda}}$)) belongs to ${N}_{\lambda}$. Furthermore, it follows from the implicit function theorem that ${N}_{\lambda}\setminus {N}_{\lambda}^{0}$ is a ${C}^{1}$ manifold and local minimizers of the restriction of ${I}_{\lambda}$ to this manifold are critical points of ${I}_{\lambda}$ (see for instance [7, Theorem 2.3]).

To analyze the structure of ${N}_{\lambda}^{\pm}$, we consider the fibering maps corresponding to ${I}_{\lambda}$, which are defined, for $u\ne 0$, as follows:

${j}_{u}(t):={I}_{\lambda}(tu)=\frac{{t}^{2}}{2}E(u)-\frac{{t}^{p}}{p}A(u)-\lambda \frac{{t}^{q}}{q}B(u),t>0.$

It is easy to see that

${j}_{u}^{\prime}(1)=0\lessgtr {j}_{u}^{\prime \prime}(1)\iff u\in {N}_{\lambda}^{\pm}$

and, more generally,

${j}_{u}^{\prime}(t)=0\lessgtr {j}_{u}^{\prime \prime}(t)\iff tu\in {N}_{\lambda}^{\pm}.$

Having this characterization in mind, we look for conditions under which ${j}_{u}$ has a critical point. Set

${i}_{u}(t):={t}^{-q}{j}_{u}(t)=\frac{{t}^{2-q}}{2}E(u)-\frac{{t}^{p-q}}{p}A(u)-\lambda B(u),t>0.$

Let $u\in {E}^{+}\cap {A}^{+}\cap {B}^{+}$. Then ${i}_{u}$ has a global maximum ${i}_{u}({t}^{*})$ at some ${t}^{*}>0$ and, moreover, ${t}^{*}$ is unique. If ${i}_{u}({t}^{*})>0$, then ${j}_{u}$ has a global maximum which is positive and a local minimum which is negative. Moreover, these are the only critical points of ${j}_{u}$.

We shall require a condition on λ that provides ${i}_{u}({t}^{*})>0$. Note that

${i}_{u}^{\prime}(t)=\frac{2-q}{2}{t}^{1-q}E(u)-\frac{p-q}{p}{t}^{p-q-1}A(u)=0$

if and only if

$t={t}^{*}:={\left(\frac{p(2-q)E(u)}{2(p-q)A(u)}\right)}^{\frac{1}{p-2}}.$

Moreover,

${i}_{u}({t}^{*})=\frac{p-2}{2(p-q)}{\left(\frac{p(2-q)}{2(p-q)}\right)}^{\frac{2-q}{p-2}}\frac{E{(u)}^{\frac{p-q}{p-2}}}{A{(u)}^{\frac{2-q}{p-2}}}-\frac{\lambda}{q}B(u)>0$

if and only if

$0<\lambda <{C}_{pq}\frac{E{(u)}^{\frac{p-q}{p-2}}}{B(u)A{(u)}^{\frac{2-q}{p-2}}},$(1.4)

where

${C}_{pq}=\left(\frac{q(p-2)}{2(p-q)}\right){\left(\frac{p(2-q)}{2(p-q)}\right)}^{\frac{2-q}{p-2}}.$

Note that

$F(u)=\frac{E{(u)}^{\frac{p-q}{p-2}}}{B(u)A{(u)}^{\frac{2-q}{p-2}}}$

satisfies $F(tu)=F(u)$ for $t>0$, i.e. *F* is homogeneous of order 0.

We then introduce

${\lambda}_{0}=inf\{E{(u)}^{\frac{p-q}{p-2}}:u\in {E}^{+}\cap {A}^{+}\cap {B}^{+},{C}_{pq}^{-1}B(u)A{(u)}^{\frac{2-q}{p-2}}=1\}.$

Note that if ${E}^{+}\cap {A}^{+}\cap {B}^{+}=\mathrm{\varnothing}$, then ${\lambda}_{0}=\mathrm{\infty}$.

We then deduce the following result, which provides sufficient conditions for the existence of critical points of ${j}_{u}$.

#### Proposition 1.5.

*The following statements hold:*

(i)

*If either *
$u\in {A}^{+}\cap {B}^{-}$
* or *
$u\in {E}^{+}\cap {A}^{+}\cap {B}_{0}$
*, then *
${j}_{u}$
* has a positive global maximum at some *
${t}_{1}>0$
*, i.e. *
${j}_{u}^{\prime}({t}_{1})=0>{j}_{u}^{\prime \prime}({t}_{1})$
* and *
${j}_{u}(t)<{j}_{u}({t}_{1})$
* for *
$t\ne {t}_{1}$
*. Moreover, *
${t}_{1}$
* is the unique critical point of *
${j}_{u}$.

(ii)

*Assume *
${\int}_{\partial \mathrm{\Omega}}b<0$
*. Then *
${\lambda}_{0}>0$
* and for any *
$0<\lambda <{\lambda}_{0}$
* and *
$u\in {E}^{+}\cap {A}^{+}\cap {B}^{+}$
* the map *
${j}_{u}$
* has a negative local minimum at *
${t}_{1}>0$
* and a positive global maximum at *
${t}_{2}>{t}_{1}$
*, which are the only critical points of *
${j}_{u}$.

#### Proof.

The first assertion is straightforward from the definition of ${j}_{u}$.
Let us assume now ${\int}_{\partial \mathrm{\Omega}}b<0$. First, we show that ${\lambda}_{0}>0$. Assume ${\lambda}_{0}=0$, so that we can choose ${u}_{n}\in {E}^{+}\cap {A}^{+}\cap {B}^{+}$ satisfying

$E({u}_{n})\to 0\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}{C}_{pq}^{-1}B({u}_{n})A{({u}_{n})}^{\frac{2-q}{p-2}}=1.$

If $({u}_{n})$ is bounded in *X*, then we may assume that ${u}_{n}\rightharpoonup {u}_{0}$ for some ${u}_{0}\in X$ and ${u}_{n}\to {u}_{0}$ in ${L}^{p}(\mathrm{\Omega})$ and ${L}^{q}(\mathrm{\Omega})$. It follows from Lemma 1.4 (i) that ${u}_{0}$ is a constant and ${u}_{n}\to {u}_{0}$ in *X*. From ${u}_{n}\in {A}^{+}\cap {B}^{+}$ we deduce that ${u}_{0}\in {A}_{0}^{+}\cap {B}_{0}^{+}$. In addition, there holds

${C}_{pq}^{-1}B({u}_{0})A{({u}_{0})}^{\frac{2-q}{p-2}}=1,$

so that ${u}_{0}\not\equiv 0$. From Lemma 1.4 we get a contradiction.

Let us assume now that $\parallel {u}_{n}\parallel \to \mathrm{\infty}$. Set ${v}_{n}=\frac{{u}_{n}}{\parallel {u}_{n}\parallel}$, so that $\parallel {v}_{n}\parallel =1$. We may assume that ${v}_{n}\rightharpoonup {v}_{0}$ and ${v}_{n}\to {v}_{0}$ in ${L}^{p}(\mathrm{\Omega})$. Since $E({v}_{n})\to 0$ and ${v}_{n}\in {A}^{+}$, we can argue as for ${u}_{n}$ to reach a contradiction.
Finally, for any $u\in {E}^{+}\cap {A}^{+}\cap {B}^{+}$ we have

${\lambda}_{0}\le {C}_{pq}\frac{E{(u)}^{\frac{p-q}{p-2}}}{B(u)A{(u)}^{\frac{2-q}{p-2}}}.$

Thus, if $0<\lambda <{\lambda}_{0}$, then ${i}_{u}({t}^{*})>0$ from (1.4).
∎

#### Proposition 1.6.

*We have the following results:*

(i)

${N}_{\lambda}^{0}$
* is empty.*

(ii)

*If *
${b}^{+}\not\equiv 0$
* and *
${\int}_{\partial \mathrm{\Omega}}b<0$
*, then *
${N}_{\lambda}^{+}$
* is non-empty for *
$0<\lambda <{\lambda}_{0}$.

(iii)

*If *
${b}^{-}\not\equiv 0$
*, then *
${N}_{\lambda}^{-}$
* is non-empty.*

#### Proof.

(i) From Proposition 1.5 it follows that there is no $t>0$ such that ${j}_{u}^{\prime}(t)={j}_{u}^{\prime \prime}(t)=0$, i.e. ${N}_{\lambda}^{0}$ is empty.

(ii) Since ${b}^{+}\not\equiv 0$, we can find $u\in {B}^{+}$. Moreover, since ${\int}_{\partial \mathrm{\Omega}}b<0$, by Lemma 1.4 we have $u\in {E}^{+}\cap {A}^{+}$. By Proposition 1.5 we infer that for $0<\lambda <{\lambda}_{0}$ there are $0<{t}_{1}<{t}_{2}$ such that ${t}_{1}u\in {N}_{\lambda}^{+}$ and ${t}_{2}u\in {N}_{\lambda}^{-}$.

(iii) Since ${b}^{-}\not\equiv 0$, we can find $u\in {B}^{-}$, so that $u\in {A}^{+}\cap {B}^{-}$. By Proposition 1.5 we infer that there exists ${t}_{1}>0$ such that ${t}_{1}u\in {N}_{\lambda}^{-}$.
∎

The following result provides some properties of ${N}_{\lambda}^{+}$.

#### Lemma 1.7.

*Assume ${b}^{\mathrm{+}}\mathrm{\not\equiv}\mathrm{0}$ and ${\mathrm{\int}}_{\mathrm{\partial}\mathit{}\mathrm{\Omega}}b\mathrm{<}\mathrm{0}$. Then, for $\mathrm{0}\mathrm{<}\lambda \mathrm{<}{\lambda}_{\mathrm{0}}$, we have the following:*

(i)

${N}_{\lambda}^{+}\subset {B}^{+}$.

(ii)

${N}_{\lambda}^{+}$
* is bounded in *
*X*.

(iii)

${I}_{\lambda}(u)<0$
* for any *
$u\in {N}_{\lambda}^{+}$.

#### Proof.

(i) Let $u\in {N}_{\lambda}^{+}$. Then $0\le E(u)<\lambda \frac{p-q}{p-2}B(u)$, i.e. $u\in {B}^{+}$.

(ii) Assume $({u}_{n})\subset {N}_{\lambda}^{+}$ and $\parallel {u}_{n}\parallel \to \mathrm{\infty}$. Set ${v}_{n}=\frac{{u}_{n}}{\parallel {u}_{n}\parallel}$. It follows that $\parallel {v}_{n}\parallel =1$, so we may assume that ${v}_{n}\rightharpoonup {v}_{0}$ in *X*, $B({v}_{n})$ is bounded and ${v}_{n}\to {v}_{0}$ in ${L}^{p}(\mathrm{\Omega})$ (implying $A(v)\to A({v}_{0})$). Since ${u}_{n}\in {N}_{\lambda}^{+}$, we see that

$E({v}_{n})<\lambda \frac{p-q}{p-2}B({v}_{n}){\parallel {u}_{n}\parallel}^{q-2},$

and thus ${lim\; sup}_{n}E({v}_{n})\le 0$. Lemma 1.4 (i) yields that ${v}_{0}$ is a constant and ${v}_{n}\to {v}_{0}$ in *X*. Consequently, $\parallel {v}_{0}\parallel =1$ and ${v}_{0}$ is a non-zero constant. On the other hand, since ${u}_{n}\in {N}_{\lambda}^{+}$, we have ${v}_{n}\in {N}_{\lambda}^{+}$, so ${v}_{n}\in {B}^{+}$. It follows that ${v}_{0}\in {B}_{0}^{+}$, a contradiction.

(iii) Let $u\in {N}_{\lambda}^{+}$, so that $u\in {B}^{+}$.
Hence *u* is not a constant and $E(u)>0$. Thus $u\in {E}^{+}\cap {A}^{+}\cap {B}^{+}$ and by Proposition 1.5 (ii) we infer that ${I}_{\lambda}(u)<0$ and $t>1$ if ${j}_{u}^{\prime}(t)>0$.
∎

#### Proposition 1.8.

*Assume ${b}^{\mathrm{+}}\mathrm{\not\equiv}\mathrm{0}$ and ${\mathrm{\int}}_{\mathrm{\partial}\mathit{}\mathrm{\Omega}}b\mathrm{<}\mathrm{0}$. Then, for any $\mathrm{0}\mathrm{<}\lambda \mathrm{<}{\lambda}_{\mathrm{0}}$, there exists ${u}_{\mathrm{1}\mathrm{,}\lambda}\mathrm{\ge}\mathrm{0}$ such that
${I}_{\lambda}\mathit{}\mathrm{(}{u}_{\mathrm{1}\mathrm{,}\lambda}\mathrm{)}\mathrm{=}{\mathrm{min}}_{{N}_{\lambda}^{\mathrm{+}}}\mathit{}{I}_{\lambda}\mathrm{<}\mathrm{0}$. In particular, ${u}_{\mathrm{1}\mathrm{,}\lambda}$ is a nontrivial non-negative solution of ((${P_{\lambda}}$)).*

#### Proof.

Let $0<\lambda <{\lambda}_{0}$. By Proposition 1.6 we know that ${N}_{\lambda}^{+}$ is non-empty. We consider a minimizing sequence $({u}_{n})\subset {N}_{\lambda}^{+}$, i.e.

${I}_{\lambda}({u}_{n})\to \underset{{N}_{\lambda}^{+}}{inf}{I}_{\lambda}<0.$

Since $({u}_{n})$ is bounded in *X*, we may assume that ${u}_{n}\rightharpoonup {u}_{0}$ in *X*, ${u}_{n}\to {u}_{0}$ in ${L}^{p}(\mathrm{\Omega})$ and ${L}^{q}(\partial \mathrm{\Omega})$. It follows that

${I}_{\lambda}({u}_{0})\le \underset{n}{lim\; inf}{I}_{\lambda}({u}_{n})=\underset{{N}_{\lambda}^{+}}{inf}{I}_{\lambda}(u)<0,$

so that
${u}_{0}\not\equiv 0$. Moreover, as ${u}_{n}\in {B}^{+}$, we have ${u}_{0}\in {B}_{0}^{+}$ and ${u}_{0}$ is not a constant. So ${u}_{0}\in {E}^{+}\cap {A}^{+}\cap {B}^{+}$. Since $0<\lambda <{\lambda}_{0}$, Proposition 1.5 yields that ${t}_{1}{u}_{0}\in {N}_{\lambda}^{+}$ for some ${t}_{1}>0$. Assume ${u}_{n}\nrightarrow {u}_{0}$. If $1<{t}_{1}$, then we have

${I}_{\lambda}({t}_{1}{u}_{0})={j}_{{u}_{0}}({t}_{1})\le {j}_{{u}_{0}}(1)<lim\; sup{j}_{{u}_{n}}(1)=lim\; sup{I}_{\lambda}({u}_{n})=\underset{{N}_{\lambda}^{+}}{inf}{I}_{\lambda},$

which is impossible.
If ${t}_{1}\le 1$, then ${j}_{{u}_{n}}^{\prime}({t}_{1})\le 0$ for every *n*, so that

${j}_{{u}_{0}}^{\prime}({t}_{1})<lim\; sup{j}_{{u}_{n}}^{\prime}({t}_{1})\le 0,$

which is a contradiction. Therefore, ${u}_{n}\to {u}_{0}$.
Now, since ${u}_{n}\to {u}_{0}$, we have ${j}_{{u}_{0}}^{\prime}(1)=0\le {j}_{{u}_{0}}^{\prime \prime}(1)$.
But ${j}_{{u}_{0}}^{\prime \prime}(1)=0$ is impossible by Proposition 1.6 (i).
Thus ${u}_{0}\in {N}_{\lambda}^{+}$ and ${I}_{\lambda}({u}_{0})={inf}_{{N}_{\lambda}^{+}}{I}_{\lambda}$. We set ${u}_{1,\lambda}={u}_{0}$.
∎

Next, we obtain a nontrivial non-negative weak solution of ((${P_{\lambda}}$)), which achieves ${inf}_{{N}_{\lambda}^{-}}{I}_{\lambda}$ for $\lambda \in (0,{\lambda}_{0})$. The following result provides some properties of ${N}_{\lambda}^{-}$.

#### Lemma 1.9.

*Assume ${\mathrm{\int}}_{\mathrm{\partial}\mathit{}\mathrm{\Omega}}b\mathrm{<}\mathrm{0}$. Then ${I}_{\lambda}\mathit{}\mathrm{(}u\mathrm{)}\mathrm{>}\mathrm{0}$ for $\mathrm{0}\mathrm{<}\lambda \mathrm{<}{\lambda}_{\mathrm{0}}$ and any $u\mathrm{\in}{N}_{\lambda}^{\mathrm{-}}$.*

#### Proof.

Let $u\in {N}_{\lambda}^{-}$.
If $u\in {B}_{0}$, then *u* is not a constant, so $u\in {E}^{+}\cap {A}^{+}$. Thus, by Proposition 1.5, ${j}_{u}$ has a positive global maximum at $t=1$. The same conclusion holds if $u\in {B}^{-}$. Finally, if $u\in {B}^{+}$, then *u* is not a constant. Hence $u\in {E}^{+}\cap {A}^{+}$, and since $0<\lambda <{\lambda}_{0}$, Proposition 1.5 yields again that ${j}_{u}$ has a positive global maximum at $t=1$.
∎

#### Proposition 1.10.

*Let ${\mathrm{\int}}_{\mathrm{\partial}\mathit{}\mathrm{\Omega}}b\mathrm{<}\mathrm{0}$. Then for any $\lambda \mathrm{\in}\mathrm{(}\mathrm{0}\mathrm{,}{\lambda}_{\mathrm{0}}\mathrm{)}$ there exists ${u}_{\mathrm{2}\mathrm{,}\lambda}\mathrm{\ge}\mathrm{0}$ such that
${I}_{\lambda}\mathit{}\mathrm{(}{u}_{\mathrm{2}\mathrm{,}\lambda}\mathrm{)}\mathrm{=}{\mathrm{min}}_{{N}_{\lambda}^{\mathrm{-}}}\mathit{}{I}_{\lambda}\mathrm{>}\mathrm{0}$. In particular, ${u}_{\mathrm{2}\mathrm{,}\lambda}$ is a non-negative solution of ((${P_{\lambda}}$)).*

#### Proof.

First of all, since ${\int}_{\partial \mathrm{\Omega}}b<0$, we have ${b}^{-}\not\equiv 0$, so that by Proposition 1.6 we know that ${N}_{\lambda}^{-}$ is non-empty. In addition, since ${I}_{\lambda}(u)>0$ for $u\in {N}_{\lambda}^{-}$, we can choose ${u}_{n}\in {N}_{\lambda}^{-}$ such that

${I}_{\lambda}({u}_{n})\to \underset{{N}_{\lambda}^{-}}{inf}{I}_{\lambda}\ge 0.$

We claim that $({u}_{n})$ is bounded in *X*. Indeed, there exists $C>0$ such that ${I}_{\lambda}({u}_{n})\le C$. Since ${u}_{n}\in {N}_{\lambda}$, we deduce

$\left(\frac{1}{2}-\frac{1}{p}\right)E({u}_{n})-\lambda \left(\frac{1}{q}-\frac{1}{p}\right)B({u}_{n})={I}_{\lambda}({u}_{n})\le C.$

Assume $\parallel {u}_{n}\parallel \to \mathrm{\infty}$ and set ${v}_{n}=\frac{{u}_{n}}{\parallel {u}_{n}\parallel}$, so that $\parallel {v}_{n}\parallel =1$. We may assume that ${v}_{n}\rightharpoonup {v}_{0}$ in *X* and ${v}_{n}\to {v}_{0}$ in ${L}^{p}(\mathrm{\Omega})$ and ${L}^{q}(\partial \mathrm{\Omega})$. Then, from

$\left(\frac{1}{2}-\frac{1}{p}\right)E({v}_{n})\le \lambda \left(\frac{1}{q}-\frac{1}{p}\right)B({v}_{n}){\parallel {u}_{n}\parallel}^{q-2}+C{\parallel {u}_{n}\parallel}^{-2},$

we infer that ${lim\; sup}_{n}E({v}_{n})\le 0$. Lemma 1.4 (i) yields that ${v}_{0}$ is a constant and ${v}_{n}\to {v}_{0}$ in *X*, which implies $\parallel {v}_{0}\parallel =1$. On the other hand, since ${u}_{n}\in {N}_{\lambda}$, we have

$E({u}_{n})=\lambda B({u}_{n})+A({u}_{n}).$

Dividing by ${\parallel {u}_{n}\parallel}^{p}$ and passing to the limit as $n\to \mathrm{\infty}$, we get $A({v}_{0})=0$, i.e. ${v}_{0}=0$, which is impossible. Hence $({u}_{n})$ is bounded.
We may then assume that ${u}_{n}\rightharpoonup {u}_{0}$ in *X* and ${u}_{n}\to {u}_{0}$ in ${L}^{p}(\mathrm{\Omega})$ and ${L}^{q}(\partial \mathrm{\Omega})$. If ${u}_{0}\equiv 0$, then we set ${v}_{n}=\frac{{u}_{n}}{\parallel {u}_{n}\parallel}$. From

$E({u}_{n})<\frac{p-q}{2-q}A({u}_{n})$

we get

$E({v}_{n})<\frac{p-q}{2-q}A({v}_{n}){\parallel {u}_{n}\parallel}^{p-2}\to 0.$

So we can assume that ${v}_{n}\to {v}_{0}$ with ${v}_{0}$ constant.
Moreover, from

$E({u}_{n})=\lambda B({u}_{n})+A({u}_{n})$

we deduce that $B({v}_{n})\to 0$, i.e. $B({v}_{0})=0$, which contradicts ${\int}_{\partial \mathrm{\Omega}}b<0$.
Thus, ${u}_{0}\not\equiv 0$. By Proposition 1.5 we infer the existence of ${t}_{2}>0$ such that ${t}_{2}{u}_{0}\in {N}_{\lambda}^{-}$.
Assume ${u}_{n}\nrightarrow {u}_{0}$. Then, since ${u}_{n}\in {N}_{\lambda}^{-}$, we get

${I}_{\lambda}({t}_{2}{u}_{0})<lim\; sup{I}_{\lambda}({t}_{2}{u}_{n})\le lim\; sup{I}_{\lambda}({u}_{n})=\underset{{N}_{\lambda}^{-}}{inf}{I}_{\lambda},$

which is a contradiction. Therefore, ${u}_{n}\to {u}_{0}$. In particular, we get ${j}_{{u}_{0}}^{\prime}(1)=0$ and ${j}_{{u}_{0}}^{\prime \prime}(1)<0$. Since ${N}_{\lambda}^{0}$ is empty for $\lambda \in (0,{\lambda}_{0})$, we infer that ${u}_{0}\in {N}_{\lambda}^{-}$ and ${I}_{\lambda}({u}_{0})={inf}_{{N}_{\lambda}^{-}}{I}_{\lambda}$. We set ${u}_{2,\lambda}={u}_{0}$.
∎

We now discuss the asymptotic profiles of ${u}_{1,\lambda}$ and ${u}_{2,\lambda}$ as $\lambda \to {0}^{+}$.

#### Lemma 1.11.

*Assume ${b}^{\mathrm{+}}\mathrm{\not\equiv}\mathrm{0}$ and ${\mathrm{\int}}_{\mathrm{\partial}\mathit{}\mathrm{\Omega}}b\mathrm{<}\mathrm{0}$. Then, for $\mathrm{0}\mathrm{<}\lambda \mathrm{<}{\lambda}_{\mathrm{0}}$, there holds*

${I}_{\lambda}({u}_{1,\lambda})<-{D}_{0}{\lambda}^{\frac{2}{2-q}}$

*for some ${D}_{\mathrm{0}}\mathrm{>}\mathrm{0}$.*

#### Proof.

Let $u\in {N}_{\lambda}^{+}$. Thus, $u\in {A}^{+}\cap {E}^{+}\cap {B}^{+}$.
Then

${I}_{\lambda}(u)\le {\stackrel{~}{I}}_{\lambda}(u):=\frac{1}{2}E(u)-\frac{\lambda}{q}B(u).$

Thus ${I}_{\lambda}(tu)\le {\stackrel{~}{I}}_{\lambda}(tu)$ for every $t>0$.
Note that ${\stackrel{~}{I}}_{\lambda}(tu)$ has a global minimum point ${t}_{0}$ given by

${t}_{0}={\left(\frac{\lambda B(u)}{E(u)}\right)}^{\frac{1}{2-q}}$

and

${\stackrel{~}{I}}_{\lambda}({t}_{0}u)=-\frac{2-q}{2q}\lambda {t}_{0}^{q}B(u)=-\frac{2-q}{2q}\frac{{(\lambda B(u))}^{\frac{2}{2-q}}}{E{(u)}^{\frac{q}{2-q}}}=-{D}_{0}{\lambda}^{\frac{2}{2-q}},$

where

${D}_{0}=\frac{2-q}{2q}\frac{B{(u)}^{\frac{2}{2-q}}}{E{(u)}^{\frac{q}{2-q}}}.$

It follows that if ${I}_{\lambda}(tu)$ has a local minimum at ${t}_{1}$, then

${I}_{\lambda}({t}_{1}u)<-{D}_{0}{\lambda}^{\frac{2}{2-q}}$

with ${D}_{0}>0$. Therefore, ${I}_{\lambda}(u)<-{D}_{0}{\lambda}^{\frac{2}{2-q}}$ for every $u\in {N}_{\lambda}^{+}$. In particular,

${I}_{\lambda}({u}_{1,\lambda})<-{D}_{0}{\lambda}^{\frac{2}{2-q}}.\mathit{\u220e}$

We now determine the asymptotic profile of ${u}_{1,\lambda}$ as $\lambda \to {0}^{+}$.

#### Proposition 1.12.

*Assume ${b}^{\mathrm{+}}\mathrm{\not\equiv}\mathrm{0}$ and ${\mathrm{\int}}_{\mathrm{\partial}\mathit{}\mathrm{\Omega}}b\mathrm{<}\mathrm{0}$.
Then ${u}_{\mathrm{1}\mathrm{,}\lambda}\mathrm{\to}\mathrm{0}$ in **X* as $\lambda \mathrm{\to}{\mathrm{0}}^{\mathrm{+}}$. Moreover, if ${\lambda}_{n}\mathrm{\to}{\mathrm{0}}^{\mathrm{+}}$, then, up to a subsequence, ${\lambda}_{n}^{\mathrm{-}\mathrm{1}\mathrm{/}\mathrm{(}\mathrm{2}\mathrm{-}q\mathrm{)}}\mathit{}{u}_{\mathrm{1}\mathrm{,}{\lambda}_{n}}\mathrm{\to}{w}_{\mathrm{0}}$ in *X*, where ${w}_{\mathrm{0}}$ is a non-negative ground state solution, i.e. a least energy solution of

$\{\begin{array}{cccc}\hfill \mathrm{\Delta}w& =0\hfill & & \hfill \mathit{\text{in}}\mathrm{\Omega},\\ \hfill \frac{\partial w}{\partial \mathbf{n}}& =b(x){|w|}^{q-2}w\hfill & & \hfill \mathit{\text{on}}\partial \mathrm{\Omega}.\end{array}$(${P_{w}}$)

#### Proof.

First we show that ${u}_{1,\lambda}$ remains bounded in *X* as $\lambda \to {0}^{+}$. Indeed, assume that $\parallel {u}_{1,\lambda}\parallel \to \mathrm{\infty}$ and set ${v}_{\lambda}=\frac{{u}_{1,\lambda}}{\parallel {u}_{1,\lambda}\parallel}$. We may then assume that for some ${v}_{0}\in X$ we have ${v}_{\lambda}\rightharpoonup {v}_{0}$ in *X*
and ${v}_{\lambda}\to {v}_{0}$ in ${L}^{p}(\mathrm{\Omega})$ and ${L}^{q}(\partial \mathrm{\Omega})$. Since ${u}_{1,\lambda}\in {N}_{\lambda}$, we have

$E({v}_{\lambda}){\parallel {u}_{1,\lambda}\parallel}^{2-p}=A({v}_{\lambda})+\lambda B({v}_{\lambda}){\parallel {u}_{1,\lambda}\parallel}^{q-p}.$

Passing to the limit as $\lambda \to {0}^{+}$, we obtain $A({v}_{0})=0$, i.e. ${v}_{0}\equiv 0$.
From ${u}_{1,\lambda}\in {N}_{\lambda}^{+}$ we have

$E({v}_{\lambda})<\lambda \frac{p-q}{p-2}B({v}_{\lambda}){\parallel {u}_{1,\lambda}\parallel}^{q-2},$

so that ${lim\; sup}_{\lambda}E({v}_{\lambda})\le 0$. By Lemma 1.4 (i) we infer that
${v}_{0}$ is a constant and ${v}_{\lambda}\to 0$ in *X*, which contradicts $\parallel {v}_{\lambda}\parallel =1$ for every λ. Thus
${u}_{1,\lambda}$ stays bounded in *X* as $\lambda \to {0}^{+}$.

Hence we may assume that ${u}_{1,\lambda}\rightharpoonup {u}_{0}$ in *X* and ${u}_{1,\lambda}\to {u}_{0}$
in ${L}^{p}(\mathrm{\Omega})$ and ${L}^{q}(\partial \mathrm{\Omega})$ as $\lambda \to {0}^{+}$.
Since ${u}_{1,\lambda}\in {N}_{\lambda}^{+}$, we observe that

$E({u}_{1,\lambda})<\lambda \frac{p-q}{p-2}B({u}_{1,\lambda}).$(1.5)

Passing to the limit as $\lambda \to {0}^{+}$, we get ${lim\; sup}_{\lambda}E({u}_{1,\lambda})\le 0$.
Lemma 1.4 (ii) provides that ${u}_{0}$ is a constant and ${u}_{1,\lambda}\to {u}_{0}$ in *X*.
Since ${u}_{1,\lambda}\in {B}^{+}$, we have ${u}_{0}\in {B}_{0}^{+}$, and ${\int}_{\partial \mathrm{\Omega}}b<0$ implies that ${u}_{0}=0$.

Let ${w}_{\lambda}={\lambda}^{-1/(2-q)}{u}_{1,\lambda}$. We claim that ${w}_{\lambda}$
remains bounded in *X* as $\lambda \to {0}^{+}$. Indeed, from (1.5)
we have

$E({w}_{\lambda})<\frac{p-q}{p-2}B({w}_{\lambda}).$

Let us assume that $\parallel {w}_{\lambda}\parallel \to \mathrm{\infty}$ and set
${\psi}_{\lambda}=\frac{{w}_{\lambda}}{\parallel {w}_{\lambda}\parallel}$. We may assume that
${\psi}_{\lambda}\rightharpoonup {\psi}_{0}$ and ${\psi}_{\lambda}\to {\psi}_{0}$ in ${L}^{p}(\mathrm{\Omega})$
and ${L}^{q}(\partial \mathrm{\Omega})$. It follows that

$E({\psi}_{\lambda})<\frac{p-q}{p-2}B({\psi}_{\lambda}){\parallel {w}_{\lambda}\parallel}^{q-2},$

so that ${lim\; sup}_{\lambda}E({\psi}_{\lambda})\le 0$. By Lemma 1.4 (i) we infer that ${\psi}_{0}$ is a constant and ${\psi}_{\lambda}\to {\psi}_{0}$ in *X*. On the other hand, from ${u}_{1,\lambda}\in {B}^{+}$ we have ${\psi}_{\lambda}\in {B}^{+}$, and consequently ${\psi}_{0}\in {B}_{0}^{+}$. From ${\int}_{\partial \mathrm{\Omega}}b<0$ we infer that ${\psi}_{0}\equiv 0$, which contradicts $\parallel {\psi}_{0}\parallel =1$. Hence ${w}_{\lambda}$ stays bounded in *X* as $\lambda \to {0}^{+}$, and
we may assume that ${w}_{\lambda}\rightharpoonup {w}_{0}$ in *X* and ${w}_{\lambda}\to {w}_{0}$ in ${L}^{p}(\mathrm{\Omega})$
and ${L}^{q}(\partial \mathrm{\Omega})$. Note that ${w}_{\lambda}$ satisfies

${\int}_{\mathrm{\Omega}}\nabla {w}_{\lambda}\nabla w-{\lambda}^{\frac{p-2}{2-q}}{\int}_{\mathrm{\Omega}}{w}_{\lambda}^{p-1}w-{\int}_{\partial \mathrm{\Omega}}b(x){w}_{\lambda}^{q-1}w=0\mathit{\hspace{1em}}\text{for all}w\in X.$

Taking $w={w}_{\lambda}-{w}_{0}$ and letting $\lambda \to 0$, we deduce that
${w}_{\lambda}\to {w}_{0}$ in *X*. Moreover, ${w}_{0}$ is a weak solution of (${P_{w}}$).
We claim that ${w}_{0}\not\equiv 0$. Indeed,
by Lemma 1.11 we have

${I}_{\lambda}({u}_{1,\lambda})<-{D}_{0}{\lambda}^{\frac{2}{2-q}},$

with ${D}_{0}>0$.
Hence

$\frac{{\lambda}^{\frac{2}{2-q}}}{2}E({w}_{\lambda})-\frac{{\lambda}^{\frac{p}{2-q}}}{p}A({w}_{\lambda})-\frac{{\lambda}^{\frac{2}{2-q}}}{q}B({w}_{n})<-{D}_{0}{\lambda}^{\frac{2}{2-q}},$

so that

$\frac{1}{2}E({w}_{\lambda})-\frac{{\lambda}^{\frac{p-2}{2-q}}}{p}A({w}_{\lambda})-B({w}_{\lambda})<-{D}_{0}.$

Letting $\lambda \to 0$, we obtain

$\frac{1}{2}E({w}_{0})-B({w}_{0})\le -{D}_{0},$

and consequently ${w}_{0}\not\equiv 0$.

It remains to prove that ${w}_{0}$ is a ground state solution of (${P_{w}}$), i.e.

${I}_{b}({w}_{0})=\underset{{N}_{b}}{\mathrm{min}}{I}_{b},$

where

${I}_{b}(u)=\frac{1}{2}E(u)-\frac{1}{q}B(u)$

for $u\in X$ and

${N}_{b}=\{u\in X\setminus \{0\}:\u3008{I}_{b}^{\prime}(u),u\u3009=0\}=\{u\in X\setminus \{0\}:E(u)=B(u)\}$

is the Nehari manifold associated to ${I}_{b}$. Since ${\int}_{\partial \mathrm{\Omega}}b<0$, it is easily seen that there exists ${w}_{b}\ne 0$ such that ${I}_{b}({w}_{b})={\mathrm{min}}_{{N}_{b}}{I}_{b}$. Note that ${w}_{0}\in {N}_{b}$, and consequently ${I}_{b}({w}_{b})\le {I}_{b}({w}_{0})$. We now prove the reverse inequality. Since ${w}_{b}$ is non-constant, we have ${w}_{b}\in {B}^{+}\cap {E}^{+}$. We set ${u}_{b}={\lambda}^{1/(2-q)}{w}_{b}$. Let ${\lambda}_{n}\to {0}^{+}$. Since ${u}_{b}\in {B}^{+}\cap {E}^{+}$ for every *n*, there exists ${t}_{n}>0$ such that ${t}_{n}{u}_{b}\in {N}_{{\lambda}_{n}}^{+}$.
Hence

${t}_{n}^{2}E({u}_{b})<{\lambda}_{n}\frac{p-q}{p-2}{t}_{n}^{q}B({u}_{b}),$

i.e.

${t}_{n}^{2-q}<\frac{p-q}{p-2}\frac{B({w}_{b})}{E({w}_{b})}=\frac{p-q}{p-2}.$

We may then assume that ${t}_{n}\to {t}_{0}$. We claim that ${t}_{0}=1$. Indeed, note that from ${t}_{n}{u}_{b}\in {N}_{{\lambda}_{n}}^{+}$ we infer that

${t}_{n}^{2}E({u}_{b})={\lambda}_{n}{t}_{n}^{q}B({u}_{b})+{t}_{n}^{p}A({u}_{b}),$

so

${t}_{n}^{2-q}E({w}_{b})=B({w}_{b})+{t}_{n}^{p-q}{\lambda}_{n}^{\frac{p-2}{2-q}}A({w}_{b}).$

From $E({w}_{b})=B({w}_{b})$ we infer that ${t}_{0}=1$, as claimed. Now, since ${t}_{n}{u}_{b}\in {N}_{{\lambda}_{n}}^{+}$, we have

${I}_{{\lambda}_{n}}({u}_{1,{\lambda}_{n}})\le {I}_{{\lambda}_{n}}({t}_{n}{u}_{b}).$

It follows that

${I}_{{\lambda}_{n}}({u}_{1,{\lambda}_{n}})\le \left(\frac{1}{2}-\frac{1}{q}\right){t}_{n}^{2}E({u}_{b})-\left(\frac{1}{p}-\frac{1}{q}\right){t}_{n}^{p}A({u}_{b}).$

Hence

$\frac{{\lambda}_{n}^{\frac{2}{2-q}}}{2}E({w}_{n})-\frac{{\lambda}_{n}^{\frac{p}{2-q}}}{p}A({w}_{n})-\frac{{\lambda}_{n}^{\frac{2}{2-q}}}{q}B({w}_{n})\le \frac{q-2}{2q}{t}_{n}^{2}{\lambda}_{n}^{\frac{2}{2-q}}E({w}_{b})-\frac{q-p}{pq}{\lambda}_{n}^{\frac{p}{2-q}}{t}_{n}^{p}A({w}_{b}),$

i.e.

$\frac{1}{2}E({w}_{n})-\frac{{\lambda}_{n}^{\frac{p-2}{2-q}}}{p}A({w}_{n})-\frac{1}{q}B({w}_{n})\le \frac{q-2}{2q}{t}_{n}^{2}E({w}_{b})-\frac{q-p}{pq}{\lambda}_{n}^{\frac{p-2}{2-q}}{t}_{n}^{p}A({w}_{b}).$

Since ${w}_{n}\to {w}_{0}$ in *X*, we obtain

${I}_{b}({w}_{0})\le \left(\frac{1}{2}-\frac{1}{q}\right)E({w}_{b})={I}_{b}({w}_{b}).$

Therefore, ${I}_{b}({w}_{0})={I}_{b}({w}_{b})$, as claimed.
∎

We now consider the asymptotic behavior of ${u}_{2,\lambda}$ as $\lambda \to {0}^{+}$. We shall prove that ${u}_{2,\lambda}\to 0$ in *X* as $\lambda \to {0}^{+}$.

#### Lemma 1.13.

*Assume ${\mathrm{\int}}_{\mathrm{\partial}\mathit{}\mathrm{\Omega}}b\mathrm{<}\mathrm{0}$. Then there exists a constant $C\mathrm{>}\mathrm{0}$ such that
$\mathrm{\parallel}{u}_{\mathrm{2}\mathrm{,}\lambda}\mathrm{\parallel}\mathrm{\le}C$ as $\lambda \mathrm{\to}{\mathrm{0}}^{\mathrm{+}}$.*

#### Proof.

First we show that there exists a constant ${C}_{1}>0$ such that ${I}_{\lambda}({u}_{2,\lambda})\le {C}_{1}$ for every
$\lambda \in (0,{\lambda}_{0})$.
To this end, we consider the eigenvalue problem

$\{\begin{array}{cccc}\hfill -\mathrm{\Delta}\phi & =\lambda \phi \hfill & & \hfill \text{in}\mathrm{\Omega},\\ \hfill \phi & =0\hfill & & \hfill \text{on}\partial \mathrm{\Omega}.\end{array}$

Let ${\lambda}_{1}$ be the first eigenvalue of this problem and ${\phi}_{1}>0$ be an eigenfunction associated to ${\lambda}_{1}$. Note that ${\phi}_{1}\in {E}^{+}\cap {A}^{+}\cap {B}_{0}$ and

${j}_{{\phi}_{1}}(t)=\frac{{t}^{2}}{2}E({\phi}_{1})-\frac{{t}^{p}}{p}A({\phi}_{1}),$

so that ${j}_{{\phi}_{1}}$
has a global maximum at some ${t}_{2}>0$, which implies ${t}_{2}{\phi}_{1}\in {N}_{\lambda}^{-}$.
Moreover, neither ${j}_{{\phi}_{1}}$ nor ${t}_{2}{\phi}_{1}$ depend on $\lambda \in (0,{\lambda}_{0})$.
Let ${C}_{1}={j}_{{\phi}_{1}}({t}_{2})={I}_{\lambda}({t}_{2}{\phi}_{1})>0$.
Since ${I}_{\lambda}({u}_{2,\lambda})={\mathrm{min}}_{{N}_{\lambda}^{-}}{I}_{\lambda}$, we have

$\left(\frac{1}{2}-\frac{1}{p}\right)E({u}_{2,\lambda})-\left(\frac{1}{q}-\frac{1}{p}\right)\lambda B({u}_{2,\lambda})={I}_{\lambda}({u}_{2,\lambda})\le {C}_{1}.$

Assume by contradiction that ${\lambda}_{n}\to 0$ and $\parallel {u}_{2,{\lambda}_{n}}\parallel \to \mathrm{\infty}$. We set ${v}_{n}=\frac{{u}_{2,{\lambda}_{n}}}{\parallel {u}_{2,{\lambda}_{n}}\parallel}$ and assume that ${v}_{n}\rightharpoonup {v}_{0}$ in *X*. Then

$\left(\frac{1}{2}-\frac{1}{p}\right)E({v}_{n})\le \left(\frac{1}{q}-\frac{1}{p}\right)\lambda B({v}_{n}){\parallel {u}_{2,{\lambda}_{n}}\parallel}^{q-2}+{C}_{1}{\parallel {u}_{2,{\lambda}_{n}}\parallel}^{-2}.$

We obtain $lim\; supE({v}_{n})\le 0$, and by Lemma 1.4 we infer that ${v}_{0}$ is a constant and ${v}_{n}\to {v}_{0}$ in *X*. In particular, $\parallel {v}_{0}\parallel =1$. Moreover, from

$E({u}_{2,{\lambda}_{n}})={\lambda}_{n}B({u}_{2,{\lambda}_{n}})+A({u}_{2,{\lambda}_{n}})$

we get $A({v}_{n})\to 0$, i.e. $A({v}_{0})=0$, which provides ${v}_{0}=0$, and we get a contradiction.
Therefore, $({u}_{2,\lambda})$ stays bounded in *X* as $\lambda \to 0$.
∎

#### Proposition 1.14.

*Assume ${\mathrm{\int}}_{\mathrm{\partial}\mathit{}\mathrm{\Omega}}b\mathrm{<}\mathrm{0}$. Then ${u}_{\mathrm{2}\mathrm{,}\lambda}\mathrm{\to}\mathrm{0}$ and ${\lambda}^{\mathrm{-}\mathrm{1}\mathrm{/}\mathrm{(}p\mathrm{-}q\mathrm{)}}\mathit{}{u}_{\mathrm{2}\mathrm{,}\lambda}\mathrm{\to}{c}^{\mathrm{*}}$ in **X* as $\lambda \mathrm{\to}{\mathrm{0}}^{\mathrm{+}}$.

#### Proof.

By Lemma 1.13, up to a subsequence, we have ${u}_{2,\lambda}\rightharpoonup {u}_{0}$ in *X* and ${u}_{2,\lambda}\to {u}_{0}$ in ${L}^{p}(\mathrm{\Omega})$ and ${L}^{q}(\partial \mathrm{\Omega})$ as $\lambda \to 0$. Since ${u}_{2,\lambda}$ is a weak solution of ((${P_{\lambda}}$)), it follows that ${u}_{2,\lambda}\to {u}_{0}$ in *X* and ${u}_{0}$ is a non-negative solution of

$\{\begin{array}{cccc}\hfill -\mathrm{\Delta}u& ={u}^{p-1}\hfill & & \hfill \text{in}\mathrm{\Omega},\\ \hfill \frac{\partial u}{\partial \mathbf{n}}& =0\hfill & & \hfill \text{on}\partial \mathrm{\Omega}.\end{array}$

But the only non-negative solution of this problem is $u\equiv 0$. Hence ${u}_{0}\equiv 0$ and ${u}_{2,\lambda}\to 0$ in *X* as $\lambda \to 0$.
We now set ${w}_{\lambda}={\lambda}^{-1/(p-q)}{u}_{2,\lambda}$. Then ${w}_{\lambda}$ is a non-negative solution of

$\{\begin{array}{cccc}\hfill -\mathrm{\Delta}w& ={\lambda}^{\frac{p-2}{p-q}}{w}^{p-1}\hfill & & \hfill \text{in}\mathrm{\Omega},\\ \hfill \frac{\partial w}{\partial \mathbf{n}}& ={\lambda}^{\frac{p-2}{p-q}}b(x){w}^{q-1}\hfill & & \hfill \text{on}\partial \mathrm{\Omega}.\end{array}$(1.6)

We claim that ${w}_{\lambda}$ stays bounded in *X* as $\lambda \to 0$. Indeed, assume that $\parallel {w}_{\lambda}\parallel \to \mathrm{\infty}$ and ${\psi}_{\lambda}=\frac{{w}_{\lambda}}{\parallel {w}_{\lambda}\parallel}\rightharpoonup {\psi}_{0}$ in *X* with ${\psi}_{\lambda}\to {\psi}_{0}$ in ${L}^{p}(\mathrm{\Omega})$ and
${L}^{q}(\partial \mathrm{\Omega})$ as $\lambda \to 0$. Let

${c}_{\lambda}={\left(\frac{-\lambda {\int}_{\partial \mathrm{\Omega}}b}{|\mathrm{\Omega}|}\right)}^{\frac{1}{p-q}}.$

We now use the fact that ${c}_{\lambda}\in {N}_{\lambda}^{-}$ for any $\lambda >0$. Hence

${I}_{\lambda}({u}_{2,\lambda})\le {I}_{\lambda}({c}_{\lambda})=D{\lambda}^{\frac{p}{p-q}},$

where

$D=\frac{p-q}{pq}\frac{{\left(-{\int}_{\partial \mathrm{\Omega}}b\right)}^{\frac{p}{p-q}}}{{|\mathrm{\Omega}|}^{\frac{q}{p-q}}}.$

Thus

$\frac{p-2}{2p}{\lambda}^{\frac{2}{p-q}}E({w}_{\lambda})-\frac{p-q}{pq}{\lambda}^{\frac{p}{p-q}}B({w}_{\lambda})\le D{\lambda}^{\frac{p}{p-q}},$

so that

$\frac{p-2}{2p}E({w}_{\lambda})-\frac{p-q}{pq}{\lambda}^{\frac{p-2}{p-q}}B({w}_{\lambda})\le D{\lambda}^{\frac{p-2}{p-q}}.$

Dividing the latter inequality by ${\parallel {w}_{\lambda}\parallel}^{2}$, we get $E({\psi}_{\lambda})\to 0$, and consequently ${\psi}_{\lambda}\to {\psi}_{0}$ in *X* as $\lambda \to 0$ and ${\psi}_{0}$ is a constant. Furthermore, integrating (1.6), we obtain

${\int}_{\mathrm{\Omega}}{w}_{\lambda}^{p-1}+{\int}_{\partial \mathrm{\Omega}}b{w}_{\lambda}^{q-1}=0,$(1.7)

so that ${\int}_{\mathrm{\Omega}}{\psi}_{\lambda}^{p-1}\to 0$, i.e. ${\psi}_{0}=0$, which is impossible since $\parallel {\psi}_{0}\parallel =1$. Therefore, ${w}_{\lambda}$ stays bounded in *X* as $\lambda \to 0$. We may then assume that ${w}_{\lambda}\rightharpoonup {w}_{0}$ in *X* and ${w}_{\lambda}\to {w}_{0}$ in ${L}^{p}(\mathrm{\Omega})$ and ${L}^{q}(\partial \mathrm{\Omega})$ as $\lambda \to 0$. It follows that

${\int}_{\mathrm{\Omega}}\nabla {w}_{0}\nabla \varphi =0\mathit{\hspace{1em}}\text{for all}\varphi \in X.$

Hence ${w}_{0}$ is a constant and ${w}_{\lambda}\to {w}_{0}$ in *X*.
It remains to show that ${w}_{0}\ne 0$. If ${w}_{0}=0$, then we set again ${\psi}_{\lambda}=\frac{{w}_{\lambda}}{\parallel {w}_{\lambda}\parallel}$. From

$E({w}_{\lambda})<\frac{p-2}{p-q}{\lambda}^{\frac{p-2}{p-q}}A({w}_{\lambda}),$

we infer that $E({\psi}_{\lambda})\to 0$, so that ${\psi}_{\lambda}\to {\psi}_{0}$ in *X* and ${\psi}_{0}$ is a constant. Moreover, from

$0\le A({w}_{\lambda})+B({w}_{\lambda})$

we have

$-{\parallel {w}_{\lambda}\parallel}^{p-q}A({\psi}_{\lambda})\le B({\psi}_{\lambda}),$

so that $B({\psi}_{0})\ge 0$. From ${\int}_{\partial \mathrm{\Omega}}b<0$ we deduce that ${\psi}_{0}=0$, which contradicts $\parallel {\psi}_{0}\parallel =1$. Therefore, we have proved that ${w}_{0}$ is a non-zero constant. Finally, letting $\lambda \to 0$ in (1.7), we obtain
${w}_{0}^{p-1}|\mathrm{\Omega}|=-{w}_{0}^{q-1}{\int}_{\partial \mathrm{\Omega}}b$, i.e. ${w}_{0}={c}^{*}$.
∎

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