Positive solutions with single and multi-peak for semilinear elliptic equations with nonlinear boundary condition in the half-space

Abstract

We consider the existence of single and multi-peak solutions of the following nonlinear elliptic Neumann problem

$$\begin{aligned} \left\{ \begin{aligned} -\Delta u+\lambda ^{2} u&=Q(x)|u|^{p-2}u \qquad&\text {in} ~~~~\mathbb {R}^{N}_{+}, \\ \frac{\partial u }{\partial n}&=f(x,u) \qquad&\text {on}~~\partial \mathbb {R}^{N}_{+}, \end{aligned}\right. \end{aligned}$$

where $\lambda$ is a large number, $p\in (2,\frac{2N}{N-2})$ for $N\ge 3$, f(x, u) is subcritical about u and Q is positive and has some non-degenerate critical points in $\mathbb {R}^{N}_{+}$. For $\lambda$ large, we can get solutions which have peaks near the non-degenerate critical points of Q.

Appendix

PROPOSITION A1

There is a positive constant $\delta$, independent of w, such that

$$\begin{aligned} \Vert w\Vert ^{2}-(p-1)\int _{\mathbf {R}^{N}_{+}}U_{\lambda a+x}^{p-2}w^{2}\mathrm{d}y\ge \delta \Vert w\Vert ^{2},~\forall w\in E_{\lambda ,x}. \end{aligned}$$

As the arguments in [4], we first consider the eigenvalue problem

$$\begin{aligned} \left\{ \begin{aligned} -\Delta \varphi + \varphi&= \mu U_{\lambda a+x}^{p-2}\varphi&\qquad&\text {in}~ \mathbb {R}^{N}_{+},\\ \frac{\partial \varphi }{\partial n}&= 0&\qquad&\text {on}~ \partial \mathbb {R}^{N}_{+} \end{aligned}\right. \end{aligned}$$

(A1)

which is equivalent to

$$\begin{aligned} \left\{ \begin{aligned} -\Delta \varphi + \varphi&= \mu U^{p-2}\varphi&\qquad&\text {in}~ \Omega _{\lambda }=\{y\in \mathbb {R}^{N}:y_{N}>-\lambda a_{N}-x_{N}\},\\ \frac{\partial \varphi }{\partial n}&= 0&\qquad&\text {on}~ \partial \Omega _{\lambda }. \end{aligned} \right. \end{aligned}$$

(A2)

Denote the j-th pair of eigenvalues and eigenfunctions of problem (A2) by ($\mu _{\lambda ,j},\phi _{\lambda ,j}$) for $j = 1, \ldots , N + 1$ with $\mu _{\lambda ,1}\le \cdots \le \mu _{\lambda ,N+1}$. As $\lambda \rightarrow \infty ,$ the limiting equation of (A2) is as follows:

$$\begin{aligned} \begin{aligned} -\Delta \varphi + \varphi = \mu U^{p-2}\varphi ~~~~~\text {in}~ \mathbb {R}^{N}. \end{aligned} \end{aligned}$$

(A3)

Denote the j-th pair of eigenvalues and eigenfunctions of problem (A3) by ($\mu _{j}, \phi _{j}$) for $j = 1, \ldots , N+1$ with $\mu _1\le \cdots \le \mu _{N+1}$. Then by [4], we have $\mu _1 = 1, \mu _2 = \mu _{N + 1} = p - 1; \phi _1=U, \phi _j\in$ span $\{\frac{\partial U_{x}}{\partial x_{i}}:~ i=1,\ldots ,N\}$ for $j=2,\ldots ,N+1$

Lemma A1

Let $\lambda \rightarrow \infty$. We have the following facts:

$$\begin{aligned} \mu _{\lambda ,j}\rightarrow \mu _{j},~~~j=1,2,\ldots ,N+1;~~~~ \phi _{\lambda ,j}\rightharpoonup \tilde{\phi _{j}}~~~in~H^{1}(\Sigma ),~~~\mathbb {R}^{N}_{+} \subset \Sigma \Subset \mathbb {R}^{N}; \end{aligned}$$

$$\begin{aligned} \tilde{\phi _{j}}\in \mathrm{span} \left\{ \frac{\partial U_{x}}{\partial x_{i}}:~ i=1,\ldots ,N\right\} ,~~j=2,\ldots ,N+1;~~ \tilde{\phi _{1}}=CU; \end{aligned}$$

$$\begin{aligned} \int _{\Omega _{\lambda }}|\nabla \phi _{\lambda ,j}-\nabla \tilde{\phi _{j}}|^{2}+(\phi _{\lambda ,j}- \tilde{\phi _{j}})^{2}\rightarrow 0,~~\lambda \rightarrow \infty ,~~j=1,\ldots ,N+1. \end{aligned}$$

Proof

The first eigenvalue of problem (A2) is defined by $\mu _{\lambda ,1}=\mathrm{inf}\{\int _{\Omega _{\lambda }}|\nabla u|^{2}+u^{2}:u\in H^{1}(\Omega _{\lambda }),\int _{\Omega _{\lambda }}U^{p-2}u^{2}=1\}$. Set  $u=\frac{U(x)}{(\int _{\Omega _{\lambda }}U^{p}\mathrm{d}x)^{1/2}}.$ Then

$$\begin{aligned} \int _{\Omega _{\lambda }}|\nabla u|^{2}+u^{2}\mathrm{d}x=\frac{\int _{\Omega _{\lambda }}|\nabla U|^{2}+U^{2}\mathrm{d}x}{\int _{\Omega _{\lambda }}U^{p}\mathrm{d}x}\rightarrow 1,~~~ \lambda \rightarrow \infty . \end{aligned}$$

Thus $\mu _{\lambda ,1}\le 1+o(1)$, so we can assume that $\mu _{\lambda ,1}\rightarrow \mu _{1}^{*}$ as $\lambda \rightarrow \infty$ and $\mu _{1}^{*}\le 1$. First, $\mu _{1}^{*}\ne 0$, else we have $\int _{\Omega _{\lambda }}U^{p-2}\phi _{\lambda ,1}^{2}\rightarrow 0$ as $\lambda \rightarrow \infty ,$ which is a contradiction with $\int _{\Omega _{\lambda }}U^{p-2}\phi _{\lambda ,1}^{2}=1$. Because $\phi _{\lambda ,1}$ is bounded in $H^{1}(\Omega _{\lambda })$ and

$$\begin{aligned}\left\{ \begin{aligned} -\Delta \phi _{\lambda ,1} + \phi _{\lambda ,1}&=\mu _{\lambda ,1} U^{p-2}\phi _{\lambda ,1}&~~~~~&\text {in}~ \Omega _{\lambda }=\{y\in \mathbb {R}^{N}:y_{N}>-\lambda a_{N}-x_{N}\}, \\ \frac{\partial \phi _{\lambda ,1}}{\partial n}&= 0&~~~~~&\text {on}~ \partial \Omega _{\lambda }, \end{aligned}\right. \end{aligned}$$

$$\begin{aligned} \int _{\Omega _{\lambda }}\nabla \phi _{\lambda ,1}\cdot \nabla \psi +\phi _{\lambda ,1}\psi =\mu _{\lambda ,1}\int _{\Omega _{\lambda }} U^{p-2}\phi _{\lambda ,1}\psi ,~~~\forall \psi \in H^{1}(\mathbb {R}^{N}), \end{aligned}$$

we have that $\phi _{\lambda ,1}\rightharpoonup \tilde{\phi _{1}}$ in $H^{1}(\Sigma )$, for $\mathbb {R}^{N}_{+}\subset \Sigma \subset \mathbb {R}^{N}$. Let $\lambda \rightarrow \infty$,

$$\begin{aligned} \int _{\mathbb {R}^{N}}\nabla \tilde{\phi _{1}}\cdot \nabla \psi +\tilde{\phi _{1}}\psi \mathrm{d}x=\int _{\mathbb {R}^{N}}\mu _{1}^{*} U^{p-2}\tilde{\phi _{1}}\psi \mathrm{d}x, \end{aligned}$$

thus

$$\begin{aligned} -\Delta \tilde{\phi _{1}}+\tilde{\phi _{1}}=\mu _{1}^{*} U^{p-2}\tilde{\phi _{1}}~~~~~\mathrm{in}~~\mathbb {R}^{N} \end{aligned}$$

and $(\mu _{1}^{*},\tilde{\phi _{1}})$ is a pair of eigenvalue and eigenfunction of (A3) and $0<\mu _{1}^{*} \le 1$. Hence, $\mu _{1}^{*}=1$ and $\tilde{\phi _{1}}=CU.$

By induction, we can get the rest of the conclusion. Suppose for $2\le k\le N+1$, we have

$$\begin{aligned}&\phi _{\lambda ,j}\rightharpoonup \tilde{\phi _{j}}~~~\mathrm{in}~H^{1}(\Sigma ),~~ \Sigma \subset \mathbb {R}^{N},~~\tilde{\phi _{j}}\in \text {span} \\&\quad \left\{ U,\frac{\partial U_{x}}{\partial x_{i}}: i=1,\ldots ,N\right\} .~~~~j=1,2,\ldots ,k-1. \end{aligned}$$

The k-th eigenvalue of (A2) is characterized by

$$\begin{aligned}&\mu _{\lambda ,k}=\mathrm{inf}\left\{ \int _{\Omega _{\lambda }}|\nabla u|^{2}+u^{2}\mathrm{d}x:u\in H^{1}(\Omega _{\lambda }), \right. \\&\quad \left. \int _{\Omega _{\lambda }}U^{p-2}u^{2}\mathrm{d}x=1, \langle u,\phi _{\lambda ,j}\rangle _{\Omega _{\lambda }}=0,j=1,\ldots ,k-1\right\} . \end{aligned}$$

Claim 1. $\lim \nolimits _{\lambda \rightarrow \infty }\mu _{\lambda ,k}=p-1$. At first, let us prove $\mu _{\lambda ,k}\le p-1+o(1)$. Choose $t_{\lambda ,j}=\frac{\langle \phi _{\lambda ,j}, \tilde{U}\rangle _{\Omega _{\lambda }}}{\mu _{\lambda ,j}}$, $j=1,\ldots ,k-1$, $\tilde{U}\in \text {span}\{U,\frac{\partial U_{x}}{\partial x_{i}}:i=1,\ldots ,N\}\backslash \text {span} \{\tilde{\phi _{i}},i=1,\ldots ,k-1\}$ and

$$\begin{aligned} \int _{\mathbb {R}^{N}}|\nabla \tilde{U}|^{2}+|\tilde{U}|^{2}\mathrm{d}x=p-1, ~~~\int _{\Omega _{\lambda }}U^{p-2}\tilde{U}^{2}\mathrm{d}x=1. \end{aligned}$$

Set $u_{\lambda }=\frac{\tilde{U}-\sum _{j=1}^{k-1}t_{\lambda ,j}\phi _{\lambda ,j}}{\big (\int _{\Omega _{\lambda }}U^{p-2}(\tilde{U}-\sum _{j=1}^{k-1}t_{\lambda ,j}\phi _{\lambda ,j})^{2}\big )^{1/2}}$. Then $u_{\lambda }$ satisfies

$$\begin{aligned} \int _{\Omega _{\lambda }}U^{p-2}u_{\lambda }^{2}\mathrm{d}x=1,~ \langle u_{\lambda },\phi _{\lambda ,j}\rangle _{\Omega _{\lambda }}=0,~j=1,\ldots ,k-1, \end{aligned}$$

hence

$$\begin{aligned} \mu _{\lambda ,k}\le ~\langle u_{\lambda },u_{\lambda }\rangle _{\Omega _{\lambda }}~\rightarrow p-1,~ j=1,\ldots ,k-1,~~\mathrm{as}~\lambda \rightarrow \infty . \end{aligned}$$

As before, we have

$$\begin{aligned} \phi _{\lambda ,k}\rightharpoonup \phi _{k}^{*}~~~\text {in}~H^{1}(\Sigma ) \end{aligned}$$

and

$$\begin{aligned} \int _{_{\mathbb {R}^{N}}}\nabla \phi _{k}^{*}\cdot \nabla \psi +\phi _{k}^{*}\psi =\int _{_{\mathbb {R}^{N}}}\mu _{k}^{*} U^{p-2}\phi _{k}^{*}\psi ,\quad \forall \psi \in H^{1}(\mathbb {R^{N}}). \end{aligned}$$

From $\langle \phi _{\lambda ,k},\phi _{\lambda ,j}\rangle _{\Omega _{\lambda }}=0$, $j=1,\ldots ,k-1$, we get that $\langle \phi ^{*}_{k},{\tilde{\phi _{j}}}\rangle =0.$ Thus

For $k=N+2,$ by similar argument we can get $\lim \nolimits _{\lambda \rightarrow \infty }\mu _{\lambda ,N+2}=\mu _{N+2}>p-1$. $\square$

Proof of Proposition A1

For all $w\in E_{\lambda ,x}$, we set $\tilde{w}(y)=w(y+\lambda a +x)$, then $\tilde{w}(y)$ satisfies $\langle \tilde{w}(y),U\rangle _{\Omega _{\lambda }}=\langle \tilde{w}(y),\frac{\partial U}{\partial y_{i}}\rangle _{\Omega _{\lambda }}=0$, $i=1,\ldots ,N$. Then the problem becomes

$$\begin{aligned} \Vert \tilde{w}\Vert _{\Omega _{\lambda }}^{2}-(p-1)\int _{\Omega _{\lambda }}U^{p-2}\tilde{w}^{2}\mathrm{d}x\ge \rho ' \Vert \tilde{w}\Vert _{\Omega _{\lambda }}^{2},~\forall w\in E_{\lambda ,x}. \end{aligned}$$

Let

$$\begin{aligned} \tilde{w}&=\sum _{i=1}^{N+1}\langle \tilde{w},\phi _{\lambda ,i} \rangle _{\Omega _{\lambda }} \phi _{\lambda ,i}+R_{\lambda }, ~~~~~ \\ \Vert \tilde{w}\Vert _{\Omega _{\lambda }}^{2}&=\sum _{i=1}^{N+1} \langle \tilde{w},\phi _{\lambda ,i}\rangle _{\Omega _{\lambda }}^{2} \Vert \phi _{\lambda ,i}\Vert _{\Omega _{\lambda }}^{2}+\Vert R_{\lambda }\Vert _{\Omega _{\lambda }}^{2}. \end{aligned}$$

So, it is not hard to derive

$$\begin{aligned} \Vert R_{\lambda }\Vert ^{2}_{\Omega _{\lambda }}\ge & {} \mu _{\lambda ,N+2}\int _{\Omega _{\lambda }}U^{p-2}R_{\lambda }^{2}\mathrm{d}x \\= & {} \mu _{\lambda ,N+2}\left[ \int _{\Omega _{\lambda }}U^{p-2}\tilde{w}^{2}\mathrm{d}x -2\sum _{i=1}^{N+1}\langle \tilde{w},\phi _{\lambda ,i}\rangle _{\Omega _{\lambda }}\right. \\&\left. \int _{\Omega _{\lambda }}U^{p-2}\tilde{w}+\sum _{i=1}^{N+1} \langle \tilde{w},\phi _{\lambda ,i}\rangle _{\Omega _{\lambda }}^{2}\right] \\= & {} \mu _{\lambda ,N+2}\left[ \int _{\Omega _{\lambda }}U^{p-2}\tilde{w}^{2}\mathrm{d}x +\sum _{i=1}^{N+1}\langle \tilde{w},\phi _{\lambda ,i}\rangle _{\Omega _{\lambda }}^{2} +o(1)\Vert \tilde{w}\Vert _{\Omega _{\lambda }}\right] . \end{aligned}$$

This implies

$$\begin{aligned} \Vert \tilde{w}\Vert _{\Omega _{\lambda }}^{2}&\ge \mu _{\lambda ,N+2} \left[ \int _{\Omega _{\lambda }}U^{p-2}\tilde{w}^{2}\mathrm{d}x +\sum _{i=1}^{N+1}\langle \tilde{w},\phi _{\lambda ,i}\rangle _{\Omega _{\lambda }}^{2}\right] \\&\quad +\sum _{i=1}^{N+1}\langle \tilde{w},\phi _{\lambda ,i}\rangle _{\Omega _{\lambda }}^{2}\mu _{\lambda ,i}+o(1). \end{aligned}$$

Because $\phi _{\lambda ,i}\rightarrow \tilde{\phi _{i}}\in \text {span}\{U,\frac{\partial U_{x}}{\partial x_{i}}:~ i=1,\cdot ,N\}$, $\langle \tilde{w},\phi _{\lambda ,i}\rangle \rightarrow 0$ as $\lambda \rightarrow \infty$;

$$\begin{aligned} \Vert \tilde{w}\Vert _{\Omega _{\lambda }}^{2}\ge \mu _{\lambda ,N+2} \int _{\Omega _{\lambda }}U^{p-2}\tilde{w}^{2}\mathrm{d}x +o(1)\Vert \tilde{w}\Vert _{\Omega _{\lambda }}^{2}. \end{aligned}$$

Hence, by $\mu _{\lambda ,N+2}>p-1$ for large $\lambda$, there exists a small constant $\rho '>0$ such that

$$\begin{aligned} \Vert \tilde{w}\Vert _{\Omega _{\lambda }}^{2}-(p-1)\int _{\Omega _{\lambda }}U^{p-2} \tilde{w}^{2}\mathrm{d}x\ge \rho '\Vert \tilde{w}\Vert _{\Omega _{\lambda }}^{2}, \end{aligned}$$

which is equivalent to

$$\begin{aligned} \Vert w\Vert ^{2}-(p-1)\int _{\mathbb {R}^{N}_{+}}U_{\lambda a+x}^{p-2}w^{2}\mathrm{d}x\ge \rho '\Vert w\Vert ^{2}. \end{aligned}$$

PROPOSITION A2

There exists a positive constant $\delta$ independent of w, such that for large $\lambda$,

$$\begin{aligned} \Vert w\Vert ^{2}-(p-1)\int _{\mathbb {R}^{N}_{+}}({U_{\lambda a^{1}+x^{1}}+U_{\lambda a^{2}+x^{2}}})^{p-2}w^{2}\mathrm{d}y\ge \delta \Vert w\Vert ^{2},~\forall w\in E_{\lambda ,x^{1},x^{2}}. \end{aligned}$$

With the help of Proposition A1, we can show this result by a similar argument in [4].

Proof

Set

$$\begin{aligned}&\Lambda _{\lambda }=\mathrm{inf}\left\{ \displaystyle \int _{\mathbb {R}^{N}_{+}}|\nabla w|^{2}+w^{2}:~w\in E_{\lambda ,x^{1},x^{2}}, \right. \\&\quad \quad \quad \quad \quad \quad \qquad \quad \left. \displaystyle \int _{\mathbb {R}^{N}_{+}}(U_{\lambda a^{1}+x^{1}}+U_{\lambda a^{2}+x^{2}})^{p-2}w^{2}\mathrm{d}y=1\right\} . \end{aligned}$$

It suffices to show that $\Lambda _{\lambda }>p-1+c_{0}$, for some positive number $c_{0}$, as $\lambda$ large enough. By contradiction, suppose there is a sequence $\lambda _{n}\rightarrow \infty$: $\lim \nolimits _{n\rightarrow \infty }\Lambda _{\lambda _{n}}=\Lambda \le p-1$. Namely, there exists $w_{n}\in H^{1}(\mathbb {R}^{N}_{+})$:

$$\begin{aligned}&\int _{\mathbb {R}^{N}_{+}}\nabla w_{n}\cdot \nabla U_{\lambda _{n} a^{i}+x^{i}}+ w_{n}\cdot U_{\lambda _{n} a^{i}+x^{i}}\mathrm{d}y=0,~~~~i=1,2; \end{aligned}$$

(A4)

$$\begin{aligned}&\int _{\mathbb {R}^{N}_{+}}\nabla w_{n}\cdot \nabla \frac{\partial U_{\lambda _{n} a^{i}+x^{i}}}{\partial x^{i}_{j}}+ w_{n}\cdot \frac{\partial U_{\lambda _{n} a^{i}+x^{i}}}{\partial x^{i}_{j}}\mathrm{d}y=0, \nonumber \\&i=1,2; ~j=1,\ldots ,N; \end{aligned}$$

(A5)

$$\begin{aligned}&\int _{\mathbb {R}^{N}_{+}}|\nabla w_{n}|^{2}+w_{n}^{2}\mathrm{d}y =\Lambda _{\lambda _{n}} \int _{\mathbb {R}^{N}_{+}} \left( U_{\lambda _{n} a^{1}+x^{1}}+ U_{\lambda _{n} a^{2}+x^{2}}\right) ^{p-2}w_{n}^{2}\mathrm{d}y; \end{aligned}$$

(A6)

$$\begin{aligned}&\int _{\mathbb {R}^{N}_{+}}\nabla w_{n}\cdot \nabla \psi +w_{n} \psi \mathrm{d}y=\Lambda _{\lambda _{n}} \int _{\mathbb {R}^{N}_{+}} (U_{\lambda _{n} a^{1}+x^{1}}+ U_{\lambda _{n} a^{2}+x^{2}})^{p-2}w_{n}\psi \mathrm{d}y,\nonumber \\&\forall \psi \in E_{\lambda _{n},x^{1},x^{2}}. \end{aligned}$$

(A7)

Set $\tilde{w}_{n}(x)=w_{n}(x+\lambda _{n} a^{1}+x^{1})$, $\Omega _{\lambda _{n}}=\mathbb {R}^{N}_{+}-\{\lambda _{n} a^{1}+x^{1}\}$. Then we have

$$\begin{aligned} \int _{\Omega _{\lambda _{n}}}|\nabla \tilde{w}_{n}|^{2}+\tilde{w}_{n}^{2}=\Lambda _{\lambda _{n}} \int _{\Omega _{\lambda _{n}}} (U+ U_{\lambda _{n} a^{2}+x^{2}-\lambda _{n} a^{1}-x^{1}})^{p-2}\tilde{w}_{n}^{2}. \end{aligned}$$

We can assume that for some $\tilde{w}\in H^{1}(\mathbb {R}^{N})$, there holds

$$\begin{aligned} \lim _{n\rightarrow \infty }\tilde{w}_{n}=\tilde{w},~~~\text {weakly}~~~\text {in}~~H^{1}(\Sigma ),~~\forall ~~ \mathbb {R}^{N}_{+}\subset \Sigma \subset \mathbb {R}^{N}. \end{aligned}$$

Then from (3) and (4) we have

$$\begin{aligned}&\int _{\mathbb {R}^{N}}\nabla \tilde{w}\cdot \nabla U+\tilde{w}U\mathrm{d}y=0, \end{aligned}$$

(A8)

$$\begin{aligned}&\quad \int _{\mathbb {R}^{N}}\nabla \tilde{w}\cdot \nabla \frac{\partial U}{\partial x_{j}}+\tilde{w}\frac{\partial U}{\partial x_{j}}\mathrm{d}y=0,~~ j=1,\ldots ,N. \end{aligned}$$

(A9)

We claim that

$$\begin{aligned} \begin{aligned} \int _{\mathbb {R}^{N}}|\nabla \tilde{w}|^{2}+|\tilde{w}|^{2}=\Lambda \int _{\mathbb {R}^{N}}U^{p-2}\tilde{w}^{2}\mathrm{d}y. \end{aligned} \end{aligned}$$

(A10)

Now we are to choose numbers $l_{\lambda _{n},1},l_{\lambda _{n},2},\ldots ,l_{\lambda _{n},2N+2}$ such that $v\in E_{\lambda _{n},x^{1},x^{2}}$, in which $w(x)=\tilde{w}(x-\lambda _{n} a^{1}-x^{1})$,

$$\begin{aligned}&v=w-l_{\lambda _{n},1}U_{\lambda _{n} a^{1}+x^{1}}-l_{\lambda _{n},2}U_{\lambda _{n} a^{2}+x^{2}} \\&\qquad -\sum \limits _{j=1}^{N}\left( l_{\lambda _{n},j+2}\frac{\partial U_{\lambda _{n} a^{1}+x^{1}}}{\partial x_{j}^{1}}+l_{\lambda _{n},N+j+2}\frac{\partial U_{\lambda _{n} a^{2}+x^{2}}}{\partial x_{j}^{2}}\right) . \end{aligned}$$

This is equivalent to the following system of linear equations

$$\begin{aligned} \left\langle w,U_{\lambda _{n} a^{1}+x^{1}}\right\rangle&=l_{\lambda _{n},1}\Vert U_{\lambda _{n} a^{1}+x^{1}}\Vert ^{2}+l_{\lambda _{n},2}\left\langle U_{\lambda _{n} a^{1}+x^{1}},U_{\lambda _{n} a^{2}+x^{2}}\right\rangle \nonumber \\&\quad +\sum _{j=1}^{N}l_{\lambda _{n},j+2}\left\langle \frac{\partial U_{\lambda _{n} a^{1}+x^{1}}}{\partial x_{j}^{1}},U_{\lambda _{n} a^{1}+x^{1}}\right\rangle \nonumber \\&\quad +\sum _{j=1}^{N}l_{\lambda _{n},N+j+2}\left\langle \frac{\partial U_{\lambda _{n} a^{2}+x^{2}}}{\partial x_{j}^{2}},U_{\lambda _{n} a^{1}+x^{1}}\right\rangle , \end{aligned}$$

(A11)

$$\begin{aligned} \left\langle w,U_{\lambda _{n} a^{2}+x^{2}}\right\rangle&= l_{\lambda _{n},2}\Vert U_{\lambda _{n} a^{2}+x^{2}}\Vert ^{2}+l_{\lambda _{n},1}\left\langle U_{\lambda _{n} a^{1}+x^{1}},U_{\lambda _{n} a^{2}+x^{2}}\right\rangle \nonumber \\&\quad +\sum _{j=1}^{N}l_{\lambda _{n},j+2}\left\langle \frac{\partial U_{\lambda _{n} a^{1}+x^{1}}}{\partial x_{j}^{1}},U_{\lambda _{n} a^{2}+x^{2}}\right\rangle \nonumber \\&\quad + \sum _{j=1}^{N}l_{\lambda _{n},N+j+2}\left\langle \frac{\partial U_{\lambda _{n} a^{2}+x^{2}}}{\partial x_{j}^{2}},U_{\lambda _{n} a^{2}+x^{2}}\right\rangle ,~~~~~~\end{aligned}$$

(A12)

$$\begin{aligned} \left\langle w,\frac{\partial U_{\lambda _{n} a^{1}+x^{1}}}{\partial x_{i}^{1}}\right\rangle&= l_{\lambda _{n},1}\left\langle U_{\lambda _{n} a^{1}+x^{1}},\frac{\partial U_{\lambda _{n} a^{1}+x^{1}}}{\partial x_{i}^{1}}\right\rangle \nonumber \\&\quad +l_{\lambda _{n},2}\left\langle \frac{\partial U_{\lambda _{n} a^{1}+x^{1}}}{\partial x_{i}^{1}},U_{\lambda _{n} a^{2}+x^{2}}\right\rangle \nonumber \\&\quad +\sum _{j=1}^{N}l_{\lambda _{n},j+2}\left\langle \frac{\partial U_{\lambda _{n} a^{1}+x^{1}}}{\partial x_{j}^{1}},\frac{\partial U_{\lambda _{n} a^{1}+x^{1}}}{\partial x_{i}^{1}}\right\rangle \nonumber \\&\quad + \sum _{j=1}^{N}l_{\lambda _{n},N+j+2}\left\langle \frac{\partial U_{\lambda _{n} a^{2}+x^{2}}}{\partial x_{j}^{2}},\frac{\partial U_{\lambda _{n} a^{1}+x^{1}}}{\partial x_{i}^{1}}\right\rangle , \end{aligned}$$

(A13)

$$\begin{aligned} \left\langle w,\frac{\partial U_{\lambda _{n} a^{2}+x^{2}}}{\partial x_{i}^{2}}\right\rangle&= l_{\lambda _{n},1}\left\langle U_{\lambda _{n} a^{1}+x^{1}},\frac{\partial U_{\lambda _{n} a^{2}+x^{2}}}{\partial x_{i}^{2}}\right\rangle \nonumber \\&\quad +l_{\lambda _{n},2}\left\langle \frac{\partial U_{\lambda _{n} a^{2}+x^{2}}}{\partial x_{i}^{2}},U_{\lambda _{n} a^{2}+x^{2}}\right\rangle \nonumber \\&\quad +\sum _{j=1}^{N}l_{\lambda _{n},j+2}\left\langle \frac{\partial U_{\lambda _{n} a^{1}+x^{1}}}{\partial x_{j}^{1}},\frac{\partial U_{\lambda _{n} a^{2}+x^{2}}}{\partial x_{i}^{2}}\right\rangle \nonumber \\&\quad + \sum _{j=1}^{N}l_{\lambda _{n},N+j+2}\left\langle \frac{\partial U_{\lambda _{n} a^{2}+x^{2}}}{\partial x_{j}^{2}},\frac{\partial U_{\lambda _{n} a^{2}+x^{2}}}{\partial x_{i}^{2}}\right\rangle . \end{aligned}$$

(A14)

To solve the system of linear equations, we need to establish estimates of the coefficients. We know that

$$\begin{aligned}&\langle U_{\lambda a^{1}+x^{1}},U_{\lambda a^{2}+x^{2}}\rangle =o(1),~~~\lambda \rightarrow \infty ; \\&\left\langle \frac{\partial U_{\lambda a^{1}+x^{1}}}{\partial x_{j}^{1}},U_{\lambda a^{1}+x^{1}}\right\rangle =\int _{\Omega _{\lambda }}\nabla \frac{\partial U}{\partial x_{j}}\cdot \nabla U+ \frac{\partial U}{\partial x_{j}}\cdot U=o(1),~~~\lambda \rightarrow \infty . \end{aligned}$$

Similarly,

$$\begin{aligned}&\quad \left\langle \frac{\partial U_{\lambda a^{i}+x^{i}}}{\partial x_{j}^{i}},U_{\lambda a^{k}+x^{k}}\right\rangle =o(1),~~\lambda \rightarrow \infty ,~~i=1,2;~~k=1,2; \\&\quad \Vert U_{\lambda a^{i}+x^{i}}\Vert ^{2}\rightarrow A^{2},~~\lambda \rightarrow \infty ,~~i=1,2;~~A^{2}=\Vert U\Vert ^{2}_{H^{1}(\mathbb {R}^{N})}, \\&\qquad \Bigg \Vert \frac{\partial U_{\lambda a^{i}+x^{i}}}{\partial x_{j}^{i}}\Bigg \Vert ^{2}\rightarrow B^{2},~~\lambda \rightarrow \infty ,~~i=1,2;~~~j=1,\ldots ,N; \\&\quad ~~~~B^{2}=\Vert \nabla U\Vert ^{2}_{H^{1}(\mathbb {R}^{N})}. \end{aligned}$$

By (6) we have

$$\begin{aligned}&\langle w,U_{\lambda _{n} a^{1}+x^{1}}\rangle =\int _{\mathbb {R}^{N}_{+}}\nabla w\cdot \nabla U_{\lambda _{n} a^{1}+x^{1}}+wU_{\lambda _{n} a^{1}+x^{1}}\mathrm{d}y \\&\quad = \int _{\mathbb {R}^{N}_{+}-\{\lambda _{n} a^{1}+x^{1}\}}\nabla w(x+\lambda _{n} a^{1}+x^{1})\cdot \nabla U+w(x+\lambda _{n} a^{1}+x^{1})U\mathrm{d}y \\&\quad \rightarrow \int _{\mathbb {R}^{N}}\nabla \tilde{w}\cdot \nabla U+\tilde{w}U\mathrm{d}y=0,~ \lambda _{n}\rightarrow \infty . \end{aligned}$$

Similarly,

$$\begin{aligned}&\left\langle w,\frac{\partial U_{\lambda _{n} a^{1}+x^{1}}}{\partial x_{j}^{1}}\right\rangle =o(1),~~~~ \lambda _{n}\rightarrow \infty ,~~j=1,\ldots ,N; \\&\quad \left\langle w,U_{\lambda _{n} a^{2}+x^{2}}\right\rangle \,=\int _{\mathbb {R}^{N}_{+}-\lambda a^{2}-x^{2}}\nabla \tilde{w}\cdot \nabla U(x+\lambda _{n} a^{1}+x^{1}-\lambda _{n} a^{2}-x^{2}) \\&\quad \quad +\,\tilde{w}U(x+\lambda _{n} a^{1}+x^{1}-\lambda _{n} a^{2}-x^{2})\mathrm{d}y. \end{aligned}$$

By (3) and $\tilde{w}_{n}\rightharpoonup \tilde{w}$, we have

$$\begin{aligned} \langle w,U_{\lambda _{n} a^{2}+x^{2}}\rangle =o(1). \end{aligned}$$

Similarly, $\langle w,\frac{\partial U_{\lambda _{n} a^{2}+x^{2}}}{\partial x_{j}^{2}}\rangle =o(1)$, $\lambda _{n}\rightarrow \infty$, $j=1,\ldots ,N.$ Owing to the above estimates, we know that the system of linear equations is equivalent to

$$\begin{aligned} \left( \begin{array}{cccc} a_{1,1} &{} a_{1,2} &{} \cdots &{} a_{1,2N+2} \\ a_{2,1} &{} a_{2,2} &{} \cdots &{} a_{2,2N+2} \\ \vdots &{} \vdots &{} \vdots &{} \vdots \\ a_{2N+2,1} &{} a_{2N+2,2} &{} \cdots &{} a_{2N+2,2N+2} \end{array} \right) \cdot \left( \begin{array}{c} l_{\lambda ,1} \\ l_{\lambda ,2} \\ \vdots \\ l_{\lambda ,2N+2} \end{array} \right) = \left( \begin{array}{c} o(1) \\ o(1) \\ \vdots \\ o(1) \end{array} \right) , \end{aligned}$$

in which $a_{ij}=o(1)$ if $i\ne j$, $a_{ii}>a_{0}>0$, for $i,j=1,\ldots ,2N+2$. Therefore, (10)–(13) has a solution and $l_{\lambda ,i}=o(1)$, $i=1,\ldots ,2N+2$. Next, we will give the proof of (9). Set

$$\begin{aligned} v_{n}= & {} w-l_{\lambda _{n},1}U_{\lambda _{n} a^{1}+x^{1}}-l_{\lambda _{n},2}U_{\lambda _{n} a^{2}+x^{2}} \\&-\sum _{j=1}^{N}\left( l_{\lambda _{n},j+2}\frac{\partial U_{\lambda _{n} a^{1}+x^{1}}}{\partial x_{j}^{1}}+l_{\lambda _{n},N+j+2}\frac{\partial U_{\lambda _{n} a^{2}+x^{2}}}{\partial x_{j}^{2}}\right) , \end{aligned}$$

such that $v_{n}\in E_{\lambda _{n},x^{1},x^{2}}$ and substitute $\psi =v_{n}$ into (6). We have

$$\begin{aligned}&\int _{\mathbb {R}^{N}_{+}}\nabla w_{n}\cdot \nabla w+w_{n}w\mathrm{d}y = \Lambda _{\lambda _{n}}\int _{\mathbb {R}^{N}_{+}}(U_{\lambda _{n} a_{1}+x^{1}}+ U_{\lambda _{n} a_{2}+x^{2}})^{p-2}w_{n}w\mathrm{d}y \\&\quad - \Lambda _{\lambda _{n}}\bigg [l_{\lambda _{n},1}\int _{\mathbb {R}^{N}_{+}}(U_{\lambda _{n} a_{1}+x^{1}}+ U_{\lambda _{n} a_{2}+x^{2}})^{p-2}w_{n}U_{\lambda _{n} a_{1}+x^{1}}\mathrm{d}y \\&\quad +l_{\lambda _{n},2}\int _{\mathbb {R}^{N}_{+}}(U_{\lambda _{n} a_{1}+x^{1}}+ U_{\lambda _{n} a_{2}+x^{2}})^{p-2}w_{n}U_{\lambda _{n} a_{2}+x^{2}}\mathrm{d}y \\&\quad +\sum _{j=1}^{N} \left( l_{\lambda _{n},j+2}\int _{\mathbb {R}^{N}_{+}}(U_{\lambda _{n} a_{1}+x^{1}}+ U_{\lambda _{n} a_{2}+x^{2}})^{p-2}w_{n}\frac{\partial U_{\lambda _{n} a^{1}+x^{1}}}{\partial x_{j}^{1}}\mathrm{d}y\right. \\&\quad \left. +l_{\lambda _{n},j+2+N}\int _{\mathbb {R}^{N}_{+}}(U_{\lambda _{n} a_{1}+x^{1}}+ U_{\lambda _{n} a_{2}+x^{2}})^{p-2}w_{n}\frac{\partial U_{\lambda _{n} a^{2}+x^{2}}}{\partial x_{j}^{2}}\right) \mathrm{d}y\bigg ]. \end{aligned}$$

Thus $\int _{\mathbb {R}^{N}_{+}-\lambda _{n} a_{1}-x^{1}}\nabla \tilde{w}_{n}\cdot \nabla \tilde{w}+\tilde{w}_{n}\tilde{w}\mathrm{d}y\rightarrow \int _{\mathbb {R}^{N}}|\nabla \tilde{w}|^{2}+\tilde{w}^{2}\mathrm{d}y$, and the right-hand side of the equation $\rightarrow \Lambda \int _{\mathbb {R}^{N}}U^{p-2}\tilde{w}^{2}$. If $\tilde{w}\not \equiv 0$, then $\Lambda =p-1$. In fact, by (7) and (8),

$$\begin{aligned} p-1\le & {} \text {inf}\left\{ \int _{\mathbb {R}^{N}}|\nabla u|^{2}+u^{2}:\,u\in H^{1}(\mathbb {R}^{N}),\,\int _{\mathbb {R}^{N}}U^{p-2}u^{2}=1,\right. \\&\left. \langle u,U\rangle _{\mathbb {R}^{N}}=\left\langle u,\frac{\partial U}{\partial x_{i}}\right\rangle _{\mathbb {R}^{N}}=0\right\} \le \Lambda \le p-1. \end{aligned}$$

Therefore $\tilde{w}$ is an eigenfunction with eigenvalue $p-1$, $\tilde{w}\in \text {Span}\{\frac{\partial U}{\partial x_{i}},i=1,\ldots ,N\}$. This is a contradiction with (8). We have $\tilde{w}=0.$ So, $\widetilde{w_{n}}\rightharpoonup 0$ in $H^{1}(\Sigma )$, $\forall ~ \Sigma :~\mathbb {R}^{N}_{+}\subset \Sigma \subset \mathbb {R}^{N}.$ Similarly, $\hat{w}_{n}=w_{n}(x+\lambda _{n}a^{2}+x^{2}),~\hat{w}_{n}\rightharpoonup 0$, in $H^{1}(\Sigma )$. Then we have

$$\begin{aligned} 1&= \int _{\mathbb {R}^{N}_{+}}(U_{\lambda _{n} a^{1}+x^{1}}+ U_{\lambda _{n} a^{2}+x^{2}})^{p-2}w_{n}^{2} \\&=\int _{\mathbb {R}^{N}_{+}}U_{\lambda _{n} a^{1}+x^{1}}^{p-2}w_{n}^{2}+ \int _{\mathbb {R}^{N}_{+}}U_{\lambda _{n} a^{2}+x^{2}}^{p-2}w_{n}^{2}+o(1) \\&=\int _{\mathbb {R}^{N}_{+}-\lambda _{n} a^{1}-x^{1}}U^{p-2}\widetilde{w_{n}}^{2}+ \int _{\mathbb {R}^{N}_{+}-\lambda _{n} a^{2}-x^{2}}U^{p-2}\widehat{w_{n}}^{2}+o(1) \\&\rightarrow 0. \end{aligned}$$

Hence it is a contradiction. Thus $\Lambda _{\lambda }>p-1+c_{0},$ for $\lambda$ large enough. $\square$