Non-degeneracy of the bubble solutions for the Hénon equation and applications

Abstract

We consider the following Hénon equation with critical growth:

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u = K(|y|)u^{\frac{N+2}{N-2}},u>0, &{}\hbox {in } B_{1}(0) \\ \displaystyle u =0,&{}\hbox {on } \partial B_{1}(0), \end{array}\right. } \end{aligned}$$

where $B_{1}(0)$ is the unit ball in ${\mathbb {R}}^{N}$, $K:[0,1] \rightarrow {\mathbb {R}}^{+}$ is a bounded function and $K''(1)$ exists. We prove a non-degeneracy result of the bubble solutions constructed in [24] via the local Pohozaev identities for $N \ge 5$. Then, as applications, by using reduction arguments combined with delicate estimates for the modified Green function and the error, we prove the new existence of infinitely many non-radial solutions, whose energy can be arbitrarily large.

Appendices

Appendix A: Pohozaev identities

Let

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle -\Delta u =K(|y|)u^{2^*-1},\quad u>0,\quad &{}\hbox {in } B_{1}(0),\\ \displaystyle u=0, &{}\hbox {on } \partial B_1(0), \end{array}\right. } \end{aligned}$$

(A.1)

and

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle -\Delta \xi =(2^*-1)K(|y|)u^{2^*-2}\xi , &{}\hbox {in }B_{1}(0), \\ \displaystyle \xi =0, &{}\hbox {on } \partial B_1(0). \end{array}\right. } \end{aligned}$$

(A.2)

Assume that $\Omega$ is a smooth bounded domain in $B_{1}(0)$.The following identities can be found in [11].

Lemma A.1

It holds

$$\begin{aligned} \begin{aligned}&-\int _{\partial \Omega }\frac{\partial u}{\partial \nu }\frac{\partial \xi }{\partial y_i} -\int _{\partial \Omega }\frac{\partial \xi }{\partial \nu }\frac{\partial u}{\partial y_i} + \int _{\partial \Omega }\langle \nabla u, \nabla \xi \rangle \nu _{i} - \int _{\partial \Omega }K(|y|)u^{2^*-1}\xi \nu _{i} \\& \quad=\,-\int _{ \Omega }u^{2^*-1}\xi \frac{\partial K(|y|)}{\partial y_{i}}, \end{aligned} \end{aligned}$$

(A.3)

and

$$\begin{aligned} \begin{aligned}&\int _{ \Omega }u^{2^*-1}\xi \langle \nabla K(|y|), y- x_0 \rangle \\ & \quad=\,\int _{\partial \Omega }K(|y|)u^{2^*-1}\xi \langle \nu , y-x_0 \rangle \\&\qquad+ \int _{\partial \Omega }\frac{\partial u}{\partial \nu }\langle \nabla \xi , y-x_0 \rangle +\int _{\partial \Omega }\frac{\partial \xi }{\partial \nu }\langle \nabla u, y-x_0 \rangle +\int _{\partial \Omega }\langle \nabla u, \nabla \xi \rangle \langle \nu , y-x_0 \rangle \\ {}&\qquad+\frac{N-2}{2}\int _{\partial \Omega }\xi \frac{\partial u}{\partial \nu } + \frac{N-2}{2}\int _{\partial \Omega } u\frac{\partial \xi }{\partial \nu }. \end{aligned} \end{aligned}$$

(A.4)

Appendix B: Some identities involving the Green’s function

Recall that

$$\begin{aligned} I_1(u,v,d) = -\int _{\partial B_{d}(x_{k_{n},1})}\frac{\partial u}{\partial \nu }\frac{\partial v}{\partial y_1} -\int _{\partial B_{d}(x_{k_{n},1})}\frac{\partial v}{\partial \nu }\frac{\partial u}{\partial y_i} + \int _{\partial B_{d}(x_{k_{n},1})}\langle \nabla u, \nabla v \rangle \nu _{1}, \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} I_2( u,v ,d ) =\,&\int _{\partial B_{d}(x_{k_{n},1})}\frac{\partial u}{\partial \nu }\langle \nabla v, y-x_{k_{n},1} \rangle +\int _{\partial B_{d}(x_{k_{n},1})}\frac{\partial v}{\partial \nu }\langle \nabla u, y-x_{k_{n},1} \rangle \\ {}&\quad-\int _{\partial B_{d}(x_{k_{n},1})}\langle \nabla u, \nabla v \rangle \langle \nu , y-x_{k_{n},1} \rangle \\ {}&\quad+\frac{N-2}{2}\int _{\partial B_{d}(x_{k_{n},1})}v\frac{\partial u}{\partial \nu } + \frac{N-2}{2}\int _{\partial B_{d}(x_{k_{n},1})} u\frac{\partial v}{\partial \nu }. \end{aligned} \end{aligned}$$

We have the following identities involving Green function.

Lemma B.1

For any $d \in (0, \frac{\delta }{k_n})$, where $\delta > 0$ is a fixed small constant, we have

$$\begin{aligned}&I_1\bigg (G(y,x_{k_{n},1}) , G(y,x_{k_{n},1}) ,d\bigg ) = -2\frac{\partial H}{\partial y_1}(x_{k_{n},1},x_{k_{n},1}), \end{aligned}$$

(B.1)

$$\begin{aligned}&I_1\bigg (G(y,x_{k_{n},1}) , \sum _{j=2}^{k_n} G(y,x_{k_{n},j}) ,d\bigg ) = \sum _{j=2}^{k_n}\frac{\partial G}{\partial y_1}(x_{k_{n},1},x_{k_{n},j}), \end{aligned}$$

(B.2)

$$\begin{aligned}&I_2\bigg (G(y,x_{k_{n},1}) , G(y,x_{k_{n},1}) ,d \bigg ) = (N-2)H(x_{k_{n},1},x_{k_{n},1}), \end{aligned}$$

(B.3)

$$\begin{aligned}&I_2\bigg (G(y,x_{k_{n},1}) ,\sum _{j=2}^{k_n} G(y,x_{k_{n},j}) ,d \bigg ) =-\frac{N-2}{2}\sum _{j=2}^{k_n}G(x_{k_{n},1},x_{k_{n},j}), \end{aligned}$$

(B.4)

$$\begin{aligned}&I_2\bigg (G(y,x_{k_{n},1}) , \frac{\partial G }{\partial x_{1}}(y,x_{k_{n},1}) ,d \bigg ) = (N-1)\frac{\partial H}{\partial y_1}(x_{k_{n},1},x_{k_{n},1}), \end{aligned}$$

(B.5)

$$\begin{aligned}&I_2\bigg (G(y,x_{k_{n},1}) , \sum _{j=2}^{k_n}\bigg (\cos \theta _{j}\frac{\partial G }{\partial x_{1}}(y,x_{k_{n},j}) +\sin \theta _{j}\frac{\partial G }{\partial x_{2}}(y,x_{k_{n},j}) \bigg ) , d \bigg ) \nonumber \\&\quad =-\frac{N-2}{2}\sum _{j=2}^{k_n}\bigg (\cos \theta _{j}\frac{\partial G }{\partial x_{1}}(x_{k_{n},1},x_{k_{n},j}) +\sin \theta _{j}\frac{\partial G }{\partial x_{2}}(x_{k_{n},1},x_{k_{n},j}) \bigg ) \nonumber \\&\quad =-\frac{N-2}{2} \sum _{j=2}^{k_n}\frac{\partial G }{\partial y_{1}}(x_{k_{n},1},x_{k_{n},j}) , \end{aligned}$$

(B.6)

and

$$\begin{aligned} \begin{aligned} I_2\bigg (\sum _{j=2}^{k_n}G(y,x_{k_{n},j}) , \frac{\partial G }{\partial x_{1}}(y,x_{k_{n},1}) , d \bigg ) =-\frac{N}{2} \sum _{j=2}^{k_n}\frac{\partial G }{\partial y_{1}}(x_{k_{n},1},x_{k_{n},j}). \end{aligned} \end{aligned}$$

(B.7)

Proof

We first proof (B.1) and (B.2).

Noting that for $j=1,\ldots ,k_n$, $G(y,x_{k_{n},j})$ are harmonic in the domain $B_{d}(x_{k_{n},1}) \setminus B_{\epsilon } (x_{k_{n},1})$, where $0<\varepsilon < d$, we have

$$\begin{aligned}&I_1\bigg (G(y,x_{k_{n},1}) ,\displaystyle G(y,x_{k_{n},1}) ,d \bigg ) - I_1\bigg (G(y,x_{k_{n},1}) ,\displaystyle G(y,x_{k_{n},1}) ,\epsilon \bigg ) \\& \quad=\,\displaystyle \int _{B_{d}(x_{k_{n},1}) \setminus B_{\epsilon } (x_{k_{n},1}) }-\Delta G(y,x_{k_{n},1}) \frac{\partial G }{\partial y_{1}}(y,x_{k_{n},j}) -\Delta G(y,x_{k_{n},j}) \frac{\partial G }{\partial y_{1}}(y,x_{k_{n},1}) =\,&0, \end{aligned}$$

and

$$\begin{aligned}&I_1\bigg (G(y,x_{k_{n},1}) ,\displaystyle \sum _{j=2}^{k_n} G(y,x_{k_{n},j}) ,d \bigg ) - I_1\bigg (G(y,x_{k_{n},1}) ,\displaystyle \sum _{j=2}^{k_n} G(y,x_{k_{n},j}) ,\epsilon \bigg ) \\ & \quad=\,\displaystyle \int _{B_{d}(x_{k_{n},1}) \setminus B_{\epsilon } (x_{k_{n},1}) }-\Delta G(y,x_{k_{n},1})\sum _{j=2}^{k_n} \frac{\partial G }{\partial y_{1}}(y,x_{k_{n},j}) -\Delta \sum _{j=2}^{k_n} G(y,x_{k_{n},j}) \frac{\partial G }{\partial y_{1}}(y,x_{k_{n},1}) =\,&0. \end{aligned}$$

Thus,

$$\begin{aligned} I_1\bigg (G(y,x_{k_{n},1}) , G(y,x_{k_{n},1}) , d \bigg ) = \lim _{\epsilon \rightarrow 0}I_1\bigg (G(y,x_{k_{n},1}) , G(y,x_{k_{n},1}) ,\epsilon \bigg ), \end{aligned}$$

$$\begin{aligned} I_1\bigg (G(y,x_{k_{n},1}) , \sum _{j=2}^{k_n} G(y,x_{k_{n},j}) , d \bigg ) = \lim _{\epsilon \rightarrow 0}I_1\bigg (G(y,x_{k_{n},1}) , \sum _{j=2}^{k_n} G(y,x_{k_{n},2}) ,\epsilon \bigg ). \end{aligned}$$

On the other hand,

$$\begin{aligned}&I_1\bigg (G(y,x_{k_{n},1}) , G(y,x_{k_{n},1}) ,\epsilon \bigg ) \\& \quad=\,I_1\bigg (\Gamma (y,x_{k_{n},1}) - H(y,x_{k_{n},1}), \Gamma (y,x_{k_{n},1}) - H(y,x_{k_{n},1}) ,\epsilon \bigg ) \\& \quad=\,I_1\bigg (\Gamma (y,x_{k_{n},1}) , \Gamma (y,x_{k_{n},1}) ,\epsilon \bigg ) - 2I_1\bigg (H(y,x_{k_{n},1}), \Gamma (y,x_{k_{n},1}) ,\epsilon \bigg ) \\&\qquad+I_1\bigg ( H(y,x_{k_{n},1}), H(y,x_{k_{n},1}) ,\epsilon \bigg ) \\& \quad =\,- 2\frac{\partial H}{\partial y_1}(x_{k_{n},1},x_{k_{n},1}) +o_{\epsilon }(1), \end{aligned}$$

and

$$\begin{aligned}&I_1\bigg (G(y,x_{k_{n},1}) , \sum _{j=2}^{k_n} G(y,x_{k_{n},j}) ,\epsilon \bigg ) \\& \quad=\,I_1\bigg (\Gamma (y,x_{k_{n},1}) , \sum _{j=2}^{k_n} G(y,x_{k_{n},j}) ,\epsilon \bigg ) - I_1\bigg (H (y,x_{k_{n},1}) , \sum _{j=2}^{k_n} G(y,x_{k_{n},j}) ,\epsilon \bigg ) \\ & \quad=\,\displaystyle \sum _{j=2}^{k_n}\frac{\partial G}{\partial y_1}(x_{k_{n},1},x_{k_{n},j}) +o_{\epsilon }(1). \end{aligned}$$

So let $\epsilon \rightarrow 0$, (B.1) and (B.2) is proved.

Now we prove (B.3). A direct calculation leads to

$$\begin{aligned}&I_2\bigg (G(y,x_{k_{n},1}) , G(y,x_{k_{n},1}) ,d \bigg ) -I_2\bigg (G(y,x_{k_{n},1}) , G(y,x_{k_{n},1}) ,\epsilon \bigg ) \\& \quad=\,\displaystyle \int _{B_{ d } (x_{k_{n},1}) \setminus B_{ \epsilon } (x_{k_{n},1})} \displaystyle \Delta G(y,x_{k_{n},1}) \langle \nabla G(y,x_{k_{n},1}), y-x_{k_{n},1} \rangle \\ {}&\qquad+ \displaystyle \int _{B_{ d } (x_{k_{n},1}) \setminus B_{ \epsilon } (x_{k_{n},1})} \Delta G(y,x_{k_{n},1}) \langle \nabla G(y,x_{k_{n},1}), y-x_{k_{n},1} \rangle \\ {}&\qquad+ \frac{N-2}{2} \displaystyle \int _{B_{ d } (x_{k_{n},1}) \setminus B_{ \epsilon } (x_{k_{n},1})} \Delta G(y,x_{k_{n},1}) \displaystyle G(y,x_{k_{n},1}) \\ {}&\qquad+ \frac{N-2}{2} \displaystyle \int _{B_{ d } (x_{k_{n},1}) \setminus B_{ \epsilon } (x_{k_{n},1})} G(y,x_{k_{n},1}) \displaystyle \Delta G(y,x_{k_{n},1}) =\,&0. \end{aligned}$$

Thus,

$$\begin{aligned}&I_2\bigg (G(y,x_{k_{n},1}) , G(y,x_{k_{n},1}) ,d \bigg ) \\& \quad=\, \lim _{\epsilon \rightarrow 0} I_2\bigg (G(y,x_{k_{n},1}) , G(y,x_{k_{n},1}) ,\epsilon \bigg ) \\& \quad= (N-2)H(x_{k_{n},1},x_{k_{n},1}). \end{aligned}$$

So (B.3) is proved. Similar to the proof of (B.3), we can prove (B.4).

Next we prove (B.5), which can be found in [8]. For completeness, we sketch the proof. It is easy to check

$$\begin{aligned}&I_2\bigg (G(y,x_{k_{n},1}) , \frac{\partial G }{\partial x_{1}}(y,x_{k_{n},1}) ,\epsilon \bigg ) \\& \quad=\,I_2\bigg (\Gamma ( y,x_{k_{n},1} ), \frac{\partial \Gamma }{\partial x_{1}}(y,x_{k_{n},1}) , \epsilon \bigg ) - I_2\bigg (H( y,x_{k_{n},1} ), \frac{\partial \Gamma }{\partial x_{1}}(y,x_{k_{n},1}) , \epsilon \bigg ) \\ {}&\qquad- I_2\bigg (\Gamma ( y,x_{k_{n},1} ), \frac{\partial H }{\partial x_{1}}(y,x_{k_{n},1}) , \epsilon \bigg ) + I_2\bigg (H( y,x_{k_{n},1} ), \frac{\partial H }{\partial x_{1}}(y,x_{k_{n},1}) , \epsilon \bigg ) \\& \quad=\,- I_2\bigg (H( y,x_{k_{n},1} ), \frac{\partial \Gamma }{\partial x_{1}}(y,x_{k_{n},1}) , \epsilon \bigg ) - I_2\bigg (\Gamma ( y,x_{k_{n},1} ), \frac{\partial H }{\partial x_{1}}(y,x_{k_{n},1}) , \epsilon \bigg ) +o_{\epsilon }(1). \end{aligned}$$

By direct computations, we have

$$\begin{aligned} \begin{aligned} I_2\bigg (\Gamma ( y,x_{k_{n},1} ), \frac{\partial H }{\partial x_{1}}(y,x_{k_{n},1}) , \epsilon \bigg ) = \frac{N-2}{2} \frac{\partial H }{\partial x_{1}}(x_{k_{n},1},x_{k_{n},1}) + o_{\epsilon }(1). \end{aligned} \end{aligned}$$

(B.8)

On the other hand, we have

$$\begin{aligned}&I_2\bigg (H( y,x_{k_{n},1} ), \frac{\partial \Gamma }{\partial x_{1}}(y,x_{k_{n},1}) , \epsilon \bigg ) \\ & \quad=\,\int _{\partial B_{\epsilon }(x_{k_{n},1})}\frac{\partial H}{\partial \nu }( y,x_{k_{n},1} )\langle \nabla \frac{\partial \Gamma }{\partial x_{1}}(y,x_{k_{n},1}), y-x_{k_{n},1} \rangle \\ {}&\qquad+\int _{\partial B_{\epsilon }(x_{k_{n},1})}\frac{\partial \frac{\partial \Gamma }{\partial x_{1}}}{\partial \nu }(y,x_{k_{n},1})\langle \nabla H( y,x_{k_{n},1} ), y-x_{k_{n},1} \rangle \\ {}&\qquad-\int _{\partial B_{\epsilon }(x_{k_{n},1})}\langle \nabla H( y,x_{k_{n},1} ), \nabla \frac{\partial \Gamma }{\partial x_{1}}(y,x_{k_{n},1}) \rangle \langle \nu , y-x_{k_{n},1} \rangle \\ {}&\qquad+\frac{N-2}{2}\int _{\partial B_{\epsilon }(x_{k_{n},1})}\frac{\partial \Gamma }{\partial x_{1}}(y,x_{k_{n},1})\frac{\partial H}{\partial \nu }( y,x_{k_{n},1} ) \\ {}&\qquad+ \frac{N-2}{2}\int _{\partial B_{\epsilon }(x_{k_{n},1})} H( y,x_{k_{n},1} )\frac{\partial \frac{\partial \Gamma }{\partial x_{1}}}{\partial \nu }(y,x_{k_{n},1}). \end{aligned}$$

Since

$$\begin{aligned} \frac{\partial \frac{\partial \Gamma }{\partial x_{1}}}{\partial \nu }(y,x_{k_{n},1})= -\frac{(N-1)( y- x_{k_{n},1} )_{1}}{\omega _{N-1}| y - x_{k_{n},1} |^{N+1}} \end{aligned}$$

we find

$$\begin{aligned} \begin{aligned}&\int _{\partial B_{\epsilon }(x_{k_{n},1})}\frac{\partial H}{\partial \nu }( y,x_{k_{n},1} )\langle \nabla \frac{\partial \Gamma }{\partial x_{1}}(y,x_{k_{n},1}), y-x_{k_{n},1} \rangle \\ {}&\qquad+\int _{\partial B_{\epsilon }(x_{k_{n},1})}\frac{\partial \frac{\partial \Gamma }{\partial x_{1}}}{\partial \nu }(y,x_{k_{n},1})\langle \nabla H( y,x_{k_{n},1} ), y-x_{k_{n},1} \rangle \\ {}&\qquad-\int _{\partial B_{\epsilon }(x_{k_{n},1})}\langle \nabla H( y,x_{k_{n},1} ), \nabla \frac{\partial \Gamma }{\partial x_{1}}(y,x_{k_{n},1}) \rangle \langle \nu , y-x_{k_{n},1} \rangle \\ & \quad=\,-2\frac{\partial H}{\partial y_1}(x_{k_{n},1},x_{k_{n},1})\int _{\partial B_{\epsilon }(x_{k_{n},1})}\frac{(N-1)( y- x_{k_{n},1} )_{1}^{2}}{\omega _{N-1}| y - x_{k_{n},1} |^{N+1}} \\ {}&\qquad-\frac{\partial H}{\partial y_1}(x_{k_{n},1},x_{k_{n},1})\int _{\partial B_{\epsilon }(x_{k_{n},1})}\bigg ( \frac{( y- x_{k_{n},1} )_{1}^{2}}{\omega _{N-1}| y - x_{k_{n},1} |^{N-1}} -\frac{N( y- x_{k_{n},1} )_{1}^{2}}{\omega _{N-1}| y - x_{k_{n},1} |^{N+1}}\bigg ) \\ {}&\qquad+ o_{\epsilon }(1) \\& \quad= -\frac{2(N-1)}{N}\frac{\partial H}{\partial y_1}(x_{k_{n},1},x_{k_{n},1})+ o_{\epsilon }(1). \end{aligned} \end{aligned}$$

(B.9)

Moreover

$$\begin{aligned} \begin{aligned} \frac{N-2}{2}\int _{\partial B_{\epsilon }(x_{k_{n},1})}\frac{\partial \Gamma }{\partial x_{1}}(y,x_{k_{n},1})\frac{\partial H}{\partial \nu }( y,x_{k_{n},1} )= \frac{N-2}{2N}\frac{\partial H}{\partial y_1}(x_{k_{n},1},x_{k_{n},1})+ o_{\epsilon }(1) \end{aligned} \end{aligned}$$

(B.10)

and

$$\begin{aligned} \begin{aligned}&\frac{N-2}{2}\int _{\partial B_{\epsilon }(x_{k_{n},1})} H( y,x_{k_{n},1} )\frac{\partial \frac{\partial \Gamma }{\partial x_{1}}}{\partial \nu }(y,x_{k_{n},1})\\ & \quad=\,-\frac{N-2}{2}\int _{\partial B_{\epsilon }(x_{k_{n},1})}\bigg (H( x_{k_{n},1},x_{k_{n},1} ) + \langle \nabla H(x_{k_{n},1},x_{k_{n},1}) , y -x_{k_{n},1} \rangle + O(\epsilon ^{2}) \bigg ) \\ {}&\qquad\times \frac{(N-1)( y- x_{k_{n},1} )_{1}}{\omega _{N-1}| y - x_{k_{n},1} |^{N+1}} \\ & \quad=\,-\frac{(N-2)(N-2)}{2N}\frac{\partial H}{\partial y_1}(x_{k_{n},1},x_{k_{n},1})+ o_{\epsilon }(1). \end{aligned} \end{aligned}$$

(B.11)

In conclusion, from (B.9), (B.10) and (B.11), we have

$$\begin{aligned} I_2\bigg (H( y,x_{k_{n},1} ), \frac{\partial \Gamma }{\partial x_{1}}(y,x_{k_{n},1}) , \epsilon \bigg ) = -\frac{N}{2}\frac{\partial H}{\partial y_1}(x_{k_{n},1},x_{k_{n},1})+ o_{\epsilon }(1) \end{aligned}$$

Taking $\epsilon \rightarrow 0$, we have

$$\begin{aligned} \begin{aligned} I_2\bigg (G(y,x_{k_{n},1}) , \frac{\partial G }{\partial x_{1}}(y,x_{k_{n},1}) , d \bigg ) = (N-1)\frac{\partial H}{\partial y_1}(x_{k_{n},1},x_{k_{n},1}). \end{aligned} \end{aligned}$$

(B.12)

Thus, we finish the proof of (B.5). Similarly, we can prove (B.6) and (B.7). $\square$

Appendix C: Green function

In this part, we give the estimate of modified Green function, which is necessary for the construction of new bubble solutions. This part is independent of interest.

In general, for any function f defined in ${\mathbb {R}}^N$, we define its corresponding function $f^*\in H_s$ as follows. We first define $A_j$ as

$$\begin{aligned} A_j z= \bigl ( r \cos (\theta +\frac{2j \pi }{k}), r \sin (\theta +\frac{2j \pi }{k}), z''\bigg ),\quad j=1, \ldots , k, \end{aligned}$$

where $z= (z', z'')\in {\mathbb {R}}^N$, $z'= (r\cos \theta , r\sin \theta )\in {\mathbb {R}}^2$, $z''\in {\mathbb {R}}^{N-2}$, while

$$\begin{aligned} B_i z= \bigl ( z_1,\ldots , z_{i-1}, -z_i, z_{i+1},\ldots , z_N),\quad i=1, \ldots , N. \end{aligned}$$

Let

$$\begin{aligned} {{\bar{f}}}(y)=\frac{1}{k}\sum _{j=1}^k f(A_j y), \end{aligned}$$

and

$$\begin{aligned} f^*(y) = \frac{1}{N-1}\sum _{i=2}^N \frac{1}{2} \bigl ( {{\bar{f}}}(y)+ \bar{f}(B_i y)\bigr ). \end{aligned}$$

Then, one can easily check that $f^*\in H_s$.

In the following, we discuss the Green’s function of $L_k$. Since $\delta _x$ is not in the space $H_s$, we consider

$$\begin{aligned} L_k u = \delta _x^*, \hbox { in } B_{1}(0), \quad u\in H_s\cap H_{0}^{1}(B_{1}(0)). \end{aligned}$$

(C.1)

The solution of (C.1) is denoted as $G_k(y, x)$, which we call it the Green function of $L_k.$ Let

$$\begin{aligned} \delta _x^* =\frac{1}{N-1}\sum _{i=2}^N \frac{1}{2} \Bigl ( \frac{1}{k}\sum _{j=1}^k \delta _{A_j x}+ \frac{1}{k}\sum _{j=1}^k \delta _{B_i A_j x}\Bigr ). \end{aligned}$$

We have

Proposition C.1

The solution $G_k(y, x)$ satisfies

$$\begin{aligned} |G_k(y, x)|\le \frac{C}{N-1}\sum _{i=2}^N \frac{1}{2} \Bigl ( \frac{1}{k}\sum _{j=1}^k \frac{C}{|y- A_j x|^{N-2}}+ \frac{1}{k}\sum _{j=1}^k \frac{C}{|y- B_i A_j x|^{N-2}}\Bigr ). \end{aligned}$$

Proof

Let $v_1= G(y,x)$ be the Green’s function of $-\Delta$ in $B_{1}(0)$ with Dirichlet boundary condition. Let $v_2$ be the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta v= (2^*-1)K(y) u_k^{2^*-2} v_1,&{} \text {in}\; B_{1}(0),\\ v= 0 ,&{} \text {on}\; \partial B_{1}(0). \end{array}\right. } \end{aligned}$$

Then, $v_2\ge 0$ and

$$\begin{aligned} \begin{array}{ll} v_2(y)&{}= \displaystyle \int _{B_{1}(0)}G(y, z) (2^*-1) u_k^{2^*-2} K(z)v_1\\ &\quad {}\le C \displaystyle \int _{B_1(0)}\frac{1}{|y-z|^{N-2}}\frac{1}{|z-x|^{N-2}}\,dz\\ &\quad {}\le \displaystyle \frac{C}{|y-x|^{N-4}}. \end{array} \end{aligned}$$

We can continue this process to find $v_i$, which is the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta v= (2^*-1) u_k^{2^*-2}K(y) v_{i-1},&{} \text {in}\; B_{1}(0),\\ v= 0 ,&{} \text {on}\; \partial B_{1}(0). \end{array}\right. } \end{aligned}$$

And satisfies

$$\begin{aligned} \begin{aligned} 0\le\,&\,v_i (y)\\ =\,&\int _{B_{1}(0)}G(y, z) (2^*-1) u_k^{2^*-2} K(z)v_{i-1}\\ \le\,&\,C \int _{B_{1}(0)}\frac{1}{|y-z|^{N-2}}\frac{1}{|z-x|^{N-2(i-1)}}\,dz\\ \le\,&\,\frac{C}{|y-x|^{N-2i}}. \end{aligned} \end{aligned}$$

Let i be large so that $v_i\in L^\infty (B_{1}(0))$. Define

$$\begin{aligned} v=\sum _{l=1}^i v_l\quad \hbox { and } w= G_k(y, x)- v^*, \end{aligned}$$

We then have

$$\begin{aligned} {\left\{ \begin{array}{ll} L_k w = f,&{}\hbox {in } B_{1}(0),\\ w = 0,&{}\hbox {on } \partial B_{1}(0), \end{array}\right. } \end{aligned}$$

(C.2)

where $f\in L^\infty \cap H_s$. By Theorem 1.1, (C.2) has a solution $w\in H_s\cap H_{0}^{1}(B_{1}(0))$.

By standard elliptic estimate, we have

$$\begin{aligned} ||w||_{L^{\infty }(B_1(0))}\le C||f||_{L^{\infty }(B_1(0))}. \end{aligned}$$

Thus, the conclusion is proved. $\square$