Bubble solutions for a supercritical polyharmonic Hénon-type equation

Abstract

We consider the following problem involving supercritical exponent and polyharmonic operator:

$$\begin{aligned} (-\Delta )^mu=K(|y|)u^{m^*-1+\varepsilon }, \;u>0, \hbox { in } B_1(0), \; u \in {\mathcal {D}}_0^{m,2}(B_1(0)), \end{aligned}$$

where $B_1(0)$ is the unit ball in ${\mathbb {R}}^{N}$, $m^*=\frac{2N}{N-2m}$ is the critical exponent,$\; N\ge 2m+2$, $m \in {\mathbb {N}}_+$, $\varepsilon > 0$, K(|y|) is a nonnegative bounded function. We prove that if $\varepsilon > 0$ is small enough, this problem has large number of bubble solutions, and the number of its bubbles varies with the parameter $\varepsilon$ at the order $\varepsilon ^{-1/(N-2m+1)}$ as $\varepsilon \rightarrow 0^+$. Moreover, all bubbles of the solutions approach the boundary of $B_1(0)$ as $\varepsilon$ goes to $0^+$.

Appendix A: Basic estimates

In this section, we will give some basic estimates used in the reduction procedure.

Lemma A.1

For each fixed i and j, $i \ne j$, let

$$\begin{aligned} g_{\varepsilon ij}=\frac{1}{(1+|y-x_{\varepsilon j}|)^{\alpha }(1+|y-x_{\varepsilon i}|)^{\beta }}, \end{aligned}$$

where $\alpha , \beta \ge 1$ are constants. Then for any constants $0 \le \sigma \le min\{\alpha , \beta \}$, there is a constant $C >0$, such that

$$\begin{aligned} g_{\varepsilon ij}(y) \le \frac{C}{|x_{\varepsilon i}-x_{\varepsilon j}|^{\sigma }}\left( \frac{1}{(1+|y-x_{\varepsilon j}|)^{\alpha +\beta -\sigma }} +\frac{1}{(1+|y-x_{\varepsilon i}|)^{\alpha +\beta -\sigma }}\right) . \end{aligned}$$

Lemma A.2

For any constant $0< \sigma < N-2m$, there is a constant $C >0$, such that

$$\begin{aligned} \int _{{\mathbb {R}}^{N}}\frac{dz}{|y-z|^{N-2m}(1+|z|)^{2m+\sigma }} \le \frac{C}{(1+|y|)^{\sigma }}. \end{aligned}$$

The proofs of the above two lemmas can be found in [20] and [43].

Lemma A.3

Suppose $N \ge 2m+2, \; \varepsilon > 0$, then there is a $\theta >0$ small, such that

$$\begin{aligned}&\displaystyle \int _{{\mathbb {R}}^{N}} \frac{1}{|y-z|^{N-2m}}Z_{r_\varepsilon , \lambda _\varepsilon }^{m^*-2+\varepsilon }(z)\displaystyle \sum _{j=1}^{k}\frac{1}{(1+|z-x_{\varepsilon j}|)^{\frac{N-2m}{2}+\tau }}dz\\&\le C\sum _{j=1}^{k}\frac{1}{(1+|y-x_{\varepsilon j}|)^{\frac{N-2m}{2}+\tau +\theta }}. \end{aligned}$$

Proof

For any $y \in \Omega _1$, we have

$$\begin{aligned} Z_{r_\varepsilon , \lambda _\varepsilon }(z) \le C\displaystyle \sum _{j=1}^{k}\frac{1}{(1+|z-x_{\varepsilon j}|)^{N-2m}} \le \frac{C}{(1+|z-x_{\varepsilon 1}|)^{N-2m-\tau _1}}, \end{aligned}$$

where $0 < \tau _1 \le \frac{N-2m}{2}$.

By Lemma A.1, we obtain

$$\begin{aligned} \begin{aligned}&\quad Z_{r_\varepsilon , \lambda _\varepsilon }^{m^*-2+\varepsilon }(z)\displaystyle \sum _{j=1}^{k}\frac{1}{(1+|z-x_{\varepsilon j}|)^{\frac{N-2m}{2}+\tau }}\\&\le \frac{C}{(1+|z-x_{\varepsilon 1}|)^{\frac{N+6m}{2}+\tau -\frac{4m\tau _1}{N-2m}+(N-2m-\tau _1)\varepsilon }}\\&\quad +\displaystyle \sum _{j=2}^{k}\frac{C}{(1+|z-x_{\varepsilon 1}|)^{4m-\frac{4m\tau _1}{N-2m}+(N-2m-\tau _1)\varepsilon }}\frac{1}{(1+|z-x_{\varepsilon j}|)^{\frac{N-2m}{2}+\tau }}\\&\le \frac{C}{(1+|z-x_{\varepsilon 1}|)^{\frac{N+6m}{2}+\tau -\frac{4m\tau _1}{N-2m}+(N-2m-\tau _1)\varepsilon }}\\ {}&\quad +\frac{C}{(1+|z-x_{\varepsilon 1}|)^{\frac{N+6m}{2}+\tau -\frac{N+2m}{N-2m}\tau _1+(N-2m-\tau _1)\varepsilon }}\displaystyle \sum _{j=2}^k\frac{1}{|x_{\varepsilon j}-x_{\varepsilon 1}|^{\tau _1}}\\&\le \frac{C}{(1+|z-x_{\varepsilon 1}|)^{\frac{N+6m}{2}+\tau -\frac{N+2m}{N-2m}\tau _1+(N-2m-\tau _1)\varepsilon }}. \end{aligned} \end{aligned}$$

By Lemma A.2, we have

$$\begin{aligned} \begin{aligned}&\quad \displaystyle \int _{\Omega _1}\frac{1}{|y-z|^{N-2m}}Z_{r_\varepsilon , \lambda _\varepsilon }^{m^*-2+\varepsilon }(z)\displaystyle \sum _{j=1}^{k}\frac{1}{(1+|z-x_{\varepsilon j}|)^{\frac{N-2m}{2}+\tau }}dz\\&\le \displaystyle \int _{\Omega _1}\frac{1}{|y-z|^{N-2m}} \frac{C}{(1+|z-x_{\varepsilon 1}|)^{\frac{N+6m}{2}+\tau -\frac{N+2m}{N-2m}\tau _1+(N-2m-\tau _1)\varepsilon }}\\&\le \frac{C}{(1+|z-x_{\varepsilon 1}|)^{\frac{N+2m}{2}+\tau -\frac{N+2m}{N-2m}\tau _1+(N-2m-\tau _1)\varepsilon }}. \end{aligned} \end{aligned}$$

Thus, choose $\tau _1$ satisfying $2m-\frac{N+2m}{N-2m}\tau _1 > 0$ and $0 < \tau _1 \le \frac{N-2m}{2}$, we obtain that

$$\begin{aligned}&\quad \displaystyle \int _{{\mathbb {R}}^{N}} \frac{1}{|y-z|^{N-2m}}Z_{r_\varepsilon , \lambda _\varepsilon }^{m^*-2+\varepsilon }(z)\displaystyle \sum _{j=1}^{k}\frac{1}{(1+|z-x_{\varepsilon j}|)^{\frac{N-2m}{2}+\tau }}dz\\&=\displaystyle \sum _{i=1}^k\displaystyle \int _{\Omega _i} \frac{1}{|y-z|^{N-2m}}Z_{r_\varepsilon , \lambda _\varepsilon }^{m^*-2+\varepsilon }(z)\displaystyle \sum _{j=1}^{k}\frac{1}{(1+|z-x_{\varepsilon j}|)^{\frac{N-2m}{2}+\tau }}dz\\&\le \displaystyle \sum _{i=1}^k\frac{C}{(1+|z-x_{\varepsilon i}|)^{\frac{N+2m}{2}+\tau -\frac{N+2m}{N-2m}\tau _1+(N-2m-\tau _1)\varepsilon }}\\ {}&\le C\sum _{j=1}^{k}\frac{1}{(1+|y-x_{\varepsilon j}|)^{\frac{N-2m}{2}+\tau +\theta }}\text {.} \end{aligned}$$

$\square$

We complete the proof of Lemma A.3.