Construction of infinitely many solutions for fractional Schrödinger equation with double potentials

Abstract

We consider the following fractional Schrödinger equation involving critical exponent:

$$\begin{aligned} (-\Delta )^su+V(y)u=Q(y)u^{2_s^*-1}, \;u>0, \; \hbox { in } \mathbb {R}^{N},\; u \in D^s(\mathbb {R}^N), \end{aligned}$$

where $2_s^*=\frac{2N}{N-2s}$, $(y',y'') \in \mathbb {R}^{2} \times \mathbb {R}^{N-2}$ and $V(y) = V(|y'|,y'')$ and $Q(y) = Q(|y'|,y'')$ are bounded nonnegative functions in $\mathbb {R}^{+} \times \mathbb {R}^{N-2}$. By using finite-dimensional reduction method and local Pohozaev-type identities, we show that if $\frac{2+N-\sqrt{N^2+4}}{4}< s <\min \{\frac{N}{4}, 1\}$ and $Q(r,y'')$ has a stable critical point $(r_0,y_0'')$ with $r_0>0,\; Q(r_0,y_0'') > 0$ and $V(r_0,y_0'') > 0$, then the above problem has infinitely many solutions, whose energy can be arbitrarily large.

Appendices

Basic estimates

In this section, we will give some basic estimates, which can be found in [15, 17] and [29].

Lemma A.1

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

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

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

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

Lemma A.2

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

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

Lemma A.3

Let $\mu > 0$, for any constants $0< \beta < N$, there exists a constant $C > 0$, independent of $\mu$, such that

$$\begin{aligned} \displaystyle \int \limits _{\mathbb {R}^N\backslash B_{\mu }(y)} \frac{1}{|y-z|^{N+2s}}\frac{1}{(1+|z|)^{\beta }} {\textrm{d}}z \le C\left( \frac{1}{(1+|y|)^{\beta +2s}}+ \frac{1}{\mu ^{2s}}\frac{1}{(1+|y|)^{\beta }} \right) . \end{aligned}$$

Lemma A.4

Suppose that $|y-x|^2 + t^2 = \rho ^2$, then there exists a constant $C > 0$ such that

$$\begin{aligned} |\tilde{Z}_{x_i,\lambda } | \le \frac{C}{\lambda ^{\frac{N-2s}{2}}}\frac{1}{(1+|y-x_i|)^{N-2s}}, \quad |\nabla \tilde{Z}_{x_i,\lambda } | \le \frac{C}{\lambda ^{\frac{N-2s}{2}}}\frac{1}{(1+|y-x_i|)^{N-2s+1}}. \end{aligned}$$

Lemma A.5

$\forall \delta > 0$, there exists $\rho = \rho (\delta ) \in (2\delta , 5 \delta )$, such that when $N>4\,s$,

$$\begin{aligned} \displaystyle \int \limits _{\partial ^{''} D_\rho ^+} t^{1-2s} |\nabla \tilde{\phi }|^2 \le \frac{Ck\Vert \phi \Vert _*^2}{\lambda ^{\tau }}. \end{aligned}$$

When $N=3=4s$,

$$\begin{aligned} \displaystyle \int \limits _{\partial ^{''} D_\rho ^+} t^{1-2s} |\nabla \tilde{\phi }|^2 \le Ck\Vert \phi \Vert _*^2, \end{aligned}$$

where C is a constant, dependent on $\delta$.

The energy expansion

Recall that, the functional corresponding to (1.1) is

$$\begin{aligned} I(u)=\frac{1}{2}\displaystyle \int \limits _{\mathbb {R}^{N}}\big (|(-\Delta )^{\frac{s}{2}}u|^2 +V(y)u^2\big )-\frac{1}{2_s^*}\displaystyle \int \limits _{\mathbb {R}^{N}} Q(y)(u)_+^{2_s^*}. \end{aligned}$$

Lemma B.1

We have the following expansion:

$$\begin{aligned} \frac{\partial I(Z_{\bar{r},\bar{y}'',\lambda })}{\partial \lambda }=k\left( -\frac{A_1}{\lambda ^{2s+1}} +\displaystyle \sum _{j=2}^{k}\frac{B_2 }{\lambda ^{N-2s+1}|x_j-x_1|^{N-2s}}+O\bigg (\frac{1}{\lambda ^{2s+1+\epsilon }}\bigg )\right) , \end{aligned}$$

where positive constants $B_1,B_2$ are defined by

$$\begin{aligned} A_1 =s V( r_0, y_0'')\displaystyle \int \limits _{\mathbb {R}^N}U_{0,1}^2, \end{aligned}$$

(B.1)

and

$$\begin{aligned} B_2 = \frac{N-2s}{2}c(N,s)\displaystyle \int \limits _{\mathbb {R}^N} U_{0,1}^{2_s^*-1}. \end{aligned}$$

(B.2)

Proof

A direct computation leads to

$$\begin{aligned} \frac{\partial I(Z_{\bar{r},\bar{y}'',\lambda })}{\partial \lambda } =\frac{\partial I(Z^*_{\bar{r},\bar{y}'',\lambda })}{\partial \lambda }+O\left( \frac{k}{\lambda ^{2s+1+\epsilon }}\right) . \end{aligned}$$

(B.3)

On the other hand, we have

$$\begin{aligned} \frac{\partial I(Z^*_{\bar{r},\bar{y}'',\lambda })}{\partial \lambda }&=\displaystyle \int \limits _{\mathbb {R}^N}V(y)Z^*_{\bar{r},\bar{y}'',\lambda }\frac{\partial Z^*_{\bar{r},\bar{y}'',\lambda }}{\partial \lambda } +\displaystyle \int \limits _{\mathbb {R}^N}(1-Q(r,y''))(Z^*_{\bar{r},\bar{y}'',\lambda })^{2_s^*-1}\frac{\partial Z^*_{\bar{r},\bar{y}'',\lambda }}{\partial \lambda } \\&\quad -\displaystyle \int \limits _{\mathbb {R}^N}\left( (Z^*_{\bar{r},\bar{y}'',\lambda })^{2_s^*-1}-\displaystyle \sum _{j=1}^kU_{x_j,\lambda }^{2_s^*-1}\right) \frac{\partial Z^*_{\bar{r},\bar{y}'',\lambda }}{\partial \lambda } \\&:=I_1+I_2-I_3. \end{aligned}$$

By symmetry, we have

$$\begin{aligned} I_1&=k\left( \displaystyle \int \limits _{\mathbb {R}^N}V(y)U_{x_1,\lambda }\frac{\partial U_{x_1,\lambda }}{\partial \lambda }+O\bigg (\frac{1}{\lambda }\displaystyle \int \limits _{\mathbb {R}^N}V(y)U_{x_1,\lambda }\displaystyle \sum _{j=2}^kU_{x_j,\lambda }\bigg )\right) \\ {}&= k\left( \displaystyle \int \limits _{\mathbb {R}^N}V(y)U_{x_1,\lambda }\frac{\partial U_{x_1,\lambda }}{\partial \lambda }+O\left( \frac{1}{\lambda ^{2s+1+\epsilon }}\right) \right) \\ {}&= k\left( -\frac{A_1}{\lambda ^{2s+1}}+O\left( \frac{1}{\lambda ^{2s+1+\epsilon }}\right) \right) , \end{aligned}$$

where $A_1 = s V( r_0, y_0'')\displaystyle \int \limits _{\mathbb {R}^N}U_{0,1}^2 > 0$ is a constant.

By symmetry, a direct computation leads to

$$\begin{aligned} |I_2|&\le Ck \displaystyle \int \limits _{B_{\lambda ^{-\frac{1}{2}+\epsilon }(x_1)}} |y-x_1|^2 U_{x_1,\lambda }^{2_s^*-1}\frac{\partial U_{x_1,\lambda }}{\partial \lambda }+ \frac{Ck}{\lambda ^{3+\epsilon }} \\ {}&\le \frac{Ck}{\lambda ^{2s+1+\epsilon }}. \end{aligned}$$

Next, we have

$$\begin{aligned} I_3&=k\displaystyle \int \limits _{\Omega _1}\left( (Z^*_{\bar{r},\bar{y}'',\lambda })^{2_s^*-1}-\displaystyle \sum _{j=1}^kU_{x_j,\lambda }^{2_s^*-1}\right) \frac{\partial Z^*_{\bar{r},\bar{y}'',\lambda }}{\partial \lambda }\\ {}&=k\left( \displaystyle \int \limits _{\Omega _1}(2_s^*-1)U_{x_1,\lambda }^{2_s^*-2}\displaystyle \sum _{j=2}^kU_{x_j,\lambda }\frac{\partial U_{x_1,\lambda }}{\partial \lambda }+O\bigg (\frac{1}{\lambda ^{2s+1+\epsilon }}\bigg )\right) \\ {}&=k\left( -\displaystyle \sum _{j=2}^k\frac{B_2 }{\lambda ^{N-2s+1}|x_j-x_1|^{N-2s}}+O\bigg (\frac{1}{\lambda ^{2s+1+\epsilon }}\bigg )\right) , \end{aligned}$$

where $B_2 = \frac{N-2s}{2}c(N,s)\displaystyle \int \limits _{\mathbb {R}^N} U_{0,1}^{2_s^*-1} > 0$ is a constant.

Thus, we obtain that

$$\begin{aligned} \frac{\partial I(Z^*_{\bar{r},\bar{y}'',\lambda })}{\partial \lambda } =k\left( -\frac{A_1}{\lambda ^{2s+1}} +\displaystyle \sum _{j=2}^{k}\frac{B_2 }{\lambda ^{N-2s+1}|x_j-x_1|^{N-2s}}+O\bigg (\frac{1}{\lambda ^{2s+1+\epsilon }}\bigg )\right) , \end{aligned}$$

(B.4)

where $A_1$, $B_2$ are defined by $A_1 =s V( r_0, y_0'')\displaystyle \int \limits _{\mathbb {R}^N}U_{0,1}^2,$ and $B_2 = \frac{N-2\,s}{2}c(N,s)\displaystyle \int \limits _{\mathbb {R}^N} U_{0,1}^{2_s^*-1}.$

Combining (B.3) and (B.4), the result follows. $\square$