Solutions concentrating around the saddle points of the potential for critical Schrödinger equations

Abstract

We consider the following singularly perturbed nonlinear elliptic problem

$$\begin{aligned} -\varepsilon ^2\Delta u+V(x)u=f(u),\ u\in H^1({\mathbb {R}}^N), \quad N\ge 3, \end{aligned}$$

where f is of critical growth. Using the variational techniques, we construct a solution $u_\varepsilon$ which concentrates around the saddle points of V as $\varepsilon \rightarrow 0$.

Appendix: A splitting Lemma

This section is devoted to the proof of Lemma 4.7. The proof is similar to that in [14]. A similar result also can be found in [3]. However, in [3, 14], only the subcritical case was considered and the decay property of u at infinity was required, i.e., $u(x)\rightarrow 0$ as $|x|\rightarrow \infty$. In this appendix, the nonlinear term f may involve the critical growth and the decay property of u at infinity is removed.

Proof of Lemma 4.7

It suffices to prove that

$$\begin{aligned} \int _{\mathbb R^N}\big (f_1(u_k)-f_1(u)-f_1(u_k-u)\big )\phi =o_k(1)\Vert \phi \Vert , \end{aligned}$$

where $f_1$ is given in Sect. 2.

In the following, we denote by C the positive constants, which are independent of $\varepsilon ,k$ and possibly different. For any fixed $\varepsilon >0$, by $(f_1)$, we can choose $s_0=s_0(\varepsilon )\in (0,1)$ such that $|f_1(t)|\le \varepsilon |t|$ for $|t|\le 2s_0$. By $(f_2)$, choosing $s_1=s_1(\varepsilon )>2$ such that $|f_1(t)|\le \varepsilon |t|^{2^{*}-1}$ for $|t|\ge s_1-1$. From the continuity of f, there exists $\delta =\delta (\varepsilon )\in (0,s_0)$ such that $|f_1(t_1)-f_1(t_2)|\le s_0\varepsilon$ for $|t_1-t_2|\le \delta , |t_1|,|t_2|\le s_1+1$. Moreover, by $(f_2)$, there exists $c(\varepsilon )>0$ such that $|f_1(t)|\le c(\varepsilon )|t|+\varepsilon |t|^{2^{*}-1}$ for $t\in \mathbb R$. Then there exists $R=R(\varepsilon )>0$ such that $\int _{\mathbb R^N{\setminus } B(0,R)}|f_1(u)\phi |\le C\varepsilon \Vert \phi \Vert$. Setting $A_k:=\{x\in \mathbb R^N{\setminus } B(0,R): |u_k(x)|\le s_0\}$, then

$$\begin{aligned} \int _{A_k\cap \{u\le \delta \}}|f_1(u_k)-f_1(u_k-u)||\phi |\le \varepsilon (\Vert u_k\Vert _2+\Vert u_k-u\Vert _2)\Vert \phi \Vert _2\le C\varepsilon \Vert \phi \Vert . \end{aligned}$$

Let $B_k:=\{x\in \mathbb R^N{\setminus } B(0,R): |u_k(x)|\ge s_1\}$, then

$$\begin{aligned} \int _{B_k\cap \{u\le \delta \}}|f_1(u_k)-f_1(u_k-u)||\phi |\le \varepsilon (\Vert u_k\Vert _{2^{*}}^{2^{*}}+\Vert u_k-u\Vert _{2^{*}}^{2^{*}})\Vert \phi \Vert _{2^{*}}\le C\varepsilon \Vert \phi \Vert . \end{aligned}$$

Setting $C_k:=\{x\in \mathbb R^N{\setminus } B(0,R): s_0\le |u_k(x)|\le s_1\}$, by $u_k\in H^1(\mathbb R^N)$, we know that $|C_k|<\infty$. Then

$$\begin{aligned} \int _{C_k\cap \{u\le \delta \}}|f_1(u_k)-f_1(u_k-u)||\phi |\le s_0\varepsilon \Vert \phi \Vert _2|C_k|^{\frac{1}{2}}\le \varepsilon \Vert u_k\Vert _2\Vert \phi \Vert _2\le C\varepsilon \Vert \phi \Vert . \end{aligned}$$

Thus,

$$\begin{aligned} \int _{(\mathbb R^N{\setminus } B(0,R))\cap \{u\le \delta \}}|f_1(u_k)-f_1(u_k-u)||\phi |\le C\varepsilon \Vert \phi \Vert \ \ \text{ for } \text{ all }\ \ k. \end{aligned}$$

Noting that for all k,

$$\begin{aligned} |f_1(u_k)-f_1(u_k-u)|\le \varepsilon (|u_k|^{2^*-1}+|u_k-u|^{2^*-1})+c(\varepsilon )(|u_k|+|u_k-u|), \end{aligned}$$

and

$$\begin{aligned}&\int _{(\mathbb R^N{\setminus }B(0,R))\cap \{u\ge \delta \}}|f_1(u_k)-f_1(u_k-u)||\phi |\\&\quad \le \int _{(\mathbb R^N{\setminus }B(0,R))\cap \{u\ge \delta \}}\varepsilon (|u_k|^{2^*-1}+|u_k-u|^{2^*-1})|\phi |\\&\qquad +c(\varepsilon )(|u_k|+|u_k-u|)|\phi |\\&\quad \le C\varepsilon \Vert \phi \Vert +\int _{(\mathbb R^N{\setminus }B(0,R))\cap \{u\ge \delta \}}c(\varepsilon )(|u_k|+|u_k-u|)|\phi |. \end{aligned}$$

By $u\in H^1(\mathbb R^N)$, we get that $|(\mathbb R^N{\setminus }B(0,R))\cap \{u\ge \delta \}|\rightarrow 0$ as $R\rightarrow \infty$. Then we can choose $R=R(\varepsilon )$ large enough, such that

$$\begin{aligned}&\int _{(\mathbb R^N{\setminus }B(0,R))\cap \{u\ge \delta \}}c(\varepsilon )(|u_k|+|u_k-u|)|\phi |\\&\quad \le c(\varepsilon )(\Vert u_k\Vert _{2^*}+\Vert u_k-u\Vert _{2^*})\Vert \phi \Vert _{2^*}|(\mathbb R^N{\setminus }B(0,R))\cap \{u\ge \delta \}|^{\frac{2}{N}}\\&\quad \le \varepsilon \Vert \phi \Vert . \end{aligned}$$

So,

$$\begin{aligned} \int _{(\mathbb R^N{\setminus }B(0,R))\cap \{u\ge \delta \}}|f_1(u_k)-f_1(u_k-u)||\phi |\le C\varepsilon \Vert \phi \Vert \ \quad \text{ for } \text{ all }\ \ k. \end{aligned}$$

Thus,

$$\begin{aligned} \int _{\mathbb R^N{\setminus }B(0,R)}|f_1(u_k)-f_1(u)-f_1(u_k-u)||\phi |\le C\varepsilon \Vert \phi \Vert \ \quad \text{ for } \text{ all }\ \ k. \end{aligned}$$

Finally, since $u_k\rightarrow u$ weakly in $H^1(\mathbb R^N)$, up to a subsequence, $u_k\rightarrow u$ strongly in $L^2(B(0,R))$ and there exists $\omega \in L^2(B(0,R))$ such that $|u_k(x)|,|u(x)|\le |\omega (x)|$ a.e. $x\in B(0,R)$. Then it is easy to check that

$$\begin{aligned} \int _{B(0,R)}|f_1(u_k-u)||\phi |\le C\varepsilon \Vert \phi \Vert \end{aligned}$$

for k large. Let $D_k:=\{x\in B(0,R): |u_k(x)-u(x)|\ge 1\}$, then

$$\begin{aligned} \int _{D_k}|f_1(u_k)-f_1(u)||\phi |&\le \int _{D_k}[c(\varepsilon )(|u|+|u_k|)+\varepsilon (|u_k|^{2^{*}-1}+|u|^{2^{*}-1})]|\phi |\\&\le C\varepsilon \Vert \phi \Vert +2c(\varepsilon )\int _{D_k}|\omega ||\phi |\\&\le C\varepsilon \Vert \phi \Vert +2c(\varepsilon )\left( \int _{D_k}|\omega |^2\right) ^{\frac{1}{2}}\Vert \phi \Vert _2. \end{aligned}$$

By $u_k\rightarrow u$ a.e. $x\in B(0,R)$, we get that $|D_k|\rightarrow 0$ as $k\rightarrow \infty$. Hence,

$$\begin{aligned} \int _{D_k}|f_1(u_k)-f_1(u)||\phi |\le C\varepsilon \Vert \phi \Vert . \end{aligned}$$

Since $u\in H^1(\mathbb R^N)$, we know $|\{u\ge L\}|\rightarrow 0$ as $L\rightarrow \infty$, then there exists $L=L(\varepsilon )>0$ such that for all k,

$$\begin{aligned}&\int _{(B(0,R){\setminus }D_k)\cap \{u\ge L\}}|f_1(u_k)-f_1(u)||\phi |\\&\quad \le \int _{(B(0,R){\setminus }D_k)\cap \{u\ge L\}}\varepsilon (|u_k|^{2^*-1}+|u|^{2^*-1})|\phi |\\&\qquad +c(\varepsilon )(|u_k|+|u|)|\phi |\\&\quad \le C\varepsilon \Vert \phi \Vert +\int _{(B(0,R){\setminus }D_k)\cap \{u\ge L\}}c(\varepsilon )(|u_k|+|u|)|\phi |\\&\quad \le C\varepsilon \Vert \phi \Vert +c(\varepsilon )(\Vert u_k\Vert _{2^*}+\Vert u\Vert _{2^*})\Vert \phi \Vert _{2^*}|(B(0,R){\setminus }D_k)\cap \{u\ge L\}|^{\frac{2}{N}}\\&\quad \le C\varepsilon \Vert \phi \Vert . \end{aligned}$$

On the other hand, by the dominated convergence theorem,

$$\begin{aligned} \int _{(B(0,R){\setminus }D_k)\cap \{u\le L\}}|f_1(u_k)-f_1(u)|^2\rightarrow 0 \end{aligned}$$

as $k\rightarrow \infty$. So,

$$\begin{aligned}\int _{(B(0,R){\setminus }D_k)\cap \{u\le L\}}|f_1(u_k)-f_1(u)||\phi |\le C\varepsilon \Vert \phi \Vert \end{aligned}$$

for k large enough. Thus,

$$\begin{aligned} \int _{B(0,R)}|f_1(u_k)-f_1(u)||\phi |\le C\varepsilon \Vert \phi \Vert . \end{aligned}$$

Then

$$\begin{aligned} \int _{B(0,R)}|f_1(u_k)-f_1(u)-f_1(u_k-u)||\phi |\le C\varepsilon \Vert \phi \Vert . \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{\mathbb R^N}|f_1(u_k)-f_1(u)-f_1(u_k-u)||\phi |\le C\varepsilon \Vert \phi \Vert , \end{aligned}$$

for k large enough. $\square$