On the prescribed scalar curvature problem with very degenerate prescribed functions

Abstract

We revisit the well known prescribed scalar curvature problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u=\big (1+\varepsilon K(x)\big )u^{2^*-1}, u(x)>0,~~ &{}{x\in \mathbb {R}^N},\\ u\in \mathcal {D}^{1,2}(\mathbb {R}^N),\\ \end{array}\right. } \end{aligned}$$

where $2^*=\frac{2N}{N-2}$, $N\ge 5$, $\varepsilon >0$ and $K(x)\in C^1(\mathbb {R}^N)\cap L^{\infty }(\mathbb {R}^N)$. It is known that there are a number of results related to the existence of solutions concentrating at the isolated critical points of K(x). However, if K(x) has non-isolated critical points with different degenerate rates along different directions, whether there exist solutions concentrating at these points is still an open problem. We give a certain positive answer to this problem via applying a blow-up argument based on local Pohozaev identities and modified finite dimensional reduction method when the dimension of critical point set of K(x) ranges from 1 to $N-1$, which generalizes some results in Cao et al. (Calc Var Partial Differ Equ 15:403–419, 2002) and Li (J Differ Equ 120:319–410, 1995; Commun Pure Appl Math 49:541–597, 1996).

Appendices

Appendix: Some estimates

Lemma A.1

(c.f. [6]) For any $k\in \mathbb {N}^+$ and $p>1$, it holds

$$\begin{aligned} \left( \sum ^k_{i=1}a_i\right) ^p- a_1^p =O\left( a_1^{p-1}\left( \sum ^k_{i=2}a_i\right) +\sum ^k_{i=2}a_i^p \right) , \end{aligned}$$

and

$$\begin{aligned} \left( \sum ^k_{i=1}a_i\right) ^p-\sum ^k_{i=1}a_i^p = {\left\{ \begin{array}{ll} O\left( \displaystyle \sum _{i\ne j}a_i^{\frac{p}{2}}a_j^{\frac{p}{2}}\right) ,~&{}1<p<2,\\ O\left( \displaystyle \sum _{i\ne j}a_i^{p-1}a_j\right) , ~&{}p\ge 2. \end{array}\right. } \end{aligned}$$

Lemma A.2

For $a,b>0$ and $p>1$, it holds

$$\begin{aligned}{} & {} (a+b)^p-a^p-b^p-pa^{p-1}b = {\left\{ \begin{array}{ll} O\big (a^{\alpha _1}b^{p-\alpha _1}\big ) ,~\text{ for } \text{ all }~\alpha _1\in (0,p-1], ~&{}1<p\le 2,\\ O\big (a^{\alpha _2}b^{p-\alpha _2}\big ) ,~\text{ for } \text{ all }~\alpha _2\in [p-2,1], ~&{}2<p\le 3. \end{array}\right. }\nonumber \\ \end{aligned}$$

(A.1)

Proof

At first, for $1<p\le 2$ and $t\le 1$, we have

$$\begin{aligned} \begin{aligned} (1+t)^p-1-t^p-pt^{p-1}=&(1+t)^p-1-O(t^{p-1})=O(t + t^{p-1})\\ =&O(t^{p-1}) =O(t^{q}),~\text{ for } \text{ all }~q\in (0,p-1], \end{aligned} \end{aligned}$$

(A.2)

and

$$\begin{aligned} (1+t)^p-1-pt-t^p=O(t^2 + t^{p})=O(t^{p})=O(t^{p-q}),~\text{ for } \text{ all }~q\in [0,p). \end{aligned}$$

(A.3)

If $a\le b$, then it holds from (A.2),

$$\begin{aligned} (a+b)^p-a^p-b^p-pa^{p-1}b=O(a^{q}b^{p-q}),~\text{ for } \text{ all }~q\in (0,p-1], \end{aligned}$$

and if $a\ge b$, from (A.3), we get

$$\begin{aligned} (a+b)^p-a^p-b^p-pa^{p-1}b=O(a^{q}b^{p-q}),~\text{ for } \text{ all }~q\in [0,p), \end{aligned}$$

which yields the first part of (A.1).

Additionally, for $2<p\le 3$ and $t\le 1$, we have

$$\begin{aligned} \begin{aligned} (1+t)^p-1-t^p-pt^{p-1}=O(t + t^{p-1})=O(t) =O(t^{q}),~\text{ for } \text{ all }~q\in (0,1], \end{aligned} \end{aligned}$$

and

$$\begin{aligned} (1+t)^p-1-t^p-pt=O(t^2 + t^{p})=O(t^{2})=O(t^{p-q}),~\text{ for } \text{ all }~q\in [p-2,p). \end{aligned}$$

Similarly, if $a\le b$,

$$\begin{aligned} (a+b)^p-a^p-b^p-pa^{p-1}b=O(a^{q}b^{p-q}),~\text{ for } \text{ all }~q\in (0,1], \end{aligned}$$

and if $a\ge b$,

$$\begin{aligned} (a+b)^p-a^p-b^p-pa^{p-1}b=O(a^{q}b^{p-q}),~\text{ for } \text{ all }~q\in [p-2,p). \end{aligned}$$

Thus (A.1) holds. $\square$

Lemma A.3

(c.f. [4]) For any $(y_\varepsilon ,\lambda _\varepsilon )\in D_{\varepsilon }$, $i,l\in \{1,2\}$ with $i\ne l$ and $j=1,\ldots ,N$, it holds

$$\begin{aligned} \int _{\mathbb {R}^N}U_{y_{\varepsilon ,i},\lambda _{\varepsilon ,i}}^{\alpha } U^{2^*-1-\alpha }_{y_{\varepsilon ,l},\lambda _{\varepsilon ,l}}\frac{\partial U_{y_{\varepsilon ,i},\lambda _{\varepsilon ,i}}}{\partial \lambda _{\varepsilon ,i}}= {\left\{ \begin{array}{ll} -\frac{N-2}{2(2^*-1)}\frac{C_NA}{\lambda _{\varepsilon ,i}^{\frac{N}{2}} \lambda _{\varepsilon ,l}^{\frac{N-2}{2}}} +O\left( \frac{1}{\overline{\lambda }_\varepsilon ^{N+1}}\right) , &{} \alpha =2^*-2,\\ O\left( \frac{1}{\overline{\lambda }_\varepsilon ^{\beta _1}}\right) , &{} \alpha \in (0,2^*-2),\\ -\frac{N-2}{2}\frac{C_NA}{\lambda _{\varepsilon ,i}^{\frac{N}{2}}\lambda _{\varepsilon ,l}^{\frac{N-2}{2}}} +O\left( \frac{1}{\overline{\lambda }_\varepsilon ^{N+1}}\right) , &{} \alpha =0, \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \int _{\mathbb {R}^N}U_{y_{\varepsilon ,i},\lambda _{\varepsilon ,i}}^{\alpha } U^{2^*-1-\alpha }_{y_{\varepsilon ,l},\lambda _{\varepsilon ,l}}\frac{\partial U_{y_{\varepsilon ,i},\lambda _{\varepsilon ,i}}}{\partial x_j}= {\left\{ \begin{array}{ll} \frac{C_0(y_{\varepsilon ,i,j}-y_{\varepsilon ,l,j})}{\lambda _{\varepsilon ,i}^{\frac{N-2}{2}}\lambda _{\varepsilon ,l}^{ \frac{N-2}{2}}} +O\left( \frac{1}{\overline{\lambda }_\varepsilon ^{N-1}}\right) , &{} \alpha =2^*-2,\\ O\left( \frac{1}{\overline{\lambda }_\varepsilon ^{\beta _2}}\right) , &{} \alpha \in (0,2^*-2),\\ \frac{C_1(y_{\varepsilon ,i,j}-y_{\varepsilon ,l,j})}{\lambda _{\varepsilon ,i}^{\frac{N-2}{2}}\lambda _{\varepsilon ,l}^{ \frac{N-2}{2}}} +O\left( \frac{1}{\overline{\lambda }_\varepsilon ^{N-1}}\right) , &{} \alpha =0, \end{array}\right. } \end{aligned}$$

where the constants $\overline{\lambda }_\varepsilon =\min \{\lambda _{\varepsilon ,1},\lambda _{\varepsilon ,2}\}$, $\beta _1=N+1-\frac{N-2}{2}|2^*-2-2\alpha |$, $\beta _2=N-\frac{N-2}{2}|2^*-2-2\alpha |$ and $C_0,C_1>0$ are constants depending on N only.

Lemma A.4

Let $A=\displaystyle \int _{{\mathbb {R}^N}}U_{0,1}^{2^*-1}$, it holds

$$\begin{aligned} A=(N-2)C_N\omega _N~~\text{ with }~~ C_N=\Big (N(N-2)\Big )^{\frac{N-2}{4}}. \end{aligned}$$

Proof

We have

$$\begin{aligned} \begin{aligned} A&=\int _{{\mathbb {R}^N}}U_{0,1}^{2^*-1}dx=\int _{{\mathbb {R}^N}}\frac{C_N^{2^*-1}}{(1+|x|^2 )^{\frac{N+2}{2}}}dx = \int _{{\mathbb {R}^N}}\frac{C_NN(N-2)}{(1+|x|^2 )^{\frac{N+2}{2}}}dx \\&= N(N-2)C_N\omega _N \int _0^{+\infty } \frac{r^{N-1}}{(1+r^2)^{\frac{N+2}{2}}}dr =(N-2)C_N\omega _N, \end{aligned} \end{aligned}$$

where we have used the following fact

$$\begin{aligned}&\int _0^{+\infty } \frac{r^{N-1}}{(1+r^2)^{\frac{N+2}{2}}}dr \mathop {=\!=\!=\!=}\limits ^{s=1+r^2} \int _1^{+\infty } \frac{(s-1)^{\frac{N-2}{2}}}{2s^{\frac{N+2}{2}}}ds \\ =&\int _1^{+\infty } \frac{1}{2s^2}\big (1-\frac{1}{s}\big )^{\frac{N-2}{2}}ds \mathop {=\!=\!=\!=}\limits ^{t=1-\frac{1}{s}} \int _0^1 \frac{1}{2}t^{\frac{N-2}{2}}dt =\frac{1}{N}. \end{aligned}$$

$\square$

Lemma A.5

(Poincaré–Miranda theorem, c.f. [16]) Consider n continuous functions of n variables, $f_1,\ldots ,f_n$. If for each variable $x_i$, the function $f_i$ is constantly negative when $x_i=-1$ and constantly positive when $x_i=1$, then there is a point in the n–dimensional cube $[-1,1]^n$ at which all functions are simultaneously equal to 0.

The nonexistence of single peak solution

Proposition B.1

Under the condition (K), problem (1.3) does not admit single peak solutions concentrating at certain point on $\Gamma$.

Proof

Suppose that problem (1.3) has a single peak solution $u_\varepsilon (x)$ concentrating at some point $b_1\in \Gamma$ as $\varepsilon \rightarrow 0$. According to Definition A, we find that there exist $y_{\varepsilon ,1}\in \mathbb {R}^N$ and $\lambda _{\varepsilon ,1}\in \mathbb {R}^+$ such that

$$\begin{aligned} \Big \Vert u_\varepsilon - U_{y_{\varepsilon ,1},\lambda _{\varepsilon ,1}}\Big \Vert =o(1), \end{aligned}$$

with

$$\begin{aligned} |y_{\varepsilon ,1}-b_1|=o(1)~\text{ and }~\lambda _{\varepsilon ,1}\rightarrow +\infty . \end{aligned}$$

When taking $\Omega =\mathbb {R}^N$ in (2.4), we have

$$\begin{aligned} \int _{\mathbb {R}^N}\Big ((x-y_{\varepsilon ,1})\cdot \nabla K(x)\Big ) u^{2^*}_\varepsilon {d}x =0. \end{aligned}$$

(B.1)

On the other hand, it follows from Proposition 2.1

$$\begin{aligned} \int _{\mathbb {R}^N}\Big ((x-y_{\varepsilon ,1})\cdot \nabla K(x)\Big ) u^{2^*}_\varepsilon {d}x =\int _{B_{2d}(y_{\varepsilon ,1})}\Big ((x-y_{\varepsilon ,1})\cdot \nabla K(x)\Big ) u^{2^*}_\varepsilon {d}x +O\Big (\frac{1}{\lambda _{\varepsilon ,1}^N}\Big ), \nonumber \\ \end{aligned}$$

(B.2)

and we can compute like (3.17) that

$$\begin{aligned}{} & {} \int _{B_{2d}(y_{\varepsilon ,1})}\Big ((x-y_{\varepsilon ,1})\cdot \nabla K(x)\Big ) u^{2^*}_\varepsilon {d}x \nonumber \\{} & {} \quad = \int _{B_{2d}(y_{\varepsilon ,1})}\Big ((x-y_{\varepsilon ,1})\cdot \nabla K(x)\Big ) \overline{U}^{2^*}_{y_{\varepsilon ,1},\lambda _{\varepsilon ,1}}dx +O\Big (\int _{B_{2d}(y_{\varepsilon ,1})}\big |x-y_{\varepsilon ,1}\big | \big ( \overline{U}^{2^*-1}_{y_{\varepsilon ,1}, \lambda _{\varepsilon ,1}}v_\varepsilon +v_\varepsilon ^{2^*}\big ) {d}x\Big )\nonumber \\{} & {} \quad = \frac{\Delta K(y_{\varepsilon ,1})}{N} \int _{B_{2d}(y_{\varepsilon ,1})} \big |x-y_{\varepsilon ,1}\big |^2 {U}^{2^*}_{y_{\varepsilon ,1},\lambda _{\varepsilon ,1}} {d}x +O\Big (\frac{\varepsilon }{\lambda _{\varepsilon ,1}^2}\Big ) +O\Big (\frac{1}{\lambda _{\varepsilon ,1}^3}\Big ) +O\Big (\frac{\Vert v_\varepsilon \Vert +\Vert v_\varepsilon \Vert ^{2^*}}{\lambda _{\varepsilon ,1}} \Big ) \nonumber \\{} & {} \quad = \frac{NB}{\lambda _{\varepsilon ,1}^2}\Big ( \Delta K(y_{\varepsilon ,1})+o(1)\Big ), \end{aligned}$$

(B.3)

where $B=\frac{C_N^{2^*}}{N^2}\displaystyle \int _{\mathbb {R}^N}\frac{|x|^2}{(1+|x|^2)^{N}}dx$. Then, combining (B.2) and (B.3), we have

$$\begin{aligned} \begin{aligned} \int _{\mathbb {R}^N}&\Big ((x-y_{\varepsilon ,1})\cdot \nabla K(x)\Big ) u^{2^*}_\varepsilon {d}x = \frac{NB}{\lambda _{\varepsilon ,1}^2}\Big ( \Delta K(y_{\varepsilon ,1})+o(1)\Big ), \end{aligned} \end{aligned}$$

(B.4)

Hence, (B.1) and (B.4) imply

$$\begin{aligned} \Delta K(b_1)=0, \end{aligned}$$

which contradicts the condition (K). $\square$