On the construction of non-simple blow-up solutions for the singular Liouville equation with a potential

Abstract

We are concerned with the existence of blowing-up solutions to the following boundary value problem

$$\begin{aligned} -\Delta u= \lambda V(x) e^u-4\pi N {\varvec{\delta }}_0 \;\hbox { in } B_1,\quad u=0 \;\hbox { on }\partial B_1, \end{aligned}$$

where $B_1$ is the unit ball in $\mathbb {R}^2$ centered at the origin, V(x) is a positive smooth potential, N is a positive integer ($N\ge 1$). Here ${\varvec{\delta }}_0$ defines the Dirac measure with pole at 0, and $\lambda >0$ is a small parameter. We assume that $N=1$ and, under some suitable assumptions on the derivatives of the potential V at 0, we find a solution which exhibits a non-simple blow-up profile as $\lambda \rightarrow 0^+$.

Appendix A

In this appendix we derive some crucial integral estimates which arise in the asymptotic expansion of the energy of approximate solution $PW_{\lambda }$.

Lemma A.1

The following holds:

$$\begin{aligned}{} & {} \int _{B_1}e^{W_{\lambda }} |x-b|dx=O(\delta ),\quad \int _{B_1}e^{W_{\lambda }} |x-b|^2dx=16\pi \delta ^2|\log \delta |+O(\delta ^2),\\{} & {} \int _{B_1}e^{W_{\lambda }} |x-b|^3 dx=O(\delta ^2) \end{aligned}$$

uniformly for b in a small neighborhood of 0.

Proof

We compute

$$\begin{aligned} \int _{B_1}e^{W_{\lambda }} |x-b|dx \le 8\delta \int _{{\mathbb {R}}^2}\frac{1}{(1+|z-\delta ^{-1} b|^2)^2}|z-\delta ^{-1}b|dz =8\delta \int _{{\mathbb {R}}^2}\frac{|z|}{(1+|z|^2)^2}dz \end{aligned}$$

and the first estimate follows. In order to show the second estimate let us observe that $B(b, 1-|b|)\subset B(0,1)\subset B(b, 1+|b|)$, so we compute

$$\begin{aligned} \int _{B_1}e^{W_{\lambda }}|x-b|^2 dx&= 8\int _{B_1 }\frac{\delta ^2|x-b|^2}{(\delta ^2+|x-b|^2)^2} dx \\&=8 \int _{B(b, 1-|b|)} \frac{\delta ^2|x-b|^2}{(\delta ^2+|x-b|^2)^2}dx\\&\quad +O\bigg (\int _{ B(b, 1+|b|)\setminus B(b,1-|b|)} \frac{\delta ^2|x-b|^2}{(\delta ^2+|x-b|^2)^2}dx\bigg )\\&= 8 \int _{B(0, 1-|b|)} \frac{\delta ^2|x|^2}{(\delta ^2+|x|^2)^3}dx +O\bigg (\int _{ B(0, 1+|b|)\setminus B(0,1-|b|)} \frac{\delta ^2|x|^2}{(\delta ^2+|x|^2)^2}dx\bigg )\\&=8\delta ^2\int _{|z|\le \frac{1-|b|}{\delta }} \frac{|z|^2}{(1+|z|^2)^2}dz+O\bigg (\delta ^2 \int _{ \frac{1-|b|}{\delta }\le |z|\le \frac{1+|b|}{\delta }}\frac{1}{|z|^2}dz\bigg ) \\&=8\delta ^2\int _{|z|\le \frac{1-|b|}{\delta }} \frac{1}{1+|z|^2}dz+O(\delta ^2)\\&=16\pi \delta ^2|\log \delta |+O(\delta ^2). \end{aligned}$$

In order to prove the third estimate, let $R>1$ so that $B(0,1) \subset B(b, R)$ if b lies in a small neighborhood of 0. Then,

$$\begin{aligned} \int _{B_1}e^{W_{\lambda }} |x-b|^3dx&=8\delta ^{3}\int _{|z| \le \frac{1 }{\delta }} \frac{1}{(1+|z-\delta ^{-1} b|^2)^2}|z-\delta ^{-1}b|^3dz\\&\le 8\delta ^{3}\int _{B(0,\frac{R}{\delta })} \frac{|z|^3}{(1+|z|^2)^2}dz \le C\delta ^{2}. \end{aligned}$$

$\square$

Since the key part in the proof of Theorem 2.1 relies in testing the Eq. (5.1) with $PZ_\lambda ^i$ in order to catch the leading terms, a crucial step consists in the evaluation of some integral estimates, as provided by the following lemma.

Lemma A.2

The following holds for $i,j=1,2$:

$$\begin{aligned}{} & {} \int _{B_1}e^{W_{\lambda }} PZ^i_\lambda dx =O(\delta ),\\{} & {} \int _{B_1}e^{W_{\lambda }} PZ^i_\lambda ( x_i-b_i)dx=2 \pi \delta +O(\delta ^{2}),\\{} & {} \int _{B_1}e^{W_{\lambda }} PZ^i_\lambda (x_j-b_j)dx=O(\delta ^{2})\quad i\ne j,\\{} & {} \int _{B_1}e^{W_{\lambda }} |PZ^i_\lambda || x-b|^2dx=O(\delta ^{2}), \\{} & {} \int _{B_1}e^{W_{\lambda }} PZ^i_\lambda (x_i-b_i)^3dx=6\pi \delta ^3 \log \frac{1}{\delta }+O(\delta ^3)\\{} & {} \int _{B_1}e^{W_{\lambda }} PZ^i_\lambda (x_j-b_j)^3dx=O(\delta ^3)\quad i\ne j,\\{} & {} \int _{B_1}e^{W_{\lambda }} PZ^i_\lambda (x_j-b_j)^2( x_i-b_i)dx =2\pi \delta ^3\log \frac{1}{\delta }+O(\delta ^3) \quad i\ne j,\\{} & {} \int _{B_1}e^{W_{\lambda }} PZ^i_\lambda (x_i-b_i)^2( x_j-b_j)dx=O (\delta ^3) \quad i\ne j,\\{} & {} \int _{B_1}e^{W_{\lambda }} |PZ^i_\lambda || x-b|^{\frac{7}{2}}dx=O(\delta ^{3}) \end{aligned}$$

uniformly for b in a small neighborhood of 0.

Proof

We compute

$$\begin{aligned} \int _{B_1}e^{W_{\lambda }} Z^i_\lambda dx&=8\int _{|z|\le \frac{1}{\delta }} \frac{1}{(1+|z-\delta ^{-1}b|^2)^3}(z_i-\delta ^{-1}b_i) dz\\&=8\int _{{\mathbb {R}}^2}\frac{1}{(1+|z-\delta ^{-1}b|^2)^3}(z_i-\delta ^{-1}b_i) dz+O(\delta ^{3})\\&=8\int _{{\mathbb {R}}^2}\frac{z_i}{(1+|z|^2)^3} dz+O(\delta ^{3}) =O(\delta ^3), \end{aligned}$$

since $\int _{{\mathbb {R}}^2}\frac{z_i}{(1+|z||^2)^3} dz=0$ by oddness. Next

$$\begin{aligned} \int _{B_1}e^{W_{\lambda }} Z^i_\lambda (x_i-b_i)dx&=8\delta \int _{|z| \le \frac{1 }{\delta }}\frac{1}{(1+|z-\delta ^{-1}b|^2)^3} (z_i-\delta ^{-1}b_i)^2 dz \\&=8\delta \int _{{\mathbb {R}}^2}\frac{1}{(1+|z-\delta ^{-1}b|^2)^3} (z_i-\delta ^{-1}b_i)^2 dz+O(\delta ^{3}) \\&= 8 \delta \int _{{\mathbb {R}}^2}\frac{z_i^2}{(1+|z|^2)^3}dz+O(\delta ^{3})=2\pi \delta +O(\delta ^{3}) \end{aligned}$$

where we have used the identity $\int _{{\mathbb {R}}^2}\frac{(z_i)^2}{(1+|z|^2)^3} =\frac{1}{2} \int _{{\mathbb {R}}^2}\frac{|z|^2}{(1+|z|^2)^3} =\frac{\pi }{4}$. Similarly for $i\ne j$

$$\begin{aligned} \int _{B_1}e^{W_{\lambda }} Z^i_\lambda (x_j-b_j)dx = 8 \delta \int _{{\mathbb {R}}^2}\frac{z_iz_j}{(1+|z|^2)^3}dz+O(\delta ^{3})=O(\delta ^{3}) \end{aligned}$$

since $\int _{{\mathbb {R}}^2}\frac{z_iz_j}{(1+|z|^2)^3}dz=0.$ Next,

$$\begin{aligned} \int _{B_1}e^{W_{\lambda }} |Z^i_\lambda || x-b|^2dx \le 8\delta ^2\int _{{\mathbb {R}}^2}\frac{|x-b|^3}{(\delta ^2+|x- b|^2)^3}dx =8\delta ^2 \int _{{\mathbb {R}}^2}\frac{|z|^3}{(1+|z|^2)^3}dz\le C\delta ^2. \end{aligned}$$

Using that $B(b, 1-|b|)\subset B(0,1)\subset B(b, 1+|b|)$, we compute

$$\begin{aligned}&\int _{B_1}e^{W_{\lambda }} Z^i_\lambda (x_i-b_i)^3dx\\&\quad = 8\delta ^3\int _{B_1 }\frac{(x_i-b_i)^4}{(\delta ^2+|x-b|^2)^3}dx\\&\quad = 8\delta ^3\int _{B(b, 1-|b|)} \frac{(x_i-b_i)^4}{(\delta ^2+|x-b|^2)^3}dx +O\bigg (\delta ^3\int _{ B(b, 1+|b|) \setminus B(b,1-|b|)}\frac{|x-b|^4}{(\delta ^2+|x-b|^2)^3}dx\bigg )\\&\quad = 8\delta ^3 \int _{B(0, 1-|b|)} \frac{( x_i)^4}{(\delta ^2+|x|^2)^3}dx +O\bigg (\delta ^3\int _{ B(0, 1+|b|)\setminus B(0,1-|b|)}\frac{|x|^4}{(\delta ^2+|x|^2)^3}dx\bigg )\\&\quad = 8\delta ^3\int _{|z|\le \frac{1-|b|}{\delta }} \frac{(z_i)^4}{(1+|z|^2)^3}dz+O(\delta ^3)\\&\quad =6\pi \delta ^3|\log \delta |+O(\delta ^3) \end{aligned}$$

where we have used the identity $\int _{|z|\le r} \frac{(z_i)^4}{(1+|z|^2)^3}dz=\frac{3}{8}\pi \log (1+r^2) +\frac{3}{4}\frac{\pi }{1+r^2}-\frac{3}{16}\frac{\pi }{(1+r^2)^2} -\frac{9}{16}\pi$.

Similarly, for $i\ne j$

$$\begin{aligned} \int _{B_1}e^{W_{\lambda }} Z^i_\lambda (x_j-b_j)^3dx =8\delta ^3\int _{|z|\le \frac{1-|b|}{\delta }} \frac{z_i(z_j)^3}{(1+|z|^2)^3}dz+O(\delta ^3) =O(\delta ^3) \end{aligned}$$

since $\int _{|z|\le r} \frac{z_i(z_j)^3}{(1+|z|^2)^3}dz=0$.

Next, for $i\ne j$

$$\begin{aligned} \int _{B_1}e^{W_{\lambda }} Z^i_\lambda (x_j-b_j)^2( x_i-b_i)dx&=8\delta ^3\int _{|z|\le \frac{1-|b|}{\delta }} \frac{(z_i)^2(z_j)^2}{(1+|z|^2)^3}dz+O(\delta ^3)\\&=2\pi \delta ^3|\log \delta |+O(\delta ^3) \end{aligned}$$

where the last equality follows by $\int _{|z|\le r} \frac{(z_i)^2(z_j)^2}{(1+|z|^2)^3}dz =\frac{\pi }{8}\log (1+r^2)+\frac{1}{4} \frac{\pi }{1+r^2}-\frac{1}{16}\frac{\pi }{(1+r^2)^2}-\frac{3}{16}\pi$. Similarly for $i\ne j$

$$\begin{aligned} \int _{B_1}e^{W_{\lambda }} Z^i_\lambda (x_i-b_i)^2( x_j-b_j)dx =8\delta ^3\int _{|z|\le \frac{1-b}{\delta }} \frac{(z_i)^3z_j}{(1+|z|^2)^3}dz+O(\delta ^3) =O(\delta ^3) \end{aligned}$$

by $\int _{|z|\le r} \frac{(z_i)^3z_j}{(1+|z|^2)^3}dz=0$. Finally

$$\begin{aligned}&\int _{B_1}e^{W_{\lambda }} |Z^i_\lambda || x-b|^{\frac{7}{2}}dx \le 8\delta ^2\int _{B(b, 1+|b|)} \frac{\delta |x-b|^{\frac{9}{2}}}{(\delta ^2+|x- b|^2)^3}dx \\&\quad =8\delta ^{\frac{7}{2}}\int _{|z| \le \frac{1+|b|}{\delta }} \frac{|z|^{\frac{9}{2}}}{(1+|z|^2)^3}dz\le C\delta ^3. \end{aligned}$$

Taking into account that $PZ^i_\lambda =Z^i_\lambda +O(\delta )$ by (3.2), and recalling Lemma A.1, the above integral estimates give the thesis. $\square$

In order to derive next integral estimate we need to expand the projections $PZ_\lambda ^i$ to a higher order with respect to (3.2).

Lemma A.3

For $i=1,2$ the following holds:

$$\begin{aligned} PZ_\lambda ^i= Z_\lambda ^i (x)-\delta (x_i-b_i)+O(\delta ^3)+O(\delta | b|)\hbox { in } B_1 \end{aligned}$$

uniformly for b in a small neighborhood of 0.

Proof

Let us consider $i=1$. Observe that

$$\begin{aligned} \hbox {if } |x|=1: \;\;Z_\lambda ^1(x)&=\frac{\delta (x_1-b_1)}{\delta ^2+|x-b|^2} =\frac{\delta (x_1-b_1)}{1+\delta ^2+O(|b|)}\\&=\delta (x_1-b_1)\Big (1+O(|b|)+O(\delta ^2) \Big )\\&=\delta (x_1-b_1)+ O(|b|\delta )+O(\delta ^3) . \end{aligned}$$

Therefore, if we set

$$\begin{aligned} \hat{Z}_\lambda ^1:= Z_\lambda ^1 (x)-\delta (x_1-b_1), \end{aligned}$$

we get

$$\begin{aligned} \hat{Z}_\lambda ^1(x)=O(\delta ^3)+O(\delta | b|) \hbox { if }|x|=1 \end{aligned}$$

and

$$\begin{aligned} -\Delta \hat{Z}_\lambda ^1 (x)=-\Delta Z_\lambda ^1(x) =\Delta PZ_\lambda ^1(x) \hbox { in }B_1. \end{aligned}$$

Hence, since by construction $PZ_\lambda ^1=0$ for $|x|=1$, the maximum principle applies and gives

$$\begin{aligned} PZ_\lambda ^1= \hat{Z}_\lambda ^1 +O(\delta ^3)+O(\delta | b|) =Z_\lambda ^1 (x)-\delta (x_1-b_1)+O(\delta ^3)+O(\delta |b|)\hbox { in } B_1. \end{aligned}$$

$\square$

Lemma A.4

Let P be a homogeneous polynomial of degree 2. Then the following holds:

$$\begin{aligned} \int _{B_1}e^{W_{\lambda }} PZ^i_\lambda P(x-b)dx=O(\delta ^3) +O(\delta ^2 |b|)\quad i=1,2 \end{aligned}$$

uniformly for b in a small neighborhood of 0.

Proof

We compute

$$\begin{aligned} \int _{B_1}e^{W_{\lambda }} Z^i_\lambda P(x-b) dx&= 8\delta ^2\int _{|z|\le \frac{1 }{\delta }} \frac{P(z-\delta ^{-1}b)}{(1+|z-\delta ^{-1} b|^2)^3}(z_i-\delta ^{-1}b_i) dz\nonumber \\&= 8\delta ^2\int _{{\mathbb {R}}^2}\frac{P(z)}{(1+|z|^2)^3}z_idz+O(\delta ^3)=O(\delta ^3) \end{aligned}$$

(A.1)

where we have used that $\int _{{\mathbb {R}}^2}\frac{P(z)}{(1+|z|^2)^3}z_idz=0$ by oddness. Taking into account of Lemma A.3 we get

$$\begin{aligned} \int _{B_1}e^{W_{\lambda }} PZ^i_\lambda P(x-b) dx&= \int _{B_1}e^{W_{\lambda }} Z^i_\lambda P(x-b) dx-\delta \int _{B_1}e^{W_{\lambda }} (x_i-b_i) P(x-b) dx\\&\quad + \Big (O(\delta ^3)+O(\delta |b|)\Big ) \int _{B_1}e^{W_{\lambda }} |x-b|^2 dx\\&= \int _{B_1}e^{W_{\lambda }} Z^i_\lambda P(x-b) dx+O(\delta ) \int _{B_1}e^{W_{\lambda }} |x -b|^3 dx\\&\quad + \Big (O(\delta ^3)+O(\delta |b|)\Big ) \int _{B_1}e^{W_{\lambda }} |x-b|^2 dx \end{aligned}$$

and the thesis follows by (A.1), and recalling Lemma A.1. $\square$

Finally we deduce some integral identities associated to the change of variable $x\mapsto x^\alpha$ which appears frequently when dealing with $\alpha$-symmetric functions.

Lemma A.5

Let $\alpha \in \mathbb {N}$, $\alpha \ge 2$, and let $f\in L^1(B_1)$. Then we have that $|x|^{2(\alpha -1)}f(x^\alpha )\in L^1(B_1)$ and

$$\begin{aligned} \int _{B_1} |x|^{2(\alpha -1)} f(x^\alpha ) dx=\frac{1}{\alpha }\int _{B_1} f(y)dy. \end{aligned}$$

Proof

It is sufficient to prove the thesis for a smooth function f. Using the polar coordinates $(\rho ,\theta )$ and then applying the change of variables $(\rho ',\theta ')=(\rho ^\alpha ,\alpha \theta )$

$$\begin{aligned} \int _{B_1} |x|^{2(\alpha -1)} f(x^\alpha ) dx&=\int _0^{+\infty } d\rho \int _0^{2\pi } \rho ^{2\alpha -1}f(\rho ^\alpha e^{\textrm{i}\alpha \theta }) d\theta \\&= \frac{1}{\alpha ^2}\int _0^{+\infty } d\rho ' \int _0^{2\alpha \pi } \rho 'f(\rho ' e^{\textrm{i}\theta '}) d\theta ' \\&=\frac{1}{\alpha }\int _0^{+\infty } d\rho '\int _0^{2\pi } \rho '|f(\rho ' e^{\textrm{i}\theta '})|^2 d\theta '\\&=\frac{1}{\alpha }\int _{B_1}f(y) dy. \end{aligned}$$