Multiple blow-up solutions for an anisotropic 2-dimensional nonlinear Neumann problem

Abstract

We study the following boundary value problem

$$\begin{aligned} \left\{ \begin{array}{ll} -div(a(x)\nabla u)+ a(x)u=0,\ u>0,\ \ \ \ &{}\quad \mathrm{in}\ \Omega ;\\ \frac{\partial u}{\partial \nu }=\lambda u^{p-1}e^{u^p},\ \ \ &{}\quad \mathrm{on}\ \partial \Omega , \end{array} \right. \end{aligned}$$

(0.1)

where $\Omega$ is a bounded domain in $\mathbb {R}^2$ with smooth boundary, $\nu$ is the unit outer normal vector of $\partial \Omega$, and a(x) is a positive smooth function in ${\overline{\Omega }}$, $\lambda >0$ is a small parameter and $0< p <2$. We construct bubbling solutions to problem (0.1), which blow-up at points near a local maximum of a(x) on $\partial \Omega$.

Appendix

Proof of Lemma 2.5

We note that, on the boundary, we have

$$\begin{aligned} \frac{\partial H_{ij} }{\partial \nu }= & {} \alpha _{ij}\left( \varepsilon e^{u_j}-\frac{\partial u_j}{\partial \nu }\right) \\= & {} \alpha _{ij} \left( 2\varepsilon \mu _j\frac{1-\nu (\xi _j)\cdot \nu (x)}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2} +2\frac{(x-\xi _j)\cdot \nu (x)}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2}\right) \end{aligned}$$

we have

$$\begin{aligned} \lim \limits _{\varepsilon \rightarrow 0}\frac{\partial H_{ij}}{\partial \nu } (x)=2\alpha _{ij}\frac{(x-\xi _j)\cdot \nu (x)}{|x-\xi _j|^2},\quad \ \forall \ x\ne \xi _j. \end{aligned}$$

On the other hand, the regular part of Green’s function H(x, y) satisfies

$$\begin{aligned} \quad \left\{ \begin{array}{l@{\quad }l@{\quad }l} -\Delta H(x,y)-\nabla \log a(x)\nabla H(x,y)+H(x,y)\\ \qquad \qquad = \nabla \log a(x)\nabla (\log \frac{1}{|x-y|^2})-\log \frac{1}{|x-y|^2}\quad &{} \mathrm{in}\ \Omega ;\\ \frac{\partial H(x,y)}{\partial \nu } = 2\frac{(x-\xi _j)\cdot \nu (x)}{|x-\xi _j|^2}\quad \ &{} \mathrm{on}\ \partial \Omega . \end{array} \right. \end{aligned}$$

(5.1)

Set $\tilde{z}_\varepsilon =H_{ij}(x)+\alpha _{ij}\log (2\mu _j\varepsilon ^2)-\alpha _{ij}H(x,\xi _j)$, then $\tilde{z}_\varepsilon$ satisfies

$$\begin{aligned} \quad \left\{ \begin{array}{ll} -\Delta \tilde{z}_\varepsilon -\nabla \log a(x)\nabla \tilde{z}_\varepsilon +\tilde{z}_\varepsilon = \bar{f} \quad &{} \mathrm{in}\ \Omega ;\\ \frac{\partial \tilde{z}_\varepsilon }{\partial \nu }= \bar{h}\quad \ &{} \mathrm{on}\ \partial \Omega , \end{array} \right. \end{aligned}$$

(5.2)

where

$$\begin{aligned} \bar{f}= & {} \alpha _{ij}\left[ \log \frac{1}{|x-\xi _j|^2}-\log \frac{1}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2}\right] \\&+\,\alpha _{ij}\nabla \log a\cdot \nabla \left( \log \frac{1}{|x-\xi _j|^2}-\log \frac{1}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2}\right) , \end{aligned}$$

and

$$\begin{aligned} \bar{h}= \frac{\partial H_{ij}}{\partial \nu }-2\alpha _{ij}\frac{(x-\xi _j)\cdot \nu (x)}{|x-\xi _j|^2}. \end{aligned}$$

First, we claim that there is a positive constant C such that

$$\begin{aligned} \left\| \bar{h}\right\| _{L^q(\partial \Omega )}\le C(\varepsilon \mu _j)^{1/q},\quad \ \forall \ q>1, \end{aligned}$$

(5.3)

In fact,

$$\begin{aligned} \bar{h}= & {} \frac{\partial H_{ij}}{\partial \nu }-2\alpha _{ij}\frac{(x-\xi _j)\cdot \nu (x)}{|x-\xi _j|^2}\\= & {} \alpha _{ij}\left( 2\varepsilon \mu _j\frac{1-\nu (\xi _j)\cdot \nu (x)}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2} +2\frac{(x-\xi _j)\cdot \nu (x)}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2}\right) -2\alpha _{ij}\frac{(x-\xi _j)\cdot \nu (x)}{|x-\xi _j|^2}\\= & {} 2\alpha _{ij}\varepsilon \mu _j\frac{1-\nu (\xi _j)\cdot \nu (x)}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2} +2\alpha _{ij}\varepsilon \mu _j\frac{(x-\xi _j)\cdot \nu (x)\left[ 2(x-\xi _j)\cdot \nu (x)-\varepsilon \mu _j\right] }{|x-\xi _j|^2|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2}. \end{aligned}$$

Now, we observe that

$$\begin{aligned} |1-\nu (\xi _j)\cdot \nu (x)|\le C|x-\xi _j|^2,\quad \ |(x-\xi _j)\cdot \nu (x)|\le C|x-\xi _j|^2,\quad \ \forall \ x\in \partial \Omega . \end{aligned}$$

Hence,

$$\begin{aligned} \left| \frac{\partial H_{ij}}{\partial \nu }-2\alpha _{ij}\frac{(x-\xi _j)\cdot \nu (x)}{|x-\xi _j|^2}\right| \le C\varepsilon \mu _j +C\frac{\varepsilon |2(x-\xi _j)\cdot \nu (\xi _j)-\varepsilon \mu _j|}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2}. \end{aligned}$$

(5.4)

For $\rho >0$ small, we have

$$\begin{aligned} \left| \frac{\partial H_{ij}}{\partial \nu }-2\alpha _{ij}\frac{(x-\xi _j)\cdot \nu (x)}{|x-\xi _j|^2}\right| \le C\varepsilon \mu _j,\ x\ \ \text{ for }\ \ \forall \ |x-\xi _j|\ge \rho ,\ x\in \partial \Omega . \end{aligned}$$

(5.5)

Now let $q>1$, we have

$$\begin{aligned}&\int _{B_\rho (\xi _j)\cap \partial \Omega }\left| \frac{\varepsilon \mu _j|2(x-\xi _j)\cdot \nu (\xi _j)-\varepsilon \mu _j|}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2}\right| ^qdx\nonumber \\&\quad =C\varepsilon \mu _j\int _{B_{\rho /\varepsilon }(0)\cap \partial \Omega _\varepsilon }\left| \frac{2y\cdot \nu (0)-\mu _j}{y-\mu _j\nu (0)}\right| ^qd y\nonumber \\&\quad \le C\varepsilon \mu _j\int _0^{\rho /\varepsilon }\frac{1}{(1+s)^q}ds\le C \varepsilon \mu _j. \end{aligned}$$

(5.6)

Combining (5.4) with (5.5) and (5.6) we conclude that (5.3) holds.

Next, we show that for any $1<q<2$,

$$\begin{aligned} \left\| \log \frac{1}{|x-\xi _j|^2}-\log \frac{1}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2}\right\| _{L^q(\Omega )} \le C\varepsilon ^\alpha . \end{aligned}$$

(5.7)

In fact, for $q\ge 1$, we write

$$\begin{aligned}&\left\| \log \frac{1}{|x-\xi _j|^2}-\log \frac{1}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2}\right\| _{L^q(\Omega )}^q\nonumber \\&\quad =\int _{B_{10\varepsilon \mu _j}(\xi _j)\cap \Omega }\ldots \ + \int _{\Omega \backslash B_{10\varepsilon \mu _j}(\xi _j)}\ldots := I_1+I_2. \end{aligned}$$

(5.8)

Next we estimate $I_1$ and $I_2$. For $I_1$, we observe that

$$\begin{aligned} \int _{B_{10\varepsilon \mu _j}(\xi _j)\cap \Omega }\left| \log \frac{1}{|x-\xi _j|^2}\right| ^qdx\le C\int _{0}^{C\varepsilon \mu _j}|\log r|^qrdr\le C(\varepsilon \mu _j)^2\left( \log \frac{1}{\varepsilon }\right) ^q, \end{aligned}$$

and the same bound is true for the integral of $|\log \frac{1}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2}|^q$ in $B_{10\varepsilon \mu _j}(\xi _j)\cap \Omega$. Hence, we have

$$\begin{aligned} |I_1|\le C(\varepsilon \mu _j)^2\left( \log \frac{1}{\varepsilon }\right) ^q. \end{aligned}$$

(5.9)

For $I_2$, if $|x-\xi _j|\ge 10\varepsilon \mu _j$, we have

$$\begin{aligned} |x-\xi _j|\le |x-\xi _j-t\varepsilon \mu _j\nu (\xi _j)|+\varepsilon \mu _jx\le |x-\xi _j-t\varepsilon \mu _j\nu (\xi _j)|+\frac{1}{10}|x-\xi _j| \end{aligned}$$

for any $t\in [0,1]$, then we have $|x-\xi _j|\le C|x-\xi _j-t\varepsilon \mu _j\nu (\xi _j)|$. Using this fact, we can obtain

$$\begin{aligned}&\left| \log \frac{1}{|x-\xi _j|^2}-\log \frac{1}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2}\right| \\&\quad \le C\sup \limits _{0\le t\le 1}\frac{C\varepsilon \mu _j}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|}\le \frac{C\varepsilon \mu _j}{|x-\xi _j|}. \end{aligned}$$

Thus for $1<q<2$,

$$\begin{aligned} |I_1|\le C\varepsilon ^q\int _{10\varepsilon \mu _j}^Dr^{1-q}dr\le C(\varepsilon \mu _j)^q, \end{aligned}$$

(5.10)

where D is the diameter of $\Omega$. Thus, combining (5.8) with (5.9) and (5.10) we obtain that (5.7) holds.

Finally, for any $1<q<2$, we show that

$$\begin{aligned} \left\| \nabla \log a\nabla \left( \log \frac{1}{|x-\xi _j|^2}-\log \frac{1}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2}\right) \right\| _{L^q(\Omega )} \le C(\varepsilon \mu _j)^{\frac{2-q}{q}}. \end{aligned}$$

(5.11)

In fact, for any $1<q<2$, we write

$$\begin{aligned}&\left\| \nabla \left( \log \frac{1}{|x-\xi _j|^2}-\log \frac{1}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2}\right) \right\| _{L^q(\Omega )}^q\nonumber \\&\quad =\int _{B_{10\varepsilon \mu _j}(\xi _j)\cap \Omega }\cdots +\int _{\Omega \backslash B_{10\varepsilon \mu _j}(\xi _j)}\cdots \nonumber \\&\quad := I_3+I_4. \end{aligned}$$

(5.12)

For $I_3$, we have

$$\begin{aligned} |I_3|\le & {} \int _{B_{10\varepsilon \mu _j}(\xi _j)\cap \Omega }\left| \nabla \log \frac{1}{|x-\xi _j|^2}\right| ^qdx\nonumber \\&+\int _{B_{10\varepsilon \mu _j}(\xi _j)\cap \Omega }\left| \nabla \log \frac{1}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^2}\right| ^qdx\nonumber \\= & {} C\int _{B_{10\varepsilon \mu _j}(\xi _j)\cap \Omega }\left| \log \frac{1}{|x-\xi _j|^q}\right| ^qdx\nonumber \\&+\, C\int _{B_{10\varepsilon \mu _j}(\xi _j)\cap \Omega }\left| \log \frac{1}{|x-\xi _j-\varepsilon \mu _j\nu (\xi _j)|^q}\right| ^qdx\nonumber \\\le & {} C(\varepsilon \mu _j)^{2-q}. \end{aligned}$$

(5.13)

Moreover, we have

$$\begin{aligned} I_4\le & {} \int _{\Omega \backslash B_{10\varepsilon \mu _j}(\xi _j)\cap \Omega }\left| \frac{2\varepsilon \mu _j}{|x-\xi _j|(|x-\xi _j|-\varepsilon \mu _j)}\right| ^qdx\nonumber \\\le & {} C\int _{10\varepsilon \mu _j}^{D}\frac{(\varepsilon \mu _j)^q}{(r-\varepsilon \mu _j)^{2q-1}}dr \le C(\varepsilon \mu _j)^{2-q}. \end{aligned}$$

(5.14)

From (5.12)–(5.14) to get (5.11).

Therefore by elliptic regularity theory, for problem (5.2) we obtain

$$\begin{aligned} \Vert z_\varepsilon \Vert _{W^{1+s,q}(\Omega )}\le C(\varepsilon \mu _j)^{\frac{2-q}{q}} \end{aligned}$$

(5.15)

By the Morrey embedding we obtain

$$\begin{aligned} \Vert z_\varepsilon \Vert _{C^\beta }(\bar{\Omega })\le C \varepsilon ^{\beta } \end{aligned}$$

for any $0<\beta <1-\frac{1}{q}$. This finishes the proof of the Lemma. $\square$

Proof of (2.21)

We shall prove

$$\begin{aligned} \alpha _{1j}=\log (2\mu _j). \end{aligned}$$

(5.16)

In fact, first we observe that $f_1$ is a symmetric function for any choice of $\alpha _{1j}$, hence

$$\begin{aligned} \int _{\partial \mathbb {R}^2_+}e^{w_{\mu _j}}f_1z_{1\mu _j}=0. \end{aligned}$$

Next, we chose parameter $\alpha _{1j}$ such that the other orthogonality condition satisfies. Since

$$\begin{aligned}&\int _{\partial \mathbb {R}^2_+}e^{w_{\mu _j}}f_1z_{0\mu _j} \\&\quad = -\frac{1}{\mu _j}\int _{\partial \mathbb {R}^2_+}e^{w_{\mu _j}}\left( \alpha _{1j}(w_{\mu _j}-1)+w_{\mu _j}+\frac{1}{2}(w_{\mu _j})^2\right) \left( y\cdot \nabla w_{\mu _j}(y)+1\right) \\&\quad =-\frac{1}{\mu _j}\alpha _{1j}\int _{-\infty }^\infty \left( e^{w_{\mu _j}(y_1,0)}\left( w_{\mu _j}(y_1,0)-1\right) \frac{\partial w_{\mu _j}}{\partial y_1}(y_1,0)y_1\right. \\&\qquad \left. +\, e^{w_{\mu _j}(y_1,0)}\left( w_{\mu _j}(y_1,0)-1\right) \right) dy_1\\&\qquad -\frac{1}{\mu _j}\int _{-\infty }^\infty \left( e^{w_{\mu _j}(y_1,0)}w_{\mu _j}(y_1,0)\frac{\partial w_{\mu _j}}{\partial y_1}(y_1,0)y_1+e^{w_{\mu _j}(y_1,0)}w_{\mu _j}(y_1,0)\right) dy_1\\&\qquad -\frac{1}{\mu _j}\int _{-\infty }^\infty \left( e^{w_{\mu _j}(y_1,0)}\frac{(w_{\mu _j})^2}{2}(y_1,0)\frac{\partial w_{\mu _j}}{\partial y_1}(y_1,0)y_1+e^{w_{\mu _j}(y_1,0)}\frac{(w_{\mu _j})^2}{2}(y_1,0)\right) dy_1\\&\quad =-\frac{1}{\mu _j}\left[ \alpha _{1j}\int _{-\infty }^\infty e^{w_{\mu _j}(y_1,0)} dy_1+\int _{-\infty }^\infty e^{w_{\mu _j}(y_1,0)}w_{\mu _j}(y_1,0) dy_1\right] . \end{aligned}$$

Thus we need to choose $\alpha _{1j}$ such that

$$\begin{aligned} \alpha _{1j}\int _{-\infty }^\infty e^{w_{\mu _j}(y_1,0)} dy_1+\int _{-\infty }^\infty e^{w_{\mu _j}(y_1,0)}w_{\mu _j}(y_1,0) dy_1=0. \end{aligned}$$

(5.17)

Since

$$\begin{aligned} \int _{-\infty }^\infty e^{w_{\mu _j}(y_1,0)} dy_1=\int _{-\infty }^\infty \frac{2\mu _j}{y_1^2+\mu _j^2}\ dy_1=2\int _{-\infty }^\infty \frac{1}{t^2+1}\ dt=2\pi , \end{aligned}$$

(5.18)

and

$$\begin{aligned}&\int _{-\infty }^\infty e^{w_{\mu _j}(y_1,0)}w_{\mu _j}(y_1,0) dy_1=\int _{-\infty }^\infty \frac{2\mu _j}{y_1^2+\mu _j^2}\log \frac{2\mu _j}{y_1^2+\mu _j^2}\ dy_1\nonumber \\&\quad =2\int _{-\infty }^\infty \frac{1}{t^2+1}\left[ \log \frac{1}{t^2+1}+\log (2\mu _j^{-1})\right] \ dt=-2\pi \log (2\mu _j). \end{aligned}$$

(5.19)

Therefore, from (5.17), (5.18) and (5.19) we get that $\alpha _{1j}$ satisfies (2.21). $\square$