Ngô Quốc Anh

May 7, 2010

The method of moving planes: An integral form

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 0:41

Today we study a beautiful version of the method of moving planes. This method just invented around 2003 by W.X. Chen, C.M. Li and B. Ou and published in Comm. Pure Appl. Math. around 2006 (for the paper, please go here). In that paper, the author proved among other things that every positive regular solution $u(x)$ of the integral equation $\displaystyle u(x) = \int_{{\mathbb{R}^n}} {\frac{1}{{{{\left| {x - y} \right|}^{n - \alpha }}}}u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}dy}$

is radially symmetric and monotone about some point and therefore assumes the form $\displaystyle u(x) = c{\left( {\frac{t}{{{t^2} + {{\left| {x - {x_0}} \right|}^2}}}} \right)^{\frac{{n - \alpha }}{2}}}$

with some constant $c=c(n,\alpha)$ and for some $t>0$ and $x_0 \in \mathbb R^n$. These solutions in case $n=3$ and $\alpha=2$ have the following shape. This integral equation is also closely related to the following family of semilinear PDEs $\displaystyle {( - \Delta )^{\frac{\alpha }{2}}}u = {u^{\frac{{n + \alpha }}{{n - \alpha }}}}$.

For a given real number $\lambda$ we define $\displaystyle \Sigma_\lambda = \{ x=(x_1,...,x_n) : x_1 \geqslant 1\}$

and let $\displaystyle x^\lambda = (2\lambda - x_1,...,x_n)$

and $\displaystyle u_\lambda (x)=u(x^\lambda)$.

We also denote by $\Omega^\lambda$ the reflection of domain $\Omega$ about the plane $x_1=\lambda$.

Lemma. For any solution $u(x)$ of the PDE, we have $\displaystyle u(x) - {u_\lambda }(x) = \int_{{\Sigma _\lambda }} {\left( {\frac{1}{{{{\left| {x - y} \right|}^{n - \alpha }}}} - \frac{1}{{{{\left| {{x^\lambda } - y} \right|}^{n - \alpha }}}}} \right)\left( {u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}} - {u_\lambda }{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}} \right)dy}$.

It is also true for $\displaystyle v(x) = \frac{1}{{|x{|^{n - \alpha }}}}u\left( {\frac{x}{{|x{|^2}}}} \right)$

the Kelvin-type transform of $u$ for any $x \ne 0$.

Set $\displaystyle \tau = \frac{n+\alpha}{n-\alpha}$

we use the method of moving planes to prove that following

Theorem. Let $u \in L_{loc}^{\tau+1}(\mathbb R^n)$ be a positive solution of the PDE. Then it must be radially symmetric and monotone decreasing about some point.

Outline of proof.

Step 1. Define $\displaystyle\Sigma _\lambda ^ - = \left\{ {x:x \in {\Sigma _\lambda }\backslash \left\{ 0 \right\},v(x) < {v_\lambda }(x)} \right\}$

and $\displaystyle\Sigma _\lambda ^C = {\mathbb{R}^n}\backslash {\Sigma _\lambda }$.

In this step, we show that the method of moving planes can be run. To this purpose, we show that there exists a sufficiently negative large value of $\lambda$ such that $v(x) \geqslant v_\lambda(x)$ in $\Sigma_\lambda$, i.e. $\Sigma _\lambda ^ -$ must be empty.

Clearly by definition of $\Sigma _\lambda ^ -$, for any $x \in \Sigma_\lambda^-$ we have $\displaystyle\begin{gathered} 0 \leqslant {v_\lambda }(x) - v(x) = \int_{{\Sigma _\lambda }} {\left( {\frac{1}{{{{\left| {x - y} \right|}^{n - \alpha }}}} - \frac{1}{{{{\left| {{x^\lambda } - y} \right|}^{n - \alpha }}}}} \right)\left( {{v_\lambda }{{(y)}^\tau } - v{{(y)}^\tau }} \right)dy} \hfill \\ \quad\leqslant \int_{\Sigma _\lambda ^ - } {\left( {\frac{1}{{{{\left| {x - y} \right|}^{n - \alpha }}}} - \frac{1}{{{{\left| {{x^\lambda } - y} \right|}^{n - \alpha }}}}} \right)\left( {{v_\lambda }{{(y)}^\tau } - v{{(y)}^\tau }} \right)dy} \hfill \\ \quad= \int_{\Sigma _\lambda ^ - } {\left( {\frac{1}{{{{\left| {x - y} \right|}^{n - \alpha }}}} - \frac{1}{{{{\left| {{x^\lambda } - y} \right|}^{n - \alpha }}}}} \right){v_\lambda }(y)\left( {{v_\lambda }{{(y)}^{\tau - 1}} - v{{(y)}^{\tau - 1}}\frac{{v(y)}}{{{v_\lambda }(y)}}} \right)dy} \hfill \\ \quad\leqslant \int_{\Sigma _\lambda ^ - } {\left( {\frac{1}{{{{\left| {x - y} \right|}^{n - \alpha }}}} - \frac{1}{{{{\left| {{x^\lambda } - y} \right|}^{n - \alpha }}}}} \right){v_\lambda }(y)\left( {{v_\lambda }{{(y)}^{\tau - 1}} - v{{(y)}^{\tau - 1}}} \right)dy} \hfill \\ \quad\leqslant \int_{\Sigma _\lambda ^ - } {\left( {\frac{1}{{{{\left| {x - y} \right|}^{n - \alpha }}}}} \right){v_\lambda }(y)\left( {{v_\lambda }{{(y)}^{\tau - 1}} - v{{(y)}^{\tau - 1}}} \right)dy} . \hfill \\ \end{gathered}$

It follows from the Hardy-Littlewood-Sobolev inequality that there is a constant $C>0$ such that $\displaystyle\begin{gathered} {\left\| {{v_\lambda } - v} \right\|_{{L^q}(\Sigma _\lambda ^ - )}} \leqslant C{\left( {\int_{\Sigma _\lambda ^ - } {v_\lambda ^{\tau + 1}(y)dy} } \right)^{\frac{\alpha }{n}}}{\left\| {{v_\lambda } - v} \right\|_{{L^q}(\Sigma _\lambda ^ - )}} \hfill \\ \qquad\leqslant C{\left( {\int_{{\Sigma _\lambda }} {v_\lambda ^{\tau + 1}(y)dy} } \right)^{\frac{\alpha }{n}}}{\left\| {{v_\lambda } - v} \right\|_{{L^q}(\Sigma _\lambda ^ - )}} \hfill \\ \qquad= C{\left( {\int_{\Sigma _\lambda ^C} {{v^{\tau + 1}}(y)dy} } \right)^{\frac{\alpha }{n}}}{\left\| {{v_\lambda } - v} \right\|_{{L^q}(\Sigma _\lambda ^ - )}} \hfill \\ \end{gathered}$

for any $q>\frac{n}{n-\alpha}$. By the assumption $u \in L_{loc}^{\tau+1}(\mathbb R^n)$ we can choose $N>0$ sufficiently large such that for $\lambda \leqslant -N$ we have $\displaystyle C{\left( {\int_{\Sigma _\lambda ^C} {{v^{\tau + 1}}(y)dy} } \right)^{\frac{\alpha }{n}}} \leqslant \frac{1}{2}$

which implies $\displaystyle \left\| {{v_\lambda } - v} \right\|_{L^q(\Sigma _\lambda ^-)}=0$.

Step 2. Now we move the plane $x_1=\lambda$ to the right as long as the following condition $\displaystyle v(x) \geqslant v_\lambda(x), \quad \forall x \in \Sigma_\lambda\backslash\{0\}$.

Suppose that at $\lambda_0$ we have $\displaystyle {v_{{\lambda _0}}}(x) \leqslant v(x) \quad \forall x \in \Sigma_{\lambda_0}\backslash\{0\} , \quad {v_{{\lambda _0}}}(x) \not\equiv v(x)$

we show that the plane can be moved further to the right. More precisely, there exists an $\varepsilon>0$ (depending on $n$, $\alpha$) such that $\displaystyle {v_\lambda }(x) \leqslant v(x), \quad \forall \lambda \in \left[ {{\lambda _0},{\lambda _0} + \varepsilon } \right)$.

It follows from our lemma that $\displaystyle {v_{{\lambda _0}}}(x) < v(x), \quad \forall x \in {\rm int} {\Sigma _{{\lambda _0}}}$.

Let $\displaystyle\overline {\Sigma _{{\lambda _0}}^ - } = \left\{ {x \in {\Sigma _{{\lambda _0}}}:v(x) \leqslant {v_{{\lambda _0}}}(x)} \right\}$.

Then obviously $\overline {\Sigma _{{\lambda _0}}^ - }$ has measure zero and $\displaystyle\mathop {\lim }\limits_{\lambda \downarrow {\lambda _0}} {\Sigma _\lambda } \subset \overline {\Sigma _{{\lambda _0}}^ - }$.

It is worth noticing that $\displaystyle {\left\| {{v_\lambda } - v} \right\|_{{L^q}(\Sigma _\lambda ^ - )}} \leqslant C{\left( {\int_{\Sigma _\lambda ^ - } {v_\lambda ^{\tau + 1}(y)dy} } \right)^{\frac{\alpha }{n}}}{\left\| {{v_\lambda } - v} \right\|_{{L^q}(\Sigma _\lambda ^ - )}} = C{\left( {\int_{{{(\Sigma _\lambda ^ - )}^\lambda }} {{v^{\tau + 1}}(y)dy} } \right)^{\frac{\alpha }{n}}}{\left\| {{v_\lambda } - v} \right\|_{{L^q}(\Sigma _\lambda ^ - )}}$.

Now the locally integrable of $v$ together with the Lebesgue Dominated Convergence theorem imply $\displaystyle C{\left( {\int_{{{(\Sigma _\lambda ^ - )}^\lambda }} {{v^{\tau + 1}}(y)dy} } \right)^{\frac{\alpha }{n}}} \leqslant \frac{1}{2}$

for all $\lambda \in [\lambda_0,\lambda_0+\varepsilon)$. Again we have $\displaystyle \left\| {{v_\lambda } - v} \right\|_{L^q(\Sigma _\lambda ^-)}=0$

which implies within $\lambda \in [\lambda_0,\lambda_0+\varepsilon)$ the method of moving planes can run.

Step 3. One the method of moving planes stops at some point $\lambda_0<0$, we get that $v_{\lambda_0}(x) \equiv v(x)$

for all $x \in \Sigma_{\lambda_0}$. Consequently, $v$ has no singularity at the origin. We are then in a position to apply the second fundamental lemma considered here to $x=0$ and $\lambda=1$ to obtain the form of solution. Otherwise, $v$ is symmetric w.r.t. the plane $x_1=0$. Since the direction $x_1$ can be chosen arbitrary, we deduce that $v$ is symmetric and monotone about the origin and thus so is $u$.

Remark. Since we do not assume any asymptotic behavior of $u$ near infinity (just assume locally integrable), we are not able to carry out the method of moving planes directly on $u$. To overcome this difficulty, we consider $v$. The advantage here is that $v$ may have singularity at the origin and no singularity at infinity.