# Ngô Quốc Anh

## June 1, 2010

### The method of moving spheres: An integral form

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 1:05

Let us now consider the following equation $\displaystyle u(x) = \int_{{\mathbb{R}^n}} {\frac{{u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy}$

where $\alpha$ is a real number satisfying $0<\alpha.

Lieb [here] proved among other things that there exist maximizing functions $f$ for the Hardy-Littlewoord-Sobolev inequality on $\mathbb R^n$ $\displaystyle {\left\| {\int_{{\mathbb{R}^n}} {\frac{{f(y)}}{{{{\left| { \cdot - y} \right|}^\lambda }}}dy} } \right\|_{{L^q}({\mathbb{R}^n})}} \leqslant {C_{p,\lambda ,n}}{\left\| f \right\|_{{L^p}({\mathbb{R}^n})}}$

When $p=\frac{2n}{n-\lambda}$ and $q=\frac{2n}{\lambda}$, the Euler-Langrange equation for a maximizing $f$ is nothing but our integral equation.

Having discussion of fractional Laplacian [here] one can easily see that this integral equation is also closely related to the following family of semilinear PDEs $\displaystyle (-\Delta)^\frac{\alpha}{2}u=u^\frac{n+\alpha}{n-\alpha}$.

The classification of solution to the integral equation was done by Chen, Li and Ou [here] published in Comm. Pure App. Math. around 2006 via the integral form of the method of moving planes. Our goal is to derive a new approach based on the integral form of the method of moving spheres. This result was due to Zhang and Hao [here] published in J. Math. Anal. Appl. around 2008.

Theorem. Let $u \in L_{loc}^\frac{2n}{n-\alpha}(\mathbb R^n)$ be a positive solution to the integral equation. Then $u(x)$ is radially symmetric and has the form $\displaystyle u(x) = {\left( {\frac{a}{{d + {{\left| {x - \overline x } \right|}^2}}}} \right)^{\frac{{n - \alpha }}{2}}}$

for some constants $a,d>0$ and $\overline x \in \mathbb R^n$.

Outline of the proof. Let $v$ be a positive function on $\mathbb R^n$, for $x \in \mathbb R^n$ and $\lambda>0$ we define $\displaystyle {v_{x,\lambda }}(y) = {\left( {\frac{\lambda }{{\left| {y - x} \right|}}} \right)^{n - \alpha }}v({y^{x,\lambda }}), \quad y \in {\mathbb{R}^n}$

where $\displaystyle {y^{x,\lambda }} = x + {\lambda ^2}\frac{{y - x}}{{{{\left| {y - x} \right|}^2}}}$.

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

Lemma 1. For any solution $u$ of the integral equation, we have $\displaystyle {u_{x,\lambda }}(y) - u(y) = \int\limits_{\left| {y - x} \right| \geqslant \lambda } {\left[ {\frac{1}{{{{\left| {y - z} \right|}^{n - \alpha }}}} - {{\left( {\frac{\lambda }{{\left| {y - x} \right|}}} \right)}^{n - \alpha }}\frac{1}{{\left| {{y^{x,\lambda }} - z} \right|^{n - \alpha }}}} \right]\left( {{u_{x,\lambda }}{{(z)}^\tau } - u{{(z)}^\tau }} \right)dz}$.

This lemma has the same form of the lemma considered in this entry. In fact, the proof is straightforward.

Lemma 2. For $x \in \mathbb R^n$, there exists $\lambda_0(x)>0$ such that $u_{x,\lambda}(y) \leqslant u(y)$

for any $0<\lambda<\lambda_0(x)$ and $|y-x| \geqslant \lambda$.

This lemma tells us that we can run the method of moving spheres. The idea is as follows: it follows from the form of our solution that our solution is indeed monotone decreasing. Thus starting from a point $x$ it must be possible to find such a $\lambda_0$ so that outside a sphere centered at $x$ with suitable radius, the attitude of function is lower that that at $x_0$. The proof is similar to the proof of step 1 in this entry.

Next for each $x\in \mathbb R^n$, we define $\displaystyle\overline \lambda (x) = \sup \left\{ {\mu > 0:{u_{x,\lambda }}(y) \leqslant u(y), \quad \forall 0 < \lambda < \mu , \quad \left| {y - x} \right| \geqslant \lambda } \right\}$.

Lemma 3. If $\overline \lambda (x_0) < \infty$ for some $x_0 \in \mathbb R^n$ then $\displaystyle u_{x_0,\overline \lambda (x_0)} \equiv u$

in $\mathbb R^n$.

The proof of this lemma is similar to the proof of step 2 in this entry. We do it by contradiction argument. Having all discussion above, we are able to complete the proof of theorem. In fact, w shall prove that $\overline\lambda(x)$ is finite for all $x$.

Proof of theorem. If there exists some $x_0$ such that $\overline \lambda (x_0) < \infty$, then by Lemma 3, $\displaystyle\mathop {\lim }\limits_{|y| \to \infty } {\left| y \right|^{n - \alpha }}u(y) = \overline \lambda {({x_0})^{n - \alpha }}u({x_0}) < \infty$.

By the definition of $\overline \lambda$, $\displaystyle {{u_{x,\lambda }}(y) \leqslant u(y), \quad \forall 0 < \lambda < \mu ,\quad\left| {y - x} \right| \geqslant \lambda }$.

Multiply the above by $|y|^{n-\alpha}$ and let $|y| \to \infty$ we get $\displaystyle\mathop {\lim \inf }\limits_{|y| \to \infty } {\left| y \right|^{n - \alpha }}u(y) \geqslant {\lambda ^{n - \alpha }}u(x),\forall 0 < \lambda < \overline \lambda (x)$.

Thus $\overline\lambda(x)<\infty, \quad \forall x$.

It now follows from the Lemma 3 that $\displaystyle u_{x,\overline \lambda (x)} \equiv u$

for any $x$. Thus gives $\displaystyle u(x) = {\left( {\frac{a}{{d + {{\left| {x - \overline x } \right|}^2}}}} \right)^{\frac{{n - \alpha }}{2}}}$

by the second fundamental lemma [here]. If $\overline \lambda(x)=\infty$ for any $x$ then $\displaystyle {{u_{x,\lambda }}(y) \leqslant u(y), \quad \forall \left| {y - x} \right| \geqslant \lambda>0 }$.

It now follows from the first fundamental lemma [here] that $u$ is constant which contradiction with the integral equation.