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<n.

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.

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