Ngô Quốc Anh

May 3, 2010

The second fundamental lemma in the method of moving spheres

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 3:32

Today we continue to discuss the second fundamental lemma in the method of moving spheres. This lemma also comes from a paper due to Y.Y. Li published in J. Eur. Math. Soc. (2004).

Recall from the previous entry where the first lemma was considered if the following

\displaystyle {\left( {\frac{\lambda }{{|y - x|}}} \right)^\nu }f\left( {x + {\lambda ^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) \leqslant f(y), \quad \forall |y - x| > \lambda > 0

holds then f is constant or \pm \infty. We now consider the equality case. Precisely,

Lemma. Let n\geqslant 1, \nu \in \mathbb R and f \in C^0(\mathbb R^n). Suppose that for every x \in \mathbb R^n there exists \lambda(x)>0 such that

\displaystyle {\left( {\frac{\lambda(x) }{{|y - x|}}} \right)^\nu  }f\left( {x + {\lambda(x) ^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) =f(y), \quad \forall |y - x| > 0.

Then for some a \geqslant 0, d>0 and \overline x \in \mathbb R^n

\displaystyle f(x) = \pm a{\left( {\frac{1}{{d + {{\left| {x - \overline x } \right|}^2}}}} \right)^{\frac{\nu }{2}}}.

Proof. It follows from the continuity of f that

\displaystyle\alpha : = \mathop {\lim }\limits_{|y| \to \infty } |y{|^\nu }f(y) = \lambda {(x)^\nu }f(x),\quad \forall x \in {\mathbb{R}^n}.

If \nu =0 then f \equiv \alpha and we are done. On the other hand, the case \nu<0 can easily be reduced to the case \nu>0 if we let

\displaystyle z=x+\lambda(x)^2 \frac{y-x}{|y-x|^2}

in the given identity. It is then sufficient to consider the only case \nu>0. If \alpha=0 then f\equiv 0 and we are also done. Otherwise, replacing f by a nonzero multiple of f, we may assume that \alpha=1. Since f(y)\to 0 as |y|\to \infty and since f is continuous and positive, f has a maximum point and we may assume that f has a maximum point at the origin.

For x \in \mathbb R^n, we have for large |y|

\displaystyle\begin{gathered} |y{|^\nu }f(y) = \lambda {(x)^\nu }{\left( {\frac{{|y|}}{{|y - x|}}} \right)^\nu }f\left( {x + \lambda {{(x)}^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) \hfill \\ \qquad= \lambda {(x)^\nu }\left( {1 + \frac{{\nu x \cdot y}}{{|y{|^2}}} + O(|y{|^{ - 2}})} \right)f\left( {x + \lambda {{(x)}^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) \hfill \\ \end{gathered}

which yields

\displaystyle |y|\left( {|y{|^\nu }f(y) - 1} \right) = |y|\lambda {(x)^\nu }\left( {f\left( {x + \lambda {{(x)}^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) - f(x)} \right) + \left( {\frac{{\nu x \cdot y}}{{|y|}} + O(|y{|^{ - 1}})} \right)f\left( {x + \lambda {{(x)}^2}\frac{{y - x}}{{|y - x{|^2}}}} \right).

Taking x=0 in the above and using the fact that f has a maximum point at the origin, we obtain

\displaystyle\mathop {\limsup }\limits_{|y| \to \infty } |y|\left( {|y{|^\nu }f(y) - 1} \right) \leqslant 0

Claim. For any \varepsilon>0 there exists M_\varepsilon such that for any |y| \geqslant M_\varepsilon there exists \widetilde x = \widetilde x(y) satisfying

\displaystyle \widetilde x + \lambda {(\widetilde x)^2}\frac{{y - \widetilde x}}{{|y - \widetilde x{|^2}}} = 0 \quad \forall |\widetilde x| \leqslant \varepsilon

Proof of claim. Remember that

\displaystyle \lambda(x)=f(x)^{-\frac{1}{\nu}}

for all x. For any \varepsilon \in (0,1) pick M_\varepsilon>1 so that

\displaystyle\frac{1}{{{M_\varepsilon } - \varepsilon }}\mathop {\max }\limits_{|x| \leqslant \varepsilon } f{(x)^{ - \frac{1}{\nu }}} < \frac{\varepsilon }{2}.

Then for all |y| \geqslant M_\varepsilon

\displaystyle\mathop {\max }\limits_{|x| \leqslant \varepsilon } \left| {\lambda {{(x)}^2}\frac{{y - x}}{{|y - x{|^2}}}} \right| = \mathop {\max }\limits_{|x| \leqslant \varepsilon } \frac{1}{{|y - x|}}f{(x)^{ - \frac{2}{\nu }}} \leqslant \frac{1}{{{M_\varepsilon } - \varepsilon }}\mathop {\max }\limits_{|x| \leqslant \varepsilon } f{(x)^{ - \frac{1}{\nu }}} < \frac{\varepsilon }{2}.

Thus by a degree argument using the continuity of f, there exists the desired \widetilde x. The proof of claim follows.

With x:=\widetilde x(y) in hand we obtain

\displaystyle\mathop {\lim \inf }\limits_{|y| \to \infty } |y|\left( {|y{|^\nu }f(y) - 1} \right) \geqslant - \varepsilon \nu f(0)\mathop {\max }\limits_{|z| \leqslant \varepsilon } f{(z)^{ - \frac{1}{\nu }}}.

Sending \varepsilon to 0 we have

\displaystyle\mathop {\lim \inf }\limits_{|y| \to \infty }  |y|\left( {|y{|^\nu }f(y) - 1} \right) \geqslant 0.

Thus

\displaystyle\mathop {\lim \inf }\limits_{|y| \to \infty }  |y|\left( {|y{|^\nu }f(y) - 1} \right) =0.

Let

\displaystyle\begin{gathered} {e_1} = \left( {1,0,0,...,0} \right), \hfill \\ {e_2} = \left( {0,1,0,...,0} \right), \hfill \\ \quad \;\vdots \hfill \\ {e_n} = \left( {0,0,0,...,1} \right). \hfill \\ \end{gathered}

For any x\in \mathbb R^n, 1\leqslant i \leqslant n and t \in \mathbb R let y=y(x,t,i) \in \mathbb R^n defined by

\displaystyle t{e_i} = \lambda {(x)^2}\frac{{y - x}}{{|y - x{|^2}}}.

Taking this y and sending t to 0, we obtain

\displaystyle\frac{{\partial f}}{{\partial {x_i}}}(x) = \mathop {\lim }\limits_{t \to 0} \frac{{f(x + t{e_i}) - f(x)}}{t} = - \frac{{\nu x \cdot {e_i}f(x)}}{{\lambda {{(x)}^2}}} = - \nu {x_i}f{(x)^{1 + \frac{2}{\nu }}}.

By the continuity of f we know that f is in C^1(\mathbb R^n) and the proof follows by writing the above system of PDEs as

\displaystyle\frac{\partial }{{\partial {x_i}}}\left( {f{{(x)}^{ - \frac{2}{\nu }}} - |x{|^2}} \right) = 0

and solving it.

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