Ngô Quốc Anh

May 4, 2010

The method of moving planes: An introduction

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 1:17

Let us go back to the following problem that we have discussed before in this topic. Denote by B_a the ball centered at the origin with radius a>0. Here we just consider our problem within a ball. We will take care the case of whole space \mathbb R^n.

Theorem. Suppose u \in C(\bar B_a) \cap C^2(B_a) is a positive solution of

-\Delta u = f(u) in B_a and u=0 on \partial B_a

where f is locally Lipschitz in \mathbb R. Then u is radially symmetric in B_a.

Since we haven’t provided any solutions to this problem, here I just figure out the idea and several steps needed during the proof.

Definition. For each unit vector k \in \mathbb R^n and each number \mu \in \mathbb R, define a hyperplane by

T_\mu(k)=\{\xi \in \mathbb R^n : \xi \cdot k = \mu\}.

The reflection in T_\mu(k) of any point x \in \mathbb R^n is

x^{\mu, k}=x+2(\mu - x \cdot k)k

and the reflection in T_\mu(k) of any function \varphi : \mathbb R^n \to \mathbb R is defined by

\varphi_{\mu,k}(x)=\varphi(x^{\mu,k})

for all x \in \mathbb R^n.

Obviously, T_\mu(k) is parallel to T_{\mu'}(k) for any \mu \ne \mu'. This is because vector k is perpendicular to T_\mu(k) for any \mu. Since vector k is chosen to be a unit vector, it is easy to check that the distance from the origin to T_\mu(k) exactly equals \mu. The following lemma plays an important role in our argument

Lemma. If v : \mathbb R^n \to \mathbb R has the property that

v(x^{0,k})=v(x)

for each unit vector k and for all x \in \mathbb R^n, then v is spherically symmetric (i.e. depends only on |x|).

Proof. We only need to prove that for any two points y and z satisfying |y|=|z| and y \ne z we have v(y)=v(z).

Indeed, if we choose k=\frac{z-y}{|z-y|} we are able to prove z=v^{0,k}. To check that,  we have

\displaystyle\begin{gathered} {y^{0,k}} = y + 2\left( {0 - y \cdot k} \right)k \hfill \\ \qquad= y + 2\left( {0 - y \cdot \frac{{z - y}}{{|z - y|}}} \right)\frac{{z - y}}{{|z - y|}} \hfill \\ \qquad= z + y - z + 2\left( {0 - y \cdot \frac{{z - y}}{{|z - y|}}} \right)\frac{{z - y}}{{|z - y|}} \hfill \\ \qquad= z + y - z - 2(z - y)y \cdot \frac{{z - y}}{{|z - y{|^2}}} \hfill \\ \qquad= z - \frac{{z - y}}{{|z - y{|^2}}}\left( {|z{|^2} - 2z \cdot y + |y{|^2} + 2\left( {y \cdot z - |y{|^2}} \right)} \right) \hfill \\ \qquad= z. \hfill \\ \end{gathered}

The proof of lemma follows.

We are now in a position to derive a preliminary sketch of the method of reflection in hyperplanes. Our aim is to show that solutions to the problem satisfy

u(x^{0,k})=u(x)

for each unit vector k and all x\in B_a.

To this end, we fix a unit vector k and contemplate all positions of the hyperplane T_\mu(k) from \mu=a to \mu=0. Let

Z(\mu,k) := \{z \in B_a :z \cdot k>\mu\}.

The procedure is to prove that (for fixed but arbitrary k)

w(z,\mu) := u(z) - u (z^{\mu,k}) < 0

if z \in Z(\mu,k) whenever \mu \in (0, a).The proof of the above estimate proceeds in two stages. The first is for small positive value a-\mu, i.e. when \mu is very closed to a, say \mu_0<\mu<a. The key ingredient used here is to apply the maximum principle in the thin domains. Once this stage can be done, we are then able to start moving the plane continuously to the left as long as the above estimate still holds. The second stage is a demonstration that one the estimate holds in an interval (\mu_0,a) it can fail only as \mu \downarrow 0. However, this is not the case. Once this whole argument has been established, a continuity argument for \mu \downarrow 0 shows that

w(z, 0) < 0

if z \in Z(0, k); in other words, that

u(x) < u(x^{0,k})

if x\cdot k > 0 and x\in B_a. But the same result for the direction -k implies that

u (x^{0,k}) < u(x) if x\cdot k > 0 and x\in B_a.

These two inequalities yield the desired result that u(x^{0,k}) = u(x) for all x.

See also: L.E. Fraenkel, An introduction to maximum principles and symmetry in elliptic problems, Cambridge, 2000.

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