Ngô Quốc Anh

May 6, 2010

The method of moving planes: The case of the whole space

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 15:06

The trick considered here can be applied successfully to any bounded set \Omega that has appropriate symmetry under the reflection in a hyperplane. But when the PDE holds in the whole space \mathbb R^n, and no boundary condition is available, one must first locate a possible center of symmetry by means of information about u(x) for |x|\to \infty and then try to derive some estimates.

Definition. Let u \in C^1(\mathbb R^n). We shall say that u has admissible assymptotic behavior iff one of the following conditions holds outside some ball, say for r=|x| \geqslant R_u. The numbers \kappa, m, \delta and \gamma are all strictly positive constants, with \delta \in (0,1] and c is a constant vector.

(A) u(x)=-\kappa r^m + (c \cdot x) r^{m-2} + h(x) where |\nabla h(x)|=O(r^{m-2-\delta} and if m-1<\delta then h(x) \to 0 as r \to \infty.

(B) u(x)=\kappa\log\frac{\gamma}{r} + (c \cdot x) r^{-2} + h(x) where |\nabla h(x)|=O(r^{-2-\delta}) and h(x)\to 0 as r \to \infty.

(C) u(x)=\kappa r^{-m} + (c \cdot x) r^{-m-2} + h(x) where |\nabla h(x)|=O(r^{-m-2-\delta}) and h(x)\to 0 as r \to \infty.

The essential features of the above definition are these.

  • First, u(x) decreases as r \to \infty.
  • Second, if c \ne 0, then c \cdot x does not depend only on r.

Therefore if u should turn out to be spherically symmetric about some point q then u(x+q) will contain no term (c\cdot x)r^\alpha. This enables us to find a priori the only point q that can be a center of symmetry of the function u.

In fact in the literature, we have the following result

Theorem. If solution u \in C^1(\mathbb R^n) of the following PDE

-\Delta u = f(u)

has admissible asymptotic behavior and u>0 in the case (C). Then under some restriction on the nonlinearity f, the following function

v(x) := u(x+q)

is spherically symmetric where q is given by

\displaystyle q = \left\{ \begin{gathered} \frac{1}{{\kappa m}}c, {\rm \; for \; cases \; (\mathbf{A}) \; and \; (\mathbf{C}) } \hfill \\ \frac{1}{\kappa }c, {\rm \; for \; case \; (\mathbf{B})}. \hfill \\ \end{gathered} \right.

Idea of the proof. In order to make use the lemma stated here we need to prove v satisfies v(x^{0,k})=v(x) for any unit vector k and all x. However, it is worth noticing that the function v still verifies the PDE, more than that, it has every good property of u and has better asymptotic behavior in that the second-order terms in the expansion of u are absent from the expansion of v.

We now use a trick to deal with the proof of v(x^{0,k})=v(x). In fact, what we need is to use a suitable rotation, precisely, by an orthogonal n \times n matrix A such that the first coordinate, say \overline x_1, of

\overline x = A x

equals x \cdot k, i.e. \overline x_1 = x \cdot k. In other words, in the new coordinates system, the first axis is parallel to vector k. We denote

\overline v(\overline x) = v(x).

Obviously, it is sufficient to show that \overline v(\overline x_1, \overline x')=\overline v(-\overline x_1, \overline x') where \overline x = (\overline x_1, \overline x') \in \mathbb R \times \mathbb R^{n-1}. For the sake of simplicity, all points are regarded in the new coordinates system and thus we omit the bar as long as there is no room for confusion. The reflection point, usually called x^{\mu, k}, will be denoted by x^\mu. Precisely, x^\mu = (2\mu-x_1, x').

The proof of this fact can be derived through the following lemmas.

Lemma 1. There is a number R(\mu) depending only on \overline v and \mu such that

if

\mu>0, \qquad y \in Y(\mu), \qquad |y| > R(\mu)

then

\overline v(y^\mu)<\overline v(y)

where Y(\mu) is defined to be the set of all points z lying in \overline x_1 such that |\overline z_1| < \mu.

The geometric meaning of this lemma can be seen from the following picture.

This lemma can be thought as the necessarily condition since once \overline v is symmetric, the statement in the lemma must hold.

Lemma 2. There is a number \mu_\star >1 such that

\mu>\mu_\star, \qquad y \in Y(\mu)

then

\overline v(y^\mu)<\overline v(y).

Next lemma is the heart of the matter.

Lemma 3. Assume for some \mu>0 we have

\overline v(y^\mu)<\overline v(y), \quad \forall y \in Y(\mu)

and there is y_0 \in Y(\mu) so that

\overline v(y_0^\mu) \ne \overline v(y_0).

Then

\overline v(y^\mu)<\overline v(y), \quad \forall y \in  Y(\mu)

The last lemma is also important.

Lemma 4. The set

\{ \mu>0: \overline v(y^\mu)<\overline v(y), \quad \forall y \in  Y(\mu)\}

is open in \mathbb R.

We are now in a position to prove the main theorem. Let (m, \infty) be the largest open interval of \mu in which the following claim

\overline v(y^\mu)<\overline v(y), \quad \forall y \in   Y(\mu)

holds. This m exists in view of lemmas 2 and 4. Assume m>0, with the help of lemma 1 and a continuity argument we are able to show that the hypotheses preceding in lemma 3 hold for \mu=m, and thus by mean of lemma 4 the interval (m, \infty) is not maximal. This proves m=0. We now have

\overline v(x_1, x')\leqslant \overline v(-x_1, x')

whenever x_1 \geqslant 0. Keeping the coordinates we repeat the argument for the unit vector -k, this yields

\overline v(x_1, x')\geqslant  \overline v(-x_1, x')

whenever x_1 \geqslant 0. Thus \overline v is an even function of x_1 for arbitrary unit vector k. Hence v is spherically symmetric by a lemma in this entry.

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

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