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

Definition. Let . We shall say that has admissible assymptotic behavior iff one of the following conditions holds outside some ball, say for . The numbers , , and are all strictly positive constants, with and is a constant vector.(

A) where and if then as .(

B) where and as .(

C) where and as .

The essential features of the above definition are these.

- First, decreases as .
- Second, if , then does not depend only on .

Therefore if should turn out to be spherically symmetric about some point then will contain no term . This enables us to find a priori the only point that can be a center of symmetry of the function .

In fact in the literature, we have the following result

Theorem. If solution of the following PDEhas admissible asymptotic behavior and in the case (

C). Then under some restriction on the nonlinearity , the following functionis spherically symmetric where is given by

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

We now use a trick to deal with the proof of . In fact, what we need is to use a suitable rotation, precisely, by an orthogonal matrix such that the first coordinate, say , of

equals , i.e. . In other words, in the new coordinates system, the first axis is parallel to vector . We denote

.

Obviously, it is sufficient to show that where . 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 , will be denoted by . Precisely, .

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

Lemma 1. There is a number depending only on and such thatif

then

where is defined to be the set of all points lying in such that .

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

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

Lemma 2. There is a number such thatthen

.

Next lemma is the heart of the matter.

Lemma 3. Assume for some we haveand there is so that

.

Then

The last lemma is also important.

Lemma 4. The setis open in .

We are now in a position to prove the main theorem. Let be the largest open interval of in which the following claim

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

whenever . Keeping the coordinates we repeat the argument for the unit vector , this yields

whenever . Thus is an even function of for arbitrary unit vector . Hence 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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