^{1}

^{*}

^{1}

In this paper, we are concerned with a positive solution of the non-homogeneous A-Laplacian equation in an open bounded connected domain. We use moving planes method to prove that the domain is a ball and the solution is radially symmetric.

In this paper, we are going to study the symmetry results for the overdetermined problem

Here is a bounded connected open subset of with boundary and is a point in. The function satisfies the regularity requirement

and the (possibly degenerate) elliptic condition

is a continuously differentiable function. is a constant and denotes the inner normal to.

J. Serrin proved the radial symmetry for positive solutions of the equation in with the same overdetermined boundary conditions as the above problem, see [

with the same boundary conditionssee [

Our main result is that for the problem (1.1)-(1.3), if u has only one critical point in, then is a ball and u is radially symmetric.

Section 2 of this paper is devoted to the main result and a more general version of this theorem. In Section 3, we will present the proof of the main theorem.

Some components, such as multi-leveled equations, graphics, and tables are not prescribed, although the various table text styles are provided. The formatter will need to create these components, incorporating the applicable criteria that follow.

In this section we give some lemma that we shall use and present our main result.

Lemma 2.1. (The boundary lemma at corner) (Lemma 2 in [^{2} boundary and be a hyperplane containing the normal to at some point. Let denote the portion of lying on some particular side of.

Suppose that is of class in the closure of and satisfies the elliptic inequality

,

where the coefficients are uniformly bounded. We assume that the matrix is uniformly definite

, and that

, where is an arbitrary real vector, is the unit normal to the plane, and is the distance from. Suppose also in and at. Let be any direction at which enters nontangentially. Then

or atunless.

Our main results are as follows:

Theorem 2.2. Let be a bounded connected open subset of with boundary and let be a point in

. Let, , be a strictly positive solution of the following overdetermined boundary value problem

Here is a continuously differentiable function, and

c is a constant and denotes the inner normal to. Assume

then is a ball and is radially symmetric.

The following remark is a general version of the theorem. It can be viewed as an extension result of p-Laplacian too. As the proof is similar to Theorem 2.2, we omit it.

Remark 2.3. Let be as in Theorem 2.2 and D be a subset of. Let be a strictly positive solution of Equation (2.1) in and verify the boundary conditions; Assume that is the critical set of, then if denotes the convex hull of1) the normal line to at an arbitrary point of intersects;

2) if is a support plane to through and is a ray from A orthogonal to which lies in the half-space determined by not containing, then intersects exactly in one point.

In what follows we assume that the origin of the coordinates system is an interior point of, and we denote with the closure of the ball centered in with radius.

Theorem 2.4. Assume that the hypotheses of Theorem 2.2 hold and furthermore assume that

for some positive. Then 1) is starshaped with respect to;

2) if

;

;

then

.

The technique we are going to use is the moving planes method. For the detailed description about moving planes method, see [

Proof. Step 1: To prove is a ball.

If we can demonstrate that for any point Q on, P lies on the normal line to at Q, then is a ball with centre P. To do this, we argue by contradiction.

Assume that there exists a point such that the normal line to at Q does not contain P. We choose a coordinate system in such that , , and the x_{n} axis coincides with r.

When we use the moving planes method, we choose a family of hyperplanes normal to the axis. Define hyperplan for any positive; Let be the infimum of such that; Define for and we denote by the reflection in. Since is, for some close to, v, we have

As decreases, condition (3.1) holds until one of the following facts happens:

1) is internally tangent to at some point of;

2) intersects at some point of.

Let be the greatest value of, , such that either condition a) or b) is true. Since is orthogonal to at, we have and then for any in. This is the crucial point of our proof. We have found a direction such that as the moving plane moves from to the critical position, it never intersects, so that the moving planes method may be applied.

Let be the reflected point of x in. We defined

for, ,

From Equation (2.1) we have for,

By the definition of v, we obtain

Differencing Equations (3.2) and (3.3) yields

Meanwhile, (3.4) can also be rewritten into

Denote, ,

.

Let

By the mean value theorem, it follows from (3.5) that

where, and c are certain functions depending on u and f. Here the matrix is uniformly positive definite, since both expressions and have this property (recall that Equation (2.1) is elliptic). So (3.6) is uniformly elliptic with bounded coefficients far from, i.e. in where is a ball centered in with radius, for any positive.

From the boundary condition (2.3) on the normal derivative of, it follows that

in (3.7)

for some sufficiently close to. Let

. We prove.

Assume, by continuity, in. On the other hand, since is not symmetric with respect to, in. By the strong version of the maximum principle, we obtain in

. Next we observe that can not be a critical point for w since while

. So as is arbitrarily small, it is

in. Since, we may apply the Hopf lemma to at each point of, we get

on (3.8)

The plane is not normal to at any point, then from inequality (3.8) and the boundary condition (2.3) on the normal derivative of, we get

By the definition of, there exists a sequence such that and

Let be a limit point for x_{n} in the closure ofby continuity, thus. But from inequality (3.10) and the mean value theorem we get and this contradicts condition (3.9).

So is proved.

Now we will prove that u must be symmetric with respect to. Assume in, so as we did for, we infer in.

Assume next that condition a) holds, then is internally tangent to at some point, where. Since P is an interior point of, , so that we can apply the Hopf lemma to w at M and we obtain

;

where is the inner normal to at M. For

we get the contradiction. Hence condition 2) must be true, i.e. is orthogonal to at some point B. From the boundary condition (2.3) and the definition of w it follows that all the first and second derivatives of w vanish at B. On the other hand, as, Equation (3.6) is uniformly elliptic with bounded coefficents in a neighborhood of B, so that the boundary lemma at corner in [

or

Then we have again a contradiction with the derivatives of w at B, so in. But this last inequality can not be true since otherwise w would be a function symmetric in whose only critical point is not on.

This completes the proof of Theorem 2.1.