Symmetry Problems for PDE

: The results of this paper allow one to derive several results of general interest: to prove the Schiffer’s conjecture, to solve the Pompeiu problem, to prove two symmetry results in harmonic analysis and to give a new method for solving an old symmetry problem.


Introduction
Let D be a bounded connected domain in R m , m ≥ 2, with a smooth boundary S, N is the outer unit normal to S, k > 0 is a constant, u N := u N | S is the limiting value of the normal derivative of u on S from D. Denote by |D| the volume of D, by |S| the area of S, and by J p (r) the Bessel function regular at r = 0. Various symmetry problems were considered by the author in [1][2][3][4][5][6].
Consider the problem ∇ 2 u + k 2 u = c 0 in D, u| S = c 1 , u N = c 2 , c j = const, j = 0, 1, 2, k = const > 0. (1) This problem is over-determined and, in general, does not have a solution. Suppose it has a solution. Our first result is the following theorem: Theorem 1. If |C 1 | + |c 2 | > 0, C 1 := c 1 − c 0 k 2 , and problem (1) is solvable, then D is a ball.
(c) two symmetry problems in harmonic analysis are reduced to Theorem 1, see [6]; these results are formulated in Theorem A in this paper in Section 4.
(d) c 0 = 1, k = 0, c 1 = 0, c 2 = |D| |S| , where |D| is the volume of D and |S| is the area of S. By different methods, this problem was treated in [2], and in [7]. The proofs in this paper are shorter than in the author's earlier works. This paper contains some new ideas.
This paper is organized as follows: In Section 2, reduction of problem (1) is given to the case when c 0 = 0. In Section 3, a proof of Theorem 1 is given. In Section 4, a proof of symmetry properties in harmonic analysis is given, see Theorem A. In Section 5, a proof of the Schiffer's conjecture is given, See Theorem B. In Section 6, the Pompeiu problem is solved, see Theorem C. In Section 7, the set of possible values of the radii of the balls in the Pompeiu problem and in the Schiffer's conjecture are given. In Section 8, a symmetry problem for the Laplace equation is considered, see Theorem 2. In Section 9, conclusions are briefly stated.

Reduction of Problem (1) to the Case c 0 = 0
Define v := u + C, C = const. Then, Choose C = − c 0 k 2 . Then, v solves the problem Let L ⊂ S be the boundary curve of a normal section P of D, that is, the plane passing through the normal N = N s 0 to S at a point s 0 .

Proof of Theorem 1
Let us start with m = 3. The proof is the same for m > 3. Let P j3 be the normal sections of D along the principal directions j = 1, 2. These directions are orthogonal; they correspond to the sections for which the curvatures at s 0 are minimal and maximal for m = 3. The section P j3 is the plane x j x 3 passing through the normal N to S at the origin s 0 . The x 3 axis is directed along N. Let L 1 be the boundary of the intersection of P 13 with S. The directions x j , 1 ≤ j ≤ 2, are the principal directions, so that the curvature k j of the normal section P j3 at the point s 0 is k j (s 0 ). If S is a smooth convex surface, then the Gauss curvature of the surface S at a point s 0 is k(s 0 ) := k 1 (s 0 )k 2 (s 0 ) and the mean curvature κ := k 1 +k 2 2 for m = 3. A point s 0 ∈ S is called umbilic if all the normal sections have the same curvatures at this point. The surface S is locally spherical at an umbilic point.

Lemma 1.
If the Gauss curvature of a smooth closed surface S ⊂ R m is a positive constant, then S is a sphere. If the mean curvature of a closed smooth surface S ⊂ R m is a positive constant, then S is a sphere. If all the points of a smooth closed surface S ⊂ R m are umbilic, then S is a sphere.
Consider now the case m = 2. Let s be the curve length of L 1 , s be the point on L 1 , corresponding to the parameter s, x(s) := x 1 (s), y(s) := x 3 (s), be the parametric representation of L 1 , s = x(s)e 1 + y(s)e 2 , where {e 1 , e 2 } is a Cartesian basis in P 13 , e 1 is directed along x 1 -axis, e 2 is directed along x 3 -axis, and s 0 := s(s 0 ) plays the role of the origin. It is known that the vector t(s) = ds ds is the unit vector, tangential to L 1 at the point s, and dt ds = k 1 (s)ν(s), where k 1 (s) ≥ 0 is the curvature of L 1 and ν(s) is the unit normal to L. Because P 13 is the normal section, one has N = −ν if S is convex. Since u N = ∇u · N = c 2 = const on S, the convexity of S does not change sign, so ν = ν(s) does not change sign. We have on L 1 . If problem (1) has a solution, then problem (3) has a solution and vice versa. We prove that, if |C 1 | + |c 2 | > 0 and (3) has a solution, then D is a ball. This is equivalent to Theorem 1.

Remark 1.
If C 1 = c 2 = 0, then v = 0 in D regardless of the shape of D. Therefore, the condition |C 1 | + |c 2 | > 0 is necessary for the solvability of problem (1) to imply that D is a ball. Theorem 1 shows that this condition and the solvability of problem (1) are sufficient for D to be a ball.
Let us prove Theorem 1 for m = 2. Let s = s 0 in (7), where s 0 is an initial value of s. If m = 2, then Equation (3) implies: Take the coordinate system in which t 2 = 0, t 1 = 1 in (7), and get: Take the coordinate system in which t 2 = 1, t 1 = 0 in (7), and get: If one uses Equation (8), then one gets ∆v = −k 2 C 1 . From (11), one obtains ∆v = 2c 2 k 1 (s 0 ). Therefore, −k 2 C 1 = 2c 2 k 1 (s 0 ), so The right side of (12) does not depend on s 0 . Therefore, k 1 (s 0 ) is a constant. If the curvature of a closed smooth curve S in R 2 is a constant, then S is a circle. Let us give a different argument. Its advantage is the choice of the equation with coefficients invariant with respect to rotations of the coordinate system. Let us write the left side of Equation (7) as where A is the 2 × 2 symmetric real-valued matrix defined by this equation One has t = h j W j , here and below summation is understood over the repeated indices, t j = (t, W j ), j = 1, 2, Remember that t 2 1 + t 2 2 = 1 and choose the coordinate system in which t 1 = 1, t 2 = 0. With this choice of t, Equation (13) yields: Take t 1 = 0, t 2 = 1. Then, Equation (13) yields: Since we obtain Formula (12)  If C 1 = 0 and c 2 = 0, then problem (3) for a ball has a solution if k 2 is the eigenvalue of the Dirichlet Laplacian in this ball, with an eigenfunction depending only on r, and if the radius of the ball has special value, see Section 7 below.
If k 2 is not a Neumann eigenvalue of the Laplacian in D, then the relation c 2 = 0 implies v = 0 in D and C 1 = 0. If k 2 is not a Dirichlet eigenvalue of the Laplacian in D, then the relation One can prove that, if problem (3) has a solution and c 2 = 0, then k 2 is simultaneously a Neumann and a Dirichlet eigenvalue of the Laplacian in D, see ( [1], p. 408).
To prove the last statement, we note that, if c 2 = 0 and (3) is solvable, then k 2 is a Neumann eigenvalue. If v = const on S, then ∇v = 0 on S because the derivative in the direction N, orthogonal to S, equals to zero, and the derivative in the direction tangential to S equals to zero because v = const on S. Since ∇v is a linear combination of derivatives along N and perpendicular to N (tangential to S), it is equal to zero on S. Therefore, each of the components of ∇v is a Dirichlet eigenfunction of the Laplacian in D with the eigenvalue k 2 . Thus, k 2 is the Dirichlet eigenvalue of the Laplacian in D. By the Laplacian, we mean here −∇ 2 .
Consider a new argument, based on the Green's formula, for proving that, if v solves problem (3), then S has to be a sphere.
Let g = g(x, y) be the Green's function of the operator ∆ + k 2 in R m , m ≥ 2, satisfying the radiation condition. Since k 2 > 0, the differential equation and the radiation condition determine g uniquely. By the Green's formula, one gets an integral representation of the solution to problem (3): Let x → t ∈ S, x ∈ D, in Formula (14). Then, using Formula (1.1.3) (a) from [10], the limiting value on S of the potential of double layer, one derives: Equation (16) is equivalent to If one proves that Equation (17) may hold only if S is a sphere, then Theorem 1 is proved. Without loss of generality, one may take C 1 = 1 if c 2 = 0, and, if C 1 = 0, then one may take c 2 = 1.
Using the known formulas for the limiting values on S of the potential of the double layer and for the normal derivative of the potential of the single layer (see, e.g., [10], p. 18, x → t means convergence in non-tangential to S directions): one derives Formulas (18) and (19) below. For c 2 = 0 and C 1 = 1, one obtains from (17) the following: For c 2 = 1 and C 1 = 0, one obtains from (14) the relations: Formula (18) implies that S is a sphere and so does Formula (19). In both cases, the proof is essentially the same. Let us give a proof for Formula (18).
Proof. If S is not a sphere, then the behavior of g N s (t, s), as s → t, depends on the location of t on S, it depends on the curvature at the point t of a normal section of S which contains the normal to S at the point t ∈ S. Only for a sphere (all the points of a sphere are umbilic) there is no such dependence and the first integral (18) can be a constant on S. A consequence of Theorem 1 is that the integral operator Qu := S g(t, s)φ(s)ds can have an eigenvalue λ = 0 with constant eigenfunction φ(s) = 1 only if S is a sphere, k 2 is the eigenvalue of the Dirichlet Laplacian in D and u N | S = const = 0, where u(x) := S g(x, s)ds, u| S = 0.
Let us give another method for proving Theorem 1.
Write Equation (7) as (At, t) = c 2 k 1 (s), where, for m = 2, t 2 1 + t 2 2 = 1 and A is a real-valued symmetric matrix, A = A , with the elements: The eigenvalues λ 1 and λ 2 of A are invariant with respect to the rotation of the coordinate axes.  where the matrix U −1 AU is diagonal. Since the matrix U describes a rotation, it preserves the length of vectors, and 1 = t 2 One may find a rotation (by the angle π/2) that sends vector (τ 1 , τ 2 ) onto (τ 2 , −τ 1 ). Then, Formula (20) takes the form: Add Equations (20) and (21) and use the formula τ 2 1 + τ 2 2 = 1 to get: The sum λ 1 + λ 2 = (u xx + u yy )| S = −k 2 C 1 is the trace of the symmetric real-valued matrix A. One has: Consequently, we obtain Formula (12) again: Since C 1 and c 2 are constants, the curvature k 1 (s 0 ) does not depend on s 0 ∈ S. It is constant on S = L 1 for m = 2. By Lemma 1, a closed smooth curve with a constant curvature is a circle (a ball in R 2 ) and its curvature k 1 > 0 is positive. This is a different method to prove Theorem 1 for m = 2. Its ideas can be used when m > 2.
Assume now that m > 2. By considering the normal sections P jm , one derives, as in the case m = 2, the formulas: Since v x m x m | s=s 0 does not depend on the normal section, it does not depend on j. Therefore, The function v x m x m | s=s 0 is a continuous function of s 0 on S. It does not depend on j. Consequently, k j (s 0 ) does not depend on j, so k j (s 0 ) = k(s 0 ), 1 ≤ j ≤ m − 1. Therefore, the point s 0 is umbilic. Since s 0 is an arbitrary point of S, all the points of S are umbilic. By Lemma 1, S is a sphere. Theorem 1 is proved for m > 2.
The quantity k(s 0 ) is equal to the mean curvature κ(s 0 ) of S: Let us derive a formula for the mean curvature κ(s) in terms of C 1 , c 2 , and m.
Put the x p -axis along the outer unit normal N at the point s 0 ∈ S and let P jp , j = p, be the normal sections. The normal sections P jp , j = p, depend on j but do not depend on p. We can use m − 1 normal sections. As in the case m = 2, one derives: Equation (3) on the boundary S yields: Using (25), one obtains (compare with Equation (11)): Therefore, where κ(s 0 ) is the mean curvature of S at the point s 0 . The right side of (28) does not depend on p. Therefore, its left side does not depend on p and from (28) we get: On the other hand, from Equations (26) and (29), we have: −k 2 C 1 = mc 2 κ(s 0 ) on S. Since c 2 and C 1 are constants independent of s 0 , we conclude, assuming that c 2 = 0, that κ(s) is constant on S: If κ(s) = const on S and S is a smooth closed surface, then, by Lemma 1, S is a sphere. Remark 2 is also valid for Formula (30). This is another proof of Theorem 1 for c 2 = 0 and C 1 = 0.

Symmetry Problems in Harmonic Analysis
One can derive from Theorem 1 the following symmetry results in harmonic analysis. Let D ⊂ R 3 be a bounded connected domain with a closed Lipschitz surface S, Ω := R 3 \ D, S 2 be the unit sphere in R 3 , k = const > 0.
Theorem A. (a) If D e ikα·s ds = 0 for all α ∈ S 2 , then D is a ball.
(b) If S e ikα·s ds = 0 for all α ∈ S 2 , then S is a sphere.

Lemma 2. From the assumption
it follows that u = 0 in Ω, where u is defined in (31).
Proof. Let us continue with the proof of Theorem A. From Lemma 2, it follows that u solves the problem: Problem (34) is problem (1) with c 0 = −1, c 1 = c 2 = 0. By Theorem 1, it follows that D is a ball. In this case C 1 = 1 k 2 , c 2 = 0. Part (a) of Theorem A is proved. (b) Let us prove part (b) of Theorem A. Define u := S g(x, s)ds. By the assumption S e ikα·s ds = 0 ∀α ∈ S 2 , it follows, as above, that u = 0 in Ω. Consequently, where the equation u N | S = 1 follows from the known relation (see [10], p. 18): , the limiting values of the normal derivative on S from Ω (D) are denoted.
Problem (35) is a particular case of problem (1). It follows that D is a ball and S is a sphere. In this case, C 1 = 0, c 2 = 1. Part (b) of Theorem A is proved.
Theorem A is proved.

Proof of the Schiffer's Conjecture
Let us prove the Schiffer's conjecture.
Theorem B. Assume that the problem is solvable. Then, D is a ball.
Proof of Theorem B. Problem (36) is a particular case of problem (1). The conclusion of Theorem B follows from Theorem 1. Theorem B is proved.

Solution to the Pompeiu Problem
Let us solve the Pompeiu problem. This problem is discussed in [10][11][12]. In modern formulation, the problem can be stated as follows: Assume that where σ is the set of all rigid motions, that is, translations and rotations, f ∈ L 1 loc (R m ) and D ⊂ R m is a bounded connected domain with a Lipschitz boundary S.
The modern formulation of the Pompeiu problem is: It is known ( see [1]) that, if D is a ball, then (37) may hold for f ≡ 0. Let us rewrite Equation (37) as where G is the group of rotations and y describes the translations. Define the Fourier transform by the formulaf := R m f (x)e ix·ξ dx. Take the Fourier transform of (38) and getf (ξ)χ(g −1 ξ) = 0, where χ(x) is the characteristic function of D and the overline stands for a complex conjugate. For the suppf to be non-void, that is, for f ≡ 0, it is necessary and sufficient that the set of zeros ofχ(g −1 ξ) be non-void. Since D is bounded, the functionχ(ξ) is an entire function of ξ. Since g ∈ G is arbitrary, the set of zeros ofχ(ξ) must be spherically symmetric. Let C k be the spherical surface of zeros ofχ(ξ), where k > 0 is the radius of the sphere C k . Therefore, if and only if f ≡ 0, there exists a k > 0 such that C k is the spherical surface of zeros ofχ(ξ). Then, by Theorem A, it follows that D is a ball. We have proved the following: Theorem C. Equation (37) can hold for f ≡ 0 if and only if D is a ball.

Possible Values of the Radii of the Balls in Theorem 1
Can the radius a of the ball in Theorem 1 be arbitrary? Let us consider the case m = 3, which is the main interest physically. The spherically symmetric solution of Equation (3) in the spherical coordinates, regular at the origin, is cj 0 (kr), where j 0 (r) := J 1/2 (kr) r 1/2 , c = const, substitute this function into the boundary conditions (3) for a sphere of radius a and get: cj 0 (ka) = C 1 , ckj 0 (ka) = c 2 . Eliminating c yields kC 1 j 0 (ka) = c 2 j 0 (ka).
Any possible radius a of the ball in Theorem 1 should solve Equation (40).
In the Pompeiu problem and Theorem A, part (a), C 1 = 1 k 2 , c 2 = 0, and Equation (40) yields equation j 0 (ka) = 0 for all possible values of the radius a.

A Symmetry Problem for the Laplace Equation
Assume that ∇ 2 u = c 0 in D, u| S = 0, u N = c 2 , c j = const, j = 0, 2, c 0 |D| = c 2 |S|, |c 0 | > 0. (41) The assumptions on D and S are the same as in the previous sections. Our result is the following.

Theorem 2. If problem (41) is solvable, then D is a ball.
Proof of Theorem 2. First, note that the condition |c 0 || > 0 is necessary: if c 0 = 0, then, for any shape of D, problem (41) has only the trivial solution u = 0 and c 2 = 0.
Secondly, integrating Equation (41) over D and using the formula one concludes that the condition c 0 |D| = c 2 |S| is necessary for the solvability of problem (41).
To prove that the solvability of problem (41) implies that D is a ball, we argue as in the proof of Theorem 1 and obtain the equation similar to (26): and the relation v x j x j (s 0 ) + v x p x p (s 0 ) = 2c 2 k j (s 0 ), j = p.
As in the proof of Theorem 1, we conclude that κ(s) = const > 0 on S, where κ is the mean curvature of S. If the mean curvature of a closed smooth surface S is a positive constant, then, by Lemma 1, S is a sphere, and D is a ball. Theorem 2 is proved.

Conclusions
In this paper, new results on symmetry problems for PDE are established. These results are formulated in Theorems 1 and 2 and Theorems A, B, and C. The symmetry results for PDE are significant and interesting.