A symmetry problem

A novel approach to an old symmetry problem is developed. A new proof is given for the following symmetry problem, studied earlier.


Introduction
Symmetry problems are of interest both theoretically and in applications.A well-known, and still unsolved, symmetry problem is the Pompeiu problem (see [9], [10], and the references therein).In modern formulation this problem consists of proving the following conjecture: If D ⊂ R n , n ≥ 2, is a domain homeomorphic to a ball, and the boundary S of D is smooth (S ∈ C 1,λ , λ > 0, is sufficient), and if the problem where c is a constant, has a solution, then S is a sphere.

Standing assumptions:
In this paper we assume that D ⊂ R 3 is a bounded domain homeomorphic to a ball, with a sufficiently smooth boundary S (S is Lipschitz suffices).
We use the following notation: H is the set of all harmonic functions in B R , R > 0 is an arbitrary large number, such that the ball B R contains D, |D| and |S| are the volume of D and the surface area of S, respectively.
In [12] it is proved that if then D is a ball.The proof in [12] is based on an idea similar to the one we are using in this paper.
In [13] a symmetry problem of interest in elasticity theory is studied by A.D.Alexandrov's method of a moving plane ( [1]), used also in [14].The result in [14], which is formulated below in Theorem 1, was proved in [15] by a method, different from the one given in [14], and discussed also in [2].The argument in [2] remained unclear to the author.
In [5] another symmetry problem of potential theory was studied.
Our goal is to give a new proof of Theorem 1.The result of Theorem 1 was obtained in [14] for R n , n ≥ 2.
Theorem 1.If D ⊂ R 3 is a bounded domain, homeomorphic to a ball, S is its Lipschitz boundary, and the problem: has a solution, then S is a sphere.This result is equivalent to the following result: then S is a sphere.The equivalence of ( 4) and ( 5) can be proved as follows.Suppose (4) holds.Multiply (4) by an arbitrary h ∈ H, integrate by parts and get (6), then one gets c = |D| |S| , so ( 6) is identical to (5).Suppose (5) holds.Then (6) holds.Let v solve the problem ∆v = 1 in D, v S = 0.This v exists and is unique.Using (6), the equation ∆h = 0 in D, and the Green's formula, one gets (7) c We will need the following lemma: Lemma A. The set of restrictions on S of all harmonic functions in D is dense in L 2 (S).
Proof of Lemma A. We give a proof for the convenience of the reader.The proof is borrowed from [12].Suppose that g ∈ L 2 (S), and S g(s)h(s Thus, a single layer potential w, with L 2 density g, vanishes in D ′ , and, by continuity, on S. Since w is a harmonic function in D vanishing on S, it follows that w = 0 in D. By the jump formula for the normal derivative of the single-layer potential across a Lipschitz boundary, one gets g = 0. ✷ Thus, (8) implies v N S = c.Therefore, (4) holds.A result, related to equation ( 5), was studied in [7] for a two-dimensional problem.The arguments in [7] were not quite clear to the author.
Our main result is a new proof of Theorem 1.The proof is simple, and the method of the proof is new.This method can be used in other problems (see [5], [10], [12], [11]).

Proofs
Proof of Theorem 1.We denote by D ′ the complement of D in R 3 , by S 2 the unit sphere, by [a, b] the cross product of two vectors, by g = g(φ) the rotation about an axis, directed along a vector α ∈ S 2 , by the angle φ, and note that if h(x) is a harmonic function in any ball B R , containing D, then h(gx) is also a harmonic function in B R .
Take h = h(g(φ)x) in ( 5), differentiate with respect to φ and then set φ = 0.This yields: Using the divergence theorem, one rewrites this as: Since α ∈ S 2 is arbitrary, one gets Denote y 0 := y/|y|.It is known (see, e.g., [3]) that where the overline stands for the complex conjugate, y 0 is the unit vector characterized by the angles θ, φ in spherical coordinates, Y nm are normalized spherical harmonics: (see [3]).The definitions of P n,m (z) in various books can differ by a factor (−1) m .Using formula (12), one obtains ( 13) Substitute this in (10), equate the terms in front of |y| −(n+1) , and define vectors (14) a nm := S [s, N]|s| n Y nm (s 0 )ds to obtain ( 15) where e θ ,e φ , and e r are orthogonal unit vectors of the spherical coordinate system, [e φ , e r ] is the cross product, [e φ , e r ] = e θ , [e θ , e r ] = −e φ , y = ry 0 , r = |y|, y 0 = (sin θ cos φ, sin θ sin φ, cos θ), ∂ θ = ∂ ∂θ .Therefore, formula (15) can be rewritten as ( 16) From (16) we want to derive that (17) If (17) is established, then it follows from ( 14) and from the completeness in L 2 (S) of the system {|s| n Y nm (s 0 )} n≥0,−n≤m≤n that [s, N] = 0 on S, and this implies that S is a sphere, as follows from Lemma 1 formulated and proved below.Consequently, Theorem 1 is proved as soon as relations (17) are established.The completeness of the system {|s| n Y nm (s 0 )} n≥0,−n≤m≤n in L 2 (S) follows from Lemma B: The functions |x| n Y nm (x 0 ), n ≥ 0, −n ≤ m ≤ n, are harmonic in any ball, centered at the origin.
Lemma B. The set of restrictions of the above functions to any Lipschitz surface homeomorphic to a sphere is complete in L 2 (S).
Proof of Lemma B. The proof is given for completeness.It is similar to the proof of Lemma A. Suppose that g ∈ L 2 (S) and S g(s)|s| n Y nm (s 0 )ds = 0, ∀n ≥ 0, |m| ≤ n.
This and (12) imply that S g(s)(4π|s − y|) −1 ds = 0 ∀y ∈ D ′ , and the argument, given in the proof of Lemma A, yields the desired conclusion g = 0. ✷ Vector a nm is written in Cartesian basis {e j } 1≤j≤3 as The relation between the components F 1 , F 2 , F 3 , of a vector F in Cartesian coordinates and its components F r , F θ , F φ , in spherical coordinates can be found, e.g., in [6], Section 6.5: Using these relations one derives from (16) the following formulas: (18)
The quantities a nm,2 and a nm,3 are proportional to c n,m .Since c n,m = 0, it follows that a nm,2 = a nm,3 = 0, −n ≤ m ≤ n, and Theorem 1 is proved.Thus, to complete the proof of Theorem 1 it is sufficient to verify the linear independence of the system of functions {f j (z)} 3 j=1 on the interval z ∈ [−1, 1].
Proof of Lemma 1.Let s = r(u, v) be a parametric equation of S. Then the vectors r u and r v are linearly independent and N s is directed along the vector [r u , r v ].Thus, the assumption [s, N s ] = 0 on S implies that [r, [r u , r v ]] = r u (r, r v ) − r v (r, r u ) = 0.
Since the vectors r u and r v are linearly independent, it follows that (r, r v ) = (r, r u ) = 0. Thus, (r, r) = R 2 , where R 2 is a constant.This means that S is a sphere.Lemma 1 is proved.✷ N]h(s)ds = S [s, c∇h(s)]ds, ∀h ∈ H, where N = N s is a unit normal to S at the point s ∈ S, pointing into D ′ .Let y ∈ B ′ R be an arbitrary point, and h(x) = 1 |x−y| ∈ H, where x ∈ B R .Then equation (9) implies that (10) v(y) := S [s, N]ds |s − y| = c[∇ ∇ s 1 |s − y| ]ds = c S [ s |s − y| 3 , y]ds = c[∇ y S ds |s − y| , y].Relation (11) actually holds for all y ∈ D ′ , because of the analyticity of its left and right sides in D ′ .Let w(y) := S |s − y| −1 ds.