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TWO SYMMETRY PROBLEMS IN POTENTIAL THEORY Tewodros Amdeberhan DeVry College of Technology, North Brunswick NJ 08902, USA Abstract. We consider two particular

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TWO SYMMETRY PROBLEMS IN POTENTIAL THEORY Tewodros Amdeberhan DeVry College of Technology, North Brunswick NJ 08902, USA Abstract. We consider two particular overdetermined boundary value problems (OBVP) as variants on J. Serrin's 1971 classical results and prove in both cases that the domains must be Euclidean balls. AMS Subject Classification: Primary 35 Key Words: Overdetermined Problems, Maximum Principles Assume throughout that R N is a bounded domain whose boundary is smooth of class C 2 and containing the origin strictly in its interior. Let ν be the outer unit normal Summation over repeated indices is in effect. In 1971, James Serrin [1] proved the following classical result. Theorem 1: Suppose there exists a function u 2 C 2 ( μ ) satisfying the elliptic differential equation a(u; jpj) u + h(u; jpj)u i u j u ij f(u; jpj) in where a; f and h; p i ;p j are continuously differentiable functions of u and p (here p (u 1 ;:::;u n ) denotes the gradient vector of u:) Suppose also that u 0in and that u satisfies the boundary conditions u 0; constant Then must be a ball and u is radially symmetric. The method of proof combines the Maximum Principles and the device (which goes back to A. D. Alexandroff: every embedded surface inr N with constant mean curvature must be a sphere) of moving planes to a critical position and then showing that the solution is symmetric about the limiting plane. In a subsequent article, H. F. Weinberger [3] gave a simplified proof for the special case of the Poisson differential equation, u 1: Our aim at present is to introduce some variants on Serrin's result and arrive at the same symmetry conclusions by employing elementary arguments. The next statement involves radial dependence on the boundary conditions. Typeset by AMS-TEX 2 TEWODROS AMDEBERHAN Proposition 1: Suppose there exists a solution u 2 C 2 ( μ ) to the overdetermined problem: u 1 in (1) u 0 cr where r p x ::: + x2 N and c is a constant. Then is an N dimensional ball. Before turning to the details we like to discuss some physical motivations for the problem itself. Consider a viscous incompressible fluid moving in straight parallel streamlines through a straight pipe of given cross-sectional form. If we fix rectangular coordinates is space with the z axis directed along the pipe, it is well known that the flow velocity u is then a function of x; y alone satisfying the Poisson differential equation (for N 2) u A in, where A is a constant related to the viscosity of the fluid and to the rate of change of pressure per unit length along the pipe. Supplementary to the differential equation one has the adherence condition u 0 Finally, the tangential stress per unit area on the pipe wall is given by the quantity μ, where μ is the viscosity. Our proposition then states that the ratio of the tangential stress on the pipe wall to its radial distance is the same at all points of the wall if and only if the pipe has a circular cross section. Exactly the same differential equation and boundary condition arise in the linear theory of torsion of a solid straight bar of cross-section ; see [2] pp In light of this, Proposition 1 states that, when a solid straight bar is subject to torsion, the ratio of the magnitude of the resulting traction which occurs at the surface of the bar to the radial distance to the surface is independent of position if and only if the bar has a circular cross-section. Lemma 1: Under the hypothesis of Proposition 1, the following holds (2) udx c 2 r 2 dx: Proof: Let us introduce the auxiliary function h 2u x i u i. Then clearly h0. Green's identities (h u u h) h dff; TWO SYMMETRY PROBLEMS IN POTENTIAL THEORY 3 the boundary conditions on u, and the harmonicity of hlead to (3) hdx c Now we compute the left and right hand sides of (3) individually. Applying the divergence theorem, we get (4) hdx (2 + N)udx (2 + div(xu)dx ux νdff (2 + N)udx as u Since u vanishes on the (therefore ν ± ru ) and kruk r i x i,we gather r that (note: the argument here could have proceeded using the so-called Pohozaev's identity, but we drop it so as not to use any heavy gun ) (5) rhdff rx i u i dff c 2 c2 4 c2 4 c 2 (N +2) Combining (3), (4) and (5) proves the Lemma. Λ r dff r 4 ) dff (r 4 )dx r 2 dx: 4 TEWODROS AMDEBERHAN Proof of Proposition 1: Consider the functional Φ u i u i c 2 r 2. Then, Φ 2u ij u ij +2u i u i 2c 2 N 2u ij u ij 2c 2 N since u is a constant 2 X i;j u ij + ffi ij N N 2c2 N (6) 2 N 2c2 N: For a moment assume that cn» 1.. Then we have, (7) Φ 0in: Note that ν ± ru, since u vanishes on the boundary. Thus wehave kruk (8) Φ 2 c 2 r 2 0: Applying Green's identities and replacing the boundary conditions results in (9) u dff udx c 2 By Lemma 1 above, equation (9) yields ru ru c 2 r 2 dx u udx c 2 r 2 dx: Φ 0: r 2 dx Standard Maximum Principles together with the properties (7)-(9) of Φ imply that Φ 0 in. This in particular forces equality in (6), i.e. u ij + ffi ij N 0: TWO SYMMETRY PROBLEMS IN POTENTIAL THEORY 5 Consequently, u takes the radial form u a r2 2N : Again, since u vanishes on the we obtain that is indeed a ball, as asserted. Λ Remarks: 1. It is not hard to see that the C 2 -smoothness assumptions, on u, made in the preceding Proposition (even Proposition 2, below) could be relaxed except the proofs would then get rather technical. 2. The assumption in the proof of Proposition 1 is not essential. To see this, notice in fact that due to the conditions (1) on u coupled with the divergence theorem verify that Therefore, cn» 1. j dff c rdff rda cn j j from geometry. N 3. Using the same argument and replacing the second boundary condition by crff, where ff 1 (note: this already implies that ff 1» 1), d diam() and 0 c» 2 ff 1 d ff 1p ffn(n +2ff 2) we still get the conclusion of Proposition Even more generally, if f(r) where g(r) f2 (r) satisfies the conditions (rg 0 2g) dx 0; g 00 + N 1 g 0 2 r N» 0 then Proposition 1 holds. Moreover, it turns out that either f c 1 or f c 2 r for some positive constants c 1 and c 2. One of the implications of which is that under these general suppostions, there can only be two possible forms for the boundary derivative: either c 1 as in Theorem 1 of Serrin, or c 2 r as in Proposition 1 of this article! 6 TEWODROS AMDEBERHAN Proposition 2: Suppose there exists a solution u 2 C 2 ( μ ) to following OBVP: u r ff in (10) u 0 x i u i + c 0 r 2+ff + c 1 0 where ff, c 0 and c 1 are constants. If fi : (ff + 2)(c 0 (ff + N) 1) is not a negative integer, then is an N dimensional ball. Proof: Introduce the functional V x i u i + c 0 r ff+2 + c 1 + fiu. Then by direct computation we obtain that V is harmonic. As can easily be seen, the boundary conditions make V Thus, V 0 in and V 0 Classical Maximum Principles show that V 0 inside, too. Hence, it follows that (11) x i u i + c 0 r ff+2 + c 1 + fiu 0 in. Rewriting equation (11) one r fi (u u(0)) + c 0r fi + ff +2 ff+2 r fi 1 fi(u u(0)) + + c 0r 0 by (11) This in turn implies that Hence, r fi (u u(0)) + c 0r fi+ff+2 fi + ff +2 G( ) for angular variables. (12) u u(0) c 0 r ff+2 fi + ff +2 +r fi G( ) Now, if fi 0 then u cannot be regular at the origin unless G( ) 0: If fi 0 is not a negative integer, then once more we must have G( ) 0 for u must satisfy u r ff and r fi G( ) is not harmonic with this value of fi. Therefore, the solution u is radial and r ff+2 u u(0) c 0 fi + ff +2 : After using (10) and the vanishing of u this proves that is a ball thereby completing the proof of the proposition. Λ TWO SYMMETRY PROBLEMS IN POTENTIAL THEORY 7 References [1] J. Serrin, A symmetry problem in potential theory, Arch. Rat. Mech. Anal., 43 (1971), [2] I.S. Sokolinikoff, Mathematical theory of elasticty, New York: McGraw Hill (1956). [3] H.F. Weinberger, Remark on the preceding paper of Serrin, Arch. Rat. Mech. Anal., 43 (1971),

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