Two Robin boundary value problems with opposite sign.

Transcription

1 Two Robin boundary value problems with opposite sign. arxiv: v [math.oc] 2 Jun 204 June 3, 204 Catherine Bandle Mathematische Institut, Universität Basel, Rheinsprung 2, CH-405 Basel, Switzerland Alfred Wagner Institut für Mathematik, RWTH Aachen Templergraben 55, D Aachen, Germany Abstract Key words: Local Rayleigh-Faber Krahn inequalities, Robin boundary conditions, domain variation, harmonic transplantation. MSC200: 49K20, 49R05, 5A42, 35J20, 35N25. Abstract In this paper we discuss a boundary value and an eigenvalue problem with Robin boundary conditions of opposite sign. The eigenvalue problem arises in the study of the reaction-diffusion equation with dynamical boundary conditions. The dependence of the energy and the principle eigenvalues of the domains is investigated. The first and second domain variations are derived for nearly spherical domains. It is shown that in contrast to the classical Robin conditions the second variation changes sign and has singularities which depend on the eigenvalues of a Steklov problem. By means of the harmonic transplantation isoperimetric inequalities are derived for the principal eigenvalues in arbitrary domains. Introduction In this paper we discuss an elliptic boundary value problem with Robin boundary conditions of the opposite sign and a corresponding eigenvalue problem with the eigenvalue in the domain and on the boundary. The motivation comes from a classical model for the heat distribution in a body R 3, expressed by the heat equation t T (x, t) T (x, t) = in R +,

2 2 where t denotes partial differentiation with respect to t, is the Laplace operator in R 3 and the right-hand side describes a source of constant density. We suppose that the body is immersed into an ice bath and that on the boundary there is a regulating system with the following property: it the temperature drops below zero heat is carried into and otherwise out of the body, according to the law ν T = αt on. Here ν denotes the outer normal derivative of and α R + is a fixed positive number. This boundary condition is of the Neumann-Robin type. Observe that this flux condition is different from the classical Newton s law where α is of opposite sign. We shall also consider the case where at the boundary a reaction takes place. Then on, T satisfies the dynamical boundary conditions ν T = σ t T where σ < 0. In the first case the stationary solutions satisfy the elliptic boundary value problem (.) u + = 0 in, ν u = αu on. Under suitable regularity assumptions it has a unique solution provided α does not coincide with a eigenvalue 0 = µ < µ 2... of the Steklov problem (.2) φ = 0 in, ν φ = µφ on. If α = µ i then the solutions are not unique. To (.) we associate the energy E() := E(u, ) where E(, v) = v 2 dx 2 v dx α v 2 ds. It is well-known that the solution of (.) is a critical point of E(v, ) in W,2 () in the sense that the Frechet derivative vanishes. However it is not a local extremum. In this paper we study the dependence of E() on the domain. In contrast to the case α < 0 the techniques based on the minimum principle E() = min W,2 () E(v, ) do not apply. Therefore, in the spirit of the previous investigations in [5] and [6], we compute the first and second order shape derivative and discuss the behavior under small perturbations of the domain which are volume preserving. This techniques fails if α is a Steklov eigenvalue. It turns out that the first variation vanishes if is a ball. The second variation for the ball depends on (v ν) 2 ds and on α. It is singular if αr = µ k and if the kth-fourier coefficient of (v ν) with repect to orthonormal system of Steklov eigenfunctions does not vanish. At this point the second variation changes sign. This is in contrast to the case α < 0 discussed in [5] where the ball is a local minimum for all α < 0. In a recent paper Bucur and Giacomini [7] have shown that it is also a global minimum for the same choice of α. This first section is a completes the investigations in our previous paper [6] where the case α < 0 is discussed. The eigenvalue problem we are dealing with has its origin in the heat equation with dynamical boundary conditions. A standard method to find solutions is the

3 3 separation of variables. In fact T (x, t) = e λt φ(x) solves the homogeneous problem if φ is an eigenfunction of (.3) φ + λφ = 0 in, ν φ = σλφ on. It was shown in [4] that there exist two sequences of eigenvalues... λ k... λ 2 λ < 0 = λ 0 < λ λ 2..., such that lim n λ n = and lim n λ n =. These eigenvalues can be characterized by a min-max principle, cf. [4]. In the second part of this paper we study the domain dependence of λ and λ. By means of the harmonic transplantation [8] a global upper bound for λ () and a global lower bound for λ is derived. They are expressed in terms of the harmonic radius which plays an essential role. Both bounds are sharp. In [3] it was shown that for small σ the ball has for all domains of given volume the smallest λ. The situation is more involved if λ. An investigation is carried out for nearly spherical domains of prescribed volume by means of the first and second domain variations. Throughout this paper we shall assume that is a bounded Lipschitz domain. This guarantees that the embedding W,2 () into L 2 () as well as the trace operator Γ : W,2 () L 2 ( ) is compact. Under this condition both problems (.) and (.3) are solvable in W,2 () in the classical sense. 2 The boundary value problem 2. First domain variation Consider problem (.) in a class of domains t which are small perturbations of. We assume that { } (2.) t = y : y = x + tv(x) + t2 2 w(x) + o(t2 ) : x. where v and w are vector fields such that v, w : R n are in C (). The solutions of (.) in t will be denoted by u(y; t) and the corresponding energy by E(t). Example If is the ball B R of radius R then for all α there exists a unique radial solution u(r) = R2 2n R αn r2 2n. The corresponding energy is ( R E(B R ) = u dx = B R B R αn R 2 ). n(n + 2)

4 4 If α is not a Steklov eigenvalue u(y, ; t) is continuous and continuously differentiable in t. Under this condition it was shown in [5] that the first domain variation is E(0) = (v ν)[ u 2 2u 2α 2 u 2 α(n )u 2 H] ds, where H is the mean curvature of and u is the solution of (.) in. Example 2 If = B R then (2.2) This leads to the following Ẽ(0) = ( (n + )R αn 2 ) R2 n 2 (v ν) ds. Corollary Let t be a family of nearly spherical domains with prescribed volume t = B R. Then E(0) = 0. Proof The volume of t is given by B R J(t) dx where J(t) is the Jacobian determinant corresponding to the transformation y : B R t.the Jacobian matrix corresponding to this transformation is up to second order terms of the form I + td v + t2 2 D w, where (D v ) ij = j v i and j = / x j. By Jacobi s formula we have for small t (2.3) J(t) := det (I + td v + t2 2 D w) = + t div v + t2 2 ( (div v) 2 D v : D v + div w ) + o(t 2 ). Here we used the notation D v : D v := i v j j v i. Hence t = J(t)dx = B R +t div vdx+ t2 ((divv) 2 D v : D v +divw)dx+o(t 2 ). B R B R 2 B R For the first variation we have only to require that y is volume preserving of the first order, that is (2.4) div v dx = B R (v ν) ds = 0. This together with (2.2) proves the assertion. A further consequence of (2.2) is the local monotonicity property Corollary 2 If 0 < αr < n + and t > B R then E(0) > 0, otherwise if αr > n + then E(0) < 0. Proof By our assumption we have (v ν) ds > 0 The sign of E(0) depends therefore on the sign of (n + )αr (αr) 2.

5 5 2.2 Second domain variation for nearly spherical domains Corollary gives rise to the following question: is E(B R ) a local extremum among the family ( t ) t of perturbed domains with the same volume as B R? The answer will be obtained from the second variation. Let u t (x) := u(y(x); t) be the solution of u + = 0 in t, ν u = αu on t transformed onto. If u t is differentiable then d dt u t(x) t=0 = u (x) + v u, where u = u 0 is the solution of (.) in B R. It was shown in [5] that formally u solves the inhomogeneous boundary value problem (2.5) (2.6) u = 0 in B R ( ) α R ν u = αu + v ν on. n Let us assume for the moment that such a u exists. This is certainly the case if α does not coincide with a Steklov eigenvalue µ i. For the next result we consider perturbations which, in addition to the condition (2.4), satisfy the volume preservation of the second order, namely (2.7) ((div v) 2 D v : D v + div w) dx = 0. B R Set Q(u ) := u 2 dx α u 2 ds. B R Then the following formula was derived in [5]. Lemma Assume α µ i, (2.4) and (2.7). Put S(t) := t. Then (2.8) Ë(0) = 2Q(u ) + 2R ( αr) (v ν) n2 BR 2 ds R2 αn S(0). 2 For a ball and for volume preserving perturbations the second variation of the surface is of the form ( ) S(0) = τ (v ν) 2 (n ) R 2 (v ν) 2 ds, where τ stands for the tangential gradient on.

6 Discussion of Ë(0) We write for short F := 2Q g (u ) + 2R (2.9) ( αr) (v ν) 2 ds. n2 In order to estimate F we consider the Steklov eigenvalue problem (.2) An elementary computation yields µ = 0, and µ k = k R (for k 2 and counted without multiplicity). The second eigenvalue µ 2 = /R has multiplicity n and its eigenfunctions are x R,..., xn R. From now on we shall count the eigenvalues µ i with their multiplicity, i.e. µ 2 = µ 3 = µ n+ = /R and µ n+2 = 2/R etc. There exists a complete system of eigenfunctions {φ i } i such that (2.0) φ i φ j ds = δ ij. In view of the completeness we can write u (x) = c i φ i and (v ν) = i= b i φ i. Note that the first eigenfunction φ constant. The condition 0 = (v ν) ds = φ (v ν) ds implies that b = 0. From (2.6) we have also c = 0. The coefficients b i for i 2 are determined from the boundary value problem (2.5), (2.6). In fact (2.) b i = n c i (µ i α) α R i= for i = 2, 3,.... From the orthonormality conditions of the eigenfunctions it follows that Inserting this into (2.9) we get (2.2) 2 Q(u ) = c 2 i (µ i α). i=2 [ F = 2 c 2 i (µ i α) 2 R α R ]. µ i α Discussion of S(0). Observe that R[χ] = τ χ 2 ds χ 2 ds is the Rayleigh quotient of the Laplace- Beltrami operator on the (n )-dimensional sphere of radius R. Its eigenvalues are k(n 2 + k)/r 2, k N +. For volume

7 7 preserving perturbations of the first order (v ν) is orthogonal to the first eigenfunction ( (v ν) ds = 0) and thus R[(v ν)] n R 2. Equality holds if and only if (v ν) belongs to the eigenspace spanned by {x i } n i=, x i = R. This does not occur if we exclude small translations. Consequently S(0) > 0. Sign of Ë(0) Let us write for short A := 2( αr)2 n 2 and d k := R α R R k αr = k 2 ( αr)(k αr). Clearly d k { > 0 if αr < or αr > k, < 0 if < αr < k. By (2.) F = 2( αr)2 n 2 2 b 2 i [ R α R ] = A µ i α b2 k d k. 3 Remark The last equality is read in the following way. Assume for some k 2 there are m identical eigenvalues µ k = µ k =... = µ km The sum to the left thus contains the finite sum m [ b 2 R k j α R ] µ k α j=. The sum on the right side abbreviates this by setting b 2 k = m j= b2 k j. Let µ be the largest eigenvalue such that µ α := max{µ i : µ i < α}. Suppose that 0 < p < αr < p. Then d k > 0 for k =,..., p and d k < 0 for k > p. Then F can be split into a positive and negative part, F = F + + F, where F + = A p b2 k d k 0, 3 F = A b2 k d k 0. p+ If 0 < αr < then d k > 0 for all k = 3,.... Hence (2.3) F = F + = A b2 k d k. 3

8 8 Moreover (2.4) Recall that F + A d p p b2 k A d p (v ν) B 2 ds R 3 F A d p+ b2 k A d p+ (v ν) B 2 ds, R p+ [ ] R F A α R (v ν) 2 ds. µ α α Ë(0) = F + + F R2 αn 2 S(0). Theorem Suppose that the perturbations y are different from translations and rotations. Under the assumptions of Lemma and if 0 < αr < ɛ for ɛ sufficiently small then Ë(0) < 0. The ball is a local maximizer for E(t). Proof By (2.3), F + is bounded from above for αr < ɛ. dominates and S(0) < 0. Hence the last term Sign changes occur if α is in a neighborhood of µ α. More precisely we have Theorem 2 Let b p 0. Suppose that α is close to µ α = p. Then there exists ɛ 0 > 0 sufficiently small and depending only on v and w such that Ë(0) > 0 if p < α < p + ɛ 0. On the other hand if b p+ 0 there exists ɛ > 0 sufficiently small such that Ë(0) < 0 if p ɛ < α < p. If b p = 0 in the first case then Ë(0) can be positive or negative depending on (v ν). The same is true in the second case if bp+ = 0. Proof The expressions containing d p in the first case and, d p+ in the second case dominate. In order to get a sharper upper bound for Ë(0) in terms of v we impose the barycenter condition (2.5) x (v(x) ν(x)) ds = 0. Setting N = (v ν) we get Thus S(0) n+ R 2 τ N 2 ds 2n R 2 N 2 ds n+ R 2 p 3 b 2 k. We then get the following upper bound for Ë(0). (2.6) { Ë(0) n + R( α R) + 2 α n2 n 2 N 2 ds. 2( α R)2 n 2 (µ α α) } p b2 k. 3

9 9 Remark 2 Observe that since (2.5) is assumed, b i = 0 for i =,..., n. By (2.) this implies c i = 0 for i =,..., n. If then Consequently (2.7) Ë(0) α n 2 { n + p < α R < p µ α = p R. 2 α R ( α R)(p 2) p α R } p b2 k. If p = 2 then it follows immediately that Ë(0) 0. Let p > 2. The second term in the brackets is monotone decreasing in αr. Thus 2 α R ( α R)(p 2) inf = 2p(p 2 3p + 2) 2. αr (p,p) p α R Hence if n 2, Ë(0) < 0 for αr close to p. Example 3 Let t R 2 be the ellipse whose boundary t is given by { } R cos(θ), ( + t)r sin(θ), + t where (r, θ) are the polar coordinates in the plane. This ellipse has the same area as the circle B R and can be interpreted as a perturbation described in (2.). We have y = x + t( x, x 2 ) + t2 2 (x, 0) + o(t 2 ). The eigenvalues and eigenfunctions of the Steklov eigenvalue problem (.3) are We have and µ = m R + α and φ = rm (a cos(mθ) + a 2 sin(mθ)). S(0) = (v ν) = R cos(2θ) = b 3 φ 3, ( τ N 2 N 2 ) ds = 3πR. A straightforward computation yields [ Ë(0) = 3 ] R( αr) + (v ν) 2 ds. 4α 2(2 αr) with (v ν) 2 ds = πr 3. From this expression it follows immediately that { > 0 if αr > 2 Ë(0) < 0 if αr < 2. In this example there is only one coefficient b 3 which does not vanish. In accordance with Theorem 2, Ë(0) has only one singularity at α = µ 3 = 2 R. R 2 3

10 0 3 Eigenvalue problem Let R n be a bounded domain with smooth boundary. We consider the eigenvalue problem (3.) u + λu = 0 in, ν u = λ σ u in. σ is a negative number. Clearly λ 0 = 0 is a eigenvalue and the corresponding eigenfunction is u 0 = const.. We define σ 0 () :=. In [4] [3] the following properties were shown. (P) There exists an infinite sequence of positive eigenvalues (λ n ) n with lim n λ n =. (P2) There exists an infinite sequence of negative eigenvalues (λ n ) n with lim n λ n =. (P3) If σ < σ 0 (), then λ is simple and the corresponding eigenfunction u is of constant sign. (P4) If 0 > σ > σ 0 (), then λ is simple and the corresponding eigenfunction u is of constant sign. (P5) Let B R be a ball such that B R =. If σ < σ 0 (B R ), then λ () λ (B R ). (P6) For any domain (with B R = ), there exist a number ˆσ (σ 0 (), 0) such that λ () λ (B R ) whenever σ (σ 0 (), ˆσ). Remark 3 For (P5) we note that the condition σ < σ 0 (B R ) is more restrictive than the condition σ < σ 0 () if = B R. This is a consequence of the isoperimetric inequality. From (3.) we obtain a representation formula for λ and λ. In fact, multiplying the equation for u (resp. u ) with u (resp. u ) and integrating over we obtain u i 2 dx λ i () = u 2 i dx + σ u 2 i ds i =,. Remark 4 Note that from (3.) we deduce u i 2 dx = 0 and hence u i = const., if the denominator u 2 i dx + σ u 2 i ds = 0.

11 From now on λ and u will denote λ i and u i in either case i = ±. Let ( t ) t be a smooth family of small perturbations of as described in (2.). In particular they are volume preserving in the sense of (2.4) and (2.7). Denote by u t the solution to (3.2) u t + λ( t )u t = 0 in t, νt u t = λ( t ) σ u t in t. Here λ( t ) has the representation u t 2 dy (3.3) λ(t) := λ( t ) = t u 2 t dy + σ t t u 2 t ds t. As in [5] we transform the integrals onto for small t and differentiate λ(t) with respect to t. Then we get ( λ(0) = u 2 λ u 2 λ 2 σ 2 u 2 (n )λ σ H ) (v ν) ds. In particular = B R implies that λ(0) = 0. We are interested in extremality properties of the ball. Thus let = B R from now on and let σ < σ 0 (B R ). We consider the following boundary value problem. (3.4) where (3.5) u + λ u = 0 in B R ν u λ σ u = k(r)(v ν) in ( ) (n ) σ k(r) := λ u(r) + + λ σ 2. R We determine the sign of k(r). In this we follow the proof of Lemma 3 in [6]. For the sake of completeness we give the details. Lemma 2 Let k(r) be given by (3.5) and let u(r) be the positive radial function in the case λ = λ or λ = λ. Then we have k(r) > 0 if λ = λ k(r) < 0 if λ = λ. Proof In the radial case either eigenfunction satisfies the differentia equation We set z = ur u u rr + n r u r + λ u(r) = 0 in (0, R), u (R) = λ σ u(r). and observe that dz dr + z2 + n z + λ = 0 in (0, R). r

12 2 At the endpoint dz dr (R) + λ2 σ 2 (n ) + R λ σ + λ = 0. We know that z(0) = 0 and z(r) = λ σ. Note that (3.6) z r (0) = λ. We distinguish two cases. The case λ = λ (B R ). In that case we have (also from (3.6)) (3.7) z(0) = 0, z(r) = λ σ < 0, z r (0) = λ < 0. Thus z(r) decreases near 0. We determine the sign of z r (R). If z r (R) 0 then because of (3.7) there exists a number ρ (0, R) such that z r (ρ) = 0, z(ρ) < 0 and z rr (ρ) 0. From the equation we get z rr (ρ) = n z(ρ) < 0 which leads to a ρ 2 contradiction. Consequently z r (R) = (λ 2 σ 2 + This implies k(r) > 0 in the case λ = λ (B R ). The case λ = λ (B R ). In that case we have (see also (3.6)) (n ) R λ σ + λ ) < 0. (3.8) z(0) = 0, z(r) = λ σ > 0, z r (0) = λ > 0. Thus z(r) increases near 0. We again determine the sign of z r (R). If z r (R) 0 then because of (3.8) there exists a number ρ (0, R) such that z r (ρ) = 0, z(ρ) > 0 and z rr (ρ) 0. From the equation we get z rr (ρ) = n z(ρ) > 0 which is contradictory. ρ 2 Consequently z r (R) = (λ 2 σ 2 + (n ) R λ σ + λ ) > 0. This also implies k(r) < 0 in the case λ = λ (B R ). To (3.4) we associate the quadratic form Q(u ) := u 2 dx λ u 2 dx λ σ u 2 ds. B R B R Computations as in [5] lead to the following formula for = B R. These are the same computations which lead to formula (2.8) in Chapter 2. λ(0) = λ 2 σ u 2 (R) S(0) + λ F

13 3 where F = 2Q(u ) 2 λ σ u(r) k(r) (v ν) 2 ds. In the remaining part of this chapter we will discuss the sign of λ(0). We modify the approach in Chapter 2.2. and consider the following Steklov eigenvalue problem (3.9) φ + λ φ = 0 in B R, ν φ λ σ φ = µ φ on. There exists an infinite number of eigenvalues µ < µ 2 µ 3... lim i µ i =. and a complete system of eigenfunctions {φ i } i. See also Chapter 7 in [5]. With the same notation we have u = i= c i φ i for the solution of (3.4) and we get (3.0) λ(0) = λ 2 σ u 2 (R) S(0) + 2 ( λ c 2 i µ 2 2 σ u(r) i k(r) i= λ µ i ). Let λ = λ (B R ). Then (3.0) is precisely the expression in [5] Chapter subsection 2. where α > 0 is now replaced by λ (B R ) σ > 0. Thus we conclude (3.) λ (0) λ 2 σ u 2 (R) S(0) > 0. This is a local version of (P5). The case λ = λ (B R ) < 0 is more involved. In that case F < 0 since k(r) < 0 by Lemma 2. It is an open problem to show that also in this case λ (0) > 0 - at least for σ close to σ 0 (B R ) = R n. This conjecture is motivated by (P6). 4 Harmonic transplantation The eigenvalues λ (resp. λ ) have a variational characterization for σ σ 0 () (see [4]). Let (4.) K := {v W,2 () : v 2 dx =, v dx + σ v ds = 0}. Then for σ < σ 0 () 0 λ () = { sup v 2 dx σ } v 2 ds : v K

14 4 has a unique minimizer (of constant sign). The same holds in the case 0 > σ > σ 0 () for 0 λ () = { inf v 2 dx σ }. v 2 ds : v K We shortly review the method of harmonic transplantation which has been deviced by Hersch[8], (cf. also [2]). In [6] it applied to some shape optimization problems involving Robin eigenvalues. To this end we need the Green s function with Dirichlet boundary condition (4.2) where (4.3) γ = G (x, y) = γ(s( x y ) H(x, y)), { 2π if n = 2 (n 2) B if n > 2 For fixed y the funcion H(, y) is harmonic. and S(t) = Definition The harmonic radius at a point y is given by { e H(y,y) if n = 2, r(y) = H(y, y) n 2 if n > 2. { ln(t) if n = 2 t 2 n if n > 2. The harmonic radius vanishes on the boundary and takes its maximum r at the harmonic center y h. It satisfies the isoperimetric inequality [8],[2] (4.4) B r. Note that G BR (x, 0) is a monotone function in r = x. Consider any radial function φ : B r R thus φ(x) = φ(r). Then there exists a function ω : R R such that φ(x) = ω(g Br (x, 0)). To φ(x) we associate the transplanted function U : R defined by U(x) = ω(g (x, y h )). Then for any positive function f(s), the following inequalities hold true (4.5) U 2 dx = φ 2 dx B r (4.6) f(u) dx f(φ) dx. B r (4.7) f(u) dx γ B n f(φ) dx, r where γ = ( ) n. B r For a proof see [8] or [2] and in particular [6] for a proof of (4.7). The following observation will be useful in the sequel.

15 5 Remark 5 Since U is constant on (U = U( )) and since φ is radial we deduce U 2 ds = U 2 ( ) = φ 2 (r ) B r B r = φ 2 ds. B r B r Let u be a positive radial eigenfunction of λ (B R ). Then U K since u K Br and (4.5) holds. The case σ < σ 0 < 0. By the variational characterization we observe that 0 λ () =: λ σ () U 2 dx σ U 2 ds. We use (4.6) for the first integral in the denominator and Remark 5 for the second. 0 λ σ () B r u 2 dx σ B r u 2 ds. B r Set σ = σ B r Then we have 0 λ σ () λ σ (B r ). Alternatively we may write B r u 2 dx σ B r B r u 2 ds = u 2 dx σ u 2 ds B r B r σ u 2 ds B r u 2 dx σ u 2 ds B r B r ( ) B r Thus (4.8) 0 λ σ () λ σ (B r ) σ u 2 ds B r λ σ (B r ) ( ) B r } {{ } =:A(,σ). Note that the multiplicative term on the right hand side is close to one if the isoperimetric defect is small. In fact for = B R we have r = R and thus A(B R, σ) =. B r

16 6 The case σ 0 < σ < 0. Again by the variational characterization we have 0 λ () U 2 dx σ U 2 ds. We apply (4.7) to the first integral in the denominator and again Remark 5 to the second. 0 λ () γ n γ n B r u 2 dx σ B r u 2 dx σ γ n B r The last inequality holds since the isoperimetric inequality was applied. Thus We may rewrite this as B r This proves the following theorem. B r = γn 0 λ () γ n λ (B r ) λ () B r λ (B r ). u 2 ds u 2 ds. Theorem 3 Let be any domain for which the trace operator W,2 () L 2 ( ) is well defined. Let λ ± () be the first positive (negative) eigenvalue of (3.) and let r be the harmonic radius of. Then the following optimality result holds. ) In the case σ < σ 0 () < 0 we have 0 λ () A(, σ)λ (B r ), where the factor A is given in (4.8) and A = for the ball. 2) In the case σ 0 () < σ < 0 we have 0 λ () B r λ (B r ). Remark 6 It is interesting to compare ) in Theorem 3 with (P5). following two sided bounds. If B R is a ball of equal volume with and if σ < σ 0 (B R ) < 0 then We get the Equality holds for the ball. λ (B R ) λ () A(, σ)λ (B r ). Acknowledgement This paper was written during a visit at the Newton Institute in Cambridge. Both authors would like to thank this Institute for the excellent working atmosphere.

17 7 References [] G. Auchmuty, Sharp boundary trace inequalities, Proc. Royal Soc. Edinbourgh A, 44 (204), -2. [2] C. Bandle and M. Flucher, Harmonic radius and concentration of energy; hyperbolic radius and Liouville s equations U = e U and U = U (n+2)/(n?2), SIAM Review 38 (996), [3] C. Bandle, A Rayleigh-Faber-Krahn inequality and some monotonicity properties for eigenvalue problems with mixed boundary conditions, Birkhäuser ISNM 57 (2008), 3-2. [4] C. Bandle, J. v. Below and W. Reichel, Parabolic problems with dynamical boundary conditions: eigenvalue expansions and blow up. Atti Accad. Naz. Lincei Cl. Sci. Fis. Mat. Natur. Rend. Lincei (9) Mat. Appl. 7 (2006), no., [5] C. Bandle and A. Wagner, Second variaion of domain functionals and applications to problems with Robin boundary conditions, arxiv (204) [6] C. Bandle and A. Wagner, Isoperimetric inequalities for the principal eigenvalue of a membrane and the energy of problems with Robin boundary conditions, to appear in Journal of Convex Analysis. [7] D. Bucur and A. Giacomini, Faber-Krahn inequalities for the Robin-Laplacian:a free discontinuity approach, [8] J. Hersch, Transplantation harmonique, transplantation par modules, et théorèmes isopérimétriques, Comment. Math. Helv. 44 (969),