Second variation of domain functionals and applications to problems

arXiv:1403.2220v1 [math.OC] 10 Mar 2014

Second variation of domain functionals and applications to problems with Robin boundary conditions March 11, 2014 Catherine Bandle Mathematische Institut, Universit¨ at Basel, Rheinsprung 21, CH-4051 Basel, Switzerland

Alfred Wagner Institut f¨ ur Mathematik, RWTH Aachen Templergraben 55, D-52062 Aachen, Germany

Abstract In this paper the first and second domain variation for functionals related to elliptic boundary and eigenvalue problems with Robin boundary conditions is computed. Minimality and maximality properties of the ball among nearly circular domains of given volume are derived. The discussion leads to the investigation of the eigenvalues of a Steklov eigenvalue problem. As a byproduct a general characterization of the optimal shapes is obtained.

MSC2010 49K20, 49J20, 34L15, 35J20, 35N25.

1

Introduction

The study of domain functionals has received in the recent years a lot of attention. New techniques have been developed to prove the existence of an optimal shape among domains which are characterized by a common geometrical property such as a fixed volume. An important question is how to describe the optimal shape analytically. In the spirit of calculus this can be done by studying the dependence of the functionals under an infinitesimal change of the domain. Hadamard [12] was the first to propose a systematic approach to this question. Let Ωt be a family of perturbations of the domain Ω ⊂ Rn of the form (1.1)

Ωt := {y = x + tv(x) +

t2 w(x) + o(t2 ) : x ∈ Ω, t small }, 2

where v = (v1 (x), v2 (x), . . . , vn (x)) and w = (w1 (x), w2 (x), . . . , wn (x)) are smooth vector 2 fields and where o(t2 ) collects all terms such that o(tt2 ) → 0 as t → 0. Consider a functional 1

2

E(t) which depends on Ωt and on a solution u˜(t) of an elliptic problem defined on Ωt . The first derivative of E(t) with respect to the parameter t is called the first domain variation and the second derivative is called the second domain variation. In modern text often the expression shape derivative is used. In their seminal paper on domain functionals Garabedian and Schiffer [7] computed the first and second domain variation for several functionals such as the first eigenvalue of the Dirichlet-Laplace operator, the virtual mass and the Green’s function. By choosing special perturbations they obtained convexity theorems. Subsequent to the work of Garabedian and Schiffer’s, D. Joseph [15] computed formally higher variations of the eigenvalues and studied the behavior of the spectrum under shear and stretching and Grinfeld [11] computed the eigenvalues of a polygon. For a long time this topic has rather been neglected. In the last years it has attracted considerable interest. New developments and new applications are found in the inspiring books by Henry [14] and Pierre and Henrot [13] where further references are given. Motivated by classical isoperimetric inequalities for domain functionals with prescribed volume, like the Rayleigh-Faber-Krahn inequality and the St. Venant-P´olya inequality for the torsional rigidity (cf. [17]) we shall focus on perturbations which are volume preserving. To our knowledge the effect of this restriction, in particular to the second variation hasn’t been explored yet. A basis for our study are the two model problems: 1. nonlinear problem (1.2) (1.3)

∆y u˜ + g(˜ u) = 0 ∂νt u˜ + α˜ u = 0

in Ωt in ∂Ωt .

Here νt is the outer unit normal to Ωt and α is a real number. This problem is the Euler-Lagrange equation corresponding to the energy functional Z Z I 2 E(t) = |∇˜ u(t)| dy − 2 G(˜ u) dy + α u˜2 (t) dS, where G0 (s) = g(s). Ωt

Ωt

∂Ωt

2. Eigenvalue problem (1.4) (1.5)

∆y u˜ + λ(Ωt )˜ u = 0 ∂νt u˜ + α˜ u = 0

in Ωt in ∂Ωt .

Like the energy functional the eigenvalue is expressed in terms of integrals Z Z I 2 2 λ(t) u˜ (t) dy = |∇˜ u(t)| + α u˜2 (t) dS. Ωt

Ωt

∂Ωt

3

We shall compute the first and second variations of E(t) and λ(t) using a change of variable approach which transforms the new domain into the original one. In fact for small t, the map y : Ω → Ωt defined in (1.1), is a diffeomorphism. Hence x can be chosen as a new variable. The first variation is a simple and elegant expression. It provides a necessary condition for extremal domains in terms of an overdetermined elliptic problem. It turns out that for the first eigenvalue the ball is a candidate for an extremal domain. The same is true for the energy E(t) if the solutions of (1.2), (1.3) are radial. This is in accordance with the Bossel-Daners inequality [5] which states that among all domains of given volume the ball yields a local minimum of the first eigenvalue and by recent results by Bucur and Giacomini [3]. As a byproduct we obtain a local monotonicity property which improves slightly the one in [9]. We then compute the second variation and study its sign in the case of the ball. For 2 this purpose we use a device by Simon [20]. The discussion of the sign of ddt2E (t)|t=0 and d2 λ (Ωt )|t=0 for volume preserving perturbations is related to an eigenvalues problem of a dt2 Steklov type problem. A theoretical approach was developed by Pierre and Novruzi [18]. In particular they found an abstract result on the structure of the second variation. However the strict positivity (coercivity) necessary for the minimality property of a domain remained a challenging open problem. In this paper we first compute the second variation for general domains and then focus on the ball which for many problems is a critical domain, i.e. the first variation vanishes. With the help of a Steklov type eigenvalue problem we are able to give an estimate for the second variation from below. It turns out that in contrast to problems with Dirichlet boundary conditions the second variation of the surface plays an crucial role, s. [4] for similar discussion. We obtain in this context an interesting result for this surface variation which to our knowledge is new. It should be pointed out that the method works for functionals which are not necessarily characterized by a variational principle, for instance E(t) with α < 0. A first attempt to tackle this problem was made in [1]. Our paper is organized as follows. First we introduce, for the reader’s convenience, the concept of the mean curvature which will play an important role and some tools concerning vector fields. We then discuss useful properties of the vector fields which are related to volume preserving perturbations. In Section 3 we describe in full details the energies and the Rayleigh quotients of the perturbed problems, expressed in the original domain Ω after the change of variables y = x + tv(x) + o(t). The first variations are derived in Section 4 from which overdetermined boundary and eigenvalue problems for optimal domains can be deduced. In Section 5 an auxiliary function related to the t-derivative of the solutions in Ωt will be discussed. It turns out that this function will play an essential role for the sign of the second variation.

4

Section 6 is devoted to the lengthly computations of the second variation. Applications to problems in nearly circular domains of fixed volume are investigated in Section 7. As a surprise we find out that the sign of the second variation for the ball depends on the sign of α.We compare our approach with Garabedian and Schiffer’s formula of the second variation of the principal eigenvalue of the Laplacian with Dirichlet boundary conditions. We show that the ball is a local minimum. For the sake of completeness we give at the end the formula for the second variation of the energy in case of Dirichlet boundary conditions.

2 2.1

Preliminaries Geometry of surfaces

In this section we collect some basic geometrical notions of surfaces needed in our study. Throughout this paper we will use the following notation. Let Ω be a bounded C 2,α -domain in Rn and let x := (x1 , x2 , . . . , xn ) denote a point in Rn . Throughout this paper x · y stands for the Euclidean scalar product of two vectors x and y in Rn and |x| = (x · x)1/2 . At every point P ∈ ∂Ω there exist therefore a neighborhood UP and a Cartesian coordinate system with the basis {e}ni=1 centered at P , such that en points n the direction of the outer normal ν and ei , i = 1, . . . n − 1 lie in the tangent space of P . The coordinates with respect to this basis will be denoted by (ξ1 , ξ2 , . . . , ξn ). Moreover we assume that Ω ∩ UP = {ξ ∈ UP : ξn < F (ξ1 , ξ2 , . . . , ξn−1 )}, F ∈ C 2,α . With this choice of coordinates clearly F (0) = 0 and Fξi (0) = 0 for i = 1, 2, . . . , n−1. For short we set ξ 0 = (ξ1 , ξ2 , · · · , ξn−1 ) which ranges in U 0 := UP ∩ {ξn = 0}. In UP ∩ ∂Ω the boundary is represented by x(ξ 0 ) = (ξ1 , ξ2 , . . . , ξn−1 , F (ξ 0 )) and the unit outer normal ν˜(ξ 0 ) = (˜ ν1 , ν˜2 . . . . , ν˜n ) with respect to the ξ 0 −coordinate system, is given by (−Fξ1 , −Fξ2 , . . . , −Fξn−1 , 1) p ν˜(ξ 0 ) = . 1 + |∇F |2 In this paper we shall use the Einstein convention where repeated indices are understood to be summed from 1 to n − 1 or from 1 to n, respectively. The vectors xξi , i = 1, 2 . . . n − 1 span the tangent space. The metric tensor of ∂Ω is denoted by gij and its inverse by g ij . We have Fξi Fξj gij = xξi · xξj = δij + Fξi Fξj and g ij = δij − , 1 + |∇0 F |2 p where ∇0 stands for the gradient in Rn−1 . The surface element of ∂Ω is dS = detgij dξ = p 1 + |∇0 F |2 dξ. Observe that any vector v can be represented in the form (2.1)

v = g js (v · xξj )xξs + (v · ν)ν, s, j = 1, 2, . . . n − 1.

5

Let f ∈ C 1 (UP ) and let f˜(ξ 0 ) := f (ξ)|∂Ω = f (ξ 0 , F (ξ 0 )). The tangential gradient of f at a boundary point is defined as ∇τ f˜ = g ij

(2.2)

∂ f˜ xξ . ∂ξj i

Let us write for short ∂i :=

∂ , ∂ξi

and ∂i∗ f := g ij ∂j f˜.

(2.3)

the tangential derivative on ∂Ω. For a smooth vector field v : ∂Ω → Rn which is not necessarily tangent to ∂Ω we define the tangential divergence by (2.4)

div ∂Ω v := g ij v˜ξi · xξj , where v˜ = v(ξ 0 , F (ξ 0 )).

By (2.3) this can also be written as div ∂Ω v = ∂j∗ v˜ · xξj .

(2.5)

If κi , i = 1, . . . , n − 1 are the principal curvatures of ∂Ω at the point P then n−1

1 X H= κi n − 1 i=1 is the mean curvature of ∂Ω at P . For a general point (ξ 0 , F (ξ 0 )) on ∂Ω it is given by ! Fξi (ξ 0 ) 0 −1 ∂ H(ξ ) := (n − 1) −p . ∂ξi 1 + |∇F |2 Observe that div ∂Ω ν = (n − 1)H.

(2.6) In particular we have H = at the origin.

1 R

if ∂Ω = ∂BR where BR denotes the ball of radius R centered

Another way of defining geometrical quantities is by projection onto the tangent space of ∂Ω. Let x ∈ ∂Ω and let Tx ∂Ω be the tangent space of ∂Ω in x. Then we define P : Rn → Tx ∂Ω

v → P (v) = v − (v · ν)ν.

From (2.1) we have for the gradient ∇f in Rn ∇f = g is (∇f · xξi )xξs + (∇f · ν)ν.

6

Notice that ∇f · xξi = ∂i f + ∂n f Fξi = ∂i f˜. Hence ∇τ f˜ = ∇f − (∇f · ν)ν = P (∇f ).

(2.7)

As in [10] some computations will be shorter if we introduce the i − th component of the tangential gradient δi f = ∂i f − νi ∂s f νs . ∗ At the origin we have δi f = ∂i f = ∂i f . In general δi f and ∂i∗ f are different, more precisely δk = (xξj · ek )∂j∗ = (∂j∗ x · ek )∂j .

(2.8)

In the same way we show that for any smooth vector field v : Ω → Rn that div ∂Ω v = div v − ν · Dv ν := ∂i vi − νj ∂j vi νi = δj v˜j .

(2.9)

At the origin we have div ∂Ω v = ∂i vi = ∂i∗ v˜i , i = 1, . . . n − 1. We will frequently use integration by parts on ∂Ω. Let f ∈ C 1 (∂Ω) and v ∈ C 0,1 (∂Ω, Rn ). The next formula is often called the Gauss theorem on surfaces. I I I τ (2.10) f (v · ν) H dS. v · ∇ f dS + (n − 1) f div ∂Ω v dS = − ∂Ω

∂Ω

∂Ω

This formula can also be written in the form I I I (2.11) vj δj f dS + (n − 1) f δj vj dS = − 2.2 2.2.1

f (v · ν) H dS.

∂Ω

∂Ω

∂Ω

Domain perturbations Volume element

The Jacobian matrix corresponding to the transformation y(t, Ω) introduced in the Introduction is up to second order terms t2 Dw , where (Dv )ij = ∂j vi and ∂j = ∂/∂xj . 2 By Jacobi’s formula we have for small t I + tDv +

(2.12)

J(t) :=

det (I + tDv +

= 1 + t div v +

t2 Dw ) 2


t2 (div v)2 − Dv : Dv + div w + o(t2 ). 2

Here we used the notation Dv : Dv := ∂i vj ∂j vi .

7

Thus y(t, Ω) is a diffeomorphism for t ∈ (−t0 , t0 ) and t0 sufficiently small. Throughout this paper we shall consider diffeomorphisms y(t, Ω) as described above. Later on we will be interested in volume preserving transformations. From Z Z Z t2 J(t) dx = |Ω| + t div v dx + ((div v)2 − Dv : Dv + div w) dx + o(t2 ) |Ωt | = 2 Ω Ω Ω it follows that y(t, Ω) is volume preserving of the first order if Z (2.13) div v dx = 0 Ω

holds and it is volume preserving of the second order if in addition to (2.13) it satisfies Z (2.14) ((div v)2 − Dv : Dv + div w) dx = 0. Ω

For volume preserving transformations of the second order we have Lemma 1 Let v ∈ C 0,1 (Ω, Rn ) Then Z
(div v)2 − Dv : Dv + div w dx = 0 Ω

is equivalent to I

I (v · ν)div v dS −

(2.15) ∂Ω

I (w · ν) dS = 0.

vi ∂i vj νj dS + ∂Ω

∂Ω

Proof Integration by parts gives I Z Z vj ∂j vi νi dS Dv : Dv dx = − vj ∂j (div v) dx + ∂Ω Ω Ω Z I I 2 = (div v) dx − (v · ν) div v dS + Ω

∂Ω

vi ∂i vj νj dS.

∂Ω

This proves (2.15).  Remark 1 The presence of w is crucial because otherwise the class of perturbations is too limited. For instance consider B1 ⊂ R2 and let Ωt be a rotation,of the type   cos(t) − sin(t) y= x. sin(t) cos(t) 2

Then for small t, y = x + t(−x2 , x1 ) − t2 (−x1 , x2 ) + o(t2 ). It is easy to see that the first order approximation x + t(−x2 .x1 ) is not volume preserving of the second order. Remark 2 For any given v we can always find a vector w such that (2.15) is satisfied. Denote by v τ = v − (v · ν)ν the tangential component of v on ∂Ω.

8

2.2.2

Surface element

In this part we shall compute the surface element of ∂Ωt . Let x(ξ 0 ), ξ 0 ∈ UP0 be local coordinates of ∂Ω introduced in Section 2.1. Then ∂Ωt is represented locally by {y(ξ 0 ) := x(ξ 0 ) + t˜ v (ξ 0 ) +

t2 w(ξ ˜ 0 ) : ξ 0 ∈ UP0 }, 2

where as before v˜(ξ 0 ) = v(ξ 0 , F (ξ 0 )) and similarly w(ξ ˜ 0 ) = w(ξ 0 , F (ξ 0 )). Setting gij := xξi · xξj , aij := xξi · v˜ξj + xξj · v˜ξi , bij := 2˜ vξi · v˜ξj + w ˜ξi · xξj + w ˜ξj · xξi we get |dy|2 := (gij + taij +

t2 bij )dξi dξj =: gijt dξi dξj . 2

Write for short G = (gij ), G−1 = (g ij ), A = (aij ), B = (bij ) and correspondingly Gt = (gijt ). Then the surface element on Ωt is
1/2 dSy = detGt dξ. Clearly √ √ t2 detGt = detG{det(I + tG−1 A + G−1 B)}1/2 . 2 {z } | k(x,t)

Set σA = trace G−1 A, σB = trace G−1 B and σA2 = trace (G−1 A)2 . For small t the Taylor expansion yields k(x, t) = 1 + tσA +


t2 σB + σA2 − σA2 + o(t2 ) 2

and (2.16)

p

t2 t k(x, t) = 1 + σA + 2 2



1 1 1 σB + σA2 − σA2 2 4 2



+ o(t2 ).

In the sequel we shall use the notation t t2 1 1 1 m(t) := 1 + σA + ( σB − σA2 + σA2 ) + o(t2 ). 2 2 2 2 4

9

Then the surface element of ∂Ωt reads as (2.17)

dSt = m(t)dS, where dS is the surface element of ∂Ω .

Our next goal is to find more explicit forms for the expressions in m(t). It follows immediately from Section 2.1 that σA = 2g ij v˜ξj · xξi = 2div ∂Ω v. The expression σA has a geometrical interpretation. We find after a straightforward computation that 1 (2.18) σA = div ∂Ω v τ + (n − 1)H(v · ν), 2 where v τ is the projection of v into the tangent space. Moreover a straightforward calculation leads to vξi · ∂k∗ x), σA2 = g is g kl ask ali = 2(∂i∗ v˜ · xξk )(∂k∗ v˜ · xξi ) + 2(∂i∗ v˜ · xξk )(˜ σB = 2g ij v˜ξi · v˜ξj + 2div ∂Ω w ∗ vξs · ν˜) + 2div ∂Ω w. x) + 2g is (˜ vξi · ν˜)(˜ vξs · ∂m = 2(∂s∗ v˜ · xξm )(˜

In the last expression we have used for v˜ξk the representation (2.1). Consequently 1 1 1 ¨ = σB − σA2 + σA2 (2.19) m(0) 2 2 4 is = g (˜ vξi · ν˜)(˜ vξs · ν˜) + div ∂Ω w − (∂i∗ v˜ · xξk )(∂k∗ v˜ · xξi ) + (div ∂Ω v)2 . This together with (2.10) implies that I H (2.20) m(0) ¨ dS = ∂Ω (∂s∗ v˜ · ν˜) (∂s v˜ · ν˜) − (∂i∗ v˜ · xξk ) (∂k∗ v˜ · xξi ) + (∂i∗ v˜ · xξi )2 dS ∂Ω R +(n − 1) ∂Ω (w · ν)H dS. 2.3

Computations for the ball

In this subsection we simplify (2.20) for the special case ∂Ω = ∂BR . For simplicity we move the Cartesian coordinate system {ei }ni=1 into the center of the ball. This transformation does not affect formula (2.20). Note that in the radial case x 1 ν= (2.21) and δi νj = (δij − νi νj ) . R R We now start with the evaluation of the different terms in (2.20). Setting N := (v · ν) we have `1 := (∂s∗ v˜ · ν˜) (∂s v˜ · ν˜) = ∂s∗ N ∂s N − 2(∂s∗ v˜ · ν˜)(˜ v · ∂s ν˜) + (˜ v · ∂s∗ ν˜)(˜ v · ∂s ν˜) τ 2 ∗ −2 τ 2 = |∇ N | − 2(∂s v˜ · ν˜)(˜ v · ∂s ν˜) − R |v | .

10

By (2.21) and (2.8) 2 2 −2(∂s∗ v˜ · ν˜)(˜ v · ∂s ν˜) = − ∂s v˜k ν˜k v˜m (∂s∗ x · em ) = − [vm δm N − vk vm ∂s νk (∂s∗ x · em )] R R 2 = − [(v · ∇τ N ) − R−1 |v τ |2 ]. R This together with the Gauss theorem on surfaces (2.10) implies (2.22) I ∂BR

(∂s∗ v˜ |

I · ν˜) (∂s v˜ · ν˜) dS = {z } `1

(|∇τ N |2 +

∂BR

2(n − 1) 2 2 1 N div BR v − N + 2 |v τ |2 ) dS. 2 R R R

It will turn out that it is convenient to eliminate the last term in (2.22). If we replace in the Gauss formula (2.10)) f by N and use (2.21) we obtain I I I 1 1 1 τ 2 |v | dS = − vj δj vi νi dS − div ∂BR v N dS R2 ∂BR R ∂BR R ∂BR I n−1 + N 2 dS. R ∂BR With this remark we rewrite (2.22).   I I (n − 1) 2 ∗ τ 2 (∂s v˜ · ν˜) (∂s v˜ · ν˜) dS = N (2.23) |∇ N | − dS {z } R2 ∂BR | ∂BR `1 I 1 (N div BR v − vj δj vi νi ) dS. + R ∂BR Next we treat the second term. Observe that `2 = −(∂i∗ v˜ · xξk ) (∂k∗ v˜ · xξi ) = −δs vj δj vs . By (2.11) we find I

n−1 `2 dS = vj δi δj vi dS − R ∂BR ∂BR I

I vj δj vi νi dS. ∂BR

At this point it is important to note that δi δj 6= δj δi . In [10] (Lemma 10.7) the following relation is proved: δi δj = δj δi + (νi δj νk − νj δi νk )δk . For the ball this gives (νi δj νk − νj δi νk )δk = R−1 (νi δj − νj δi ).

11

Hence n−2 `2 dS = vj δj δi vi dS − R ∂BR ∂BR I 1 − N div ∂BR v dS. R ∂BR

I (2.24)

I

I vj δj vi νi dS ∂BR

We apply (2.11) to the first integral on the right hand side of (2.24) and obtain I I I n−1 2 N div ∂BR v dS. vj δj δi vi dS = − (δi vi ) dS + R ∂BR ∂BR ∂BR Introducing this expression into (2.24) we find I I I n−2 2 vj δj vi νi dS `2 dS = − (δi vi ) dS − R ∂BR ∂BR ∂BR I n−2 N div ∂BR v dS. + R ∂BR Thus I (`2 + ∂BR

(∂i∗ v˜

n−2 · xξi ) ) dS = − R 2

n−2 vj δj vi νi dS + R ∂BR

I

I N div ∂BR v dS. ∂BR

This identity together with (2.23) and the fact that N div BR v − vj δj vi νi = N div v − vj ∂j vi νi implies the following lemma. Lemma 2 For an arbitrary vector field v = v τ +N ν the second variation assumes the form   I I (n − 1) 2 τ 2 m(0) ¨ dS = |∇ N | − N dS R2 ∂BR ∂BR I n−1 + (N div v − vj ∂j vi νi + w · ν) dS. R ∂BR Let us now consider vector fields which are volume preserving of the second order (cf. Lemma 1). We observe that in view of (2.15) the second integral on the right-hand side in Lemma 2 vanishes. Therefore n−1 m(0) ¨ dS = |∇ N | dS − R2 ∂BR ∂BR

I (2.25)

I

τ

2

I

N 2 dS.

∂BR

H Remark 3 It is interesting to observe that the second variation ∂BR m(0) ¨ dS does not depend on w nor on the tangential components of the vector field v.

12

Let us introduce the following notation. I S(t) := m(t) dS : surface of ∂Ωt . ∂BR

Next we determine all volume preserving vector fields of first and second order for which ¨ the second variation of S(0) vanishes. They will be called the kernel of S(0). For this purpose we recall the eigenvalue problem (2.26)

∆Sn−1 φ + µφ = 0 on Sn−1 .

It is well-known that the eigenfunctions are the spherical harmonics of order k and the . corresponding eigenvalues are k(k +n−2), k ∈ N+ with the multiplicity (2k +n−2) k+n−3)! (n−2)!k! H If the v is volume preserving of the first order then ∂BR N dS = 0 and by the variational characterization of the eigenvalues I I I µ2 n−1 τ 2 2 |∇ N | dS ≥ 2 N dS = N 2 dS. 2 R R ∂BR ∂BR ∂BR Equality holds if and only if the projection of v onto the normal (v · ν) is an element of the eigenspace corresponding to µ = (n − 1)/R2 . It is generated by the basis {ei · ν}ni=1 or if N = 0. Example Suppose that on ∂BR the vector field v points only in tangential direction. Then v = g ij (v · xξj )xξi . The vector y = x(ξ 0 ) + t˜ v (ξ 0 ) is orthogonal to ∂BR . Its length is |y|2 = R2 + t2 g ks (v · xξk )(v · xξs ). The boundary ∂Ωt can therefore be represented by 2 y + t2 g0 ν + o(t2 ) where g0 = g ks (v · xξk )(v · xξs ) and ν = Ry . The domain Ωt is therefore a second order perturbation of BR . Definition 1 A perturbation of the form y = x + tN ν +

t2 w + o(t2 ) on ∂BR 2

is called a Hadamard perturbation. From the previous consideration it follows immediately that every small perturbation of the ball can be described by a Hadamard perturbation. Consequently we have ¨ > 0. Lemma 3 Assume N 6= ai xi on ∂BR . Then for every Hadamard perturbation S(0)

3

Energies

Let (Ωt )t be a family of domains described in the previous chapter and let G : R → R 2 denote a smooth function i.e. G ∈ Cloc (R) at least. We denote by g its derivative: G0 = g. Consider the energy functional

13

Z E(Ωt , u) :=

(3.1)

Z

2

I

|∇y u| dy − 2 Ωt

G(u) dy + α Ωt

u2 dSt .

∂Ωt

A critical point u˜ ∈ H 1 (Ωt ) of (3.1) satisfies the Euler Lagrange equation (3.2) (3.3)

∆y u˜ + g(˜ u) = 0 ∂νt u˜ + α˜ u = 0

in Ωt in ∂Ωt ,

where νt stands for the outer normal of ∂Ωt . A special case is the torsion problem (1.2) and (1.3) with G(w) = w. Assume that u˜ solves (3.2) - (3.3). We set E(t) := E(Ωt , u˜). 2

In a first step we transform the integrals onto Ω and ∂Ω. Let y = x + tv(x) + t2 w(x) be defined as in (1.1) and let x(y) be its inverse. Then after change of variables we get    Z Z ∂xi ∂xj E(t) = ∂i u˜(t) ∂j u˜(t) (3.4) J(t) dx − 2 G(˜ u(t)) J(t) dx ∂yk ∂yk Ω Ω I +α u˜2 (t) m(t) dS ∂Ω

where u˜(t) := u˜(x + tv(x) +

t2 w(x), 2

t ∈ (−, ). We set

Aij (t) :=

(3.5)

∂xi ∂xj J(t). ∂yk ∂yk

The expression (3.4) assumes now the concise form Z Z I (3.6) E(t) = ∇˜ uA∇˜ u dx − 2 G(˜ u)J dx + α Ω

Ω

u˜2 m dx.

∂Ω

Thus in the domain Ω the solution u˜(t) solves the transformed equation (3.7) (3.8)

LA u˜(t) + g(˜ u(t)) J(t) = 0 ∂νA u˜(t) + α m(t) u˜(t) = 0

in Ω in ∂Ω,

where (3.9)

LA = ∂j (Aij (t)∂i )

and

∂νA = νi Aij (t)∂j .

It turns out to be convenient to write the equations (3.7) - (3.8) for u˜ in the weak form Z I Z (3.10) ∇φA∇˜ u dx + α φ u˜ m dS = g(˜ u) φ J dx, ∀φ ∈ W 1,2 (Ω). Ω

∂Ω

Ω

14

The eigenvalue λ(t) in (1.4) and (1.5) is characterized by the Rayleigh quotient R H |∇y u˜|2 dy + α ∂Ωt u˜2 dS Ωt R (3.11) . λ(t) = R(t) := 2 dy u ˜ Ωt The change of variable (1.1) yields R H A (t) ∂ u ˜ (t) ∂ u ˜ (t) dx + α u˜(t)2 m(t) dS ij i j ∂Ω R R(t) = Ω (3.12) . u˜2 (t) J(t) dx Ω Thus in the domain Ω the solution u˜(t) solves the transformed equation (3.13) (3.14)

LA u˜(t) + λ(t)J(t) u˜(t) = 0 ∂νA u˜(t) + α m(t) u˜(t) = 0

in Ω in ∂Ω.

Testing the above equation with u˜ we obtain the identity Z I Z 2 (3.15) u˜ m dS = λ(t) u˜2 J dx ∇˜ uA∇˜ u dx + α Ω

∂Ω

Ω

which will be used later. 3.1

Expansions

In this subsection we expand formally all relevant quantities with respect to t about the origin. Under suitable regularity assumption on Ωt such processes can be justified. We start with the energy (3.6) I Z Z u˜2 m dx, u)J dx + α ∇˜ uA∇˜ u dx − 2 G(˜ E(t) = ∂Ω

Ω

Ω

where u˜ is a weak solution of (3.10). Recall that u˜(t) = u˜(x + tv(x) + following expansion is valid

t2 w(x), 2

t ∈ (−, ). Under sufficient regularity the

t2 u˜(t) = u˜(0) + t u˜˙ (0) + u¨˜(0) + o(t2 ). 2 We set u0 (x) := ∂t u˜(x + tv(x) + cients of this expansion: (3.16) u˜(0) = u(x)

t2 w(x), t)|t=0 2

and get the following formulas for the coeffi-

and

t2 t2 (3.17) u˜˙ (0) = ∂t u˜(x + tv(x) + w(x), t)|t=0 + v(x) · ∇˜ u(x + tv(x) + w(x), t)|t=0 2 2 = u0 (x) + v(x) · ∇u(x).

15

We also expand Aij (t) with respect to t: (3.18)

Aij (t) = Aij (0) + t A˙ ij (0) +

t2 ¨ Aij (0) + o(t2 ). 2

˙ ¨ Later we will compute E(0) and E(0). For this purpose we shall need the explicit terms in (3.18). A lengthy but straightforward computation gives Lemma 4 Aij (0) = δij ; A˙ ij (0) = div v δij − ∂j vi − ∂i vj ;
A¨ij (0) = (div v)2 − Dv : Dv δij + 2 (∂k vi ∂j vk + ∂k vj ∂i vk ) +2 ∂k vi ∂k vj − 2 div v (∂j vi + ∂i vj ) + div w δij − ∂i wj − ∂j wi . Finally we recall from (2.12) (3.19) (3.20) (3.21) 3.2 3.2.1

J(0) = 1 ˙ J(0) = div v ¨ J(0) = (div v)2 − Dv : Dv + div w. Differentiation of the energy and the eigenvalue First and second variation

Direct computation gives I Z Z ˙ ˙ ˙ u˜2 m ˙ dS u)J dx + α ∇˜ uA∇˜ u dx − 2 G(˜ E(t) = ∂Ω Ω Ω Z Z I +2 ∇u˜˙ A∇˜ u dx − 2 g u˜˙ J dx + 2α u˜˙ u˜m dS. Ω

Ω

∂Ω

We now eliminate the terms containing u˜˙ by means of (3.10) with φ = u˜˙ and obtain Z I Z 2 ˙ ˙ (3.22) E(t) = ∇˜ uA∇˜ u dx + α u˜ m ˙ dS − 2 GJ˙ dx. Ω

∂Ω

Ω

˙ Notice that E(t) is independent of u˜˙ . Next we want to find an expression for the second derivative. Differentiation of (3.10) implies Z h I i ˙ ˙ (3.23) ∇φA∇u˜ + ∇φA∇˜ u dx + α (φu˜˙ m + φ˜ um) ˙ dS Ω ∂Ω Z ˙ dS, = (g 0 (˜ u)u˜˙ φJ + g(˜ u)φJ) Ω

16

for all φ ∈ W 1,2 (Ω). Differentiation of (3.22) yields I Z h i ¨ ˙ ˙ ¨ ¨ ˙ ˙ (3.24) E(t) = ∇˜ uA∇˜ u + 2∇u˜A∇˜ u − 2g u˜J − 2GJ dx + α (2˜ uu˜˙ m ˙ + u˜2 m) ¨ dS. Ω

∂Ω

By means of (3.23) with φ = u˜˙ we get Z I Z 2 ¨ ¨ (3.25) E(t) = ∇˜ uA∇˜ u dx + α u˜ m ¨ dS − 2 GJ¨ dx Ω ∂Ω Ω Z I Z 2 −2 ∇u˜˙ A∇u˜˙ dx − 2α u˜˙ m dS + 2 g 0 u˜˙ 2 J dx. Ω

∂Ω

Ω

In accordance with the first derivative which does not depend on u˜˙ , the second derivative does not depend on u¨˜. In order to compute the variations of the eigenvalue we first recall that u˜ solves (3.15). We impose the normalization Z u˜2 (t) J(t) dx = 1. Ω

This implies (3.26)

d dt

Z

Z

2

u˜ (t) J(t) dx = 2 Ω

u˜(t) u˜˙ (t) J(t) dx +

Ω

Z

˙ dx = 0. u˜(t)2 J(t)

Ω

We differentiate (3.11), use the normalization and then we choose φ = u˜˙ as a test function in the weak formulation of (3.13) and (3.14). With (3.26) we get Z I Z 2 ˙ ˙ (3.27) λ(t) = ∇˜ uA∇˜ u dx + α u˜ m ˙ dS − λ(t) u˜2 J˙ dx. Ω

∂Ω

Ω

˙ Thus λ(t) does not depend on u˜˙ (t). We differentiate (3.27) with respect to t. Then we differentiate (3.15) with respect to t and choose φ = 2u˜˙ in the weak formulation of (3.13) and (3.14). With (3.26) and under ˙ the assumption λ(0) = 0 we get for t = 0 Z Z I I 2 ¨ ¨ ˙ ˙ (3.28) λ(0) = ∇˜ uA∇˜ u dx − 2 ∇u˜A∇u˜ dx + α u˜ m ¨ dS − 2α u˜˙ 2 m dS Ω Ω ∂Ω ∂Ω Z Z 2 ¨ −λ(0) u˜ J dx + 2λ(0) u˜˙ 2 J dx. Ω

¨ Thus λ(0) does not depend on u¨˜(0).

Ω

17

3.2.2

Third variation

... In order to compute the third variation E (s) we proceed exactly in the same way as before. We differentiate (3.25). Z I Z   ... ... ¨ u · ( A ∇˜ u) dx + 2α u˜ u˜˙ m ¨ dS u dx + 2 ∇˜ E (t) = 2 ∇u˜˙ · A∇˜ ΩI Ω ∂Ω Z Z Z
... 2 ... 0 ¨ ˙ +α u˜ m dS − 2 G (˜ u) u˜ J dx − 2 G(˜ u) J dx − 4 ∇u¨˜ · A∇u˜˙ dx Ω Ω Ω I Z ∂Ω I   2 ˙ u˜˙ dx − 4α u˜˙ u¨˜ m dS − 2α u˜˙ m ˙ dS −2 ∇u˜˙ · A∇ Ω ∂Ω ∂Ω Z Z Z +2 G000 (˜ u) u˜˙ 3 J dx + 4 G00 (˜ u) u˜˙ u¨˜ J dx + 2 G00 (˜ u) u˜˙ 2 J˙ dx. Ω

Ω

Ω

Differentiation of (3.23) gives Z ˙ + g(˜ ¨ dx (g 0 (˜ u)u¨˜Jφ + g 00 (˜ u)u˜˙ 2 Jφ + 2g 0 (˜ u)u˜˙ Jφ u)Jφ ΩZ I ¨ ˙ ˙ ¨ (˜ uφm ¨ + 2u˜˙ φm ˙ + u¨˜φm) dS u + 2∇φA∇u˜ + ∇φA∇u˜) dx + α = (∇φA∇˜ ∂Ω

Ω

If φ = 4˜ u(t) then Z Z I
u) u˜˙ u¨˜ J dx −4 ∇u¨˜ · A∇u˜˙ dx + 4 u˜˙ ∂νA u¨˜ m dS + 4 G00 (˜ ΩZ ΩZ ∂ΩI     ˙ u˜˙ dx ¨ u dx + 4 (3.29) u˜˙ ∂νA¨ u˜ m dS − 8 ∇u˜˙ · A∇ −4 ∇u˜˙ · A∇˜ Ω IΩ Z ∂Ω +8 u˜˙ ∂νA˙ u˜˙ m dS + 4 G000 (˜ u) u˜˙ 3 J dx ∂Ω Ω Z Z 00 2 u) u˜˙ 2 J¨ dx = 0. u) u˜˙ J˙ dx + 4 G0 (˜ +8 G (˜ Ω

Ω

... ... Notice that only three integrals in (3.29) contain u˜¨. They also appear in E (t). Thus E (t) does not depend on u¨˜. Hence Z Z Z     ... ... ˙ ¨ ˙ ˙ ˙ u dx + 2 ∇˜ u · ( A ∇˜ u) dx + 8 ∇u˜ · A∇u˜ dx E (t) = 2 ∇u˜ · A∇˜ ΩZ Ω Z Ω Z ... (3.30) −6 G0 (˜ u) u˜˙ J¨ dx − 2 G(˜ u) J dx − 6 G00 (˜ u) u˜˙ 2 J˙ dx Ω I Ω I ZΩ I ... 000 3 2 −2 G (˜ u) u˜˙ J dx + 6α u˜ u˜˙ m ¨ dS + 6α u˜˙ m ˙ dS + α u˜2 m dS. Ω

∂Ω

∂Ω

∂Ω

18

¨ with respect to t. Then Similarly we compute the third variation of λ. We differentiate λt we differentiate the weak formulation of (3.13) and (3.14) twice with respect to t and choose φ = −4u˜˙ . With (3.26) we get Z Z Z ... ... ¨ u dx + 6 ∇u˜˙ A∇ ˙ u˜˙ dx (3.31) ∇˜ u A ∇˜ u dx + 6 ∇u˜˙ A∇˜ λ (t) = Ω I ΩI ΩI ... +α u˜2 m dS + 6α u˜ u˜˙ m ¨ dS + 6α u˜˙ 2 m ˙ dS ∂Ω ∂Ω ∂Ω Z Z Z ... 2 2 ¨ ˙ ˙ −λ(t) u˜ J dx − 6λ(t) u˜ u˜ J dx − 3λ(t) u˜2 J¨ dx Ω Z ZΩ ZΩ ˙ ˙ u˜˙ 2 J dx − 6λ(t) u˜˙ 2 J˙ dx − 12λ(t) u˜ u˜˙ J˙ dx −6λ(t) Ω

Ω

Ω

... Thus λ (t) does not depend on u¨˜. A direct consequence is Corollary 1 The derivatives of E(t) and of λ(t) of order greater than two are expressed in terms of the derivatives of u˜ of two orders lower. This phenomenon was observed by D. D. Joseph [15] for the eigenvalues.

4 4.1

First variation Energies

˙ The goal of this section is to represent E(0) as a boundary integral. By (3.22) we have Z I Z 2 ˙ ˙ ˙ dx . E(0) = ∂i uAij (0)∂j u dx +α u m(0) ˙ dS − 2 G(u)J(0) ∂Ω |Ω {z } |Ω {z } E˙1

E˙2

From Lemma 4 we conclude after integration by parts that Z I 2 ˙ {|∇u| (v · ν) − 2(v · ∇u)(ν · ∇u)} dS − 2 g(u)(v · ∇u) dx. E1 = ∂Ω

Ω

Moreover E˙2 =

I

Z G(u)(v · ν) dS −

∂Ω

g(u)(v · ∇u) dx. Ω

Hence by (2.18) and the boundary condition (1.3) for u I ˙ E(0) = {|∇u|2 − 2G(u))(v · ν) + 2α(v · ∇u)u ∂Ω

+αu2 (div ∂Ω v τ + (n − 1)(v · ν)H)} dS.

19

Observe that v · ∇u = (v τ + (v · ν)ν) · (∇τ u + (ν · ∇u)ν) = v τ · ∇τ u − α(v · ν)u. Thus I

˙ E(0) =

∂Ω

(v · ν){|∇u|2 − 2G(u) − 2α2 u2 + α(n − 1)Hu2 } dS I (2v τ u∇τ u + u2 div ∂Ω v τ ) dS. +α ∂Ω

The last integral vanishes by (2.10). Finally we have I ˙ (4.1) E(0) = (v · ν){|∇u|2 − 2G(u) − 2α2 u2 + α(n − 1)Hu2 } dS. ∂Ω

˙ In particular we observe that E(0) = 0 for all purely tangential deformations. From the expression (4.1) above we deduce Theorem 1 Let Ωt be a family of volume preserving perturbations of Ω as described in ˙ (1.1). Then Ω is a critical point of the energy E(t), i.e. E(0) = 0, if and only if |∇u|2 − 2G(u) − 2α2 u2 + α(n − 1)u2 H = const.

(4.2)

on ∂Ω.

Proof Write for short 2

2 2

2

−1

I

z(x) := |∇u| − 2G(u) − 2α u + α(n − 1)u H and z := |∂Ω|

z dS. ∂Ω

Then, since

H ∂Ω

(v · ν) dS = 0, I

I (v · ν)(z − z) dS.

(v · ν)z dS = ∂Ω

∂Ω

Put Z ± =max{0, ±(z − z)}. Hence I I (v · ν)z dS = (v · ν)(Z + − Z − ) dS. ∂Ω

∂Ω ±

Suppose that z 6=const. Then Z 6= 0 and we can construct a volume preserving perturba˙ tion such that (v · ν) > 0 in suppZ + and (v · ν) < 0 in suppZ − . In this case we get E(0) >0 which is obviously a contradiction.  ˙ Example If Ω = BR and u(x) = u(|x|) then E(0) = 0. The question arises: are there domains other than the ball for which we can find a solution u : Ω → R of the overdetermined

20

problem ∆u + g(u) = 0 in Ω ∂ν u + αu = 0 in ∂Ω 2 2 2 2 |∇u| − 2G(u) − 2α u + α(n − 1)u H = const. in ∂Ω? Such overdetermined problems cannot be treated with the technique proposed by Serrin in [19] 4.2

Eigenvalues

The same arguments as in Section 4.1 imply that I ˙ (4.3) λ(0) = (|∇u|2 − λ(0)u2 − 2α2 u2 + α(n − 1)Hu2 )(v · ν) dS. ∂Ω

In analogy to Theorem 1 we get by the same arguments Theorem 2 Let Ωt be a family of volume preserving perturbations of Ω as described in ˙ (1.1). Then Ω is a critical point of the principal eigenvalue λ(t), i.e. λ(0) = 0, if and only if (4.4)

|∇u|2 − λu2 − 2α2 u2 + α(n − 1)u2 H = const.

in

∂Ω.

Application Let us now determine the sign of the constant in (4.4) for the Ball BR . We set z = uur and observe that n−1 dz + z2 + + λ = 0 in (0, R). dr r At the endpoint dz n−1 (R) + α2 − α + λ = 0. dr R We know that z(0) = 0 and z(R) = −α. Assume α > 0. If zr (R) > 0 then there exists a number ρ ∈ (0, R) such that zr (ρ) = 0, z(ρ) < 0 and zrr (ρ) ≤ 0. From the equation we get zrr (ρ) = n−1 > 0 which leads to a contradiction. Consequently ρ α(n − 1) + λ > 0. R Similarly we prove that A < 0 if α < 0. Consequently we have for α > 0 (< 0) (4.5)

A := α2 −

˙ λ(0) < 0 (< 0) H for all volume increasing perturbations ∂BR v · ν dS > 0. Notice that this observation extends partly the result of Giorgi and Smits [9] who proved that λ(Ω) > λ(BR ) for any Ω ⊂ BR . The result for negative α was observed in [2].

21

An equation for u0

5

In this section we derive a boundary value problem for the function u0 defined in (3.17). Let u˜(t) solve (3.7) - (3.8). If we differentiate with respect to t and evaluate the derivative at t = 0 we get (5.1) (5.2)

˙ LA(0) u˜˙ (0) + LA(0) u˜(0) + g 0 (˜ u(0))u˜˙ (0) J(0) + g(˜ u(0))J(0) = 0 ˙ ∂νA(0) u˜˙ (0) + ∂νA(0) u˜(0) + α m(0) u˜˙ (0) + α m(0) ˙ u˜(0) = 0 ˙

in Ω in ∂Ω.

From Lemma 4 we then get ∆u0 + g 0 (u) u0 = 0 in Ω. The computation for the boundary condition for u0 is more involved. ∂νA(0) u˜˙ (0) = ∂ν (v · ∇u) + ∂ν u0 ∂νA(0) u˜(0) = div v ∂ν u − νj ∂j vi ∂i u − ∂i u ∂i vj νj ˙ = div v ∂ν u − ν · Dv ∇u − ∇u · Dv ν ˙ α m(0) u˜(0) = α v · ∇u + α u0 α m(0) ˙ u˜(0) = α(n − 1)(v · ν)H u + α u div ∂Ω v τ . Inserting these expressions into (5.2) and taking into account (2.18) and the boundary condition ∂uν + αu = 0, we obtain ∂ν u0 + αu0 = −∂ν (v · ∇u) + ∇u · Dv ν + ν · Dv ∇u − αv · ∇u +αu(div v − (n − 1)(v · ν)H − div ∂Ω v τ ). We observe that since div ∂Ω ν = (n − 1)H, div v = div ∂Ω v τ + (n − 1)(v · ν) H + νi ∂i vj νj

on ∂Ω.

Thus ∂ν u0 + αu0 = −∂ν (v · ∇u) + ∇u · Dv ν + ν · Dv ∇u −α v · ∇u + α u ν · Dv ν. In view of (2.7) and the boundary condition for u we have ∇u · Dv ν = −α u ν · Dv ν + ∇τ u · Dv ν. Hence ∂ν u0 + αu0 = −∂ν (v · ∇u) + ∇τ u · Dv ν + ν · Dv ∇u − α v · ∇u. Thus u0 solves (5.3) (5.4)

∆u0 + g 0 (u) u0 = 0 in Ω ∂ν u0 + αu0 = −∂ν (v · ∇u) + ∇τ u · Dv ν + ν · Dv ∇u − α v · ∇u in ∂Ω.

22

Analogously we get for the eigenvalue problem ˙ ∆u0 + λ(0)u0 + λ(0)u = 0 in Ω 0 0 ∂ν u + αu = −∂ν (v · ∇u) + ∇τ u · Dv ν + ν · Dv ∇u − α v · ∇u in ∂Ω.

(5.5) (5.6)

Examples 1. Of special interest will be the case where Ω is the ball BR of radius R centered at the origin and u is a radial solution of ∆u + g(u) = 0 in BR with ∂ν u + αu = 0 on ∂BR . Then (5.4) becomes   α(n − 1) 2 0 0 (5.7) u(R) + α u(R) v · ν. ∂ν u + αu = g(u(R)) − R For the torsion problem g(u) = 1 we have u(x) =

(5.8) Inserting u(R) = (5.9) (5.10)

R αn


R 1 + R2 − |x|2 . α n 2n

and g 0 (u) = 0 into (5.3) and (5.4) we obtain ∆u0 = 0 in BR   1+αR 0 0 ∂ν u + αu = v·ν n

in ∂BR .

2. Similarly we get for the eigenvalue problem in BR 0 ˙ (5.11) ∆u0 + λ(0)u0 + λ(0)u = 0

(5.12)

6

in BR

∂ν u0 + αu0 = ([1 + α R − n]α + λ R)

u(R) (v · ν) R

in ∂BR .

The second domain variation

¨ The aim of this section is to find a suitable form of E(0) in order to determine its sign. ¨ Recall that E(t) is given by (3.25) and that consequently Z I Z 2 ¨ ¨ (6.1) E(0) = ∇uA∇u dx + α um ¨ dS − 2 G(u)J¨ dx Ω ∂Ω Ω Z I Z 2 ˙ ˙ ˙ −2 ∇u˜A∇u˜ dx − 2α u˜ dS + 2 g 0 (u)u˜˙ 2 dx. Ω

∂Ω

Ω

For the moment we do not assume that Ω is a critical domain. This enables us to give a rather general formula.

23

The following integrals which appear in (6.1), will be expanded with respect to t. Z (6.2) F1 (t) := A¨ij (t) ∂i u˜(t) ∂j u˜(t) dx Ω I (6.3) u˜2 (t) m(t) ¨ dS F2 (t) := α ∂Ω Z ¨ dx (6.4) F3 (t) := −2 G(˜ u(t)) J(t) Ω Z (6.5) F4 (t) := −2 Aij (t) ∂i u˜˙ (t) ∂j u˜˙ (t) dx Ω I (6.6) F5 (t) := −2α u˜˙ 2 (t)m(t) dS ∂Ω Z (6.7) F6 (t) := 2 g 0 (˜ u)u˜˙ 2 (t)J(t) dx. Ω

6.1

The expression F1 (0) + F4 (0)

From Lemma 4 we have Z Z
2 2 F1 (0) = (div v) − Dv : Dv |∇u| dx + 2 (∂k vi ∂j vk + ∂k vj ∂i vk ) ∂i u ∂j u dx Ω Z Ω Z +2 ∂k vi ∂k vj ∂i u ∂j u dx − 2 div v (∂j vi + ∂i vj ) ∂i u ∂j u dx + D, Ω

Ω

where

Z

Z D=−

(∂i wj + ∂j wi )∂i u∂j u dx +

div w|∇u|2 dx.

Ω

Ω

Using our notation (Dv )ij = ∂j vi we rewrite this as Z Z
2 2 F1 (0) = (div v) − Dv : Dv |∇u| dx + 4 (∇u · Dv ) · (Dv ∇u) dx Ω Z Ω Z +2 (Dv ∇u) · (Dv ∇u) dx − 4 div v ∇u · Dv ∇u dx + D. Ω

Ω

From Lemma 4 and (3.17) we also have Z Z Z F4 (0) := −2 ∂i u˜˙ (0) ∂i u˜˙ (0) dx = −2 ∂i vk ∂k u ∂i vl ∂l u dx − 2 vk ∂k ∂i u vl ∂l ∂i u dx Ω Ω Ω Z Z Z −2 |∇u0 |2 dx − 4 vl ∂l ∂i u ∂i vk ∂k u dx − 4 ∂i u0 ∂i vk ∂k u dx Ω Ω ZΩ −4 vk ∂k ∂i u ∂i u0 dx. Ω

24

Moreover in terms of matrices we have, setting (D2 u)ij = ∂i ∂j u, Z Z F4 (0) := −2 ∂i u˜˙ (0) ∂i u˜˙ (0) dx = −2 (Dv ∇u) · (Dv ∇u) dx Ω Ω Z Z Z 2 2 0 2 −2 (D u v) · (D u v) dx − 2 |∇u | dx − 4 (D2 u v) · (Dv ∇u) dx Ω Z Ω ZΩ −4 ∇u0 · (Dv ∇u) dx − 4 (D2 u v) · ∇u0 dx. Ω

Ω

R For the sum F1 (0) + F4 (0) we observe that the integral 2 Ω (Dv ∇u) · (Dv ∇u) dx cancels: Z Z
2 2 (div v) − Dv : Dv |∇u| dx + 4 (∇u · Dv ) · (Dv ∇u) dx F1 (0) + F4 (0) = Ω Z Ω Z 2 −4 div v ∇u · Dv ∇u dx − 2 (D u v) · (D2 u v) dx ZΩ ZΩ Z 2 0 2 −4 (D u v) · (Dv ∇u) dx − 2 |∇u | dx − 4 ∇u0 · (Dv ∇u) dx Ω Ω ZΩ −4 (D2 u v) · ∇u0 dx + D. Ω

Observe that the last two integrals can be written as Z Z 0 −4 ∇u · (Dv ∇u) dx − 4 (D2 u v) · ∇u0 dx Ω Ω Z I 0 = 4 v · ∇u ∆u dx − 4 v · ∇u ∂ν u0 dS. Ω

∂Ω

We will show that F1 (0) + F4 (0) can be written as a sum of boundary integrals and two domain integrals involving the Laplace operator. The computations are done in three steps. Step 1 We observe that Z Z I := −4 div v ∇u · (Dv ∇u) dx − 4 (v · D2 u) · (Dv ∇u) dx Ω ZΩ = −4 ∂j (vj ∂i u) ∂i vk ∂k u dx Z Ω Z = 4 vj ∂i u ∂i ∂j vk ∂k u dx + 4 vj ∂i u ∂i vk ∂j ∂k u dx ΩI Ω −4 (v · ν) ∇u · (Dv ∇u) dS. ∂Ω

We integrate again the integral 4

R Ω

vj ∂i u ∂i ∂j vk ∂k u dx by parts. This gives a term with

25

∆u: Z

Z

I := −4

∆u v · (Dv ∇u) dx − 4

(∇u · Dv ) · (Dv ∇u) dx Z 2 −4 (v · Dv ) · (D u∇u) dx + 4 (∇u · Dv ) · (D2 u v) dx IΩ I Ω +4 ∂ν u v · (Dv ∇u) dS − 4 (v · ν) ∇u · (Dv ∇u) dS. ZΩ

Ω

∂Ω

∂Ω

Then Z F1 (0) + F4 (0) =

2

2




Z

(div v) − Dv : Dv |∇u| dx − 4 ∆u v · (Dv ∇u) dx Ω Z Z −4 (v · Dv ) · (D2 u∇u) dx + 4 (∇u · Dv ) · (D2 u v) dx ZΩ I Ω −2 (D2 u v) · (D2 u v) dx + 4 ∂ν u v · (Dv ∇u) dS Ω ∂Ω I Z −4 (v · ν) ∇u · (Dv ∇u) dS − 2 |∇u0 |2 dx Ω I Z∂Ω 0 v · ∇u ∂ν u0 dS + D. +4 v · ∇u ∆u dx − 4 Ω

∂Ω

Ω

Step 2 Again by partial integration we get Z Z 2 −2 (v · Dv ) · (D u∇u) dx − 2 (D2 u v) · (D2 u v) dx Ω Z Ω Z 2 = 2 div v v · (D u∇u) dx + 2 vi vj ∂k u ∂i ∂j ∂k u dx Ω Ω I (v · ν) v · (D2 u∇u) dS. −2 ∂Ω

Moreover Z Z Z 2 2 4 (∇u · Dv ) · (D u v) dx = −2 ∆u v · (D u v) dx − 2 vi vj ∂k u ∂i ∂j ∂k u dx Ω Ω IΩ 2 +2 ∂ν u v · (D u v) dS. ∂Ω

26

Thus Z F1 (0) + F4 (0) =

2

2




Z

(div v) − Dv : Dv |∇u| dx − 4 ∆u v · (Dv ∇u) dx Ω Z Z −2 (v · Dv ) · (D2 u∇u) dx + 2 div v v · (D2 u∇u) dx Ω ZΩ I −2 ∆u v · (D2 u v) dx − 2 (v · ν) v · (D2 u ∇u) dS ∂Ω I IΩ 2 ∂ν u v · (D u v) dS + 4 ∂ν u v · (Dv ∇u) dS +2 ∂Ω ∂Ω I Z −4 (v · ν) ∇u · (Dv ∇u) dS − 2 |∇u0 |2 dx Ω Z∂Ω I +4 v · ∇u ∆u0 dx − 4 v · ∇u ∂ν u0 dS + D. Ω

Ω

∂Ω

Step 3 Finally we note that
div [v div v − v · Dv ]|∇u|2
= (div v)2 − Dv : Dv |∇u|2 + 2(D2 u∇u) · (v div v − v · Dv ). In addition straightforward partial integration implies I I I 2 (6.8) (w · ν)G(u) dS (w · ν)|∇u| dS − 2 wi ∂i u∂ν u dS + D =−2 ∂Ω ∂Ω ∂Ω Z + 2 G(u)div w dx. Ω

In summary we have proved Proposition 1 A formal computation without any further assumption on v yields Z Z
2 F1 (0) + F4 (0) = div [v div v − v · Dv ]|∇u| dx − 4 ∆u v · (Dv ∇u) dx Ω Z Ω I −2 ∆u (D2 u v) · v dx + 4 ∂ν u v · (Dv ∇u) dS Ω ∂Ω I I −4 (v · ν) ∇u · (Dv ∇u) dS + 2 (D2 u v) · v ∂ν u dS I∂Ω Z ∂Ω −2 (v · ν) v · (D2 u∇u) dS + 4 v · ∇u ∆u0 dx Ω I∂Ω Z −4 v · ∇u ∂ν u0 dS − 2 |∇u0 |2 dx + D. ∂Ω

Ω

27

6.2

The expression F3 (0) + F6 (0)

From (6.4), (3.19) and (3.16) we have Z Z
¨ u(0)) J(0) dx = −2 G(u(x)) (div v)2 − Dv : Dv + div w dx. F3 (0) = −2 G(˜ Ω

Ω

Using again the fact that (div v)2 − Dv : Dv = div (v div v − v · Dv ) we get Z F3 (0) = 2

g(u(x)) (v · ∇u div v − v · (Dv ∇u)) dx ΩI

−2

G(u(x)) ((v · ν) div v − v · (Dv ν)) dS Z∂Ω

−2

G(u)div w dx. Ω

From (6.7), (3.21) and (3.16) we have Z Z 2 0 2 ˙ F6 (0) = 2 g (˜ u(0)) u˜ (0) J(0) dx = 2 g 0 (u(x)) (v · ∇u + u0 ) dx Ω Z ZΩ 2 0 = 2 g (u(x)) (v(x) · ∇u(x)) dx + 4 g 0 (u(x)) v(x) · ∇u(x) u0 (x) dx Ω ΩZ 0 02 +2 g (u(x)) u (x) dx. Ω

We note that Z Z 2 0 2 g (u(x)) (v(x) · ∇u(x)) dx = 2 v · ∇g(u) v · ∇u dx Ω Ω Z Z = −2 g(u(x))div v(x) v(x) · ∇u dx − 2 g(u(x)) v(x) · (Dv ∇u(x)) dx ΩZ Ω I 2 g(u(x)) v(x) · ν v · ∇u(x) dS. −2 g(u(x)) v(x) · (D u(x)v(x)) dx + 2 Ω

From this we easily deduce the following proposition.

∂Ω

28

Proposition 2 A formal computation without any further assumption on v yields Z F3 (0) + F6 (0) = −4 g(u(x)) v(x) · (Dv ∇u(x)) dx ZΩ −2 g(u(x)) v(x) · (D2 u(x) v(x)) dx IΩ −2 G(u) (v(x) · ν div v − v(x) · (Dv ν)) dS Z I∂Ω g(u(x)) v(x) · ν v · ∇u(x) dS + 2 g 0 (u(x)) u02 (x) dx +2 Ω Z∂Ω +4 g 0 (u(x)) v(x) · ∇u(x) u0 (x) dx ZΩ −2 G(u)div w dx. Ω

6.3

The expression F2 (0) + F5 (0)

From (6.3) and (3.16) we deduce I I 2 F2 (0) := α u˜ (0) m(0) ¨ dS = α ∂Ω

u2 (x) m(0) ¨ dS.

∂Ω

We will not use the explicit for of m(0). ¨ From (3.17) and the fact that m(0) = 1 we obtain I u˜˙ 2 (0)m(0) dS F5 (0) := −2α I∂Ω I I 2 0 = −2α (v · ∇u) dS − 4α v · ∇u u dS − 2α u02 dS. ∂Ω

∂Ω

∂Ω

Thus I (6.9) F2 (0) + F5 (0) = −2α I +α

2

I ∂Ω

u2 (x) m(0) ¨ dS.

∂Ω

6.4

I

v · ∇u u dS − 2α

(v · ∇u) dS − 4α ∂Ω

0

Main result

Adding up all these contributions we arrive at our final result.

∂Ω

u02 dS

29

Theorem 3 Assume that ∆u + g(u) = 0 in Ω and ∂ν u + αu = 0 on ∂Ω. Let u0 satisfy ¨ (5.3) and (5.4). Then the second variation E(0) can be expressed in the form I
¨ (6.10) E(0) [(v · ν) div v − v · (Dv ν) + w · ν] |∇u|2 − 2G(u) dS = −2Qg (u0 ) + ∂Ω I +4 (∂ν u v · (Dv ∇u) − (v · ν) ∇u · (Dv ∇u)) dS I∂Ω
∂ν u v · (D2 u v) − (v · ν) v · (D2 u∇u) dS +2 I∂Ω I +2 g(u) (v · ν) v · ∇u dS − 4 v · ∇u (∂ν u0 + αu0 ) dS ∂Ω ∂Ω I I I 2 (v · ∇u) dS − 2 w · ∇u ∂ν u dS + α u2 (x) m(0) ¨ dS, −2α ∂Ω

∂Ω

∂Ω

where (6.11)

Z

0

0 2

Z

|∇u | dx −

Qg (u ) := Ω

0

02

I

g (u)u dx + α Ω

u02 dS

∂Ω

is a form in u0 . This formula is very general because no volume constraint is used. It could for instance be used to study problems with a prescribed perimeter.

7 7.1

Applications to nearly spherical domains Second variation

We evaluate (6.10) if Ω = BR , u = u(|x|) and when the domain perturbations preserve the volume and satisfy (2.15) and (2.14). Then 1 ∂i ∂j u(|x|)|∂BR = ∂ν2 u(R) νi νj + ∂ν u(R) (δij − νi νj ) . R 2 Since |∇u| −2G(u) = const. on ∂BR the contribution in the first integral of (6.10) vanishes by (2.15). Keeping in mind the Robin boundary condition for u we get I I 0 2 2 2 ¨ (7.1) m(0) ¨ dS + 4α u (R) v τ · Dv ν dS E(0) = −2Qg (u ) + α u (R) ∂BR ∂BR I I 2α2 2 + u (R) (v τ )2 dS − 2α u(R) g(u) (v · ν)2 dS R ∂BR I ∂BR I 0 0 3 2 +4α u(R) (v · ν) (∂ν u + αu ) dS − 2α u (R) (v · ν)2 dS ∂BR I ∂BR −2ur (R)2 w · ν dS. ∂BR

30

We need the following technical lemma for v τ . Lemma 5 For volume preserving perturbations there holds I I 2α2 2 τ 2 2 2 u (R) (v ) dS = −4α u (R) v τ · Dv ν dS R ∂BR ∂BR I I 2α2 (n − 1) 2 2 2 2 + u (R) (v · ν) dS + 2α u(R) w · ν dS. R ∂BR ∂BR Proof At first observe that I I 1 τ 2 (v ) dS = v · Dν v dS. R ∂BR ∂BR On the other hand I I v · Dν v dS = ∂BR

viτ (∇τi νk )vk dS

∂BR

I

I

dS − div ∂BR v τ (v · ν) dS ∂BR I∂BR I = − v · Dv ν dS − div v (v · ν) dS ∂BR ∂BR I I +2 (v · ν) ν · Dv ν dS + div ∂BR ν (v · ν)2 dS.

= −

viτ (∇τi vk )νk

∂BR

Next we use (2.15). Then I I v · Dν v dS = −2

n−1 v · Dv ν dS + R ∂BR

∂BR

τ

∂BR

I

2

I

(v · ν) dS + ∂BR

w · ν dS. ∂Ω

t u

This proves the claim.

This lemma together with the Robin condition ur (R) + αu(R) = 0 implies that (7.1) can be written as (7.2)

¨ = −2Qg (u0 ) E(0) I I 2 +α u (R) m(0) ¨ dS + 4α u(R) (v · ν) (∂ν u0 + αu0 ) dS ∂BR ∂BR I α(n − 1) 2 −2α u(R) g(u) − u(R) − α u(R) (v · ν)2 dS. R ∂BR

We rewrite (5.4) for the radial situation. Recall that ∂ν u0 + αu0 = −∂ν (v · ∇u) + ∇τ u · Dv ν + ν · Dv ∇u − α v · ∇u

in ∂Ω.

31

Thus in the radial case we get ∆u0 + g 0 (u)u0 = 0 in BR ∂ν u0 + αu0 = kg (u(R))(v · ν) in ∂BR ,

(7.3) (7.4) with

α(n − 1) u(R) + α2 u(R). R ¨ We can insert (7.4) into (7.2) and obtain E(0) as a quadratic functional in u0 alone. I I 2α u(R) 2 0 2 ¨ E(0) = −2Qg (u ) + α u (R) m(0) ¨ dS + (7.6) (∂ν u0 + αu0 ) dS. k (u(R)) g ∂BR ∂BR (7.5)

kg (u(R)) := g(u(R)) −

Further simplification is possible if we use the equation (5.3) for u0 . We multiply this equation with u0 and integrate over BR . This leads to I 0 (∂ν u0 + αu0 )u0 dS. Qg (u ) = ∂BR

Lemma 6 For every volume preserving perturbation of the ball and for radially symmetric solutions u we have I I 0 0 0 2 ¨ (7.7) E(0) = −2 (∂ν u + αu )u dS + α u (R) m(0) ¨ dS ∂BR ∂BR I 2α u(R) 2 (∂ν u0 + αu0 ) dS. + kg (u(R)) ∂BR Remark 4 The second variation is independent of v τ and w. We can therefore restrict ourselves to Hadamard perturbations y = x + tN ν + O(t2 ). Consider the case where (v · ν) = 0 on ∂BR . Then by (7.3),(7.4) and (6.11) we have ¨ Qg (u0 ) = 0. Moreover by (2.25), Lemma 3 (7.6) it follows that E(0) = 0. Consequently ¨ perturbations which preserve the volume and with (v · ν) = 0 lie in the kernel of E(0). 7.2 7.2.1

¨ Discussion of the sign of E(0) in the radial case General strategy

Recall that by (7.6) and (7.3) ¨ = αu2 (R)S(0) ¨ +F E(0)

(7.8) where

0

I

F := −2Qg (u ) + 2α u(R)kg (u(R)) ∂BR

(v · ν)2 dS.

32

¨ > 0. In order to estimate F we consider the following Steklov eigenvalue By Lemma 3 S(0) problem ∆φ + g 0 (u)φ = 0 in BR , ∂ν φ + αφ = µφ on ∂BR .

(7.9)

If g 0 (u) is bounded there exists an infinite number of eigenvalues µ1 < µ2 ≤ µ3 ≤ ... lim µi = ∞. i→∞

and a complete system of eigenfunctions {φi }i≥1 . Testing (7.9) with φj we find Z I I 0 [−∇φi · ∇φj + g (u)φi φj ] dx − α φi φj dS + µi φi φj dS = 0. BR

∂BR

∂BR

If we interchange i and j we see immediately that the system of eigenfunctions {φi }i can be chosen such that I (7.10) φi φj dS = δij , ∂BR I Z Z 0 φi φj dS = µi δij . g (u)φi φj dx + α ∇φi · ∇φj dx − and q(φi , φj ) := ∂BR

BR

BR

We write u0 (x) =

∞ X

ci φi

and

(v · ν) =

i=1

∞ X

bi φ i .

i=1

Note that the first eigenfunction φ1 is radially symmetric and does not change. The condition I I 0= (v · ν) dS = φ1 (v · ν) dS ∂BR

∂BR

implies that b1 = 0. It gives a condition on c1 µ1 if we take into account (7.3) - (7.4): I 0 = (∂ν u0 + αu0 ) φ1 dS Z∂BR Z I 0 0 = ∆u φ1 dx + ∇u ∇φ1 dx + α u0 φ1 dS BR ∂BR IBR I = u0 ∂ν φ1 dS + α u0 φ1 dS ∂BR ∂BR I = µ1 u0 φ1 dS = c1 µ1 . ∂BR

33

Thus c1 µ1 = 0. The coefficients bi for i ≥ 2 are determined from the boundary value problem (5.3), (5.4). In fact ci µ i for i = 2, 3 . . . . bi = kg (u(R)) By means of the orthonormality conditions of the eigenfunctions we find ∞ ∞ X X Qg (u0 ) = q(u0 , u0 ) = c2i q(φi , φi ) = c2i µi . i=1

i=2

Inserting this into (7.8) we find (7.11)

F =2

∞ X 2

c2i

µ2i



 αu(R) 1 − , kg (u(R)) µi

where kg is defined in (7.5). Let µp = min{µi : µi > 0} be the smallest positive eigenvalue. Then    I ∞ X αu(R) 1 αu(R) 1 2 2 2 (v · ν)2 dS. F ≥2 ci µ i − = 2kg (u(R)) − k (u(R)) µ k (u(R)) µ g p g p ∂BR 2 The expression F vanishes if • u0 = 0 • µi = kgαu(R) for some i, and (ν · v) = di φi . (u(R)) The first case occurs only in the case of translations. This together with Lemma 3 implies ¨ Lemma 7 The kernel of E(0) consists only on first order translations (ν · v) = ai xi . ¨ In order to get an estimate of E(0) in terms of v we impose the ”barycenter” condition I (7.12) x (v(x) · ν(x)) dS = 0, ∂BR

By (2.25) and (2.26) it then follows that I I 2n τ 2 |∇ N | dS ≥ 2 N 2 dS R ∂BR ∂BR H ¨ ≥ n+1 and thus S(0) N 2 dS. Observe that bi = 0 for i = 1, . . . , n. Hence the estimate R2 ∂BR (7.13) can be improved by replacing µp by µp0 = min{µk > 0, k > n}. This together with the estimate for F given above implies I  2kg2 (u(R)) n + 1 2 ¨ ≥ αu (R) (7.13) E(0) + 2kg (u(R))αu(R) − (v · ν)2 dS. 2 0 R µp ∂BR In summary we have

34

Theorem 4 The second variation of E for volume preserving perturbations of the first and second order is of the form ¨ = αu2 (R)S(0) ¨ + F. E(0) (i) If α > 0 it is bounded from below by  I αu(R) 1 2 2 ¨ + 2kg (u(R)) αu (R)S(0) − (v · ν)2 dS. kg (u(R)) µp ∂BR (ii) Under the additional assumption (7.12) we have for α > 0 I  2kg2 (u(R)) n+1 2 ¨ (v · ν)2 dS. E(0) ≥ αu (R) 2 + 2kg (u(R))αu(R) − R µp 0 ∂BR 7.2.2

Applications

1. The torsion problem g = 1 The problem is well-posed provided α 6= 0. From (5.8) we have u(R) = k1 =

R αn

and by(5.10)

k1 (u(R)) 1 + αR 1 + αR and = . n αu(R) R

The Steklov problem (7.9) is in this case −∆φ = 0 in BR ,

∂ν φ + αφ = µφ on ∂BR .

An elementary computation yields µ1 − α = 0 and µk − α = k−1 (for k ≥ 2 and counted R without multiplicity). The second eigenvalue µ2 = 1/R + α has multiplicity n and its ¨ ≥ S(0) ¨ ≥ 0. eigenfunctions are xR1 , . . . , xRn . If α > 0 then µ2 > 0 and by Theorem 4 (i) E(0) Equality holds only for translations. The estimate can be improved by assuming (7.12). Notice that this condition implies in addition to c1 = 0 also also that c2 = · · · = cn = 0. Hence we can take µp0 = R2 + α. By Theorem 7.13 (ii)  I (1 + αR)R n+1 ¨ +2 2 (v · ν)2 dS > 0. E(0) ≥ αn2 n (2 + αR) ∂BR Next consider the case where − R1 < α < 0. Then µ2 > 0 and thus Q(u0 ) > 0. Moreover ¨ < 0. F < 0 and consequently by (7.8) the second variation becomes negative, E(0) This property is not longer true if α ≤ −1/R. In fact we can always find ci or equivalently ¨ bi such that F > 0 orF < 0 and E(0) is positive or negative, respectively. For the torsion problem we have proved the following ¨ > 0 for all volume preserving perturbations. Theorem 5 (i) Assume α > 0. Then E(0) 1 ¨ < 0 for all volume preserving perturbations. (ii) If − R < α < 0 then E(0) 1 ¨ (ii) If α ≤ − R then the sign of E(0) can change depending on he particular perturbation.

35

2. Principal eigenvalue g(u) = λu In [2] it was shown that for α0 < α < 0 the ball yields the maximal principal eigenvalue among all nearly spherical solutions of the same volume. In this section we therefore restrict ourselves to the discussion of the case α > 0. √ n−2 The first eigenfunction is of the form u = u(r) = J n−2 ( λr)r− 2 . Furthermore 2

kλu (u(R)) := (α2 R − α(n − 1) + λR)

u(R) . R

¨ Then by (4.5), A := α2 R − α(n − 1) + λR is positive. In order toR prove that λ(0) is non 2 negative we shall use the form (7.6). Under the assumption that BR u dx = 1 it follows that I 0 ¨ λ(0) = −2Qλu (u ) + 2αu(R)kλu (7.14) (v · ν)2 dS. ∂BR

The corresponding Steklov eigenvalue problem is (7.15) (7.16)

∆φ + λφ = 0 ∂ν φ + αφ = µφ

in BR in ∂BR .

Notice that φ1 = u and therefore φ1 =const. on ∂BR . Moreover µ1 = 0, therefore µp = µ2 . − µ12 in Theorem 4 . This is equivalent Next we want to check the sign of the expression αu(R) kλu to the sign of n−1 λ L := µ2 − α + − . R α For this purpose we need the eigenvalues of (7.16). The eigenfunctions of (7.15) − (7.16) are of the form X φ(x) = cs,i as (r) Ys,i (θ), θ ∈ S n−1 . s,i

Here s ∈ N ∪ {0} and i = 1, . . . , ds for ds = (2s + n − 2) (s+n−3)! ∈ N. The function Ys,i (θ) s!(n−2)! denotes the i - th spherical harmonics of order s. In particular ∆∗ Ys,i + s(s + n − 2)Ys,i = 0

in S n−1 ,

where ∆∗ is the Laplace Beltrami operator on the sphere. As a consequence of this Ansatz we get from (7.15) √ 2−n as (r) = r 2 Js+ n2 −1 ( λ r). The corresponding eigenvalue follows from (7.16), namely a0s (R) = (µ − α)as (R).

36

Since the first eigenfunction does not change sign u = r−

n−2 2

√ J n−2 ( λr). 2

It follows from the well-known Bessel identity (z −ν Jν (z))z = −r−ν Jν+1 (z) and from ur (R) + αu(R) = 0 that √ (7.17)

α=

√ Jn/2 ( λR) √ λ . J(n−2)/2 ( λR)

The eigenfunctions corresponding to µ2 span the n-dimensional linear space (s = 1) φ(r, θ) =

n X

c2,i r

2−n 2

√ J n2 ( λ r)Y,i (θ).

i=1

The boundary condition gives by means of the same identity as before √ √ Jn/2+1 ( λR) 1 √ +α− λ = µ2 . R Jn/2 ( λR) If we replace α and µ2 we obtain

√ √ √ n λ √ L= − (J n2 +1 ( λR) + J n2 −1 ( λR)). R J n ( λR) 2

From the identity (7.18)

nJn/2 (z) = z(Jn/2+1 (z) + Jn/2−1 (z)

it follows that L = 0. Consequently for all v 6= const. ¨ ¨ > 0. λ(0) ≥ αu2 (R)S(0)

(7.19)

As for the torsion problem the inequality can be improved by imposing the barycenter conditions (2.15). The positivity of the decond variation is in accordance with DanersBossel’s inequality [5].

8

The ball is optimal

By the Taylor expansion t2 ˙ E(t) = E(0) + t E(0) + 2

¨ + E(0)

Z 0

tˆ

... E (s) ds

!

37

... for some tˆ ∈ (−t, t). If | E | ≤ c then for the critical domain E(t) ≥ E(0) +

t2 (E(0) − c t) 2

which shows that ... for small t that E(0) is minimal. In the first step we find upper bounds ... ... for m, J and A . The main tool is the formula !k ∞ ∞ X X 1 (−1)j j det (Id + tA)) = − tr((t A) ) k! j j=1 k=0 where tr(A) denotes the trace of the matrix A. In our case the matrix A will depend on t as well: t ˜ ˜ = Dw . A := A˜ + B where A˜ = Dv , B 2 We now assume (8.1)

kDv kL∞ (Ω) + kDw kL∞ (Ω) ≤ 1 and 0 ≤ t < 1/2,

then c

det (Id + tA)) ≤ e 1−t ≤ c and c does not depend on v or d det (Id + tA) , dt

w. With these assumptions we also get the estimates 2 3 d d dt2 det (Id + tA) , dt3 det (Id + tA) ≤ c

and again c does not depend on v or w. With this we can easily prove the following lemma Lemma 8 Let J (resp. m and A) be defined as in (2.12) (resp. (2.16) and (3.5)). Moreover we assume (8.1) for the vector fields v, w and the parameter t. Then the following estimates hold: ... m(t) + m(t) ˙ + m(t) ¨ + m(t) ≤ c0 ˙ + J(t) ¨ + ... J(t) + J(t) J (t) ≤ c1 ... ˙ ¨ kA(t)k + kA(t)k + kA(t)k + k A (t)k ≤ c1 , where c0 and c1 do not depend on v and w. k · k denotes any matrix norm. In a final step we assume that for some number c ∈ R we have (8.2)

|G0 (u)| ≤ c.

38

Then from (3.7) and (3.8) and the corresponding equation for u˜˙ we get Z I 2 |∇˜ u| dx + α u˜2 dS ≤ c IΩ ZΩ |∇u˜˙ |2 dx + α u˜˙ 2 dS ≤ c. Ω

Ω

Theorem 6 Let t ∈ R and let v and w be two smooth vector fields satisfying (8.1). Then there exists a number c ∈ R which is independent of v and w such that ... | E (t)| ≤ c

1 ∀0≤t< . 2

Consequently ¨ ≥ E(0) ¨ −ct E(t)

1 ∀0≤t< . 2

¨ For t sufficiently small we thus get the uniform positivity of E(t). ¨ does not depend on u¨˜ it is also independent of the tangential component of v, Since E(t) and w.

9

Back to Garabedian and Schiffer’s second variation

In [7] the authors computed the second domain variation of the first Dirichlet eigenvalue of the Laplace operator for the ball. Since the Krahn - Faber inequality holds one would ¨ D (0). However, from the formula Garabedian and Schiffer expect the strict positivity of λ obtained, namely Z I
2 2 ¨ (∂ν u) (v · ν) H dS − 2 |∇u˜˙ (0)|2 − λ(u˜˙ (0))2 dx λD (0) = − ∂Ω

Ω

¨ it seems R to2 be difficult to show that λD (0) ≥ 0. Throughout this section we shall assume that BR u dx = 1. Following the device of our paper we find Z I 1¨ n−1 0 2 02 λD (0) = |∇u | − λD u dx + u02 dS. 2 R BR ∂BR Here u0 satisfies the equation ∆u0 + λD u0 = 0 in BR 0 u = (v · ν)∂ν u

in ∂BR .

39

In this computation λ˙ D (0) = 0 is already taking into account. We define R |∇φ|2 − λD φ2 dx B . Rs (φ) = R H φ2 dS ∂BR and we set 

 φ dS = 0 ,

I

µ := inf Rs (φ), ∂BR

From the previous considerations we observe that µ = µ2 − α where µ2 is as in the previous subsection. As before p 2−n u = u(r) = c r 2 J n−2 ( λD r). 2

λD is determined by the boundary condition u(R) = 0, i.e. p (9.1) J n−2 ( λD R) = 0. 2

By the same arguments as in the previous section ( )I √ √ λD J n2 +1 ( λD R) 1¨ n √ u02 dS. λD (0) ≥ − 2 R J n2 ( λD R) ∂BR The identity (7.18) and (9.1) imply that ¨ λ(0) ≥ 0. As in the last section the equality sign can be excluded if v satisfies (7.12). 9.1

The Case of Dirichlet data

In case of Dirichlet data u = 0 on ∂Ω the energy E(t) has the form Z Z (9.2) E(t) = ∇˜ uA∇˜ u dx − 2 G(˜ u) J dx. Ω

Ω

As in (3.7) the function u˜ solve LA u˜(t) + g(˜ u(t)) J(t) = 0

in Ω

with the boundary condition (3.8) replaced by (9.3)

u˜(t) = 0

in ∂Ω.

The function u0 solves (9.4) (9.5)

∆u0 + g 0 (u) u0 = 0 in Ω 0 u = −v · ∇u in ∂Ω.

40

In complete analogy with Chapter 4.1 we get I  ˙ (9.6) E(0) = (v · ν) |∇u|2 − 2G(u) dS ∂Ω

˙ Thus any critical domain for which E(0) = 0 satisfies the overdetermined boundary condition (9.7)

|∇u|2 − 2G(u) = const.

in ∂Ω,

thus (9.8)

|∇u| = c0

in ∂Ω.

Note that by a result of Serrin for positive solutions this would already imply that Ω is a ball. For the second variation we observe that only S and F1 + F4 + F3 + F6 contribute. Hence ¨ = S(0) ¨ + F1 (0) + F4 (0) + F3 (0) + F6 (0), E(0) Computations very similar to those in Chapter 6 lead to the following lemma. Lemma 9 Let Ω be a smooth domain and let E(t) be as in (9.2). Let u be a solution of ∆u + g(u) = 0 in Ω and u = const. on ∂Ω. Let u0 be a solution of (9.4) - (9.5). For any ˙ critical domain Ω in the sense that E(0) = 0 we have I I 2g(0) 0 02 2 ¨ (9.9) E(0) = 2Qg (u ) − (v · ν)2 H dS, u dS + 2(n − 1) c0 c0 ∂Ω ∂Ω where c0 is given by (9.8) and H denotes the mean curvature of ∂Ω.. Acknowledgement The authors would like to thank J. Arrieta E. Harrell and M. Pierre for drawing our attention to some of the more recent literature on the subject.

References [1] C. Bandle and A. Wagner, Domain derivatives for energy functionals with boundary integrals; optimality and monotonicity, Inequalities and Applications 10, Birkh¨auser (2012). [2] M. Bareket, On an isoperimetric inequality for the first eigenvalue of a boundary value problem, SIAM J. Math.Anal. 8 (1977), 280-287. [3] D. Bucur and A. Giacomini,Faber-Krahn inequalities for the Robin-Laplacian: a free discontinuity approach, to appear. [4] M. Dambrine and M. Pierre, About stability of equilibrium shapes M2AN Math. Model. Numer. Anal. 34 (2000), 811-834.

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[5] D. Daners, Faber- Krahn inequality for Robin problems in any space dimension, Math. Ann. 335 (2006), 767-785. [6] G. Faber, Beweis, dass unter allen homogenen Membranen von gleicher Fl¨ache und gleicher Spannung, die kreisf¨ormige den tiefsten Grundton gibt. Sitz. Ber. Bayer. Akad. Wiss. (1923), p 162 - 172. [7] P. R. Garabedian, M. Schiffer,Convexity of domain functionals, J. Anal. Math. 2 (1953), 281-368. [8] D. Gilbarg and N. Trudinger, Elliptic partial differential equations of second order. Second edition. Grundlehren der Mathematischen Wissenschaften, 224. Springer-Verlag, Berlin, 1983. [9] T. Giorgi and R. Smits, Monotonicity results for the principal eigenvalue of the generalized Robin problem, Illinois J. Math. 49 (4) (2005), 1133-1143. [10] E. Giusti,Minimal Surfaces and Functions of Bounded Variation, Birkh¨auser (1984). [11] P. Grinfeld, Hadamard’s formula Inside and Out, J. Optim. Theory and Appl. 146 (2010), 654-690. [12] J. Hadamard, M´emoire sur le probl´eme d’analyse relatif l’´equilibre des plaques ´ ´elastiques encastr´ees, M´emoires des Savants Etrangers, 33 (1908). [13] A. Henrot, M. Pierre,Variation et optimisation de formes, Springer (2005). [14] D. Henry, Pertubation of the boundary in boundary value problems of partial differential equations, London Math. Soc. Lecture Notes Series, 318, Cambridge University Press (2005). [15] D. D. Joseph, Parameter and Domain Dependence of Eigenvalues of Elliptic Partial Differential Equations, Arch. Rat. Mech. 24, (1967), 325-351. ¨ [16] E. Krahn, Uber eine von Rayleigh formulierte Minimaleigenschaft des Kreises. (German) Math. Ann. 94 (1925), no. 1, 97-100. [17] G. P´olya and G. Szeg¨o,Isoperimetric inequalities in mathematical physics, Princeton (1951). [18] A. Novruzi and M. Pierre, Structure of shape derivatives J. evol. equ. 2 (2002), 365-382. [19] J. Serrin,A symmetry problem in potential theory, Arch. Rat. Mech. Anal. 43 (1971), 304-318. [20] J. Simon, Differentiation with respect to the domain in boundary value problems, Num. Funct. Anal. Optimiz. 2 (1980), 649-687.