A surgery result for the spectrum of the Dirichlet Laplacian
aa r X i v : . [ m a t h . O C ] O c t A SURGERY RESULT FOR THE SPECTRUM OF THE DIRICHLETLAPLACIAN
DORIN BUCUR AND DARIO MAZZOLENI
Abstract.
In this paper we give a method to geometrically modify an open set such that thefirst k eigenvalues of the Dirichlet Laplacian and its perimeter are not increasing, its measureremains constant, and both perimeter and diameter decrease below a certain threshold. Thekey point of the analysis relies on the properties of the shape subsolutions for the torsion energy. Keywords: shape optimization, eigenvalues, Dirichlet Laplacian1.
Introduction and statement of the main results
The results of this paper are motivated by spectral shape optimization problems for theeigenvalues of the Dirichlet Laplacian, e.g.min (cid:8) λ k (Ω) , Ω ⊂ R N , | Ω | = 1 (cid:9) , (1.1)where λ k denotes the k th eigenvalue of the Dirichlet Laplacian and | · | the N dimensionalLebesgue measure ( N ≥ k eigenvalues of theDirichlet Laplacian are not increasing, its measure remains constant and its diameter decreasebelow a certain threshold. This result together to the Buttazzo-Dal Maso existence theorem [11](which has a local character) gives a proof of global existence of solutions. By a different method,based on the so called shape subsolutions (see the definition in Section 2), in [7] is proved theexistence of solutions and moreover that all minimizers have finite diameter and finite perimeter.Recently, Van den Berg has studied in [4] a minimum problem with both a measure and aperimeter constraint: min (cid:8) λ k (Ω) , Ω ⊂ R N , | Ω | ≤ , Per(Ω) ≤ C (cid:9) . (1.2)An existence result for this problem cannot be deduced from the results [7, 17]. The surgerymethod of [17] can hardly control the perimeter since the procedure generates new pieces ofboundary which may have a large surface area. As well, in the presence of two simultaneousconstraints, the notion of shape subsolution can not be used in a direct manner due to the lackof suitable Lagrange multipliers which can take into account both geometric constraints. Theresults of this paper are also intended to provide a tool allowing to prove existence of a solutionfor (1.2).In this paper we give a result which follows the main objectives of [17], but with the newrequirement on the control of the perimeter. For this purpose, the “surgery” is done in a different manner, using some of the key ideas of the shape subsolutions. Roughly speaking, we look atthe local behavior of the torsion function and prove that if this function is small enough in someregion, then one can cut out a piece of the domain controlling simultaneously the variation ofthe low part of the spectrum, of the measure and of the perimeter.Throughout the paper, by ˜Ω we denote an open set of finite measure. For simplicity, andwithout restricting the generality, we shall assume that its measure is equal to 1. Here is ourmain result which, for clarity, is stated in a simplified way: Theorem 1.1.
For every
K > , there exists D, C > depending only on K and the dimension N , such that for every open set ˜Ω ⊂ R N with | ˜Ω | = 1 there exists an open set Ω satisfying (1) | Ω | = 1 , diam (Ω) ≤ D and Per(Ω) ≤ min { Per( ˜Ω) , C } , (2) if λ k ( ˜Ω) ≤ K , then λ k (Ω) ≤ λ k ( ˜Ω) . The set Ω is essentially obtained by removing some parts of ˜Ω and rescaling it to satisfy themeasure constraint. In case the measure of ˜Ω is not equal to 1, the constants D and C abovedepend also on | ˜Ω | , following the rescaling rules of the eigenvalues, measure and perimeterWe shall split the main result stated above in two distinct theorems, Thereoms 3.3 and 4.1.The construction of Ω differs depending on which kind of control of the perimeter is desired. Ifthe perimeter of ˜Ω is infinite (or larger than C ), it is convenient to use an optimization argumentrelated to the shape subsolutions to directly construct the set Ω satisfying all the requirementsabove on eigenvalues, measure and diameter, but with a perimeter less than C (Theorem 3.3).If the perimeter of ˜Ω is finite (for example smaller than C ), we produce a different argument,by cutting in a suitable way the set ˜Ω with hyperplanes, and removing some strips, decreasingin this way the perimeter (Thereom 4.1) and of course satisfying all the requirements aboveon eigenvalues, measure and diameter. In this last case, the control of the perimeter is donethrough a De Giorgi type argument. We point out that the assertions of the two theorems areslightly stronger than the unified formulation stated in Theorem 1.1.We note that the results of this paper hold true in exactly the same way if instead of“open” sets one works with “quasi-open” or “measurable” sets (see the precise definitions of thespectrum for this weaker settings in [10]). In general, if ˜Ω is quasi-open or measurable, then theconstructed set Ω is of the same type. In some situations in which the diameter of ˜Ω is large, Ωcould be chosen open and smooth.2. The spectrum of the Dirichlet Laplacian and the torsion function
Let Ω ⊂ R N be an open set of finite measure. Denoting by H (Ω) the usual Sobolev space,the eigenvalues of the Dirichlet Laplacian on Ω are defined by λ k (Ω) := min S k max u ∈ S k \{ } R Ω |∇ u | dx R Ω u dx , (2.1)where the minimum ranges over all k -dimensional subspaces S k of H (Ω). SURGERY RESULT FOR THE SPECTRUM OF THE DIRICHLET LAPLACIAN 3
The torsion function of Ω is the function denoted w Ω which minimizes the torsion energy E (Ω) := min u ∈ H (Ω) Z R N | Du | dx − Z R N udx, and satisfies in a weak sense − ∆ w Ω = 1 in Ω , w Ω ∈ H (Ω) . Note that the torsion energy is negative if Ω = ∅ and E (Ω) = − Z R N w Ω dx < . We recall (see for instance [11]) that if one extends the torsion function by zero on R N \ Ω, thenit satisfies − ∆ w Ω ≤ R N .A fundamental property of the torsion function is the Saint Venant inequality, which statesthat among all (open) sets of equal volume, the ball maximizes the L -norm of the torsionfunction. This leads to the following inequality Z Ω w Ω dx ≤ | Ω | N +2 N ω − /NN N ( N + 2) , (2.2)where ω N is the volume of the ball of radius R N . A similar inequality between the L ∞ norms was proved by Talenti in [18] and leads to : k w Ω k L ∞ ≤ (cid:16) | Ω | ω N (cid:17) N N . (2.3)We recall the following bound on the ratio between eigenvalues of the Dirichlet Laplacian,which can be found in [2]. For all k ∈ N there exists a constant M k , depending only on k andthe dimension N , such that 1 ≤ λ k (Ω) λ (Ω) ≤ M k . (2.4)Another fundamental inequality, proved in [3] (see also [5]), relates the L ∞ norm of the torsionfunction with the first eigenvalue and reads1 λ (Ω) ≤ k w Ω k ∞ ≤ N log 2 λ (Ω) . (2.5)We also recall the following inequality due to Berezin, Li and Yau (see [16]), which asserts thatfor some constant C N depending only on the dimensions of the space, we have ∀ k ∈ N λ k (Ω) ≥ C N (cid:16) k | Ω | (cid:17) N . The way we shall use this inequality is the following: if one fixes
K >
0, then the number ofeigenvalues of Ω below K , is at most of (cid:16) KC N (cid:17) N | Ω | .The γ -distance between two open sets with finite measure Ω , Ω is defined by: d γ (Ω , Ω ) := Z R N | w Ω − w Ω | dx. DORIN BUCUR AND DARIO MAZZOLENI
For sets satisfying Ω ⊆ Ω , the following inequality was proved in [7] : for every k ∈ N (cid:12)(cid:12)(cid:12) λ k (Ω ) − λ k (Ω ) (cid:12)(cid:12)(cid:12) ≤ k e / π λ k (Ω ) N/ d γ (Ω , Ω ) , (2.6)and we notice that there is a strong relation between the γ -distance and the torsion energy: d γ (Ω , Ω ) = 2( E (Ω ) − E (Ω )).Let c >
0. It is said that ˜Ω ⊂ R N is a shape subsolution for the energy if for all Ω ⊂ ˜Ω E (Ω) + c | Ω | ≥ E ( ˜Ω) + c | ˜Ω | . It is proved in [7] that, if ˜Ω is a shape subsolution for the energy, then it is bounded (withcontrolled diameter) and has finite perimeter.We conclude this Section with a result relating a pointwise value of the torsion function toits integral on some neighborhood.
Lemma 2.1.
Let Ω ⊂ R N be an open set and w = w Ω be its torsion function. For every θ > ,there exists δ > depending only on N, θ such that if w ( x ) ≥ θ for some x ∈ R N , then Z B δ ( x ) wdx ≥ θω N δ N , ∀ δ ∈ (0 , δ ) . Proof.
Since for every x ∈ R N the function x w ( x ) + | x − x | N is subharmonic in R N , we havethat, for all δ > θ ≤ w ( x ) ≤ | B δ | Z B δ ( x ) ( w ( x ) + | x − x | N ) dx = 1 | B δ | Z B δ ( x ) wdx + δ N + 2) . For some δ sufficiently small (e.g equal to p θ ( N + 2)), we have ∀ < δ ≤ δ Z B δ ( x ) wdx ≥ θω N δ N . (cid:3) Control of the spectrum by subsolutions
Before stating our first result, we outline the main ideas. Let ˜Ω ⊂ R N be a given open setof finite measure. Assume that for some set Ω ⊂ ˜Ω and for some constant c > E (Ω) + c | Ω | ≤ E ( ˜Ω) + c | ˜Ω | . (3.1)Then, we shall observe that a certain number of low eigenvalues of the rescaled set (cid:16) | ˜Ω || Ω | (cid:17) N Ω arenot larger than the corresponding eigenvalues on ˜Ω, provided that c is small enough. Smaller isthe constant c , more eigenvalues satisfy this property. Indeed, from (2.6), we get λ k (Ω) − λ k ( ˜Ω) ≤ k e / π λ k (Ω) λ k ( ˜Ω) ( N +2) / [ E (Ω) − E ( ˜Ω)] . (3.2)Setting K Ω , ˜Ω = 4 k e / π λ k (Ω) λ k ( ˜Ω) ( N +2) / , using inequality (3.1) we get λ k (Ω) − λ k ( ˜Ω) ≤ cK Ω , ˜Ω ( | ˜Ω | − | Ω | ) ≤ cK Ω , ˜Ω | ˜Ω | N − N N | ˜Ω | N − | Ω | N ) . (3.3) SURGERY RESULT FOR THE SPECTRUM OF THE DIRICHLET LAPLACIAN 5
Then, for every Λ such that cK Ω , ˜Ω | ˜Ω | N − N N ≤ Λ (3.4)we get λ k (Ω) − λ k ( ˜Ω) ≤ Λ( | ˜Ω | N − | Ω | N ) , so λ k (Ω) + Λ | Ω | N ≤ λ k ( ˜Ω) + Λ | ˜Ω | N . If c is small enough so that we can choose Λ satisfying λ k ( ˜Ω) = Λ | ˜Ω | N , we get λ k (Ω) | Ω | N ≤ λ k ( ˜Ω) | ˜Ω | N . The construction above can be carried out provided that one has control on an upper bound of K Ω , ˜Ω in (3.4). We shall prove that this is the case, if c is small enough. Lemma 3.1.
Let k ∈ N , K > and ˜Ω ⊂ R N be an open set of unit measure, satisfying λ k ( ˜Ω) ≤ K . There exist two constants c, β > depending only on K and N such that for all Ω ⊂ ˜Ω satisfying E (Ω) + c | Ω | ≤ E ( ˜Ω) + c | ˜Ω | (3.5) we have λ i (Ω) | Ω | /N ≤ λ i ( ˜Ω) | ˜Ω | /N , ∀ i = 1 , . . . , k (3.6) and | Ω | ≥ β | ˜Ω | .Proof. We divide the proof in several steps.
Step 1.
The constant c can be chosen such that E ( ˜Ω) + c | ˜Ω | is negative. In order to find theright information on c , we start by proving the following inequality: Z ˜Ω w ˜Ω dx ≥ C ( N ) 1(2 λ ( ˜Ω)) N +22 , (3.7)with C ( N ) := (2 N ) N +22 ω N N ( N +2) . We first note that ( w ˜Ω − λ (˜Ω) ) + is the torsion function of the set { w ˜Ω > λ (˜Ω) } and that k w ˜Ω − λ (˜Ω) k ∞ ≥ / λ ( ˜Ω), thanks to (2.5). As a consequence of theTalenti inequality (2.3), the measure of the set { w ˜Ω > λ (˜Ω) } is controlled from below by λ ( ˜Ω),precisely we have Z ˜Ω w ˜Ω dx ≥ Z n w ˜Ω > λ o (cid:18) w ˜Ω − λ ( ˜Ω) (cid:19) dx + Z n w ˜Ω > λ o λ ( ˜Ω) dx ≥ C ( N ) 1(2 λ ( ˜Ω)) N +22 . Then it is clear that we have E ( ˜Ω) + c | ˜Ω | ≤ − C ( N )2(2 λ ( ˜Ω)) N +22 + c | ˜Ω | . The right hand side above is negative, as soon as we choose c ≤ C ( N )2 | ˜Ω | (2 K ) N +22 , (3.8)since λ ( ˜Ω) ≤ K . DORIN BUCUR AND DARIO MAZZOLENI
Step 2.
The constant c can be chosen such that for every Ω satisfying (3.5), we have k w Ω k ∞ ≥ k w ˜Ω k ∞ . Indeed, denote h := k w ˜Ω k ∞ and assume that k w Ω k ∞ < k w ˜Ω k ∞ . Then0 ≤ c | Ω | ≤ c | ˜Ω | + 12 Z w Ω dx − Z w ˜Ω dx = c | ˜Ω | + 12 Z w Ω dx − Z min (cid:8) w ˜Ω , h/ (cid:9) dx − Z { w ˜Ω >h/ } (cid:0) w ˜Ω − h/ (cid:1) dx ≤ c | ˜Ω | − Z { w ˜Ω >h/ } (cid:0) w ˜Ω − h/ (cid:1) dx. Thanks to the fact that ( w ˜Ω − h/ + = w { w ˜Ω >h/ } using the same argument as in Step 1, wehave that C ( N ) h N +22 ≤ Z { w ˜Ω >h/ } ( w ˜Ω − h/ + ≤ c | ˜Ω | . Consequently, if c ≤ C ( N ) K − N +22 then k w Ω k ∞ ≥ k w ˜Ω k ∞ , since by inequality (2.5) k w ˜Ω k ∞ ≥ λ (˜Ω) ≥ K .We note that, using the inequalities (2.4) and (2.5) together with the fact that k w Ω k ∞ ≥k w ˜Ω k ∞ /
2, one can easily deduce that for every i ∈ N the corresponding eigenvalues on Ω and ˜Ωare comparable λ i ( ˜Ω) ≤ λ i (Ω) ≤ (8 + 6 N log 2) M i λ i ( ˜Ω) . (3.9) Step 3.
Proof of inequality (3.6). Choosing c satisfying Steps 1 and 2, and c ≤ λ ( B )2 M k k (8 + 6 N log 2) e / π k K N +2 , (3.10)where B is the ball of volume equal to 1, from the Faber-Krahn inequality we have c M k k (8 + 6 N log 2) e / π k K N +2 ≤ λ ( ˜Ω) | ˜Ω | /N , (3.11)or this precisely gives, in view of (3.2)-(3.3), λ i (Ω) + Λ | Ω | /N ≤ λ i ( ˜Ω) + Λ | ˜Ω | /N . (3.12)Thanks to Step 2, c is such that Λ ≤ λ (˜Ω) | ˜Ω | /N . Consequently, (3.12) also holds for all values largerthan λ (˜Ω) | ˜Ω | /N , thus this inequality holds for all i = 1 , . . . , k with constant λ i (˜Ω) | ˜Ω | /N .With the choice of λ i (˜Ω) | ˜Ω | /N , using the arithmetic geometric inequality, we note that λ i (Ω) | Ω | /N ≤ λ i ( ˜Ω) | ˜Ω | /N , ∀ i = 1 , . . . , k. Step 4.
In order to control the diameter of the rescaled set, we prove the existence of β > | Ω | ≥ β | ˜Ω | . Indeed, we have the chain of inequalities (the last one being a consequence of(2.3)) 12 K ≤ λ ( ˜Ω) ≤ k w ˜Ω k ∞ ≤ k w Ω k ∞ ≤ (cid:16) | Ω | ω N (cid:17) N N , which gives the estimate for β , depending only on K, N . (cid:3) SURGERY RESULT FOR THE SPECTRUM OF THE DIRICHLET LAPLACIAN 7
Remark 3.2.
Inequality (3.9) in Step 2 could also be obtained in a different way, as a conse-quence of inequality (2.6) by choosing c small enough. Indeed, if c is such that k e / π λ k ( ˜Ω) N/ d γ (Ω , ˜Ω) ≤ λ k ( ˜Ω) , (3.13) then λ k ( ˜Ω) ≤ λ k (Ω) ≤ λ k ( ˜Ω) . Since by the hypothesis of Lemma 3.1 we have E (Ω) − E ( ˜Ω) ≤ c, inequality (3.13) holds as soon as c ≤ k e / π λ k ( ˜Ω) N/ . Now, we are in position to prove the first result. Below, the diameter of a disconnected setis referred as the sum of the diameters of each connected component.
Theorem 3.3.
For every
K > , there exists D, C > depending only on K and the dimension N such that for every open set ˜Ω ⊂ R N with | ˜Ω | = 1 there exist an open set Ω with diam (Ω) ≤ D , | Ω | = 1 , Per(Ω) ≤ C and if λ k ( ˜Ω) ≤ K then λ k (Ω) ≤ λ k ( ˜Ω) .Proof. From the Berezin-Li-Yau inequality, the maximal index k for which it is possible that λ k ( ˜Ω) ≤ K is lower than a constant depending only on K and N .Let us consider the minimum problemmin Ω ⊂ ˜Ω { E (Ω) + c | Ω |} , with c the constant given by Lemma 3.1 and k = k . This problem has at least one solution,denoted Ω ∗ , which is an open set (see for instance [14]) and it is also a shape subsolution ofthe energy. The results from [7] give that diam (Ω ∗ ) ≤ D ( c ) and that Per(Ω ∗ ) ≤ C ( c ) and weremind that c depends only on K and the dimension N . Moreover, using Step 4 of Lemma 3.1,we have that the set Ω := | Ω ∗ | − /N Ω ∗ has still diameter and perimeter bounded by constantsdepending only on K, N , thanks to the fact that | Ω | ≥ β | ˜Ω | . Moreover we have that ∀ i = 1 , . . . , k, λ i (Ω) ≤ λ i ( ˜Ω) , since λ k ( ˜Ω) ≤ K . (cid:3) Control of the perimeter
In order to give precise statements, we introduce a suitable notion of diameter in a prescribeddirection. In the coordinate direction e ∈ R N we setdiam e (Ω) := H (cid:0) t ∈ R : H N − (Ω ∩ { x = t } ) > (cid:1) . Theorem 4.1.
For every
K, P > , there exist D > depending only on K, P and the dimension N , such that for every open set ˜Ω ⊂ R N with | ˜Ω | = 1 , Per( ˜Ω) ≤ P , there exists an openset Ω of unit measure with diam e (Ω) ≤ D , Per(Ω) ≤ Per( ˜Ω) such that if λ k ( ˜Ω) ≤ K then λ k (Ω) ≤ λ k ( ˜Ω) . DORIN BUCUR AND DARIO MAZZOLENI
For every x ∈ R and r >
0, we define the strip centered in x of width 2 r orthogonal to R e by S r ( x ) := [ − r + x , r + x ] × R N − . Its topological boundary is ∂S r := {− r + x , r + x } × R N − . If x = 0, we simply denote S r instead of S r (0).The main idea of the following lemma is inspired from [1] and was also used in [12], [7],and [8] under different settings. We point out that here we do not use optimality, but only aninequality between two fixed domains. Lemma 4.2.
For all c > , there exist C , r > , with C r ≤ min (cid:8) c , K (cid:9) such that if forsome r ≤ r the function w is not identically zero in S r and E ( ˜Ω) + c | ˜Ω | ≤ E ( ˜Ω \ S r ) + c | ˜Ω \ S r | , (4.1) then max S r w ˜Ω ≥ C r. (4.2) Proof.
Below, we denote w := w ˜Ω and ε := max S r w and introduce the function η : R N → R + : η = 0 in S r , η = ε in R N \ S r , − ∆ η = 1 in S r \ S r ,η = 0 on ∂S r ,η = ε on ∂S r . (4.3)Since the function min { w, η } := w ∧ η belongs to H ( ˜Ω \ S r ) we get E ( ˜Ω \ S r ) ≤ Z | D ( w ∧ η ) | dx − Z w ∧ ηdx. Hypothesis (4.1) gives12 Z | Dw | dx − Z wdx + c | S r ∩ ˜Ω | ≤ Z | D ( w ∧ η ) | dx − Z w ∧ ηdx. Since ε ≤ C r ≤ c we get w ∧ η = w in ˜Ω \ S r . Denoting the outer unit normal to a set by ν ,12 Z S r | Dw | dx + c | S r ∩ ˜Ω | ≤ Z S r | Dw | dx − Z S r wdx + c | S r ∩ ˜Ω |≤ Z S r \ S r ( | D ( w ∧ η ) | − | Dw | ) dx − Z S r \ S r ( w ∧ η − w ) dx = 12 Z S r \ S r ∩{ w>η } ( | Dη | − | Dw | ) dx − Z S r \ S r ( w − η ) + dx ≤ Z S r \ S r ∩{ w>η } − Dη · D ( w − η ) dx − Z S r \ S r ( w − η ) + dx = − Z ∂S r ∂η∂ν ( w − η ) + dx = | η ′ ( r ) | Z ∂S r w d H N − . SURGERY RESULT FOR THE SPECTRUM OF THE DIRICHLET LAPLACIAN 9
The following trace inequality holds: Z ∂S r w d H N − ≤ C ( N ) (cid:18) r Z S r wdx + Z S r | Dw | dx (cid:19) . By using hypothesis (4.2) and the Cauchy-Schwarz inequality on the gradient term, we get to Z ∂S r w d H N − ≤ C ( N ) (cid:18) ( C + 12 ) | S r ∩ ˜Ω | + 12 Z S r ∩ ˜Ω | Dw | dx (cid:19) . If R S r | Dw | dx + | S r ∩ ˜Ω | = 0 then w = 0 in the “strip” S r . Otherwise R S r | Dw | dx + | S r ∩ ˜Ω | > (cid:26) , c (cid:27) ≤ | η ′ ( r ) | C ( N )( C + 1) . Since | η ′ ( r ) | = | C − r/ | , choosing C and r small enough we get a contradiction. We noticethat the choice of these constants depends only on N and on c . (cid:3) The following corollary can be proved in the very same way as Lemma 4.2.
Corollary 4.3.
For all c > there exist C , r > with C r ≤ min (cid:8) c , K (cid:9) such that if forsome r ≤ r and x , . . . , x n ∈ R such that S r ( x i ) ∩ S r ( x j ) = ∅ for all i = j, it holds max (cid:8) w ˜Ω ( x ) : x ∈ ∪ i S r ( x i ) (cid:9) ≤ C r , then we have that E ( ˜Ω \ ∪ i S r ( x i )) + c | ˜Ω \ ∪ i S r ( x i ) | ≤ E ( ˜Ω) + c | ˜Ω | . (4.4)Here we outline the main idea for proving Theorem 4.1. Let c be as in Lemma 3.1 and r , C be the constants from Lemma 4.2, for that particular choice of c . We shall remove a finitenumber of strips S r ( x i ) from the region where w ˜Ω ( x ) ≤ C r thus, following inequality (4.4) andLemma 3.1, we can control the eigenvalues after rescaling. The control of the perimeter, willbe done by a suitable choice of the position of the strips. Contrary to the construction in [17],the new perimeter introduced by sectioning with hyperplanes does not depend on the H N − measure of the boundary of the sections.Let l > n ∈ N . The value of l will be precised below, in Lemma 4.4. Assume x i ∈ R N and L i > l , i = 1 , . . . , n are such that S L i ( x i ) ∩ S L j ( x j ) = ∅ if i = j . For every t ∈ [0 , l ] we define: S ( t ) := ∪ ni =1 S L i − t ( x i ) . For an open set of unit measure ˜Ω, we denote m ( t ) := | S ( t ) ∩ ˜Ω | the mass of the union of stripsin ˜Ω and σ ( t ) := n X i =1 H N − ( ˜Ω ∩ { L i − t, L i + t } × R N − ) , the new perimeter introduced by the sections with the hyperplanes and p ( t ) = n X i =1 Per( ˜Ω ∩ ( L i − t, L i + t ) × R N − ) − σ ( t ) , the perimeter of ˜Ω inside the strips. We denote the rescaled set,Ω( t ) := (1 − m ( t )) − /N ( ˜Ω \ S ( t )) . Lemma 4.4.
Given
P > and an open set ˜Ω of unit measure, with Per( ˜Ω) ≤ P , there existtwo constants l and b m , depending only on P and the dimension N , such that if m ( l ) ≤ b m then there exists t ∈ [0 , l ] such that Per(Ω( t )) ≤ Per( ˜Ω) .Proof.
First of all, we notice that, by definition, t m ( t ) is a nonincreasing function and for a.e. t ∈ (0 , l ), we have that σ ( t ) = − m ′ ( t ). If for every t ∈ [0 , l ] we would have Per(Ω( t )) > Per( ˜Ω),we get: Per( ˜Ω) − p ( t ) + σ ( t ) ≥ Per( ˜Ω)(1 − m ( t )) N − N . There exists a constant b m (depending only on P, N ), such that if m ( t ) ≤ b m , then(1 − m ( t )) N − N ≥ − m ( t ) N − N P ≥ − m ( t ) N − N . Putting the above inequalities together and using the isoperimetric inequality for the set S ( t ),Per( ˜Ω) + 2 σ ( t ) ≥ Per( ˜Ω) − m ( t ) N − N p ( t ) + σ ( t ) ≥ Per( ˜Ω) − m ( t ) N − N N ω /NN m ( t ) N − N . Since 2
N ω /NN − >
0, we obtain: − m ′ ( t ) ≥ (2 N ω /NN − m ( t ) N − N . By integrating on [0 , l ] we get m /N (0) − m /N ( l ) ≥ (2 ω /NN − l N .
Since m (0) = b m and m ( l ) ≥
0, choosing l > N ω /NN − b m /N we get a contradiction. (cid:3) Remark 4.5.
If we denote by A a subset of ˜Ω with max A w ˜Ω ≤ C r then, having in mind (2.5) ,if b m is small enough (depending only on C and r ) we get λ ( A )(1 − b m ) /N ≥ C r ≥ K. Remark 4.6.
Thanks to the choice of C , r made in Lemma 4.2, we deduce that if A ⊂ ˜Ω issuch that max A w ˜Ω ≤ C r , then E ( A ) + c | A | ≥ . Indeed, using the monotonicity of the torsionfunction: E ( A ) + c | A | = − Z w A dx + c | A | ≥ − Z A w ˜Ω dx + c | A | ≥ − C r | A | c | A | ≥ , since C r ≤ c from the hypotheses of Lemma 4.2. We are now in position to prove the main result of this section.
SURGERY RESULT FOR THE SPECTRUM OF THE DIRICHLET LAPLACIAN 11
Proof of Theorem 4.1.
We fix the constant c such that Lemma 3.1 is satisfied, we get C , r fromLemma 4.2 and we fix a constant b m that works both for Lemma 4.4 and for Remark 4.5. Forsimplicity we rename w = w ˜Ω . The region where w ( x ) ≥ C r is contained in a finite union ofstrips with width 4 r . Indeed, we define X := (cid:26) x ∈ R : max S r ( x ) w ≥ C r (cid:27) , e X := [ n S r ( t ) : t ∈ X o . From Lemma 2.1 and the Saint Venant inequality (2.2) the set e X is contained in the union ofat most n = n ( r , N ) of disjoint strips (each of width at least 4 r ). Let us call X the projectionof e X on R e .The set R \ X is a finite union of disjoint segments and of the infinite intervals at ±∞ , say R \ X = ( −∞ , b ) ∪ " n [ i =1 ( a i , b i ) ∪ ( a n +1 , ∞ ) . If a segment ( a i , b i ) has a length less than or equal to 8 r + 2 l , we shall ignore it in our furtherconstruction and just add the corresponding strip to the set ˜ X and renumber the index i ifnecessary. The total length of those such segments is at most n (8 r + 2 l ).Therefore, we shall assume in the sequel that all segments ( a i , b i ) have a length greater than8 r + 2 l . We denote a i = a i + (4 r + l ), b i = b i − (4 r + l ) and Y = " n +1 [ i =1 ( a i , a i ) ∪ " n [ i =0 ( b i , b i ) . In order to highlight the main idea, let us assume in a first instance that | (cid:0) Y × R N − (cid:1) ∩ ˜Ω | ≤ b m. (4.5)If we are in this situation, we perform a simultaneous “cut” as in Lemma 4.2, removing thefollowing union of strips: S t := S r ( b − r − t ) [ S r ( a i + 2 r + t ) [ S r ( b i − r − t ) [ S r ( a n +1 + 2 r + t ) , for every t ∈ [0 , l ].Following the assumption (4.5) and Lemma 4.4, there exists a value t such that the perimeterof the rescaled set | ˜Ω \ S t | − N ( ˜Ω \ S t ) is at most Per( ˜Ω). Moreover, from the choice of c andLemma 4.2, all the eigenvalues less than K of the rescaled set are not greater than the ones on˜Ω. In order to handle the diameter of the rescaled set, we replace all the connected componentshaving a projection on R e disjoint from X by one ball, such that the volume remains unchanged.In this way, the perimeter does not increase, while the low part of the spectrum (below K ) canonly decrease, since the first eigenvalue of every such a connected component is not smaller than1 / ( C r ) ≥ K (see Remark 4.5).It is clear that the new set satisfies the diameter bound:diam e (Ω) ≤ diam e ( b Ω)(1 − b m ) − /N ≤ (cid:16) H ( X ) + n (8 r + 2 l ) + 2 r ( n + 2) (cid:17) + 2 ω − N N . If assumption (4.5) does not hold, we can not apply directly Lemma 4.4. Let p ∈ N dependingonly on P and the dimension, be such that p ≤ b m < p − . If a i + p (4 r + l ) > b i − p (4 r + l ) , we ignore this strip and add it to X , renumbering the index i if necessary. There exists s ∈ [0 , p −
1] such that replacing simultaneously all a i with a i + s (4 r + l ) and b i with b i − s (4 r + l )the assumption (4.5) is satisfied and so we finish the proof, adding at worst 4 np (4 r + l ) to thediameter. (cid:3) Remark 4.7.
Since the choice of the direction e was arbitrary, we can repeat all the processof the proof of Theorem 4.1 for all the coordinate direction, finding a set which has diameterbounded in all directions, unit measure, better eigenvalues than ˜Ω up to level k and perimeterlower than ˜Ω . Acknowledgments.
This work was initiated during the first author stay at the Isaac NewtonInstitute for Mathematical Sciences Cambridge in the programme “Free boundary problems andrelated topics”, as an answer to a question raised by Michiel Van den Berg. The first author alsoacknowledges the support of ANR–12–BS01–0007 Optiform. The work of the second author hasbeen supported by the ERC Starting Grant n. 258685 “AnOptSetCon”.
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