Introduction to spectral geometry

Transcription

1 Introduction to spectral geometry Bruno Colbois, University of Neuchâtel Beirut, February 6-March 7, 018 Preamble. These notes correspond to a six hours lecture given in the context of the CIMPA School 018 Elliptic problems and applications in geometry, Beirut. They are intended for participants of the School and not for publication. A good part of this lecture is issue from [1] where you will find more references and other developments, in particular people interested to the geometric spectral theory on Riemannian manifolds. In this present lecture, in order to make things easier, I will stay at the level of Euclidean domains, where it exits already a lot of interesting and open questions. I would just mention a very interesting new book which appeared in May 017 and is on open access: Shape optimization and spectral theory, by Antoine Henrot, with the contribution of a lot of experts about recent results on the field: https : // Also the interested students may look at the notes in progress Spectral Theory of PDEs on the webpage of Richard Laugesen https : //f aculty.math.illinois.edu/ laugesen/ which are of course much more complete of this lecture. 1 Lecture 1: a first approach of the spectral geometry After Wikipedia, under Laplace operator, one can find the following explanations: In mathematics, the Laplace operator or Laplacian is a differential operator given by the divergence of the gradient of a function on Euclidean space. The Laplace operator is named after the French mathematician Pierre-Simon de Laplace ( ), who first applied the operator to the study of celestial mechanics, where the operator gives a constant multiple of the mass density when it is applied to a given gravitational potential. Solutions of the equation f = 0, now called Laplace s equation, are the so-called harmonic functions. The Laplacian occurs in differential equations that describe many physical phenomena, such as electric and gravitational potentials, the diffusion equation for heat and fluid flow, wave propagation, and quantum mechanics. For these reasons, it is extensively used in the sciences for modeling all kinds of physical phenomena. The Laplacian is the simplest elliptic operator, and is at the core of Hodge theory as well as the results of de Rham cohomology. In one sentence, one can say that the Laplacian occurs in a lot of domain of applied sciences. However, the goal of this lecture is to understand the relation between the spectrum of the Laplacian on a domain and the geometry of the domain. (See also the introduction of the notes on line of Yaiza Canzani, Analysis on Manifolds via the Laplacian (http : //canzani.web.unc.edu/f iles/016/08/laplacian.pdf, Chapter 1 What makes the Laplacian special?, relation with heat diffusion, wave propagation, etc.). Let us present this more formally in a particular situation, the Dirichlet problem. 1

2 1.1 The Laplacian for Euclidean domains with Dirichlet boundary conditions Let Ω R n be a bounded, open, connected domain with Lipschitz boundary. The Laplacian we will consider is given by n f f = x i where f C (Ω). We investigate the spectrum of the Laplacian on Ω with the Dirichlet boundary condition, that is we study the eigenvalue problem under the Dirichlet condition This problem has a discrete and real spectrum i=1 f = λf (1) f Ω = 0. () 0 < λ 1 (Ω) < λ (Ω) λ 3 (Ω)..., with the eigenvalues repeated according to their multiplicity. Example 1. If Ω is the interval ]0, L[ R, the Laplacian is given by f = f and the spectrum, for the Dirichlet boundary condition, is given by λ k = k π L, k = 1,,... The eigenfunction corresponding to λ k is f k (x) = sin kπ L x. This very simple example illustrate a simple but important fact coming directly from the definition: if aω is the image of Ω by an homothety of ratio a, then we have λ k (aω) = 1 a λ k (Ω). However, it is exceptional to be able to calculate explicitly the spectrum. In dimensions greater than one, it is generally not possible (apart from a few exceptions like the ball or a product of intervals that we will look at below) to calculate the spectrum of a domain. The calculation of the spectrum of a ball is classic, but not easy. It is done with enough details, for example, in the introduction of the PhD. D. thesis of Amandine Berger [5] that we can find online in https : //tel.archives ouvertes.f r/tel /document. Given that, physically, the Dirichlet problem is a modelization of the vibration of a fixed string in dimension 1 and of the vibration of a fixed membrane in dimension, it can be hoped to have connections between the geometry (or shape) of a domain Ω and its spectrum. In dimension 1, the relation is quite poor: it depends only on the length L of the string (or of the interval in Example 1). In dimension, things are much more interesting. By fixed membrane is understood a membrane whose boundary is fixed, like a drum. This leads to the following very classical image for two types of problems in the context of geometrical spectral theory: The direct problems. When we see a drum, we have an idea of how it would sound. The mathematical translation is: if we have some information about the geometry (or the shape) of a domain Ω, we can hope to get information (or estimates) about its spectrum. By information about the geometry we understand for example that we have information about the volume, the diameter of Ω, the mean curvature of its boundary, the inner radius. By estimates of the spectrum, we understand lower or upper bounds for some or all of the eigenvalues, with respect to the geometrical information we have.

3 Figure 1: Can we guess how sound these drums? Another way to get an intuitive understanding of the spectrum as vibration of a membrane is to look at the Chladni plates on YouTube, for example https : // = wvjagru BF 4w What you see in the video are the nodal lines of the eigenfunctions, that is the points where the eigenfunction takes the value 0, that is the membrane stay fixed at these points. Less realistic but closer from our problem, one can look at the vibration of a circular membrane on wikipedia. The inverse problems. If we hear a drum, even if we do not see it, we have an idea of its shape. The mathematical translation is: If we have information about the spectrum of a domain, we can hope to get information about its geometry. The most famous question is whether the shape of a domain is determined by its spectrum (question of M. Kac in 1966: can one hear the shape of a drum?). In other words: if we know all the spectrum of a domain, are we able to reconstruct the domain? The formal answer is no: There are a lot of domains having the same spectrum without being isometric. However, in some sense, this is often true, which makes things complicated. I will not go further in this 3

4 direction in these notes, see [17] for a survey of the subject. Figure : Example of two isospectral domains Example. Let us consider the rectangle ]0, a[ ]0, b[ and consider the equation f(x, y) = λf(x, y) with Dirichlet boundary condition. We use the method of separation of variable and look at solutions of the type f(x, y) = g(x)h(y). The equation becomes which gives g (x)h(y) g(x)h (y) = λg(x)h(y) (3) g (x) g(x) h (y) h(y) = λ (4) and g(0) = g(a) = 0; h(0) = h(b) = 0. This implies that g (x) g(x) = p; h (y) h(y) = q or g (x) = pg(x); h (y) = qg(y). We are in the situation of the interval, and the solutions are f(x, y) = sin kπ a x sin lπ b y, for the corresponding eigenvalue λ k,l = k π a + l π b, k, l 1. In order to be complete, we need to show that we have found all the eigenvalue: this is a consequence of the Fourier series decomposition. The family f k,l (x, y) = sin kπ a x sin lπ b y, h of eigenfunctions forms a Hilbert basis for the L functions on the rectangle. This rather simple calculation leads to some elementary and less elementary questions: 1. What is the smallest eigenvalue? We see that we have to choose k = 1 and l = 1, and the first eigenvalue is λ 1 = π a + π b. We also see immediately that this first eigenvalue becomes very large if a and b are small and very small if a and b are large. But if we look at all the rectangle of given area, say equal to 1 (that is b = 1 a ), what can be said? Again, if a becomes small, λ 1 becomes large. But if a becomes large, 1 a becomes small and, again, λ 1 is large. It is not possible to make λ 1 arbitrarily small if we consider rectangles of fixed area. 3. But how is the smallest λ 1 among all the rectangle of area 1? We have to minimize the expression π(a + 1 a ): this is an easy problem of high school: the answer is that the minimum is reached for a = 1. 4

5 4. We can ask the same question for λ : what is the smallest λ? Without lost of generality, we can suppose a 1, and in order to make λ small, we have to choose l = 1. But we need to take k = in order to find the second eigenvalue which becomes λ = π ( 4 a + a ). Again, it is easy to calculate that the minimum is reached for a =. These a priori simple consideration are indeed not so easy and lead to the question Open question 1: for rectangles of area 1, what is the minimal k the eigenvalue and how is the corresponding rectangle? In [1], it is shown that, for large k, the shape tends to be a square. Exercise. Try to solve this question for k = 3 and for k = 10. For more information about this non trivial question, one can look at the original paper where it was investigated (by P. Antunes and P. Freitas) [1] or on the recent paper of K. Gittins and S. Larson (and the reference therein) where a much more general question is solved [16] Exercise: counting the eigenvalues. For the square (that is a = b = 1) let N(γ) be the number of eigenvalues less or equal to γ. Prove that asymptotically, N(γ) γ 4π (that is lim γ N(γ) γ = 1 4π ). 1. The Faber-Krahn inequality and its developments This is an old but very enlightening result found independently by Faber and Krahn in the 190 s: More explanations and a sketch of the proof can be found in [3] p The proof uses the isoperimetric inequality and is not easy. For a new proof of this inequality, see also [8]. Theorem 3. Let Ω R n be a bounded open domain in R n and B R n, a ball with the same volume as Ω. If λ 1 denotes the first eigenvalue for the Dirichlet boundary conditions, then λ 1 (B) λ 1 (Ω), with equality if and only if Ω is equal to B up to a displacement. Firstly, the theorem gives a lower bound for the first Dirichlet eigenvalue of a domain Ω which is a typical direct problem. If the volume of Ω is known, then one gets immediately information about the spectrum: a lower bound for the first eigenvalue. But the equality case can be understood as an inverse result. If Ω and B have the same volume and if λ 1 (Ω) = λ 1 (B), then Ω is a ball. Remark 4. This result has to be related to another celebrated result in geometry: the isoperimetric inequality. The question consists in comparing the volume of a bounded domain Ω R n with the volume of its boundary Σ = Ω. The isoperimetric ratio of Ω is I(Ω) = V ol n 1 (Σ) V ol n+1 (Ω) (n 1)/n. Note that this ratio is invariant by homothety. One can ask which domain among all the reasonably regular domains of given volume has the smallest isoperimetric ratio. The answer is the ball! So, if Ω is a domain in R n, B a ball such that V ol( B) = V ol( Ω), we have V ol(b) V ol(ω) 5

6 with equality if and only if Ω is a ball. The interpretation correspond to the Dido problem and the Founding Myth of Carthage (see http : //math.arizona.edu/ dido/welcome.html and http : //math.arizona.edu/ dido/elissa.html): with a string of given length, the domain of largest area we can bound is a disc ao area L 4π. Some of the possible proofs of the Faber-Krahn inequality use the isoperimetric inequality. This result is also very interesting because it allows very modern questions to be asked. We will mention a few of them, without going into much detail. The same type of questions could also be asked in other situations (other boundary conditions, Laplacian on Riemannian manifolds, other differential operators, etc.). 1. The stability. If a domain Ω R n has the same volume as a ball B and if, moreover, λ 1 (Ω) is close to λ 1 (B), can one say that Ω is close to the ball B? Of course, a difficulty is to state what is meant by close. There are a lot of results around this problem that I will not describe in detail, but see [14] for a recent and important contribution and [7] for a very recent survey about this question. B Figure 3: The domain Ω is close to the ball B in the sense of the measure Note that it is not too difficult to see that one can do a small hole in any domain or add a thin hair without affecting the spectrum too much (easy part in Example 4.1 and more explanations in [6]). Therefore, it cannot be hoped for example, that Ω will be homeomorphic to a ball, or will be Hausdorff close to a ball. However, essentially, one can show that Ω is close to a ball in the sense of the measure, see Figure 4. and [14, 7].. The spectral stability. The next natural question is to see what this stability will imply: in [6], we have shown that if Ω and B have the same volume, and if λ 1 (Ω) is close to λ 1 (B) then, for all k, λ k (Ω) is close to λ k (B) (but, of course, not uniformly in k). The difficulty is that the proximity of λ 1 (Ω) and λ 1 (B) implies only that the domains are close in the sense of the measure, and we have to show that this is enough to control all the eigenvalues. 3. The extremal domains. We have seen that, among all domains having the same volume, the ball is the domain having the minimal λ 1. We can of course ask for domain(s) having the minimal λ k. 6

7 The various aspects of this problem are very complicated and most are far from being solved. For λ, it is well known that the solution is the union of two disjoint balls, see [18] thm (A proof will be given in lecture 4). Note that this domain is not regular in the sense that was expected. But in general, it is already difficult to find a right context where it can be shown that an extremal domain exists, see [8, 10] 4. Numerical estimates. Knowing about the existence of extremal domains from a theoretical viewpoint, one can try to investigate the shape of such domains. It is almost impossible to do this from a formal point of view, and this was investigated from a numerical point of view, in particular for domains of R. But this is also a difficult numerical problem, in particular due to the fact that there are a lot of local extrema. This approach was initiated in the thesis of E. Oudet. Note that it gives us a candidate to be extremal, but we have no proof that the domain appearing as extremal is the right one. It is important to apply different numerical approaches, and to see if they give the same results. See the thesis of A. Berger [5] for a description of results. Lecture : the Laplacian for Euclidean domains with Neumann boundary conditions, variational characterization of the spectrum.1 The Neumann problem Let Ω R n be a bounded, connected domain with a Lipschitz boundary. The Laplacian is given by n f f = x i where f C (Ω). What is investigated is the spectrum of the Laplacian on Ω with the Neumann boundary condition, that is the eigenvalue problem under the Neumann condition where n denotes the exterior normal to the boundary. This problem has a discrete and real spectrum i=1 f = λf (5) ( f n ) Ω = 0. (6) λ 0 = 0 < λ 1 (Ω) < λ (Ω) λ 3 (Ω)..., with the eigenvalues repeated according to their multiplicity. Notation. Sometimes, we will write λ D k (Ω) and λn k (Ω) to avoid confusion between the eigenvalues with respect to the Dirichlet boundary conditions and Neumann boundary conditions. Example 5. If Ω is the interval ]0, L[ R, the Laplacian is given by f = f and the spectrum, for the Neumann boundary condition, is given by λ k = k π L, k = 0, 1,,... The eigenfunction corresponding to λ k is f k (x) = cos kπ L x. 7

8 We also refer to [5] for the calculation of the spectrum of the ball in dimensions and 3. For rectangles, we can do the same calculations and ask similar open question as in Example for Neumann boundary condition (see [16] for the general problem). Some similarities between the spectrum for the Neumann boundary condition and Dirichlet boundary condition can be noticed. However, in this introduction, I will mainly focus on the differences. A first obvious difference is that zero is always an eigenvalue, corresponding to the constant eigenfunction. This means that the first interesting eigenvalue is the second, or the first nonzero eigenvalue. In Example 10, for the Dirichlet problem, we will see that if Ω 1 Ω, then we have for each k that λ k (Ω 1 ) λ k (Ω ) (we say that we have monotonicity). On the contrary, in Example 15, we will see that we have no monotonicity for the Neumann problem. Indeed, in this case, the fact that Ω 1 Ω has no implication on the spectrum. We will also see that the spectrum for the Neumann conditions is very sensitive to perturbations of the boundary. At this stage, I must clarify that, because the goal of these lectures was to present a geometric approach, I have chosen bounded domains with a Lipschitz boundary. This guarantees the discreteness of the spectrum, but this hypothesis may be relaxed for the Dirichlet problem..1.1 The Szegö-Weinberger inequality For further details, please refer to [3] p It is similar to the Faber-Krahn inequality, but is more delicate to prove, precisely because it concerns the second eigenvalue λ 1. Theorem 6. Let Ω be an open bounded domain in R n with Lipschitz boundary and B R n a ball with the same volume of Ω. If λ 1 denotes the first eigenvalue for the Neumann boundary conditions, then λ 1 (Ω) λ 1 (B), with equality if and only if Ω is equal to B up to a displacement. Remark In contrast with the case of the Dirichlet boundary condition, λ 1 cannot be very large for a domain of fixed volume when we take the Neumann boundary condition.. We will see that there are domains of fixed volume with as many eigenvalues close to 0 as we want. 3. There is no spectral stability in the Neumann problem. Indeed, we can perturb a ball by pasting a thin full cylinder of convenient length L: It does not affect the first nonzero eigenvalue but all the others. Using results in the spirit of [An1], the spectrum converges to the union of the spectrum of the ball and of the spectrum of [0, L], with the Dirichlet boundary condition on 0 and the Neumann condition on L. It suffices to choose L in such a way that the first eigenvalue of the interval lies between the two first eigenvalues (without multiplicity) of the ball. Exercise. Calculate the spectrum of a rectangle ]0, a[ ]0, b[ and show that there exists rectangle of area 1 with as many small eigenvalues as we want.. Variational characterization of the spectrum and simple applications In this section, I give a variational characterization of the spectrum: Roughly speaking, this allows to use geometry in order to investigate the spectrum: more precisely, this allows us to investigate the spectrum 8

9 of the Laplacian without looking at the equation u = λu itself, but rather by considering some test functions. Moreover, we have only to take into account the gradient of these test functions, and not their second derivative, which is much easier. One can say that this variational characterization is one of the fundamental tools allowing us to use the geometry in order to investigate the spectrum. To illustrate this, I will explain with some detail very elementary constructions of small eigenvalues: These constructions are well known by the specialists of the topic, and, in general, they are mentioned in one sentence. Although these are used here at an elementary level, they allow us to understand principles that will be used later in a more difficult context. Please, refer to the book of P. Bérard [4] for more details about the theory. Let Ω R n be a bounded domain with Lipschitz boundary Ω. Let f, h C (Ω). Then we have the well-known Green s Formula fhdx = f, h dx h df dn da Ω Ω where df dn denotes the derivative of f in the direction of the outward unit normal vector field n on Ω. In particular, if one of the following conditions h Ω = 0 or ( df dn ) Ω = 0 is satisfied, then we have the relation ( f, h) = ( f, h) = (f, h) where (, ) denotes the L -scalar product. The following standard result about the spectrum can be seen in [4] p. 53. Theorem 8. Let Ω R n be a bounded domain with Lipschitz boundary Ω Then, for both eigenvalue problems (Dirichlet and Neumann) 1. The set of eigenvalues consists of an infinite sequence 0 < λ D,N 1 λ D,N λ D,N 3..., where 0 is not an eigenvalue in the Dirichlet problem;. Each eigenvalue has finite multiplicity and the eigenspaces E(λ i ) corresponding to distinct eigenvalues are L (M)-orthogonal; Ω 9

10 3. The direct sum of the eigenspaces E(λ i ) is dense in L (M) for the L -norm. Furthermore, each eigenfunction is C -smooth. As already mentioned, the spectrum cannot be computed explicitly, with a few exceptions. In general, it is only possible to get estimates of the spectrum, and these estimations are related to the geometry of the domain Ω we consider. To introduce some geometry on the study of the Laplacian, it is very relevant to use the variational characterization of the spectrum. To this aim, let us introduce the Rayleigh quotient. In order to do this, we have to consider functions in the Sobolev space H 1 (Ω) or H 1 0 (Ω). It is possible to find the definition of H 1 (M) or H 1 0 (M) in every book on PDE, but what we will really consider are functions which are of class C 1 by part and glue them continuously along a submanifold. Moreover, for H 1 0 (Ω), we consider only functions taking the value 0 on Ω. If a function f lies in H 1 (Ω) in the closed and Neumann problems, and in H0 1 (M) for the Dirichlet problem, the Rayleigh quotient of f is Ω R(f) = f dx ( f, f) Ω f =. dx (f, f) Note that in the case where f is an eigenfunction for the eigenvalue λ k, then Ω R(f) = f dx Ω f dx = f fdx ΩΩ f dx = λ k. Theorem 9. (Variational characterization of the spectrum, [4] p ) Let us consider one of the eigenvalues problems. Min-Max formula: we have λ k = inf V k sup{r(u) : u 0, u V k } (7) where V k runs through k + 1-dimensional subspaces of H 1 (Ω) for the Neumann problem and the k- dimensional subspaces of H 1 0 (Ω) for the Dirichlet eigenvalue problem. Another variational characterization for the Neumann problem is the following: Let us denote by {f i } an orthonormal system of eigenfunctions associated to the eigenvalues {λ i }. We have λ k (Ω) = inf{r(u) : u 0; u f 0,.., f k 1 } (8) where u H 1 (Ω) and R(u) = λ k if and only if u is an eigenfunction for λ k. In particular, we have the classical useful facts (recalling that f 0 is constant and that to be orthogonal to f 0 is equivalent of being of integral 0) λ 1 (Ω) = inf{r(u) : u 0; u dx = 0}, (9) and, if u dx = 0 we have the upper bound Ω Ω λ 1 (Ω) R(u). (10) 10

11 For the Dirichlet problem, we have a similar characterization λ k = inf{r(u) : u 0; u f 1,.., f k 1 } (11) where u H 1 0 (Ω) and R(u) = λ k if and only if u is an eigenfunction for λ k. In particular λ 1 (Ω) = inf{r(u) : u H 1 0 (Ω); u 0}. (1) This min-max formula is useless to calculate λ k, but it is very useful to find an upper bound because of the following fact: For example, in the case of the Neumann (resp. Dirichlet) problem, for any given (k + 1) dimensional (resp. k dimensional) vector subspace V of H 1 (Ω), we have λ k (Ω) sup{r(u) : u 0, u V }. (13) This gives immediately an upper bound for λ k (Ω) if it is possible to estimate the Rayleigh quotient R(u) of all the functions u V k. Note that there is no need to calculate the Rayleigh quotient, it suffices to estimate it from above. Of course, this upper bound is useful if the vector space V is conveniently chosen. Let us sketch the proof of (11): let f be a smooth function taking the value 0 on Ω. By the third point of Theorem 8, one can write u = a i f i where {f i } i=1 denotes an orthonormal basis of eigenfuctions. If u f 1,.., f k 1, we have So i=k u = i=1 a i f i. i=k u = a i ; u = λ i a i λ k and R(u) λ k. We already know that R(f k ) = λ k, and this allows to conclude. The proof of (8) is similar. Exercise. The Laplacian is a self-adjoint operator acting on functions, that is on a space of infinite dimension. In finite dimension, we have what we learn in a classical linear algebra lecture: let consider a symmetric matrix A. It is well know that it has n real eigenvalues. Give and prove a min-max characterization as in (7) of these eigenvalues in this simple case. Then do the proof of (7) using similar ideas as in the proof of (11). 3 Lecture 3: applications of the variational characterization 3.1 Comparison between Dirichlet and Neumann problem, monotonicity. Example 10. Let us consider the Dirichlet problem. Use the min-max characterization of the spectrum to show that, if Ω 1 Ω, then we have for each k that λ k (Ω 1 ) λ k (Ω ) (we say that we have monotonicity). i=k i=k a i, 11

12 Note that for the Neumann problem the situation is completely different: in Example 15, we show that given a domain Ω, we can construct Ω 1 Ω with as many eigenvalues as small as we will! Let us prove the monotony result using the min-max characterization of the spectrum. Each eigenfunction of Ω 1 may be continuously extended by 0 on Ω and may be used as a test function for the Dirichlet problem on Ω. Let us construct an upper bound for λ k (Ω ): for V k, we choose the vector subspace of H 1 0 (Ω ) generated by an orthonormal basis f 1,..., f k of eigenfunctions of Ω 1 extended by 0 on Ω. Clearly, these functions vanish on Ω and they are C on Ω 1 and Ω Ω 1 c. They are continuous on Ω1. Let f = α 1 f α k f k. We have (f, f) = α α k. We have ( f i, f j ) = f i, f j dx = f i, f j dx = f i, f j dx = λ i f i f j dx, Ω Ω 1 Ω 1 Ω 1 and it follows that ( f i, f i ) = λ i (Ω 1 )(f i, f i ) and ( f i, f j ) = 0 if i j. We have ( f, f) = α 1λ 1 (Ω 1 ) α kλ k (Ω 1 ) λ k (Ω 1 )(α α k). We conclude that R(f) λ k (Ω 1 ), and we have λ k (Ω ) λ k (Ω 1 ). As this is true for each k, we have the result. Example 11. Note that, in Example 10, a test function for the Dirichlet problem is also a test function for the Neumann problem. If we denote by λ D k and λn k the spectrum for respectively the Dirichlet and Neumann boundary condition, we deduce from the proof that and, in particular, taking Ω 1 = Ω = Ω, we have λ N k (Ω ) λ D k+1(ω 1 ) λ N k (Ω) λ D k+1(ω). 1

13 Exercise. Let Ω R n, and Ω 1,..., Ω p Ω a family of p disjointly supported domains. We denote by {µ i } i=1 the union of the Dirichlet spectrum of Ω 1,..., Ω p. Then, the k th eigenvalue for the Dirichlet problem satisfies λ k (Ω) µ k. Example 1. We consider now Ω R n, and Ω 1,..., Ω p Ω a family of p disjointly supported domains such that p i=1 Ω i = Ω and we denote by {µ i } i=1 the union of the Neumann spectrum of Ω 1,..., Ω p. Then the k th eigenvalue for the Neumann problem satisfies λ k (Ω) µ k+1. Let h 1,..., h k an orthogonal basis corresponding to the k first eigenvalues µ 1,..., µ k and V the vector space generated by the first k+1 eigenfunction f 0,..., f k of the Neumann problem of Ω. We consider the linear application L : V R k given by L(f) = ((f, h 1 ),..., (f, h k ). As the dimension of V is k + 1, the kernel of L is not trivial, and there exists f V such that (f, h i ) = 0 for i = 1,..., k. This implies µ k+1 R(f). But, by definition R(f) λ k and we have the conclusion. Observe however that µ 1 =... = µ p = 0. spectrum. So the method is not relevant for the beginning of the These examples and theorems may be understood as follows: in general, it is difficult to study the spectrum of a domain Ω because of the complexity of the domain. It is sometimes a good idea to write Ω as a disjoint union of smallest domains Ω i that we understand better and deduce information on the spectrum of Ω form the spectrum of the Ω i. 3. Construction of small eigenvalues for the Neumann problem: the Cheeger dumbbell construction. Example 13. The idea is to consider two n-balls of fixed volume A connected by a small cylinder C of length L and radius ϵ. We denote by Ω ϵ this domain. The first nonzero eigenvalue of Ω ϵ converges to 0 as ϵ goes to 0. It is even possible to estimate very precisely the asymptotic of λ 1 in term of ϵ, but here, let us just shows that it converges to 0. Let f be a function with value 1 on the first ball, 1 on the second and decreasing linearly along the cylinder. The maximum norm of its gradient is 1 L. By construction (and because we can suppose for simplicity that the manifold is symmetric), we have M ϵ fdvol gϵ = 0. First method. We introduce a specific vector space, the vector space V generated by f and by the constant function 1. If h V, we can write h = a + bf, a, b R and h dx = a V ol(ω ϵ ) + b f dx, Ω ϵ Ω ϵ h dx = b f dx. Ω ϵ Ω ϵ By the min-max characterization, we have λ 1 sup{r(h) : h V },and we get λ 1 (Ω ϵ ) R(f). The function f varies only on the cylinder C and its gradient has norm 1 L. This implies df dvol gϵ = 1 M ϵ L V ol(c). 13

14 f=-1 f=1 f=x/l Sphere Volume V Cylinder Length L Sphere Volume V Moreover, because f takes the value 1 on both balls of volume A, we have Ω ϵ f dx A. This implies that the Rayleigh quotient of f is bounded above by which goes to 0 as ϵ does. V olc/l Second method (easiest, but specific to λ 1 ). By the min-max characterization, we have by (10), A λ 1 (Ω ϵ ) R(f), where f is orthogonal to the first eigenfunction which is constant. So, we can use the function f defined above, because it satisfies Ω ϵ f = 0 and we obtain the same conclusion with the same calculations. Exercise. Prove a similar result if the two balls joined by the thin cylinder have not the same volume. More generaly, prove the result if we join two fixed domains by a thin cylinder. Example 14. A similar construction with k balls connected by thin cylinders shows that there exist examples with k arbitrarily small eigenvalues. Observe that we can easily fix the volume and the diameter in all these constructions: Thus to fix the volume and diameter is not enough to have a lower bound on the spectrum. 14

15 Exercise. Write precisely Example 14. Example 15. Let Ω R n a domain. There exists a domain Ω 1 Ω with a smooth boundary such that the k first eigenvalues of Ω 1 for the Neumann problem are arbitrarily small. To do this, choose Ω 1 to be a like a Cheeger dumbbell, with k+1 balls related by very thin cylinders. We see immediately that the k first eigenvalues can be made as small as we wish, with the same calculations as above. Note that we have no monotonicity for the Neumann problem even when we restrict ourselves to convex domains. For example, in dimension, we can consider a square [0, π] [0, π]. The first nonzero eigenvalue is 1. But we can construct inside this square a thin rectangle around the diagonal of the square of length almost π: the first eigenvalue is close to 1. Note, however, that we cannot find convex subsets of [0, π] [0, π] with arbitrarily small first nonzero eigenvalue. By a result of Payne and Weinberger [1], we have for Ω R n, convex, the inequality λ 1 (Ω) π diam(ω). Example 16. This example shows that a small, local perturbation of a domain may strongly affect the spectrum of the Laplacian for the Neumann conditions. The text gives only qualitative arguments, the calculations are left to the reader if he needs them to understand the examples. We begin with a domain Ω (a ball in the picture) and show that a very small perturbation of the boundary may change drastically the spectrum for the Neumann boundary condition: A nonzero eigenvalue arbitrarily close to zero may appear as a consequence of a very small perturbation. Figure 4: We want to perturb locally a domain Ω We add a small ball close to Ω (but at a positive distance) and link it to Ω using a thin cylinder. The important point is not the size of the ball or of the cylinder, but the ratio. As on the picture, the radius of the cylinder must be much smaller than the radius of the ball. This creates a Cheeger dumbbell: the only difference with the above calculations is that the two thick parts do not have the same size, but it is easy to adapt the calculations and show that if the radius of the cylinder goes to zero, then the same is true for the first nonzero eigenvalue. The ball can be very small and very close to Ω, so the perturbation may be confined in an arbitrarily small region. We can iterate this first deformation, and get a small local deformation given as many eigenvalues as small as we want. 15

16 Figure 5: Add a small ball and link it to Ω with a thin cylinder 4 Lecture 4: perturbations, nodal domains, Courant and Pleijel theorems 4.1 Elementary surgery. We will explain in detail the following fact (in the context of the Dirichlet boundary conditions): A small hole in a domain does not affect the spectrum too much (see (14) below). Let Ω R n a domain with smooth boundary. Let x Ω, and B(x, ϵ) the ball of radius ϵ centered at the point x (ϵ is chosen small enough such that B(x, ϵ) Ω). We denote by Ω ϵ the subset Ω B(x, ϵ), that is we make a hole of radius ϵ on Ω. Theorem 17. We consider the two domains Ω and Ω ϵ with the Dirichlet boundary conditions, and denote by {λ k } k=1 and by {λ k(ϵ)} k=1 their respective spectrum. Then, for each k, we have Before showing this in detail, let us make a few remarks. Remark The convergence is not uniform in k. lim ϵ 0 λ k(ϵ) = λ k (14). This result is a very specific part of a much more general facts. We get this type of results on manifolds, with subset much more general than balls, for example tubular neighborhood of submanifolds of codimension greater than 1. For the interested reader, we refer to the paper of Courtois [13] and to the book of Chavel [11]. 3. We can get very precise asymptotic estimates of λ k (ϵ) in terms of ϵ. Again, we refer to [13] and references therein. 4. We have also the convergence of the eigenspace associated to λ k (ϵ) to the eigenspace associated to λ k, but this has to be defined precisely: a problem occurs if the multiplicity of λ k is not equal to the multiplicity of λ k (ϵ), see [13]. 5. As a consequence of the monotonicity, the same result is true if we excise a family of domains V ϵ contained in a ball of radius ϵ 0: because Ω ϵ Ω V ϵ we have λ k (Ω V ϵ ) λ k (Ω ϵ ) λ k (Ω) and also λ k (Ω V ϵ ) λ k (Ω). 16

17 6. A similar result is true for a hole and Neumann boundary condition. The proof is more difficult and cannot be extended to the domains contained in a ball. Exercise. For the Neumann boundary condition, show that, given a ball B Ω, we can excise a domain V ϵ B such that Ω V ϵ has as many small eigenvalues as one wish. To this aim, such ideas similar as in Example 15. Proof of Theorem 17 In order to see the convergence, we first observe that by monotonicity (10), for all k λ k λ k (ϵ). Our goal is to show that with c(ϵ) 0 as ϵ 0. λ k (ϵ) λ k + c(ϵ), In order to use the min-max construction, we need to consider a family of test-functions as explained in (13). This family is constructed using a perturbation of the eigenfunctions of Ω. Note that we do the proof for n 3, but the strategy is the same if n =. Let {f i } i=1 be an orthonormal basis of eigenfuctions on Ω. We fix a C -plateau function χ : R R defined by 0 if r 3 χ(r) = 1 if r 0 χ(r) 1 if 3 r The test function f i,ϵ associated to f i is defined by where d denotes the distance. f i,ϵ (p) = χ( d(p, x) )f i. ϵ By construction, the function f i,ϵ (p) takes the value 0 on the ball B(x, ϵ) and satisfies the Dirichlet boundary condition (for ϵ small enough). So, it can be used as a test function in order to estimate λ k,ϵ. Let f be a function on the vector space generated by {f i,ϵ } k i=1. In order to estimate its Rayleigh quotient, it makes a great simplification if the basis is orthonormal and the ( f i ) are orthogonal. This is not the case here. However, the basis is almost orthonormal and the ( f i ) are almost orthogonal: this is what we will first show and use. Note that this is a very common situation. The basis {f i,ϵ } k i=1 is almost orthonormal. Let us denote by a i (resp. b i ) the maximum of the function f i (resp. f i ) on Ω. Then fi a i n ω n ϵ n, f i b i n ω n ϵ n (15) B(x,ϵ) where ω n is the volume of the unit ball in R n. Let us show that where C k depends on a 1,..., a k. B(x,ϵ) δ ij C k ϵ n/ (f i,ϵ, f j,ϵ ) δ ij + C k ϵ n/, 17

18 (f i,ϵ, f j,ϵ ) = (χ ϵ f i, χ ϵ f j ) = ((χ ϵ 1)f i + f i, (χ ϵ 1)f j + f j ) = = (f i, f j ) + ((χ ϵ 1)f i, f j ) + ((χ ϵ 1)f j, f i ) + ((χ ϵ 1)f i, (χ ϵ 1)f j ). By (15), (χ ϵ 1)f i a i n ω n ϵ n, and by Cauchy-Schwarz where C k depends on a 1,.., a k. δ ij C k ϵ n/ (f i,ϵ, f j,ϵ ) δ ij + C k ϵ n/, The set { f i,ϵ } k i=1 is almost orthogonal. Let us observe that and f i,ϵ = χ ϵ f i + χ ϵ f i, ( f i,ϵ, f j,ϵ ) = ( χ ϵ f i, χ ϵ f j ) + ( χ ϵ f i, χ ϵ f j ) + (χ ϵ f i, χ ϵ f j ) + (χ ϵ f i, χ ϵ f j ). We have also χ ϵ f i C 1 ϵ fi B(x,ϵ) Ca i n ω n ϵ n where C depends only on the derivative χ of χ. This is the place where n 3 is used in order to have a positive exponent to ϵ. With the same considerations as before where C k depends on a 1,..., a k, b 1,..., b k, C. ( f i,ϵ, f j,ϵ ) ( f i, f j ) + C i ϵ n/ = λ i δ ij + C k ϵ n/. Now, we can estimate the Rayleigh quotient of a function f [f 1,ϵ,..., f k,ϵ ]. This is done exactly in the same spirit as the proof of the monotonicity, but we have to deal with the fact that the basis of test functions is only almost-orthonormal. Let f = α 1 f 1,ϵ α k f k,ϵ. Thanks to the above considerations, we have f = α α k + O(ϵ n/ ), This implies and gives the result. f = α 1λ α kλ k + O(ϵ n ). R(f) λ k + O(ϵ n ) 18

19 4. Nodal domains On different pictures or video, it seems to appear that the eigenfunction(s) of the eigenvalue λ k on any domain Ω becomes more and more complicated as k increase The simplest example is the interval [0, L] where we have seen that the k th eigenvalue of the Dirichlet problem was k π L, and the eigenfunction corresponding to λ k was f k (x) = sin kπ L x. This is the object of this part of the lecture to elaborate a little around this. I follow mainly the book of Chavel [11], p Definition 19. Let Ω R n a domain and f : Ω R a continuous function. The nodal set of f is the set {f 1 (0)} and a nodal domain of f is one connected component of Ω {f 1 (0)}. In the sequel, I will describe a situation for the Dirichlet eigenvalues. The situation for the Neumann problem is often similar, but not always. I can be more precise about what we see on the video around Chladni plates: the sound oblige the plate to vibrate and the vibration of the plate correspond to an eigenfunction. What we see is the place where the plate does not vibrate, and this correspond to the nodal set of the corresponding eigenfuctions. The impression is that for large frequence (large eigenvalue) the nodal line is more and more complicated and tends to be everywhere on the plate. moreover, we have the impression that the number of nodal domains tends to increase with k. This is what we want to clarify thanks to Theorem 0 and 1. A first observation is that if f is an eigenfunction for the k th eigenvalue of a domain Ω, then the restriction of f to each of its nodal domains is an eigenfunction for the Dirichlet boundary condition. As it does not change its sign, one can even say that this is an eigenfunction for the first eigenvalue of this nodal domain that we call G. Now, if p G and r is small enough so that the ball B(p, r) of center p and radius r stays in G, then by monotonicity property (Example 10), λ 1 (B(p, r)) λ 1 (G) = λ k (Ω). Moreover, we have λ 1 (B(p, r)) = 1 r λ 1 (B) where B is the unit ball in R n. In summary, we have and we have prove the following result: 1 r λ 1(B) λ k (Ω), Theorem 0. There exist a constant C(n) = λ 1 (B) (B unit ball in R n ) such that the following is C true: if p Ω and if r, then the ball B(p, r) intersect the nodal set the any eigenfunction f of λ k (Ω). λk (Ω) This gives an measure of the complexity of the nodal set which becomes more and more dense for large λ k! However, the number of nodal domains cannot increase too quickly. This is the theorem of Courant: Theorem 1. Let Ω be a domain in R n and f 1, f,.. be an orthonormal basis of eigenfunctions for the eigenvalues λ 1, λ,... Then, the number of nodal domains of f k is less or equal to k, for all k. In particular, the first eigenfunction does not change its sign. So, it has to be of multiplicity 1. Note that in dimension 1, there are exactly k nodal domains for λ k. For the proof, we suppose that there are k + 1 nodal domains that we denote by G 1,..., G k+1. We consider the function h j equal to f k on G j and 0 outside, for j = 1,...k. We consider the vector space V 19

20 generated by h 1,..., h k. The dimension of V is k because the functions h 1,..., h k are clearly independent. We consider the linear map H : V R n 1 given as follow: if h = k j=1 a jh j, we take by H(h) = ((h, f 1 ),...(h, f k 1 )). Because of the respective dimensions of these spaces, the kernel of H is not reduced to 0 and there exists h = k j=1 a jh j orthogonal to f 1,..., f k 1. By the min-max characterization (11), we have λ k (Ω) R(h) and h is an eigenfunction of λ k (Ω) in case of equality. In order to calculate R(h), because the functions h j are disjointly supported, we have and h = k k a j h j λ k ( a j), j=1 h = k a j. It follows that R(h) λ k so that R(h) = λ k (Ω). But this is not possible because of a very general principle (unique continuation principle of Aronszajn, see []), telling that an eigenfunction cannot be constant in an open set. Here, it turns out that h is equal to 0 in the open set G k+1. After the theorem of Courant, a new natural question appears: does a k th eigenfunction f associated to λ k (Ω) have exactly k nodal domains as it is the case in dimension 1. Examples show that this is not the case, but there a lot of tricky and interestiong problems around this question. Let us just mention j=1 j=1 0

21 the theorem of Pleijel (1956) which says the following: if n k denote the number of nodal set of λ k (Ω), then n k lim sup γ(n) < 1, k k where γ n depends only on the dimension n. 5 Lecture 5: The Weyl formula. It is in general not possible to calculate the spectrum of a domain. However, asymptotically, it is possible to say something: the eigenvalues satisfy the Weyl law. Weyl law: For each of the eigenvalue problems (Dirichlet and Neumann) on a domain Ω, λ k (Ω) (π) ω /n n as k, where ω n denotes the volume of the unit ball of R n. This also means λ k (Ω) lim k k /n = (π) ω /n n In particular, the spectrum of Ω determines the volume of Ω. ( ) /n k (16) V ol(ω) ( ) /n 1 (17) V ol(ω) It is important to stress that the result is asymptotic! We do not know in general for which k the asymptotic estimate becomes good! It depends in fact on the domain we consider. However, this formula is a guide when trying to get upper bounds. Let us see how to establish the Weyl formula for the Dirichlet problem in dimension. Recall that, for a given domain Ω R n and for the Neumann or Dirichlet problem, we have to show that λ k (Ω) = (π) ω /n n ( ) /n k + ϵ(k)k /n (18) V ol(ω) with ϵ(k) 0 as k. Note that finding precisely the second term in the asymptotic is a very interesting question, but we will not look at it. In order to prove this, it is better to have another equivalent formulation of this equality. For γ > 0, let N(Ω, γ) = {k : λ k (Ω) γ}, (19) that is the number of eigenvalues less or equal to γ. The formula N(Ω, γ) = ω nv ol(ω) (4π ) n/ γn/ + γ n/ ϵ(γ) (0) with ϵ(γ) 0 as γ implies the Weyl law. Again, the interest of this formula is only asymptotic and means that N(Ω, γ) lim = ω nv ol(ω) γ γ n/ (4π ). n/ 1

22 Let us suppose that Formula 0 is verified and write c = ω nv ol(ω) (4π ) n/. This mean that for a given ϵ > 0 and for γ large enough, we have (1 ϵ)cγ n/ N(γ) (1 + ϵ)cγ n/. If we take γ = λ k, we have N(λ k ) k (one cannot say = k because of the possible multiplicity), and we get (1 + ϵ)cλ n/ k k. Then, choose δ > 0 and take γ = λ k δ. We have N(λ k δ) < k and (1 ϵ)c(λ k δ) n/ < k. As it is true for all δ > 0, we deduce (1 ϵ)cλ n/ k k. In summary, we have for each ϵ > 0 and k large enough and, as ϵ is arbitrary, we have 1 (1 + ϵ) /n c /n λ k k /n 1 (1 ϵ) /n c /n, lim k λ k k /n = 1 c /n. Formula 0 in dimension for the Dirichlet problem. This uses mainly Examples 10, 11, 1 and the associate exercises. Step 1: a particular case. We do first the proof for a very special domain Ω which is a union of adjacent squares R j of edge of length ϵ. Intuitively, think that ϵ is small, but fixed: Ω = m R j=1 j. We also introduce Ω 0 = m j=1 R j, which is a disjoint union of squares. We denote λ D k and λn k the Dirichlet and Neumann eigenvalues and, for the convenience of the proof, we denote by λ N 1 the first eigenvalue (before we called it λ N 0 ). Let us summary in the present situation what say the examples 10, 11, 1. We have Ω 0 Ω, Ω = Ω 0 and Ω 0 is the disjoint union of the R j. It follows λ N k (Ω 0 ) λ N k (Ω) λ D k (Ω) λ D k (Ω 0 ). (1)

23 Recall that the spectrum of a disjoint union is simply the ordered union of the spectrum of the different components. This implies in particular N D (Ω 0, γ) N D (Ω, γ) N N (Ω 0, γ). () So, in order to estimate N D (Ω, γ), which could be very tricky, we are lead to estimate N D (Ω 0, γ) and N N (Ω 0, γ), which is more tractable: the spectrum of Ω 0 is just m copies of the spectrum of a square R of radius ϵ. This means that N D (Ω 0, γ) = mn D (R, γ); N N (Ω 0, γ) = mn N (R, γ) and, regarding the asymptotic as γ (but ϵ fixed) and We have to calculate that is the asymptotic for a square. N D (Ω 0, γ) (R, γ) lim = lim γ γ γ mnd γ N N (Ω 0, γ) (R, γ) lim = lim γ γ γ mnn. γ lim γ Proposition. For a square R, we have N D (R, γ) N N (R, γ) ; lim γ γ γ N D (R, γ) N N (R, γ) lim = lim γ γ γ γ = V ol(r). 4π If we admit this proposition, because mv ol(r j ) = V ol(ω), we deduce that N D (Ω, γ) lim γ γ = V ol(ω). 4π This prove the Weyl law for the Dirichlet problem in dimension for these very particular domains which are union of squares. Second step: general case. Let Ω R a bounded domain with smooth boundary. We write the plane R as a natural union of squares R j of radius ϵ. We define We have Ω ϵ 1 Ω Ω ϵ, which implies again Ω ϵ 1 = R j : R j Ω; Ω ϵ = R j : R j Ω. λ D k (Ω ϵ ) λ D k (Ω) λ D k (Ω ϵ 1) 3

24 and From Step 1, we have N D (Ω ϵ 1, γ) N D (Ω, γ) N D (Ω ϵ, γ). and N D (Ω ϵ lim 1, γ) γ γ This implies that for each ϵ > 0, = V ol(ωϵ 1) N D (Ω ϵ ; lim, γ) 4π γ γ lim sup γ lim inf γ N D (Ω, γ) γ N D (Ω, γ) γ V ol(ωϵ ) 4π V ol(ωϵ 1) 4π = V ol(ωϵ ). 4π Moreover V ol(ω ϵ ) V ol(ω ϵ 1) 0 as ϵ 0 (this difference corresponds to the area of the squares touching Ω; note that at this stage, we need some regularity of the boundary!), and this implies that N D (Ω, γ) lim γ γ = V ol(ω). 4π Proof of Proposition. We prove it for a square of edges of length 1. By homothety, we get the result for length ϵ. The eigenvalues of a unit square for the Dirichlet boundary condition were calculated in Example. We have λ k,l = π (k + l ) : k, l 1. We have N(γ) = {(k, l) : k + l γ ; k, l 1}. π 4

25 Intuitively, this is more or less the couple of positive integer contained in a square of radius γ π. Always intuitively, this must be related to the area of the part of the disc of radius γ π contained in the positive quadrant, which is γ 4π as expected in Proposition. Let us see this more rigorously. To each couple (k, l) such that k + l γ π one can associate a unit square of center (k, l). Some of these squares may intersect the circle of radius γ π contained in the first quadrant, but they are contained in the circle of radius ( γ π + ). So there are at most π γ 4 π γ π of such points. At the other side, this union of squares cover the disc of radius γ π, with a possible exception of a neighborhood of radius along the different boundaries which total length is ( + π ) γ π. This means that there are at least γ 4π ( + π ) γ π such points. (Goemetrically, we have used the idea that a ϵ tubular neighbourhood of a curve of length L has an area bounded by Lϵ; it would be more complicated in higher dimension). and So, we have as expected. N D (R, γ) = γ 4π + O( γ), N D (R, γ lim γ γ = 1 4π The proof for the Neumann case is essentially the same. We have to take into account the couple of points (k, l) with k 0, l 0. 5

26 6 Lecture 6: extremal problems, open questions 6.1 Minimization of the second eigenvalue of the Dirichlet problem: Theorem of Szegö. I will explain the proof of the fact announced in Lecture 1: we have to show that the infimum of λ for the Dirichlet boundary conditions on planar domains (or more generally in R n ) of given volume is given the union of two disjoint balls (see [18] thm ). Note that this domain is not regular in the sense that was expected. I sketch the proof as it is done in [18]. First, recall what we know: 1. The first eigenvalue for the Dirichlet problem on a regular domain (connected, Lipschitz boundary) is of multiplicity 1 and the associate eigenfunction does not change its sign.. The theorem of Courant: the number of nodal domains for any eigenfunction associated to λ is less or equal to. 3. The Faber-Krahn inequality: for a domain Ω, λ 1 (Ω) λ 1 (B) where B is a ball of the same volume as Ω. Let go to the proof: The first implication of the theorem of Courant is that any eigenfunction f 1 associated to the first eigenvalue λ 1 of a domain Ω has a constant sign (say it is positive) and has multiplicity one. The second eigenvalue λ of Ω could have multiplicity. If f is an eigenfunction associated to λ, it has to be orthogonal to f 1, and, as a consequence, changes its sign. Let A 1 = {x Ω : f (x) > 0}; A = {x Ω : f (x) < 0}. The Courant Nodal Theorem says that A 1 and A are connected. The restriction of f to A 1 and to A satisfies the Laplace equation f = λ (Ω)f, and f is an eigenfunction of eigenvalue λ. But on A 1 and A, f does not change its sign. This mean that f is an eigenfunction for the first eigenvalue of A 1 and A, and it follows λ 1 (A 1 ) = λ 1 (A ) = λ (Ω). Let B 1, B two balls with V ol(b 1 ) = V ol(a 1 ) and V ol(b ) = V ol(a ). By Faber-Krahn, we have λ 1 (A i ) λ 1 (B i ), i = 1,, with equality if A i is congruent to B i, so that λ (Ω) max(λ 1 (B 1 ), λ 1 (B )). In summary, we have shown that for any domain Ω, there exist two balls B 1, B with V ol(b 1 ) + V ol(b ) = V ol(ω) and λ (Ω) max(λ 1 (B 1 ), λ 1 (B )). The difficulty is that we do not know how these balls look like. Recall that, if we consider now the disjoint union B of B 1 and B, the spectrum of B is the union of both spectrum, and V ol(b ) = V ol(ω), so that λ (B ) max(λ 1 (B 1 ), λ 1 (B )). So, we deduce λ (Ω) max(λ 1 (B 1 ), λ 1 (B )) = λ (B ) and we just have to minimize the number max(λ 1 (B 1 ), λ 1 (B )) when V ol(b 1 ) + V ol(b ) = V ola 1 + V ola = V ol(ω). 6