Nodal domains of Laplacian eigenfunctions

Transcription

1 Nodal domains of Laplacian eigenfunctions Yoni Rozenshein May 20 & May 27, 2014 Contents I Motivation, denitions, Courant's theorem and the Faber-Krahn inequality 1 1 Motivation / entertainment The vibration problem The 1-dimensional case Higher dimensions Denitions and basic results 5 3 Courant's nodal domain theorem A brief encounter with weak derivatives Rayleigh's variational method Nodal domains of eigenfunctions Isoperimetric properties The Coarea Formula The Faber-Krahn inequality II Weyl's law and Pleijel's theorem 15 5 Variational methods on eigenvalues and Weyl's law Neumann eigenvalues The variational method for Neumann eigenvalues Domain monotonicity of eigenvalues The asymptotic growth of eigenvalues Improvements to Courant's theorem 24 1

2 Part I Motivation, denitions, Courant's theorem and the Faber-Krahn inequality 1 Motivation / entertainment 1.1 The vibration problem We start by describing the vibration problem: A string (1D) or membrane (2D) is held static at the edges (or boundary glued to the drum), and plucked or beaten. How does it vibrate? What sounds can you hear? The physics behind this problem concern the wave equation. Suppose our domain of vibration is Ω (interval or planar domain). Describe by u (x, t) the string or membrane's displacement at point x and at time t. The wave equation is: 2 u t 2 = c2 u (1) where the Laplacian is taken in the spatial variables, and c 2 is a positive physical constant related to the material and tension of the string or membrane. We add a static boundary condition u (x, t) = 0 whenever x Ω and search for solutions u C 2 (Ω R). These solutions describe how Ω can vibrate. How to solve this? We try to nd simple modes of vibration by separation of variables. (It is, in fact, true that all complex modes of vibration are representable as a series of simpler ones, by a Fourier-like expansion. But we will not go into this.) Let u (x, t) = U (x) T (t) (U will be the spatial part and T the temporal part of the solution). We assume c = 1. Plugging this into (1) gives which means U (x) T (t) = U (x) T (t), U (x) U (x) = T (t) T (t) This leads to the existence of a constant λ R that satises U + λu = 0 with U = 0 on Ω T + λt = 0 2

3 T is therefore given by a well-known and easy-to-solve ODE for which the solutions are exponential if λ < 0, linear if λ = 0 and oscillating if λ > 0. Our physical intuition therefore requires λ > 0 (that's why the equations are written in this way, and not U = λu...). Anyway, solving for T is no problem. U takes the form of a Laplacian eigenfunction, and there is much interest in what functions U are possible and what values λ are possible. 1.2 The 1-dimensional case In dimension 1, the analysis is very easy, and will serve as a model for what we should expect in higher dimensions. There is only one shape of a domain - an open interval. Let Ω is a bounded interval in R with two endpoints. We assume the length of the interval is l and that Ω = (0, l). Our problem is to solve: U + λu = 0 U (0) = U (l) = 0 Following our physical intuition, we assume that λ > 0. Then ( the ODE λx ) (without boundary conditions) has two independent solutions: sin and ( λx ) cos. The rst boundary condition, U (0) = 0, eliminates the cosine ( λl ) solution. The second boundary condition adds the constraint sin = 0. This means not every λ > 0 gives a nontrivial solution to our problem! We have a very simple constraint: λ > 0 solves our problem if and only if λl πz. To summarize, we have proven: Proposition 1 (spatial part of vibration of a string). 1. The eigenvalues of a vibrating string of length l form a discrete sequence 0 < λ 1 < λ 2 <..., where λ n = ( ) πn 2. l 2. Each eigenvalue is simple, with eigenfunction φ n (x) = sin ( λ n x ). 3. The n'th eigenfunction has exactly n 1 nodes (zeroes) in the interior of the string, and thus exactly n nodal domains (nonzero connected components). Remark. This was known to the Pythagoreans, and is known to anyone who plays the guitar. Corollary 2. You can hear the length of a string. Proof. Pluck the string and listen to the frequencies. Arrange them in order and divide the n'th frequency by n 2. The result is a constant ( π l ) 2. 3

4 1.3 Higher dimensions In one dimension, we completely solved the problem easily thanks to two things: 1. Our PDE became an ODE; 2. Domains are very simple - only intervals are domains. In higher dimensions (for example - two-dimensional membranes stretched over drums), we no longer have these easy conditions. It is of much interest to generalize what's easy in 1 dimension to higher dimensions. Here are a few questions: Question 3. In the 1-dimensional case, the eigenvalues grow quadratically. How do they grow in higher dimensions? Question 4. Are the eigenvalues simple like in the 1-dimensional case, or can they have multiplicities? The next question concerns the nodal domains. For example, in dimension 2, the nodal set is composed of lines rather than points, and the nodal domains are the areas between the lines. An experiment by E. Chladni (Chladni plates) allows one to reveal the nodal lines, by vibrating a plate at an eigenfrequency and spilling sand over it. The nodal lines remain stationary and other points vibrate, so the sand naturally moves to the nodal lines. This experiment can be seen in several YouTube videos, for example [17]. Question 5. How many nodal domains can the n'th eigenfunction have? Question 6 (Kac, 1966 [13]). Can you hear the shape of a drum? (How much do the eigenvalues tell us about the shape of the domain?) Question 6 is very hard. A partial positive answer (which also answers Question 3) is given by Weyl's law, which we will prove in the second part of this talk: Theorem 7 (Weyl, 1911). In the two-dimensional 1 case, λ n 4πn particular, you can hear the area of a drum. Area Ω. However, for more complicated notions of shape, the answer is negative. Theorem 8 (Milnor, 1964 [14]). There are two 16-dimensional isospectral 2 non-congruent tori, and this can be proven in less than a page. 1 Results for other dimensions are also given by Weyl's law, but 2D is the simplest. Note ( ) 2 πn that this is very similar to the formula λ n = Length Ω that we found in the 1-dimensional case. But why isn't there a second power? To understand why this makes sense, notice that if a function u measures somethings, then the Laplacian (which is really a second spatial derivative) measures somethings per centimeter squared. Since the PDE is u + λu = 0, the eigenvalues λ must have units cm 2, and this is for any dimension. This is why the powers make sense; in the d-dimensional case, we expect the value to be proportional to (Vol Ω) 2/d. 2 Two domains are isospectral if they have the same eigenvalues. In 4

5 Theorem 9 (Gordon, Webb, Wolpert, 1992 [15]). The following drumheads are isospectral: Question 5 also has a more complicated answer than in the rst dimensional case. In Figure 1, we see how a small perturbation can wildly aect the number of nodal domains. Figure 1: The nodal lines of sin 12x sin y + µ sin x sin 12y, for µ = 1 (left image; 12 nodal domains) and for µ 1 (right image; 2 nodal domains). Image reproduced from Courant & Hilbert's book [1]. 2 Denitions and basic results Problem (the Dirichlet eigenvalue problem). Let Ω R d be a domain. Describe all values λ C and all (complex-valued) functions u C 2 (Ω) C ( Ω) that satisfy: { u + λu = 0 on Ω u = 0 on Ω 5

6 From this point on, we'll assume Ω is bounded and has a suciently regular boundary (for example, piecewise smooth). Exercise 10. Any eigenvalue must be real and positive. To prove this, we recall: Theorem (Green's rst identity). Let f C 1 (Ω) C ( Ω) and g C 2 (Ω) C ( Ω) 1. Then ( ) f g + f, g dv = f g ν ds, Ω where ν is the outward normal vector, and the inner product is taken in C. Solution of exercise. Applying Green's rst identity with f = g = u, we get ( u u + u 2) dv = 0 Plugging in the PDE and rearranging gives us Ω Ω λ u 2 = u 2 (2) (where the norms are understood as the L 2 norms), which shows λ = λ 0. Finally, if λ = 0 then u 2 = 0, which means u = 0, and u is constant, but then u is zero by the boundary condition. (Another argument is that if λ = 0 then u is harmonic, but then it must be zero by the maximum principle.) Hereafter we will no longer deal with complex numbers. The equation (2) plays a central role, and we will return to it later. Theorem 11 (spectral theorem for - 19th century). Suppose Ω is bounded and Ω is nice (for example, smooth). Then: 1. The set of eigenvalues is discrete. We will write them in the following way: 0 < λ 1 λ 2... (Eigenvalues with multiplicity are repeated.) 2. There is an orthonormal sequence of real-valued eigenfunctions φ 1, φ 2,... such that φ i corresponds to λ i. This sequence is a complete orthonormal basis for the Hilbert space L 2 (Ω). Remark. The boundedness of Ω is absolutely necessary. For example, if Ω = R d ((relaxing the boundary condition to bounded at innity), the function λx1 ) sin is an eigenfunction for any λ > 0. 6

7 We shall not dene nice boundary nor prove Theorem 11. The idea of the proof: 1. There exists z C such that (zi ) 1 is a compact operator. Its spectrum is therefore discrete, except maybe 0 as an accumulation point. 2. is self-adjoint, so the spectral theorem applies. Theorem 12 (regularity of eigenfunctions). Each eigenfunction is C (Ω) and has the unique continuation property: it cannot vanish in an open set. 3 Remark. Many of the results in this talk are still valid when replacing Ω with a compact Riemannian manifold with nice boundary, where is understood as the Laplace-Beltrami operator. 3 Courant's nodal domain theorem 3.1 A brief encounter with weak derivatives In order to prove Courant's nodal domain theorem, we will need to calculate gradients of functions that are not C 1. To do this, we need the notion of a weak gradient. In this talk, we will not provide the most general denition of this notion. We will content ourselves with a denition that works only on continuous functions that are zero on the boundary. The motivation for dening a weak gradient is the desire to be able to integrate by parts. We rst recall the divergence theorem. Theorem (divergence theorem). Let X C 1 (Ω) C ( Ω) be a vector eld. Then div X dv = X, ν da where ν is the outward normal vector. Ω Exercise 13 (integration by parts). Let f C 1 (Ω) C ( Ω) with f = 0 on Ω. Let X C 1 (Ω) C ( Ω) be a test vector eld. Then 4 Ω f, X L 2 (Ω) = f, div X L 2 (Ω) Solution of exercise. By the Leibniz rule, div (fx) = f div X + f, X, so the result immediately follows from the divergence theorem on fx (remember f = 0 on Ω). 3 In fact, the eigenfunctions are real-analytic, which immediately guarantees the unique continuation property. However, in the generalization of the problem to smooth manifolds with a non-analytic Riemannian metric, they may be non-analytic, but the unique continuation property still holds (this is Aronszajn's theorem). 4 There is some abuse of notation here: The inner product f, X L 2 (Ω) isn't in the L2 space of functions, but the L 2 space of vector elds (i.e. vector elds with L 2 components). This inner product is dened as Ω f, X dv. For simplicity we will continue using the same notation for both, and the subscript L 2 (Ω) should clear up any confusion (without the subscript it will be the Euclidean inner product). 7

8 This exercise motivates the following denition: Denition 14 (weak gradient). Dene by D (Ω) the space of functions f C ( Ω) with f = 0 on Ω. Let f D (Ω). We say that a vector eld G f L 2 (Ω) is a weak gradient of f, if for any test vector eld X C 1 (Ω) C ( Ω), we have G f, X L2 (Ω) = f, div X L 2 (Ω) Remark. This is not the most general denition of the weak gradient: We only handle the case where f is continuous with zero on the boundary. This is the only case needed in this talk, but there are more general denitions: f is allowed to be L 2 (Ω), but then the test vector eld X must be compactly supported. The space of L 2 functions possessing a weak gradient is the Sobolev space, denoted W 1,2 (Ω) or H 1 (Ω). A distributional denition allows the weak gradient to have point masses and other irregular behavior. Unfortunately, a more general denition complicates things 5, so we avoid it in this talk. Exercise 15. If G f exists, it is unique, and we denote it by f. (This notation is justied by Exercise 13.) Solution of exercise. Assume G 1, G 2 are two weak gradients for some f, and let G = G 1 G 2. Then G, X L 2 (Ω) = 0 for any X C1 (Ω) C ( Ω). But C 1 is dense in L 2, so G = 0. Denition 16. Dene by W (Ω) D (Ω) the space of functions that have a weak gradient. Notice that this is a vector space. The following lemma will be useful later. Lemma 17 (patching lemma). Let f D (Ω) C 1 (Ω). Let Ω be subdomain of Ω. Let g D (Ω) be the function { f on Ω g = 0 elsewhere If f = 0 on Ω, then g W (Ω) (g has a weak gradient), and { f on Ω g = 0 elsewhere 5 For example, it is harder to require f = 0 on the boundary when f is only L 2. The boundary is of measure zero, so this doesn't make sense. Instead, one must say f belongs to the completion of C0 inside the Sobolev space. But I want to avoid such nonsense in my talk. 8

9 Proof. We will verify that g dened above is indeed a weak gradient, by the denition. Let X C 1 (Ω) C ( Ω) be a vector eld. By Exercise 13 applied to f and Ω, we have f, X L 2 (Ω ) = f, div X L 2 (Ω ). Replacing f by g and increasing the domain from Ω to Ω does not change the value of either side, so we are done. Remark. The requirement that f = 0 on Ω is crucial to the patching lemma. If it doesn't hold, then when using the divergence theorem in Exercise 13, we have the integral Ω fx, ν da, which has no good reason to be zero. 3.2 Rayleigh's variational method Denition 18 (Rayleigh quotient). Let w W (Ω), w 0. quotient of w is where the norms are taken in L 2 (Ω). Q (w) := w 2 w 2, The Rayleigh Remark. To understand this quotient, think about functions normalized with w = 1. Then Q (w) = w 2. The bigger Q is, the more w uctuates, where we don't measure the values but just how quickly they change. We already know that Q (φ n ) = λ n, so bigger eigenfunctions represent higher energy modes. Theorem 19 (Rayleigh's theorem). Let w W (Ω), w Q (w) λ 1, with equality if and only if w is an eigenfunction of λ Suppose w is orthogonal to φ 1,..., φ n 1 in L 2 (Ω). Then Q (w) λ n, with equality if and only if w is an eigenfunction of λ n. Proof. We prove the second part, which implies the rst by taking n = 1. Let α j := w, φ j L 2 (Ω). (We know that α j = 0 for 1 j n 1.) We make some auxiliary calculations using integration by parts: w, φ j L2 (Ω) = w, φ j L2 (Ω) = w, λ jφ j L2 (Ω) = λ jα j φ i, φ j L2 (Ω) = φ i, φ j L2 (Ω) = λ jδ ij Let r n be an integer. 0 2 r w α j φ j = j=n r = w 2 2 α j w, φ j L2 (Ω) + = w 2 2 j=n r λ j αj 2 + j=n n i,j r r λ j αj 2 = w 2 j=n 9 α i α j φ i, φ j L2 (Ω) = r λ j αj 2 j=n

10 Letting r, this means On the other hand, λ j αj 2 w 2. j=n λ j αj 2 λ n j=n j=n α 2 j = λ n w 2, where the last step is Parseval's identity. This gives w 2 λ n w 2, or Q (w) λ n. In the case of equality Q (w) = λ n, we get λ j αj 2 = λ n j=n which implies α j = 0 for all j with λ j λ n, so w is an eigenfunction of λ n. Corollary 20 (max-min theorem). The eigenvalues are given by: { } λ n = max min Q (w) v 1,...,v n 1 L 2 (Ω) w W (Ω) orthogonal to v 1,...,v n 1 Proof (sketch). Whatever the functions v 1,..., v n 1 are, we can always pick w to be a linear combination of φ 1..., φ n, because there are more unknowns (coecients) than constraints. A calculation like in the previous proof shows that then Q (w) λ n. On the other hand, choosing v j = φ j for 1 j n 1, by Rayleigh's theorem we force the constraint Q (w) λ n. j=n α 2 j. 3.3 Nodal domains of eigenfunctions Denition 21 (nodal domain). A nodal domain of a function f D (Ω) is a connected component of the set {f 0} Ω. Theorem 22 (Courant, 1923). Any eigenfunction of λ n has at most n nodal domains. Proof. Assume that u is an eigenfunction of λ n with at least n+1 nodal domains, and let Ω 1,..., Ω n+1 be distinct nodal domains of u. For each j = 1,..., n, dene the following function: { u on Ω j ψ j = 0 elsewhere By the patching lemma (Lemma 17), ψ j W (Ω), with ψ j = u on Ω j and 0 elsewhere. 10

11 Notice that ψ j (and similarly ψ j ) are clearly pairwise orthogonal, as they are dened in distinct subdomains. For any linear combination w = α 1 ψ α n ψ n, we therefore get: w 2 = n αj 2 ψ j, ψ j = j=1 so Q (w) = λ n. Introducing the linear constraints n αjλ 2 n ψ j, ψ j = λ n w 2, j=1 w, φ j = 0 for each 1 j n 1, we see that a nontrivial linear combination w exists, since there are more unknowns (coecients) than constraints. By Rayleigh's theorem, w must be an eigenfunction. However, w must vanish on Ω n+1, which contradicts the unique continuation property. Exercise 23 (immediate corollaries from Courant's nodal domain theorem). 1. λ 1 is simple (no multiplicity). Therefore φ 1, called the rst eigenfunction, is unique (up to a constant factor). 2. φ 1 has constant sign. Eigenfunctions of other eigenvalues change sign, and thus have at least 2 nodal domains. 3. Eigenfunctions of λ 2 have exactly 2 nodal domains. 4. Let Ω be a nodal domain of an eigenfunction of λ n. Then the rst eigenvalue of the Dirichlet eigenvalue problem on Ω is λ n. Solution of exercise. Everything follows immediately from the following observation: φ 1 does not change sign by Courant's theorem, and any function orthogonal to φ 1 must therefore change sign - otherwise the inner product would not be zero. The last part follows from the fact that in the nodal domain, the eigenfunction does not vanish, and so it must be the rst eigenfunction. 4 Isoperimetric properties We will show another geometric property arising from the eigenvalues. Recall the classical isoperimetric inequality: Theorem 24 (isoperimetric inequality). A bounded domain Ω with nice boundary has S ( Ω) S ( B), where B is a ball of the same volume and S () is hypersurface area, with equality if and only if Ω is a ball. Amazingly, the same is true if we replace hypersurface area of the boundary with rst eigenvalue. This is the Faber-Krahn inequality, which we prove in this section. 11

12 4.1 The Coarea Formula We denote by dv the d-dimensional volume element and by da the (d 1)- dimensional hypersurface area element. The coarea formula is simply a change of variables, which allows integration over a volume element by integrating over level sets of an auxiliary function. (See, for example, [7] for more on this subject.) Lemma 25 (coarea formula). Let f C 1 (Ω) be positive in Ω, and extend to zero in Ω. Suppose that Vol {f = t} = 0 for all t. Denote by Ω t = {f > t} Ω and V (t) = Vol Ω t. 1. V is continuous everywhere, dierentiable almost everywhere (in the regular values of f), and V 1 (t) = {f=t} f da. 2. For any g L 1 (Ω), we have Ω g dv = max f 0 g dt {f=t} f da. Sketch of the proof. Let R f (0, max f) be the regular values (the values t for which f 0 whenever f = t). R f is clearly open, and by Sard's theorem, it is of full measure. Let (a, b) R f and µ (a, b). We shall construct a dieomorphism ψ : f 1 (µ) (a, b) f 1 ((a, b)) that takes a point on the µ level and returns a point on the given level. Let X be the vector eld f f 2. Let φ t (q) be the ow of X starting at q (that is, the solution to the ODE φ t (q) = X (φ t (q)) with initial value φ 0 (q) = q). Dene: ψ (q, t) = φ t µ (q) We check that the conditions are satised: d f (ψ (q, t)) f (ψ (q, t)) = f (ψ (q, t)) dt f (ψ (q, t)) 2 = 1 Also, f (ψ (q, µ)) = µ, so f (ψ (q, t)) = t. Additionally we saw that ψ t 1 ψ f(ψ(q,t)), and t is parallel to f (ψ (q, t)), so is orthogonal to the level set {f = t}. Therefore one can compute the Jacobian (which I will not do here) and get dv = 1 f da dt on f 1 (a, b), and the results follow from this. is Example 26. Let Ω R 2 be the unit disk. Let f (x, y) = 1 x 2 y 2. This is valued between 0 and 1, with zero boundary values. We have f (x, y) = 12

13 (x,y) x2 +y 2 and therefore f (x, y) = 1. This gives Ω g dv = 1 0 dt { g da, x2 +y =1 t} 2 which is equivalent to the well-known change of coordinates dx dy = r dθ dr. 4.2 The Faber-Krahn inequality Lemma 27. Let φ be the rst eigenfunction of Ω. Then Vol {φ = t} = 0 for any t. Proof. For any point with φ 0, the level set is locally (n 1)-dimensional by the implicit function theorem. So we only need to show the critical points have no volume. Let x be a point for which φ (x) = 0. We have φ (x) 0 in Ω by Courant's nodal domain theorem, so φ (x) = λ 1 φ (x) 0 as well. Therefore there is a neighborhood of x for which the j'th second derivative 2 φ is nonzero. Inside this x 2 j neighborhood, critical points satisfy φ = 0. By the implicit function theorem, φ x j x j = 0 is locally (n 1)-dimensional, which in particular shows critical points have no volume. Theorem 28 (Rayleigh's conjecture (1877); Faber (1923), Krahn (1924)). Let B be a ball with the same volume as Ω. Then λ 1 (Ω) λ 1 (B), with equality if and only if Ω is a ball. Proof. Let φ be the rst eigenfunction of Ω, which we take to be positive inside Ω, and normalize it so that max Ω φ = 1. The idea of the proof: We know that Q Ω (φ) = λ 1 (Ω). We will use φ to construct a function φ on B with Q B (φ ) λ 1 (Ω). From Rayleigh's theorem we know that Q B (φ ) λ 1 (B), and the result will follow. The equality case will arise from the calculations. We assume B is centered at the origin, and dene Ω t = {φ > t} Ω, and B t B a ball with the same volume as Ω t, centered at the origin. (Note Ω 0 = Ω and B 0 = B.) We now dene φ : B [0, 1] to be a radially symmetric function that satises B t = {φ > t}. Notice that φ (0) = 1, and it can be shown that φ is C 1 as well as φ. (See Figure 2 for an illustration of these sets.) Since Vol Ω t = Vol B t, by the coarea formula we have, for almost all t (0, 1), {φ=t} 1 φ da = {φ =t} 1 φ da (3) 13

14 Figure 2: Illustration of Ω and B and their t = 1/2 level sets. Again by the coarea formula, this gives Ω φ 2 dv = = In other words, {φ=t} t 2 On the other hand, dene {φ =t} φ 2 da dt = φ 1 1 φ da = 0 t 2 B {φ=t} (φ ) 2 dv 1 da dt = φ φ 2 L 2 (Ω) = φ 2 L 2 (B) (4) ψ (t) := φ 2 L 2 (Ω t) = Ω t φ 2 dv ψ (t) := φ 2 L 2 (B t) = B t φ 2 dv By denition, ψ (1) = ψ (1) = 0. Also, by the coarea formula: ψ (t) = 1 t dτ {φ=τ} φ 2 φ da = 1 t dτ {φ=τ} φ da Therefore ψ (t) = {φ=t} φ da, and similarly (ψ ) (t) = {φ =t} φ da. We have Vol Ω t = Vol B t, so by the isoperimetric inequality (Theorem 24), Area {φ = t} Area {φ = t}. This means: ( {φ =t} da) 2 ( {φ=t} da ) 2 14

15 For the left hand side, we use the fact that φ is constant on {φ = t} (since φ is radial), to get: ( {φ =t} 2 ( ) ( da) = φ da {φ =t} {φ =t} ) 1 φ da For the right hand side, we use the Cauchy-Schwarz inequality: ( 2 ( ) ( ) 1 da) φ da {φ=t} {φ=t} {φ=t} φ da Now, using (3), we get: {φ =t} φ da {φ=t} φ da This means (ψ ) ψ, but since ψ (1) = ψ (1) = 0, we have ψ (0) ψ (0), which means φ 2 L 2 (B) φ 2 L 2 (Ω) (5) By (4) and (5), we have Q Ω (φ) Q B (φ ), which by Rayleigh's theorem gives λ 1 (Ω) λ 1 (B). Finally, the equality case implies an equality case in the isoperimetric inequality, which means Ω t = B t for almost all t, and therefore Ω = B. (Furthermore, the equality case in the Cauchy-Schwarz inequality means φ is constant on level sets of φ, which means that φ is radially symmetric and φ = φ, so the rst eigenfunction on the ball is radially symmetric.) Part II Weyl's law and Pleijel's theorem 5 Variational methods on eigenvalues and Weyl's law 5.1 Neumann eigenvalues We begin by dening the Neumann eigenvalue problem: Problem (the Neumann eigenvalue problem). Let Ω R d be a domain. Describe all values λ C and all (complex-valued) functions u C 2 (Ω) C 1 ( Ω) that satisfy: { u + λu = 0 on Ω u ν = 0 on Ω where ν is the outward normal vector to Ω. 15

16 Like before, we'll always assume Ω is bounded and has a nice boundary. Example 29. In the one-dimensional case, we have a vibrating string, Ω = (0, l). Finding the Neumann solutions like we did in Proposition 1, we nd that ) they are cosines: φ n (x) = cos ( λn x, where λ ) 2. n = ( π(n 1) l Exercise 30. Any eigenvalue must be real and nonnegative. The rst eigenvalue is zero, it is simple, and the rst eigenfunction is a constant function. Solution of exercise. The proof of nonnegativity is exactly like the solution of Exercise 10, where we used Green's rst identity to get Ω ( u u + u 2) dv = u u ds = 0, Ω ν yielding by the PDE and by rearrangement λ u 2 = u 2 which shows nonnegativity (we will no longer deal with complex numbers). If λ = 0 then u = 0, yielding u = const (and one checks that a constant function indeed satises the problem with λ = 0). Like in the Dirichlet problem, the Neumann problem yields a spectral decomposition theorem (cf. Theorem 11) and a regularity result (cf. Theorem 12): Theorem 31 (spectral theorem for - 19th century). Suppose Ω is bounded and Ω is nice (for example, smooth). Then: 1. The set of eigenvalues is discrete. We will write them in the following way: 0 = λ 1 < λ 2 λ 3... (Eigenvalues with multiplicity are repeated.) 2. There is an orthonormal sequence of real-valued eigenfunctions φ 1, φ 2,... such that φ i corresponds to λ i. This sequence is a complete orthonormal basis for the Hilbert space L 2 (Ω). Theorem 32 (regularity of eigenfunctions). Each eigenfunction φ i is in C (Ω). Remark. We make a convention where λ, φ appearing without a tilde refer to Dirichlet eigenvalues and eigenfunctions, and λ, φ refer to Neumann eigenvalues and eigenfunctions. 16

17 5.2 The variational method for Neumann eigenvalues In this section we revisit Rayleigh's theorem, and delve deeper into its corollaries. We show the maximum-minimum principle, and the results on domain monotonicity of eigenvalues: If we decrease the domain (i.e. consider a subdomain), the Dirichlet eigenvalues increase. For Neumann eigenvalues this monotonicity is not true, but we will get a dierent kind of monotony which is good enough for what we need. We rst recall Rayleigh's theorem with Dirichlet eigenvalues. Recall the necessary denitions: W (Ω) is the set of C ( Ω) functions that are zero on Ω and have a weak gradient in Ω; the Rayleigh quotient is dened as Q (w) = w 2 w 2, where the norms are taken in L 2 (Ω). Theorem (Rayleigh's theorem for Dirichlet eigenvalues - Theorem 19). Let w W (Ω), w 0 (note zero boundary values). 1. Q (w) λ 1, with equality if and only if w is an eigenfunction of λ Suppose w is orthogonal to φ 1,..., φ n 1 in L 2 (Ω). Then Q (w) λ n, with equality if and only if w is an eigenfunction of λ n. We now present a similar theorem for Neumann eigenvalues. Theorem 33 (Rayleigh's theorem for Neumann eigenvalues). Let w C 1 (Ω) C ( Ω), w 0 (with arbitrary boundary values). 1. Q (w) λ 1, with equality if and only if w is an eigenfunction of λ Suppose w is orthogonal to φ 1,..., φ n 1 in L 2 (Ω). Then Q (w) λ n, with equality if and only if w is an eigenfunction of λ n. Remark. By Exercise 30, λ 1 = 0 with only the constant function as an eigenfunction, so the rst part is completely trivial. This was not the case in the Dirichlet eigenvalue problem, where we had to do some work even to get the result for λ 1! Proof. The proof of the second part is exactly like the Dirichlet case. We only outline the dierence, which is in the integration by parts is slightly dierent. Let α j := w, φ j. (We know that α j = 0 for 1 j n 1.) By L 2 (Ω) Green's rst identity, ( w φ j + w, φ ) j dv = w φ j ds = 0, Ω Ω ν therefore (replacing φ j by λ j φj ) we get w, φ j = λ j α j. (Here we L 2 (Ω) used the fact that we're integrating against an eigenfunction - integration by parts would not work with some arbitrary function instead.) By plugging in w = φ i we then get φ i, φ j = λ j δ ij. The rest of the proof is exactly the same as the proof of Theorem

18 Corollary 34 (max-min theorems). The Dirichlet eigenvalues 0 < λ 1 < λ 2 λ 3... and the Neumann eigenvalues 0 = λ 1 < λ 2 λ 3... are given by { } λ n = max min Q (w) v 1,...,v n 1 L 2 (Ω) w W (Ω) orthogonal to v 1,...,v n 1 { } λ n = max v 1,...,v n 1 L 2 (Ω) min Q (w) w C 1 (Ω) C( Ω) orthogonal to v 1,...,v n 1 Proof (sketch). Whatever the functions v 1,..., v n 1 are, we can always pick w (orthogonal to them) to be a linear combination of the rst n eigenfunctions, because there are more unknowns (coecients) than constraints. A calculation like in the proof of Theorem 19 shows that Q (w) is at most the n'th eigenvalue. On the other hand, choosing v 1,..., v n 1 to be the rst n 1 eigenfunctions, by Rayleigh's theorem we force the constraint Q (w). 5.3 Domain monotonicity of eigenvalues As a result of the above, we get that the Neumann eigenvalues for a given domain are less than the Dirichlet eigenvalues. The following lemma generalizes this. Lemma 35 (Neumann monotonicity of eigenvalues). Let Ω be a domain with disjoint subdomains Ω 1,..., Ω m such that Ω = Ω 1 Ω 2... Ω m. Let 0 < λ 1 < λ 2... be the Dirichlet eigenvalues for Ω, as above. Let 0 = µ 1 µ 2... be the Neumann eigenvalues for Ω 1,..., Ω m, combined into one increasing sequence. Then µ n λ n for any n. Proof. Let φ n be the Dirichlet eigenfunction for λ n, as above. Let ψ n be the L 2 (Ω) function that is the Neumann eigenfunction for µ n on the appropriate subdomain Ω j, and zero outside this subdomain. Suppose f C 1 (Ω) C ( Ω) is orthogonal in L 2 (Ω) to ψ 1,..., ψ n 1. Then by Rayleigh's theorem for the Neumann eigenfunctions, we have Q Ωj (f) µ n (note the restriction to Ω j ) for every j. Using the fact that the subdomains make up the entire domain, we have m f 2 L 2 (Ω) = m f 2 L 2 (Ω µ j) n f 2 L 2 (Ω = µ j) n f 2 L 2 (Ω), j=1 which gives Q Ω (f) µ n. On the other hand, we can choose such f to be a linear combination of φ 1,..., φ n (because we have more unknowns than equations for this choice), and then we have Q Ω (f) λ n as we've seen last time: Recall that φ 1,..., φ n are pairwise orthogonal, and so are their gradients (by integration by j=1 18

19 parts), so opening the parentheses we have 2 f 2 n L 2 (Ω) = α j φ j j=1 L 2 (Ω) n = αjλ 2 j φ j 2 L 2 (Ω) j=1 n λ n αj 2 φ j 2 L 2 (Ω) = λ n f 2 L 2 (Ω). j=1 Lemma 36 (Dirichlet monotonicity of eigenvalues). Let Ω be a domain with disjoint subdomains Ω 1,..., Ω m (with no requirement on them). Let 0 < λ 1 < λ 2... be the Dirichlet eigenvalues for Ω, as above. Let 0 < µ 1 µ 2... be the Dirichlet eigenvalues for Ω 1,..., Ω m, combined into one increasing sequence. Then λ n µ n for any n. Proof. Let φ n be the Dirichlet eigenfunction for λ n, and let ψ n be the function that is the Dirichlet eigenfunction for µ n on the appropriate subdomain Ω j, and zero outside this subdomain. We assume that {ψ 1,..., ψ n } are orthogonal in L 2 (Ω): Functions from dierent subdomains are obviously orthogonal, and on each subdomain we require a sequence of orthonormal functions as eigenfunctions. By the patching lemma (Lemma 17), ψ n W (Ω) (it has a weak gradient). Their gradients are also orthogonal (as we've shown last time, using integration by parts, i.e. Green's rst identity). Therefore any f that is a linear combination of ψ n is also in W (Ω), and satises Q Ω (f) µ n, just like in the previous proof (opening parentheses). On the other hand, we can choose such f to be orthogonal to φ 1,..., φ n 1 (as before - more unknowns than equations), and then by Rayleigh's theorem for Dirichlet eigenvalues, we have Q Ω (f) λ n. Corollary 37 (Dirichlet simple monotonicity of eigenvalues). All Dirichlet eigenvalues of a domain increase as the domain decreases. In other words: Let Ω Ω. Then λ n (Ω ) λ n (Ω), for any n. Proof. This is a direct result of Lemma 36, with m = 1. Remark. This is not true for Neumann eigenvalues. Notice that Lemma 35 requires a complete partition of the domain, so it cannot be used for a single subdomain. A simple counterexample can be constructed for the Neumann case (see Figure 3). We consider rectangular domains. The rst Neumann eigenvalue is always zero, so consider the second eigenvalue. Suppose a rectangle's sides are A < B. We will later see that λ 2 = const B, i.e. it depends only on the longer 2 side of the rectangle and is inversely proportional to it. But now take two subrectangles, one whose long side is less than B and one whose long side is more than B. This demonstrates that there is no monotonicity (at least in λ 2 ). 19

20 Figure 3: Couterexample to subdomain monotonicity in the Neumann eigenvalue problem. 5.4 The asymptotic growth of eigenvalues We are ready to formulate and prove Weyl's law. Below, ω d denotes the volume of the d-dimensional unit ball. Theorem 38 (Weyl, 1911). 1. In dimension 2: λ n 4πn, as n. Area ( Ω ) 2/d In dimension d: λ n 4π 2 n, as n. ω d Vol Ω 2. Let N (λ) denote the number of Dirichlet eigenvalues λ. Then: N (λ) In dimension 2: lim = 1 Area Ω. λ λ 4π N (λ) In dimension d: lim λ λ = ω d Vol Ω. d/2 d (2π) A historical note is told in [13]. In 1910, Lorentz (who had already received a Nobel prize in physics) was invited to give a series of lectures in physics at Göttingen. During these lectures, he mentioned a problem which might interest the mathematicians in the audience - a conjecture that was essentially Weyl's law. As the story goes, Hilbert and his student Weyl were both among the audience. Hilbert made a comment that he doesn't believe this problem will be solved within his life time. The next year, Weyl published his solution. In a more advanced paper of Weyl, he conjectured a stronger result - known today as Weyl's conjecture. This is still unsolved, although a lot of progress has been made since. Conjecture 39 (Weyl's conjecture, 1913, two-dimensional version). N (λ) = Area Ω 4π λ Length Ω 4π λ + o ( λ ), as λ. 20

21 In our proof of Theorem 38, we will only show the case of d = 2 - higher dimensions follow exactly the same proof, but staying in dimension 2 makes some of the details easier to present. Also, the rst part of the theorem follows immediately from the second, as shown in Figure 4, so we will only prove the second part. Figure 4: Picture-proof that the rst part of Theorem 38 follows from the second. Proof of Weyl's law (second part). We split the proof to three increasingly more general cases: 1. The case where Ω is a rectangle, which we will prove directly by computing the eigenvalues. In this case we will also show Weyl's law for Neumann eigenvalues. 2. The case where Ω is a nite union of rectangles, which we will prove using the variational results on monotonicity. 3. The case where Ω is a general nice domain, which we will prove by approximating it by nite unions of rectangles. Case 1: Suppose Ω is a rectangle, Ω = (0, A) (0, B). We will simply solve the PDE u + λu = 0 and nd the eigenfunctions, and thus the eigenvalues. We will do this both in the Dirichlet case and the Neumann case. We will use separation of variables to solve 6 the PDE. 6 By general PDE theory which we will not go into, the general solutions are given by series in solutions in separated variables. In the case of a xed λ, this series must be nite, 21

22 Let u (x, y) = X (x) Y (y). The PDE becomes: X (x) Y (y) + X (x) Y (y) + λx (x) Y (y) = 0 Dividing by X (x) Y (y) gives: X (x) X (x) + Y (y) Y (y) + λ = 0 This means there are nonnegative constants α, β such that λ = α 2 + β 2 and X + α 2 X = 0 Y + β 2 Y = 0 Ignoring boundary conditions, the solutions are X Span {cos (αx), sin (αx)} and Y Span {cos (βx), sin (βx)}. Now we solve the PDE introducing Dirichlet boundary conditions: X (0) = X (A) = Y (0) = Y (B) = 0 This gives eigenfunctions u (x, y) = sin (αx) sin (βy), where αa, βb πn (zero is not allowed because this gives the zero eigenfunction). The eigenvalues λ = α 2 + β 2 are ( ) n λ n,m = π 2 2 A 2 + m2 B 2, where n, m {1, 2,...}. We solve the PDE again, introducing Neumann boundary conditions: X (0) = X (A) = Y (0) = Y (B) = 0 This gives eigenfunctions u (x, y) = cos (αx) cos (αy), where αa, βb πz (this time, zero is allowed). The eigenvalues λ = α 2 + β 2 are therefore given by the same λ n,m formula as above, but this time, n, m {0, 1, 2,...}. We are ready to prove Weyl's law for both the Dirichlet and Neumann eigenvalues in the rectangle. Notice that if λ n,m λ, then the lattice point ) 2 = λ π 2. Therefore, the (n, m) R 2 is in the interior of the ellipse ( x 2 ( A) + y B number N (λ) of eigenvalues that satisfy λ n,m λ is precisely the number of lattice points in the rst quadrant of this ellipse, where we don't count points on the axis in the Dirichlet case, and do count them in the Neumann case. (See Figure 5.) The area of the rst quadrant in this ellipse is λab 4π, and an elementary calculation 7 therefore shows N (λ) = λab 4π Area Ω + o (λ) = 4π λ + o (λ), and therefore a general eigenfunction is simply a linear combination of eigenfunctions with separated variables. Therefore, separation of variables gives us the basis for the eigenspace. 7 Scaling by λ/π, we count the number of n, m λ 1/2 πn such that (n, m) is in the interior of the ellipse (x/a) 2 + (y/b) 2 = 1. As λ, this resembles the Riemann-integral denition of the area of this ellipse. 22

23 Figure 5: The lattice points in the rst quadrant of our ellipse. Image reproduced from Strauss's book [4]. where N counts either Dirichlet or Neumann eigenvalues. Dividing by λ and letting λ gives us what we need. Case 2: Now suppose Ω is a nite union of rectangles Ω 1,..., Ω m. Let 0 < λ 1 < λ 2... be the Dirichlet eigenvalues of Ω. Let 0 < µ 1 µ 2... be the Dirichlet eigenvalues of the rectangles that compose Ω, combined together into one increasing sequence. Let 0 = µ 1 µ 2... be the same, but with Neumann eigenvalues. According to Lemma 35 and Lemma 36, we have for every n: µ n λ n µ n (6) Let N j (λ) be the number of Dirichlet eigenvalues of Ω j less than λ, and let M j (λ) be the same with Neumann eigenvalues. According to (6), we have for every λ: m m N j (λ) N (λ) M j (λ) j=1 According to the rectangle case, we have: j=1 M j (λ) N j (λ) lim = lim λ λ λ λ which by the sandwich rule, gives N (λ) lim = Area Ω λ λ 4π. 23 = Area Ω j, 4π

24 Case 3: Finally, suppose Ω is a general (nice) domain. Suppose that R 1 Ω R 2, where R 1, R 2 are domains composed of a nite union of rectangles. By Corollary 37, for any λ, N R1 (λ) N (λ) N R2 (λ). Dividing by λ and taking λ, we get Area R 1 4π lim inf λ N (λ) λ lim sup λ N (λ) λ Area R 2. 4π Finally, taking the supremum over all R 1 and the inmum over all R 2, using the assumed regularity of the domain Ω, we get our wanted result. 6 Improvements to Courant's theorem We focus only on the two-dimensional case, but most of what we'll say is true in higher dimensions as well (with correctly modied formulas). Let Ω be a planar domain with nice boundary and compact closure, 0 < λ 1 < λ 2... its Dirichlet eigenvalues and φ 1, φ 2,... its Dirichlet eigenfunctions. We know by Courant's nodal domain theorem (Theorem 22) that the number of nodal domains of φ n is at most n. But in fact, more is true: Theorem 40 (Pleijel's nodal domain theorem, 1956 [10]). lim sup n # nodal domains of φ n n ( ) , j where j is the rst zero of the Bessel function J 0. The case of Neumann eigenvalues is more dicult, and it was settled only recently. Theorem 41 (I. Polterovich, 2009 [16]). If Ω is piecewise analytic, then lim sup n # nodal domains of φ n n ( ) , j where φ n are the Neumann eigenfunctions. ( ) 2 2 The constant j that appears in the theorem is known not to be sharp. For the case of the rectangle, for example, it is known that the true bound is This bound is conjectured to hold for all domains: 2 π Conjecture 42 (I. Polterovich, 2009 [16]). lim sup n # nodal domains of φ n n 2 π

25 However, this bound was explicitly improved only very recently: Theorem 43 (Bourgain, 2013 [12]). lim sup n # nodal domains of φ n n ( ) ,000,003. j Proof of Pleijel's theorem in d = 2. Let Ω Ω be some nodal domain of φ n. Let A = Area Ω and A = Area Ω. By Courant's nodal domain theorem (see Exercise 23), By the Faber-Krahn inequality (Theorem 28), λ 1 (Ω ) = λ n. (7) λ 1 (Ω ) λ 1 (disk of area A ) = πj2 A. (8) (Where the constant comes from: The eigenvalues of the disk can be computed explicitly by just solving the PDE, using separation of variables - radius and angle. The radial part becomes Bessel's equation, and imposes the condition that the square root of the eigenvalue is a Bessel zero.) Combining (7) and (8) gives A πj2 λ n. Suppose φ n has exactly N nodal domains with areas A 1,..., A N. Then Rearranging this gives A = By Weyl's law (Theorem 38), n i=1 N n Finally, plugging (10) into (9) gives as we wanted to show. A i Nπj2 λ n. ( ) 2 2 λ n A j 4πn. (9) λ n lim n n = 4π A. (10) lim sup n N n ( ) 2 2, j 25

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