SPECTRAL PROPERTIES OF THE LAPLACIAN ON BOUNDED DOMAINS TSOGTGEREL GANTUMUR Abstract. After establishing discrete spectra for a large class of elliptic operators, we present some fundamental spectral properties of the Dirichlet and Neumann Laplace op- erators on bounded domains, including eigenvalue comparison theorems, Weyl's asymptotic law, and Courant's nodal domain theorem. Note: This is an incomplete draft. Contents 1. Introduction1 2. Spectral theory of compact self-adjoint operators3 3. Application to the Laplace eigenproblems5 4. Eigenvalue comparison theorems7 1. Introduction In these notes, we will be concerned with the eigenvalue problem − ∆u = λu, (1) n in a bounded domain Ω ⊂ R , with either the Dirichlet u = 0 or the Neumann @νu = 0 condition on the boundary @Ω. The unknown in the problem is the pair (u; λ) where u is a function and λ is a number. If (u; λ) is a solution then u is called an eigenfunction, and λ is called the eigenvalue associated to u. Let us note the following. • Since the right hand side involves λu, the problem is not linear. • If (u; λ) is a solution then so is (αu; λ) for any number α. • We exclude the trivial solution u = 0 from all considerations. We have studied the problem −∆u + tu = f with the Dirichlet or Neumann boundary con- 2 ditions, where t 2 R and f 2 L (Ω) are given. Since (1) is equivalent to −∆u + tu = 0 with 1 1 t = −λ, we can give the following weak formulation for (1). Let V be either H0 (Ω) or H (Ω), depending on the boundary condition we wish to impose. Then the problem is to find u 2 V and λ 2 R satisfying ru · rv = λ uv for all v 2 V: (2) ˆΩ ˆΩ We know that a unique weak solution u 2 V to −∆u+tu = f exists if t is larger than a certain threshold value t0 that depends on the type of the boundary condition and the geometry of the domain Ω. This shows that if (2) has a nontrivial solution then −λ ≤ t0, that is, λ cannot ! be less than −t0. If u 2 V satisfies (2) then the regularity results imply that u 2 C (Ω), and so in particular u is a classical solution of (1) in Ω, and moreover that u satisfies the desired boundary condition in the classical sense provided the boundary is regular enough. Date: December 3, 2013. 1 2 TSOGTGEREL GANTUMUR Now we want to write (2) as an abstract operator eigenvalue problem. We introduce the linear 0 1 operator A : V ! V and the bilinear form a : V × V ! R by hAu; vi = a(u; v) = ru · rv; u; v 2 V; (3) ˆΩ where h·; ·i is the duality pairing between V 0 and V . Recall that a is continuous ja(u; v)j ≤ kukH1 kvkH1 ; u; v 2 V; (4) symmetric a(u; v) = a(v; u); u; v 2 V; (5) and satisfies 2 a(u; u) + thu; uiL2 ≥ αkukH1 ; u 2 V; (6) for all t > t0, with α > 0 possibly depending on t. The operator A is called the energy extension of −∆ with the given boundary condition, in the sense that it is an extension of the classical Laplacian acting on a dense subset of V . We can check that it is bounded: hAu; vi a(u; v) kAukV 0 = sup = sup ≤ kukH1 : (7) v2V kvkH1 v2V kvkH1 In terms of the operator A, the problem (2) can be written as Au = λJu; (8) where the inclusion map J : L2(Ω) ! V 0 is defined by hJf; vi = fv; v 2 V: (9) ˆΩ Obviously, J is injective because Jf = 0 implies f = 0 for f 2 L2(Ω) by the du Bois-Reymond lemma. It is also continuous: hJf; vi kfkL2 kvkL2 kJfkV 0 = sup ≤ sup ≤ kfkL2(Ω); (10) v2V kvkH1 v2V kvkH1 and hence J defines a continuous embedding of L2(Ω) into V 0. In what follows we will identify L2(Ω) with a subspace of V 0 through J. So for instance, we write (8) simply as Au = λu. (11) −1 0 By the Riesz representation theorem, A + tI is invertible for t > t0, and (A + tI) : V ! V is bounded. In what follows, we fix some t > t0. Then adding tu to both sides of (11), and applying (A + tI)−1, we get u = (t + λ)(A + tI)−1u: (12) At this point, we introduce the resolvent2 −1 2 2 Rt = (A + tI) jL2(Ω) : L (Ω) ! L (Ω); (13) −1 2 which is the restriction of (A + tI) to L (Ω). Hence if u 2 V and λ 2 R satisfy (11), then (t + λ)Rtu = u: (14) 2 Conversely, if u 2 L (Ω) and λ 2 R satisfy (14), then by applying A + tI on both sides, we derive (11), proving the equivalence of the two formulations. Let us derive some straightforward properties of the resolvent. 1Note that A in these notes corresponds to A~ from class. In class we denoted by A the Friedrichs extension, but I realized later that we don't really need it and removing it makes the presentation a bit simpler. 2The usual definition is (A − tI)−1, but we are using the plus sign for convenience. SPECTRAL PROPERTIES OF THE LAPLACIAN ON BOUNDED DOMAINS 3 2 • The resolvent is bounded as an operator Rt : L (Ω) ! V , because kRtfkV ≤ ckfkV 0 ≤ ckfkL2(Ω); (15) where the constant c may have different values at its different occurrences. 2 • The resolvent is positive, in the sense that hRtf; fi > 0 for f 6= 0, where h·; ·i is the L 2 inner product on Ω. To see this, let f 2 L (Ω) and let u = Rtf, so that f = (A+tI)u. Then the strict coercivity property (6) gives 2 hRtf; fi = hu; (A + tI)ui = a(u; u) + thu; ui ≥ αkukH1 : (16) • The resolvent is injective: If Rtf = 0 then f = 0. 2 • The resolvent is symmetric, in the sense that hRtf; gi = hf; Rtgi for f; g 2 L (Ω). With u = Rtf and v = Rtg, we have hRtf; gi = hu; (A + tI)vi = a(u; v) + thu; vi; (17) which clearly shows the claim. 2 2 • A function u 2 L (Ω) is in the range of Rt if and only if Au 2 L (Ω). To show this, 2 2 first let u be such that Au 2 L (Ω). Then (A + tI)u 2 L (Ω), hence u = Rt(A + tI)u, which means that u 2 RanRt. Second, let u 2 RanRt, i.e., let u = Rtf for some f 2 L2(Ω). It is obvious that f = (A+tI)u. From this, we have Au = f −tu 2 L2(Ω). Apart from these simple properties, a crucial property we would like to have for the resolvent is compactness. Recall Rellich's lemma, which says that if Ω is a bounded domain then the unit 1 2 1 2 ball in H0 (Ω) is relatively compact in L (Ω), i.e., the embedding H0 (Ω) ,! L (Ω) is compact. 1 2 1 Since in the Dirichlet case, i.e., when V = H0 (Ω), the resolvent as a map Rt : L (Ω) ! H0 (Ω) 2 1 is bounded, the image of the unit ball of L (Ω) under Rt is a bounded set in H0 (Ω). Hence by Rellich's lemma this image is relatively compact in L2(Ω), showing that the resolvent as a 2 2 map Rt : L (Ω) ! L (Ω) is a compact operator. Therefore in the Dirichlet case, boundedness of Ω is sufficient for the compact resolvent. In the Neumann case, it is known that the resolvent is not compact without additional assump- tions on the boundary regularity of Ω. n 1 Definition 1. An open set Ω ⊂ R is said to have the H -extension property, if there exists 1 1 n a bounded linear map E : H (Ω) ! H (R ). For example, it is known that Lipschitz domains have the H1-extension property. n 1 Exercise 2. Show that in the Neumann case, if Ω ⊂ R is bounded and has the H -extension 2 2 property, then the resolvent as a map Rt : L (Ω) ! L (Ω) is compact. Our strategy to solve the Laplace eigenvalue problem (11) is through the equivalent formu- lation (14) in terms of the resolvent. The main feature that makes this formulation attractive is the fact that the resolvent is compact under some very mild assumptions on Ω. 2. Spectral theory of compact self-adjoint operators In this section, we will prove the spectral theorem for compact, symmetric, positive opera- tors on a real Hilbert space H. This theorem will then be applied to the resolvent in the next section. We start with some preliminary results. Lemma 3. Let B : H ! H be a bounded symmetric operator. Then we have the following. a) If B is positive, and if u 2 H n f0g and λ 2 R satisfy Bu = λu, then λ > 0. b) If u; v 2 H n f0g and λ, µ 2 R satisfy Bu = λu, Bv = µv, and λ 6= µ, then hu; vi = 0. c) Let fung be a complete orthonormal basis of H, such that Bun = λnun for each n. Suppose that u 2 H satisfies Bu = λu with λ 62 fλng. Then u = 0. 4 TSOGTGEREL GANTUMUR Proof. a) Recall that positivity of B : H ! H means that hBv; vi > 0 for v 2 H n f0g.