Complex Analytic Functions and Differential Operators. Robert Carlson

Transcription

1 Complex Analytic Functions and Differential Operators Robert Carlson

2 Some motivation Suppose L is a differential expression (formal operator) N L = p k (z)d k, k=0 D = d dz (0.1) with p k (z) = j=0 b jz j analytic on the unit disk D C. Think of L as a (big) matrix with eigenvalues and eigenvectors. Use these (N = 2) to understand du dt = Lu, du dt = ilu, d 2 u dt 2 = Lu, (heat equation) (Schrodinger equation) (wave equation)

3 Some hair splitting This plan works best in a Hilbert space H with inner product f, g, and when L is (nearly) self adjoint. Distinguish an adjoint pair L, L + Lf, g = f, L + g, f, g D min and (formal) symmetry L = L + from the adjoint L (graph) L = {(g, h) Lf, g = f, h, all f in domain of L} with self adjoint L = L. Typically, L is only densely defined, and a symmetric L may have no self-adjoint extensions, exactly one, or many.

4 Previous work Often for real variables, f, g = b a f (x)g(x) dx. What if L acts on a Hilbert space of analytic functions on D? When is L symmetric/self adjoint? A. Villone (E.A. Coddington) initiated just such a project in his dissertation (1966), and in subsequent papers ( ). Hilbert space was the Bergman space A 2 of analytic functions square integrable with respect to area measure on D. f, g = f (z)g(z) dxdy D This project was extended in the dissertation of W. Stork (J. Weidmann), pubs (1975).

5 Extend to weighted Hardy spaces On D min = polynomials f = K n=0 a nz n, use a sequence β = {β n > 0, n = 0, 1, 2,... } to define an inner product f 1, f 2 β = a n c n βn, 2 n=0 f 1 2 β = a n 2 βn. 2 n=0 The Hilbert space H β is g = n=0 a nz n such that g 2 β = a n 2 βn 2 <. n=0 H β has an orthonormal basis e n = z n /β n, n = 0, 1, 2,.... Examples include the classical Hardy space H 2, β n = 1, and the Bergman space A 2, β n = π/(n + 1).

6 Linking β to D Assume (M z is bounded) C β = sup β n+1 /β n <, (0.2) n and lim n β1/n n = r, r > 0. (0.3) Proposition If (0.3) holds, then every f H β is analytic in the open disk D r. (RKHS) For j = 0, 1, 2,... and z 1 D r, the linear functional f (j) (z 1 ) given by derivative evaluation is uniformly bounded in the H β norm on compact subsets of D r. For any R > r, every function analytic in D R belongs to H β, and some functions in H β do not extend analytically to D R.

7 Formal adjoints are rare Villone has the next result in A 2 when L = L +. Theorem Suppose p k (z) and q k (z) are in H β for k = 0,..., N, with β satisfying (0.2) and (0.3). If L = N k=0 p k(z)d k has a formal adjoint L + = N k=0 q k(z)d k, then L has polynomial coefficients, with deg(p k (z)) N + k. Idea of proof: Use e n = z n /β n to form matrices A mn = Le m, e n for L and L +. Since D k e m = Ce m k while multiplication by p k (z) raises degree, A mn = 0 for n < m N. Matrix for L + should be (i) same form, and (ii) the conjugate transpose of A. So the nonzero entries of A must be located near the diagonal.

8 Interesting β are rare Theorem Let {β n } be a positive sequence. Suppose L 1 = [a 2 z 2 + a 0 ]D + b 1 z has a formal adjoint L + 1 = [c 2z 2 + c 1 z + c 0 ]D + [d 1 z + d 0 ] on H β. If a 2 0, then b 1 = σa 2 for some σ > 0. The weights satisfy βm+1 2 = βm 2 β1 2 (m + 1)σ β0 2 m + σ, (0.4) and H β is a Hilbert space of functions analytic on D r with r 2 = β 2 1 σ/β2 0. For convenience assume restricted weight sequences β with σ > 0, β 0 = 1, β 2 1 = 1/σ, β 2 m+1 = β 2 m m + 1 m + σ, m 1. (0.5) D is the natural domain for H β.

9 Existence of adjoint pairs Theorem Let β be a weight sequence satisfying (0.5). On H β every has a formal adjoint L 1 = [a 2 z 2 + a 1 z + a 0 ]D + σa 2 z + b 0 L + 1 = [a 0z 2 + a 1 z + a 2 ]D + σa 0 z + b 0. Let A β be expressions with a formal H β adjoint L +. Theorem As an algebra, A β is generated by its expressions with order at most one. If L = N k=0 p k(z)d k A β, with p N (z) = 2N j=0 c jz j, then the leading coefficient of L + is q N (z) = 2N j=0 c jz 2N j. If z 1 0, then p N (z 1 ) = 0 if and only if q N (1/z 1 ) = 0.

10 Classical examples in A β Suppose N = 2. Then A β includes the Jacobi (Gauss) operators (1 z 2 )D 2 y + {(β α) (α + β + 2)z}Dy, (0.6) and other operators with polynomial eigenfunctions. It is easy to show that L A β of order 2 can have any nonzero polynomial of order 4 with distinct roots as leading coefficient. This produces many equations Ly = λy with L A β with three (Riemann-Papperitz), four (Heun), and five regular singular points ( being one). Starting from the classical (Riemann-Papperitz, Heun) operators, the adjoint map (typically?) increases the number of regular singular points.

11 When does L = L +? Villone and Stork consider characterizing formally symmetric expressions L on A 2. Villone succeeds with a fairly complex recursive technique. Applied when N = 1!! Stork shows that for c 0, n = 1, 2, 3,..., and 0 r n the examples l n,r = (cz n+r +cz n r )D n + r ( ) r (n + 1)! c k (n + 1 k)! zn+r k D n k, k=1 are formally symmetric in A 2 without characterizing formal symmetry. Of course you can take real linear combinations of the l n,r.

12 Adjoint formulas give a characterization in H β Let B 0,0 = 1 and B n,r = (zd) n r D r, n = 1, 2,..., 0 r n. Taking the adjoint factors in reverse order gives B + n,r = (z 2 D + σz) r (zd) n r. Since L + L + is formally symmetric, Lemma For n = 0, 1, 2,..., 0 r n, and c n,r C, the expressions c n,r B n,r + c n,r B + n,r are formally symmetric, with highest order term (c n,r z n r + c n,r z n+r )D n when written in the standard form (0.1).

13 Main result 1 Theorem An differential expression L is formally symmetric in H β if and only if it can be written in the form L = N n=0 r=0 n [c n,r B n,r + c n,r B n,r + ]. (0.7) Idea of proof: Show that these examples allow you to match any possible leading coefficient. Subtract and use induction.

14 Passing from symmetric to self-adjoint in A 2. Stork showed that L = L + and p N (z) 0 for z = 1 implies L with polynomial domain is essentially self-adjoint. Also characterizes the domain. (When z = exp(iθ) you get a regular periodic problem, so the condition is not surprising.) Villone later found conditions for a variety of equal (nonzero) deficiency indices, meaning self-adjoint extensions exist. ( Boundary conditions?)

15 Some definitions If the differential expression L has coefficients p k (z) H β, the minimal operator L min may be extended to a maximal operator L max with the domain, D max = {f H β Lf H β }. Say that L is D regular if the coefficients p k (z) are analytic on the closed unit disk D, and p N (z) 0 when z = 1; in particular the set {z k z k < 1, p N (z k ) = 0} is finite. For k = 0, 1, 2,..., define additional (Sobolev) Hilbert spaces Hβ k of functions f H β whose k-th derivative is also in H β. For f = n=0 c nz n, and g = n=0 b nz n, f, g k = (1 + n 2k )c n b n βn. 2 n=0

16 Main result 2: Regular case After some analysis (zeros in p N (z) are singularities ), Theorem Suppose L = N k=0 p k(z)d k is D regular and formally symmetric on H β, with order N 1. Then L max is the closure of L min, and L max is self adjoint. The resolvent R(λ) : H β H β is compact, implying the spectrum is a sequence {λ n } of eigenvalues, λ n, eigenfunctions orthogonal and complete. A closed operator T is Fredholm if T has a finite dimensional null space and a closed range of finite codimension. The index is ind(t ) = dim(null T ) codim(range T ). Theorem Suppose σ 1, and L = N k=0 p k(z)d k is D regular of order N 1. If p N (z) has K roots in D, counted with multiplicity, then L with domain Hβ N is Fredholm with index N K.

17 Main result 3: eigenvalue location More analysis starting with a periodic problem on R with good leading coefficient. Theorem Suppose L = N k=0 p k(z)d k is a D regular formally symmetric expression of order N 2 on H β, with self-adjoint maximal operator L. The eigenvalues of L can be enumerated as a sequence {λ n, n = 0, 1, 2,... } with the following description: for 0 < ɛ < 1 there are real nonzero constants C 1 and τ > 0, and a C 2 C such that λ n /C 1 = (n/τ + C 2 ) N + O(n N 2+ɛ ). In particular, with at most finitely many exceptions, the eigenvalues are either all positive or all negative, and have multiplicity 1.

18 Comments on monodromy Suppose p N (z 1 ) = 0, p N (z) 0. Solutions of Ly = λy have analytic continuations around z 1, but may not return to the same value. The local analysis near z 1 is usually easy (regular singular point, Frobenius series, indicial equation). Suppose p(z k ) = 0 for k = 1,..., K and z k D. Then having an eigenfunction y in H β means y has trivial monodromy around each z k. Instead of a simple local problem, we re solving a tricky global one. Why should this be linked to self-adjointness/leading coefficient symmetry? Hilbert s 21st problem (1900): To show that there always exists a linear differential equation of the Fuchsian class, with given singular points and monodromic group. Significant work Generalizations in algebraic geometry (Deligne).