Outline of the Weyl-Titchmarsh m Function A Qualifying Exam April 27, 21 The goal for this talk is to present the asymptotic expansion for the Weyl-Titchmarsh m-function as described in Levitan and Danielyan, 1991 [3. Agenda: Define the m-function Outline the asymptotic expansion problem Derive the expansion formula Describe how to compute the first few exact terms of the expansion. The Half-Line Sturm-Liouville Problem A Brief Overview of I Consider the singular Sturm-Liouville problem d 2 u(x) + q(x)u(x) = λu(x), x [, ) dx 2 u () = where q is real valued and locally integrable and λ is a complex parameter. Notation: L q := d 2 dx 2 + q(x). Weyl s 191 dissertation describes a rigorous framework for analyzing singular Sturm-Liouville problems. His method is to regard singular problems as a limit of non-singular problems: { Lq u(x) = λu(x) x [, b u () = u (b) = There exists a sequence of real λ n and functions ψ n = ψ(x, λ n ) satisfying L q ψ n = λ n ψ n and ψ n() = ψ n(b) =. Every f in L 2 [, b can be expanded as f = n=1 (ψ n,f )ψ n ψ n 2 L 2 [,b
A Brief Overview of II Let ρ b (λ) = λ n λ f L 2 [, b as where 1 ψ n 2 L 2 [,b f =. Rewrite the expansion of F (λ)ψ(x, λ)dρ b (λ) F (λ) = (ψ(x, λ), f ). The function ρ b is the spectral function of the finite interval problem on [, b. Now we take a limit as b Want: Unique limiting value for ρ b and a well-defined L 2 [, ) expansion. A Brief Overview of III For a broad class of operators L q, Weyl theory produces a unique spectral function ρ for the half-line problem and an expansion of the form f = F (λ)ψ(x, λ) dρ(ρ). Operators L q for which this happens are said to be in the limit point case at. This classification depends only on the nature of q. Sufficient condition: q(x) cx 2. Example: More For the case q(x) = and ρ b ( λ) 2 λ/π. A Brief Overview of IV Consider the problem The m-function ( ) { Lq y = λy x [, ) y L 2 [, ). In the limit-point case, the Weyl-Titchmarsh m-function, m(λ), is uniquely determined by the condition that The functions L q φ = λφ φ(, λ) = φ (, λ) = 1 L q ψ = λψ ψ(, λ) = 1 ψ (, λ) = form a basis for the solution space of (*). Weyl theory shows that for Iλ, up to a multiplicative constant: y(x, λ) = φ(x, λ) + m(λ)ψ(x, λ). y(x, λ) = φ(x, λ) + m(λ)ψ(x, λ) L 2 [, ). An important, technical result of the Weyl analysis is that the m function is the Stieltjes transform of spectral function: dρ(λ) λ z.
Theorem Theorem For L q in the limit point case, q(x) C n [, δ), and θ outside tan θ < ɛ, the m function admits the asymptotic expansion Linearizing q m Special case: Let q L 1 [, ). Then the solution u(x, z; q) of L q u(x, z) = zu(x, z) lim x + u(x, z) e i zx = lim x + u (x, z) i ze i zx = i n+1 + a k ( z) (k+2)/2 + o z k=1 ( z (n+2)/2). where z has its branch cut on [, ) is called the Jost solution. Let us motivate this theorem in a special case. Our goal is to understand the potential to m-function map, q m. One way to do this is to linearize this map. In the following special case, we can show that the linearization of the q to m map is strikingly similar to the expansion of the theorem. The asymptotic conditions imply that this solution is L 2 [, ). Linearizing q m Linearizing q m Define m(z; q) = u (, z; q) u(, z; q). Let q(t) := q(, t) be a family of real-valued, integrable, measurable functions depending smoothly on t, with q(x, ) = and q(x, ) = h(x). We can compute that the linearization of q m near q = is: d m(z, q(t)) dt = e 2i z h(x) dx. t= Integrating by parts twice, for example, the linearization takes the form: L q= (m) = ih() 2 z h () 1 4z (2i z) 2 e 2i zx h (x). Linearization is a Fourier transform. Thus, by studying the mapping q m we are engaged in a kind of non-linear Fourier analysis.
Preliminaries The first term Assume that L q has no negative spectrum. Set More λ = µ 2, σ(µ) = ρ(λ) = ρ(µ 2 ), and τ(µ) = σ(µ) 2µ π Regarding L q as a perturbation of the free Sturm-Liouville operator L, we note that τ(µ) is the spectral function of L q with the spectral function of L removed. Use the identity to write Apply the residue formula: Integrate by parts n + 3 times: i z + (n + 3) dρ(λ) λ z dτ(µ) µ 2 z + 2 dµ π µ 2 z. i dτ(µ) + z µ 2 z. 2µ (µ2 t 2 ) n+2 dτ(t) (µ 2 z) n+4 dµ. Hypotheses of the Theorem Theorem (Levitan, 1973) Assume 1 τ(µ) is an odd function for µ R 2 The total variation of τ(µ) is uniformly bounded on [a, a + 1 as a. 3 The Fourier transform of τ(µ) = for some interval ( ɛ, ɛ). That is, the map from S(R) S(R) given by φ φ(x) dτ(µ) vanishes on C ( ɛ, ɛ) for some ɛ >. Then for any s : R Use the theorem to control the term expansion. (µ 2 t 2 ) n+2 dτ(t) (µ 2 z) n+4 in the Variation hypothesis: Regard τ as the limit of spectral measures for a sequence of finite interval problems, establish a bound in the finite case, and taking a limit. Fourier transform hypothesis: Study the Fourier transform of τ(µ) through an associated wave equation. (µ 2 t 2 ) s dτ(t) = o(µ s ) µ.
A Spectral Solution Representation Consider the wave problem u tt = u xx q(x)u x [, ) u(x, ) = f (x) f is even and compactly supported u t (x, ) = u x (, t) = Notation: Denote solutions of this problem with initial data f by u(x, t; f ). We will present two representation formulas for the solution, one spectral and one classical, and compare them to obtain the necessary information to apply the theorem. Motivated by the Fourier transform method for solving this problem with q(x) =, define the generalized Fourier transform of u S(R) by Fu(λ) = and its inverse transform where F 1 û(x) = ψ(x, λ)u(x) dx ψ(x, λ)û(λ, t) dρ(λ) L q ψ(x, λ) = λψ(x, λ), ψ(, λ) = 1, ψ (, λ) = A Spectral Solution Representation A Spectral Solution Representation Apply F to (L q u)(x, t) and integrate by parts to find that Transform the PDE with F F(L q u)(λ, t) = λf(u)(λ, t). (Fu) tt (λ, t) = λ(fu)(λ, t) The general solution to the transformed PDE is: (Fu)(λ, t) = F (λ) cos λt + G(λ) sin( λt) λ. Since F(u(x, )) = Ff (λ) and F(u t (x, )) = we have that F (λ) = Ff (λ) and G(λ) =. Thus and so u(x, t; f ) = (Fu)(λ, t) = Ff (λ) cos ( λt). cos ( [ λt) ψ(x, λ) f (s)ψ(s, λ) ds dρ(λ).
A Classical Solution Representation Now we construct a classical representation for the solution. To reduce some of the computation, assume that q(x) admits a smooth, even extension to R. More Then the solution to our problem is the solution to the Cauchy problem: restricted to [, ). U tt = U xx q(x)u U(x, ) = f (x) U t (x, ) = x R A standard energy argument establishes that the solution to the half-line wave problem is unique. Thus u(x, t, s) = U(x, t, s) x. A Classical Solution Representation For x t, the solution of the Cauchy problem has the form U(x, t; f ) = 1 2 [f (t + x) + f (t x) + 1 2 t+x t x w(x, t, s)f (s); ds The term 1 2 [f (t + x) + f (t x) is the d Alembert solution to the homogeneous wave equation with initial profile f (x). The function w(x, t, s) is a kind of Green s kernel for the Cauchy problem. An elegant method of Riemann shows that solutions to the inhomogeneous problem u tt u xx = g(x, t), u(x, ) = u t (x, ) = can be written in the form u(x, t) = K(x, t, ξ, τ)g(ξ, τ) dξdτ for some region in the plane. A Classical Solution Representation Comparing Representations Taking g(x, t) = q(x)u(x, t) in the inhomogeneous problem, a careful iteration argument shows that there exists a function w(x, t, s) so that U(x, t; f ) = 1 2 [f (t + x) + f (t x) + 1 2 t+x t x w(x, t, s)f (s); ds By the remarkable fact that L q u(x, t; f ) = u(x, t; L q f ) we see that w must satisfy the Goursat problem: { wtt = w ss q(s)w w(x, t, x ± t) = 1 t. 2 q(x ± τ) dτ We thus have the two representations for the solution to our wave problem. Spectral: u(x, t; f ) = cos ( [ λt) ψ(x, λ) f (s)ψ(s, λ) ds dρ(λ). Classical (x t): where u(x, t) = 1 2 [f (t + x) + f (t x) + 1 2 t+x t x { wtt = w ss q(s)w w(x, t, x ± t) = 1 t 2 q(x ± τ) dτ. Use uniqueness to equate them. w(x, t, s); ds.
Comparing Representations Comparing Representations Equate and subtract from both sides the representations for the free (q(x) = ) problem: cos [ λt ψ(x, λ) f (s)ψ(s, λ) ds dρ(λ) cos λt cos [ λx f (s) cos ( λs ds d = 1 2 t+x w(x, t, s)f (s) ds + 1 2 t x w(x, t, s)f (s) ds. Let g(t) be an even, smooth function of compact support and ĝ(µ) its Fourier cosine transform. Multiply both sides by g(t), integrate in t, and interchange the order of integration. 2 ) λ π Then [ f (s) = 1 2 ĝ(µ) (ψ(x, µ 2 )ψ(s, µ 2 ) dρ(µ 2 ) 2π ) cos µx cos µs [ x + 1 2 + 1 2 f (s) x w(x, t, s)g(t) dt ds [ f (s) w(x, t, s)g(t) dt ds s x [ f (s) w(x, t, s)g(t) dt ds. s+x dµ ds Comparing Representations Applying the Theorem Since f (s) is arbitrary: ĝ(µ) [ψ(x, µ 2 )ψ(s, µ 2 ) dρ(µ 2 ) 2π cos µx cos µs dµ 1 w(x, t, s)g(t) dt + 1 w(x, t, s)g(t) dt, (s x) 2 2 s+x = 1 w(x, t, s)g(t) dt + 1 w(x, t, s)g(t) dt, (s x) 2 s x 2 s+x Set τ(µ) = ρ(µ 2 ) 2 π µ and a(t) = w(, t, ). For s = x = we obtain the trace-type formula ĝ(µ) dτ(µ) = a(t)g(t) dt. Now let g(t) have support in ( ɛ, ɛ) and set By Parseval s equality A(µ) = ɛ a(t)g(t) dt = 2 π Then by the trace formula: ĝ(µ) d µ [ τ(µ) 2 π a(t) cos(µt) dµ. A(µ)ĝ(µ) dµ. A(ν) dν =
Applying the Theorem The Leading Terms By Taylor s Theorem: So [ apply the Tauberian theorem with the measure d µ τ(µ) 2 µ π A(ν) dν and s = n + 2: where (µ 2 ν 2 ) n+2 dτ(ν) = 2 π (µ 2 ν 2 ) n+2 A(ν) d(ν) + O(µ n+2 ) A(ν) = ɛ a(t) cos (νt) dt a(t) = W (, t, ) so that a() = and a(t) C n+1 [, δ). n+1 a(t) = α k t k + r n (t) k=1 where α k = a (k) ()/k! and r n (t) = o(t n+1 ). Insert into: obtain: (µ 2 ν 2 ) n+2 dτ(ν) = 2 π k= (µ 2 ν 2 ) n+2 A(ν) d(ν) + O(µ n+2 ). n+1 (µ 2 ν 2 ) n+2 dτ(ν) = C n,k α k µ 2n k+4 + o(µ n+3 ) C n,k = 1 2 k Γ((k + 1)/2) (n + 2)! π Γ(n k/2 + 3). The Leading Terms The Remainder Term Now put this expression for (µ2 ν 2 ) n+2 dτ(ν) into and simplify: where i z + (n + 3) 2µ (µ2 t 2 ) n+2 dτ(t) (µ 2 z) n+4 dµ i n+1 + a (k) ()( z) (k+2)/2 + R n (z) z k=1 R n (z) = o(µ n+4 ) (µ 2 dµ. z) n+4 To prove the theorem, we must show that R n (z) = o ( z (n+2)/2). Estimate R n (z) C and write R n (z) = I 1 + I 2 where I 1 = I 2 = µ n+4 dµ z = x + iy. [(µ 2 x) 2 + y 2 (n+4)/2 2x µ n+4 dµ [(µ 2 x) 2 + y 2 (n+4)/2 µ n+4 dµ. 2x [(µ 2 x) 2 + y 2 (n+4)/2 We estimate I 1. The I 2 estimate is similar.
The Remainder Term Suppose that x, y > and assume that < tan θ α 1 : Then y > α 1 x so that z/y = O(1). Since µ [, 2x we have: z = x 2 + y 2 = y 1 + x 2 /y 2 y cot θ µ 2 x = 1 ( ) 1 2 µ2 + 2 µ2 x 1 2 µ2. The Remainder Term We then estimate Similarly, 2x µ n+4 I 1 dµ y n+4 C x n+5 y n+4 = C y (n+5)/2 y (n+3)/2 x n+5 Cy (n+3)/2 = O( z (n+3)/2 ) I 2 = O( z (n+3)/2 ) and the also estimates hold for x <, y >, and z outside < tan θ α 1. The Expansion Expand w The w-function has an asymptotic expansion: All together, we have that for z C + /R: i n+1 + a (k) ()( z) (k+2)/2 + o( z (n+3)/2 ) z k=1 where a(t) = w(, t, ). Next, we will show that the coefficients of the expansion can be determined exactly through a set of recurrence relations. k= l= w(x, t, s) = A k,l (x)t k (s x) l. k= l= Inserting w into the PDE w tt = w ss q(s)w: (k + 2)(k + 1)A k+2,l (x)t k (s x) l = (l + 2)(l + 1)A k,l+2 (x)t k (s x) l k= l= q(s)a k,l t k (s x) l. k= l=
Expand q Taylor expand q(s) = q (r) r= (s x)r then: k= l= (k + 2)(k + 1)A k+2,l (x)t k (s x) l = k= l= (l + 2)(l + 1)A k,l+2 (x)t k (s x) l l k= l= r= (k + 1)(k + 2)A k+2,l (x) = (l + 2)(l + 1)A k,l+2 (x) A k,l r (x) q(r) (x) t k (s x) l. l r= A k,l r (x) q(r) (x) Use the Boundary Conditions Now by the boundary conditions for the w-problem: Expand t w(x, t, x t) = 1 2 w(x, t, t + x) = 1 2 t t q(x ± τ) dt in a Taylor series: w(x, t, x t) = 1 2 w(x, t, x + t) = 1 2 r=1 r=1 q(x τ) dτ q(x + τ) dτ. q (r 1) (x) ( 1) r t r q (r 1) (x) t r. Equate the Expansions Also, by the asymptotic expansion for w: w(x, t, x t) = r= l= A r l,l (x)( 1) l t r, w(x, t, x+t) = Equate the two expansions for w(x, t, x ± t): 1 2 r= 1 2 q (r 1) (x) ( 1) r t r = q (r 1) (x) ( 1) r = 1 q (r 1) (x) = 2 r= l= A r l,l (x)( 1) l t r A r l,l (x)( 1) l l= A r l,l (x). l= r= l= A r l,l (x)t r The Recurrence Relations In total, we can compute the coefficients of our asymptotic expansion from the relations A, (x) = 1 q (r 1) (x) = ( 1) l A r l,l (x) 2 l= 1 q (r 1) (x) = A r l,l (x) 2 l= (k + 1)(k + 2)A k+2,l (x) = (l + 2)(l + 1)A k,l+2 (x) l r= A k,l r (x) q(r) (x)
Computing the Coefficients Sample Computation I We now use the recurrence relations to generate the first few coefficients of the asymptotic expansion i n+1 + a (k) ()( z) (k+2)/2 + o( z (n+3+α)/2 ). z k=1 Note that a(t) = W (, t, ) and differentiating, we see that a (k) () = k!a k, (). Determine the coefficients A k, (). k =, l = : A, () = k =, l = 1 : A 2, () = A,2 () A,1 () + A 1, () = 1 2 q() A,1 () A 1, () = 1 2 q() A,1 () = ; A 1, () = 1 2 q() A 2,1 () = 3A,3 () k =, l = 2 : A,2 () + A 1,1 () + A 2, () = 1 4 q () A,2 () A 1,1 () + A 2, () = 1 4 q () A,2 () + A 2, () = A,2 () = ; A 2, () = A 1,1 () = 1 4 q () A 2,2 () = 6A,4 () ll Sample Computation II I k = 1, l = 6A 3, = 2A 1,2 A 1, ()q() A 1,2 () = 3A 3, () 1 4 q()2 k =, l = 3 A,3 () + A 1,2 () + A 2,1 () + A 3, () = 1 A,3 () A 1,2 () + A 2,1 () A 3, () = 1 A,3 () + A 2,1 () = A,3 () = A 2,1 () = A 1,2 () + A 3, () = 1 12 q () A 3, () = 1 48 q () + 1 16 q()2 12 q () 12 q () Earl A. Coddington and Norman Levinson, Theory of ordinary differential equations, McGraw-Hill Book Company, Inc., New York-Toronto-London, 1955. MR MR69338 (16,122b) R. Courant and D. Hilbert, Methods of mathematical physics. Vol. II, Wiley Classics Library, John Wiley & Sons Inc., New York, 1989, Partial differential equations, Reprint of the 1962 original, A Wiley-Interscience Publication. MR MR11336 (9k:351) A. A. Danielyan and B. M. Levitan, Asymptotic behavior of the Weyl-Titchmarsh m-function, Izv. Akad. Nauk SSSR Ser. Mat. 54 (199), no. 3, 469 479. MR MR172691 (92d:3456)
II Appendix I Appendix 1 In this case, we are considering the sequence of problems B. M. Levitan, On a special Tauberian theorem, Izvestiya Akad. Nauk SSSR. Ser. Mat. 17 (1953), 269 284. MR MR5868 (15,316a) B. M. Levitan and I. S. Sargsjan, to spectral theory: selfadjoint ordinary differential operators, American Mathematical Society, Providence, R.I., 1975, Translated from the Russian by Amiel Feinstein, Translations of Mathematical Monographs, Vol. 39. MR MR369797 (51 #626) This leads to the eigenpairs λ n = n2 π 2 b 2, ψ n(x) = and the expansion { utt = λu x [, b u () = u (b) = f (x) = 2 b cos( λx) n =, 1, 2,... (f, ψ k )ψ k. k=. Appendix Appendix Appendix II Appendix III for every f L 2 [, b. Set ψ(x, λ) = 2 b cos( λx) and let ρ b (λ) be the nondecreasing step function which jumps by 2 b when λ passes through kπ b. Then we have f (x) = (f, ψ(, λ))ψ(x, λ)dρ b ( λ). Taking b, dρ b (λ) 2 π d λ and the expansion becomes the Fourier cosine transform. Return 2 If L q has negative spectra, write d ρ(λ) m(z) λ z = dρ(λ) λ z. We expand the right hand side, as above, with slight modifications as in [3. Then expanding the integral on the left hand side in powers of 1/z and comparing coefficients, we can obtain the same expansion as in the theorem. Return
Appendix Appendix IV 3 Without the assumption that q admits a smooth, even extension, the classical solution representation formula becomes more complicated. In this case, additional integral terms, which are associated to a second Goursat problem, will appear. The analysis of this general situation, which is not much more difficult, is fully detailed in [3. Return