A Smooth Operator, Operated Correctly

Transcription

1 Clark Department of Mathematics Bard College at Simon s Rock March 6, 2014

2 Abstract By looking at familiar smooth functions in new ways, we can make sense of matrix-valued infinite series and operator-valued integrals. We can then use them to solve partial differential equations, both stochastic and classic. Slides available at: Clark.weebly.com

3 Initial-value Problem Consider the following initial-value problem (IVP). { d dt x(t) = Ax(t) x(0) = x 0 Your choice, A is: a real or complex number (scalar) a square matrix (bounded operator) an unbounded operator If A is a scalar, then the solution is x(t) = e At x 0. Does the above solution make sense if A is not a scalar?

4 e to the what? In other words, can we give a meaning to the expression e At when A is more complicated? Will this new conception of the exponential behave as expected? d dt eat = Ae At? In any context that e At makes sense and behaves well, x(t) = e At x 0 is a solution to the IVP. Therefore, many IVP s reduce to understanding e At.

5 Bounded Operators Suppose the following: (B, ) is a Banach Space (complete, normed vector space). A : B B is a bounded linear operator: A(cx + y) = cax + Ay A := sup Ax < x =1 For example, could be max-row sum : If A = then A = 6 and A is bounded If U = then U = and U is unbounded.

6 Bounded Operators In Quantum Mechanics, A represents the Schrödinger operator. A = 1 ( ) 2 ı 2m 2 + V Alternatively, A could be the convolution with a function f. Ax(t) = f x(t) = f (t s)x(s) ds

7 The Exponential of a Bounded Operator For any bounded operator A on a vector space, I + A A ! A n! An is another bounded operator. Since the domain of A is a Banach Space (complete, normed, vector space), then e A := n=0 1 n! An = I + A A ! A n! An + is also a bounded operator.

8 The Exponential of a Bounded Operator Why does e A = e A = n=0 n=0 1 n! An make sense? 1 n! A n is a convergent Taylor series. The tail of a convergent series is always small: N n=m 1 n! A n < ɛ if M and N are large. The partial sums are a Cauchy sequence: N 1 N n! An 1 n! A n < ɛ n=m n=m

9 First Conclusion If A is a bounded operator on a Banach space B, then: e At is well-defined and x(t) = e At x 0 is a solution to the initial-value problem. Moreover, if f is any function with a Taylor series, f (x) = c k (x a) k, then f (A) := k=0 is a bounded operator on B. c k (A a) k k=0

10 Cauchy s Integral Theorem If f : C C is analytic (i.e. has a power series expansion) and C is a loop (winding number one) around a complex number w, then f (w) = 1 2πi C f (z) z w dz. Im(z) C w Re(z)

11 Cauchy s Integral Theorem If f : C C is analytic (i.e. has a power series expansion) and C is a loop (winding number one) around a complex number w, then f (w) = 1 2πi C f (z) z w dz. Some implications: A quick formula for evaluating certain integrals An analytic function can be reconstructed entirely from its values around C. f (A) = 1 f (z) dz??? 2πi C z A

12 Spectral Theory Suppose H is a Hilbert Space (complete vector space with an inner product) and A : H H is an (unbounded) operator. If (z, x) is an eigenvalue/eigenvector pair for A, then zi A has a nullspace. ρ(a) := {z in C : zi A has a bounded inverse} is the resolvent set. σ(a) := {z in C : zi A does not have a bounded inverse} is the spectrum. R(z) = 1 is the resolvent, defined on ρ(a). z A

13 Analytic Functional Calculus C f (z) dz??? z A f (z) dz is an operator-valued integral, defined C z A analogously to the Riemann integral. The integral converges because R(z) = 1 z A is a bounded operator. We can safely define f (A) := 1 2πi C f (z) z A dz.

14 Initial-value Problem In particular, the exponential is an analytic function. e At := 1 2πi d dt eat = 1 2πi C C e zt z A dz ze zt dz = AeAt z A

15 Second Conclusion If A is an (unbounded) operator on a Hilbert space H, then: e At is well-defined (analytic functional calculus) and x(t) = e At x 0 is a solution to the initial-value problem. Moreover, if f is any analytic function on C, f (At) := 1 f (zt) 2πi z A dz is a bounded operator on H. C

16 Example: Diffusion for Markov Wave Equations If ψ t l 2 (Z d ) is a solution to the discrete Schrödinger initial-value problem { t ψ t (x) = i θ ω(t) ψ t (x), t > 0, x Z d ψ 0 (x) = δ 0 (x), x Z d, then there exists a symmetric matrix D such that ( lim ηk x E ψ t/η (x) 2) = e t k,dk for k T d. e i η 0 + x Z d

17 Example: Diffusion for Markov Wave Equations

18 Thank You & References Thank you! Slides available at: Clark.weebly.com John B. Conway. Functions of One Complex Variable I. Number 11 in Graduate Texts in Mathematics. Springer, C. and J. Schenker. Diffusive scaling for all moments of the Markov Anderson model. ArXiv e-prints, December Klaus-Jochen Engel & Rainer Nagel. One-Parameter Semigroups for Linear Evolution Equations, volume 194 of Graduate Studies in Mathematics. Springer, 2000.