Chapter 24 SturmLiouville problem Speaker LungSheng Chien Reference

Chapter 24 Sturm-Liouville problem Speaker: Lung-Sheng Chien Reference: [1] Veerle Ledoux, Study of Special Algorithms for solving Sturm-Liouville and Schrodinger Equations. [2] 王信華教授, chapter 8, lecture note of Ordinary Differential equation

Existence and uniqueness [1] on interval Sturm-Liouville equation: Assumptions: 1 is continuous differentiable on closed interval , and 2 3 is continuous on closed interval , or say , and Proposition 1. 1: Sturm-Liouville initial value problem is unique on under three assumptions. proof: as before, we re-formulate it as integral equation and apply “contraction mapping principle” Let be continuous space equipped with norm

Existence and uniqueness [2] 1 2 is complete under norm define a mapping by extract supnorm is a contraction mapping if Existence and uniqueness

Fundamental matrix Sturm-Liouville equation: on interval Transform to ODE system by setting has two fundamental solutions satisfying Solution of initial value problem is Definition: fundamental matrix of is satisfying Question: Can we expect that we have two linear independent fundamental solutions, say

Abel’s formula [1] Consider general ODE system of dimension two: with ( from product rule) First order ODE implies are linearly independent

Abel’s formula [2] with fundamental matrix Abel’s formula: since Definition: Wronskian , then fundamental matrix of Sturm-Liouville equation can be expressed as In our time-independent Schrodinger equation, we focus on eigenvalue problem Definition: boundary condition is called Dirichlet boundary condition Question: What is “solvability condition” of Sturm-Liouville Dirichlet eigenvalue problem?

Dirichlet eigenvalue problem [1] Dirichlet eigenvalue problem: Observation: 1 If , in fact, is eigen-pair of Dirichlet eigenvalue problem, then is continuous on closed interval Exercise: 2 Under assumption , and , we can define inner-product where 1 is complex conjugate of If w is not positive, then such definition is not an “inner-product” 2 is Dirac Notation, different from conventional form used by Mathematician Thesis of Veerle Ledoux: Matrix computation: Dirac Notation:

Dirichlet eigenvalue problem 3 Define differential operator 4 Green’s identity: [2] , then it is linear and The same hence Dirichlet boundary condition:

Dirichlet eigenvalue problem 5 [3] Definition: operator L is called self-adjoint on inner-product space If Green’s identity is zero, say Matrix computation (finite dimension): Matrix A is Hermitian (self-adjoint): Functional analysis (infinite dimension): operator L is self-adjoint: Matrix A is Hermitian, then 1 A is diagonalizable and has real eigenvalue 2 eigenvectors are orthogonal : orthonormal operator L is Hermitian, then 1 L is diagonalizable and has real eigenvalue 2 eigenvectors are orthogonal

Dirichlet eigenvalue problem Suppose and are two eigen-pair of are two eigen-pair Theorem 1: all eigenvalue of Sturm-Liouville Dirichlet problem are real <proof> Hence is eigen-pair if and only if is eigen-pair [4]

Dirichlet eigenvalue problem Theorem 2: if If eigenvalue and [5] are two eigen-pair of , then are orthogonal in <proof> from Theorem 1, we know E 1 and E 2 are real Theorem 3 (unique eigen-function): eigenfunction of Sturm-Liouville Dirichlet problem is unique, in other words, eignevalue is simple. <proof> Abel’s formula suppose Let and , then for some constant c since

Dirichlet eigenvalue problem [6] Theorem 4 (eigen-function is real): eigenfunction of Sturm-Liouville Dirichlet problem can be chosen as real function. <proof> suppose is eigen-function with eigen-value and Hence , satisfying and are both eigen-pair, from uniqueness of eigen-function, we have we can choose real function u as eigenfunction So far we have shown that Sturm-Liouville Dirichlet problem has following properties 1 Eigenvalues are real and simple, ordered as 2 Eigen-functions are orthogonal in 3 Eigen-functions are real and twice differentiable Exercise (failure of uniqueness): consider with inner-product , find eigen-pair and show “eigenvalue is not simple”, can you explain this? (compare it with Theorem 3)

Dirichlet eigenvalue problem Theorem 5 (Sturm’s Comparison theorem): let suppose , then be eigen-pair of Sturm-Liouville Dirichlet problem. . Precisely speaking is more oscillatory than Between any consecutive two zeros of [7] , there is at least one zero of <proof> Let suppose are consecutive zeros of and on are eigen-pair, then define IVT on as left figure

Dirichlet eigenvalue problem [8] on and Question: why on is more oscillatory than implies , any physical interpretation? Sturm-Liouville equation Definition: average quantity of operator on Time-independent Schrodinger equation over interval by

Dirichlet eigenvalue problem Express operator L as [9] where neglect boundary term where and Heuristic argument for Theorem 5 Average value of is more oscillatory than

Prufer method [1] Sturm-Liouville Dirichlet eigenvalue problem: Idea: introduce polar coordinate in the phase plane Exercise: check it Objective: find eigenvalue E such that Advantage of Prufer method: we only need to solve when solution then is found with condition

Prufer method [2] Observation: 1 Suppose has unique solution with initial condition 2 fixed is a strictly increasing function of variable is Hook’s law has general solution and implies

Prufer method [3] numerically Question: Given energy E, how can we solve Forward Euler method: consider first order ODE 1 Uniformly partition domain 2 Let as and approximate by one-side finite difference continuous equation: discrete equation: Exercise : use forward Euler method to solve angle equation for different E = 1, 4, 9, 16 as right figure It is clear that is a strictly increasing function of variable

Prufer method [4] 3 Although such that Example: for model problem has a unique solution , that is . But we want to find energy E is constraint.

Prufer method [5] 4 never decrease in a point where number of zeros of in number of multiples of Ground state in has no zeros except end points.

Prufer method [6] First excited state has one zero in model problem: second excited state conjecture: k-th excited state has two zeros in has k zeros in

Prufer method [7] Theorem 6 : consider Prufer equations of regular Sturm-Liouville Dirichlet problem and Let boundary values satisfy following normalization: Then the kth eigenvalue satisfies Moreover the kth eigen-function has k zeros in Remark: for detailed description, see Theorem 2. 1 in Veerle Ledoux, Study of Special Algorithms for solving Sturm-Liouville and Schrodinger Equations.

Prufer method [8] Scaled Prufer transformation: a generalization of simple Prufer method where scaling function, continuous differentiable 1 2 Theorem 6 holds for scaled Prufer equations Exercise: use Symbolic toolbox in MATLAB to check scaled Prufer transformation.

Prufer method [9] Scaled Prufer transformation Recall time-independent Schrodinger’s equation dimensionless form reduce to 1 D Dirichlet problem Simple Prufer transformation

Exercise 1 [1] Consider 1 D Schrodinger equation with Dirichlet boundary condition 1 use standard 3 -poiint centered difference method to find eigenvalue number of grids = 50, (n = 50) 1 2 all eigenvalues ae positive, can you explain this? Ground state has no zeros in 1 st excited state has 1 zero in 2 nd excited state has 2 zero in Check if k-th eigen-function has k zeros in

Exercise 1 2 [2] solve simple Prufer equation for first 10 eigenvalues Forward Euler method: N = 200 is consistent with that in theorem 6 1 2 increases gradually “staircase” shape with “plateaus” at and steep slope around Question: what is disadvantage of this “staircase” shape?

Exercise 1 Forward Euler method: N = 100 [3] Forward Euler method: N = 200 under Forward Euler method with number of grids, N = 100, this is wrong! Question: Can you explain why is not good? hint: see page 21 in reference Veerle Ledoux, Study of Special Algorithms for solving Sturm-Liouville and Schrodinger Equations.

Exercise 2 [1] Scaled Prufer transformation 1 D Schrodinger: where Suppose we choose is continuous Question: function f is continuous but not differentiable at x = 1. How can we obtain Although we have known on open set and

Exercise 2 [2] solve simple Prufer equation for first 10 eigenvalues Choose scaling function Forward Euler method: N = 100 scaling function S NO scaling function S Compare both figures and interpret why “staircase” disappear when using scaling function