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

Sturm-Liouville equation: Assumptions: 1 on interval is continuous differentiable on closed interval, or say, and 2 is continuous on closed interval, or say 3, 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” Existence and uniqueness [1] Let be continuous space equipped with norm

Existence and uniqueness [2] is complete under norm define a mapping by 1 2 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] since with fundamental matrix Abel’s formula: 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 is eigen-pair of Dirichlet eigenvalue problem, then,in fact, is continuous on closed interval Exercise: 2 Under assumption, and, we can define inner-product where is complex conjugate of 1 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 [2] 3 Define differential operator, then it is linear and 4 Green’s identity: The same hence Dirichlet boundary condition:

Dirichlet eigenvalue problem [3] 5 Definition: operator L is called self-adjoint on inner-product space If Green’s identity is zero, say Matrix computation (finite dimension):Functional analysis (infinite dimension): Matrix A is Hermitian (self-adjoint):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 [4] Suppose and are two eigen-pair of and are two eigen-pair Theorem 1: all eigenvalue of Sturm-Liouville Dirichlet problem are real Hence is eigen-pair if and only if is eigen-pair

Dirichlet eigenvalue problem [5] If eigenvalue, then are orthogonal in Theorem 2: if and are two eigen-pair of from Theorem 1, we know E1 and E2 are real Theorem 3 (unique eigen-function): eigenfunction of Sturm-Liouville Dirichlet problem is unique, in other words, eignevalue is simple. since Abel’s formula suppose and Let, then for some constant c

Dirichlet eigenvalue problem [6] Theorem 4 (eigen-function is real): eigenfunction of Sturm-Liouville Dirichlet problem can be chosen as real function. suppose is eigen-function with eigen-value and, satisfying and Hence 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 with inner-product 3 Eigen-functions are real and twice differentiable Exercise (failure of uniqueness): consider, find eigen-pair and show “eigenvalue is not simple”, can you explain this? (compare it with Theorem 3)

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

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

Dirichlet eigenvalue problem [9] Express operator L as 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 Objective: find eigenvalue E such that Advantage of Prufer method: we only need to solve when solution is found with condition then Exercise: check it

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

Prufer method [3] Question: Given energy E, how can we solve numerically Forward Euler method: consider first order ODE 1 Uniformly partition domain as 2 Let 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 has a unique solution. But we want to find energy E such that, that is is constraint. Example: for model problem

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

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

Prufer method [7] Theorem 6 : consider Prufer equations of regular Sturm-Liouville Dirichlet problem Let boundary values satisfy following normalization: Then the kth eigenvalue satisfies Moreover the kth eigen-function has k zeros in and 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 scaling function, continuous differentiable where 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 Simple Prufer transformation Recall time-independent Schrodinger’s equation dimensionless form reduce to 1D Dirichlet problem

Exercise 1 [1] Consider 1D Schrodinger equation with Dirichlet boundary condition use standard 3-poiint centered difference method to find eigenvalue 1 1 all eigenvalues ae positive, can you explain this? 2 number of grids = 50, (n = 50) 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] solve simple Prufer equation for first 10 eigenvalues 2 Forward Euler method: N = is consistent with that in theorem 6 2 increases gradually “staircase” shape with “plateaus” at and steep slope around Question: what is disadvantage of this “staircase” shape?

Exercise 1 [3] Forward Euler method: N = 100Forward 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.

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

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