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Chapter 24 Sturm-Liouville problem

Chapter 24 Sturm-Liouville problem. Speaker: Lung-Sheng Chien. Reference: [1] Veerle Ledoux, Study of Special Algorithms for solving Sturm-Liouville and Schrodinger Equations.

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Chapter 24 Sturm-Liouville problem

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  1. 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

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

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

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

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

  6. 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?

  7. Dirichlet eigenvalue problem [1] Dirichlet eigenvalue problem: Observation: 1 ,in fact, If is eigen-pair of Dirichlet eigenvalue problem, then 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:

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

  9. 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 operator L is Hermitian, then eigenvectors are orthogonal 2 1 L is diagonalizable and has real eigenvalue eigenvectors are orthogonal 2 : orthonormal

  10. Dirichlet eigenvalue problem [4] and are two eigen-pair of Suppose and 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

  11. Dirichlet eigenvalue problem [5] and are two eigen-pair of Theorem 2: if If eigenvalue , then are orthogonal in <proof> 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. <proof> Abel’s formula suppose and Let , then for some constant c since

  12. 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 , satisfying and and are both eigen-pair, from uniqueness of eigen-function, we have Hence 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 Eigen-functions are orthogonal in with inner-product 2 Eigen-functions are real and twice differentiable 3 , find eigen-pair and Exercise (failure of uniqueness): consider show “eigenvalue is not simple”, can you explain this? (compare it with Theorem 3)

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

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

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

  16. 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 Objective: find eigenvalue E such that Advantage of Prufer method: we only need to solve when solution is found with condition then

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

  18. Prufer method [3] numerically Question: Given energy E, how can we solve Forward Euler method: consider first order ODE Uniformly partition domain as 1 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

  19. 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

  20. 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.

  21. 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

  22. 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.

  23. 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.

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

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

  26. 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 increases gradually “staircase” shape with 2 “plateaus” at and steep slope around Question: what is disadvantage of this “staircase” shape?

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

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

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

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