Green's function for Sturm-Louiville ODE

  • #1

member 428835

Hi PF!

Given the following ODE $$(p(x)y')' + q(x)y = 0$$ where ##p(x) = 1-x^2## and ##q(x) = 2-1/(1-x^2)## subject to $$y'(a) + \sec(a)\tan(a)y(a) = 0$$ and $$|y(b)| < \infty,$$ where ##a = \sqrt{1-\cos^2\alpha} : \alpha \in (0,\pi)## and ##b = 1##, what is the Green's function?

This is the associated Legendre ODE, which admits two linearly independent solutions: $$y_1 = P_1^1, \,\,\, y_1 = Q_1^1$$ where ##P_1^1## and ##Q_1^1## are associated Legendre polynomials first and second kind respectively. Things now get a little murky: notice ##y_1## automatically satisfies ##y_1'(a) + \sec(a)\tan(a)y_1(a) = 0##, but ##y_2(b) = \infty##. If ##y_2(b)## was bounded, the Green's function would be very simple to calculate. But it's not bounded, so I'm confused how to proceed?

Any help is greatly appreciated.
 

Answers and Replies

  • #2
If you want a solution bounded at 1, then you are limited to scalar multiples of [itex]P_1^1[/itex]: [tex]
G(x;x_0) = \begin{cases}
AP_1^1(x) & x \leq x_0 \\
BP_1^1(x) & x > x_0\end{cases}.[/tex] The conditions you need to satisfy are [tex]
\begin{split}
(A - B)P_1^1(x_0) &= 0 \\
(A - B){P_1^1}'(x_0) &= \frac{1}{p(x_0)}\end{split}[/tex] which cannot be done unless [itex]P_1^1(x_0) = 0[/itex] or you are prepared to relax the constraint that [itex]G(\cdot;x_0)[/itex] be continuous at [itex]x_0[/itex].
 
  • #3
I appreciate your reply. So it looks like a Green's function doesn't exist for this problem, right?
 
  • #4
I appreciate your reply. So it looks like a Green's function doesn't exist for this problem, right?

Yes.

The way to solve the non-homogenous boundary value problem [tex]
L(y) = (py')' + qy = -g[/tex] on the interval [itex][a,b][/itex] subject to arbitrary boundary conditions on [itex]y[/itex] is to first solve the eigenvalue problem [tex]
L(\phi) = -\lambda w \phi[/tex] for a convenient weight function [itex]w[/itex] which is strictly positive on [itex](a,b)[/itex] with [itex]\phi[/itex] subject to a convenient self-adjoint boundary condition. This gives you a discrete spectrum of eigenvalues [itex]\lambda_n \neq 0[/itex] with corresponding eigenfunctions [itex]\phi_n[/itex] which are orthogonal with respect to the inner product [itex](f,g)_w = \int_a^b f(x)g(x) w(x)\,dx[/itex].

(In this case, choosing a suitable boundary condition and setting [itex]w = 1[/itex] will give [itex]\phi_n = P_n^1[/itex] with [itex]\lambda_n = n(n+1) - 2[/itex] for [itex]n \geq 2[/itex].)

Next, look for a particular solution [tex]y_P = \sum_n a_n \phi_n[/tex] so that [tex]
L(y_P) = -\sum_na_n\lambda_n w \phi_n = -g[/tex] and then multiplying by [itex]\phi_m[/itex] and integrating you should find[tex]
a_m = \frac{1}{\lambda_m \|\phi_m\|_w^2}\int_a^b g(x)\phi_m(x)\,dx.[/tex] A complimentary function [itex]y_C \in \ker L[/itex] can then be added to [itex]y_P[/itex] in order to satisfy the boundary conditions on [itex]y[/itex].
 

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