A Green's function for Sturm-Louiville ODE

[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.

 

[pasmith]

If you want a solution bounded at 1, then you are limited to scalar multiples of P_1^1: <br /> G(x;x_0) = \begin{cases}<br /> AP_1^1(x) &amp; x \leq x_0 \\<br /> BP_1^1(x) &amp; x &gt; x_0\end{cases}. The conditions you need to satisfy are <br /> \begin{split}<br /> (A - B)P_1^1(x_0) &amp;= 0 \\<br /> (A - B){P_1^1}&#039;(x_0) &amp;= \frac{1}{p(x_0)}\end{split} which cannot be done unless P_1^1(x_0) = 0 or you are prepared to relax the constraint that G(\cdot;x_0) be continuous at x_0.

 

[member 428835]

I appreciate your reply. So it looks like a Green's function doesn't exist for this problem, right?

 

[pasmith]

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 <br /> L(y) = (py&#039;)&#039; + qy = -g on the interval [a,b] subject to arbitrary boundary conditions on y is to first solve the eigenvalue problem <br /> L(\phi) = -\lambda w \phi for a convenient weight function w which is strictly positive on (a,b) with \phi subject to a convenient self-adjoint boundary condition. This gives you a discrete spectrum of eigenvalues \lambda_n \neq 0 with corresponding eigenfunctions \phi_n which are orthogonal with respect to the inner product (f,g)_w = \int_a^b f(x)g(x) w(x)\,dx.

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

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