# Green's function for Sturm-Louiville ODE

• A
• member 428835

#### 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.

If you want a solution bounded at 1, then you are limited to scalar multiples of $P_1^1$: $$G(x;x_0) = \begin{cases} AP_1^1(x) & x \leq x_0 \\ BP_1^1(x) & x > x_0\end{cases}.$$ The conditions you need to satisfy are $$\begin{split} (A - B)P_1^1(x_0) &= 0 \\ (A - B){P_1^1}'(x_0) &= \frac{1}{p(x_0)}\end{split}$$ 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 $x_0$.

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

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 $$L(y) = (py')' + qy = -g$$ on the interval $[a,b]$ subject to arbitrary boundary conditions on $y$ is to first solve the eigenvalue problem $$L(\phi) = -\lambda w \phi$$ 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 $$L(y_P) = -\sum_na_n\lambda_n w \phi_n = -g$$ and then multiplying by $\phi_m$ and integrating you should find$$a_m = \frac{1}{\lambda_m \|\phi_m\|_w^2}\int_a^b g(x)\phi_m(x)\,dx.$$ A complimentary function $y_C \in \ker L$ can then be added to $y_P$ in order to satisfy the boundary conditions on $y$.