Finding Green's Function for Linear DE

Question

I'm trying to find the Green's function for

$x^3\frac{d^2y(x)}{dx^2}+x^2\frac{dy(x)}{dx}-xy(x)=d(x)$,

with boundary conditions $y(0)=y(1)=0$.

I said the Green's function equation was

$x^3\frac{d^2g(x|\xi)}{dx^2}+x^2\frac{dg(x|\xi)}{dx}-xg(x|\xi)=\delta(x-\xi)$,

with $g(0|\xi)=g(1|\xi)=0$.

The solution I thought could be found by:

$x^3\frac{d^2g(x|\xi)}{dx^2}+x^2\frac{dg(x|\xi)}{dx}-xg(x|\xi)=0$,

giving

$g(0<x<\xi|\xi) = A_1\frac{x^2+1}{2x}+iB_1\frac{x^2-1}{2x}$,

$g(\xi<x<1|\xi) = A_2\frac{x^2+1}{2x}+iB_2\frac{x^2-1}{2x}$.

But when I plug in $g(0|\xi)$, I get $A_1, B_1 = 0$, which screws the rest of it up.

Did I miss something or take a wrong turn somewhere?

Answer 1

Just wondering if you have solved this problem. It is still worth some discussion.

You shall test for the existence of Green's function first, by solving the completely homogeneous problem (with $d\left(x\right)=0$). If it only admits the trivial zero solution, then there exists a Green's function, which can be constructed explicitly.

The homogeneous problem here is of the Euler-Cauchy type. By guessing $y\left(x\right) = x^{r}$, we arrive at the characteristic equation $r^{2}-1=0$. Therefore, the solution to the homogeneous problem takes the form \begin{equation} y\left(x\right)=c_{1}x+\frac{c_2}{x} \end{equation}

Set up the Green's function as the following, \begin{equation} G\left(x,\xi\right)=\begin{cases} G_{-}\left(x,\xi\right), & 0\leq x<\xi\leq1\\ G_{+}\left(x,\xi\right), & 0\leq\xi<x\leq1 \end{cases} \end{equation} where $G_{\pm}\left(x,\xi\right)$ both solve the homogeneous problem, with \begin{equation} G{-}\left(0,\xi\right)=0 \\ G_{+}\left(0,\xi\right)=1 \end{equation}

Two more conditions also need to be satisfied by the Green's function, namely,

1. Continuity at some point $x=\xi$
2. Jump discontinuity at $x=\xi$ such that \begin{equation} a_{0}\left(\xi\right)\left[\frac{dG}{dx}\right]_{x=\xi^{-}}^{x=\xi^{+}}=1 \end{equation} for which $a_{0}\left(\xi\right) = \xi^{3}$ in this particular case. (it is the function multiplying the highest order derivative)

More precisely, \begin{equation} G_{-}\left(x,\xi\right) = Ax+\frac{B}{x} \\ G_{+}\left(x,\xi\right) = Cx+\frac{D}{x} \\ \end{equation}

With the four conditions (two boundary conditions and the two conditions above), you can solve for the four unknown coefficients with some algebra