Find the Green's function that satisfies 2nd order ODE

Question

Find the Green's function that satisfies:

$ x \frac {d^2G}{dx^2}-(2x+1)\frac {dG}{dx}+(x+1)G=\delta (x-s)$.

I have found it to be: $G= \begin {cases} e^x(\frac{1}2-\frac{x^2}{2}) & 0\leq x < s\leq 1 \\ e^x(\frac{1}2-\frac{x^2}{2}) & 0\leq s < x\leq 1 \end{cases}$.

With boundary conditions $G(0)=0, G(1)=0$

I'm not entirely happy with this answer as I would have thought that the function would differ for $x<s$ and $s<x$. As this is my first time doing Green's functions, I'm not sure if this is correct

Thanks

Answer 1

In general the way these things go is as follows. Suppose $f_1,f_2$ are the general solutions to the homogeneous ODE. For fixed $s$, $G$ needs to have the following properties:

$$G(0)=0 \\ \lim_{x \to s^-} G(x)=\lim_{x \to s^+} G(x) \\ \lim_{x \to s^+} G'(x)=\lim_{x \to s^-} G'(x)+1/s \\ G(1)=0.$$

The third equation is where the Dirac delta is actually entering, and it appears as $1/x$ because the equation says that $xG''$ is the derivative of a thing with a jump of size $1$ at $x=s$, and integrating by parts tells you that $xG''$ is the derivative of $xG'-G$. For reasons relating to distribution theory the entire jump must be in $G'$, so you get these equations.

Now

$$G(x)=\begin{cases} a_1 f_1 + a_2 f_2 & x \leq s \\ b_1 f_1 + b_2 f_2 & x \geq s \end{cases}$$

and you are left to solve a system of linear equations for the coefficients. These equations read

$$a_1 f_1(0) + a_2 f_2(0) = 0 \\ b_1 f_1(1) + b_2 f_2(1) = 0 \\ a_1 f_1(s) + a_2 f_2(s) = b_1 f_1(s) + b_2 f_2(s) \\ b_1 f_1'(s) + b_2 f_2'(s) = a_1 f_1'(s) + a_2 f_2'(s) + 1/s.$$

This system can be solved to give you the coefficients, which specify $G$.

Allowing $s$ to be non-fixed then gives you $G(x,s)$, with basically everything going through the same except that all the derivatives and integrals that happened above are with respect to $x$ specifically now.

In your situation with $f_1(x)=e^x,f_2(x)=x^2 e^x$ the equations read

$$a_1 = 0 \\ eb_1 + eb_2 = 0 \\ a_1 e^s + a_2 s^2 e^s = b_1 e^s + b_2 s^2 e^s \\ b_1 e^s + b_2 (2s + s^2) e^s = a_1 e^s + a_2 (2s+s^2) e^s + 1/s.$$

Replace $b_2=-b_1$ and $a_2=\frac{b_1(1-s^2)}{s^2}$ and substitute into the fourth equation to find $b_1$, then you're done.

As Lutz suggested, you can clean this up somewhat by choosing $f_1$ and $f_2$ such that the first two equations read, for example, $a_1=0,b_2=0$, which makes the third and fourth equations less complicated after substitution.

Answer 2

You know from earlier computations that the general solution is $e^x(Ax^2+B)$. Now determine two solutions that satisfy one of the boundary conditions each. These are $G_0(x)=e^xx^2$ and $G_1(x)=e^x(1-x^2)$. The Green function is a multiple of $G_0$ for $x<s$ and of $G_1$ for $x>s$. You need continuity at $x=s$, so you get that $G(s,x)$ is a multiple of $$ e^x[x^2(1-s^2)1_{x\le s}+s^2(1-x^2)1_{x>s}]. $$ The derivative of this expression for $x\ne s$ is $$ e^x[(x^2+2x)(1-s^2)1_{x\le s}+s^2(1-x^2-2x)1_{x>s}] $$ This gives at $x=s$ a jump of the derivative of $e^s(s^2(-2s)-2s(1-s^2))=-2se^s$. For the Green function this jump needs to be $\frac1s$, so multiply with the factor $-\frac{e^{-s}}{2s^2}$ to get $$ G(s,x)=\frac1{2}e^{x-s}\left[x^2-\min\left(1,\frac{x^2}{s^2}\right)\right] $$