# Thread: 3rd and Final Green Function Question

1. ## 3rd and Final Green Function Question

(Q4) Derive the Green's function for

$\displaystyle \frac{d^2y}{dx^2} - a^2y = \delta(x-x_0)$

with boundary conditions $y(x) \rightarrow 0$ as $x \pm \infty$

2. First note that the solution is almost already is its self adjoint form.

$\frac{d}{dx}\left(-1\cdot y' \right)+ay=-\delta(x-x_0)$

So we need to find two different solutions such that each one satisfies one of the Boundary Conditions.

$v_0(t)=c_1e^{-ax}+c_2e^{ax}$
As $x \to -\infty$ this gives $c_1=0$ and

$v_0(t)=c_2e^{ax}$

$v_1(t)=c_3e^{-ax}+c_4e^{ax}$

As $x \to \infty$ this gives $c_4=0$

$v_1(t)=c_3e^{-ax}$

So the Green's function has the form

$g(x,s)=\begin{cases} Ae^{as}\cdot e^{-ax} \text{ for } s \le x \\ Be^{ax}\cdot e^{-as} \text{ for } x \le s\end{cases}$

Now we use the Wronskian and the and the condiditon that

$p(x)W(x)=-1$ so from above we have that $p(x)=-1$ and

$\begin{vmatrix} c_1e^{a(s-x)} && c_2e^{a(x-s)} \\ c_1(-a)e^{a(s-x)} && c_2(a)e^{a(x-s)} \end{vmatrix}=c_1c_2a+c_1_c2a$

Now using this we get

$2c_1c_2a(-1)=-1 \iff c_1c_2=\frac{1}{2a}$

Now we can pick $c_1$ and $c_2$ to be any real numbers that satisfy the above equation for example

$c_1=c_2=\frac{1}{\sqrt{2a}}$

This gives

$g(x,s)=\begin{cases} \frac{1}{\sqrt{2a}}e^{as}\cdot e^{-ax} \text{ for } s \le x \\ \frac{1}{\sqrt{2a}}e^{ax}\cdot e^{-as} \text{ for } x \le s\end{cases}$

3. Thank-you so much for this.