# Help with Greens functions

1. Jun 21, 2011

### hunt_mat

I have to solve the following PDE:
$$\frac{\partial^{2}u}{\partial t^{2}}+2\frac{\partial^{2}u}{\partial t\partial x}+\frac{\partial u}{\partial x}+\frac{\partial u}{\partial t}+k^{2}u=f$$
I use the Greens function method and examine the equation:
$$\frac{\partial^{2}G}{\partial t^{2}}+2\frac{\partial^{2}G}{\partial t\partial x}+\frac{\partial G}{\partial x}+\frac{\partial G}{\partial t}+k^{2}G=4\pi\delta (x-x')\delta (t-t')$$
I then write:
$$G=\frac{1}{2\pi}\int_{-\infty}^{\infty}g(X|X')e^{i\omega (T-T')}d\omega\quad \delta (T-T') =\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega (T-T')}d\omega$$
The equation then becomes:
$$(1+2\omega i)\frac{\partial g}{\partial X}+(k^{2}-\omega^{2}+i\omega)g=4\pi\delta (X-X')$$
Take the Fourier transform to obtain:
$$i\xi (1+2\omega i)\hat{g}+(k^{2}-\omega^{2}-\omega i)\hat{g}=4\pi e^{i\xi X'}$$
Rearrange and take the inverse Fourier transform to obtain:
$$g=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{4\pi e^{-i\xi (X-X')}}{i\xi (1+2\omega i)+k^{2}-\omega^{2}-\omega i}d\xi$$
Am I on the right track here?

2. Jun 21, 2011

### hunt_mat

I think that I can find g by using contour integration. Write:
$$\frac{1}{2\pi}\oint_{\gamma}\frac{4\pi e^{-iz(X-X')}}{iz(1+2\omega i)+k^{2}-\omega^{2}-\omega i}dz$$
Which can then be evaluated via Cauchy's integral formula:
$$g=4\pi ie^{i(X-X')h(\omega )},\quad h(\omega )=\frac{k^{2}-\omega^{2}-\omega i}{1+2\omega i}$$

Thoughts?