- #1
arpon
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Homework Statement
Find out the Green's function, ##G(\vec{r}, \vec{r}')##, for the following partial differential equation:
$$\left(-2\frac{\partial ^2}{\partial t \partial x} + \frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} \right) F(\vec{r}) = g(\vec{r})$$
Here ##\vec{r} = (t,x,y,z)## and ##\vec{r}'=(t',x',y',z')##
The boundary conditions are:
i) When ##|(x-x') + (t-t')| < 0##, ##~~G(\vec{r}, \vec{r}') = 0##
ii) When ##|(x-x') - (t-t')| \rightarrow \infty##, ##~~G(\vec{r}, \vec{r}') = 0##
iii) When ##|y-y'| \rightarrow \infty##, ##~~G(\vec{r}, \vec{r}') = 0##
iv) When ##|z-z'| \rightarrow \infty##, ##~~G(\vec{r}, \vec{r}') = 0##
2. Homework Equations
Fourier transformation:
$$\tilde{G}(\vec{k}) = \int d^4 R~~G(\vec{R}) e^{-i \vec{k} \cdot \vec{R}}$$
Inverse Fourier transformation:
$$G(\vec{R}) = \frac{1}{(2\pi)^4} \int d^4 k~~\tilde{G}(\vec{k}) e^{i \vec{k} \cdot \vec{R}}$$
Here ##\vec{k} = (k_0,k_1,k_2,k_3)##
The Attempt at a Solution
The Green's function satisfies this equation:
$$\left(-2\frac{\partial ^2}{\partial t \partial x} + \frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} \right) G(\vec{r}, \vec{r}') = \delta(\vec{r} - \vec{r}')$$
Let ##\vec{R} = \vec{r} - \vec{r}' = (T,X,Y,Z)##
The boundary conditions and the differential operator of the PDE are translation invariant. So ##G## is a function of ##\vec{r} - \vec{r}'##
Now we have
$$\left(-2\frac{\partial ^2}{\partial T \partial X} + \frac{\partial^2}{\partial Y^2} +\frac{\partial^2}{\partial Z^2} \right) G(\vec{R}) = \delta(\vec{R})$$
Fourier transform:
$$(2k_0k_1 -k_2^2 - k_3 ^2)~\tilde{G}(\vec{k}) = 1$$
$$\implies \tilde{G}(\vec{k}) = \frac{1}{2k_0k_1 -k_2^2 - k_3 ^2}$$
Now inverse Fourier transform:
$$G(\vec{R}) = \frac{1}{(2\pi)^4} \int d^4 k~~\frac{e^{i \vec{k} \cdot \vec{R}}}{2k_0k_1 -k_2^2 - k_3 ^2} $$
Consider the integral over ##k_0##. Apart from some constants, this is:
$$\int^{\infty}_{-\infty} \frac{e^{ik_0T}}{2k_0k_1 -k_2^2 - k_3 ^2} ~dk_0$$
$$= \frac{1}{2k_1} \int^{\infty}_{-\infty} \frac{e^{ik_0T}}{k_0 - \frac{k_2^2 + k_3 ^2}{2k_1}} ~dk_0$$
Considering these contour integral and the definition of Cauchy Principal Value,
we get:
$$sgn(T)~~\frac{ \pi i}{2k_1} e^{iT\frac{k_2^2+k_3^2}{2k_1}}$$
Now let us consider the integral over ##k_1##. Apart from some constants we get:
$$\int^{\infty}_{-\infty} dk_1 \frac{e^{i(Xk_1+T\frac{k_2^2+k_3^2}{2k_1})}}{k_1}$$
There is an essential singularity at ##k_1 = 0##. The integral does not converge.
Any help would be appreciated.