Green's Function for a Partial Differential Equation

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SUMMARY

The forum discussion focuses on finding the Green's function, ##G(\vec{r}, \vec{r}')##, for the partial differential equation defined by the operator $$\left(-2\frac{\partial ^2}{\partial t \partial x} + \frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} \right) F(\vec{r}) = g(\vec{r})$$. The boundary conditions specify that ##G(\vec{r}, \vec{r}') = 0## under certain conditions of the variables. The solution involves applying Fourier transformation techniques, leading to the expression for the Fourier transform $$\tilde{G}(\vec{k}) = \frac{1}{2k_0k_1 - k_2^2 - k_3^2}$$ and the inverse Fourier transform for ##G(\vec{R})##. The discussion highlights the challenges in evaluating integrals, particularly due to singularities in the variable ##k_1##.

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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,

drawing.png

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.
 

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May I suggest a change of variables to ##\tau = (x+t)/\sqrt 2## and ##\xi = (x-t)/\sqrt 2##?
 
Orodruin said:
May I suggest a change of variables to ##\tau = (x+t)/\sqrt 2## and ##\xi = (x-t)/\sqrt 2##?
This will transform the PDE into a wave equation. But this exercise asks to solve this problem not using this coordinate transformation.
Thanks for your suggestion anyway.
 
Last edited:

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