Green's Function for a Partial Differential Equation

AI Thread Summary
The discussion focuses on finding the Green's function for a specific partial differential equation involving mixed derivatives. The equation is given by a combination of second derivatives with respect to time, space, and boundary conditions that dictate the behavior of the Green's function at infinity and certain limits. The Fourier transform method is employed to derive the Green's function, leading to a complex integral that ultimately does not converge due to an essential singularity. A suggestion is made to change variables to simplify the equation into a wave equation, but the original problem constraints discourage this approach. The thread highlights the challenges in solving the problem directly while adhering to the specified conditions.
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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.
 
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