# Green's Function for a Partial Differential Equation

• arpon
In summary, the conversation discusses finding the Green's function for a specific partial differential equation. The boundary conditions and equations for Fourier and inverse Fourier transformations are also provided. A change of variables is suggested, but the exercise requires solving the problem without this transformation.
arpon

## 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.

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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.

Last edited:

## 1. What is a Green's function of a PDE?

A Green's function of a PDE (partial differential equation) is a mathematical concept used to solve certain types of PDEs. It is a type of integral transform that maps a linear differential operator into an integral operator.

## 2. How is a Green's function used to solve a PDE?

A Green's function is used to solve a PDE by representing the solution as a linear combination of the Green's function and the boundary conditions. This allows for the PDE to be transformed into an integral equation, which can then be solved to find the solution.

## 3. What are the properties of a Green's function?

A Green's function has several important properties, including linearity, symmetry, and orthogonality. It also satisfies the boundary conditions of the PDE and is unique for a given PDE.

## 4. What types of PDEs can be solved using a Green's function?

A Green's function can be used to solve linear PDEs with constant coefficients, such as the heat equation, wave equation, and Laplace's equation. It can also be used for nonhomogeneous PDEs with variable coefficients, but the calculations may be more complex.

## 5. How does a Green's function relate to the fundamental solution of a PDE?

The fundamental solution of a PDE is the solution for a point source at the origin with zero initial conditions. A Green's function is closely related to the fundamental solution as it represents the response of the system to a point source at any location. In other words, the Green's function is the fundamental solution for a PDE with non-zero boundary conditions.

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