Green's Functions Explained

General Idea

Suppose you want to find $f(x)$ that satisfies some linear, differential equation

$L\,f(x) = u(x),$

where $L$ is a linear differential operator and $u(x)$ some given function (the inhomogeneous term of the differential equation). The Green's function $G(x, x_0)$ can help you find the solution. $G$ is a function of two variables, and it satisfies the equation

$L\,G(x, x_0) = \delta(x - x_0).$

If you happen find $G(x, x_0)$, then you can multiply this equation by $u(x_0)$ and integrate over $x_0$:

$\int dx_0\,L\,G(x, x_0) u(x_0) = \int dx_0\,\delta(x-x_0)u(x_0).$

Rearrange the terms and you end up with

$L \int dx_0 G(x, x_0) u(x_0) = u(x).$

This solves your initial problem, because you can identify

$f(x) = \int dx_0 G(x, x_0) u(x_0).$

In the real world, you often deal with differential equations that have boundary conditions, for instance the electric potential must be 0 on the boundary. Green's functions are a great tool for these problems, because if the Green's function is 0 on the boundary, then any integral over Green's functions will also be 0 on the boundary. Conceptually, this is very similar to finding the image charge for any single charge within your boundary.

Completely Useless Example

Typical problem in electrostatics: Point charge in vacuum. (I left out all units). I'm including this because I first made a mistake when assigning a name to the location of the point charge. You must not name this $x_0$ and then integrate over $x_0$!

$\Delta \Phi(x) = \rho(x) = -\frac{1}{4\pi} \delta(x - x_p)$

Everyone "knows" that the solution is

$\Phi(x) = \frac{1}{\|x - x_p\|}$.

To find it with Green's function: Solve

$\Delta G(x,x_0) = \delta(x - x_0)$

Obviously, we know

$G(x,x_0) = \frac{1}{\|x - x_0\|}.$

To find the solution, calculate

$\Phi(x) = \int dx_0 G(x, x_0) \rho(x_0)\\ = \int dx_0 \delta(x - x_0) \frac{-1}{4\pi} \delta(x_0 - x_p)\\ = \frac{1}{\|x - x_p\|}$

Simple, useful example: Undamped, harmonic oscillator

Consider the harmonic oscillator forced by a function $u(x)$:

$f''(x) + \omega^2 f(x) = u(x)\\ \Rightarrow L = \frac{d^2}{dx^2} + \omega^2$

So we are looking for a function $G(x, x_0)$ that satisfies

$L\,G(x,x_0) = \delta(x-x_0)\quad.$

The corresponding physics case to this equation is a harmonic oscillator that is "bumped" at $x = x_0$. One solution that should work is

$G(x, x_0) = \begin{cases} 0 &\text{if } x < x_0 \\ A \sin(\omega x - \omega x_0) &\text{if } x \geq x_0 \end{cases}$

Or, using the Heaviside step function $\Theta(x)$, which I find easier to handle when doing derivatives:

$G(x,x_0) = A\,\Theta(x-x_0)\, \sin(\omega x - \omega x_0)$

We can prove that this is the right equation by showing that it holds true for $x < x_0$ (trivial), for $x > x_0$ (almost as trivial), and for:

$\int_{x_0 - a}^{x_0+a}\! G(x, x_0)\,dx = 1 \quad \forall \, a \quad .$

The last part is the only one that is a bit tricky, but you can quickly work out that the second derivative of $\Theta(x) \sin(x)$ is basically $\delta(x)$. In our case, the factor $A$ works out to $A = 1/\omega$.

It is quite easy to go to the damped harmonic oscillator from here. Just ask yourself "What does my system do when I poke it at $x = x_0$ with a $\delta$-pulse?", use that as a guess for $G(x,x_0)$, and find the factor $A$ of the solution.