Green's Function Method

The idea of Green’s function is very simple. To solve a general solution of equation

\begin{equation} \frac{d^2}{d x^2} y(x) + y(x) = f(x),\label{eqn-green-function-example} \end{equation}

where $f(x)$ is the source and some given boundary conditions. To save ink we define

\begin{equation} \hat L_x = \frac{d^2}{dx^2} + 1, \end{equation}

which takes a function $y(x)$ to $f(x)$, i.e.,

\begin{equation} \hat L_x y(x) = f(x).\label{eqn-green-function-example-1} \end{equation}

Now we define the Green’s function to be the solution of equation \ref{eqn-green-function-example-1} but replacing the source with delta function $\delta(x-z)$

\begin{equation} \hat L_x G(x,z) = \delta(z-x). \end{equation}

Why do we define this function? The solution to equation \ref{eqn-green-function-example} is given by

\begin{equation} y(x) = \int G(x,z) f(z) dz. \end{equation}

To verify this conclusion we plug it into the LHS of equation \ref{eqn-green-function-example}

\begin{align} & \left(\frac{d^2}{dx^2} +1 \right) \int G(x,z) f(z) dz \\ =& \int \left[ \left(\frac{d^2}{dx^2} +1 \right) G(x,z) \right] f(z) dz \\ =& \int \delta(z-x) f(z) dz \\ =& f(x), \end{align}

in which we used one of the properties of Dirac delta distribution

\begin{equation} \int f(z) \delta(z-x) dz = f(x). \end{equation}

Also note that delta function is even, i.e., $\delta(-x) = \delta(x)$.

So all we need to do to find the solution to a standard second differential equation

\begin{equation} \left( \frac{d^2}{dx^2} + p(x) \frac{d}{dx} + q(x) \right)y(x) = f(x) \end{equation}

is do the following.

1. Find the general form of Green’s function (GF) for operator for operator $\hat L_x$.
2. Apply boundary condition (BC) to GF. This might be the most tricky part of this method. Any ways, for a BC of the form $y(a)=0=y(b)$, we can just choose it to vanish at $a$ and $b$. Otherwise we can move this step to the end when no intuition is coming to our mind.
3. Continuity at $n-2$ order of derivatives at point $x=z$, that is

\begin{equation} G^{(n-2)}(x,z) \vert_{x<z} = G^{(n-2)} \vert _{x>z} ,\qquad \text{at } x= z. \end{equation}

1. Discontinuity of the first order derivative at $x=z$, i.e.,

   \begin{equation} G^{(n-1)}(x,z)\vert_{x>z} - G^{(n-1)}(x,z)\vert_{x<z} = 1, \qquad \text{at } x= z. \end{equation}

   This condition comes from the fact that the integral of Dirac delta distribution is Heaviside step function.

2. Solve the coefficients to get the GF.

3. The solution to an inhomogeneous ODE $y(x)=f(x)$ is given immediately by

   \begin{equation} y(x) = \int G(x,z) f(z) dz. \end{equation}

   If we haven’t done step 2 we know would have some unkown coefficients which can be determined by the BC.

So we are bound to find Green’s function. Solving a nonhonogeneous equation with delta as source is as easy as solving homogeneous equations.

We do this by demonstrating an example differential equation. The problem we are going to solve is

\begin{equation} \left(\frac{d^2}{dx^2} + \frac{1}{4}\right) y(x) = f(x), \end{equation}

with boundary condition

\begin{equation} y(0) = y(\pi) = 0. \label{eqn-green-function-example2-bc} \end{equation}

For simplicity we define

\begin{equation} \hat L_x = \frac{d^2}{dx^2} + \frac{1}{4}. \end{equation}

First of all we find the GF associated with

\begin{equation} \hat L_x G(x,z) = \delta(z-x). \end{equation}

We just follow the steps.

The general solution to

\begin{equation} \hat L_x G(x,z) = 0 \end{equation}

is given by

\begin{equation} G(x,z) = \begin{cases} A_1\cos (x/2) + B_1 \sin(x/2), & \qquad x \leq z, \\ A_2\cos (x/2) + B_2 \sin(x/2), & \qquad x \geq z. \end{cases} \end{equation}

Continuity at $x=z$ for the 0th order derivatives,

\begin{equation} G(z_-,z) = G(z_+,z), \end{equation}

which is exactly

\begin{equation} A_1\cos(z/2) + B_1 \sin(z/2) = A_2 \cos(z/2) + B_2\sin(z/2). \label{eqn-green-function-example2-continuity} \end{equation}

Discontinuity condition at 1st order derivatives,

\begin{equation} \left.\frac{d}{dx} G(x,z) \right\vert_{x=z_+} - \left.\frac{d}{dx} G(x,z) \right\vert_{x=z_-} = 1, \end{equation}

which is

\begin{equation} -\frac{A_2}{2}\sin\frac{z}{2} + \frac{B_2}{2} \cos\frac{z}{2} - \left( -\frac{A_1}{2}\sin\frac{z}{2} + \frac{B_1}{2}\cos\frac{z}{2} \right) = 1 \label{eqn-green-function-example2-discontinuity} \end{equation}

Now we combine \ref{eqn-green-function-example2-continuity} and \ref{eqn-green-function-example2-discontinuity} to eliminate two degrees of freedom. For example, we can solve out $A_1$ and $B_1$ as a function of all other coefficients. Here we have

\begin{align} B_1 =& \frac{ - 2/\sin(z/2) }{\tan(z/2) + \cot(z/2)} + B_2 , \\ A_1 =& A_2 + B_2(\tan(z/2)-1) + \nonumber \\ & \frac{2}{\sin(z/2) + \cot(z/2)\cos(z/2)}. \end{align}

Write down the form solution using $y(x) = \int G(x,z) f(z) dz$. Then we still have two unknown free coefficients $A_2$ and $B_2$, which in fact is to be determined by the BC equation \ref{eqn-green-function-example2-bc}.

The corresponding homogeneous equation is

\begin{equation} \frac{d^2}{dx^2} u(x) - u(x) = 0. \end{equation}

To find the general solution we assume it has the form

\begin{equation} u(x) = A e^{\omega x}, \end{equation}

which is then plugged back into the equation,

\begin{equation} (\omega^2 - 1) u(x) = 0. \end{equation}

We require $\omega^2-1=0$ to make the solution most general, which leads to

\begin{equation} \omega = \pm 1. \end{equation}

Finally we write down the general solution to this homogeneous equation,

$$ \begin{equation} u(x) = C e^{x} + D e^{-x}. \end{equation} $$