---
title: Green function
tags: Applied_math_method
GA: G-77TT93X4N1
---

# Green's function

Consider the boundary value problem
$$
-T g'' = \delta(x-\xi), \quad g(x=0)=g(x=\ell)=0.
$$
The solution is denoted by $g(x|\xi)$ and is known as the Green's function for the linear operator $-T\frac{d^2}{dx^2}$ with vanishing boundary conditions at $x=0,\ell$.

## Construction of Green's function

Since $g''=0$ for $0\le x<\xi$ and $\xi<x\le \ell$, we have, after applying the boundary conditions,
$$
g(x|\xi) = Ax, \, 0\le x< \xi, \quad g(x|\xi)= C(x-\ell), \, \xi<x\le \ell.
$$

### Continuity condition
The Green's function is continuous at $x=\xi$ that gives rise to the condition
$$
A\xi = C(\xi-\ell).
$$

### Jump condition
By integrating over a small region covering the jump location, we have
$$
\int^{\xi^+}_{\xi^-}-T g''\,dx = \int^{\xi^+}_{\xi^-}\delta(x-\xi)dx.
$$
We therefore obtain
$$
g'(\xi^+) - g'(\xi^-) = -\frac{1}{T},
$$

that is, $C - A = -\frac{1}{T}$.

### Green's function

With the two conditions one can solve $A$ and $C$. Finally, we obtain the Green's function:
$$
g(x|\xi)=
\begin{cases}
\frac{\ell - \xi}{\ell T}\,x, \quad 0\le x<\xi,\\[.1cm]
\frac{\ell-x}{\ell T}\,\xi, \quad \xi<x\le \ell.
\end{cases}
$$

### Remark

There is an interesting symmetry: $g(x|\xi) = g(\xi | x)$.