# Chapter 12: Green's Function

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The material covered in this chapter is partially presented in Boas Chapter 8, Section 12.

## 12.1 Introduction

Poisson's equation

\begin{equation} \nabla^2 V = -f, \tag{12.1} \end{equation}

where $f$ is a prescribed function of position $\boldsymbol{r}$, is a generalization of Laplace's equation considered in Chapter 10. The equation is relevant in many areas of physics, including electrostatics, where $f$ is related to the charge density $\rho$ by $f = \rho/\epsilon_0$. We would like to find solutions to this equation, but the presence of a source term on the right-hand side creates a substantial difficulty. A little thought will convince you, for example, that an approach based on factorized solutions will fail. And if we happened to find a successful approach to solve Eq.(12.1) for one particular $f$, chances are that the approach would fail as soon as a different $f$ is inserted in the equation. What are we to do?

The answer is that we follow the British mathematician George Green (1793--1841), and consider a specific case of Eq.(12.1) in which the arbitrary $f$ is replaced by the three-dimensional delta function $\delta(\boldsymbol{r}-\boldsymbol{r'})$, where $\boldsymbol{r'}$ is an arbitrary reference position. The solution to this version of Poisson's equation will depend on $\boldsymbol{r}$ and $\boldsymbol{r'}$, and to indicate this we denote it by $G(\boldsymbol{r},\boldsymbol{r'})$. This shall be called a Green's function, and it shall be a solution to Green's equation,

\begin{equation} \nabla^2 G(\boldsymbol{r},\boldsymbol{r'}) = -\delta(\boldsymbol{r}-\boldsymbol{r'}). \tag{12.2} \end{equation}

The good news here is that since the delta function is zero everywhere except at $\boldsymbol{r} = \boldsymbol{r'}$, Green's equation is everywhere the same as Laplace's equation, except at $\boldsymbol{r} = \boldsymbol{r'}$. This observation suggests that we should be able to rely on the techniques presented in Chapter 10 to find solutions to Eq.(12.2). A way forward therefore presents itself.

But why should we be interested in the special case of Green's equation when we really want to solve Eq.(12.1) for a very different $f$? The answer is that if we know $G(\boldsymbol{r},\boldsymbol{r'})$, we can obtain $V$ for any $f$ from simple operations. The link is provided by

\begin{equation} V(\boldsymbol{r}) =\int G(\boldsymbol{r},\boldsymbol{r'}) f(\boldsymbol{r'})\, dV', \tag{12.3} \end{equation}

in which $f$ is expressed as a function of the variables $(x', y', z')$ contained in $\boldsymbol{r'}$, and $dV' := dx' dy' dz'$ is the volume element associated with these variables. It is easy to see that the $V$ of Eq.(12.3) is a solution to Poisson's equation. Acting with the Laplacian operator, we have that

\begin{align} \nabla^2 V &= \nabla^2 \int G(\boldsymbol{r},\boldsymbol{r'}) f(\boldsymbol{r'})\, dV' \nonumber \\ &= \int \bigl[ \nabla^2 G(\boldsymbol{r},\boldsymbol{r'}) \bigr] f(\boldsymbol{r'})\, dV' \nonumber \\ &= \int \bigl[ -\delta(\boldsymbol{r}-\boldsymbol{r'}) \bigr] f(\boldsymbol{r'})\, dV' \nonumber \\ &= -f(\boldsymbol{r}), \tag{12.4} \end{align}

and we find that indeed, $V$ satisfies Eq.(12.1). In the second step we observed that since $\nabla^2$ contains derivatives with respect to $(x, y, z)$, the variables contained in $\boldsymbol{r}$, it does not act on $f(\boldsymbol{r'})$, which depends on the independent set of variables $(x', y', z')$. In the third step we used Eq.(12.2), and in the last step we imported the defining property of the delta function, as expressed by Eq.(6.19).

The miracle of Green's method is that once $G(\boldsymbol{r},\boldsymbol{r'})$ is known, the solution to Poisson's equation for any $f$ can be obtained at the mere cost of evaluating a three-dimensional integral. And since Green's equation is the same as Laplace's equation except at $\boldsymbol{r} = \boldsymbol{r'}$, the task of solving it should be relatively straightforward.

The power of Green's method was illustrated here in the context of Poisson's equation. The method, however, can easily be adapted to other contexts, and it has become one of the most powerful tools of mathematical physics. We shall run through a number of examples in the following sections. The solution to Eq.(12.2) will be revealed in Sec.12.4.

## 12.2 One-Dimensional Poisson Equation

To see how this works in practice, we begin with the simple case of the one-dimensional Poisson equation,

\begin{equation} \frac{d^2 V}{dx^2} = -f, \tag{12.5} \end{equation}

with $V$ and $f$ depending on $x$ only. To get a unique solution to Eq.(12.5) we impose the boundary conditions

\begin{equation} V(x=0) = 0, \qquad V(x=1) = 0. \tag{12.6} \end{equation}

These are chosen for simplicity; the calculation can easily be adapted to allow for more general conditions.

To solve Eq.(12.5) we look for a Green's function $G(x,x')$ that satisfies the one-dimensional version of Green's equation,

\begin{equation} \frac{\partial^2}{\partial x^2} G(x,x') = -\delta(x-x'), \tag{12.7} \end{equation}

together with the same boundary conditions, $G(0,x') = 0 = G(1,x')$. We take $x'$ to be somewhere between $0$ and $1$. Then the solution to Eq.(12.5) for any $f$ shall be

\begin{equation} V(x) = \int_0^1 G(x,x') f(x')\, dx', \tag{12.8} \end{equation}

as can be verified by inserting the integral within Eq.(12.5) and using Eq.(12.7). It can also be verified that the $V$ of Eq.(12.8) satisfies the boundary conditions of Eq.(12.6).

Exercise 12.1: Check that the $V$ of Eq.(12.8) satisfies the one-dimensional Poisson equation and has the correct boundary values.

To find $G(x,x')$ we note first that when $x < x'$ or $x > x'$, the Green's function satisfies $\partial^2 G/\partial x^2 = 0$, the one-dimensional Laplace equation. This implies that $G(x,x') = G_<(x,x') = a_< x + b_<$ when $x < x'$, and that $G(x,x') = G_>(x,x') = a_> x + b_>$ when $x > x'$, where $a_<$, $b_<$, $a_>$, and $b_>$ are constants. The Green's function is required to satisfy boundary conditions at $x=0$ and $x=1$, and these determine some of the constants. It must vanish at $x = 0$, where $x$ is smaller than $x'$, and this implies that $G_<(0, x') = b_< = 0$. It must also vanish at $x = 1$, where $x$ is larger than $x'$; this implies that $G_>(1,x') = a_> + b_> = 0$, so that $b_> = -a_>$. At this stage we have obtained

\begin{equation} G(x,x') = \left\{ \begin{array}{ll} G_<(x,x') = a_< x & \quad x < x' \\ G_>(x,x') = a_> (x-1) & \quad x > x' \end{array} \right. ,\tag{12.9} \end{equation}

and we have two constants left to determine.

We demand that the Green's function be continuous at $x = x'$, so that $G_<(x',x')$ should agree with $G_>(x',x')$. From this we obtain $a_< x' = a_> (x'-1)$. To implement this condition we write $a_< = c\, (x' - 1)$ and $a_> = c\, x'$, where $c$ is another constant. The Green's function becomes

\begin{equation} G(x,x') = \left\{ \begin{array}{ll} G_<(x,x') = c\, (x'-1) x & \quad x < x' \\ G_>(x,x') = c\, x'(x-1) & \quad x > x' \end{array} \right. , \tag{12.10} \end{equation}

and we have one final constant to determine.

Equation (12.7) implies that the first derivative of the Green's function must be discontinuous at $x=x'$. To see this, we integrate the equation with respect to $x$, from $x' - \epsilon$ to $x' + \epsilon$, where $\epsilon$ is some positive number. We write

\begin{equation} \int_{x'-\epsilon}^{x'+\epsilon} \frac{\partial^2 G}{\partial x^2}\, dx = -\int_{x'-\epsilon}^{x'+\epsilon} \delta(x-x')\, dx, \tag{12.11} \end{equation}

and get

\begin{equation} \frac{\partial G}{\partial x} \biggl|^{x'+\epsilon}_{x'-\epsilon} = -1 \tag{12.12} \end{equation}

after integration. In the limit $\epsilon \to 0$ we obtain

\begin{equation} \frac{\partial G_>}{\partial x}(x',x') - \frac{\partial G_<}{\partial x}(x',x') = -1, \tag{12.13} \end{equation}