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}

the statement that the discontinuity in the first derivative of the Green's function at $x=x'$ is equal to $-1$. From Eq.(12.10) we find that $\partial G_>/\partial x = c\, x'$ and $\partial G_

Our final expression for the Green's function is

\begin{equation} G(x,x') = \left\{ \begin{array}{ll} G_<(x,x') = x(1-x') & \quad x < x' \\ G_>(x,x') = x'(1-x) & \quad x > x' \end{array} \right. . \tag{12.14} \end{equation}

The Green's function is a straight line with positive slope $1-x'$ when $x < x'$, and another straight line with negative slope $-x'$ when $x > x'$.

Exercise 12.2: With the notation $x_< := \text{min}(x,x')$ and $x_> := \text{max}(x,x')$, show that the Green's function can be written as $G(x,x') = x_<(1-x_>)$.

We may now insert the Green's function within Eq.(12.8). To account for the change of identity of $G(x,x')$ at $x = x'$, we break the integration domain into two pieces, one going from $x' = 0$ to $x' = x$, the other going from $x' = x$ to $x' = 1$. We write

\begin{equation} V(x) = \int_0^{x} G(x,x') f(x')\, dx' + \int_{x}^1 G(x,x') f(x')\, dx', \tag{12.15} \end{equation}

and note that since $x'$ is smaller than $x$ in the first interval, $G(x,x') = G_>(x,x')$ in the first integral, while $G(x,x') = G_<(x,x')$ in the second integral. We therefore have

\begin{equation} V(x) = \int_0^{x} x'(1-x) f(x')\, dx' + \int_{x}^1 x(1-x') f(x')\, dx', \tag{12.16} \end{equation}

and pulling constant factors out of the integrals, we obtain

\begin{equation} V(x) = (1-x) \int_0^{x} x' f(x')\, dx' + x \int_{x}^1 (1-x') f(x')\, dx' \tag{12.17} \end{equation}

for the final solution to Eq.(12.5) with the boundary conditions of Eq.(12.6). The solution applies to any source function $f$, and the integrals can be evaluated as soon as this function is specified.

As an example, we might consider a source term given by $f(x) = \sin x$. Inserting this within Eq.(12.17) and evaluating the integrals returns $V(x) = \sin x - x \sin 1$.

Exercise 12.3: Verify that the solution to $d^2V/dx^2 = -\sin x$ with boundary conditions $V(0) = 0$ and $V(1) = 0$ is given by $V(x) = \sin x - x\sin 1$. First show that this is the solution returned by Eq.(12.17). Then check explicitly that this $V(x)$ satisfies the differential equation and the boundary conditions.

Exercise 12.4: Generalize Eq.(12.17) so that $V(x)$ satisfies the modified boundary conditions $V(0) = V_0$ and $V(1) = V_1$. Do this by reconstructing the Green's function so that it enforces the new boundary conditions.

12.3: Driven Harmonic Oscillator

Green's method is not restricted to the Poisson equation. As a second example we examine a harmonic oscillator of natural frequency $\omega$ driven by an external force $f(t)$. We assume that the force begins to act at $t = 0$, when the oscillator was in a state of rest. The differential equation that governs the motion of this oscillator is

\begin{equation} \frac{d^2 X}{dt^2} + \omega^2 X = f, \tag{12.18} \end{equation}

with $X$ measuring the oscillator's displacement from its equilibrium position. The initial conditions are

\begin{equation} X(0) = 0, \qquad \frac{dX}{dt}(0) = 0. \tag{12.19} \end{equation}

We wish to find the solution to Eq.(12.18) for any force $f$.

We need a Green's function for Eq.(12.18), and we adapt the general strategy detailed previously by replacing the arbitrary $f(t)$ by the specific $\delta(t-t')$, where $t'$ is a reference time. Green's equation becomes

\begin{equation} \frac{\partial^2}{\partial t^2} G(t,t') + \omega^2 G(t,t') = \delta(t-t'), \tag{12.20} \end{equation}

and we wish to solve it with the initial conditions

\begin{equation} G(0,t') = \frac{\partial G}{\partial t}(0,t') = 0. \tag{12.21} \end{equation}

With $G(t,t')$ identified, the solution to Eq.(12.8) with the initial conditions of Eq.(12.19) is

\begin{equation} X(t) = \int_0^\infty G(t,t') f(t')\, dt'. \tag{12.22} \end{equation}

Exercise 12.5: Verify that the $X(t)$ of Eq.(12.22) is the solution to Eq.(12.18) with the initial conditions of Eq.(12.19).

When $t < t'$ or $t > t'$, the Green's function satisfies the homogeneous equation $\partial^2 G/\partial t^2 + \omega^2 G = 0$, and the solution to this is a superposition of sine and cosine functions. We therefore write

\begin{equation} G(t,t') = \left\{ \begin{array}{ll} G_<(t,t') = a_< \cos(\omega t) + b_< \sin(\omega t) & \quad t < t' \\ G_>(t,t') = a_> \cos(\omega t) + b_> \sin(\omega t) & \quad t > t' \end{array} \right. \tag{12.23} \end{equation}

where $a_<$, $b_<$, $a_>$, and $b_>$ are constants. The initial conditions are imposed at $t = 0$, so that $t < t'$ and $G(t,t') = G_<(t,t')$. They imply that $a_< = 0$ and $b_< = 0$, so that $G_<(t,t') = 0$. At this stage we have that

\begin{equation} G(t,t') = \left\{ \begin{array}{ll} G_<(t,t') = 0 & \quad t < t' \\ G_>(t,t') = a_> \cos(\omega t) + b_> \sin(\omega t) & \quad t > t' \end{array} \right. \tag{12.24} \end{equation}

and we still have $a_>$ and $b_>$ left to determine.

Continuity of the Green's function at $t = t'$ provides the condition

\begin{equation} 0 = G_>(t',t') = a_> \cos(\omega t') + b_> \sin(\omega t'), \tag{12.25} \end{equation}

and we solve this for $a_>$ and $b_>$ by writing $a_> = -c\, \sin(\omega t')$ and $b_> = c\, \cos(\omega t')$, where $c$ is another constant. Making the substitution in $G_>(t,t')$ yields

\begin{equation} G_>(t,t') = c\, \bigl[ \sin(\omega t) \cos(\omega t') - \cos(\omega t) \sin(\omega t') \bigr] = c\, \sin[ \omega(t-t') ]; \tag{12.26} \end{equation}

the last form makes it clear that $G_>(t,t')$ goes to zero at $t=t'$. To find $c$ we must account for the delta function in Green's equation, which gives rise to a discontinuity in the first derivative of $G(t,t')$ at $t = t'$. We can derive this condition as we did previously, by integrating Green's equation from $t = t' - \epsilon$ to $t = t' + \epsilon$. We have

\begin{equation} \int_{t'-\epsilon}^{t'+\epsilon} \frac{\partial^2 G}{\partial t^2}\, dt + \omega^2 \int_{t'-\epsilon}^{t'+\epsilon} G\, dt = \int_{t'-\epsilon}^{t'+\epsilon} \delta(t - t')\, dt, \tag{12.27} \end{equation}

and we see that we can deal easily with the right-hand side as well as the first term on the left-hand side, but that the second term poses more of a challenge. To deal with it we recall that we are interested in the limit $\epsilon \to 0$, and that the Green's function is continuous at $t = t'$. This implies that the second term can be approximated by $2\epsilon \omega^2 G(t',t')$, and that this contribution to the equation vanishes when $\epsilon \to 0$. In the end we are left with

\begin{equation} \frac{\partial G_>}{\partial t}(t',t') - \frac{\partial G_<}{\partial t}(t',t') = 1. \tag{12.28} \end{equation}

Examining this discontinuity condition, we see that the second term vanishes, that the first evaluates to $c\, \omega$, and we arrive at $c = \omega^{-1}$.

Our final expression for the Green's function is

\begin{equation} G(t,t') = \left\{ \begin{array}{ll} G_<(t,t') = 0 & \quad t < t' \\ G_>(t,t') = \omega^{-1} \sin[ \omega(t-t') ] & \quad t > t' \end{array} \right. . \tag{12.29} \end{equation}

Substitution within Eq.(12.22) gives

\begin{align} X(t) &= \int_0^\infty G(t,t') f(t')\, dt' \nonumber \\ &= \int_0^{t} G(t,t') f(t')\, dt' + \int_t^\infty G(t,t') f(t')\, dt' \nonumber \\ &= \int_0^{t} G_>(t,t') f(t')\, dt' + \int_t^\infty G_<(t,t') f(t')\, dt' , \tag{12.30} \end{align}


\begin{equation} X(t) = \frac{1}{\omega} \int_0^t \sin[ \omega(t-t')] f(t')\, dt'. \tag{12.31} \end{equation}

This is the solution to Eq.(12.18) with the initial conditions of Eq.(12.19). The solution applies to any forcing function $f$.

As an example we may choose a forcing function that grows linearly with time. We write $f(t) = f_0\, \omega t$, where $f_0$ is a constant, and substitute this within Eq.(12.31). A quick calculation returns $X(t) = f_0\, \omega^{-2} [\omega t - \sin(\omega t)]$.

Exercise 12.6: Verify that the solution to $d^2 X/dt^2 + \omega^2 X = f_0\, \omega t$ with boundary conditions $X(0) = 0$ and $(dX/dt)(0) = 0$ is given by $X(t) = f_0\, \omega^{-2} [\omega t - \sin(\omega t)]$. First show that this is the solution returned by Eq.(12.31). Then check explicitly that this $X(t)$ satisfies the differential equation and the boundary conditions.

12.4 Three-Dimensional Poisson Equation

We may now return to the discussion initiated in Sec.12.1 and ask: what is the Green's function for the three-dimensional Poisson equation? In general the answer to this question is complicated, and it depends on the boundary conditions that we wish to impose on $V$. To keep the discussion simple we shall assume that there are no boundaries, and therefore no need to impose boundary conditions. We merely assume that the potential vanishes at infinity.

To recapitulate, we have Poisson's equation $\nabla^2 V = -f$, Green's equation

\begin{equation} \nabla^2 G(\boldsymbol{r},\boldsymbol{r'}) = -\delta(\boldsymbol{r}-\boldsymbol{r'}), \tag{12.32} \end{equation}

and the formal solution to Poisson's equation,

\begin{equation} V(\boldsymbol{r}) = \int G(\boldsymbol{r},\boldsymbol{r'}) f(\boldsymbol{r'})\, dV'. \tag{12.33} \end{equation}

We want the Green's function to vanish at infinity, so that the potential will behave in the same way.

You might expect that to get the Green's function would require a lot of detailed work. In fact there is no need: we've done the work before. If you refer back to Sec.6.6 and consult Eq.(6.38), you will recall the identity

\begin{equation} \nabla^2 \frac{1}{|\boldsymbol{r}-\boldsymbol{r'}|}= -4\pi \delta(\boldsymbol{r}-\boldsymbol{r'}), \tag{12.34} \end{equation}

which is identical to Green's equation except for the numerical factor on the right-hand side. This, without further ado, allows us to identify

\begin{equation} G(\boldsymbol{r},\boldsymbol{r'}) = \frac{1}{4\pi} \frac{1}{|\boldsymbol{r}-\boldsymbol{r'}|} \tag{12.35} \end{equation}

as the Green's function for Poisson's equation in the absence of boundaries. As required, $G(\boldsymbol{r},\boldsymbol{r'})$ vanishes at infinity.

Inserting the Green's function within the formal solution for $V$ returns

\begin{equation} V(\boldsymbol{r}) = \frac{1}{4\pi} \int \frac{f(\boldsymbol{r'})}{|\boldsymbol{r}-\boldsymbol{r'}|}\, dV', \tag{12.36} \end{equation}

an expression that should be familiar from your course on electrostatics. It should be even more recognizable when $f$ is written as $\rho/\epsilon_0$, so that Eq.(12.36) is the solution to $\nabla^2 V = -\rho/\epsilon_0$.

These results take a familiar turn, and they can be given a compelling physical interpretation. Forgetting about the annoying factors of $4\pi\epsilon_0$, Green's equation is identical to Poisson's equation for a point charge of unit strength ($q=1$) situated at position $\boldsymbol{r'}$. Its solution, the Green's function $G(\boldsymbol{r},\boldsymbol{r'})$, is then the potential of this point charge, and we certainly recognize it in Eq.(12.35). To obtain the potential $V$ of a distribution of charge density $f$, we invoke the superposition principle and add up contributions $f(\boldsymbol{r'}) dV'/|\boldsymbol{r}-\boldsymbol{r'}|$ from all the elements of charge in the distribution. It is this idea that is expressed mathematically by Eq.(12.36).

So it turns out that the Green's function was a familiar entity all this time!

12.5 Wave Equation

We saw back in Chapter 11 that the propagation of a dispersionless wave is governed by the wave equation $\Box \psi = 0$. The wave, however, is generated by some kind of physical process, and the wave equation should be modified to account for this. In the same way that Laplace's equation $\nabla^2 V = 0$ becomes Poisson's equation $\nabla^2 V = -f$ to account for the presence of a source, the wave equation becomes

\begin{equation} \Box \psi = -f, \tag{12.37} \end{equation}

with $f(t,\boldsymbol{r})$ representing the source of the wave, imagined to be contained in some bounded region of space. We recall that

\begin{equation} \Box := -\frac{1}{v^2} \frac{\partial^2}{\partial t^2} + \nabla^2 \tag{12.38} \end{equation}

is the wave (d'Alembertian) operator. The wave equation reduces to its previous form, $\Box \psi = 0$, outside the region occupied by the source. An example of Eq.(12.37) is provided by an electromagnetic wave generated by a time-dependent distribution of charge; in this case $\psi$ is the scalar potential $V$, and $f = \rho/\epsilon_0$, where $\rho$ is the time-dependent charge density. Another example involves a sound wave generated by a loud speaker.

We wish to integrate Eq.(12.37) for any source function $f$, and for this we require a Green's function. In the case of Poisson's equation, which involves spatial variables only, the Green's function was seen to depend on two positions, $\boldsymbol{r}$ and $\boldsymbol{r'}$. But the wave equation involves time in addition to spatial variables, and we should expect that in this case, the Green's function will depend on two times, $t$ and $t'$, in addition to the two positions $\boldsymbol{r}$ and $\boldsymbol{r'}$. We also saw that Green's equation is obtained by replacing the generic source term by a delta function. Because the wave equation involves time, the replacement must also include a delta function in the time variable.

We are therefore looking for a Green's function $G(t,\boldsymbol{r};t',\boldsymbol{r'})$ that satisfies

\begin{equation} \Box G(t,\boldsymbol{r};t',\boldsymbol{r'}) = -\delta(t-t') \delta(\boldsymbol{r}-\boldsymbol{r'}). \tag{12.39} \end{equation}

With this object at our disposal, the solution to Eq.(12.37) is

\begin{equation} \psi(t,\boldsymbol{r}) = \int G(t,\boldsymbol{r};t',\boldsymbol{r'}) f(t',\boldsymbol{r'})\, dt' dV'. \tag{12.40} \end{equation}

In this equation $f$ is expressed as a function of $t'$ and $\boldsymbol{r'}$, and integration is carried out over all time and space. Equation (12.39) is the specific form of Green's equation that is associated with $\Box \psi = -f$, and that Eq.(12.40) is a solution to the wave equation can be verified by direct substitution.

Exercise 12.7: Verify that Eq.(12.40) is a solution to Eq.(12.37).

To calculate $G(t,\boldsymbol{r};t',\boldsymbol{r'})$ we take advantage of the fact that the origin of time and the origin of the spatial coordinates can be chosen arbitrarily. For the purpose of solving Eq.(12.39), it turns out to be convenient to place the origin of time at $t'$, and the origin of the spatial coordinates at $\boldsymbol{r'}$. This amounts to setting $t' = 0$ and $\boldsymbol{r'} = \boldsymbol{0}$ in Green's equation, which allows us to find $G(t,\boldsymbol{r}) := G(t,\boldsymbol{r}; 0, \boldsymbol{0})$. This choice will no longer be convenient when it is time to insert the Green's function in Eq.(12.40), but it will be easy, at this point, to restore the dependence on $t'$ and $\boldsymbol{r'}$. We shall have to remember that $t$ measures time from $t'$, and that $\boldsymbol{r}$ measures a displacement from $\boldsymbol{r'}$.

We wish to solve

\begin{equation} \Box G(t,\boldsymbol{r}) = -\delta(t) \delta(\boldsymbol{r}) \tag{12.41} \end{equation}

for the Green's function $G(t,\boldsymbol{r})$. The equation indicates that physically, $G(t,\boldsymbol{r})$ represents a wave created by a sudden and momentary disturbance localized at $\boldsymbol{r} = \boldsymbol{0}$. We would expect such a disturbance to create a wave that travels outward with speed $v$. And because the disturbance is localized at a point at the origin, we expect that this wave will be spherically symmetric, so that $G(t,\boldsymbol{r})$ will depend on $t$ and $r$, but will be independent of $\theta$ and $\phi$. So we are dealing with the kind of spherical wave that we studied back in Sec.11.9, and Eq.(11.91) suggests that our Green's function should take the form

\begin{equation} G(t,\boldsymbol{r}) = \frac{1}{r} A(r-vt), \tag{12.42} \end{equation}

where $A$ is a function that remains to be determined.

To find $A$ we integrate Eq.(12.41) over a three-dimensional sphere of radius $R$ centred at the origin of the coordinate system. We denote this region $V$, and the surface of the sphere is denoted $S$. Working as we are in spherical coordinates, the volume element is $dV = r^2\sin\theta\, drd\theta d\phi$, and the area element on the surface of the sphere is $d\boldsymbol{a} = \boldsymbol{\hat{r}} R^2\sin\theta\, d\theta d\phi$. Integration gets rid of $\delta(\boldsymbol{r})$, and we obtain

\begin{equation} \int_V \Box G\, dV = -\delta(t), \tag{12.43} \end{equation}


\begin{equation} \frac{1}{v^2} \int_V \frac{\partial^2 G}{\partial t^2}\, dV - \int_V \nabla^2 G\, dV = \delta(t). \tag{12.44} \end{equation}

We can re-express the second integral as a surface integral by exploiting the divergence theorem. We have that $\nabla^2 G = \boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} G)$, so that $\int \nabla^2 G\, dV = \oint \boldsymbol{\nabla} G \cdot d\boldsymbol{a}$, and the equation becomes

\begin{equation} \frac{1}{v^2} \int_V \frac{\partial^2 G}{\partial t^2}\, dV - \oint_S \boldsymbol{\nabla} G \cdot d\boldsymbol{a} = \delta(t). \tag{12.45} \end{equation}

We will be able to evaluate the volume and surface integrals, and the outcome of this exercise will be the precise identity of $A(r-vt)$.

We begin with the volume integral. With $G$ given by Eq.(12.42), we have that $\partial G/\partial t = -v r^{-1} A'$, in which a prime indicates differentiation with respect to $r-vt$, the function's argument. Similarly, $\partial^2 G/\partial t^2 = v^2 r^{-1} A''$, and we get

\begin{equation} \frac{1}{v^2} \int_V \frac{\partial^2 G}{\partial t^2}\, dV = \int \frac{1}{r} A''\, r^2\sin\theta\, drd\theta d\phi = 4\pi \int_0^R r A''\, dr \tag{12.46} \end{equation}

for the volume integral; the factor of $4\pi$ comes from performing the angular integration. Now $A'' = \partial A^2/\partial r^2$, and we can express the integrand as

\begin{equation} r A'' = \frac{\partial}{\partial r} \biggl( r \frac{\partial A}{\partial r} \biggr) - \frac{\partial A}{\partial r} = \frac{\partial}{\partial r}\biggl( r \frac{\partial A}{\partial r} - A \biggr). \tag{12.47} \end{equation}

This allows us to evaluate the integral, and we obtain

\begin{align} \frac{1}{v^2} \int_V \frac{\partial^2 G}{\partial t^2}\, dV &= 4\pi \biggl( r \frac{\partial A}{\partial r} - A \biggr)\biggr|^R_0 \nonumber \\ &= 4\pi \bigl[ R A'(R-vt) - A(R-vt) + A(-vt) \bigr]. \tag{12.48} \end{align}

Next we deal with the surface integral. Because $G$ is independent of $\theta$ and $\phi$, we have that $\boldsymbol{\nabla} G = \boldsymbol{\hat{r}}\, \partial G/\partial r = \boldsymbol{\hat{r}} (r^{-1} A' - r^{-2} A)$. Evaluating this at $r = R$ and taking the scalar product with $d\boldsymbol{a}$ gives

\begin{equation} \boldsymbol{\nabla} G \cdot d\boldsymbol{a} = \bigl[ R A'(R-vt) - A(R-vt) \bigr] \sin\theta\, d\theta d\phi, \tag{12.49} \end{equation}

and integrating over the angles returns

\begin{equation} \oint_S \boldsymbol{\nabla} G \cdot d\boldsymbol{a} = 4\pi \bigl[ R A'(R-vt) - A(R-vt) \bigr]. \tag{12.50} \end{equation}

If we now insert Eqs.(12.48) and (12.50) within Eq.(12.45), we obtain the remarkably simple equation

\begin{equation} 4\pi A(-vt) = \delta(t), \tag{12.51} \end{equation}

which informs us that the function $A$ is nothing but a delta function.

Writing $p := -vt$, so that $t = -p/v$, we have obtained that $4\pi A(p) = \delta(-p/v) = v \delta(p)$ --- recall the scaling property of delta functions expressed by Eq.(6.14). Evaluating this at $p = r - vt$, we have

\begin{equation} A(r-vt) =\frac{v}{4\pi} \delta(r-vt), \tag{12.52} \end{equation}

so that

\begin{equation} G(t,\boldsymbol{r}) = \frac{v}{4\pi r} \delta(r-vt). \tag{12.53} \end{equation}

It is useful to rewrite the delta function as $\delta[-v(t-r/v)] = v^{-1} \delta(t-r/v)$, so that the Green's function becomes

\begin{equation} G(t,\boldsymbol{r}) = \frac{1}{4\pi} \frac{\delta(t-r/v)}{r}. \tag{12.54} \end{equation}

We may now restore the dependence on $t'$ and $\boldsymbol{r'}$. We merely observe that since $t$ measures time from $t'$, it can safely be replaced by $t - t'$ when $t' \neq 0$. Similarly, because $r$ measures the distance from $\boldsymbol{r'}$, it can be replaced by $|\boldsymbol{r}-\boldsymbol{r'}|$ when $\boldsymbol{r'} \neq \boldsymbol{0}$. With this we finally arrive at

\begin{equation} G(t,\boldsymbol{r}; t',\boldsymbol{r'}) = \frac{1}{4\pi} \frac{ \delta(t-t' - |\boldsymbol{r}-\boldsymbol{r'}|/v) }{|\boldsymbol{r}-\boldsymbol{r'}|} \tag{12.55} \end{equation}

for the Green's function associated with the wave equation.

Inserting the Green's function within Eq.(12.40) gives

\begin{equation} \psi(t,\boldsymbol{r}) = \frac{1}{4\pi} \int \frac{f(t',\boldsymbol{r}')}{|\boldsymbol{r}-\boldsymbol{r'}|}\, \delta(t-t' -|\boldsymbol{r}-\boldsymbol{r'}|/v)\, dt' dV'. \tag{12.56} \end{equation}

The delta function can be rewritten as $\delta(t' - t + |\boldsymbol{r}-\boldsymbol{r'}|/v)$, and integration with respect to $dt'$ produces

\begin{equation} \psi(t,\boldsymbol{r}) = \frac{1}{4\pi} \int \frac{f(t-|\boldsymbol{r}-\boldsymbol{r'}|/v,\boldsymbol{r}')}{|\boldsymbol{r}-\boldsymbol{r'}|}\, dV', \tag{12.57} \end{equation}

our final expression for the wave function.

Figure 12.1: The wave at $\boldsymbol{r}$ is made up of elementary contributions originating from all positions $\boldsymbol{r'}$ within the source. An element of wave originating at $\boldsymbol{r'}$ requires a time $\Delta t = |\boldsymbol{r}-\boldsymbol{r'}|/v$ to travel to $\boldsymbol{r}$.

Notice the remarkable fact that Eq.(12.57) looks almost identical to the solution to the three-dimensional Poisson equation, as expressed by Eq.(12.36). As we had in the case of Poisson's equation, the wave at $\boldsymbol{r}$ is built up from elementary contributions $f(\boldsymbol{r'})\, dV'/|\boldsymbol{r}-\boldsymbol{r'}|$ that are added together as we sample the region occupied by the source --- see Fig.12.1. The only difference is that $\psi$ and $f$ now depend on time.

It is another remarkable fact that inside the integral of Eq.(12.57), the source function $f$ is evaluated at the time $t - \Delta t$, where $\Delta t := |\boldsymbol{r}-\boldsymbol{r'}|/v$. This shift in time is evidently a consequence of the delta function in Eq.(12.55). Notice that the shift is negative: $f$ is sampled at earlier times than $t$, the time at which the wave function is evaluated. What is the meaning of this?

A clue is provided by the fact that $|\boldsymbol{r}-\boldsymbol{r'}|$ is the distance between $\boldsymbol{r}$, where the wave function is evaluated, and $\boldsymbol{r'}$, where the source is sampled as we add up the contributions. This divided by $v$ is the time $\Delta t$ required for an elementary wave to travel from $\boldsymbol{r'}$ to $\boldsymbol{r}$. The source function, then, is sampled at time $t - \Delta t$ precisely because it takes a time $\Delta t$ for the elementary wave to leave $\boldsymbol{r'}$ and reach $\boldsymbol{r}$. The picture that emerges is that the total wave at position $\boldsymbol{r}$ and time $t$ is built up from elementary contributions that leave $\boldsymbol{r'}$ at time $t-\Delta t$ so that they can all arrive at $\boldsymbol{r}$ at the same time $t$. This is exactly what we would expect of elementary waves that travel at a speed $v$, as implied by the wave equation.

12.6 Practice Problems

  1. (Boas Chapter 8, Section 12, Problem 2) Solve $\ddot{X} + \omega^2 X = f_0 \sin(\omega t)$ with initial conditions $X(0) = \dot{X} = 0$. An overdot indicates differentiation with respect to $t$.

  2. (Boas Chapter 8, Section 12, Problem 3) Solve $\ddot{X} + \omega^2 X = f_0 e^{-\kappa t}$ with initial conditions $X(0) = \dot{X} = 0$. An overdot indicates differentiation with respect to $t$.

  3. (Boas Chapter 8, Section 12, Problems 6 and 7) Obtain the Green's function $G(t,t')$ for the differential equation $\ddot{X} - \kappa^2 X = f(t)$, with initial conditions $X(0) = \dot{X}(0) = 0$. An overdot indicates differentiation with respect to $t$. Then find the solution to the differential equation when $f = f_0 e^{-t/t_0}$.

  4. (Boas Chapter 8, Section 12, Problem 8) Solve the differential equation $y'' + 2y' + y = f(x)$ with the boundary conditions $y(0) = y'(0) = 0$, for any function $f$. A prime indicates differentiation with respect to $x$. Then specialize your solution to the case in which $f(x)$ is defined to be equal to $1$ when $0 < x < a$ and to $0$ when $x > a$. To express your solution it is helpful to consider the cases $x < a$ and $x > a$ separately.

  5. (Boas Chapter 8, Section 12, Problem 11). Find the Green's function for the differential equation $y'' + y = f$, with boundary conditions $y(0) = 0$ and $y(\pi/2) = 0$. A prime indicates differentiation with respect to $x$. Then construct the solution for $f(x) = \sin(2x)$.

12.7 Challenge Problems

  1. Our main goal with this problem is to derive the identity

    \begin{equation} \frac{1}{|\boldsymbol{r}-\boldsymbol{r'}|} = \sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell \frac{4\pi}{2\ell+1} \frac{r_<^\ell}{r_>^{\ell+1}}\bigl[ Y^m_\ell(\theta',\phi') \bigr]^* Y^m_\ell(\theta,\phi) \tag{12.58}\end{equation}

    where $(r,\theta,\phi)$ are the spherical coordinates associated with $\boldsymbol{r}$, $(r',\theta',\phi')$ are those associated with $\boldsymbol{r'}$, $r_< := \text{min}(r,r')$, $r_> := \text{max}(r,r')$, and an asterisk indicates complex conjugation. We shall also derive a corollary,

    \begin{equation} P_\ell(\cos\gamma) = \frac{4\pi}{2\ell+1}\sum_{m=-\ell}^\ell\bigl[ Y^m_\ell(\theta',\phi') \bigr]^* Y^m_\ell(\theta,\phi)\tag{12.59}\end{equation}

    known as the addition theorem for spherical harmonics, where $\cos\gamma := \boldsymbol{r}\cdot\boldsymbol{r'}/(rr') = \cos\theta\cos\theta' + \sin\theta\sin\theta'\cos(\phi - \phi')$. These identities have many applications, for example in the theory of multipole expansions. To obtain these results you may proceed as follows.

    a) Express Green's equation $\nabla^2 G(\boldsymbol{r},\boldsymbol{r'}) = -\delta(\boldsymbol{r}-\boldsymbol{r'})$ in spherical coordinates. Don't forget to use the rules of Sec.6.5 --- see in particular Eq.(6.27) --- to write the three-dimensional delta function in this coordinate system.

    b) Establish the identity

    \[\frac{\delta(\theta-\theta') \delta(\phi - \phi')}{\sin\theta}= \sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell\bigl[ Y^m_\ell(\theta',\phi') \bigr]^* Y^m_\ell(\theta,\phi).\]

    Hint: this is an expansion in spherical harmonics, and it might be a good idea to refer back to Sec.8.3.

    c) Expand also the Green's function in spherical harmonics. Write the expansion as

    \[G(\boldsymbol{r},\boldsymbol{r'}) = \sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell g^m_\ell(r,r') \bigl[ Y^m_\ell(\theta',\phi') \bigr]^*Y^m_\ell(\theta,\phi),\]

    with factors $[Y^m_\ell(\theta',\phi')]^*$ inserted for convenience, to match the expansion of the delta function. Show that with this expansion, Green's equation reduces to

    \[r^2 \frac{\partial^2 g^m_\ell}{\partial r^2}+ 2r \frac{\partial g^m_\ell}{\partial r}- \ell(\ell+1) g^m_\ell = -\delta(r-r').\]

    d) The solution to the reduced Green's equation must be nonsingular at both $r=0$ and $r=\infty$, and it must account for the delta function at $r = r'$. Show that it is given by

    \[g^m_\ell = \frac{1}{2\ell + 1} \left\{\begin{array}{ll}r^\ell/r^{\prime \ell+1} & \quad r < r' \\ r^{\prime \ell}/r^{\ell+1} & \quad r > r'\end{array} \right. .\]

    With the notation introduced previously, this is

    \[g^m_\ell = \frac{1}{2\ell+1} \frac{r_<^\ell}{r_>^{\ell+1}}.\]

    e) Collect results, make use of Eq.(12.35), and arrive at Eq.(12.58).

    f) To derive the addition theorem of Eq.(12.59), relate $|\boldsymbol{r}-\boldsymbol{r'}|^{-1}$ to the generating function for the Legendre polynomials, just as we did back in Sec.3.8. Be careful to consider both cases $r < r'$ and $r > r'$. Then make use of Eq.(3.4).