Chapter 11: Wave Equation

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Matterial covered in this chapter is also presented in Boas Chapter 13, Sections 4 and 6.

11.1 Introduction

The wave equation,

\begin{equation} 0 = \Box \psi := -\frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2} + \nabla^2 \psi, \tag{11.1} \end{equation}

is a partial differential equation that implicates four independent variables, the three spatial variables $x$, $y$, $z$, and the time variable $t$. The dependent variable $\psi$ is the wave function, and $v$ is the speed at which the wave travels. The wave operator $\Box$ is a four-dimensional generalization of the Laplacian operator $\nabla^2$, with a minus sign and factor of $v^{-2}$ inserted in front of the time derivatives; it is sometimes referred to as the d'Alembertian operator. The wave equation governs the propagation of waves in the absence of dispersion, which means that all wavelengths travel at the same speed.

Waves are produced in many different physical processes, and they travel on many different media. An example is a wave traveling on a stretched string. In this case the medium is the string, which is one-dimensional, and the wave equation reduces to

\begin{equation} 0 = -\frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2} + \frac{\partial^2 \psi}{\partial x^2}, \tag{11.2} \end{equation}

with $\psi$ measuring the lateral displacement of the string; the wave speed is related to the string tension $T$ and its mass per unit length $\lambda$ by $v = \sqrt{T/\lambda}$. Another example is a wave on a stretched membrane. Here the medium is the two-dimensional membrane, the wave equation becomes

\begin{equation} 0 = -\frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2} + \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2}, \tag{11.3} \end{equation}

the function $\psi$ describes the membrane's lateral displacement, and $v = \sqrt{\gamma/\sigma}$, with $\gamma$ denoting the surface tension and $\sigma$ the membrane's mass per unit area.

In three dimensions, Eq.(11.1) becomes

\begin{equation} 0 = -\frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2} + \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2}, \tag{11.4} \end{equation}

and it governs, for example, the propagation of sound waves. In this case, the medium is air (or water, or any other fluid), $\psi$ describes the perturbation in the air pressure, and $v = \sqrt{\gamma p/\rho}$, where $p$ is the average pressure, $\rho$ is the air's mass density, and $\gamma := c_P/c_V$ is the ratio of heat capacities. The three-dimensional wave equation also governs the propagation of electromagnetic waves. In this case $\psi$ represents a component of the electric or magnetic field, and $v$ is the speed of light. It even governs the propagation of gravitational waves, with a speed $v$ that happens to be identical to the speed of light.

There are many more examples, but the point is made that the wave equation is associated with a plethora of phenomena involving physics from many different fields. Whenever the physics produces a dispersionless wave, the propagation of the wave is governed by Eq.(11.1). Given the abundance of wave phenomena in physics, the wave equation is clearly one of the most important equations of mathematical physics. Our task in this chapter is to find ways to solve the wave equation, building on the techniques that proved so successful in Chapter 10, in the case of Laplace's equation.

11.2 Vibrating String

To get us started we consider a traveling wave on a one-dimensional medium, a stretched string. We take the string to have a fixed end at $x=0$ and another fixed end at $x=L$; see Fig.11.1. The wave equation reduces to the one-dimensional form given by Eq.(11.2),

\begin{equation} 0 = \Box \psi = -\frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2} + \frac{\partial^2 \psi}{\partial x^2}, \tag{11.5} \end{equation}

with $\psi$ representing the string's lateral displacement and $v$ the wave speed. We wish to find solutions $\psi(t,x)$ to this partial differential equation. Apart from the minus sign and the factor of $v^{-2}$ in front of the time derivatives, the wave equation is similar in form to the two-dimensional Laplace equation, and we should be able to adapt the techniques introduced in Chapter 10 to find solutions.

 

The y-axis corresponds to the wave function (psi) and the x axis corresponds to the length of the string (x). The plot depicts a wave-like form with two peaks. At x=0 and x=L the wave function is equal to 0.
Figure 11.1: Stretched string with fixed ends at $x=0$ and $x=L$. The wave function $\psi(t,x)$ measures the lateral displacement of the string with respect to the $x$-axis.

The intuition gained from Laplace's equation suggests that Eq.(11.5) will admit a unique solution once a suitable number of boundary conditions are imposed. Two such conditions come automatically from the fact that the string has fixed ends. The lateral displacement must vanish at these ends, and we have that

\begin{equation} \psi(t,x=0) = 0 = \psi(t,x=L). \tag{11.6} \end{equation}

We must also pay attention to the time boundaries. Suppose that we wish to solve the wave equation between the times $t = 0$ and $t = t_1$. From a mathematical point of view it would be natural to impose, in addition to Eq.(11.6), boundary conditions at $t = 0$ and $t = t_1$ to obtain a unique solution to the wave equation. From a physics point of view, however, this would be a strange thing to do. To impose a condition at $t = 0$ is natural enough: to specify $\psi(t=0, x)$ is to give a mathematical description of the initial shape of the wave. But to impose a condition at $t = t_1$ is far from natural, because $\psi(t = t_1, x)$ would describe the shape of the wave at the final time $t_1$. Isn't this what we would like to predict?

The central goal of physics is indeed to predict what will happen next once we know the initial conditions. The prototypical example is ballistic motion in mechanics. We have a particle of mass $m$, a force $\boldsymbol{F}$ acting on it, we know the particle's initial state, and we let $\boldsymbol{F} = m \boldsymbol{a}$ tell us where the particle will be at a later time. The initial state at $t = 0$ is specified with two pieces of information: the particle's initial position $\boldsymbol{r}_0$ and its initial velocity $\boldsymbol{v}_0$. We need both pieces of information to get a unique solution to the equations of motion. For example, when $\boldsymbol{F} = \boldsymbol{0}$ the unique solution is $\boldsymbol{r}(t) = \boldsymbol{r}_0 + \boldsymbol{v}_0\, t$.

In the case of ballistic motion, the conditions that ensure a unique solution are initial conditions; we specify the particle's position and velocity at $t = 0$. We could, if we wanted to, change the nature of the problem and replace one of the initial conditions by a final condition. For example, we could choose to solve the equations of motion by specifying $\boldsymbol{r}_0 := \boldsymbol{r}(t=0)$ and $\boldsymbol{r}_1 := \boldsymbol{r}(t=t_1)$ instead of $\boldsymbol{r}_0$ and $\boldsymbol{v}_0$. This is perfectly acceptable from a mathematical point of view, and we still get a unique solution. In the simple example with $\boldsymbol{F} = \boldsymbol{0}$, setting $\boldsymbol{r}_1 = \boldsymbol{r}_0 + \boldsymbol{v}_0\, t_1$ and solving for $\boldsymbol{v}_0$ yields $\boldsymbol{r}(t) = \boldsymbol{r}_0 + (\boldsymbol{r}_1 - \boldsymbol{r}_0) (t/t_1)$, a perfectly acceptable form of solution. It is unique, and entirely equivalent to the one obtained previously.

From a mathematical point of view, the specification of one initial condition and one final condition is just as acceptable as the specification of two initial conditions. From the point of view of a physicist, however, the specification of a final condition is unnatural, because we do not know the future. The specification of initial conditions is far better, because we do know the present state of the system, and we can exploit this knowledge together with the laws of physics to predict its future.

Returning to the wave equation, we would like to replace the final condition $\psi(t=t_1, x)$ with a second initial condition, the first being $\psi(t=0, x)$, the string's initial lateral displacement. By analogy with the ballistic problem, we could think of specifying the string's initial lateral velocity, measured by $\partial \psi/\partial t$ evaluated at $t = 0$. (The lateral velocity is not to be confused with the wave speed $v$. The lateral velocity describes how fast a given point on the string moves up and down; the wave speed describes how fast the wave as a whole travels from left to right or from right to left.) The analogy with ballistic motion suggests that the specification of

\begin{align} & \text{initial lateral displacement} = \psi(t=0, x) = f(x), \tag{11.7a}\\ & \text{initial lateral velocity} = \frac{\partial \psi}{\partial t}(t=0, x) = g(x), \tag{11.7b} \end{align}

should be as mathematically acceptable as the specification of $\psi(t=0, x)$ and $\psi(t=t_1, x)$, and should ensure that the solutions to the wave equation is unique. The statement is in fact true.

11.3 Solution to the 1D Wave Equation

We adopt the tried and true method of looking for factorized solutions. We write

\begin{equation} \psi = T(t) X(x\tag{11.8}\end{equation}

insert within Eq.(11.5), and get

\begin{equation} \frac{1}{v^2 T} \frac{d^2 T}{dt^2} = \frac{1}{X} \frac{d^2 X}{dx^2}. \tag{11.9} \end{equation}

Both sides of the equation must be equal to a constant, which we write as $-k^2$. The solutions for $X$ are $e^{\pm i k x}$, or

\begin{equation} X(x) = \left\{ \begin{array}{l} \cos(kx) \\ \sin(kx) \end{array} \right. \tag{11.10} \end{equation}

The solutions for $T$ are $e^{\pm i k v t}$, or

\begin{equation} T(t) = \left\{ \begin{array}{l} \cos(k v t) \\ \sin(k v t) \end{array} \right. \tag{11.11} \end{equation}

The factorized solutions are therefore

\begin{equation} \psi_k(t,x) = \left\{ \begin{array}{l} \cos(kx) \\ \sin(kx) \end{array} \right\} \left\{ \begin{array}{l} \cos(k v t) \\ \sin(k v t) \end{array} \right\}, \tag{11.12} \end{equation}

and they depend on the arbitrary constant $k$. This quantity has the interpretation of a wave number, related to the wavelength $\lambda$ by $k = 2\pi/\lambda$. The quantity $\omega := k v$, which appears in the functions of time, has the interpretation of an angular frequency.

We must now impose the boundary conditions of Eq.(11.6). The wave function must vanish at $x = 0$, and to enforce this we must eliminate $\cos(kx)$ from the factorized solutions. The wave function must also vanish at $x = L$, and to achieve this we set $k = n\pi/L$, with $n = 1, 2, 3, \cdots$. The factorized solutions become

\begin{equation} \psi_n(t,x) = \sin \Bigl( \frac{n \pi x}{L} \Bigr) \left\{ \begin{array}{l} \cos(\omega_n t) \\ \sin(\omega_n t) \end{array} \right\}, \tag{11.13} \end{equation}

where

\begin{equation} \omega_n := \frac{n \pi v}{L} \tag{11.14} \end{equation}

are the string's natural frequencies. The functions $\psi_n(t,x)$ are known as the normal modes of vibration of the one-dimensional string.

The final solution to the wave equation will be a superposition of normal modes, which we write as

\begin{equation} \psi(t,x) = \sum_{n=1}^\infty \sin \Bigl( \frac{n \pi x}{L} \Bigr) \bigl[ a_n \cos(\omega_n t) + b_n \sin(\omega_n t) \bigr], \tag{11.15} \end{equation}

where $a_n$ and $b_n$ are expansion coefficients. These are determined by the initial conditions. Setting $t = 0$ in Eq.(11.15) and incorporating Eq.(11.7) gives

\begin{equation} f(x) = \sum_{n=1}^\infty a_n \sin \Bigl( \frac{n \pi x}{L} \Bigr), \tag{11.16} \end{equation}

which is recognized as a sine Fourier series for the function $f(x)$. From Eq.(7.47) we get that the expansion coefficients are given by

\begin{equation} a_n = \frac{2}{L} \int_0^L f(x) \sin \Bigl( \frac{n \pi x}{L} \Bigr)\, dx. \tag{11.17} \end{equation}

With $\psi(t=0,x)$ specified by the function $f(x)$, and $a_n$ obtained from Eq.(11.17), we have determined half the terms in Eq.(11.15).

To get the second half we must relate $b_n$ to the wave's lateral velocity at $t=0$. Differentiating Eq.(11.15) with respect to $t$ gives

\begin{equation} \frac{\partial \psi}{\partial t} = \sum_{n=1}^\infty \sin \Bigl( \frac{n \pi x}{L} \Bigr) \bigl[ -a_n \omega_n \sin(\omega_n t) + b_n \omega_n \cos(\omega_n t) \bigr], \tag{11.18} \end{equation}

and evaluating this at $t=0$ returns

\begin{equation} g(x) = \sum_{n=1}^\infty b_n \omega_n \sin \Bigl( \frac{n \pi x}{L} \Bigr), \tag{11.19} \end{equation}

a sine Fourier series for the function $g(x)$. Inverting this yields

\begin{equation} b_n \omega_n = \frac{2}{L} \int_0^L g(x) \sin \Bigl( \frac{n \pi x}{L} \Bigr)\, dx. \tag{11.20} \end{equation}

With the initial lateral velocity specified by the function $g(x)$, and $b_n$ obtained from Eq.(11.20) together with $\omega_n = n\pi v/L$, we have determined the second set of terms in Eq.(11.15).

To summarize, the unique solution to Eq.(11.5) with the boundary conditions of Eq.(11.6) and the initial conditions of Eq.(11.7) is given by Eq.(11.15), with $a_n$ determined from $f(x)$ by Eq.(11.17) and $b_n$ determined from $g(x)$ by Eq.(11.20).

11.4 Triangle Wave

As an example of a wave on a string with ends fixed at $x=0$ and $x=L$, we consider the triangle wave described by the initial profile

\begin{equation} f(x) =\frac{8h}{L} \left\{ \begin{array}{ll} 0 & \quad 0 \leq x < \frac{3}{8} L\\ x - \frac{3}{8} L & \quad \frac{3}{8} L \leq x < \frac{1}{2} L \\ \frac{5}{8} L - x & \quad \frac{1}{2} L \leq x < \frac{5}{8} L \\ 0 & \quad \frac{5}{8} L \leq x < L \end{array} \right. |tag{11.21} \end{equation}

The function describes a triangle of height $h$ and width $\frac{1}{4} L$ centred at $x = \frac{1}{2} L$. We choose the string to be momentarily at rest at $t=0$, so that the initial lateral velocity vanishes, giving $g(x) = 0$ for the second initial condition. With this initial data, Eq.(11.20) informs us that $b_n = 0$, and Eq.(11.15) becomes

'\begin{equation} \psi(t,x) = \sum_{n=1}^\infty a_n \sin\Bigl( \frac{n \pi x}{L} \Bigr) \cos\Bigl( \frac{n \pi v t}{L} \Bigr). \tag{11.22} \end{equation}

We next use Eq.(11.17) to calculate $a_n$ from $f(x)$, and a quick calculation returns

\begin{equation} a_n = \frac{16h}{\pi^2 n^2} \biggl[ 2 \sin\Bigl(\frac{n\pi}{2} \Bigr) - \sin\Bigl( \frac{3n\pi}{8} \Bigr) - \sin\Bigl( \frac{5n\pi}{8} \Bigr) \Biggr]. \tag{11.23} \end{equation}

Insertion within Eq.(11.22) allows us to calculate the wave profile at any time $t$.


Exercise 11.1: Verify the preceding expression for the expansion coefficients.


Snapshots of the wave are displayed in Fig.11.2. The figure reveals that the wave immediately splits up into two half waves, one traveling to the right, the other traveling to the left. When each half wave meets the end of the string, it reverses direction as well as reverses sign. The half waves eventually meet again, momentarily, as they continue their journey.

The y axis of the plot corresponds to height and the x axis to the length of the string. At t equal to 0 there is a single triangle-shaped wave of amplitude 1. When t is equal to 3 the wave splits into two waves of height 0.5 that moving away from each other. When t equals 8, the half waves reach the end of the string and reverse direction, moving towards each other. The sign of their amplitudes change and the height of the waves is now negative 0.5. At t equal to 10 the two half waves have recombined to form a single wave of amplitude negative 0.5. At t equal 13 the wave has, again, split into two half waves that mpve into opposite directions. By t equal 18, these waves have reached the end of the string and have reversed amplitudes, moving toward each other once again.
Figure 11.2: Triangle wave at various times. For the computation we set $h = 1$, $L = 10$, and $v = 1$.

To explain why we get both right-moving and left-moving waves in our solution, we observe that the wave function of Eq.(11.22) is built from functions of the form $\sin(kx) \cos(kvt)$, where $k = n\pi/L$. The trigonometric identity $\sin(a) \cos(b) = \frac{1}{2} \sin(a-b) + \frac{1}{2} \sin(a+b)$ implies that this can be written as

\begin{equation} \sin(kx) \cos(kvt) = \frac{1}{2} \sin\bigl[ k(x-vt) \bigr] + \frac{1}{2} \sin\bigl[ k(x+vt) \bigr]. \tag{11.24} \end{equation}

This equation describes the split of the initial wave into the two half waves. The first term, which has $k(x - vt)$ inside the sine function, describes a right-moving wave. This can be seen from the fact that the maximum of the sine function occurs at $kx = \pi/2$ at $t = 0$, but that it occurs at $kx = \pi/2 + kvt$ at later times; the peak has moved to the right by a distance $vt$, which confirms that the wave travels with a speed $v$. On the other hand, the second term, which has $k(v+vt)$ inside the sine function, describes a left-moving wave. We therefore see that each term in Eq.(11.22) contains an equal mixture of right-moving and left-moving waves, and the total wave has no choice but to respect this decomposition. The fact that we have an equal mixture is a consequence of the zero initial condition for the lateral velocity. A nonzero $g(x)$ would give rise to unequal mixtures, and initial conditions can be concocted so that the solution describes, before reflection at the two ends of the string, a right-moving wave only, or a left-moving wave only.

11.5 Infinite String

We continue to study the one-dimensional wave equation of Eq.(11.5), but now take the string to be infinite in length. We continue to impose the initial conditions of Eq.(11.7), but the boundary conditions of Eq.(11.6) must now be abandoned. Our intuition, based on the example of the preceding section, suggests that a right-moving wave will just keep on traveling to the right, because there is no longer a fixed end to provoke a reflection. Similarly, we would expect that a left-moving wave will just keep on traveling to the left. These expectations will be confirmed by a re-analysis of the wave equation.

With the absence of boundary conditions, the factorized solutions to the wave equation revert back to Eq.(11.12), which we rewrite as

\begin{equation} \psi_k(t,x) = \left\{ \begin{array}{l} e^{ik x} \\ e^{-ik x} \end{array} \right\} \left\{ \begin{array}{l} e^{i kv t} \\ e^{-i kv t} \end{array} \right\}, \tag{11.25} \end{equation}

in terms of complex exponentials; here $k$ is again an arbitrary parameter. The four combinations contained in the factorized solutions are \[ e^{ik(x+vt)}, \qquad e^{ik(x-vt)}, \qquad e^{-ik(x-vt)}, \qquad e^{-ik(x+vt)}, \] and with $k$ allowed to be both positive and negative, we see that the last two entries are redundant. The general solution to the wave equation can therefore be expressed as a superposition of the first two combinations only. Because $k$ is a continuous variable, this superposition must be written as an integral, and we have

\begin{equation} \psi(t,x) = \int_{-\infty}^\infty a(k)\, e^{ik(x-vt)}\, dk + \int_{-\infty}^\infty b(k)\, e^{ik(x+vt)}\, dk, \tag{11.26} \end{equation}

where the functions $a(k)$ and $b(k)$ play the role of expansion coefficients. The first integral defines a function of $x-vt$, which we denote $A$, while the second integral defines a function of $x+vt$, which we denote $B$. Comparison with Eq.(9.6) reveals that $a(k)$ is in fact the Fourier transform of $A(x-vt)$, while $b(k)$ is the Fourier transform of $B(x+vt)$.

The general solution to the one-dimensional wave equation in the absence of boundaries can therefore be expressed as

\begin{equation} \psi(t,x) = A(x-vt) + B(x+vt), \tag{11.27} \end{equation}

where $A(p)$ and $B(q)$ are arbitrary functions of their arguments, $p := x-vt$ and $q := x+vt$. The first term, $A(x-vt)$, is recognized as a right-moving wave of arbitrary shape, while the second term, $B(x+vt)$, is a left-moving wave, also of arbitrary shape.

It is easy to verify that Eq.(11.27) is a solution to Eq.(11.5). We simply make the substitution and evaluate the partial derivatives. We have

\begin{equation} \frac{\partial \psi}{\partial x} = \frac{dA}{dp} \frac{\partial p}{\partial x} + \frac{dB}{dq} \frac{\partial q}{\partial x} =\frac{dA}{dp} + \frac{dB}{dq}, \tag{11.28} \end{equation}

and similarly,

\begin{equation} \frac{\partial^2 \psi}{\partial x^2} =\frac{d^2A}{dp^2} + \frac{d^2B}{dq^2}. \tag{11.29} \end{equation}

On the other hand,

\begin{equation} \frac{\partial \psi}{\partial t} = \frac{dA}{dp} \frac{\partial p}{\partial t} + \frac{dB}{dq} \frac{\partial q}{\partial t} = -v\frac{dA}{dp} + v\frac{dB}{dq}, \tag{11.30} \end{equation}

and similarly,

\begin{equation} \frac{\partial^2 \psi}{\partial t^2} = v^2 \frac{d^2A}{dp^2} + v^2 \frac{d^2B}{dq^2}. \tag{11.31} \end{equation}

Collecting these results, we have that

\begin{equation} \Box \psi = -\frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2}+ \frac{\partial^2 \psi}{\partial x^2} = -\frac{d^2A}{dp^2} - \frac{d^2B}{dq^2} + \frac{d^2A}{dp^2} + \frac{d^2B}{dq^2} = 0, \tag{11.32} \end{equation}

and find that indeed, the $\psi(t,x)$ of Eq.(11.27) satisfies the wave equation for any function $A(p)$ and any function $B(q)$.

In applications we may wish to relate $A(p)$ and $B(q)$ to the initial conditions of Eq.(11.7). At $t=0$ we have

\begin{align} f(x) &= \psi(0,x) = A(x) + B(x) \tag{11.33a} \\ g(x) &= \frac{\partial \psi}{\partial t}(0, x) = -v A'(x) + v B'(x), \tag{11.33b} \end{align}

where a prime indicates differentiation of the function with respect to its argument. Thus, $A'(p) := dA/dp$ and $B'(q) := dB/dq$, but because the derivatives are evaluated at $t = 0$, so that $p = q = x$, the prime can also be taken to represent a derivative with respect to $x$. Differentiating Eq.(11.33a) with respect to $x$ and dividing Eq.(11.33b) by $v$ yields

\begin{equation} A' + B' = f', \qquad A' - B' = -g/v. \tag{11.34} \end{equation}

Adding and subtracting the equation produces

\begin{equation} A' = \frac{1}{2} \Bigl( f' - g/v \Bigr), \qquad B' = \frac{1}{2} \Bigl( f' + g/v \Bigr). \tag{11.35} \end{equation}

Integration yields

\begin{align} A(x) &= \frac{1}{2} f(x) - \frac{1}{2v} \int^x g(x')\, dx' + c_1, \tag{11.36a}\\ B(x) &= \frac{1}{2} f(x) + \frac{1}{2v} \int^x g(x')\, dx' + c_2, \tag{11.36b} \end{align}

where $c_1$ and $c_2$ are constants of integration. These are related to each other, because according to Eq.(11.33a), $f = A + B = f + c_1 + c_2$, so that $c_2 = -c_1$. Our final expressions for the functions $A(p)$ and $B(q)$ are therefore

\begin{align} A(p) &= \frac{1}{2} f(p) - \frac{1}{2v} \int^p g(x')\, dx' + c, \tag{11.37a} \\ B(q) &= \frac{1}{2} f(q) + \frac{1}{2v} \int^q g(x')\, dx' - c, \tag{11.37b} \end{align}

where $c := c_1$ is an arbitrary constant.

Substitution of Eq.(11.37) into Eq.(11.27) gives

\begin{equation} \psi = \frac{1}{2} f(x-vt) - \frac{1}{2v} \int^{x-vt} g(x')\, dx' + \frac{1}{2} f(x+vt) +\frac{1}{2v} \int^{x+vt} g(x')\, dx', \tag{11.38} \end{equation}

or

\begin{equation} \psi(t,x) = \frac{1}{2} \Bigl[ f(x-vt) + f(x+vt) \Bigr] + \frac{1}{2v} \int_{x-vt}^{x+vt} g(x')\, dx', \tag{11.39} \end{equation}

and we see that the final expression for $\psi(t,x)$ is independent of the constant $c$. Equation (11.39) is known as d'Alembert's formula. It gives $\psi(t,x)$ directly in terms of the initial conditions $f(x)$ and $g(x)$. When, for example, the string is momentarily at rest at $t=0$, so that $g(x) = 0$, the solution reduces to $\psi = \frac{1}{2} f(x-vt) + \frac{1}{2} f(x+vt)$, which represents an equal mixture of right-moving and left-moving waves, just as was seen in Sec.11.4.


Exercise 11.2: Reproduce the steps that lead to Eq.(11.39). In particular, make sure that in the end, the integral terms in $A(p)$ and $B(q)$ can be written as the single integral from $x-vt$ to $x+vt$.


11.6 Vibrating Rectangular Membrane

A stretched membrane can support a wave that satisfies the two-dimensional wave equation

\begin{equation} 0 = \Box\psi = -\frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2} + \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial x^2}, \tag{11.40} \end{equation}

where $\psi(t,x,y)$ is the lateral displacement of the membrane, and $v$ is the wave speed. In this section we suppose that the membrane has a rectangular shape, with fixed ends at $x = 0$, $x = a$, $y = 0$, and $y = b$. We wish to solve the wave equation with boundary conditions

\begin{equation} \psi(t,x=0,y) = \psi(t,x=a,y) = \psi(t,x,y=0) = \psi(t,x,y=b) = 0 \tag{11.41} \end{equation}

and initial conditions

\begin{equation} \psi(t=0, x, y) = f(x,y), \qquad \frac{\partial \psi}{\partial t}(t=0, x, y) = g(x,y), \tag{11.42} \end{equation}

where $f(x,y)$ and $g(x,y)$ are arbitrary functions.

The starting point is the factorized solution $\psi = T(t) X(x) Y(y)$, which gives rise to

\begin{equation} 0 = -\frac{1}{v^2 T} \frac{d^2 T}{dt^2} + \frac{1}{X} \frac{d^2 X}{dx^2} + \frac{1}{Y} \frac{d^2 Y}{dy^2} \tag{11.43} \end{equation}

after substitution into the wave equation. Each term must be a constant, and we write

\begin{equation} \frac{1}{X} \frac{d^2 X}{dx^2} = -k_x^2, \qquad \frac{1}{Y} \frac{d^2 Y}{dy^2} = -k_y^2, \tag{11.44} \end{equation}

where $k_x$ and $k_y$ are two of the constants. The solutions to these equations are

\begin{equation} X(x) = \left\{ \begin{array}{l} \cos(k_x x) \\ \sin(k_x x) \end{array} \right., \qquad Y(y) = \left\{ \begin{array}{l} \cos(k_y y) \\ \sin(k_y y) \end{array} \right. . \tag{11.45} \end{equation}

The equation for $T(t)$ becomes

\begin{equation} \frac{1}{T} \frac{d^2 T}{dt^2} = -v^2 (k_x^2 + k_y^2), \tag{11.46} \end{equation}

and the solutions are

\begin{equation} T(t) = \left\{ \begin{array}{l} \cos(\omega t) \\ \sin(\omega t) \end{array} \right. , \tag{11.47} \end{equation}

where $\omega := v (k_x^2 + k_y^2)^{1/2}$. Collecting results, the factorized solutions to the wave equation are

\begin{equation} \psi_{k_x, k_y}(t,x,y) = \left\{ \begin{array}{l} \cos(k_x x) \\ \sin(k_x x) \end{array} \right\} \left\{ \begin{array}{l} \cos(k_y y) \\ \sin(k_y y) \end{array} \right\} \left\{ \begin{array}{l} \cos(\omega t) \\ \sin(\omega t) \end{array} \right\}, \tag{11.48} \end{equation}

and they depend on the two arbitrary constants $k_x$ and $k_y$.

We may now impose the boundary conditions of Eq.(11.41). With $\psi(t,0,y) = \psi(t,x,0) = 0$ we eliminate $\cos(k_x x)$ and $\cos(k_y y)$ from the factorized solutions, and with $\psi(t,a,y) = \psi(t,x,b) = 0$ we find that $k_x = n\pi/a$ and $k_y = m\pi/b$, where $n$ and $m$ are positive integers. The factorized solutions become

\begin{equation} \psi_{nm}(t,x,y) = \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{b} \Bigr) \left\{ \begin{array}{l} \cos(\omega_{nm} t) \\ \sin(\omega_{nm} t) \end{array} \right\}, \tag{11.49} \end{equation}

where $n = 1, 2, 3, \cdots$, $m = 1, 2, 3, \cdots$, and

\begin{equation} \omega_{nm} = \pi v \sqrt{ \frac{n^2}{a^2} + \frac{m^2}{b^2} }. \tag{11.50} \end{equation}

The functions $\psi_{nm}(t,x,y)$ are the normal modes of vibration of the rectangular membrane, and $\omega_{nm}$ are the membrane's natural frequencies. The modes are illustrated in Fig.11.3.

Modes of vibration
Figure 11.3: Modes of vibration of a rectangular membrane with $a=1$ and $b = 2$, plotted as functions of $x$ and $y$. The modes are labelled with two positive integers, $n$ and $m$, and they are evaluated at $t=0$.

As usual the general solution to the wave equation will be a superposition of normal modes. We write this as

\begin{equation} \psi(t,x,y) = \sum_{n=1}^\infty \sum_{m=1}^\infty \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{b} \Bigr) \Bigl[ a_{nm} \cos(\omega_{nm} t) + b_{nm} \sin(\omega_{nm} t) \Bigr], \tag{11.51} \end{equation}

where $a_{nm}$ and $b_{nm}$ are expansion coefficients. These are determined by the initial conditions of Eq.(11.42), which give

\begin{align} f(x,y) &= \sum_{n=1}^\infty \sum_{m=1}^\infty a_{nm} \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{b} \Bigr) \tag{11.52a}\\ g(x,y) &= \sum_{n=1}^\infty \sum_{m=1}^\infty b_{nm} \omega_{nm} \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{b} \Bigr). \tag{11.52b} \end{align}

These are double sine Fourier series for the functions $f(x,y)$ and $g(x,y)$, and inversion yields

\begin{align} a_{nm} &= \frac{4}{ab} \int_0^a \int_0^b f(x,y)\, \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{b} \Bigr)\, dxdy, \tag{11.53a}\\ b_{nm} &= \frac{4}{ab\, \omega_{nm}} \int_0^a \int_0^b g(x,y)\, \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{b} \Bigr)\, dxdy. \tag{11.53b} \end{align}

The unique solution to Eq.(11.40) with the boundary conditions of Eq.(11.41) and the initial conditions of Eq.(11.42) is therefore given by Eq.(11.51) with the expansion coefficients of Eq.(11.53). The membrane's natural frequencies are given by Eq.(11.50).


Exercise 11.3: Make sure that you can reproduce the steps that go from Eq.(11.51) to Eqs.(11.53).


As an example we consider the initial conditions

\begin{align} f(x,y) &= 16 h\, (x/a)(1-x/a)(y/b)(1-y/b), \tag{11.54a}\\ g(x,y) &= 0, \tag{11.54b} \end{align}

which describe a parabolic displacement of a membrane initially at rest. The displacement is maximum at the centre of the membrane, at $x = \frac{1}{2} a$ and $y = \frac{1}{2} b$, where it is equal to $h$. If we insert these initial data into Eqs.(11.53) we obtain $b_{nm} = 0$, as well as

\begin{equation} a_{nm} = \frac{1024h}{\pi^6} \frac{1}{n^3 m^3} \tag{11.55} \end{equation}

when $n$ and $m$ are both odd; otherwise $a_{nm} = 0$. Inserting these coefficients into Eq.(11.51), we find that the wave at any time is described by

\begin{equation} \psi(t,x,y) = \frac{1024h}{\pi^6} \sum_{n=1, 3, 5, \cdots}^\infty \sum_{m=1, 3, 5, \cdots}^\infty \frac{1}{n^3 m^3} \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{b} \Bigr) \cos(\omega_{nm} t). \tag{11.56} \end{equation}

Snapshots of this wave are shown in Fig.11.4. In this example the traveling nature of the wave is not apparent; we are in fact dealing with a standing wave. In other situations the behaviour of the wave would be more complicated. In general, a wave on a two-dimensional membrane will be split into waves that travel in all directions, and all these partial waves will be repeatedly reflected by the fixed ends of the membrane; this can lead to a lot of wonderful complexity.

Parabolic wave on a rectangular membrane
Figure 11.4: Parabolic wave on a rectangular membrane with $a=1$ and $b = 2$. For the computation we set $h = 1$ and $v = 1/3$.

Exercise 11.4: Verify that the coefficients $a_{nm}$ associated with the initial conditions of Eq.(11.54) are given by Eq.(11.55).


11.7 Vibrating Circular Membrane

When the stretched membrane has a circular shape, the two-dimensional wave equation and the boundary conditions are best expressed in polar coordinates $(s, \phi)$. Taking our cue from Eq.(1.75), which gives the Laplacian operator in cylindrical coordinates, we have that the wave equation becomes

\begin{equation} 0 = \Box \psi = -\frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2} + \frac{1}{s} \frac{\partial}{\partial s} \biggl( s \frac{\partial\psi}{\partial s} \biggr) + \frac{1}{s^2} \frac{\partial^2 \psi}{\partial \phi^2} \tag{11.57} \end{equation}

when it is written in polar coordinates. We wish to solve this equation for the wave function $\psi(t,s,\phi)$, assuming that the circular edge of the membrane, at $s = R$, is held fixed. This gives rise to the boundary condition

\begin{equation} \psi(t, s=R, \phi) = 0. \tag{11.58} \end{equation}

We take the initial conditions to be given by

\begin{equation} \psi(t=0, s, \phi) = f(s,\phi), \qquad \frac{\partial \psi}{\partial t}(t=0, s, \phi) = g(s,\phi), \tag{11.59} \end{equation}

where $f(s,\phi)$ and $g(s,\phi)$ are arbitrary functions.

The starting point, as usual, is the factorized solution $\psi = T(t) S(s) \Phi(\phi)$, which we insert within the wave equation. We obtain \begin{equation} \frac{1}{v^2 T} \frac{d^2 T}{dt^2} = \frac{1}{sS} \frac{d}{ds} \biggl( s \frac{dS}{ds} \biggr) + \frac{1}{s^2 \Phi} \frac{d^2 \Phi}{d\phi^2}, \tag{11.60} \end{equation}

and conclude that both sides of the equation must be equal to a constant, which we write as $-k^2$. The solutions for $T(t)$ are

\begin{equation} T(t) = \left\{ \begin{array}{l} \cos(\omega t) \\ \sin(\omega t) \end{array} \right. , \tag{11.61} \end{equation}

where $\omega := k v$. The equation becomes

\begin{equation} \frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2} = -k^2 s^2 -\frac{s}{S} \frac{d}{ds} \biggl( s \frac{dS}{ds} \biggr), \tag{11.62} \end{equation}

and again we have that both sides must be equal to a constant, which we write as $-m^2$. The solutions for $\Phi(\phi)$ are

\begin{equation} \Phi(\phi) = \left\{ \begin{array}{l} \cos(m\phi) \\ \sin(m\phi) \end{array} \right. ,\tag{11.63} \end{equation}

and single-valuedness of the wave function requires that $m = 0, 1, 2, 3, \cdots$ --- recall the discussion following Eq.(10.48) in Sec.10.5. With this the wave equation becomes

\begin{equation} s \frac{d}{ds} \biggl( s \frac{dS}{ds} \biggr) + (k^2 s^2 - m^2) S = 0, \tag{11.64} \end{equation}

or

\begin{equation} s^2 \frac{d^2 S}{ds^2} + s \frac{dS}{ds} + (k^2 s^2 - m^2) S = 0, \tag{11.65} \end{equation}

which is the same form of Bessel equation that we encountered back in Sec.10.5 --- refer to Eq.(10.56). The solutions are

\begin{equation} S(s) = \left\{ \begin{array}{l} J_m(ks) \\ N_m(ks) \end{array} \right., \tag{11.66} \end{equation}

and we have obtained the factorized solutions to the two-dimensional wave equation in polar coordinates. They are given by

\begin{equation} \psi_{k,m}(t,s,\phi) = \left\{ \begin{array}{l} J_m(ks) \\ N_m(ks) \end{array} \right\} \left\{ \begin{array}{l} \cos(m\phi) \\ \sin(m\phi) \end{array} \right\} \left\{ \begin{array}{l} \cos(\omega t) \\ \sin(\omega t) \end{array} \right\} , \tag{11.67} \end{equation}

where $m = 0, 1, 2, 3, \cdots$, $k$ is an arbitrary parameter, and $\omega := k v$.

Modes of vibration of a circular membrane
Figure 11.5: Modes of vibration of a circular membrane with $R=1$. The modes are labelled with two integers, $m$ and $p$, and they are evaluated at $t=0$.

The factorized solutions may now be refined to enforce the boundary conditions. An implicit condition is that the wave function should be finite at $s=0$, and this requires the elimination of $N_m(ks)$. The condition of Eq.(11.58) implies that $kR$ must be a zero of $J_m$, so that $k = \alpha_{mp}/R$, where $\alpha_{mp}$ is the $p^{\rm th}$ zero of the $m^{\rm th}$ Bessel function. The factorized solutions are now restricted to

\begin{equation} \psi_{mp}(t,s,\phi) = J_m(\alpha_{mp} s/R) \left\{ \begin{array}{l} \cos(m\phi) \\ \sin(m\phi) \end{array} \right\} \left\{ \begin{array}{l} \cos(\omega_{mp} t) \\ \sin(\omega_{mp} t) \end{array} \right\} , \tag{11.68} \end{equation}

where

\begin{equation} \omega_{mp} := \alpha_{mp} v/R. \tag{11.69} \end{equation}

The functions of Eq.(11.68) are the normal modes of vibration of the circular membrane, and $\omega_{mp}$ are the membrane's natural frequencies. A sample of modes are plotted in Fig.11.5.

The general solution to the wave equation will be a superposition of normal modes. Because there are four possible choices among Eq.(11.68), the general superposition must be written as

\begin{align} \psi(t,s,\phi) &= \sum_{m=0}^\infty \sum_{p=1}^\infty J_m(\alpha_{mp} s/R) \Bigl[ a_{mp} \cos(m\phi) \cos(\omega_{mp} t) + b_{mp} \cos(m\phi) \sin(\omega_{mp} t) \nonumber \\ & \quad \mbox{} + c_{mp} \sin(m\phi) \cos(\omega_{mp} t) + d_{mp} \sin(m\phi) \sin(\omega_{mp} t) \Bigr], \tag{11.70} \end{align}

where $a_{mp}$, $b_{mp}$, $c_{mp}$, and $d_{mp}$ are expansion coefficients that must be determined from the initial conditions of Eq.(11.59).

Let us examine an example. We consider the initial conditions

\begin{equation} f(s,\phi) =h \bigl[ 1 - (s/R)^2 \bigr], \qquad g(s,\phi) = 0, \tag{11.71} \end{equation}

which describe a parabolic deformation of a membrane initially at rest. The parabola has its maximum at the centre of the membrane, $s = 0$, and the initial height of the wave is $h$. Because the initial conditions do not depend on $\phi$, the membrane possesses a rotational symmetry, and the solution to the wave equation will also be independent of $\phi$. This implies that only terms with $m=0$ survive in Eq.(11.70), and we get

\begin{equation} \psi(t,s) = \sum_{p=1}^\infty J_0(\alpha_{0p} s/R) \Bigl[ a_{0p} \cos(\omega_{0p} t) + b_{0p} \sin(\omega_{0p} t) \Bigr]. \tag{11.72} \end{equation}

This can be refined further. The membrane is initially at rest, which means that $\partial \psi/\partial t$ vanishes at $t = 0$. The cosine terms in $\psi$ give rise to sine terms after differentiation with respect to $t$, and these will all properly vanish at $t = 0$. The sine terms, however, give rise to cosine terms after differentiation, and these will not vanish at $t = 0$, in violation of the initial conditions. We must therefore eliminate the sine terms by setting $b_{0p} = 0$.

We are left with

\begin{equation} \psi(t,s) = \sum_{p=1}^\infty a_{0p} J_0(\alpha_{0p} s/R)\,\cos(\omega_{0p} t), \tag{11.73} \end{equation}

and evaluating this at $t = 0$ while incorporating the initial condition of Eq.(11.71) gives

\begin{equation} h ( 1 - u^2 ) = \sum_{p=1}^\infty a_{0p} J_0(\alpha_{0p} u), \tag{11.74} \end{equation}

where $u := s/R$. This is a Bessel series for the parabolic function, and we learned how to invert this series back in Sec.8.4. According to Eq.(8.35), the expansion coefficients are given by

\begin{equation} a_{0p} = \frac{8 h}{ (\alpha_{0p})^3 J_1(\alpha_{0p}) }. \tag{11.75} \end{equation}

The unique solution to the two-dimensional wave equation for the initial conditions of Eq.(11.71) is therefore

\begin{equation} \psi(t,s) = 8 h \sum_{p=1}^\infty \frac{1}{ (\alpha_{0p})^3 J_1(\alpha_{0p}) }\, J_0(\alpha_{0p} s/R)\,\cos(\omega_{0p} t), \tag{11.76} \end{equation}

where $\omega_{0p} := \alpha_{0p} v/R$. Snapshots of this wave are shown in Fig.11.6.

Parabolic wave on a circular membrane
Figure 11.6: Parabolic wave on a circular membrane with $R=1$. For the computation we set $h = 1$ and $v = 1/2$.

11.8 Infinite Membrane

In Sec.11.5 we considered an infinite string, and discovered that thanks to the absence of boundaries, the solution to the one-dimensional wave equation took the simple form displayed in Eq.(11.27), $\psi(t,x) = A(x-vt) + B(x+vt)$. This describes the superposition of a right-moving wave and a left-moving wave. The solution is simple because the absence of fixed ends implies that the wave will not undergo multiple reflections as it travels along the string.

In view of these observations it might be expected that waves on an infinite membrane would admit a simpler description than what was provided in Secs.11.6 and 11.7. The absence of reflections from a fixed edge should, after all, produce a substantial simplification. While this is true, there are other factors that ensure that the description is a far cry from the simple superposition of right-moving and left-moving waves obtained in the one-dimensional case. One complicating factor is that the wave can now travel in different directions on the two-dimensional medium. Another is that a two-dimensional wave will not keep its original shape as it travels on the membrane; the wave tends to spread out.

If we adopt Cartesian coordinates, the absence of boundaries on the membrane implies that the factorized solutions revert back to the form given by Eq.(11.48), which we rewrite as

\begin{equation} \psi_{k_x, k_y}(t,x,y) = e^{\pm i k_x x} e^{\pm i k_y y} e^{\pm i\omega t}, \tag{11.77} \end{equation}

where $k_x$ and $k_y$ are arbitrary parameters, and $\omega := v (k_x^2 + k_y^2)^{1/2}$. With $k_x$ and $k_y$ allowed to be both positive and negative, the choice of factorized solutions can be restricted to $e^{i(k_x x + k_y y \pm \omega t)}$, and the general solution to the two-dimensional wave equation can be written as the superposition

\begin{equation} \psi(t,x,y) = \int_{-\infty}^\infty \int_{-\infty}^\infty \Bigl[ a(k_x, k_y) e^{i(k_x x + k_y y - \omega t)} + b(k_x, k_y) e^{i(k_x x + k_y y + \omega t)} \Bigr]\, dk_x dk_y, \tag{11.78} \end{equation}

where $a(k_x, k_y)$ and $b(k_x, k_y)$ play the role of expansion coefficients. This solution is in the form of a two-dimensional Fourier transform. It cannot be expressed more simply than this.

We could also attempt to solve the wave equation in polar coordinates, but this would lead to a form of solution that is even more complicated. Simplification occurs when the wave is independent of the angle $\phi$. In this case the wave is traveling either toward larger values of $s$ --- an outgoing wave --- or toward smaller values of $s$ --- an incoming wave. But even in this simple case the solution has the rather complicated expression

\begin{equation} \psi(t,s) = \int_0^\infty \frac{A(\beta + s - vt)}{\sqrt{\beta^2+2\beta s}}\, d\beta + \int_0^\infty \frac{B(\beta + s + vt)}{\sqrt{\beta^2+2\beta s}}\, d\beta, \tag{11.79} \end{equation}

where $A$ and $B$ are arbitrary functions of their arguments. The first integral describes an outgoing wave, while the second describes an incoming wave.

We won't work through a derivation of Eq.(11.79), but we can verify that any integral of the form

\begin{equation} \Psi(t,s) := \int_0^\infty \frac{C(\beta + s \pm vt)}{\sqrt{\beta^2+2\beta s}}\, d\beta, \tag{11.80} \end{equation}

where $C$ is an arbitrary function, is a solution to

\begin{equation} 0 = \Box \Psi = -\frac{1}{v^2} \frac{\partial^2 \Psi}{\partial t^2} + \frac{1}{s} \frac{\partial}{\partial s} \biggl( s \frac{\partial\Psi}{\partial s} \biggr). \tag{11.81} \end{equation}

The identity

\begin{equation} \Box \biggl[ \frac{C(\beta + s \pm vt)}{\sqrt{\beta^2+2\beta s}} \biggr] = \frac{d}{d\beta} \biggl[ \frac{\beta^2 C(\beta+s \pm vt)}{s(\beta^2 + 2\beta s)^{3/2}} \biggr] \tag{11.82} \end{equation}

is the key input in this exercise. We write

\begin{align} \Box \Psi &= \Box \int_0^\infty \frac{C(\beta + s \pm vt)}{\sqrt{\beta^2+2\beta s}}\, d\beta \nonumber \\ &= \int_0^\infty \Box \biggl[ \frac{C(\beta + s \pm vt)}{\sqrt{\beta^2+2\beta s}} \biggr]\, d\beta \nonumber \\ &= \int_0^\infty \frac{d}{d\beta} \biggl[ \frac{\beta^2 C(\beta+s \pm vt)}{s(\beta^2 + 2\beta s)^{3/2}} \biggr]\, d\beta \nonumber \\ &= \frac{\beta^2 C(\beta+s \pm vt)}{s(\beta^2 + 2\beta s)^{3/2}} \biggr|^\infty_{0} \nonumber \\ &= 0, \tag{11.83} \end{align}

and obtain that $\Psi$ does indeed satisfy the wave equation. In the last step of the calculation, the boundary term at $\beta = \infty$ vanishes because with $C$ assumed to be bounded in the limit, the function behaves as $\beta^{-1}$ when $\beta$ is large. The boundary term at $\beta = 0$ vanishes also, because the function behaves as $\beta^{1/2}$ when $\beta$ is small.


Exercise 11.5: Verify the identity of Eq.(11.82) by evaluating the derivatives on both sides of the equation.


11.9 Three-Dimensional Wave

In three dimensions the wave equation is

\begin{equation} 0 = \Box \psi = -\frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2} + \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2}, \tag{11.84} \end{equation}

and it can be written in alternative forms by adopting different coordinates. We can think of finding solutions to this equation by relying once more on the good old technique of constructing factorized solutions. These would depend on the coordinate system, and on the boundary conditions that we might wish to impose on $\psi$.

For example, we might consider sound waves traveling in a sound-insulated room. In this case we would require $\psi$ to vanish on each wall of the room (including floor and ceiling), and this would make a straightforward generalization of the two-dimensional problem examined in Sec.11.6. Another example would be a study of electromagnetic waves confined to an optical cavity, or to a wave guide. We shall not pursue these topics here, as it would just be more of the same; the old methods work just as well in three dimensions.

We can learn something new by examining a spherical wave in an infinite three-dimensional medium. This is a wave that is independent of the angles $\theta$ and $\phi$ when we describe it in terms of spherical coordinates. The wave function depends on $t$ and $r$ only, and it moves either toward larger values of $r$ --- an outgoing wave --- or toward smaller values of $r$ --- an incoming wave. Writing the Laplacian operator in spherical coordinates, as in Eq.(1.86), and eliminating the angular derivatives, we find that the wave equation becomes

\begin{equation} 0 = \Box \psi = -\frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2} +\frac{1}{r^2}\frac{\partial}{\partial r} \biggl( r^2 \frac{\partial \psi}{\partial r} \biggr). \tag{11.85} \end{equation}

We wish to find the general solution to this equation.

The equation looks complicated, but it simplifies dramatically when we define a new variable $u := r\psi$. With $\psi = r^{-1} u$ we find that $\partial \psi/\partial r = r^{-1} \partial u/\partial r - r^{-2} u$, so that $r^2 \partial \psi/\partial r = r \partial u/\partial r - u$. Another round of differentiation gives

\begin{equation} \frac{\partial}{\partial r} \biggl( r^2 \frac{\partial \psi}{\partial r} \biggr) = r \frac{\partial^2 u}{\partial r^2} + \frac{\partial u}{\partial r} - \frac{\partial u}{\partial r} = r \frac{\partial^2 u}{\partial r^2}, \tag{11.86} \end{equation}

and we arrive at

\begin{equation} \frac{1}{r^2}\frac{\partial}{\partial r} \biggl( r^2 \frac{\partial \psi}{\partial r} \biggr) = \frac{1}{r} \frac{\partial^2 u}{\partial r^2} \tag{11.87} \end{equation}

for $\nabla^2 \psi$ in spherical coordinates. On the other hand, differentiation with respect to $t$ yields

\begin{equation} \frac{\partial^2 \psi}{\partial t^2} = \frac{1}{r}\frac{\partial^2 u}{\partial t^2}, \tag{11.88} \end{equation}

and all this gives

\begin{equation} \Box \psi = \frac{1}{r} \biggl( -\frac{1}{v^2} \frac{\partial^2 u}{\partial t^2} + \frac{\partial^2 u}{\partial r^2} \biggr)\tag{11.89} \end{equation}

for the wave operator acting on $\psi(t,r)$.

We see that the wave equation becomes

\begin{equation} -\frac{1}{v^2} \frac{\partial^2 u}{\partial t^2} + \frac{\partial^2 u}{\partial r^2} = 0 \tag{11.90} \end{equation}

when it is written in terms of the new variable $u = r \psi$. We recognize this as the one-dimensional form of the wave equation, presented in a different notation, with $r$ playing the role of $x$. The general solution to this equation is $u = A(r-vt) + B(r+vt)$, and this immediately implies that the general solution to Eq.(11.85) is

\begin{equation} \psi(t,r) = \frac{1}{r} A(r-vt) + \frac{1}{r} B(r+vt). \tag{11.91} \end{equation}

As expected, this represents the superposition of an outgoing wave described by $r^{-1} A(r-vt)$ and an incoming wave described by $r^{-1} B(r+vt)$. The functions $A$ and $B$ are arbitrary, just as in the case of the infinite string, and they can be related to the initial conditions imposed on the wave. The factor of $r^{-1}$ in front of $A(r-vt)$ indicates that the wave's amplitude decreases as the wave expands toward $r=\infty$; the wave becomes increasingly attenuated, a familiar phenomenon in the context of sound waves. On the other hand, the factor of $r^{-1}$ in front of $B(r+vt)$ reveals that the amplitude increases as the wave converges toward $r=0$.

11.10 Practice Problems

  1. Find the solution $\psi(t,x)$ to the one-dimensional wave equation that describes a wave on a string of length $L$ with initial conditions $\psi(0,x) = f(x)$ and $(\partial \psi/\partial t)(0,x) = 0$. The function $f(x)$ is equal to $2hx/L$ when $0 \leq x < \frac{1}{2}L$ and $2h(L-x)/L$ when $\frac{1}{2}L \leq x < L$.

  2. (Boas Chapter 13, Section 4, Problem 5) Find the solution $\psi(t,x)$ to the one-dimensional wave equation that describes a wave on a string of length $L$ with initial conditions $\psi(0,x) = 0$ and $(\partial \psi/\partial t)(0,x) = g(x)$. The function $g(x)$ is equal to $2hx/L$ when $0 \leq x < \frac{1}{2}L$ and $2h(L-x)/L$ when $\frac{1}{2}L \leq x < L$.

  3. (Boas Chapter 13, Section 4, Problem 6) Find the solution $\psi(t,x)$ to the one-dimensional wave equation that describes a wave on a string of length $L$ with initial conditions $\psi(0,x) = 0$ and $(\partial \psi/\partial t)(0,x) = g(x)$. The function $g(x)$ is equal to $0$ when $0 \leq x < \frac{1}{2} L - w$, $h$ when $\frac{1}{2} L - w \leq x < \frac{1}{2} L + w$, and $0$ when $\frac{1}{2} L + w \leq x < L$.

  4. Find the solution $\psi(t,x)$ to the one-dimensional wave equation that describes a wave on a string of length $L$ with initial conditions $\psi(0,x) = h\sin(\pi x/L)$ and $(\partial \psi/\partial t)(0,x) = 0$.\

  5. (Boas Chapter 13, Section 6, Problem 2) Find the six lowest natural frequencies $\omega_{mp}$ of a circular membrane, and express them as multiples of the fundamental frequency.

  6. (Boas Chapter 13, Section 6, Problem 8) Consider the Schr\"odinger equation $-\hbar^2 \nabla^2 \Psi = 2m E\, \Psi$ for a quantum particle of mass $m$ confined to a two-dimensional, circular region of radius $R$. Find the allowed values for the particle's energy $E$. Hint: You are meant to solve the partial differential equation in polar coordinates $(s,\phi)$, and to impose the boundary condition $\Psi(s=R, \phi) = 0$.

 

11.11 Challenge Problems

  1. A vibrating string of length $L = \pi$ has an initial displacement and initial velocity described by

    \[\psi(0,x) = 0, \qquad\frac{\partial \psi}{\partial t}(0,x) = \sin(2x) + \sin(4x).\]

    What is $\psi(t,x)$ for these initial conditions?

  2. A vibrating, circular membrane of radius $R$ has an initial displacement described by

    \[\psi(0,s,\phi) = J_1(\alpha_{13} s/R) \cos\phi,\]

    where $\alpha_{13}$ is the third zero of $J_1(x)$. The initial velocity vanishes. What is $\psi(t,s,\phi)$ for these initial conditions?