Wave Equation

The wave equation is a linear second-order partial differential equation which describes the propagation of oscillations at a fixed speed in some quantity $y$:

A solution to the wave equation in two dimensions propagating over a fixed region [1].

$$\frac{1}{v^2} \frac{\partial^2 y}{\partial t^2} = \frac{\partial^2 y}{\partial x^2},$$

where $v$ is the velocity of the wave.

The equation is a good description for a wide range of phenomena because it is typically used to model small oscillations about an equilibrium, for which systems can often be well approximated by Hooke's law. Solutions to the wave equation are of course important in fluid dynamics, but also play an important role in electromagnetism, optics, gravitational physics, and heat transfer. Especially important are the solutions to the Fourier transform of the wave equation, which define Fourier series, spherical harmonics, and their generalizations.

The derivation of the wave equation varies depending on context. A particularly simple physical setting for the derivation is that of small oscillations on a piece of string obeying Hooke's law. Consider the below diagram showing a piece of string displaced by a small amount from equilibrium:

Small oscillations of a string (blue). On a small element of mass contained in a small interval $dx$, tensions $T$ and $T^{\prime}$ pull the element downwards.

Consider the forces acting on a small element of mass $dm$ contained in a small interval $dx$. If the displacement is small, the horizontal force is approximately zero. The vertical force is

$$\sum F_y = -T^{\prime} \sin \theta_2 - T \sin \theta_1 = (dm) a = \mu dx \frac{\partial^2 y}{\partial t^2},$$

where $\mu$ is the mass density $\mu = \frac{\partial m}{\partial x}$ of the string.

On the other hand, since the horizontal force is approximately zero for small displacements, $T \cos \theta_1 \approx T^{\prime} \cos \theta_2 \approx T$. Therefore,

$$-\frac{\mu dx \frac{\partial^2 y}{\partial t^2}}{T} \approx \frac{T^{\prime} \sin \theta_2+ T \sin \theta_1}{T} =\frac{T^{\prime} \sin \theta_2}{T} + \frac{ T \sin \theta_1}{T} \approx \frac{T^{\prime} \sin \theta_2}{T^{\prime} \cos \theta_2}+ \frac{ T \sin \theta_1}{T \cos \theta_1} = \tan \theta_1 + \tan \theta_2.$$

However, $\tan \theta_1 + \tan \theta_2 = -\Delta \frac{\partial y}{\partial x}$, where the difference is between $x$ and $x + dx$. This is because the tangent is equal to the slope geometrically. Dividing over $dx$, one finds

$$-\frac{\mu \frac{\partial^2 y}{\partial t^2}}{T} = \frac{\tan \theta_1 + \tan \theta_2}{dx} = -\frac{ \Delta \frac{\partial y}{\partial x}}{dx}.$$

The rightmost term above is the definition of the derivative with respect to $x$ since the difference is over an interval $dx$, and therefore one has

$$\frac{\mu}{T} \frac{\partial^2 y}{\partial t^2} = \frac{\partial^2 y}{\partial x^2},$$

which is exactly the wave equation in one dimension for velocity $v = \sqrt{\frac{T}{\mu}}$.

Many derivations for physical oscillations are similar. Below, a derivation is given for the wave equation for light which takes an entirely different approach.

Another derivation can be performed providing the assumption that the definition of an entity is the same as the description of an entity.

So, a wave is a squiggly thing, with a speed, and when it moves it does not change shape:

The squiggly thing is $f(x)$, the speed is $v$, and the red graph is the wave after time $t$ given by a graph transformation of a translation in the $x$-axis in the positive direction by the distance $vt$ (the distance travelled by the wave travelling at constant speed $v$ over time $t$): $f(x-vt)$.

Now, since the wave can be translated in either the positive or the negative $x$ direction, I do not think anyone will mind if I change $f(x-vt)$ to $f(x\pm vt)$. Now, I am going to let $u = x \pm vt$, so differentiating with respect to $x$, keeping $t$ constant,

$$\partial u = \partial x,$$

and differentiating with respect to $t$, keeping $x$ constant,

$$\partial u = \pm v \partial t.$$

Now, I have a $\pm$ sign, which I do not like, so I think I am going to take the second derivative later, which will introduce a square value of $v^2$.

So, let me take the second derivative of $f$ with respect to $u$ and substitute the various $\partial u$:

$$\frac{\partial}{\partial u} \left( \frac{\partial f}{\partial u} \right) = \frac{\partial}{\partial x} \left(\frac{\partial f}{\partial x} \right) = \pm \frac{1}{v} \frac{\partial}{\partial t} \left(\pm \frac{1}{v} \frac{\partial f}{\partial t}\right) \implies \frac{\partial^2 f}{\partial u^2} = \frac{\partial^2 f}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 f}{\partial t^2}.$$

So we finally have the wave equation:

$$\frac{\partial^2 f}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 f}{ \partial t^2}.$$

All solutions to the wave equation are superpositions of "left-traveling" and "right-traveling" waves, $f(x+vt)$ and $g(x-vt)$. These are called left-traveling and right-traveling because while the overall shape of the wave remains constant, the wave translates to the left or right in time. Furthermore, any superpositions of solutions to the wave equation are also solutions, because the equation is linear.

The fact that solutions to the wave equation are superpositions of "left-traveling" and "right-traveling" waves is checked explicitly in this wiki. Here a brief proof is offered:

Prove that all solutions to the wave equation are superpositions of "left-traveling" and "right-traveling" waves.

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Define new coordinates $a = x - vt$ and $b=x+vt$ representing right and left propagation of waves, respectively. Then the partial derivatives can be rewritten as

$$\begin{aligned} \frac{\partial}{\partial x}&= \frac12 (\frac{\partial}{\partial a} + \frac{\partial}{\partial b}) \implies \frac{\partial^2}{\partial x^2} = \frac14 \left(\frac{\partial^2}{\partial a^2}+2\frac{\partial^2}{\partial a\partial b}+\frac{\partial^2}{\partial b^2}\right) \\ \frac{\partial}{\partial t} &=\frac{v}{2} (\frac{\partial}{\partial b} - \frac{\partial}{\partial a}) \implies \frac{\partial^2}{\partial t^2} = \frac{v^2}{4} \left(\frac{\partial^2}{\partial a^2}-2\frac{\partial^2}{\partial a\partial b}+\frac{\partial^2}{\partial b^2}\right). \end{aligned}$$

Since the wave equation is

$$\frac{\partial^2 y}{\partial x^2} - \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2} = 0,$$

substituting in for the partial derivatives yields the equation in the coordinates $a$ and $b$:

$$\frac{\partial^2 y}{\partial a \partial b} = 0.$$

This is solved in general by $y = f(a) + g(b) = f(x-vt) + g(x+vt)$ as claimed. $_\square$

The most commonly used examples of solutions are harmonic waves:

$$y(x,t) = A \sin (x-vt) + B \sin (x+vt) ,$$

where $y_0$ is the amplitude of the wave and $A$ and $B$ are some constants depending on initial conditions.

A superposition of left-propagating and right-propagating traveling waves creates a standing wave when the endpoints are fixed [2].

If the boundary conditions are such that the solutions take the same value at both endpoints, the solutions can lead to standing waves as seen above. These take the functional form

$$y(x,t) = y_0 \sin (x-vt) + y_0 \sin (x+vt) = 2y_0 \sin(x) \cos (vt),$$

where $y_0$ is the amplitude of the wave. The solution has constant amplitude and the spatial part $\sin (x)$ has no time dependence.

Formally, there are two major types of boundary conditions for the wave equation:

• Dirichlet boundary conditions: The amplitude is fixed at the boundary point $x^{\ast}$. This can be written as $$y(x^{\ast}) = y^{\ast},$$ where $y^{\ast}$ is some fixed constant. A more useful way of writing this condition is $$\biggl.\frac{dy}{dt}\biggr|_{x=x^{\ast}} = 0,$$ which says that the amplitude at the endpoint does not change over time.
• Neumann boundary conditions: The derivative of the amplitude is specified at the endpoints. Often, this derivative is taken to be zero. For an approximately massless (very light) string, this means the endpoints are free rather than fixed; for a pipe with pressure waves inside of it, this means that the endpoints are open to the atmosphere: $$\biggl.\frac{dy}{dx}\biggr|_{x=x^{\ast}} = 0.$$ A string with Dirichlet boundary conditions at the left end, where the string is fixed to a wall, and Neumann boundary conditions at the right end, where the string is attached to a freely sliding ring

A string of tension $T$ and mass density per unit length $\mu$ is attached to a small massless ring which slides on a slippery rod. The rod exerts a damping force $F_d= - b\frac{\partial y}{\partial t}$ on the ring. Find the boundary condition on the oscillations of the string at the end attached to the ring.

A string attached to a ring sliding on a slippery rod.

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The ring is free to slide, so the boundary conditions are Neumann and since the ring is massless the total force on the ring must be zero. Consider the following free body diagram:

All vertically acting forces on the ring at the end of the oscillating string.

Using the fact that the wave equation holds for small oscillations only, $dx \gg dy$. Balancing the forces in the vertical direction thus yields

$$-T \frac{\partial y}{\partial x} - b \frac{\partial y}{\partial t} = 0 \implies \frac{\partial y}{\partial x} = -\frac{b}{T} \frac{\partial y}{\partial t}.$$

This slope condition is the Neumann boundary condition on the oscillations of the string at the end attached to the ring. $_\square$

One way of writing down solutions to the wave equation generates Fourier series which may be used to represent a function as a sum of sinusoidals. This method uses the fact that the complex exponentials $e^{-i\omega t}$ are eigenfunctions of the operator $\frac{\partial^2}{\partial t^2}$. Using this fact, ansatz a solution for a particular $\omega$:

$$y(x,t) = e^{-i\omega t} f(x),$$

where the exponential has essentially factored out the time dependence. Plugging into the wave equation, one finds

$$\frac{\partial^2 y}{\partial t^2} = -\omega^2 y(x,t) = v^2 \frac{\partial^2 y}{\partial x^2} = v^2 e^{-i\omega t} \frac{\partial^2 f}{\partial x^2}.$$

The function $f$ therefore satisfies the equation

$$\frac{\partial^2 f}{\partial x^2} = -\frac{\omega^2}{v^2} f.$$

This is solved by the plane waves

$$f(x) = f_0 e^{\pm i \omega x / v}.$$

Therefore, the general solution for a particular $\omega$ can be written as

$$y(x,t) = f_0 e^{i\frac{\omega}{v} (x \pm vt)} .$$

This is consistent with the assertion above that solutions are written as superpositions of $f(x-vt)$ and $g(x+vt)$ for some functions $f$ and $g$. By the linearity of the wave equation, an arbitrary solution can be built up in terms of superpositions of the above solutions that have $\omega$ fixed. This is exactly the statement of existence of the Fourier series.

The wave equation governs a wide range of phenomena, including gravitational waves, light waves, sound waves, and even the oscillations of strings in string theory. Depending on the medium and type of wave, the velocity $v$ can mean many different things, e.g. the speed of light, sound speed, or velocity at which string displacements propagate.

Prove that light obeys the wave equation directly from Maxwell's equations.

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Begin by taking the curl of Faraday's law and Ampere's law in vacuum:

$$\begin{aligned} \vec{\nabla} \times (\vec{\nabla} \times \vec{E}) &= - \frac{\partial}{\partial t} \vec{\nabla} \times \vec{B} = -\mu_0 \epsilon_0 \frac{\partial^2 E}{\partial t^2} \\ \vec{\nabla} \times (\vec{\nabla} \times \vec{B}) &= \mu_0 \epsilon_0 \frac{\partial}{\partial t} \vec{\nabla} \times \vec{E} = -\mu_0 \epsilon_0 \frac{\partial^2 B}{\partial t^2}. \end{aligned}$$

Now using the vector identity:

$$\vec{\nabla} \times (\vec{\nabla} \times A) = \vec{\nabla} (\vec{\nabla} \cdot A)-\vec{\nabla}^2 A,$$

the left-hand sides can also be rewritten. Since $\vec{\nabla} \cdot \vec{E} = \vec{\nabla} \cdot \vec{B} = 0$ according to Gauss' laws for electricity and magnetism in vacuum, this reduces to

$$\vec{\nabla} \times (\vec{\nabla} \times \vec{E}) = -\vec{\nabla}^2 \vec{E}, \qquad \vec{\nabla} \times (\vec{\nabla} \times \vec{B}) = -\vec{\nabla}^2 \vec{B}.$$

Equating both sides above gives the two wave equations for $\vec{E}$ and $\vec{B}$

$$\vec{\nabla}^2 E = \mu_0 \epsilon_0 \frac{\partial^2 E}{\partial t^2}, \qquad \vec{\nabla}^2 B = \mu_0 \epsilon_0 \frac{\partial^2 B}{\partial t^2}.$$

Since it can be numerically checked that $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, this shows that the fields making up light obeys the wave equation with velocity $c$ as expected. $_\square$

A plasma is an ionized gas, typically very hot. Ripples in a plasma, in the form of perturbations $\rho$ to the plasma density, satisfy a modified wave equation

$$v^2 \frac{\partial^2 \rho}{\partial x^2} - \omega_p^2 \rho = \frac{\partial^2 \rho}{\partial t^2},$$

where $v$ is the speed at which the perturbations propagate and $\omega_p^2$ is a constant, the plasma frequency.

What is the frequency of traveling wave solutions for small velocities $v \approx 0?$

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Ansatz a solution $\rho = \rho_0 e^{i(kx - \omega t)}$. Plugging in, one finds the equation

$$-v^2 k^2 \rho - \omega_p^2 \rho = -\omega^2 \rho,$$

i.e. the dispersion relation:

$$\omega^2 = \omega_p^2 + v^2 k^2 \implies \omega = \sqrt{\omega_p^2 + v^2 k^2}.$$

For small velocities $v \approx 0$, the binomial theorem gives the result

$$\omega \approx \omega_p + \frac{v^2 k^2}{2\omega_p}.$$

The size of the plasma frequency $\omega_p$ thus sets the dynamics of the plasma at low velocities. $_\square$

[1] By BrentHFoster - Own work, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=38870468.

[2] Image from https://upload.wikimedia.org/wikipedia/commons/7/7d/Standing_wave_2.gif under Creative Commons licensing for reuse and modification.