The (one-dimensional) wave equation is the second-order linear partial differential equation $$\frac{\partial^2 f}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2 f}{\partial t^2}\tag{second order PDE}$$ that admits as its solutions functions $f$ of the form $$f=f(x\pm vt),\tag{solution}$$ as can be verified in a straight-forward manner. These solutions have a convenient interpretation that justifies the phrase wave equation.

I noticed there are first-order partial differential equations which have as solutions functions of the form $f(x\pm vt)$: $$\frac{\partial f}{\partial x}=\pm\frac{1}{v}\frac{\partial f}{\partial t}\tag{first order PDE}.$$

A quick Google search shows this is indeed called the first-order wave equation, but it usually shows up in the context of math classes.

So now the question:

Why is the usual second-order PDE favored over these first-order ones if both admit the same solutions? Is there a physical reason? Are this first-order equations useful in their own right?

Perhaps there other solutions that one admits that isn't desired, or maybe it just looks cleaner since one doesn't have to carry around the $\pm$ symbol in the differential equation.

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    $\begingroup$ It will have not the same solutions if e.g. $v(x)\ne \text{const}$. Also, you can't set two boundary conditions for a first-order PDE. $\endgroup$
    – Ruslan
    Commented Mar 30, 2014 at 15:47
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    $\begingroup$ The Dirac equation is a first order wave equation. $\endgroup$ Commented Mar 30, 2014 at 15:48
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    $\begingroup$ @Raskolnikov it's not just an equation, it's a system of equations. Usual (second-order) wave equation could also be rewritten as a system of first order PDEs. $\endgroup$
    – Ruslan
    Commented Mar 30, 2014 at 15:49
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    $\begingroup$ It's still first order. $\endgroup$ Commented Mar 30, 2014 at 15:50
  • $\begingroup$ As is often the case, posting a question leads me to magically stumble upon relevant information. One of these first-order PDEs is known as the Advection Equation, though I'd still appreciate insight as to why the second-order form is preferred to the first-order form when dealing with waves. $\endgroup$
    – BMS
    Commented Mar 30, 2014 at 15:55

4 Answers 4


There's nothing wrong with the first order wave equation mathematically, but it's just a little boring. If you want to use this equation to describe waves, it basically amounts to having a 1d solid with speed of sound $v$ for left moving waves (say) and speed of sound $0$ for right moving waves. It wouldn't surprise me if such a thing could be constructed (you would have to introduce some external fields to break time reversal invariance) but it is a very special system that we are not generically interested in.

Let's take the Fourier transforms of both equations to get the dispersion relationships. The normal second order equation gives \begin{equation} \omega^2=v^2 k^2 \end{equation} So for each frequency $\omega$ there are two allowed values of $k$, corresponding to right and left moving waves. Note that if we generalize the second order equation to include more spatial directions, there would be an infinite number of allowed $k$ values.

The first order equation meanwhile always has one allowed solution for a given frequency \begin{equation} \omega=v k \end{equation} So we get either right moving or left moving waves but not both. This restricts the allowed behavior, you can't have standing waves for example. If I try to generalize to higher dinensions, this equation picks out a single allowed $k$ for each frequency, so waves will only propagate along one very special direction.

Physically this is not what we would normally call a wave bc I only need one initial condition, not two. Usually dynamical systems can only be evolved given their initial position and velocity, but the first order equation needs only the initial position. (Or if you like, your equation is not a Hamiltonian system bc the phase space is odd dimensional).

Last but not least the first order equation necessarily picks out a preferred frame. By doing a boost I can change the sign of $v$, Thus the equation is not a good starting point for dealing with relativistic waves, which is one major application for the wave equation. (Of course you can have waves in materials that do pick out a preferred frame, and that is fine, but there you run into the problems above that you are looking at something with a preferred direction of motion as well).

(The Dirac equation gets around this by using spinor reps of the Lorentz group, but from your question I am supposing $f$ is a scalar).

Edit: rereading your question I see you want to have the $\pm$. Then you aren't looking at solutions of one single equation, you are looking at solutions to two equations and saying both are allowed. This is a little ugly for a few reasons. First,philosophically there should be one single equation for any system. Second, super positions don't solve either first order equation separately but do solve the second order equation. Third, the analogue of your idea for more spatial dimensions is to have an infinite set of first order equations, one for each direction.

On the other hand there is a way to rewrite the second order equation as two first order equations in a way which generalizes to any dimension, this is the way of the Hamiltonian and it is indeed a very useful thing to do in many situations.

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    $\begingroup$ This is what I call an answer that cuts to the chase. I thoroughly enjoyed reading it. $\endgroup$ Commented Mar 31, 2014 at 23:05
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    $\begingroup$ It’s perhaps interesting to note that while the first-order wave equation in itself isn’t interesting (and even more boring in the comoving frame of the wave), it’s what’s you use as a starting point for slightly nonlinear variations on the wave equation: dependence of speed on amplitude (Hopf equation), dissipation (Burgers equation), dispersion (Korteweg–de Vries equation), etc. $\endgroup$ Commented Apr 7, 2018 at 3:00

There are a few reasons I can think of:

(1) The second order system is that it is time-reversible. If you let $t\to-t$, you get $$ \frac{\partial^2f}{\partial(-t)^2}=\frac{\partial^2f}{\partial t^2}=v^2\frac{\partial^2f}{\partial x^2} $$ whereas the first order system has $$ \frac{\partial f}{\partial(-t)}=-\frac{\partial f}{\partial t}=\pm v\frac{\partial f}{\partial x} $$ which is now a different equation because of the minus sign.

(2) Superposition. If $f(x+vt)$ and $f(x-vt)$ are both solutions, then $F(x,t)=af(x+vt)+bf(x-vt)$ is also a solution. Inserting this into the second order equation, $$ \frac{\partial^2F}{\partial t^2}=av^2f(x+vt)+bv^2f(x-vt)\\ v^2\frac{\partial^2F}{\partial x^2}=v^2\left(af(x+vt)+bf(x-vt)\right) $$ and the first order equation $$ \frac{\partial F}{\partial t}=avf(x+vt)-bvf(x-vt) \\ \pm v\frac{\partial F}{\partial x}=\pm v\left(af(x+vt)+bf(x-vt)\right) $$ which are different results (i.e., super-position is not held here except maybe for particular cases).

However, in terms of solving such equations numerically, it is far easier to use the first order equation because less memory is required. The two equations can be approximated via the finite differences $$ \frac{f(x,t+dt)+f(x,t-dt)-2f(x,t)}{\Delta t^2}=v^2\frac{f(x+dx,t)+f(x-dx,t)-2f(x,t)}{\Delta x^2} $$ $$ \frac{f(x,t+dt)-f(x,t)}{\Delta t}=\pm v\frac{f(x+dx,t)-f(x-dx,t)}{2\Delta x} $$ The first equation (second order) requires storing the previous, current, and future time steps whereas the second equation (first order) requires only the previous time steps.

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    $\begingroup$ I kind of disagree with this answer. 1) time-reversal not only requires $t \to -t$ but also $v\to -v$; as such the two minus signs cancel each other and the equation still retain the same form. 2) Superposition always holds for linear equations (of any order): why are you looking at solutions of the form $f(x\pm vt)$? Those are solutions only of the second order equations and they obviously need not hold for the first order one. $\endgroup$
    – gented
    Commented Nov 11, 2016 at 22:59

One very useful application of the first-order wave equation is in describing the propagation of the slowly varying envelope of a nearly sinusoidal wave.

Suppose we have a wave satisfying the second-order wave equation that is known to be nearly sinusoidal, with an envelope $a$ that varies slowly both in space (with a small variation over a wavelength) and in time (with a small variation over an oscillation period). In the case of light waves, for instance, this applies in many situations, even for light fields considered to be "pulsed," or otherwise time-varying. In complex notation we can write the positive- or negative-propagating wave as $$f=ae^{i(\pm kx-\omega t)},$$ with $k$ positive.

Substituting into the second-order wave equation and taking $k=\omega/v$, we have $$\pm\frac{2i\omega}{v}\frac{\partial a}{\partial x}+\frac{\partial^2a}{\partial x^2}=-\frac{2i\omega}{v^2}\frac{\partial a}{\partial t}+\frac{1}{v^2}\frac{\partial^2a}{\partial t^2}.$$ Under the slowly varying envelope approximation (SVEA) we neglect the second-order derivatives relative to the first-order, and are left with $$\frac{\partial a}{\partial x}=\mp\frac{1}{v}\frac{\partial a}{\partial t},$$ i.e., the first-order wave equation.

This greatly simplifies numerical computations, especially since the relatively high-frequency and high-wavenumber oscillation at $\omega$ and $k$ has been removed from explicit consideration.


First order wave equations are common, it's just that the first order wave equation over the real numbers is very boring. The Schrodinger equation, Maxwell's equations, and the Dirac equation are all first order in time.

The next simplest example is $\frac{\partial f}{\partial t} = i \frac{\partial f}{\partial x}$. Taking the time derivative on both sides and rewriting using the original equation, $\frac{\partial^2 f}{\partial t^2} = i \frac{\partial^2 f}{\partial t \partial x} = i^2 \frac{\partial^2 f}{\partial x \partial x} = - \frac{\partial^2 f}{\partial x \partial x}$. So the first order wave equation with complex coefficient is equivalent to the second order wave equation $\frac{\partial^2 f}{\partial t^2} = - \frac{\partial^2 f}{\partial x^2}$. The second order equation needs 2 initial conditions ($f(0,x)$ and $\frac{\partial f}{\partial t}(0,x))$, and the complex first order equation also needs 2 initial conditions (real and imaginary parts).

The Maxwell equations are directly analogous to this but with multiple space dimensions, and the Dirac equation is closely related too.


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