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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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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. – Ruslan Mar 30 '14 at 15:47
The Dirac equation is a first order wave equation. – Raskolnikov Mar 30 '14 at 15:48
@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. – Ruslan Mar 30 '14 at 15:49
It's still first order. – Raskolnikov Mar 30 '14 at 15:50
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. – BMS Mar 30 '14 at 15:55
up vote 9 down vote accepted

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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This is what I call an answer that cuts to the chase. I thoroughly enjoyed reading it. – WetSavannaAnimal aka Rod Vance Mar 31 '14 at 23:05

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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