Why is the wave equation a second order differential?

[thegirl]

I don't know if this is a silly question? Am I missing simple math? How does a wave depending on amplitude and frequency make it's equation a second order differential equation?

 

[blue_leaf77]

Actually it's the other way around, every system which is described by a second order type differential equation in space and time supports the propagation of waves.

 

[UncertaintyAjay]

I don't know much about second order differential equations, only about the equations for harmonic oscillators. So if any second order diff. eq describes a system that supports wave propagation, that means that the solution or solutions to any second order differential equation must have a bit that's sinusoidal right?

 

[boneh3ad]

Actually it's the other way around, every system which is described by a second order type differential equation in space and time supports the propagation of waves.

More correctly, shouldn't it be any system described by a hyperbolic differential equation supports wave propagation?

 

[vanhees71]

Actually it's the other way around, every system which is described by a second order type differential equation in space and time supports the propagation of waves.

Mh, I'd say, it should be a hyperbolic partial differential equation, because, e.g., the Laplace equation doesn't describe waves. The sign pattern in the corresponding differential operator of the linear case is important, i.e., for waves it should be a D'Alembertian rather than a Laplacian to lead to wave propgation solutions.

 

[blue_leaf77]

Thanks boneh3ad, I missed the requirement on the coefficients.

 

[nasu]

Actually what we call usually "wave equation" it has that form because we make a lot of approximations. You can have much more complicated equations (with higher order differential terms) describing waves that do not satisfy various approximations implicit in the simple wave equation.
So there is nothing special about second order. Is maybe the lowest order that allow for a wave solution.
Same as there is nothing special about a lot of "laws" being linear. Just the first order approximation,

 

[blue_leaf77]

because we make a lot of approximations

For mechanical waves, may be. But Maxwell equations are no approximation.

 

[nasu]

I did not see Maxwell's equations mentioned in the OP. Or EM waves only.
But this is a good (interesting) point.

 

[blue_leaf77]

I mean the EM wave equation is build from Maxwell's equations.

But this is a good (interesting) point.

I can agree with you, this shows that the EM waves are an example of a truly sinusiodal disturbance in nature provided they propagate in free space.

 

[boneh3ad]

Actually what we call usually "wave equation" it has that form because we make a lot of approximations.

Well, not really. What we call the wave equation has that form because it is a simple model of a wave (or really a pair of waves) and is readily solvable and a good analog for all other hyperbolic equations. We certainly make a lot of approximations to other equations to make it resemble the wave equation more closely, but the wave equation itself is not an approximation. It's really just more of a model equation for others.

Is maybe the lowest order that allow for a wave solution.

No, there are first order equations that admit wave-like behavior. The wave equation itself can be factored into two first-order advection equations (below) which represent an left-running and right-running wave individually.

\dfrac{\partial \phi}{\partial t} \pm c\dfrac{\partial \phi}{\partial x} = 0