# PDE from dispersion relation?

Suppose I have knowledge of a system's dispersion relation $f(\omega,k)$. Is it possible to recover the underlying PDE describing the system? Can I simply use the replacement $k=-i\nabla$, $\omega=i\frac{d}{dt}$ to go back?

I came across one source which claimed that the original PDE could only be recovered to a certain extent. An example given was the Dirac and Klein Gordon equations which both satisfy the same dispersion relation. But I didn't quite follow.

Example: I have a polynomial dispersion relation of the form $\omega^2+c^2k^2=1$. Can I automatically say that the underlying PDE is $\frac{d^2u(x,t)}{dt^2}+c^2\nabla^2u(x,t)+u(x,t)=0$? Or is there a subtlety I'm missing?

Thanks

• Can you give an example with equations and stuff? – Your Majesty Feb 23 '14 at 14:31
• Ok, added an example. – user2053563 Feb 23 '14 at 20:35

Of course you can restore the original equation from dispersion relation, at least sometimes. But I think your equations of $k = -i\nabla$ is not correct, because $k$ is a number the other is an operator, you cannot put = between them. Be careful when you abuse the notation and know what you are doing.

But you are on the right track.

The first condition your equation should obey is the superposition principle. That is the linear combinations of the solutions should be also a solution. In order to do that you need to bring your equation in the form like this:

$$A\Psi = 0$$

in the entire domain of $\Psi$ (see why). There $A$ is an expression that evaluates to 0, and $\Psi(x, t)$ is your wave function. $x$ is a spatial coordinate, $t$ is time.

The wave function is a linear combination of elementary plane waves which can be expressed in the form (and is a solution to the wave equation in itself):

$$\Psi(x, t) = e^{i(kx - \omega t)}$$

There $k$ is the wave number $\omega$ is the frequency.

The second ingredient is seeing how this function react to the differential operators.

Wave equations usually contain a Laplacian operator. Since that connects the points in the space and allow the disturbance to spread. If you differentiate $\Psi$ with respect to $x$ twice you get:

$$\nabla^2 \Psi = -k^2\Psi$$

And time derivatives will be in the form:

$$\partial_t^n \Psi = \left( -i \omega \right)^n \Psi$$

(There $\partial_t^n$ is a shorthand for $\frac{\partial^n}{\partial t^n}$).

Now all you need to do is, combine these two equations, substitute the dispersion relation for $\omega$, and fiddle with it till the right side equals to zero. One important thing that $k$ shouldn't appear in the left side, otherwise, your equation will be true for a single particular $k$ which is obviously not what you want.

There is no standard method how to fiddle with these equations. It's a mathematical puzzle which probably doesn't always have a solution.

My examples are in 1 dimension but it's straightforward to generalize it to multiple dimensions.

Example

Consider light which have a very simple dispersion relation:

$$\omega = ck$$

In that case the time derivatives are:

$$\partial_t^n \Psi = \left( -i c k \right)^n \Psi$$

The second derivative looks promising:

$$\partial_t^2 \Psi = -c^2 k^2 \Psi$$

Combining them into an equation:

$$\nabla^2 \Psi + \partial_t^2 \Psi = -k^2 \Psi - c^2 k^2 \Psi$$

Multiplying the second time derivative with $-1/c^2$ we get:

$$\nabla^2 \Psi - \frac{1}{c^2}\partial_t^2 \Psi = -k^2 \Psi + k^2 \Psi = 0$$

Reordering we just arrive at the wave equation:

$$\partial_t^2 \Psi = c^2 \nabla^2 \Psi$$

And the same thing can be done to obtain the Schrödinger or Klein-Gordon equations, from their respective dispersion relations.