# Diffusion and dispersion relation

I'm looking at some dispersion relations for some complex systems and realised I actually don't have a clear understanding of what physics I can get from a dispersion relation from equations that produce complex relations between frequencies and wavevectors. For instance, if you take a diffusion equation

$$\dot u=\nabla^2u$$

for an initial condition $$u_0(x)=u(x,t=0)$$, then this equation can be simply solved in Fourier space, and the solution is

$$u(x,t) = \int \frac{d^dq}{(2\pi)^d}\tilde u_0(q)e^{-q^2t}e^{i q\cdot r}$$

If the initial equation is transformed to Fourier space also in time we get a complex relation between frequency $$\omega$$ and wavevector $$k$$, $$i\omega+q^2=0.$$

Which will be generaly the case for a scalar field $$u$$ for any overdamped dynamics $$\dot u= F[u]$$ for any $$F$$.

My question is what do we make of a relation of that kind? I'm guessing something along the lines of, since this is overdamped and just a scalar field, you don't expect wave-like solutions, so since an initial condition won't propagate like a wave, there's no real correspondence between $$\omega$$ and $$q$$. I know this is extremely vague, but I'm looking for some physical insight on this.

To expand the question a bit, if $$F[u]$$, for instance $$F[u]=\nabla^2u +A[u]$$ for a nonlinear operator $$A$$, we can 'solve' for $$u$$ through its linear propagator $$\tilde u = \frac{1}{i\omega+q^2}\tilde A[\tilde u]$$

How do we interpret this propagator function physically?

Thanks!

• yes you can solve it in terms of perturbation theory. you can assume $A=0$ then order by order correct the solution. Commented Nov 5, 2020 at 13:35