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I know that for the wave equation $$\frac{\partial^2\psi}{\partial t^2}=c^2\frac{\partial^2\psi}{\partial x^2} \;, $$ we could always plug in the ansatz $A\cos(kx-\omega t)$, since equations of this form always satisfy the wave equation. However, for the diffusion equation, $$\frac{\partial T}{\partial t}=D\frac{\partial^2 T}{\partial x^2} \;, $$ where $T$ is temperature and $D$ is thermal diffusivity, does $A\cos(kx-\omega t)$ still serve as a valid guess for the solution?

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    $\begingroup$ Have you plugged it in to see if it actually solves the equation? $\endgroup$ – ACuriousMind Jun 24 at 16:39
  • $\begingroup$ well I'd get a cosine on one side and a sine on the other side of the equal sign. So does that mean no? If the answer is no, then how come using $Ae^{ikz-i\omega t}$ seems to solve the equation then? $\endgroup$ – Houndbobsaw Jun 24 at 16:47
  • $\begingroup$ What relationship did you get between $\omega$ and $k$? $\endgroup$ – G. Smith Jun 24 at 17:01
  • $\begingroup$ using the guess $Ae^{ikx-i\omega t}$ I get a relation of $k=\pm\sqrt{\frac{\omega}{D}}e^{i\pi/4}$. I'm not sure if I use the guess $A\cos(kx-\omega t) though. $\endgroup$ – Houndbobsaw Jun 24 at 17:11
  • $\begingroup$ When you solve for $k$ it’s messy. Solve for $\omega$ and then put that into your exponential ansatz. $\endgroup$ – G. Smith Jun 24 at 17:19
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Well, you can always put in any trial solution into any differential equation, and see if it works. In this case, however, as you indicated,

well I'd get a cosine on one side and a sine on the other side of the equal sign.

Since sines and cosines are linearly independent, that means that the function you've given will not work as a solution.


If the answer is no, then how come using $Ae^{ikz-i \omega t}$ seems to solve the equation then?

Because $\cos(kx-\omega t)$ and $e^{ikz-i\omega t}$ are different functions $-$ "related" does not mean "identical". For the specific case of the imaginary-exponential trial solution, the correct procedure is to use the standard matching procedure to obtain the dispersion relationship, i.e. $\omega$ as a function of $k$ (and not the other way around), which will give you $\omega(k) = -iDk^2$.

(Why $\omega=\omega(k)$ and not $k=k(\omega)$? Essentially, because the former generalizes well to 2D and 3D problems, whereas the latter doesn't.)

To find out what happens to the initial condition $T(x,0) = A\cos(kx)$, add together your initial trial $e^{ikz-i\omega(k) t}$ with its counter-propagating counterpart $e^{-ikz-i\omega(-k) t}$, and see what the time dependence simplifies to. You will see that the time dependence is not oscillatory, as would be implied by a trial function of the type $\cos(kx-\omega t)$, which means that it's, as they say, right out.

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