In particular, for a paper, I am looking to compare this equation

to the heat equation in one dimension, where $\omega$ is a constant and $i$ is the familiar imaginary unit. My problem is, anytime I multiply by $i$, it can look like either the forward or backward heat equation, and I can't just have it look like either one arbitrarily...

I tried rewriting $\Phi$ in terms of parts with real and imaginary coefficients:

$\Phi(y,t) = N \text{exp}(\frac{i\epsilon}{4\omega}e^{-4\omega t})sin(\sqrt{\epsilon}y)$

$= N [cos(\frac{\epsilon}{4\omega}e^{-4\omega t})+isin(\frac{\epsilon}{4\omega}e^{-4\omega t})]sin(\sqrt{\epsilon}y)$

$= Ncos(\frac{\epsilon}{4\omega}e^{-4\omega t})sin(\sqrt{\epsilon}y) + iNsin(\frac{\epsilon}{4\omega}e^{-4\omega t})sin(\sqrt{\epsilon}y) $

I could then write this as:

$= \Phi_{re} + i\Phi_{im}$

However, when I plug it back into the PDE, I get a system of PDEs with mixed functions...

$e^{4\omega t}\frac{\partial\Phi_{im}}{\partial t} -\frac{\partial^2\Phi_{re}}{\partial y^2} = 0$

$e^{4\omega t}\frac{\partial\Phi_{re}}{\partial t} + \frac{\partial^2\Phi_{im}}{\partial y^2} = 0$

How I can I still compare to the forward and/or backward heat equation? Please help soon, this is due by Friday April 28 for a 15-page paper. I am almost done, except for this.

  • $\begingroup$ Why would you try and compare a wave equation to the heat equation, this is basically a Schrödinger wave equation $\endgroup$ – Triatticus Apr 22 '17 at 19:36
  • $\begingroup$ Because when I classified the heat equation and this equation via the Delta discriminant, I got that both were parabolic. $\endgroup$ – timaeus222 Apr 22 '17 at 19:53
  • $\begingroup$ Well a Schrödinger wave equation IS a diffusion equation like the heat equation but the imaginary unit makes it a wave equation $\endgroup$ – Triatticus Apr 22 '17 at 20:29
  • $\begingroup$ I thought the wave equation had a second-order time derivative? What is it about the imaginary unit that makes it a wave equation PDE instead of a normal diffusion equation? I understand that it functions as a wave equation from a physical point of view, but I don't get why from a mathematical point of view. $\endgroup$ – timaeus222 Apr 22 '17 at 20:31
  • 1
    $\begingroup$ Applied Asymptotic Analysis by P Miller $\endgroup$ – Bananach Apr 22 '17 at 22:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.