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I just started to learn a little bit of fourier analysis in solving PDEs. I want to find a solution $u(x,t)$ to $u_t+u_{xxxx}+u_{xx}=0$.

My attempt: Applying the fourier transform to both sides gives $\frac {d}{dt}\hat u(\omega,t)+(i\omega)^4\hat u(\omega,t)+(i\omega)^2\hat u(\omega,t)$

$\implies \frac {d}{dt}\hat u(\omega,t)=(\omega^2-\omega^4)\hat u(\omega,t)$

This is a first order, autonomous ODE in $\hat u(x,t)$. With integrating factor $e^{\int (\omega^4-\omega^2)d\omega}$, we have $\frac {d}{d\omega}(e^{1/5\omega^5-1/3\omega^3}\hat u(\omega,t))=0$

$\implies e^{1/5\omega^5-1/3\omega^3}\hat u(\omega,t)=C$

$\implies \hat u(\omega,t)=Ce^{1/3\omega^3-1/5\omega^5}$

Does this look alright so far? If so, I'm not sure how I should proceed, as I couldn't find the inverse transform of $e^{1/3\omega^3-1/5\omega^5}$ in any tables.

user153582
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    Normally you use k vector as the transform for spatial (quite an important concept in physics) and $\omega$ for the transform to frequency space. So a Fourier transform tends to look like $\exp(i\omega t +i\textbf{k}\cdot\textbf{x})$. Then you solve a dispersion relation $f(\omega,k)$=0. – Chinny84 Jul 05 '14 at 08:43
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    To be honest I was going to give a full answer, but why bother when there is a great answer here. Give some love there and thumbs up. If you can follow the related link then solve it yourself and fill out the answer part below and everyone benefits, but mainly yourself :). – Chinny84 Jul 05 '14 at 08:54
  • @Chinny84 Thanks for the link. :) I was able to follow the first response, so I guess I should be looking at $\hat u(k,t)=e^{(k^2-k^4)t}$, right? And just use the straight definition for the inverse transform. I was hoping there'd be a short cut, but I'll see what I can do with the integral and post the results – user153582 Jul 05 '14 at 09:10
  • No worries. I will look out for the answer you post :). If you have no success comment here and no doubt one of the fellow members will help :). – Chinny84 Jul 05 '14 at 09:39
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    In the integrating factor, you should have $dt$ instead of $d\omega$. – Hans Lundmark Jul 05 '14 at 10:59

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