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How do I solve:

$$u_t = -u_{xxxx} + \pi^2u_{xx},$$ with BCs: $u_x(0,t)=u_x(1,t)=u_{xxx}(0,t)=u_{xxx}(1,t)=0$ and initial condition $u(x,0)=\cos(\pi x)$.

We have been told that we can use separation of variables however I can't seem to get the required solution $\cos(\pi x)\exp(-2\pi^4t)$.

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  • $\begingroup$ Enclose your math in \$$\dotsc$\$ to make your post more readable. Use $\{\dotsc\}$. In other words learn some basic TeX. $\endgroup$ – Oskar Limka Apr 10 '15 at 10:02
  • $\begingroup$ You may want to do some Fourier transforms, to get an ordinary differential equations before attempting separation of variables. Alternatively (which is basically the same), you write $u(x,t)$ as a time-dependent coefficients Fourier series in $x$. $\endgroup$ – Oskar Limka Apr 10 '15 at 10:07
  • $\begingroup$ pea91, the separation of variables thing is huge I think since it means the solution "u is a product in which the dependence of u on x, t is separated". @KittyL shows how it's done. en.wikipedia.org/wiki/… $\endgroup$ – user198044 Apr 10 '15 at 10:32
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Hint: Assume $u=X(x)T(t)$, you will get

$$XT'=-X^{(4)}T+\pi^2X''T=T(-X^{(4)}+\pi^2X'')$$ $$\frac{X}{-X^{(4)}+\pi^2X''}=\frac{T}{T'}$$

Then you can continue from there.

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  • $\begingroup$ Equate each side to $- \lambda$ or something? $\endgroup$ – user198044 Apr 10 '15 at 10:17
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    $\begingroup$ @Jack: Yes. They would be constant. $\endgroup$ – KittyL Apr 10 '15 at 10:30
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    $\begingroup$ I did do that however I struggled solving the differential equations for each case of lambda. For example lambda less than zero trying to solve the characteristic equation of the DE I get a square root inside a square root! $\endgroup$ – pea91 Apr 10 '15 at 10:41
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    $\begingroup$ @pea91: If you mean the $X$ part, you don't need to solve for that. Solve the $T$ part first and the initial condition will give you the $X$ part. $\endgroup$ – KittyL Apr 10 '15 at 10:51
  • $\begingroup$ Solving for $T$ I get $T$=cos(pi*x)exp(lambda*t)/$X$. How do I then use my $X$ Equation to calculate that lambda=-2pi^4, so I get the correct solution? $\endgroup$ – pea91 Apr 10 '15 at 11:38

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