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Solve the equation $$u_t=17u_{xx}, \ 0<x<\pi, \ t>0,$$ with the boundary conditions $$u(0,t)=u(\pi,t)=0, \ t\ge 0,$$ and the initial conditions $$u(x,0)=\left\{ \begin{array}{l l} 0 & \quad \text{if} \ 0\le x\le \pi/2\\ 2 & \quad \text{if} \ \pi/2<x\le\pi \end{array} \right.$$

How will I be able to solve this PDE? My book didn't provide an example of how to solve a PDE of the form $u_t=Au_{xx}$ and I am not sure on how to go about solving it?

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  • $\begingroup$ Which book are you following ? $\endgroup$
    – creative
    Sep 23, 2014 at 6:13
  • $\begingroup$ @Abstraction I am reading, "An introduction of PDE", by Pinchover. $\endgroup$
    – Robben
    Sep 23, 2014 at 6:15
  • $\begingroup$ Also note that the way you have written your initial condition is wrong. You cannot write two inequalities in a same expression. $\endgroup$
    – creative
    Sep 23, 2014 at 6:35
  • $\begingroup$ @Abstraction yes, you're right. It was a latex issue. I dont know how to do brackets. Which is why I had to write it that way. $\endgroup$
    – Robben
    Sep 23, 2014 at 6:37

1 Answer 1

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Consider $u(x,t)=X(x)T(t)$. Then $u_{xx}=X''(x)T(t)$ and $u_t=X(x)T'(t)$. Substitute in the original equation to get

$X(x)T'(t)=17X''(x)T(t).$ So

$\dfrac{T'(t)}{T(t)}=17\dfrac{X''(x)}{X(x)}=-\lambda$. So this gives

$T'(t)+\lambda T=0$ and $X''(x)+\dfrac{\lambda}{17}X=0.$ Can you proceed from here. I think its trivial now.

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  • $\begingroup$ Oh, wow, I didn't know I could use the method of $u(x,t)=X(x)T(t)$ here. I thought it was only a strict form of two second order PDE. I am pretty sure I can proceed thank you very much for the help! $\endgroup$
    – Robben
    Sep 23, 2014 at 6:23
  • $\begingroup$ So, $T'(t)+\lambda T=0$ will have a general solution of the form $T(t)=ce^{kt}$ and $X''(x)+\dfrac{\lambda}{17}X=0,$ will have the form $X(x)=a\cos\frac{\lambda}{17}x +b\sin\frac{\lambda}{17}x$. Is that correct? $\endgroup$
    – Robben
    Sep 23, 2014 at 6:28
  • $\begingroup$ No, $T'(t)+\lambda T(t)=0$ will have $T=ce^{-\lambda t}$ while the second one will have $X(x)=a\cos \sqrt{\frac{\lambda}{17}}x +b\sin\sqrt{\frac{\lambda}{17}}x$ $\endgroup$
    – creative
    Sep 23, 2014 at 6:32
  • $\begingroup$ Forgot square roots. Thank for clarifying! $\endgroup$
    – Robben
    Sep 23, 2014 at 6:35
  • $\begingroup$ Don't forget to find $\lambda$. This is a Sturm-Liouville system. Check the cases $\lambda > 0$, $\lambda < 0$ and $\lambda=0$. Substitute the $\lambda$ obtained from the X equation into the T equation $\endgroup$
    – creative
    Oct 6, 2014 at 6:56

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