I need some help on solving this heat PDE :

Question : Consider a bar length L. The face at x=0 is insulated so that the heat flow across is zero, and the face at x=L is held at temperature u=0. The temperature distribution is governed by heat equation

$$ {\partial u \over \partial t} - k {\partial ^2 u \over \partial x^2} =0$$

Show that the normal modes of u(x,t) are

$$\large U_n(x,t)=B_n\cos[(2n-1)\pi x/2L]e^{[-(2n-1)^2\pi^2k^2t]/4L^2}$$

Given BCs are : u(0,t)=0 and u(L,t)=0

I already solve this PDE

and I got my answer in this form :

$$\large U_n(x,t)=B_n\sin[n\pi x/2L] e^{[-(n)^2\pi^2k^2t]/4L^2}$$

How am I need to change my answer into the form they asking ??? Anyone willing to help me ?

  • $\begingroup$ Isn't the B.C. at x=0 expressed as $u_t (0,t) = 0$, as the temperature at $x=0$ isn't changing due to the insulation? $\endgroup$ – Ron Gordon Jan 8 '13 at 17:52
  • $\begingroup$ No it cant. Because boundary condition is given in interval 0<x<L. The one you said only applies for ICs only. But in this case ICs are not given $\endgroup$ – Garett Jan 8 '13 at 17:55
  • $\begingroup$ I'm pretty sure rlgordonna is correct: if the heat flow across x=0 is 0, that means $u_t (0,t) = 0$. This should lead to the right eigenfunction, too. $\endgroup$ – AndrewG Jan 8 '13 at 18:06
  • $\begingroup$ Ok let me try it out first and then I confirm back to you again $\endgroup$ – Garett Jan 8 '13 at 18:11
  • $\begingroup$ I cant get it. I think it mucg be easier if you could could show me the steps on how you done it Andrew Gibson. Please help me out $\endgroup$ – Garett Jan 8 '13 at 18:18

You need to express the problem as

$$u_t = k u_{xx} $$

$$ u_x(0,t) = 0 ,\;\;\; u(L,t) = 0 $$

Use separation of variables, i.e. $u(x,t) = X(x)T(t)$ and get

$$X'' + \frac{\lambda}{k} X = 0 $$

where $-\lambda$ is the separation constant. The solution to this equation is

$$ X(x) = A \cos{\left (\sqrt{\frac{\lambda}{k}} x \right)} + B \sin{\left (\sqrt{\frac{\lambda}{k}} x \right)}$$

The condition at $x=0$ implies that $B=0$:

$$ X'(x) = -\sqrt{\frac{\lambda}{k}} A \sin{\left (\sqrt{\frac{\lambda}{k}} x \right)} + B \sqrt{\frac{\lambda}{k}}\cos{\left (\sqrt{\frac{\lambda}{k}} x \right)}$$

$$X'(0) = B \sqrt{\frac{\lambda}{k}} = 0$$

The condition at $x=L$ implies that

$$\cos{\left (\sqrt{\frac{\lambda}{k}} L \right)} = 0$$

so that

$$\lambda = \left [ (2 n-1) \frac{\pi}{2 L} \right ]^2 k $$

You should be able to take it from there with the time equation.

  • $\begingroup$ I just cant get the part where you do X'(x). Shouldnt the BCs $u_t(0,t)=O$ which is $X(0)T'(t)=0$?? Why you did X'(x) rlgordonma? Could you explain $\endgroup$ – Garett Jan 8 '13 at 18:37
  • $\begingroup$ My mistake - it should be $u_x(0,t) = 0$. I will fix it above. $\endgroup$ – Ron Gordon Jan 8 '13 at 18:40
  • $\begingroup$ For some reason, I mixed up holding a fixed temp., and heat flow, with is a temp. flux, i.e. a gradient, i.e., a spatial derivative. I hope that clears it up. $\endgroup$ – Ron Gordon Jan 8 '13 at 18:43
  • $\begingroup$ Ok now I understand. But just one thing could you teach me how you indentify it is $u_x(0,t)=0$ rather than $u_t(0,t)=0$ or $u(0,t)=0$. Tell me some tips so that I can understand clearly about this part $\endgroup$ – Garett Jan 8 '13 at 18:46
  • $\begingroup$ My comment above should explain some of it. The B.C.'s are described using certain terms. If the temp is held fixed to zero at one end, then $u=0$ at that end. If the temp is held fixed over time in general, then $u_t = 0$, which obviously returns a similar result to just setting $u=0$ at that end. (My mistake in being so glib before and confusing you.) If there is no heat flow, or the end is insulated, then $u_x = 0$ at that end. $\endgroup$ – Ron Gordon Jan 8 '13 at 18:52

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