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Consider a metal rod, with $K=1$ and longitude $\pi$ .

On its extremes, there is heat transmission with the ambient, according to the differential equation: $$u_t(x,t)=u_{xx}(x,t)-hu(x,t),$$ with $h>0$.

The extremes of the rod have fixed temperatures of $0^\circ C$ on the left, and $1^\circ C$ on the right. The initial temperature of the rod is the function: $$f(x)=x(\pi-x).$$

Find the temperature distribution $u(x,t)$ on the rod.

Could someone please explain me how to start? I've never seen something like that (with an $hu(x,t)$ added or taken from the equation). I'm really stuck.

Thank you beforehand, and sorry for my english, it's not my natural language.

JohnD
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Federico
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1 Answers1

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It sounds like you know how to solve the "usual" heat equation, $u_t=u_{xx}$ by separation of variables. Is that correct? If so, proceed in exactly the same way as you did before, i.e. letting $u(x,t)=X(x)T(t)$.

The eigenvalue problem will be just like before, but the temporal problem will look like ${T'\over T}+h=-\lambda$.

JohnD
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  • Oh, thank you for your answer, i really appreciate it. But I have a little problem:

    When i do that, i say: $u(x,t)=X(x)T(t)$, so i replace it on the equation:

    $$-T(t)/T(t)=X''(x)/X(x)-h = -\lambda²$$

    The problem is that, when i solve the temporal problem:

    $$-T'(t)/T(t) + h = - \lambda²$$ $$-T'(t)=(-\lambda²-h)*T(t)$$

    Is that right? Then it is homogeneous...

    Sorry for asking again :(

     – Federico Dec 16 '12 at 18:45
  • Looks good. Keep in mind that you would solve the $T$ problem after you've finished solving the $X$ problem, so the values of $\lambda^2$ would be known, and thus the $T$ problem is easily solvable (in terms of $\lambda$ and $h$). – JohnD Dec 16 '12 at 18:58
  • Thank you very, very much! One more question and i'll stop bothering you: why did you say

    $$T′/T+h=0$$ instead of $$T′/T+h=\lambda$$ ?

    Am i missing something?

    Sorry again for asking, and thank you very much.

     – Federico Dec 16 '12 at 19:00
  • No, I just mistyped. Fixed now. – JohnD Dec 16 '12 at 19:44