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I was just wondering how do I use change of variables to obtain a more suitable equation to solve for the following PDE? If I know how to do that then I am sure I can solve the rest.

$$u_t=Du_{xx}+\alpha u, \ 0<x<1, t>0$$ $$u(0,x)=x(1-x), \ 0<x<1$$ $$u(t,0)=0, \ u_x(t,1)=0, \ t>0$$

I see that if I use $u(x,t)=e^{\alpha t}v(x,t)$ then $v(x,t)$ satsifies $v_t=Dv_{xx}$ but I am not sure how that was found?

Robben
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1 Answers1

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If we write $u(x,t) = g(x,t)v(x,t)$, then your equation becomes

\begin{equation} g_t(x,t)v(x,t) + g(x,t)v_t(x,t) = D\left(g_{xx}(x,t)v(x,t) + 2 g_x(x,t)v_x(x,t) + g(x,t)v_{xx}(x,t)\right) + \alpha g(x,t)v(x,t) \end{equation}

And we want to find $g(x,t)$ such that the equation is simplified. In particular, if we want to recover the heat equation, we can do it by setting $g(x,t) = g(t)$ and then noting that we can make the extraneous terms go away if $g_t(t) = \alpha g(t)$, which has solution $g(t) = \exp(\alpha t)$, so that $v_t(x,t) = Dv_{xx}(x,t)$.

You might need to be a bit careful with your boundary conditions, as now we want them to apply to $v(x,t) = \exp(-\alpha t)u(x,t)$, however in this case we have that $v(x,0) = 1*u(x,0) = x(1-x)$ and $v(0,t) = \exp(-\alpha t)u(0,t) = 0$, $v_x(1,t) = \exp(-\alpha t)u_x(1,t) = 0$ so everything is as nice as can be. In general, though, the boundary conditions will need to be transformed as well.

Baron Mingus
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  • Baron, can you elaborate please? – Robben Nov 22 '14 at 18:48
  • Ah, now I've reread your answer and edit I think what I've written was probably a red herring - as you observe, once you transform your equation in that way so that you can rewrite the equation as the heat equation. You can solve this using Fourier series and then transform back to have a solution in terms of a series to the original equation. – Baron Mingus Nov 22 '14 at 20:52
  • How was that obtained? Where we can rewrite the equation as the heat equation? – Robben Nov 22 '14 at 21:18
  • You wrote it yourself? If $u(x,t) = \exp(\alpha t)v(x,t)$ then $u_t(x,t) = \alpha\exp(\alpha t)v(x,t) + \exp(\alpha t)v_t(x,t)$ and the first term on in that expressions cancels with the second one on the right hand side. As for how it was found - play around with setting $u(x,t) = g(x,t)v(x,t)$ and finding what kinds of $g(x,t)$ do what to the equation satisfied by the transformed variable $v(x,t)$. – Baron Mingus Nov 22 '14 at 21:38
  • I see. Thank you for elaborating! If you want, you can edit your answer with your comment so that I can accept it. – Robben Nov 22 '14 at 21:41
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    Hope that helped. Can you solve it from there? – Baron Mingus Nov 22 '14 at 21:58