I am trying to solve the following exercise:

$u_t=4u_{xx}, 0,x<x<\pi, t>0$

$u(x,0)=g(x), 0\le x\le\pi$

$-u_x(0,t)=u_x(\pi,t)=0, t>0$,

considering that $u(x,t) \to U$ as $t \to \pm \infty$.

I tried so solve it using separation of variables (suppose $u(x,t)=X(x)T(t)$) and using the Neumann boundary conditions and the initial condition in order to find the constants. I came up with the following:

$u(x,t)= \sum_{k=1}^ \infty a_k \cos (kx) e^{-4k^2t}$

$a_0 = \frac{1}{\pi} \int_0^\pi g(x)dx$

$a_k = \frac{2}{\pi} \int_o^\pi f(x) \cos(kx)dx$.

Is this correct? What does it mean that $u(x,t) \to U$ as $t \to \pm \infty$?


From my solution $u(x,t)$ I use the initial condition and get:

$g(x) = \sum_{k=1}^\infty a_k \cos(kx)$.

From this I should use Fourier series to find $a_k$ right?

  • $\begingroup$ as far as I can see, $u(x,t)\to U$ means $u(x,t)$ as a function of $x$ tends to a constant function $U$ as $t\to \infty$. Natural interpretation would be that heat diffuses over the domain indefinitely so heat distribution converges to uniform distribution, as you would expect with diffusion $\endgroup$ – PLE Oct 26 '16 at 16:06
  • $\begingroup$ Thank you @PLE but do I have to do something \textit{particular} in order to take into account this information? Thank you. $\endgroup$ – wrong_path Oct 26 '16 at 16:08
  • $\begingroup$ Something is wrong here: first your $a_k$ should be integrals involving $g(x)$ not $f(x)$. Also, your $u(x,t)$ seems to be wrong - given your solution, as $t\to \infty$ $u\to 0$ for every $x$, so $U=0$ BUT because your boundary conditions imply zero flux the heat cannot escape through boundary (ie it is insulated) so total amount of heat must be conserved, which would eqaul $U\pi$. To this end I think you are missing a term involving $a_0$ in your expression of $u(x,t)$. Once you find that, you know $u(x,t)\to \text{some term involving $a_0$}=U$, so you can in fact find $a_0$ in terms of $U$. $\endgroup$ – PLE Oct 26 '16 at 16:23
  • $\begingroup$ Thank you @PLE yes it is a typo, it is $g(x)$. For the other problem I will try to go through the problem again. Thank you. $\endgroup$ – wrong_path Oct 26 '16 at 16:27

I think this is the correct procedure/answer:



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.