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Given the heat equation $u_t$=$u_{xx}$ with period boundary conditions u(x) = u(x+2$\pi$), and u(x,0) = $\sum_{n=1}^{10}$cos(nx),

  1. Find an expression for the maximum value of u at time t > 0?

  2. What happens to the solution as t → ∞?

So I attempted separation of variables and got u(x,0) = $\sum_{k=1}^{\infty}$$B_k$sin($\frac{xk\pi}{l}$), but I have no idea how to proceed. Any hints/links would be helpful.

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As you mentioned by separation of variables $u(x,t)=X(x)T(t)$, the heat equations decouples to $\dot{T} = -ET$ and $X''=-EX$ with $E$ an unknown positive real number (for now) and at this point we have no idea what it is. We might end up finding multiple $E$'s satisfying our conditions (and indeed this is the case). So assuming there is some $E$ like that: $$T_E(t)= e^{-Et}, \quad X_E=A_E \cos (\sqrt{E} x) + B_E \sin (\sqrt{E} x)$$ note that I chose $E$ positive since $E$ negative does not have any periodic solution. Let say a solution corresponding to a value $E$ is "good", if $u(x+2\pi,t)=u(x,t)$. In other words we say a solution is "good" if it is indeed consistent with periodic boundary condition. Hence must have $X_E(x+2\pi)=X_E(x)$ for a good solution. This means $$ \cos (\sqrt{E}(x+2\pi))=\cos (\sqrt{E}x), \quad \sin (\sqrt{E}(x+2\pi))=\sin (\sqrt{E}x) $$ This can only happen if $\sqrt{E}=n$ for some integer $n$, or if $E =n^2$ for "good" solutions. The full solution, before applying initial conditions, would be linear combination of all "good" solutions: $$ u(x,t)=\frac{A_0}{2}+\sum_{n=1}^\infty [A_n \cos (nx)+B_n\sin (nx)]e^{-n^2t} $$ Now comes the initial conditions that $u(x,0)=\sum_{n=1}^{10}\cos (nx)$. By matching this to the solution above, can you finish the job?

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  • $\begingroup$ How did you get the value for $E_n$? Also, I attempted setting the u(x,0) equal to each other, I decided that $B_n$ = 0 and $A_n$ = 1 , and $A_0$ = 2 as well (?). I'm not sure if my steps were logical or not. $\endgroup$ – Tesuji Apr 7 '17 at 3:26
  • $\begingroup$ Yes $B_n=0$ for all $n$, $A_0=0$ (note that your sum runs starts $n=1$) and for $n=1,\cdots, 10$ you have $A_n=1$, for the rest of $n$, you have $A_n=0$. $\endgroup$ – Hamed Apr 7 '17 at 3:31
  • $\begingroup$ Whoops oh yea. How did you determine $E_n$ I tried comparing the U(x+2$\pi$) and u(x) but the sin functions and cos functions just didn't change at all. $\endgroup$ – Tesuji Apr 7 '17 at 3:34
  • $\begingroup$ I edited the solution slightly to make it clearer where the values of $E$ are coming from. $\endgroup$ – Hamed Apr 7 '17 at 3:39
  • $\begingroup$ Ok that makes a lot of sense. Thank you so much. $\endgroup$ – Tesuji Apr 7 '17 at 3:47

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