Heat Equation with Period Boundary Condition. 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), 


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

*What happens to the solution as t → ∞?
So I attempted separation of variables and got u(x,0) = $\sum_{k=1}^{\infty}$$B_k$sin($\frac{xk\pi}{l}$), but I have no idea how to proceed. Any hints/links would be helpful. 
 A: As you mentioned by separation of variables $u(x,t)=X(x)T(t)$, the heat equations decouples to $\dot{T} = -ET$ and $X''=-EX$ with $E$ an unknown positive real number (for now) and at this point we have no idea what it is. We might end up finding multiple $E$'s satisfying our conditions (and indeed this is the case). So assuming there is some $E$ like that:
$$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?
