Heat equation asymptotic behavior This was a question on one of my analysis finals, but I was unable to answer it. It seemed very interesting though.

Suppose that $f(x, t)$ is a solution of the heat equation $\displaystyle \frac{\partial f}{\partial t} = k \frac{\partial^2 f}{\partial x^2},$
  where $k > 0$, with initial condition $f(x, 0) = f_0(x)$ that is continuous
  in [0, 1] and $f_0(0) = f_0(1)$.
Supposing also that $\displaystyle \int^1_0 f_0(x)dx = 0$, explain what happens asymptotically to the function $e^{kt}f(x, t)$ as $t \to \infty$?

Can anyone help me solve this?

Separate variables, let $f(x,t) = X(x)T(t)$. 
We find $T' = -k\omega^2 T$ and $X'' = -\omega^2 X$ where $\omega^2$ is the separation constant.
Thus, $T = e^{-k\omega^2 t}$ and 
$X(x) = A \cos \omega x + B \sin\omega x$. 
To get the asymptotic behavior we need the eigenvalues. 
The boundary conditions $X(0) = X(1) = 0$ and $\int_0^1 dx\, X(x) = 0$ give 
$$\begin{eqnarray*}
-A\sin\omega + B(1-\cos\omega) &=& 0 \hspace{5ex} \textrm{and} \\
\frac{1}{\omega}\left(A(1-\cos\omega) + B \sin\omega\right) &=& 0,
\end{eqnarray*}$$
respectively. 
For a general $\omega$ this implies $A=B=0$. 
However, if $\omega = 2n\pi$ for $n=1,2,\ldots$ these equations will be satisfied for any $A$ and $B$.
(Notice that $n = 0$ is excluded due to the second boundary condition.) 
Therefore, 
$$\begin{eqnarray*}
f(x,t) &=& \sum_{n=1}^\infty \left(a_{n} \cos 2 n \pi x + b_{n} \sin 2n\pi x\right)e^{-4k n^2\pi^2 t} \\
&\sim& (a_1\cos 2\pi x + b_1 \sin 2\pi x)e^{-4k\pi^2t},
\end{eqnarray*}$$
and so $e^{k t}f(x,t) \sim e^{-k(4\pi^2-1)t}$.
That is, $\lim_{t\to\infty} e^{k t}f(x,t) = 0$.
If we generalize the problem slightly and let $x\in[0,l]$ be the region of interest we find 
$e^{k t}f(x,t) \sim e^{k(1-(2\pi/l)^2)t}$. 
There are three distinct scenarios where $f(x,t)$ decays faster than, at the same rate, or slower than $e^{-k t}$ depending on whether $l<2\pi$, $l=2\pi$, or $l>2\pi$. 
Aside: 
Of course, we can get the coefficients from the condition $f(x,0) = f_0(x)$, knowing the form of $f_0$. 
Any $f_0$ satisfying the boundary conditions imposed on the eigenfunctions can be expanded in that basis. 
The coefficients are
$$\begin{eqnarray*}
a_n &=& 2\int_0^1 dx\, f_0(x) \cos 2 n \pi x \\
b_n &=& 2\int_0^1 dx\, f_0(x) \sin 2 n \pi x.
\end{eqnarray*}$$
