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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?

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Express $f$ in a Fourier series in space and plug it into the PDE to get an ODE in time for each Fourier coefficient. The $j$'th mode will decay exponentially as $\hat{f_j}(t)=C e^{-kt(j\pi)^2/2}$. – Nick Alger May 7 '12 at 3:27
up vote 2 down vote accepted

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
$$\begin{eqnarray*} a_n &=& 2\int_0^1 dx\, f_0(x) \cos 2 n \pi x \\ b_n &=& 2\int_0^1 dx\, f_0(x) \sin 2 n \pi x. \end{eqnarray*}$$

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Great answer, thanks very much. – Julien Clancy May 7 '12 at 16:23
@JulienClancy: Glad to help. Thank you for the interesting question. – user26872 May 7 '12 at 22:46

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