From Example 5, here:
\begin{align*}& \frac{{\partial u}}{{\partial t}} = k\frac{{{\partial ^2}u}}{{\partial {x^2}}}\\ & u\left( {x,0} \right) = f\left( x \right)\hspace{0.25in}u\left( { - L,t} \right) = u\left( {L,t} \right)\hspace{0.25in}\frac{{\partial u}}{{\partial x}}\left( { - L,t} \right) = \frac{{\partial u}}{{\partial x}}\left( {L,t} \right)\end{align*}
The solution to the heat equation system on a thin circular ring is given by
$$u\left( {x,t} \right) = \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}} + \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}}$$
Is there a standard way to show that this solution $u(x, t)$ is smooth? It of course seems obvious due to the decaying exponential coefficients.
I suspect that there should be some simple theorem to directly conclude that each higher-order derivative exists and is continuous, but I couldn't find one right away.
