Motivation

Consider the following boundary value problem:

\[u_t = k u_{xx}, \quad u(0,t) = u(l,t) = 0,\]

Separate the variables by taking $u(x,t) = X(x) T(t)$, this becomes

\[T'(t)/kT(t) = X''(x)/X(x)\]

along with $X(0) = X(l) = 0$. Since the LHS depends solely on $t$ and the RHS on $x$, both must be equal to a constant, $A$:

\[T'(t) = AkT(t), X''(x) = AX(x).\]

The general solution for the equation for $T$ is

\[T(t) = C_0 e^{Akt}\]

, while for $X$ is

\[X(x) = C_1 \cos\lambda x + C_2 \sin \lambda x\]

where $\lambda = - \sqrt{-A}$. The boundary condition $X(0) = 0$ forces $C_1 = 0$, hence $C_2 \sin\lambda l =0$. For nontrivial solutions, $\lambda l = n\pi$ for some $n\in \mathbb{Z}$, in other words $A = -(n\pi/l)^2$. We take $n>0$ since $n=0$ gives zero solution and $n\mapsto -n$ amounts to $C_2 \mapsto -C_2$. Hence, taking $C_0 = C_2 = 1$, for every $n\in\mathbb{N}^+$ we have a solution $u_n(x,t)$ for the boundary value problem, namely,

\[u_n(x,t) = \exp\left(\frac{-n^2\pi^2 k t}{l^2}\right)\sin\frac{n\pi x}{l}.\]

Linear combinations of the $u_n$'s are also solutions. Hence we can take an infinite linear combination $u = \sum a_n u_n$. Does it converge?

Now with an initial condition $u(x,0) = f(x)$, provided that $f(x) = \sum a_n \sin\frac{n\pi x}{l}$, the solution $u$ shall do the job. The problem now becomes the determination of $a_n$. Now the problem becomes: how to expand $f(x)$ in terms of $\sin\frac{n\pi x}{l}$? Can it be done?

If the boundary condition is not Dirichlet but Neumann, i.e., $u_x(0,t) = u_x(l,t) = 0$, then the solution $u$ becomes the previous $u$ with $\sin$ replaced by $\cos$ The problem becomes how to expand $f(x)$ in terms of $\cos\frac{n\pi x}{l}$. Can it be done?

To simplify the issue, we note that $f$ is a function on an interval $(0,l)$. A function on an interval $(0,l)$ is a periodic function with period $l$ restricted to $(0,l)$ (with some subtleties on $nl$'s; which we for the moment ignore), thus we can without loss of generality study periodic functions $f(\theta + 2\pi) = f(\theta)$. The expansion we want can be written as

\[f(\theta) = \frac{1}{2}a_0 + \sum (a_n \cos n\theta + b_n \sin n\theta),\]

or even better,

\[f(\theta) = \sum c_n e^{in\theta}\]

, but this time with $n\in\mathbb{Z}$.