$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$

In this Chapter we consider simplest separation of variables problems, arising simplest eigenvalue and eigenfunction problems, and the corresponding Fourier series.

Consider IBVP for homogeneous 1D-wave equation on the *finite interval* $(0,l)$:
\begin{align}
& u_{tt}-c^2 u_{xx}=0,&& 0< x< l, \label{eq-4.1.1}\\[3pt]
& u|_{x=0}=u|_{x=l}=0, \label{eq-4.1.2}\\[5pt]
& u|_{t=0}=g(x), \quad u_t|_{t=0}=h(x).\label{eq-4.1.3}
\end{align}
Note, that boundary conditions are also homogeneous. So inhomogeneous are only initial conditions.

Let us skip temporarily initial conditions (\ref{eq-4.1.3}) and consider only (\ref{eq-4.1.1})--(\ref{eq-4.1.2}) and look for a solution in a special form \begin{equation} u(x,t)= X(x) T(t) \label{eq-4.1.4} \end{equation} with unknown functions $X(x)$ on $(0,l)$ and $T(t)$ on $(-\infty,\infty)$.

**Remark 1.**
We are looking for *non-trivial solution* $u(x,t)$ which means that $u(x,t)$ is not identically $0$.

Therefore neither $X(x)$ nor $T(t)$ could be identically $0$ either.

Plugging (\ref{eq-4.1.4}) into (\ref{eq-4.1.1})--(\ref{eq-4.1.2}) we get \begin{align*} & X(x) T''(t)=c^2 X''(x)T(t), \\[3pt] & X(0)T(t)=X(l)T(t)=0, \end{align*} which after division by $X(x)T(t)$ and $T(t)$ respectively become \begin{align} & \frac{ T''(t)}{T(t)}=c^2 \frac{X''(x)}{X(x)}, \label{eq-4.1.5}\\[3pt] & X(0)=X(l)=0. \label{eq-4.1.6} \end{align} Recall, neither $X(x)$ nor $T(t)$ are identically $0$.

In (\ref{eq-4.1.5}) the l.h.e. does not depend on $x$ and the r.h.e. does not depend on $t$ and since we have an identity we conclude that

**Remark 2.**
Both expressions do not depend on $x,t$ and therefore they are constant.

This is a **crucial conclusion of the separation of variables method**. We rewrite We rewrite the (\ref{eq-4.1.5}) as two equalities
\begin{equation*}
\frac{T''(t)}{T(t)}=-c^2\lambda,\qquad \frac{ X''(x)}{X(x)}=-\lambda,
\end{equation*}
with \alert<2>{unknown constant $\lambda$},
which in turn we rewrite as (\ref{eq-4.1.7}) and (\ref{eq-4.1.8}):
\begin{align}
&X'' +\lambda X=0, \label{eq-4.1.7}\\[3pt]
& X(0)=X(l)=0, \tag{\ref{eq-4.1.6}}\\[5pt]
& T''+ c^2\lambda T=0.\label{eq-4.1.8}
\end{align}

Consider BVP (for ODE) (\ref{eq-4.1.7})--(\ref{eq-4.1.6}). It is usually called *Sturm-Liouville problem*. We need to find its solution $X(x)$ which is not identically $0$.

**Definition 1.**
Such solutions are called *eigenfunctions* and corresponding numbers $\lambda$ *eigenvalues* (compare with *eigenvectors* and *eigenvalues*).

**Proposition 1.**
Problem (\ref{eq-4.1.7})--(\ref{eq-4.1.6}) has eigenvalues and eigenfunctions
\begin{align}
& \lambda_n = \frac{\pi^2n^2}{l^2}&& n=1,2,\ldots, \label{eq-4.1.9}\\[3pt]
&X_n(x)=\sin (\frac{\pi n x}{l}). \label{eq-4.1.10}
\end{align}

*Proof.* Note that (\ref{eq-4.1.7}) is a 2-nd order linear ODE with constant coefficients and to solve it one needs to consider *characteristic equation*
\begin{equation}
k^2+\lambda =0 \label{eq-4.1.11}
\end{equation}
and therefore $k_{1,2}=\pm \sqrt{-\lambda}$ and
$X= Ae^{\sqrt{-\lambda}x} + Be^{-\sqrt{-\lambda}x}$ (provided
$\lambda\ne 0$). So far $\lambda \in \mathbb{C}$.

Plugging into $X(0)=0$ and $X(l)=0$ we get \begin{align*} &A \qquad +B\qquad = 0,\\ &Ae^{\sqrt{-\lambda}l} + B e^{-\sqrt{-\lambda}l}=0 \end{align*} and this system has a non-trivial solution $(A,B)\ne 0$ if and only if its determinant is $0$: \begin{multline*} \left| \begin{matrix} 1 & 1\\ e^{\sqrt{-\lambda}l} & e^{-\sqrt{-\lambda}l}\end{matrix}\right|= e^{-\sqrt{-\lambda}l}-e^{\sqrt{-\lambda}l}=0\iff e^{2\sqrt{-\lambda}l}=1 \\ \iff 2\sqrt{-\lambda}l=2\pi ni \end{multline*} with $n=1,2,\ldots$. Here we excluded $n=0$ since $\lambda \ne 0$, and also $n=-1,-2,\ldots$ since both $n$ and $-n$ lead to the same $\lambda$ and $X$. The last equation is equivalent to (\ref{eq-4.1.9}). Then $k_{1,2}=\pm \frac{\pi n i}{l}$.

Meanwhile $B=-A$ anyway and we get $X=2Ai \sin (\frac{\pi n x}{l})$ i.e. (\ref{eq-4.1.10}) since *the constant factor does not matter*.

So far we have not covered $\lambda=0$. But then $k_{1,2}=0$ and $X=A+Bx$ and plugging into (\ref{eq-4.1.6}) we get $A=A+Bl=0 \implies A=B=0$ and $\lambda=0$ is not an eigenvalue.

After eigenvalue problem has been solved we plug $\lambda=\lambda_n$ into (\ref{eq-4.1.8}): \begin{equation} T''+ (\frac{c\pi n}{l})^2 T=0 \label{eq-4.1.12} \end{equation} which is also a 2-nd order linear ODE with constant coefficients.

Characteristic equation $k^2+ (\frac{c\pi n}{l})^2=0$ has solutions $k_{1,2}=\pm \frac{c\pi n i}{l}$ and therefore
\begin{equation}
T_n(t)=A_n \cos(\frac{c\pi nt}{l} ) + B_n \sin (\frac{c\pi nt}{l}) \label{eq-4.1.13}
\end{equation}
and finally we get a *simple solution*
\begin{multline}
u_n(x,t)=
\underbrace{\bigl( A_n \cos(\frac{c\pi nt}{l} )+ B_n \sin (\frac{c\pi nt}{l} )\bigr)}_{=T_n(t)} \cdot
\underbrace{\sin (\frac{\pi n x}{l})}_{=X_n(x)}\\
\text{with }\ n=1,2,\ldots
\label{eq-4.1.14}\qquad\end{multline}

This *simple solution* (\ref{eq-4.1.14}) can be rewritten as
\begin{equation*}
u_n(x,t) =C_n \cos(\frac{c\pi nt}{l}+\phi_n)\sin (\frac{\pi n x}{l})
\end{equation*}
and represents a *standing wave*

which one can decompose into a sum of running waves

and the general discussion of standing waves could be found in wikipedia.

Points which do not move are called *nodes* and points with maximal amplitude are called *antinodes*.

The sum of solutions of (\ref{eq-4.1.1})-(\ref{eq-4.1.2}) is also a solution: \begin{equation} u(x,t)= \sum_{n=1}^\infty \bigl( A_n \cos(\frac{c\pi nt}{l} )+ B_n \sin (\frac{c\pi nt}{l} )\bigr) \cdot \sin (\frac{\pi n x}{l}). \label{eq-4.1.15} \end{equation} We have an important question to answer:

Have we covered all solutions of (\ref{eq-4.1.1})-(\ref{eq-4.1.2})? -- Yes, we did, but we need to justify it.

Plugging (\ref{eq-4.1.15}) into initial conditions (\ref{eq-4.1.3})_{1,2} (that is $u|_{t=0}=g(x)$ and $u_t|_{t=0}=h(x)$) we get respectively
\begin{align}
&\sum_{n=1}^\infty A_n \sin (\frac{\pi n x}{l})=g(x),\label{eq-4.1.16}\\
&\sum_{n=1}^\infty \frac{c\pi n}{l} B_n \sin (\frac{\pi n x}{l} )=h(x). \label{eq-4.1.17}
\end{align}

How to find coefficients $A_n$ and $B_n$? Do they exist? Are they unique?

What we got are *Fourier series* (actually $\sin$-Fourier series).
And we consider their theory in the several next sections.

More coming! Including multidimensional: