$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ #[Chapter 4. Separation of variables and Fourier Series](id:chapter-4) In this Chapter we consider simplest separation of variables problems, arising simplest eigenvalue problems and corresponding Fourier series. ##[4.1. Separation of variables (the first blood)](id:sect-4.1) ----------------------------------------- > 1. [Separation of variables](#sect-4.1.1) > 2. [Eigenvalue problem](#sect-4.1.2) > 2. [Simple solutions](#sect-4.1.3) > 3. [General solutions](#sect-4.1.4) Consider IBVP for 1D-wave equation on $(0,l)$: \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) &&u\_t|\_{t=0}=h(x).\label{eq-4.1.3} \end{align} ###[Separation of variables](id:sect-4.1.1) 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.](id:remark-4.1.1)** 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 $T(t)$ 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.](id:remark-4.1.2)** 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 it as two equalities \begin{equation\*} \frac{T''(t)}{T(t)}=-c^2\lambda,\qquad \frac{ X''(x)}{X(x)}=-\lambda \end{equation\*} 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} ###[Egenvalue problem](id:sect-4.1.2) Consider BVP (for ODE) (\ref{eq-4.1.7})--(\ref{eq-4.1.6}). We need to find its solution $X(x)$ which is not identically $0$. **[Definition 1.](id:definition-4.1.1)** Such solutions are called *eigenfunctions* and corresponding numbers $\lambda$ *eigenvalues* (compare with *eigenvectors* and *eigenvalues*.) **[Proposition 1.](id:prop-4.1.1)** (\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.* (\ref{eq-4.1.9}) 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 $\lambda\ne 0$). 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{equation\*} \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} \iff 2\sqrt{-\lambda}l=2\pi in \end{equation\*} with $n=1,2,\ldots$. Here we excluded $n=0$ since $\lambda \ne 0$ and both $n$ and $-n$ lead to the same $\lambda$ and $X$. The last equation is equivalent to (\ref{eq-4.1.9}). Then $k\_{1,2}=\frac{\pi n}{l} i$. Meanwhlie $B=-A$ anyway and we get $X=2Ai \sin (\frac{\pi n x}{l})$ i.e. (\ref{eq-4.1.10}) as 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. ###[Simple solutions](id:sect-4.1.3) 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 $k\_{1,2}=\pm \frac{c\pi n}{l}i$ and therefore \begin{equation} T\_n(t)=A\_n \cos(\frac{c\pi n}{l}t ) + BA\_n \sin (\frac{c\pi n}{l}t ) \label{eq-4.1.13} \end{equation} and finally \begin{multline} u\_n(x,t)= \underbrace{\bigl( A\_n \cos(\frac{c\pi n}{l}t )+ B\_n \sin (\frac{c\pi n}{l}t )\bigr)}\_{=T\_n(t)} \cdot \underbrace{\sin (\frac{\pi n x}{l})}\_{=X\_n(x)}\\\\ n=1,2,\ldots \label{eq-4.1.14}\qquad\end{multline} This *simple solution* (\ref{eq-4.1.14}) 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](http://en.wikipedia.org/wiki/Standing_wave) ###[General solutions](id:sect-4.1.4) 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 (\ref{eq-4.1.3})_1,2 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 $A\_n$ 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. --------------- [$\Leftarrow$](../Chapter3/S3.2.html)  [$\Uparrow$](../contents.html)  [$\Rightarrow$](./S4.2.html)