4.1. Separation of variables (the first blood)

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

Chapter 4. Separation of variables and Fourier Series

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)

  1. Separation of variables
  2. Eigenvalue problem
  3. Simple solutions
  4. General solutions

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

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 $T(t)$ 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 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

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

Proposition 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

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

General solutions

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.

$\Leftarrow$  $\Uparrow$  $\Rightarrow$