Solving the wave equation with Neumann conditions

Question

I have some difficulties solving this Wave Equation under Neumann BC. Here is what I have so far.

$u_{tt} = 4u_{xx}$ for $0<x<\pi, t>0$

$u_x(0,t) = u_x(\pi$, t) = 0 for t>0

$u(x,0) = 0, u_t(x,0) = \sin(x), 0 <= x <= \pi$

I used the separation of variables method to solve this problem.

$u(x,t) = X(x)T(t)$

$\frac{-X''(x)}{X(x)} = \frac{-1}{4}\frac{T''(t)}{T(t)} = \lambda$

BC: $X'(0) = 0$ and $X'(\pi) = 0$

IC: $T(0) = 0$ and $T'(0) = \sin (x)$

I get $u(x,t) = \frac{1}{2}A_0t + \frac{1}{2}B_0$ + $\sum_{n=1}^\infty(A_n \cos (4nt) + B_n \sin (4nt))\cos(nx)$

I think everything here so far is correct.

Here is where I am stuck:

I need $\psi(x) = \sin (x)$ = $\frac{1}{2}B_0$ + $\sum_{n=1}^\infty$$4n B_n \cos(nx)$

Do I calculate the coefficients $B_n$ using

$4nB_n$ = $\frac{2}{\pi} \int_0^\pi \sin(x) \cos(nx) dx$ ?

or

$B_n$ = $\frac{2}{\pi} \int_0^\pi \sin(x) \cos(nx) dx$

Thanks for any help.

Answer 1

Indeed, everything correct until you stuck. What about $A_n$? You don't mention that $A_n=0\,$. From your expansion $$ \sin{(x)}=\frac{1}{2}B_0+\sum_{n=1}^{\infty}4nB_n\cos{(nx)}, $$ it is absolutely clear that Fouries coefficients of $\sin{(x)}$ are $$ \frac{1}{2}B_0\,,4B_1\,,8B_2\,,\dots, 4nB_n\,,\dots $$ Hence you have $$ \begin{align*} \frac{1}{2}B_0=\frac{2}{\pi}\int\limits_0^{\pi}\sin{(x)}\,dx,\\ 4B_n=\frac{2}{\pi}\int\limits_0^{\pi}\sin{(x)}\cos{(nx)}\,dx,\;n\geqslant 1. \end{align*} $$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\rm u}_{tt}\pars{x,t} = 4{\rm u}_{xx}\pars{x,t}\,,\quad x \in \pars{0,\pi}\,, \quad t > 0.\qquad \color{#00f}{{\rm u}_{x}\pars{0,t} = {\rm u}_{x}\pars{\pi,t} = 0}.\quad \color{#c00000}{{\rm u}\pars{x,0}=0\,,\ {\rm u}_{t}\pars{x,0} = \sin\pars{x}}}$

The general solution has the form $\ds{\color{#c00000}{% {\rm u}\pars{x,t} \equiv \fermi\pars{x - 2t} + {\rm g}\pars{x + 2t}}}$ where $\ds{\fermi}$ and $\ds{\rm g}$ are functions to be determined. First, we'll find the solution for a general $\ds{{\rm u}_{t}\pars{x,0} \equiv \phi\pars{x}}$. Later on, $\ds{\phi\pars{x}}$ is chosen to agree with the original condition and in such way it satisfies the remaining boundary conditions.

$$ \begin{array}{rclcrcrcl} {\rm u}\pars{x,0} & = & 0 & \imp & \fermi\pars{x} & + & {\rm g}\pars{x} & = & 0 \\ {\rm u}_{t}\pars{x,0} & = & 0 & \imp & -2\fermi'\pars{x} & + & 2{\rm g}'\pars{x} & = & \phi\pars{x} \end{array} $$ $$ \mbox{which yields}\quad {\rm g}\pars{x} = -\fermi\pars{x}\,,\quad \fermi\pars{x} = \fermi\pars{0} - {1 \over 4}\int_{0}^{x}\phi\pars{\xi}\,\dd\xi $$

Then \begin{align} {\rm u}\pars{x,t}&= \bracks{\fermi\pars{0} - {1 \over 4}\int_{0}^{x - 2t}\phi\pars{\xi}\,\dd\xi} +\bracks{-\fermi\pars{0} + {1 \over 4}\int_{0}^{x + 2t}\phi\pars{\xi}\,\dd\xi} \\[3mm]&={1 \over 4}\int_{x - 2t}^{x + 2t}\phi\pars{\xi}\,\dd\xi \end{align} Whenever $\ds{\phi\pars{x}}$ is defined in $\pars{0,\pi}$, this solution becomes invalid whenever $\ds{x \pm 2t}$ 'leaves' $\pars{0,\pi}$. That's is the reason we extend the initial condition.

$$ \begin{array}{rclcrcl} {\rm u}_{x}\pars{0,t} & = & 0 & \imp & \phi\pars{2t} - \phi\pars{-2t} & = & 0 \\ {\rm u}_{x}\pars{\pi,t} & = & 0 & \imp & \phi\pars{\pi + 2t} - \phi\pars{\pi - 2t} & = & 0 \end{array} $$ Those conditions are equivalent to: $$ \phi\pars{-\xi} = \phi\pars{\xi}\,,\qquad\phi\pars{\xi + 2\pi} = \phi\pars{\xi}. \quad\mbox{Also,}\ \left.\phi\pars{\xi}\right\vert_{\xi\ \in\ \pars{0,\pi}}\ =\sin\pars{\xi} $$

which yields \begin{align} &\color{#00f}{\large{\rm u}\pars{x,t}= {1 \over 4}\int_{x - 2t}^{x + 2t}\phi\pars{\xi}\,\dd\xi} \\[3mm]& \mbox{where}\quad\color{#c00000}{\large% \phi\pars{\xi} \equiv \left\lbrace\begin{array}{ccrcccl} \sin\pars{\xi} & \mbox{if} & 0 & \leq & \xi & \leq & \pi \\[1mm] \phi\pars{-\xi} & \mbox{if} & -\pi & \leq & \xi & < & 0 \\[1mm] \phi\pars{\xi + 2\pi} & \mbox{if} & \xi & < & -\pi&& \\[1mm] \phi\pars{\xi - 2\pi} & \mbox{if} & \xi & > & \pi&& \end{array}\right.} \end{align}