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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.

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  • $\begingroup$ Oops that was a typo. Thanks. $\endgroup$ Apr 28, 2014 at 10:03

2 Answers 2

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

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}

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  • $\begingroup$ To be logically correct, your definition of extension is to be treated as recursive. You'd better extend $\phi$ from $[0,\pi]$ to $\mathbb{R}$ periodically to make it a $\pi$-periodic function, that equals just taking $\phi(\xi)=|\sin{(\xi)}|$. For a sine function, such extension is even on each pair of adjacent periods, which is enough to satisfy the boundary conditions. With Neumann boundary conditions, this nice trick works for all initial data $u_t|_{t=0}=u_1(x)$ such that $u_1(0)=u_1(\pi),\;u'_1(0)=-u'_1(\pi)$. Solution will be only semi-classical, just like with $u_1=\sin$. $\endgroup$
    – mkl314
    Apr 28, 2014 at 13:47
  • $\begingroup$ @mkl314 I knew that but I wrote in that way to avoid the temptation to move to Fourier series which is quite bad given the derivative discontinuity. Thanks. $\endgroup$ Apr 29, 2014 at 0:19
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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 $$ \begin{align*} \frac{1}{2}B_0=\frac{2}{\pi}\int\limits_0^{\pi}\sin{(x)}\,dx,\\ 4B_n=\frac{2}{\pi}\int\limits_0^{\pi}\sin{(x)}\cos{(nx)}\,dx,\;n\geqslant 1. \end{align*} $$

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