Wave Equation BC's

Question

So, I'm trying to solve the wave equation with the Fourier transform, and I'm struggling to figure out how to apply the BC's. Here's the problem I considered:

$$\frac{d^2u}{dt^2}=c^2\frac{d^2u}{x^2}$$ $$u(x,0)=g(x)$$ $$\frac{du}{dt}=0$$ at t = 0

Running through computations, I find that the Fourier transform solution is as follows:

$$F(u(\lambda,t))=A(\lambda)e^{ic\lambda t}+B(\lambda)e^{-ic \lambda t}$$

How would I apply boundary conditions to this and then transform back to u? ANy detailed explanation on this would be appreciated (I'm still learning this stuff). I'm thinking it may be easier to work with cosines and sines.

Answer 1

You basically fourier the initial condition to get:

$\hat{u}(\lambda,0)=\hat{g}(\lambda)=A(\lambda)+B(\lambda)\Rightarrow A(\lambda)=\hat{g}(\lambda)-B(\lambda)$

$\frac{\partial\hat{u}}{\partial t}=ic\lambda A(\lambda)e^{ic\lambda t}-ic\lambda B(\lambda)e^{-ic\lambda t}=0\Rightarrow A(\lambda)=B(\lambda)e^{-2ic\lambda t}$

$\Rightarrow \hat{g}(\lambda)-B(\lambda)=B(\lambda)e^{-2ic\lambda t}\Rightarrow B(\lambda)=\large\frac{\hat{g}(\lambda)}{(1+e^{-2ic\lambda t})}$

$\Rightarrow A(\lambda)=\large\hat{g}(\lambda)-\frac{\hat{g}(\lambda)}{(1+e^{-2ic\lambda t})}$

Thus $\large\hat{u}(\lambda,t)=(\hat{g}(\lambda)-\frac{\hat{g}(\lambda)}{(1+e^{-2ic\lambda t})})e^{ic\lambda t}+(\frac{\hat{g}(\lambda)}{(1+e^{-2ic\lambda t})})e^{-ic\lambda t}$

$\large=\frac{e^{-ic\lambda t}}{1+e^{-2ic\lambda t}}+(\frac{\hat{g}(\lambda)}{(1+e^{-2ic\lambda t})})e^{-ic\lambda t}=\frac{1+\hat{g}(\lambda)e^{-ic\lambda t}}{1+e^{-2ic\lambda t}}$

Now you inverse fourier to find $u(x,t)$

Hope that helps.

Inverse fourier: (you are correct, this seems hard)

Write $\large\frac{1+\hat{g}(\lambda)e^{-ic\lambda t}}{1+e^{-2ic\lambda t}}=\frac{1}{1+e^{-2ic\lambda t}}+\frac{\hat{g}(\lambda)e^{-ic\lambda t}}{1+e^{-2ic\lambda t}}=\frac{1}{1+e^{-2ic\lambda t}}+\hat{g}(\lambda)\hat{f}(\lambda,t)$

Then $u(x,t)=F^{-1}(\frac{1}{1+e^{-2ic\lambda t}})+(\sqrt{2\pi})g(x)f(x,t)$

This simplifies things, so you need to calculate:

$\large F^{-1}(\frac{1}{1+e^{-2ic\lambda t}})$, and $\large F^{-1}(\frac{e^{-ic\lambda t}}{1+e^{-2ic\lambda t}})$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\rm u}\pars{x,t} = \fermi\pars{x - ct} + {\rm h}\pars{x + ct}}$ is a general solution.

\begin{align} \left.\partiald{{\rm u}\pars{x,t}}{t}\right\vert_{t\ =\ 0} = 0\quad\imp\quad 0=-c\fermi'\pars{x} + c{\rm h}'\pars{x}\quad\imp\quad{\rm h}\pars{x} = f\pars{x} + A\,,\quad A\ \mbox{is a constant} \end{align}

Then, $\ds{{\rm u}\pars{x,t} = \fermi\pars{x - ct} + \fermi\pars{x + ct} + A}$

$$ {\rm u}\pars{x,0} = {\rm g}\pars{x}\quad\imp\quad 2\fermi\pars{x} + A = {\rm g}\pars{x}\quad\imp\quad\fermi\pars{x} = {{\rm g}\pars{x} - A \over 2} $$

$$ {\rm u}\pars{x,t}={{\rm g}\pars{x - ct} - A \over 2} +{{\rm g}\pars{x + ct} - A \over 2} + A $$

$$\color{#00f}{\large% {\rm u}\pars{x,t} = \half\,\bracks{{\rm g}\pars{x - ct} + {\rm g}\pars{x + ct}}} $$