Solving a particular partial differential equation

Question

Need help solving this PDE: \begin{equation} \frac{\partial^2u(x,t)}{\partial t^2} + \frac{2\tau}{m}\frac{\partial{u}}{\partial x} = 0. \end{equation}

Context:
I don't know how to specify the boundary conditions, but I can give some context. This is a mass centered between two strings. We need to determine the equation of motion of the mass so that we can use it to specify the boundary conditions (??) of the two strings. I.e., the left end of the right string moves with the mass, and vice versa. I mean, presumably it has some oscillatory solution (in time). I don't really get how the equation works, because it really shouldn't have an x dependence. I.e., it only moves vertically (u axis), not left and right (x axis).

We arrive at this equation by looking at some differential region centered about the mass, extending dx in the left right direction. The strings attached to the mass exert some tension $\tau$ on it and make an angle $\theta$ with the horizontal. To get the $\frac{\partial{u}}{\partial x} $ term, we take $\sin(\theta)$, use the small angle approximation to set $\sin(\theta) = \theta = \tan(\theta)$ which is equal to du/dx. This is kinda the standard thing we do in grad text books when going from discrete masses connected to eachother to a continuous string.

This is part of what is, IMO, a really really hard physics problem and I'm not even entirely sure that this is the correct way to approach it. However, I would still really appreciate it if someone knew how to solve this, or could even offer any guidance otherwise.

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Lets assume the mass is at $\ds{x = 0}$. The boundary condition at $\ds{x = 0}$ is given by the Newton Second Law: \begin{align} m\,\left.\partiald[2]{\Psi\pars{x,t}}{t}\right\vert_{\,x\ =\ 0} =\left. T\,\partiald{\Psi\pars{x,t}}{x}\right\vert_{\,x\ =\ 0^{+}} -\left. T\,\partiald{\Psi\pars{x,t}}{x}\right\vert_{\,x\ =\ 0^{-}}\tag{1} \end{align}

such that $\ds{\Psi\pars{x,t}}$ is continuos at $\ds{x = 0}$ but $\ds{\partiald{\Psi\pars{x,t}}{x}}$ is not, in general, continuos at $\ds{x = 0}$.

$\ds{\Psi\pars{x,t}}$ satisfies the Classical Wave equation (CWE) $\ds{\pars{\partiald[2]{}{x} - {1 \over v_{\varphi}^{2}}\, \partiald[2]{}{t}}\Psi\pars{x,t} = 0}$ for $\ds{x \not= 0}$. In general, the constant $\ds{v_{\varphi}}$ $\ds{\pars{~\it\mbox{the phase velocity}~}}$ can be different at both sides of the mass.

So, the general procedure is:

1. Write two solutions for the CWE ( for $x < 0$ and $x > 0$ ).
2. You have two boundary conditions:
   1. One is given by expression $\pars{1}$.
   2. The other one is $\ds{\Psi\pars{0^{-},t} = \Psi\pars{0^{+},t}}$
3. It's likely you have some other boundary conditions when $\ds{x_{<\atop >} 0}$.

This program determines the possible modes of oscillations of the whole system.

Answer 2

That is a linear partial differential equation with constant coefficients so it should be solved by any linear combination of exponentials and separation of variables (however there's a little bit more behind it not yet far enough in distribution theory sry)

Answer 3

The exponential comes about when you separate the variables you will have a second linear ode in time and a first order ode in x .. So if we choose a certain value for the constant that ties the two equations together (see the method of separation of variables as that last sentence will make sense) we will get oscillatory in time and exponential in x. As for why you will get any explicit x dependence I have not really read the problem, and instead helping you try solve the pde :) .