# 1D wave equation with non-stationary boundary conditions

Consider a string with natural length $L$ with initial displacement $u(x, 0) = p(x) = 0$, and velocity $u'(x,0) = v(x) = 0$. Let the boundary conditions be $u(0, t) = f(t)$. This describes a string that is initially stationary, with its left end being moved.

The string is governed by the wave equation: $$c^2 \frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial t^2}$$ Using D'Alembert's solution yields: $$u(x, t) = \frac{p(x - ct) + p(x + ct)}{2} + \frac{1}{2c}\int_{x - ct}^{x + ct} v(s) \, \mathrm{d}s = 0$$ But this equation cannot satisfy the boundary conditions, i.e. $u(0, t) = f(t) = 0$. Am I doing something wrong here? It seems the wave equation only describes a string when the boundary conditions are stationary.

• And what, pray tell, is $p(x-ct)$ for times $t>x/c$? ;-). – Emilio Pisanty Feb 17 '17 at 16:54
• @EmilioPisanty ok, I see how this is a problem, but how do I fix it? I think $p(x)$ can have any value for $x \notin [0, L]$ – user110971 Feb 17 '17 at 21:06
• Well, you can set $v(s)\equiv0$ and ignore the term in $p(x+ct)$, so what does $u(0,t)$ tell you about the relationship between $f(t)$ and $p$ at negative arguments? (This assuming that you're solving on a half-infinite interval; if you have a fixed end on the other side then you obviously need to introduce reflections via the $p(x+ct)$ term, as well as $v(x)$ outside your interval.) – Emilio Pisanty Feb 17 '17 at 21:09
• @EmilioPisanty I guess you have $u(0, t) = f(t) = p(-ct)/2 + p(ct)/2$. For the interval $ct \in [0, L]$ $f(t) = p(-ct)/2$. The question then is: what is $p(a)$ for $a>L$. Because the other end does not move until the wave propagates: $u(L, t) = 0$ until $t = L/c$, i.e. $p(a) = 0$ for $a\in [0, 2L]$. After that I am not sure. I guess the wave will reflect, return, reflect again etc. $p(a)$ cannot remain zero after that because then $u(L, t)$ will just be a delayed version of $f(t)$. – user110971 Feb 17 '17 at 22:03
• @EmilioPisanty What about the velocity component? Wouldn't the same arguments apply? – user110971 Feb 17 '17 at 22:07