# For a wave that follows the d'Alembert solution, fixed at one end upon a table, why should it bounce back instead of inverting itself?

Jun 19, 2017

Well, you've cut off the wave so that only positive values exist (assuming you mean you've propagated the wave perpendicular to a horizontal plane).

In technical terms, the trough has gone into the imaginary domain (underneath the ground, but not actually in the real domain). Only the crest remains in the real domain.

The wave equation describes a propagating wave. Let's assume the wave is traveling with one end position fixed (by your hand).

Then the problem data are (with arbitrary amplitudes, because we are focusing on the $x$ direction):

$\frac{{\partial}^{2} u}{\partial {t}^{2}} - {c}^{2} \frac{{\partial}^{2} u}{\partial {x}^{2}} = 0$, $\text{ } 0 < x < \infty , t > 0$

$u \left(x , 0\right) = \phi \left(x\right)$, $\text{ } 0 < x < \infty$

$\frac{\partial u \left(x , 0\right)}{\partial t} = \psi \left(x\right)$, $\text{ } 0 < x < \infty$

where $u \left(x , t\right)$ describes the wave amplitude as a function of position and time. $\phi \left(x\right)$ is then the initial position, and $\psi \left(x\right)$ is the initial velocity. $c = \sqrt{\frac{T}{\rho}}$ is a constant, $T$ is the tension, and $\rho$ is the mass density of the string.

Since the wave is fixed at one end, the solution has odd symmetry with respect to the origin, i.e. $u \left(x , t\right) = - u \left(- x , t\right)$.

However, $u \left(x , t\right)$ for negative $x$ and amplitudes below the $x$ axis don't exist here, because positions can only be positive (and you've further restricted the amplitude to be above the $x$ axis).

Your wave is traveling in one dimension (perpendicular to the horizontal plane), so it is described by the one-dimensional d'Alembert solution:

$u \left(x , t\right) = \frac{1}{2} \left[\phi \left(x - c t\right) + \phi \left(x + c t\right)\right] + \frac{1}{2 c} {\int}_{x - c t}^{x + c t} \psi \left(\tau\right) d \tau$

By placing it on a table, let's say, you've cut the wave off so that the amplitude is only above the $x$ axis.

As seen below, its position is given by $v {t}_{0}$, where $v$ is another notation for our $c$ and ${t}_{0}$ is the particular time we have taken a snapshot of the wave.

The real wave travels forward, whereas an imaginary component, with odd symmetry with respect to $\left(u , x\right) = \left(0 , 0\right)$, is traveling backwards in forward time.