Longitudinal wave in a falling elastic body

Question

Consider an elastic rod hung from a high point with density $\rho$ and Young's modulus $Y$, subject to gravitational acceleration $g$. The coordinate from the hanging point is $x$, while the displacement from the original position is $\xi$.

The equation of a longitudinal wave travelling through the rod is (I will omit detailed derivations for brevity, but am happy to add them as an addendum if that's helpful):

$$ \boxed{ \frac{\partial^2 \xi}{\partial t^2} = c^2 \frac{\partial^2\xi}{\partial x^2} + g, \quad c=\sqrt{\frac{Y}{\rho}}.\; } $$

We solve this partial differential equation using Fourier series. The initial conditions at $t=0$ for $\xi\left(x, 0\right)$ and its time derivative $\frac{\partial \xi}{\partial t}\left(x,0\right)$ are

$$ \xi\left(x, 0\right) = \frac{g}{c^2}\left(Lx-\frac{1}{2}x^2\right) , \qquad \frac{\partial \xi}{\partial t}\left(x,0\right) = 0, $$

where the former condition is derived by considering the stationary case without free fall, so $\frac{\partial^2 \xi}{\partial t^2} = 0$. The boundary conditions are

$$ \frac{\partial\xi}{\partial x}\left(0,t\right)=0, \qquad \frac{\partial \xi}{\partial x}\left(L,t\right) = 0, $$

since the endpoints are free during the fall.

Fit the time-independent part of the solution to $\xi\left(x,0\right)$ using the usual Fourier series method with $t=0$, we have:

$$ \boxed{ \xi\left(x,t\right) = \frac{1}{2}gt^2 + \frac{g L^2}{c^2}\left[ \frac{1}{3} - \frac{2}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}\cos\left(k_n x\right) \cos\left(\omega_n t\right) \right], \quad k_n = \frac{n\pi}{L}, \quad \omega_n = ck_n. \;} $$

This is a reasonable solution, where the time averaged motion is that of a free fall, with a longitudinal wave travelling back and forth with period $\Delta T = \frac{2L}{c}$.

---

But we can find the solution to the wave equation above with another method, using the generalised d'Alembert formula:

For a solution to $$\frac{\partial^2 u}{\partial t^2} - c^2 \frac{\partial^2u}{\partial x^2} = h\left(x, t\right), $$

subject to the boundary conditions $$ u\left(x,0\right) = \phi(x), \quad \frac{\partial u}{\partial t}\left(x, 0\right)=\psi(x), $$

the solution is given by $$ u\left(x,t\right) = \frac{1}{2}\left[ \phi\left(x+ct\right) + \phi\left(x-ct\right) \right] + \frac{1}{2c}\int_{x-ct}^{x+ct}\psi\left(\zeta\right) \mathrm{d}\zeta + \frac{1}{2c}\int_0^t \int_{x-c\left(t-\tau\right)}^{x+c\left(t-\tau\right)} h\left(\zeta, \tau\right) \mathrm{d}\zeta\mathrm{d}\tau. $$

However, plugging in

$$h\left(x,t\right) = g,$$

$$\xi\left(x,0\right) = \phi\left(x\right) = \frac{g}{c^2}\left(Lx-\frac{1}{2}x^2\right),$$

and

$$\frac{\partial \xi}{\partial t}\left(x,0\right) = \psi\left(x\right) = 0, $$

I am unable to recover the falling solution from the Fourier method, instead only recovering the static solution

$$\xi\left(x,t\right) = \xi\left(x,0\right) = \frac{g}{c^2}\left(Lx-\frac{1}{2}x^2\right).$$

Where have I gone wrong in the second solution?

---

Edit:

As @hyportnex pointed out, the boundary condition $$\psi(x) = \frac{\partial\xi}{\partial t}\left(x,0\right) $$ is ambiguous. I will assume that $g$ is not turned on at $t=0$, such that $\psi\left(x\right)=0$. From hindsight, by differentiating the solution found using Fourier's method and plugging in $t=0$, we also get $\psi\left(x\right)=0$.

Answer 1

The generalized solution does not have any boundary conditions imposed. In particular, note that we have in general $$ \frac{\partial u}{\partial x}\left(0,t\right) = \frac{1}{2}\left[ \phi'\left(ct\right) + \phi'\left(-ct\right) \right] + \frac{1}{2c} \left[ \psi(ct) - \psi(-ct)\right] + \frac{1}{2c}\int_0^t \left[ h(c(t-\tau),\tau) - h(-c(t-\tau),\tau) \right] \mathrm{d}\tau. $$ In your case you have $$ \phi'(x) = \frac{g}{c^2}\left(L-x\right) \qquad \psi(x) = 0 \qquad h(x,t) = g $$ and so the general solution for the given initial conditions will have $$ \frac{\partial u}{\partial x}\left(0,t\right) = \frac12 \left[\phi'(ct) + \phi'(-ct)\right] = \frac{gL}{c^2} \neq 0. $$ One can also find a similar expression for $\frac{\partial u}{\partial x}$ at $x=L$, and it will not generally vanish either.

To use the generalized solution while maintaining boundary conditions, I think (I haven't worked through all the details explicitly) that one can use a method-of-images technique. For your initial conditions, use a period-$L$ version of your original initial displacement function, defined for all $-\infty < x < \infty$: $$ \phi(x) = \begin{cases} \frac{g}{c^2}\left(Lx-\frac12x^2\right) & 0 \leq x \leq L \\ \phi(2L - x) & n L < x \leq (n+1) L \quad \text{($n$ odd)} \\ \phi(x + 2L) & n L < x \leq (n+1) L \quad \text{($n$ even & $n \neq 0$)} \end{cases} $$ Effectively, this "periodic version" is the function on your original domain $0 \leq x \leq L$ repeated over the real numbers, with the function "flipped" left-to-right in the "odd" domains. (This doesn't matter in your case, since this "flipping" doesn't change the parabola, but it would matter in a more general case.) The periodicity & symmetry of this function means that we will have $\phi'(x) = -\phi'(-x)$ for all $x$, and so the boundary condition $$ \frac{\partial u}{\partial x}\left(0,t\right) = 0 $$ will always be maintained.

Finally, the above periodic version of $\phi(x)$ can be easily written as a Fourier series—in fact, I believe that it will be the Fourier series you found in your original solution. Plug this function into the general d'Alembert solution, simplify it using standard trigonometric identities, and I expect that you'll end up with your original solution.