Say, a thin long rod is occupying the space $[0,L]$. It's isotropic, linear elastic, homogeneous. The partial differential equations for stress $\sigma(x,t)$ and displacement $u(x,t)$ are as follows ($E$ denotes the Young's Modulus and $\rho$ the density):
$$\frac{\partial^2 u}{\partial t^2} = \frac{1}{\rho} \frac{\partial\sigma}{\partial x}$$ $$ \frac{\partial \sigma}{\partial t} = E \frac{\partial^2u}{\partial x\partial t}$$
The bar is initially in rest and free of any stresses. It is fixed to the "left" (at $x = 0$) and a constant velocity is applied to the right. Formally:
$$ \sigma(x,0) = 0$$ $$ u(x,0) = 0$$ $$ u(0,t) = 0$$ $$ \frac{\partial u\left(L,t\right)}{\partial t} = v_{bc}$$
Now my question is simply: Is there an analytical solution for this system? I already tried to find one in Maple, but to now avail.
The equations can be combined for a solution, if initial conditions are known. Strain is $\epsilon = \frac{\partial u}{\partial x}$, stress $\sigma = E \epsilon$ and the balance of forces yields
$$ \frac{\partial \sigma}{\partial x} = E \frac{\partial \epsilon}{\partial x} = E \frac{\partial^2 u}{\partial x^2} = \rho \frac{\partial^2 u}{\partial t^2} $$
and with $c^2=\frac{E}{\rho}$ the wave equation is
$$ \frac{\partial^2 u}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} $$
with initial conditions
$$ u(x,0) = 0 $$ $$ \dot{u}(x,0) = \frac{x}{L} u_{bc} $$ (for smooth velocity function)
The general solution has the form:
$$ u(x,t) = A_0 + B_0 t + \sum_{n=0}^{\infty} \cos\left(\frac{n \pi x}{L}\right) \left(A_n \sin \left( \frac{n \pi c}{L} t \right) + B_n \left( \frac{n \pi c}{L} t \right) \right) $$ $$ \dot{u}(x,t) = \frac{\partial}{\partial t} u(x,t) $$
with the coefficients $A_n$ and $B_n$ derived from the initial conditions ($u(x,0)$ and $\dot{u}(x,0)$)
$$ A_n = \frac{2}{n \pi c} \int_0^L \cos \left( \frac{n \pi x}{L} \right) \dot{u}(x,0)\,{\rm d} x $$ $$ B_n = \frac{2}{L} \int_0^L \cos \left( \frac{n \pi x}{L} \right) u(x,0)\,{\rm d} x $$
For your case
$$ A_0 = 0 $$ $$ B_0 = \frac{1}{2} v_{bc} $$ $$ B_n = 0 $$ $$ A_n = \frac{L \left(\cos(\pi n)+\pi n \sin(\pi n)-1\right)}{c \pi^3 n^3} v_{bc} $$
with example output: