Riemann Problem for linear system of second-order PDEs I've been trying to solve the following system of partial differential equations,
\begin{equation}
u(x,t)_{t}+\frac{D}{2}v(x,t)_{xx}=0 
\qquad 
\text{and} 
\qquad 
v(x,t)_{t}-\frac{D}{2}u(x,t)_{xx}=0,
\end{equation}
in the domain $x\in[-\infty +\infty]$ and $t\in[-\infty +\infty]$, with initial conditions,
\begin{equation}
  u(x,0)=\begin{cases}
            u_L \qquad x<0\\
            u_R \qquad x>0
            \end{cases}
\qquad
  v(x,0)=\begin{cases}
            v_L \qquad x<0\\
            v_R \qquad x>0
            \end{cases}
\end{equation}
and no boundary conditions, except that the functions must remain bounded. My first question is if the problem is well posed without any boundary conditions? Here $D$ is a constant, and so are $u_L$, $u_R$, $v_L$ and $v_R$. 
I'm trying to solve this in the context of the Finite Volume Method, specifically, I need to solve a Riemann Problem (the problem above is a Riemann Problem, I think) in order to derive a numerical flux for my scheme.
I tried using Laplace Transforms to turn the system into two inhomogeneous ODEs in the variable $x$. My second question is if this approach is even valid?
To continue, I solve for the homogeneous ODE's for the Laplace Transforms $\mathcal{V}(x,s)$ and $\mathcal{U}(x,s)$, and I get the fundamental system of solutions for both to be $FSS=\{e^{wx},e^{-wx},\cos{wx},\sin{wx}\}$, where $w^2=\frac{2s}{D}$. 
Now I don't know how to proceed any further, and this is my third question. It might help to awnser to say that, for the FV scheme, I only need $u(0,0+)$ and $v(0,0+)$.
 A: Introduce the spatial Fourier transformation operator $\hat\cdot = \int_{\Bbb R} \cdot\, e^{-\text i k x}\text d x$. Thus, we have
$\hat u_t = k^2 \tfrac{D}{2} \hat v$ and $\hat v_t = -k^2 \tfrac{D}{2} \hat u$,
which gives
$$
\hat w_{tt} = -k^4 \tfrac{D^2}{4} \hat w
$$
where $w$ equals either $u$ or $v$.
At $t=0$, we have $w(x,0) = w_L+ (w_R-w_L)\,\text{u}(x)$ where $\text u$ is the Heaviside unit step function. The Fourier transform of the initial data is obtained straightforwardly by looking to the tables of Fourier transforms.
Thus, we are left with the resolution of
$$
\hat w_{tt} = -k^4 \tfrac{D^2}{4} \hat w
\qquad\text{with}\qquad
\begin{aligned}
\hat w(k,0) &= \pi (w_L+w_R)\delta(k) + \frac{w_R-w_L}{\text i k} \\
\hat w_t(k,0) &= \pm k^2 \tfrac{D}{2} \hat {\bar w}(k,0)
\end{aligned}
$$
where $\bar w$ is the remaining unknown ($\neq w$). The "$+$" sign corresponds to $(w,\bar w) = (u,v)$ and the "$-$" sign corresponds to $(w,\bar w) = (v,u)$.
Solutions read
$$\hat w(k,t) = \hat w(k,0)\cos(k^2 \tfrac{D}{2} t) \pm \hat {\bar w}(k,0)\sin(k^2 \tfrac{D}{2} t) \, ,$$
which inverse spatial Fourier transform provides the integral representation
$$
w(x,t) = \frac{1}{2\pi} \int_{\Bbb R} \hat w(k,t) \, e^{\text i k x}\text d k
$$
of the solution. It may be simplified further by looking at the expression of $k\mapsto \hat w(k,t)$ above. Hope that helps.

Note: If we differentiate the first PDE w.r.t. $t$ and the second PDE w.r.t. $x$ twice (and vice versa), we get in particular
$$
u_{tt} = -\tfrac{D}{2} v_{xxt}
\qquad\text{and}\qquad
v_{txx} = \tfrac{D}{2} u_{xxxx}
$$
provided that $u$, $v$ are sufficiently smooth.
Using the equality of mixed derivatives, one can then show that $w_{tt} + \tfrac14 D^2 w_{xxxx} = 0$. This is the dynamic Euler-Bernoulli beam equation.
A: *

*Yes, this system is well-posed, as seen in Harry's answer, at least for initial data and regularity classes where all of the steps in his solution continue to make sense. Periodic boundary conditions will likely not change this.

*I'm not sure if Laplace transforms are a good approach in this case, especially considering how simple the problem becomes if you use a Fourier transform

*It is not clear to me that there is a simple way to get this without just explicitly performing the inverse Fourier transform on Harry's solution 
