At a fixed point in time, consider the equation of motion $$ \nabla \cdot \boldsymbol \sigma(u) + \boldsymbol f = \rho \ddot{\boldsymbol u} \quad \text{in $\Omega \subset \mathbb R^d$} $$ for a displacement $\boldsymbol u$, a stress tensor $\boldsymbol\sigma(\boldsymbol u)$, an outer force $\boldsymbol f$ and a density $\rho$. Testing this equation with functions $\boldsymbol v \in X = H^1(\Omega)^d$ yields the equation $$ \int_\Omega \langle \rho \ddot{\boldsymbol u}, \boldsymbol v\rangle + \int_\Omega \langle \boldsymbol \sigma(u), \nabla\boldsymbol v\rangle = \int_\Omega \langle \boldsymbol f, \boldsymbol v \rangle - \int_{\partial \Omega} \langle \boldsymbol \sigma \boldsymbol n, \boldsymbol v\rangle \quad \forall \boldsymbol v \in X $$ To derive a weak formulation now, in particular, one needs to determine what space $\ddot{\boldsymbol u}$ should lie in. If we simply set $\rho = 1$, then the answer is trivial and given by $\ddot{\boldsymbol u} \in X^*$, where the corresponding integral is reinterpreted as a duality bracket.
Q: What if we are dealing with a density $\rho$ that varies in space, though?
Positive results: If we assume $\rho X \subset X$, which should e.g. be the case whenever $\rho \in W^{1,\infty}(\Omega)$, then we can define $$\langle \rho \ddot{\boldsymbol u}, \boldsymbol v\rangle_{(X, X^*)} := \langle \ddot{\boldsymbol u}, \rho \boldsymbol v\rangle_{(X, X^*)}$$ Negative results: The regularity of $\rho \boldsymbol v$ can be as bad as the regularity of $\rho$ (since $\boldsymbol v$ can be a constant function), so that we must have $\rho \in H^1(\Omega)$.
In summary, the way I see it, we can choose
- $\ddot{\boldsymbol u} \in X^*$, which restricts $\rho$ to a space like $W^{1,\infty}(\Omega)$ or
- $\ddot{\boldsymbol u} \in L^2(\Omega)^d$, which allows for arbitrary $\rho \in L^\infty(\Omega)$.
Is this issue discussed anywhere?