Consider the Cauchy momentum equation for a steady flow: $$ \rho\mathbf{u}\cdot\nabla\mathbf{u} = \rho\mathbf{g} + \nabla\cdot\mathbf{\sigma} $$ Here, $\mathbf{\sigma}$ is the stress tensor and $\mathbf{u}$ is the velocity field. In $d$ dimensions, the equation may be written in coordinates (and using the summation convention), viz. $$ \frac{\partial}{\partial x_j} \left(\sigma_{ij} + \frac{\rho}{d+1} (g_i x_j + g_j x_i) - \rho u_i u_j \right) = 0. $$ My question is: Is it possible to recover $\mathbf{\sigma}$ in terms of $\mathbf{u}$?
This tensor differential equation (TDE) needs to be supplemented with boundary conditions. One is the free-surface boundary condition that $\mathbf{\sigma}\cdot\mathbf{n} = \mathbf{0}$ on a free surface. However, the free surface may form only part of the boundary of a fluid domain, so I think this doesn't fully determine the solution. I think we will need an extra 'normalisation' condition on the total force on the fluid: $$ \int_V \left( \frac{\partial \sigma_{ij}}{\partial x_j} + \rho g_i \right) \mathrm{d} V = F_i$$ for a prescribed $F_i$.
The TDE is of the form $$ \frac{\partial S_{ij}}{\partial x_j} = 0 \text{ for } i=1,\dots,d,$$ where $\mathbf{S}$ is a symmetric tensor. Therefore, my question boils down to: What do solenoidal symmetric tensor fields look like? For a solenoidal vector field $\bf{u}$, with $\nabla\cdot\mathbf{u} = \partial u_i / \partial x_i = 0$, the Helmholtz decomposition tells us that $\mathbf{u}$ may be written as the curl of a vector potential, $\mathbf{u} = \nabla\times\mathbf{A}$. Hence, regarding each 'column' of the tensor $\mathbf{S}$ as a vector field in its own right, we can write $$ S_{ij} = \epsilon_{jkl}\frac{\partial A_{il}}{\partial x_k} $$ for a 'tensor potential' $A_{il}$. I don't know what the fact that $S_{ij} = S_{ji}$ means for $A_{il}$.