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In this paper, equations of rigid body motion (eq 4 and 5 in the paper) are written in Eulerian form (eq 12 in the paper). The rigid body is submerged in a viscous incompressible fluid. $$m\frac{dw_G(t)}{dt}=\int_{\partial\Omega(t)}\sigma n\,ds+\int_{\Omega(t)}\rho f\,dx,$$

$$I\frac{dR(t)}{dt}=R\times(IR)+\int_{\partial\Omega(t)}r\times\sigma n \,ds+\int_{\Omega(t)}\rho r\times f\,dx,$$

are transformed into

$$\partial_t(\rho w)+\mathrm{div}(\rho w\otimes w)=\rho f+\frac{1}{\rho}\mathrm{div}(\rho\Sigma)-\frac{1}{\rho}\sigma\nabla\rho,$$ where $\rho$ is the density of the solid, $\sigma$ is the stress tensor of the fluid in which the solid is submerged, $\Sigma$ is being called an internal rigidity force and $f$ is an external force.

Could someone help me to understand how the 'Eulerian' equation has been obtained from the earlier equations?

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  • $\begingroup$ Looks like stokes theorem was used to get sine volume integrals. Are you familiar with the conversion from integral formulation to differential for the normal hydro equations? $\endgroup$ – Kyle Kanos Jun 9 '15 at 12:47
  • $\begingroup$ DOI dx.doi.org/10.1007/s002050050136 $\endgroup$ – Qmechanic Jun 9 '15 at 14:36
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Giving a full answer would take long, but here are a few steps to help you. The easiest is probably to start from the result, eq. (11). Multiply by a vector test function and integrate it over a given volume $V(t)$ within the solid phase, which is advected by the velocity $w$. Replace $w(x,t)$ by its value, defined in (3) -- note that $x_G$ has for derivative $w_G$.

Then taking the test function equal in turn to each direction of space $e_i$, or to $x \times e_i$, you will obtain the six scalar equations corresponding to (4) and (5).

Note that $\nabla \rho/\rho$ is a way to obtain the outward-pointing normal vector.

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