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I'm trying to solve the coupled Navier-Stokes/elasto-dynamics equations using a discontinuous Galerkin approach in order to propagate waves through a fluid-solid interface. I was wondering how to properly impose the coupling between the solid and the fluid as one has two conditions to fulfill at the interface: $$ \mathbf{v}_f.\mathbf{n} = \mathbf{v}_s.\mathbf{n} \quad \mbox{and }\quad\Sigma_f.\mathbf{n} = \Sigma_s.\mathbf{n} $$ where $\mathbf{v}_f, \mathbf{v}_s$ are respectively the fluid and solid velocity and $\Sigma_f, \Sigma_s$ are respectively the fluid and solid stress tensors. $\mathbf{n}$ is the outer normal.

In the discontinuous Galerkin framework, at an interface, one defines the interior and exterior states for constitutive variables (velocity, density and total energy) to solve the Riemann problem. For example, for the fluid domain, at a hard wall boundary one sets $$ \rho_f^- = \rho_f^+ \quad \mathbf{v}_f^- = \mathbf{v}_f^+-2(\mathbf{v}_f^+.\mathbf{n}).\mathbf{n}\quad p^- = p^+ $$ where $\rho_f$ is the fludi density and $p$ the fluid pressure. The sign $-$ denotes the exterior state (to be set) and $+$ the interior state. The fluxes at the interface are then computed from these states and you can find many different ways to do so, especially if you have shocks.

Then my question is, how do you do the same for a fluid-solid boundary where both the normal stress and the normal velocity are imposed ?

Thanks a lot

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  • $\begingroup$ Not only do the normal components of velocity match, but so also are the tangential components. Also, the pressures don't have to match; only the total tractions. $\endgroup$ – Chet Miller Sep 15 '17 at 20:53
  • $\begingroup$ You're right ! But in the simple case of an inviscid fluid, one only imposes the normal velocity and the fluid stress is equivalent to the pressure ($\Sigma_f = -p\mathbf{Id}$, where $\mathbf{Id}$ is the identity matrix) $\endgroup$ – Amzocks Sep 18 '17 at 14:09
  • $\begingroup$ Yes. That's right. But, when you indicate Navier-Stokes, the implication is that the fluid is not inviscid. $\endgroup$ – Chet Miller Sep 18 '17 at 14:11
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Im assuming your solving the inviscid compressible navier stokes due to your discussion of the Riemann problem, and using ghost cells due to the way your implementing boundary conditions.

The boundary condition is essentially the imposition of a shock at the boundary of the fluid. In the case when the boundary is against a solid wall we can consider the boundary as a shock with speed $\boldsymbol{s}=0$ and zero mass flux through it, which gives values for the ghost cell by constructing some fluid exterior to the domain at each time step that produces the correct behaviour at the boundary.

For a moving boundary we do the same thing, only the 'shock' moves withe the boundary so $\boldsymbol{s}=s \hat{\boldsymbol{n}}$ and the computation of the shock condition to obtain the correct values for the ghost cells is more subtle, but the same principle applies.

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  • $\begingroup$ Thanks a lot ! Have you any reference for shock waves formula at the boundary ? $\endgroup$ – Amzocks Sep 18 '17 at 18:42
  • $\begingroup$ Simply the Rankine-Hugoniot shock condition for whatever system of equations your using. Slick in your b.c. and thats the boundary shock condition. I like LeVerques book on hyperbolic PDEs, but thats mainly based around methods you're not using. $\endgroup$ – Eddy Sep 18 '17 at 18:46
  • $\begingroup$ Finding references for numerical boundary conditions is really hard, usually best to figure it out, explain it clearly and become the reference! $\endgroup$ – Eddy Sep 18 '17 at 18:48
  • $\begingroup$ That's true and thanks again ! But when you use the relationships for a stationary shock and consider $\mathbf{v}_f = \mathbf{v}_s$ (meaning that the total velocity is continuous at the boundary), you obtain $p_f = \gamma p_s$ (subscript $1$ on the wikipedia page en.wikipedia.org/wiki/Rankine%E2%80%93Hugoniot_conditions being the solid and $2$ the fluid and $\gamma$ is the heat capacity ratio). How do you set the pressure in the solid ? Is it straightforward from the normal stress ? $\endgroup$ – Amzocks Sep 19 '17 at 10:04
  • $\begingroup$ The stress should be continuous across the boundary (unless the boundary has finite mass), so just set solid stress equal to fluid stress. $\endgroup$ – Eddy Sep 19 '17 at 10:07

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