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I am trying to solve FSI problems with finite elements and using a projection scheme (I am taking as reference the review of Guermond:

Guermond, J. L.; Minev, P.; Shen, Jie, An overview of projection methods for incompressible flows, Comput. Methods Appl. Mech. Eng. 195, No. 44-47, 6011-6045 (2006). ZBL1122.76072.), for the flow equations, for reasons of computational efficiency.

These methods are based on the Helmoltz decomposition as I understand: $$[L^2(\Omega)]^d = H \oplus \nabla H^1 (\Omega),$$ where $$H=\{ v\in [L^2(\Omega)]^d; \nabla\cdot v = 0; v\cdot n|_{\Gamma} = 0\}.$$ So to solve the Navier-Stokes equations $$\frac{\partial\mathbf{u}}{\partial t} + (\mathbf{u}\cdot \nabla)\mathbf{u} = -\nu \Delta\mathbf{u} + \nabla p,\\ \nabla\cdot \mathbf{u} = 0,$$ first one solves a problem for velocity only, where the pressure is approximated in some way, and then solves a problem for the pressure to enforce the incompressibility constraint $$ \frac{\partial\mathbf{u}}{\partial t} -\nabla p = 0, \qquad \frac{\partial p}{\partial n}\bigg|_{\Gamma} = 0,$$ where the boundary condition should come from the Helmoltz decomposition. My doubt is that I usually see this framework for standard Navier-Stokes problems, where boundaries are fixed and velocity is imposed with a Dirichlet condition. If that is the case, taking the product of the NS equations with the normal to the boundary $n|_{\Gamma}$, one sees that the condition respected by pressure at the boundary is $$ \frac{\partial p}{\partial n}\bigg|_{\Gamma} = \nu \Delta \mathbf{u} \cdot n|_{\Gamma}.$$

Now, if I want to apply this to a FSI problem, one of the boundaries is moving and therefore the velocity is assigned but changing in time, that is an acceleration is imposed. My question is, does the classic framework of pressure projection still work? And how does one derive rigorously the two subproblems to be solved (velocity and pressure), and more importantly, which are the boundary conditions to be imposed on pressure and velocity at the boundaries and at the FSI interfaces? To me, it doesn't look like that the Helmoltz decomposition in its basic form works if the velocity on Dirichlet boundaries is time dependent. Intuitively, if again I take the product of NS equations with the boundary normal $n|_{\Gamma}$, the $\frac{\partial \mathbf{u}}{\partial t}$ does not cancel if the boundary is accelerating, therefore the pressure satisfies now $$ \frac{\partial p}{\partial n}\bigg|_{\Gamma} = \left(-\frac{\partial \mathbf{u}}{\partial t} + \nu\Delta\mathbf{u} \right)\cdot n|_{\Gamma}.$$ But do I simply add the acceleration as boundary condition to the pressure problem? And how do I rigorously derive it? Also, is there a good reference on this that I am missing out?

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  • $\begingroup$ I don't think these boundary conditions make sense. Mathematically, $p$ doesn't even have a trace, let alone $\nabla p \cdot n$. Nor does $\Delta \mathbf u$. $\endgroup$ – Wolfgang Bangerth May 13 '20 at 14:49
  • $\begingroup$ @WolfgangBangerth True, I guess I was looking at the NS equations directly to see what the pressure is expected to satisfy in strong form, to get an hint. So from the last equation I am led to expect that pressure has to satisfy an additional term with respect to the fixed boundaries situation, but I don't see how to translate this in weak form, and furthermore in the framework of a projection scheme. $\endgroup$ – gc11 May 13 '20 at 16:32
  • $\begingroup$ You often can't impose these additional conditions. For the Laplace equation, you might infer from $\Delta u=0$ that $\partial_n^2 u = \partial_t^2 u$, but that's an unenforceable condition -- it just comes out that way after solving the equation, there is nothing you need or can do with the condition. $\endgroup$ – Wolfgang Bangerth May 13 '20 at 18:51
  • $\begingroup$ @WolfgangBangerth Therefore, the projection scheme as is with no modification already takes into account for the modified pressure condition? Would I be correct in just using it in the standard form for FSI applications? $\endgroup$ – gc11 May 13 '20 at 21:02
  • $\begingroup$ I'm not sufficiently knowledgeable about this subject to say for sure one way or the other. $\endgroup$ – Wolfgang Bangerth May 15 '20 at 15:41

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