2
$\begingroup$

""we aim for a consistent variational-monolithic coupling scheme in which we need all equations defined in the same domain; therefore, $\mathrm{ALE}_{f x}$ was introduced. In variational monolithic coupling, Neumann-like interface conditions, like the continuity of normal stresses are fulfilled exactly in a weak sense on the continuous level: $$ \left\langle\hat{J} \widehat{\sigma}_f \widehat{F}^{-T} \hat{n}_f, \varphi\right\rangle+\left\langle\widehat{F} \widehat{\Sigma} \hat{n}_s, \varphi\right\rangle=0 \quad \forall \varphi \in V, $$ "" this is a part from Thomas Wick's lecture on FSI, then he gives the general variational formulation, but how to deal with this condition in ALE FSI, could you please help me, thank you

$\endgroup$

2 Answers 2

3
$\begingroup$

@Wolfgang Bangerth already gave a really good answer to your question and I just want to make some additions that are specific to monolithic ALE-FSI. In fluid-structure interaction, we have the interface conditions $$ u_f = u_s, \\ v_f = v_s, \\ \sigma_f n_f + \sigma_s n_s = 0, $$ where for simplicity I omitted all ALE mappings which you included in your question. The first two conditions on the continuity of $u$ and $v$ are enforced through the function spaces, since we just define $u$ and $v$ over the entire domain $\Omega_f \cup \Omega_s$ and in the subdomains, e.g. $u = u_f$ in $\Omega_f$. So you see that the first two conditions are directly baked into the function spaces. Now we just need to deal with the last interface condition, which as @Wolfgang Bangerth pointed out will be enforced through the weak form. In the fluid equations you have a term like $-\nabla \cdot \sigma_f$ and in the solid equations you have a term like $-\nabla \cdot \sigma_s$. Multiplying from the right with a test function $\varphi_s$ or $\varphi_f$ and integrating by parts, we get $$ (-\nabla \cdot \sigma_s, \varphi_s)_{\Omega_s} = (\sigma_s, \nabla \varphi_s)_{\Omega_s} - (\sigma_s n_s, \varphi_s)_{\Gamma}, \\ (-\nabla \cdot \sigma_f, \varphi_f)_{\Omega_f} = (\sigma_f, \nabla \varphi_f)_{\Omega_f} - (\sigma_f n_f, \varphi_f)_{\Gamma}, $$ where I used the notation $$ (f,g)_{\Omega_\Box} := \int_{\Omega_\Box}f \cdot g\ \mathrm{d}x, \qquad \Box \in \{f,s\}, \\ (f,g)_{\Gamma} := \int_{\Gamma}f \cdot g\ \mathrm{d}x =: \langle f,g \rangle. $$ If you add up the boundary terms that you get from the integration by parts, you get $$ - (\sigma_s n_s, \varphi_s)_{\Gamma} - (\sigma_f n_f, \varphi_f)_{\Gamma} = - (\sigma_s n_s + \sigma_f n_f, \varphi)_{\Gamma} = 0, $$ where we used that $\varphi_s = \varphi_f =: \varphi$ on $\Gamma$ and used the continuity of normal stresses interface condition $\sigma_f n_f + \sigma_s n_s = 0$. So you see that this Neumann-like interface condition just cancels out some interface terms in the monolithic discretization of FSI and you don't need to enforce this anywhere else. The same thing happens if you look at monolithic ALE-FSI, I just didn't want to write down all the ALE mappings.

$\endgroup$
1
  • 2
    $\begingroup$ Thanks for taking the time to work out what I was too lazy to put into formulas :-) $\endgroup$ Mar 7 at 20:30
2
$\begingroup$

You deal with it in the usual way: In the derivation of the weak form, at some point or other you integrate by parts and obtain an integral over the interface between the two parts of the domain. In that integral, you then use the equation you show to replace whatever term you have there.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.