Assume we have two domains $\Omega_a$ a acoustic domain with boundary $\Gamma_a$ and $\Omega_s$ a domain of a solid body with boundary $\Gamma_s$. $\Omega_a$ and $\Omega_s$ have the common interface $\Gamma_{as}$. In $\Omega_s$ we have the equation of motion for a solid: \begin{equation} \rho_s \frac{\partial^2 u}{\partial t^2} - \nabla \cdot \sigma = 0 \end{equation} And in $\Omega_a$ the wave equation: \begin{equation} \frac{1}{c_a^2} \frac{\partial p^2}{\partial t^2} - \nabla \cdot \nabla p =0 \end{equation} where $u$ is the displacement of the solid and $p$ is the acoustic pressure. Further $\rho_s$ is the density in $\Omega_s$ , $\sigma$ the stress-tensor and $c_a$ the speed of sound in $\Omega_a$. The coupling at $\Gamma_{as}$ can be formulated by the following equations: \begin{align} -\frac{1}{\rho_a} \nabla p \cdot n = \frac{\partial^2 u}{\partial t^2} \cdot n\\ -pn = \sigma \cdot n \end{align} where $\rho_a$ denotes the density in $\Omega_a$ and $n$ the normal vector at $\Gamma_{as}$.

Now I want to derive a finite element formulation such that I can have a non-matching grid at $\Gamma_{as}$ and I want to do this by the Mortar method.

The first step is to derive the weak form of the two equations on their domains with the test functions $p'$ and $u'$:

\begin{align} \int_{\Omega_s} u' \cdot \rho_s \frac{\partial^2 u}{\partial t^2} dx + \frac{1}{2} \int_{\Omega_s} \nabla u' : C : (\nabla u + (\nabla u)^T ) dx -\int_{\Gamma_{as}} (u' \cdot \sigma) \cdot n \ ds =0 \\ \int_{\Omega_a} p' \frac{1}{c_a^2} \frac{\partial p^2}{\partial t^2} dx + \int_{\Omega_a} \nabla p' \cdot \nabla p dx - \int_{\Gamma_{as}} p' \nabla p \cdot n \ ds =0 \end{align}

Here I'm stuck. What I think I should do next is:

For the Mortar method one needs to define a lagrange multiplier as $t=-pn=\sigma\cdot n$. How to incooperate it into both equations?

The second interface condidtion is then added in the weak form with the test function $\mu'$: \begin{align} \int_{\Gamma_{as}} \mu'(\frac{1}{\rho_a} \nabla p \cdot n +\frac{\partial^2 u}{\partial t^2} \cdot n) ds = 0 \end{align} Is this correct?

Edit: Here are two papers about the subject that I found so far:

[1] Triebenbacher, S., Kaltenbacher, M., Wohlmuth, B., & Flemisch, B. (2010). Applications of the mortar finite element method in vibroacoustics and flow induced noise computations. Acta Acustica united with Acustica, 96(3), 536-553.

[2] Walsh, T., Reese, G., Dohrmann, C., & Rouse, J. (2009). Finite element methods for structural acoustics on mismatched meshes. Journal of Computational Acoustics, 17(03), 247-275.

They use the coupling conditions that I have given above. A pity is that they dont show how a mortar formulation is done for the interface between the structural and acoustic domain.


1 Answer 1


After some digging I could finally find a paper that more or less answered my question:

Bermúdez, A., Rodrıguez, R., & Santamarina, D. (2003). Finite element approximation of a displacement formulation for time-domain elastoacoustic vibrations. Journal of computational and applied mathematics, 152(1-2), 17-34.

It turns out that the equation for the acoustic pressure from above might not be suited for a Mortar coupling. Instead, we can alternatively describe the behaviour of the fluid in the domain $\Omega_a$ by a displacement based formulation. The governing equation are then:

\begin{equation} \rho_s \frac{\partial^2 u_s}{\partial t^2} - \nabla \cdot \sigma = 0 \quad \text{on } \Omega_s\\ \rho_a \frac{\partial^2 u_a}{\partial t^2} -\nabla(\rho_a c^2 \nabla \cdot u_a) =0 \quad \text{on } \Omega_a \end{equation} Here we have now the displacement-vector of the solid by $u_s$ and of the fluid in the acoustic domain by $u_a$. For this new formulation we get the coupling conditions on $\Gamma_{as}$: \begin{align} u_s\cdot n = u_a \cdot n\\ \sigma \cdot v = \rho_a c^2 \nabla \cdot u_a v \end{align} where $v$ is any vector at the interace $\Gamma_{as}$. This coupling conditions allow us to use the Mortar method as usual, which is to define a vectorial-lagrange multiplier as $tv:=\sigma \cdot v = \rho_a c^2 \nabla u_a v$ and then derive the weak formulation :

\begin{align} \int_{\Omega_s} u_s' \cdot \rho_s \frac{\partial^2 u_s}{\partial t^2} dx + \frac{1}{2} \int_{\Omega_s} \nabla u' : C : (\nabla u_s + (\nabla u_s)^T ) dx +\int_{\Gamma_{as}} (u_s' \cdot \sigma) \cdot n \ ds =0 \\ \int_{\Omega_a} u_a' \cdot \rho_a \frac{\partial^2 u_a}{\partial t^2} dx + \int_{\Omega_a} \rho_a c^2 \nabla \cdot u_a\nabla \cdot u_a' dx -\int_{\Gamma_{as}} \rho_a c^2 \nabla \cdot u_a u_a' \cdot n \ ds =0 \end{align} In these equations we can now easily insert the Lagrange multiplier $t$ and enforce the other coupling condition in a weak sense by a third equation. The final weak formulation is then:

\begin{align} \int_{\Omega_s} u_s' \cdot \rho_s \frac{\partial^2 u_s}{\partial t^2} dx + \frac{1}{2} \int_{\Omega_s} \nabla u' : C : (\nabla u_s + (\nabla u_s)^T ) dx +\int_{\Gamma_{as}} t u_s' \cdot n \ ds =0 \\ \int_{\Omega_a} u_a' \cdot \rho_a \frac{\partial^2 u_a}{\partial t^2} dx + \int_{\Omega_a} \rho_a c^2 \nabla \cdot u_a\nabla \cdot u_a' dx -\int_{\Gamma_{as}} t u_a' \cdot n \ ds =0 \\ \int_{\Gamma_{as}} \mu'(u_s - u_a )\cdot n ds = 0 \end{align} Note that $u_a'$ and $u_s'$ are the test functions for the two domains $\Omega_a$ and $\Omega_s$ and we have to choose their Sobolev space such that $u_a'\cdot v = u_s'\cdot v$ for an vector $v$ on $\Gamma_{as}$.


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