# Coupled Finite Elements ⤶

Dec 12, 2017

In this post, I’ll introduce the FE formulation of a generalized linear and coupled weak form. Said weak formulation has the form

Find $u\in V_1$, $y\in V_2$ such that

\begin{alignedat}{3} a(u, v) &+ b(y, v) &&= c(v) \\ d(u, w) &+ e(y, w) &&= f(w) \\ \end{alignedat} \label{eq:coupledweakform1}

for all $v\in V_1$, $w \in V_2$ where $a(\cdot, \cdot): V_1\times V_1 \to \IR$, $b(\cdot, \cdot): V_2\times V_1 \to \IR$, $d(\cdot, \cdot): V_1\times V_2 \to \IR$, $e(\cdot, \cdot): V_2\times V_2 \to \IR$ are bilinear forms and $c(\cdot): V_1\to \IR$, $f(\cdot): V_2\to \IR$ are linear forms.

Here, the objective is to solve for the two unknown functions $u$ and $y$. One can also imagine an arbitrary degree of coupling between $n$ variables with $n$ equations.

We introduce the following discretizations

\begin{alignedat}{3} u_h &= \suml{J=1}{n_n^1} u^J N^J \qquad\qquad & v_h &= \suml{I=1}{n_n^1} u^I N^I \qquad\qquad & u_h, v_h\in V_{h1} \\ y_h &= \suml{L=1}{n_n^2} u^L N^L & w_h &= \suml{K=1}{n_n^2} u^K N^K & y_h, w_h\in V_{h2} \\ \end{alignedat}

where the corresponding number of shape functions are $n_n^1$ and $n_n^2$, respectively.

Substituting the discretizations in \eqref{eq:coupledweakform1}, we obtain two linear systems of equations

\begin{alignedat}{3} \suml{J=1}{n_n^1} a(N^J, N^I) \,u^J &+ \suml{L=1}{n_n^2} b(N^L, N^I) \, y^L &&= c(N^I) \\ \suml{J=1}{n_n^1} d(N^J, N^K) \,u^J &+ \suml{L=1}{n_n^2} e(N^L, N^K) \, y^L &&= f(N^K) \\ \end{alignedat}

for $I=1,\dots,n_n^1$ and $K=1,\dots,n_n^2$.

We write this system as

\boxed{ \begin{alignedat}{3} \BA \Bu &+ \BB\By &&= \Bc \\ \BD \Bu &+ \BE\By &&= \Bf \\ \end{alignedat} \eqor \begin{bmatrix} \BA & \BB \\ \BD & \BE \end{bmatrix} \begin{bmatrix} \Bu \\ \By \end{bmatrix} = \begin{bmatrix} \Bc \\ \Bf \end{bmatrix} } \label{eq:coupledsystem1}

where the components of given matrices and vectors are defined as

\begin{alignedat}{6} A^{I\!J} &:= a(N^J, N^I) \qquad & B^{I\!L} &:= b(N^L, N^I) \qquad & c^{I} &:= c(N^I)\\ D^{K\!J} &:= d(N^J, N^K) & E^{K\!L} &:= e(N^L, N^K) & f^{K} &:= f(N^K)\\ \end{alignedat}

Solution of \eqref{eq:coupledsystem1} yields the unknown vectors $\Bu$ and $\By$.