$$
\newcommand{\x}{\boldsymbol{x}}
\renewcommand{\u}{\boldsymbol{u}}
\renewcommand{\v}{\boldsymbol{v}}
\newcommand{\V}{\boldsymbol{V}}
\newcommand{\basphi}{\varphi}
\newcommand{\dx}{\, \mathrm{d}x}
$$
    
Study guide: Computing with variational forms for systems of PDEs
  
Hans Petter Langtangen [1, 2]
[1] Center for Biomedical Computing, Simula Research Laboratory
[2] Department of Informatics, University of Oslo
Oct 16, 2015
 
Table of contents
 Systems of differential equations 
       Variational forms: treat each PDE as a scalar PDE 
       Variational forms: treat the PDE system as a vector PDE 
       A worked example 
       Identical function spaces for the unknowns 
       Variational form of each individual PDE 
       Compound scalar variational form 
       Alternative inner product notation 
       Decoupled linear systems 
       Coupled linear systems 
       Alternative notation for coupled linear system 
       Different function spaces for the unknowns 
Systems of differential equations
Consider \( m+1 \) unknown functions: \( u^{(0)},\ldots, u^{(m)} \) governed
by \( m+1 \) differential equations:
$$
\begin{align*}
\mathcal{L}_0(u^{(0)},\ldots,u^{(m)}) &= 0\\ 
&\vdots\\ 
\mathcal{L}_{m}(u^{(0)},\ldots,u^{(m)}) &= 0,
\end{align*}
$$
Goals.
 -  How do we derive variational formulations of systems of differential
   equations?
 
 -  How do we apply the finite element method?
 
 
 -  First approach: treat each equation as a scalar equation
 
 -  For equation no. \( i \), use test function \( v^{(i)}\in V^{(i)} \)
 
$$
\begin{align*}
\int_\Omega \mathcal{L}^{(0)}(u^{(0)},\ldots,u^{(m)}) v^{(0)}\dx &= 0\\ 
&\vdots\\ 
\int_\Omega \mathcal{L}^{(m)}(u^{(0)},\ldots,u^{(m)}) v^{(m)}\dx &= 0
\end{align*}
$$
Terms with second-order derivatives may be integrated by parts, with
Neumann conditions inserted in boundary integrals.
$$ V^{(i)} = \hbox{span}\{\basphi_0^{(i)},\ldots,\basphi_{N_i}^{(i)}\},$$
$$ u^{(i)} = B^{(i)}(\x) + \sum_{j=0}^{N_i} c_j^{(i)} \basphi_j^{(i)}(\x),
$$
Can derive \( m \) coupled linear systems for the unknowns
\( c_j^{(i)} \), \( j=0,\ldots,N_i \),
\( i=0,\ldots,m \).
Variational forms: treat the PDE system as a vector PDE 
 -  Second approach: work with vectors (and vector notation)
 
 -  \( \u = (u^{(0)},\ldots,u^{(m)}) \)
 
 -  \( \v = (u^{(0)},\ldots,u^{(m)}) \)
 
 -  \( \u, \v \in  \V = V^{(0)}\times \cdots \times V^{(m)} \)
 
 -  Note: if \( \boldsymbol{B} = (B^{(0)},\ldots,B^{(m)}) \) is needed for
   nonzero Dirichlet conditions, \( \u - \boldsymbol{B}\in \V \) (not \( \u \) in \( \V \))
 
 -  \( \boldsymbol{\mathcal{L}}(\u ) = 0 \)
 
 -  \( \boldsymbol{\mathcal{L}}(\u ) = (\mathcal{L}^{(0)}(\u),\ldots, \mathcal{L}^{(m)}(\u)) \)
 
The variational form is derived by taking the inner product of
\( \boldsymbol{\mathcal{L}}(\u ) \) and \( \v \):
$$
\begin{equation*}
\int_\Omega \boldsymbol{\mathcal{L}}(\u )\cdot\v = 0\quad\forall\v\in\V
\end{equation*}
$$
 -  Observe: this is a scalar equation (!).
 
 -  Can derive \( m \) independent equation by choosing \( m \) independent \( \v \)
 
 -  E.g.: \( \v = (v^{(0)},0,\ldots,0) \) recovers \eqref{fem:sys:vform:1by1a}
 
 -  E.g.: \( \v = (0,\ldots,0,v^{(m)} \) recovers \eqref{fem:sys:vform:1by1b}
 
A worked example
$$
\begin{align*}
\mu \nabla^2 w &= -\beta\\ 
\kappa\nabla^2 T &= - \mu ||\nabla w||^2 \quad (= \mu \nabla w\cdot\nabla w)
\end{align*}
$$
 -  Unknowns: \( w(x,y) \), \( T(x,y) \)
 
 -  Known constants: \( \mu \), \( \beta \), \( \kappa \)
 
 -  Application: fluid flow in a straight pipe, \( w \) is velocity, \( T \) is
   temperature
 
 -  \( \Omega \): cross section of the pipe
 
 -  Boundary conditions: \( w=0 \) and \( T=T_0 \) on \( \partial\Omega \)
 
 -  Note: \( T \) depends on \( w \), but \( w \) does not depend on \( T \) (one-way coupling)
 
Identical function spaces for the unknowns 
Let \( w, (T-T_0) \in V \) with test functions \( v\in V \).
$$ V = \hbox{span}\{\basphi_0(x,y),\ldots,\basphi_N(x,y)\}, $$
$$
\begin{equation*}
w = \sum_{j=0}^N c^{(w)}_j \basphi_j,\quad T = T_0 +
\sum_{j=0}^N c^{(T)}_j\basphi_j
\end{equation*}
$$
Variational form of each individual PDE 
Inserting \eqref{fem:sys:wT:ex:sum}
in the PDEs, results in the residuals
$$
\begin{align*}
R_w &= \mu \nabla^2 w + \beta\\ 
R_T &= \kappa\nabla^2 T + \mu ||\nabla w||^2
\end{align*}
$$
Galerkin's method: make residual orthogonal to \( V \),
$$
\begin{align*}
\int_\Omega R_w v \dx &=0\quad\forall v\in V\\ 
\int_\Omega R_T v \dx &=0\quad\forall v\in V
\end{align*}
$$
Integrate by parts and use \( v=0 \) on \( \partial\Omega \) (Dirichlet conditions!):
$$
\begin{align*}
\int_\Omega \mu \nabla w\cdot\nabla v \dx &= \int_\Omega \beta v\dx
\quad\forall v\in V\\ 
\int_\Omega \kappa \nabla T\cdot\nabla v \dx &= \int_\Omega \mu
\nabla w\cdot\nabla w\, v\dx \quad\forall v\in V
\end{align*}
$$
Compound scalar variational form 
 -  Test vector function \( \v\in\V = V\times V \)
 
 -  Take the inner product of \( \v \) and the system of PDEs (and integrate)
 
$$ \int_{\Omega} (R_w, R_T)\cdot\v \dx = 0\quad\forall\v\in\V
 $$
With \( \v = (v_0,v_1) \):
$$ \int_{\Omega} (R_w v_0 + R_T v_1) \dx = 0\quad\forall\v\in\V
 $$
$$
\begin{equation*}
\int_\Omega (\mu\nabla w\cdot\nabla v_0 + \kappa\nabla T\cdot\nabla v_1)\dx
= \int_\Omega (\beta v_0 + \mu\nabla w\cdot\nabla w\, v_1)\dx,
\quad\forall \v\in\V
\end{equation*}
$$
Choosing \( v_0=v \) and \( v_1=0 \) gives the variational form
\eqref{fem:sys:wT:ex:w:vf1}, while \( v_0=0 \) and \( v_1=v \) gives
\eqref{fem:sys:wT:ex:T:vf1}.
Alternative inner product notation 
$$
\begin{align*}
\mu (\nabla w,\nabla v) &= (\beta, v)
\quad\forall v\in V\\ 
\kappa(\nabla T,\nabla v) &= \mu(\nabla w\cdot\nabla w, v)\quad\forall v\in V
\end{align*}
$$
Decoupled linear systems 
$$
\begin{align*}
\sum_{j=0}^N A^{(w)}_{i,j} c^{(w)}_j &= b_i^{(w)},\quad i=0,\ldots,N\\ 
\sum_{j=0}^N A^{(T)}_{i,j} c^{(T)}_j &= b_i^{(T)},\quad i=0,\ldots,N\\ 
A^{(w)}_{i,j} &= \mu(\nabla \basphi_j,\nabla\basphi_i)\\ 
b_i^{(w)} &= (\beta, \basphi_i)\\ 
A^{(T)}_{i,j} &= \kappa(\nabla \basphi_j,\nabla\basphi_i)\\ 
b_i^{(T)} &= (\mu\nabla w_{-}\cdot (\sum_k
c^{(w)}_k\nabla\basphi_k), \basphi_i)
\end{align*}
$$
Matrix-vector form (alternative notation):
$$
\begin{align*}
\mu K c^{(w)} &= b^{(w)}\\ 
\kappa K c^{(T)} &= b^{(T)}
\end{align*}
$$
where
$$
\begin{align*}
K_{i,j} &= (\nabla \basphi_j,\nabla \basphi_i)\\ 
b^{(w)} &= (b_0^{(w)},\ldots,b_{N}^{(w)})\\ 
b^{(T)} &= (b_0^{(T)},\ldots,b_{N}^{(T)})\\ 
c^{(w)} &= (c_0^{(w)},\ldots,c_{N}^{(w)})\\ 
c^{(T)} &= (c_0^{(T)},\ldots,c_{N}^{(T)})
\end{align*}
$$
First solve the system for \( c^{(w)} \), then solve the system for \( c^{(T)} \)
Coupled linear systems 
 -  Pretend two-way coupling, i.e., need to solve for \( w \) and \( T \) simultaneously
 
 -  Want to derive one system for \( c_j^{(w)} \) and \( c_j^{(T)} \), \( j=0,\ldots,N \)
 
 -  The system is nonlinear because of \( \nabla w\cdot\nabla w \)
 
 -  Linearization: pretend an iteration where \( \hat w \) is computed
   in the previous iteration and set \( \nabla w\cdot\nabla w
   \approx \nabla\hat w\cdot\nabla w \) (so the term becomes linear in \( w \))
 
$$
\begin{align*}
\sum_{j=0}^N A^{(w,w)}_{i,j} c^{(w)}_j + \sum_{j=0}^N A^{(w,T)}_{i,j} c^{(T)}_j
&= b_i^{(w)},\quad i=0,\ldots,N,
\\ 
\sum_{j=0}^N A^{(T,w)}_{i,j} c^{(w)}_j + \sum_{j=0}^N A^{(T,T)}_{i,j} c^{(T)}_j &= b_i^{(T)},\quad i=0,\ldots,N,\\ 
A^{(w,w)}_{i,j} &= \mu(\nabla \basphi_j,\basphi_i)\\ 
A^{(w,T)}_{i,j} &= 0\\ 
b_i^{(w)} &= (\beta, \basphi_i)\\ 
A^{(w,T)}_{i,j} &= \mu(\nabla w_{-}\cdot\nabla\basphi_j), \basphi_i)\\ 
A^{(T,T)}_{i,j} &= \kappa(\nabla \basphi_j,\basphi_i)\\ 
b_i^{(T)} &= 0
\end{align*}
$$
Alternative notation for coupled linear system 
$$
\begin{align*}
\mu K c^{(w)} &= b^{(w)}\\ 
L c^{(w)} + \kappa K c^{(T)} & =0
\end{align*}
$$
\( L \) is the matrix from the \( \nabla w_{-}\cdot\nabla \) operator:
\( L_{i,j} = A^{(w,T)}_{i,j} \).
Corresponding block form:
$$
\left(\begin{array}{cc}
\mu K & 0\\ 
L & \kappa K
\end{array}\right)
\left(\begin{array}{c}
c^{(w)}\\ 
c^{(T)}
\end{array}\right) =
\left(\begin{array}{c}
b^{(w)}\\ 
0
\end{array}\right)
$$
Different function spaces for the unknowns 
 -  Generalization: \( w\in V^{(w)} \) and \( T\in V^{(T)} \),
   \( V^{(w)} \neq V^{(T)} \)
 
 -  This is called a mixed finite element method
 
$$
\begin{align*}
V^{(w)} &= \hbox{span}\{\basphi_0^{(w)},\ldots,\basphi_{N_w}^{(w)}\}\\ 
V^{(T)} &= \hbox{span}\{\basphi_0^{(T)},\ldots,\basphi_{N_T}^{(T)}\}
\end{align*}
$$
$$
\begin{align*}
\int_\Omega \mu \nabla w\cdot\nabla v^{(w)} \dx &= \int_\Omega \beta v^{(w)}\dx
\quad\forall v^{(w)}\in V^{(w)}\\ 
\int_\Omega \kappa \nabla T\cdot\nabla v^{(T)} \dx &= \int_\Omega \mu
\nabla w\cdot\nabla w\, v^{(T)}\dx \quad\forall v^{(T)}\in V^{(T)}
\end{align*}
$$
Take the inner product with \( \v = (v^{(w)}, v^{(T)}) \) and integrate:
$$
\begin{equation*}
\int_\Omega (\mu\nabla w\cdot\nabla v^{(w)} +
\kappa\nabla T\cdot\nabla v^{(T)})\dx
= \int_\Omega (\beta v^{(w)} + \mu\nabla w\cdot\nabla w\, v^{(T)})\dx,
\end{equation*}
$$
valid \( \forall \v\in\V = V^{(w)}\times V^{(T)} \).