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\begin{document}
\hbox to\hsize{\rlap{MATH 8446, Spring '18}\hfil Homework 4 \hfil\llap{Arnold}}
\bigskip
\parskip=10pt plus 2pt minus 2pt
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One way to derive a mixed formulation is to minimize a quadratic energy function (the dual or complementary energy)
subject to a linear constraint. The ingredients needed to carry this out are
\begin{itemize}
 \item a pair of Hilbert spaces $V$ and $W$
 \item a quadratic functional $J:V\to\R$
 \item a bounded linear operator $L:V\to W^*$
 \item an element $f\in W^*$.
\end{itemize}
The problem  is then to find a minimizer $\sigma$ of $J$ over $V$ subject to the constraint $L\sigma=f$.  That is, we seek
$$
\sigma \in V \text{ such that $L\sigma=f$ and } J(\sigma) =\min_{\substack{\tau\in V\\L\tau=f}} J(\tau).
$$

To solve the constrained minimization problem, we introduce a Lagrange multiplier $u\in W$ which converts the
problem to that of finding a stationary point of a different functional, without any constraint.  Specifically, we seek
a pair $(\sigma, u) \in V \x W$ which is a stationary point of the quadratic functional
$$
L:V\x W\to \R, \quad L(\tau,v) = J(\tau) + \langle L\tau -f , u\rangle.
$$
We may then compute the corresponding Euler--Lagrange equations to get a system of linear equations that determine
$\sigma$ and $u$.


{\bf Example:}  For example, let $\Omega$ be a domain in $\R^2$ and take 
$$
V = H(\div,\Omega),\quad W = L^2(\Omega), \quad L\tau = \div\tau,\quad f\in L^2(\Omega).
$$
Note that, in this case, $W^*=L^2$, the same as $W$.
Then the constrained minimization problem is to find $\sigma\in H(\div)$ such that $\div\sigma=f$ and
$$
\frac 12\|\sigma\|_{L^2}^2 =\min_{\substack{\tau\in H(\div)\\\div\tau=f}} \frac12\|\tau\|_{L^2}^2.
$$
The Lagrange multiplier $u$ belongs to $L^2$ and the pair $(\sigma,u)\in H(\div)\x L^2$ is a stationary point of
$$
L(\tau, v) =  \frac12\|\tau\|_{L^2}^2 + (\div\tau -f, v).
$$
The Euler--Lagrange equations are then
$$
(\sigma,\tau) + (\div\tau, u) = 0, \quad \tau\in H(\div), \qquad
(\div\sigma, v) = (f, v), \quad v\in L^2.
$$
This is a weak formulation of the boundary value problem
$$
\sigma -\grad u = 0,\quad \div\sigma = f, \quad u=0 \text{ on $\partial\Omega$}.
$$
In this case we can eliminate $\sigma$ from the system and write a boundary
value problem for the
Lagrange multiplier variable $u$ alone: it satisfies the boundary value problem
$$
\Delta u = f \text{ in $\Omega$}, \quad u=0 \text{ on $\partial\Omega$},
$$
which is the Dirichlet problem for the Poisson equation.

\bigskip
The homework assignment is to work out a few more examples in a way similar to the example above.
Present each example carefully and clearly, defining the spaces and quadratic functionals involved, writing
out the Euler--Lagrange equations in weak and strong form, etc.

\bigskip

1.  Let $H(\div\div)$ be the space of $2\x 2$ symmetric-matrix-valued functions $M=M_{ij}$ for which
$$
\div\div M:= \sum_{i,j=1}^2 \frac{\partial^2 M_{ij}}{\partial x^i\partial x^j}
$$
belongs to $L^2$.  Discuss (as above) the 
minimization of
$$
J(M) = \frac12 \int_\Omega \sum_{i,j=1}^2 |M_{ij}|^2\, dx
\text{  subject to $\div\div M = f$}.
$$
Be sure to state (among the other things) the boundary value problem satisfied by the Lagrange multiplier variable alone.



\bigskip

2. Consider minimizing the energy functional on $H^1$
$$
J(v) = \frac12 \int_\Omega|\grad v|^2\,dx -\int_\Omega fv\,dx
$$
subject to the constraint $\tr v = 0$.  Here $\tr v = v|_{\partial\Omega}$ is the usual
trace operator which maps $H^1(\Omega)$ to $H^{1/2}(\partial\Omega)$ so the space $W$ should be
taken to be $H^{-1/2}(\partial\Omega)$, the dual space of $H^{1/2}(\partial\Omega)$.
This is an approach to solve the Dirichlet problem without building the boundary
condition into the space (instead enforcing it through a Lagrange multiplier).  In this case, it
is not natural to reduce the problem to the Lagrange multiplier alone, but be sure
to say how the Lagrange multiplier variable is related to the solution of the Dirichlet problem.

\bigskip

3. The non-mixed form of the biharmonic problem minimizes
$$
\frac12 \int_\Omega |\grad\grad w|^2\,dx - \int_\Omega fw\,dx
$$
over $w\in \0H^2$.  One way to get a mixed method is to minimize
$$
J(\tau,w) = \frac12\int_\Omega |\grad \tau|^2\,dx -\int_\Omega fw\,dx,
$$
over $V=\0H^1(\Omega;\R^2)\x \0H^1(\Omega)$ subject to the constraint
$\tau-\grad w=0$.  Analyze this method in the framework above.


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