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Subsections


17 Minimal Surface with Obstacle

Find the surface with minimal area, given boundary conditions, and above an obstacle.

Formulation

Plateau's problem is to determine the surface of minimal area with a given closed curve in $ \mbox{${\mathbb{R}}$}^3 $ as boundary. We assume that the surface can be represented in nonparametric form $ v : \mbox{${\mathbb{R}}$}^2 \mapsto \mbox{${\mathbb{R}}$}$, and we add the requirement that $ v \ge v_L $ for some obstacle $ v_L $. The solution of this obstacle problem [13] minimizes the function $f:K \mapsto \mbox{${\mathbb{R}}$}$,

\begin{displaymath}
\ f(v)=\int_{{\cal D}} \left ( 1+\Vert \nabla v(x) \Vert^2 \right )^{1/2} \; dx,
\end{displaymath}

over the convex set $K$, where

\begin{displaymath}
\ K=\left\{v \in H^1 ({\cal D}):
v(x) = v_D (x) \mbox{ for }...
... \ \
v(x) \geq v_L (x) \mbox{ for } x \in {\cal D} \right\} ,
\end{displaymath}

$ H^1 ({{\cal D}}) $ is the space of functions with gradients in $ L^2 ( {\cal D} ) $, the function $ v_D: \partial {{\cal D}} \mapsto \mbox{${\mathbb{R}}$}$ defines the boundary data, and $ v_L: {\cal D} \mapsto \mbox{${\mathbb{R}}$}$ is the obstacle. We assume that $ v_L \le v_D $ on the boundary $ \partial {\cal D} $.

A finite element approximation to the minimal surface problem is obtained by triangulating ${{\cal D}}$ and minimizing $f$ over the space of piecewise linear functions with values $ v_{i,j} $ at the vertices of the triangulation. We set $ {\cal D} = [0,1] \times [0,1] $ and use a triangulation with, respectively, $ n_x $ and $ n_y $ internal grid points in the coordinate directions. Data for this problem appears in Table 16.1.



Table 17.1: Minimal surface problem data
Variables $ n_x n_y$
Constraints 0
Bounds $ n_x n_y $
Linear equality constraints 0
Linear inequality constraints 0
Nonlinear equality constraints 0
Nonlinear inequality constraints 0
Nonzeros in $ \nabla ^2 f(x) $ $5n_x n_y -2(n_x+n_y)$
Nonzeros in $ c'(x) $ 0

Performance

We provide results for the AMPL formulation in Table 17.2. For these results we fix $ n_x = 50 $ and vary $ n_y $. The starting guess is the function $1-(2x-1)^2$ evaluated at the grid nodes. We used boundary data

\begin{displaymath}
v_D (x,y)=
\left \{
\begin{tabular}{rcl}
$ 1-(2x-1)^2 $, & & $ y = 0, 1 $\ \\
0, & & otherwise, \\
\end{tabular}\right .
\end{displaymath}

and the obstacle

\begin{displaymath}
v_L (x,y)=
\left \{
\begin{tabular}{rcl}
1 & & if
$\vert x...
... \frac{1}{4}$\ \\
0, & & otherwise. \\
\end{tabular}\right .
\end{displaymath}

Figure 17.1 shows the minimal surface for this data.



Table 17.2: Performance on minimal surface area with obstacle problem
Solver $n_y = 25$ $n_y = 50$ $n_y = 75$ $n_y = 100$
LANCELOT 2.77 s 5.9 s 10.34 s 16.33 s
$f$ 2.51948e+00 2.51488e+00 2.50568e+00 2.50694e+00
$c$ violation 0.00000e+00 0.00000e+00 0.00000e+00 0.00000e+00
iterations 8 9 10 13
LOQO 2.98 s 9.76 s 23.32 s $\ddagger$
$f$ 2.51948e+00 2.51488e+00 2.50568e+00 $\ddagger$
$c$ violation 2.4e-15 3.8e-15 3.4e-15 $\ddagger$
iterations 20 28 46 $\ddagger$
MINOS 103.76 s 984.81 s $\ddagger$ $\ddagger$
$f$ 2.51948e+00 2.51488e+00 $\ddagger$ $\ddagger$
$c$ violation 0.0e+00 0.0e+00 $\ddagger$ $\ddagger$
iterations 1 1 $\ddagger$ $\ddagger$
SNOPT 137.88 s $\ddagger$ $\ddagger$ $\ddagger$
$f$ 2.51948e+00 $\ddagger$ $\ddagger$ $\ddagger$
$c$ violation 0.0e+00 $\ddagger$ $\ddagger$ $\ddagger$
iterations 171 $\ddagger$ $\ddagger$ $\ddagger$
$\dagger$ Errors or warnings. $\ddagger$ Timed out.

Figure 17.1: Minimal surface problem with a plate obstacle
\includegraphics[width=4.5in]{ps/minsurf.eps}


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Liz Dolan
2001-01-02