Minimal Surface Problem on the Unit Disk

This example shows how to solve a nonlinear elliptic problem numerically, using the pdenonlin function in Partial Differential Equation Toolbox™.

In many problems, the coefficients not only depend on spatial coordinates, but also on the solution itself. In toolbox wording, this kind of problem is called nonlinear. An example of this is the minimal surface equation

$$-\nabla \cdot \left( \frac{1}{\sqrt{1 + \left|\nabla u\right|^2}} \nabla u \right) = 0$$

on the unit disk, with

$u(x,y) = x^2$

on the boundary. The PDE coefficient c is the multiplier of $\nabla u$, namely

$$\frac{1}{\sqrt{1 + \left|\nabla u\right|^2}}$$

c is a function of the solution $u$, so the problem is nonlinear.

Problem Definition

The following variables define the problem:

  • g: A geometry specification function used by initmesh. For more information, see the documentation page for circleg and pdegeom.

  • c, a, f: The coefficients of the PDE. Note that c is a character array. For more information on passing coefficients into pdenonlin, see the documentation page for assempde.

  • rtol: Tolerance for nonlinear solver.

g = @circleg;
c = '1./sqrt(1+ux.^2+uy.^2)';
a = 0;
f = 0;
rtol = 1e-3;

Boundary Conditions

Plot the geometry and display the edge labels for use in the boundary condition definition.

pdegplot(g, 'edgeLabels', 'on');
axis equal
title 'Geometry With Edge Labels Displayed';
% Create a pde entity for a PDE with a single dependent variable
numberOfPDE = 1;
pb = pde(numberOfPDE);
% Create a geometry entity
pg = pdeGeometryFromEdges(g);
bcFunc = @(thePde, loc, state) loc.x.^2;
pb.BoundaryConditions = pdeBoundaryConditions(pg.Edges(1:4), 'u', bcFunc);

Generate Mesh

[p,e,t] = initmesh(g);
[p,e,t] = refinemesh(g,p,e,t);
axis equal

Solve PDE

Because the problem is nonlinear, we solve it using the pdenonlin function.

u = pdenonlin(pb,p,e,t,c,a,f,'tol',rtol);

Plot Solution


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