examples.levelSet.distanceFunction.square

Here we solve the level set equation in two dimensions for a square. The equation is given by:

\[\begin{split}\abs{\nabla \phi} &= 1 \\ \phi &= 0 \qquad \text{at} \qquad \begin{cases} x = \left( L / 3, 2 L / 3 \right) & \text{for $L / 3 \le y \le 2 L / 3$} \\ y = \left( L / 3, 2 L / 3 \right) & \text{for $L / 3 \le x \le 2 L / 3$} \end{cases}\end{split}\]

Do the tests:

>>> var.calcDistanceFunction(order=1) 

>>> def evalCell(phix, phiy, dx, dy):
...     aa = dy**2 + dx**2
...     bb = -2 * ( phix * dy**2 + phiy * dx**2)
...     cc = dy**2 * phix**2 + dx**2 * phiy**2 - dx**2 * dy**2
...     sqr = numerix.sqrt(bb**2 - 4. * aa * cc)
...     return ((-bb - sqr) / 2. / aa,  (-bb + sqr) / 2. / aa)
>>> val = evalCell(-dy / 2., -dx / 2., dx, dy)[0]
>>> v1 = evalCell(val, -3. * dx / 2., dx, dy)[0]
>>> v2 = evalCell(-3. * dy / 2., val, dx, dy)[0]
>>> v3 = evalCell(v2, v1, dx, dy)[0]
>>> v4 = dx * dy / numerix.sqrt(dx**2 + dy**2) / 2
>>> arr = numerix.array((
...     v3, v2, -3. * dy / 2., v2, v3,
...     v1, val, -dy / 2., val, v1,
...     -3. * dx / 2., -dx / 2., v4, -dx / 2., -3. * dx / 2.,
...     v1, val, -dy / 2., val, v1,
...     v3, v2, -3. * dy / 2., v2, v3           ))
>>> print(var.allclose(arr)) 
1

Last updated on Feb 14, 2025. Created using Sphinx 7.1.2.