Laplace’s equation

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2024-11-26

Table of contents

1 How to solve a maze without AI

See these LinkedIn posts to see some comments and discussions:

Other people’s maze-related posts:

Say you are Homer Simpson and you want to solve a maze drawn in a restaurant’s placemat, one where both the start and end are known beforehand. In order to avoid falling into the alligator’s mouth, you can exploit the ellipticity of the Laplacian operator to solve any maze (even a hand-drawn one) without needing any fancy AI or ML algorithm. Just FeenoX and a bunch of standard open source tools to convert a bitmapped picture of the maze into an unstructured mesh.

Bitmapped maze from https://www.mazegenerator.net (left) and 2D mesh (right) 
PROBLEM laplace 2D  # pretty self-descriptive, isn't it?
READ_MESH maze.msh

# boundary conditions (default is homogeneous Neumann)
BC start  phi=0 
BC end    phi=1

SOLVE_PROBLEM

# write the norm of gradient as a scalar field
# and the gradient as a 2d vector into a .msh file
WRITE_MESH maze-solved.msh \
    sqrt(dphidx(x,y)^2+dphidy(x,y)^2) \
    VECTOR dphidx dphidy 0 
$ gmsh -2 maze.geo
[...]
$ feenox maze.fee
$
Solution to the maze found by FeenoX (and drawn by Gmsh)

1.1 Transient top-down

Instead of solving a steady-state en exploiting the ellipticity of Laplace’s operator, let us see what happens if we solve a transient instead.

PROBLEM laplace 2D
READ_MESH maze.msh

phi_0(x,y) = 0              # inital condition
end_time = 100              # some end time where we know we reached the steady-state
alpha = 1e-6                # factor of the time derivative to make it advance faster
BC start   phi=if(t<1,t,1)  # a ramp from zero to avoid discontinuities with the initial condition
BC end     phi=0            # homogeneous BC at the end (so we move from top to bottom)

SOLVE_PROBLEM
PRINT t

WRITE_MESH maze-tran-td.msh phi    sqrt(dphidx(x,y)^2+dphidy(x,y)^2) VECTOR -dphidx(x,y) -dphidy(x,y) 0 
$ feenox maze-tran-td.fee
0
0.00433078
0.00949491
0.0170774
0.0268599
[...]
55.8631
64.0819
74.5784
87.2892
100
$ gmsh maze-tran-td-anim.geo
# all frames dumped, now run
ffmpeg -y -framerate 20 -f image2 -i maze-tran-td-%03d.png maze-tran-td.mp4
ffmpeg -y -framerate 20 -f image2 -i maze-tran-td-%03d.png maze-tran-td.gif
$ ffmpeg -y -framerate 20 -f image2 -i maze-tran-td-%03d.png maze-tran-td.mp4
[...]
$ ffmpeg -y -framerate 20 -f image2 -i maze-tran-td-%03d.png maze-tran-td.gif
[...]
Transient top-bottom solution to the maze found by FeenoX (and drawn by Gmsh)

1.2 Transient bottom-up

Now let us see what happens if we travel the maze from the exit up to the inlet. It looks like the solver tries a few different paths that lead nowhere until the actual solution is found.

PROBLEM laplace 2D
READ_MESH maze.msh

phi_0(x,y) = 0
end_time = 100
alpha = 1e-6
BC end     phi=if(t<1,t,1)
BC start   phi=0 

SOLVE_PROBLEM
PRINT t

WRITE_MESH maze-tran-bu.msh phi    sqrt(dphidx(x,y)^2+dphidy(x,y)^2) VECTOR -dphidx(x,y) -dphidy(x,y) 0 
$ feenox maze-tran-bu.fee
0
0.00402961
0.00954806
0.0180156
0.0285787
[...]
65.3715
72.6894
81.8234
90.9117
100
$ gmsh maze-tran-bu-anim.geo
# all frames dumped, now run
ffmpeg -y -framerate 20 -f image2 -i maze-tran-bu-%03d.png maze-tran-bu.mp4
ffmpeg -y -framerate 20 -f image2 -i maze-tran-bu-%03d.png maze-tran-bu.gif
$ ffmpeg -y -framerate 20 -f image2 -i maze-tran-bu-%03d.png maze-tran-bu.mp4
[...]
$ ffmpeg -y -framerate 20 -f image2 -i maze-tran-bu-%03d.png maze-tran-bu.gif
[...]
Transient bottom-up solution. The first attempt does not seem to be good.

2 Potential flow around an airfoil profile

The Laplace equation can be used to model potential flow, as illustrated with this example from Prof. Enzo Dari for his course “Fluid Mechanics” at Instituto Balseiro.

For the particular case of a airfoil profile, the Dirichlet condition at the wing has to satisfy the Kutta condition.

This example

  1. Creates a symmetric airfoil-like Joukowsky profile using the the Gmsh Python API
  2. Solves the steady-state 2D Laplace equation with a different Dirichlet value at the airfoil until the solution \phi evaluated at the continuation of the wing tip matches the boundary value c. It then computes the circulation integral of the velocities
    1. over the profile itself
    2. over a circle around the airfoil, computing the unitary tangential vector
      1. from the internal normal variables nx and ny
      2. from two functions tx and ty using the circle’s equation
    3. around the original rectangular domain
    It also computes the drag and the lift scalars as the integral of the pressure (computed from Bernoulli’s principle) times nx and nz over the circle, respectively.
Full domain
Zoom over the airfoil profile 
import sys
import math
import cmath
import gmsh

gmsh.initialize()

# size of the domain [-a,+a]x[-b,+b]
a = 20
b = 10

# characteristic element sizes
lc0 = 0.6       # near the boundary
lc = 0.1*lc0    # near the wing

# number of points of definition
nperf = 50
# attack angle
alpha = (float(sys.argv[1]) if len(sys.argv) > 1 else 10) * math.pi/180

# Joukowsky parameters
x0 = (float(sys.argv[2]) if len(sys.argv) > 2 else -0.11)
y0 = 0
r = math.sqrt((x0-1)**2+y0**2)

# append points
pts = []
coords = []
for i in range(nperf):
  theta = i * 2*math.pi/nperf
  dz = complex(x0 + r*math.cos(theta), y0 + r*math.sin(theta))
  z = (dz + 1/dz)*cmath.exp(complex(0, -alpha))
  # remember the coords of these three points
  if i == 0 or i == 1 or i == nperf-1:
    coords.append([z.real, z.imag])
  pts.append(gmsh.model.occ.addPoint(z.real, z.imag, 0, lc))

# create a closed loop of points
pts.append(pts[0])

# add a spline
spl = []
spl.append(gmsh.model.occ.addSpline(pts))
wing = gmsh.model.occ.addCurveLoop(spl)

# compute the location where to evaluate the Kutta condition 
# this is ugly and not pythonic at all!
vsup = [-(coords[1][0] - coords[0][0]),
        -(coords[1][1] - coords[0][1])]  
norm = math.sqrt(vsup[0]**2 + vsup[1]**2)
vsup[0] /= norm
vsup[1] /= norm

vinf = [-(coords[0][0] - coords[2][0]),
        -(coords[0][1] - coords[2][1])] 
norm = math.sqrt(vinf[0]**2 + vinf[1]**2)
vinf[0] /= norm
vinf[1] /= norm

vmed = vsup + vinf
norm = math.sqrt(vmed[0]**2 + vmed[1]**2)
vmed[0] /= norm
vmed[1] /= norm

eps = 0.04
p = [coords[0][0] + eps * vmed[0],
     coords[0][1] + eps * vmed[1]]
# this is just to show the point in case one opens the .brep
kutta = gmsh.model.occ.addPoint(p[0], p[1], 0, lc)

# f = open("p.fee", "w")
# f.write( "VECTOR p[2] DATA %g %g\n" % (p[0], p[1]))
# f.close()


# add the external boundary
p1 = gmsh.model.occ.addPoint(-a, -b, 0, lc0)
p2 = gmsh.model.occ.addPoint(+a, -b, 0, lc0)
p3 = gmsh.model.occ.addPoint(+a, +b, 0, lc0)
p4 = gmsh.model.occ.addPoint(-a, +b, 0, lc0)
bottom   = gmsh.model.occ.addLine(p1, p2)
right    = gmsh.model.occ.addLine(p2, p3)
top      = gmsh.model.occ.addLine(p3, p4)
left     = gmsh.model.occ.addLine(p4, p1)
boundary = gmsh.model.occ.addCurveLoop([bottom, right, top, left])
air      = gmsh.model.occ.addPlaneSurface([boundary, wing])

# create a circle to compute the circulation
r_circle = 3
circle = gmsh.model.occ.addCircle(0, 0, 0, r_circle)


# we could embed the kutta point into the mesh but
# then the geometrical entities 
ov, ovv = gmsh.model.occ.fragment([(2, air)], [(0, kutta), (1,circle)])
print(ov)
print(ovv)

gmsh.model.occ.synchronize()

# physical groups
# after the boolean fragment we lost the ids
# gmsh.model.addPhysicalGroup(1, [bottom], name="bottom")
# gmsh.model.addPhysicalGroup(1, [right],  name="right")
# gmsh.model.addPhysicalGroup(1, [top],    name="top")
# gmsh.model.addPhysicalGroup(1, [left],   name="left")
# gmsh.model.addPhysicalGroup(1, [wing],   name="wing")
# gmsh.model.addPhysicalGroup(2, [air],    name="air")

gmsh.model.addPhysicalGroup(0, [kutta], name="kutta")
gmsh.model.addPhysicalGroup(1, [6],     name="circle")
gmsh.model.addPhysicalGroup(1, [7],     name="bottom")
gmsh.model.addPhysicalGroup(1, [9],     name="right")
gmsh.model.addPhysicalGroup(1, [10],    name="top")
gmsh.model.addPhysicalGroup(1, [8],     name="left")
gmsh.model.addPhysicalGroup(1, [11],    name="wing")
gmsh.model.addPhysicalGroup(2, [1,2],   name="air")

# dump the brep (or xao) just in case
gmsh.write("airfoil.brep")
gmsh.write("airfoil.xao")


# mesh
gmsh.option.setNumber("Mesh.RecombineAll", 1)
gmsh.option.setNumber("Mesh.MeshSizeFromCurvature", 2*math.pi * 0.25 * r_circle/lc)
gmsh.option.setNumber("Mesh.Algorithm", 6)
gmsh.model.mesh.setAlgorithm(2, 1, 8);
gmsh.model.mesh.setAlgorithm(2, 2, 6);
gmsh.option.setNumber("Mesh.Smoothing", 10)

gmsh.model.mesh.generate(2)
gmsh.model.mesh.optimize("Netgen")
gmsh.model.mesh.optimize("Laplace2D")
gmsh.model.mesh.setOrder(2)
gmsh.model.mesh.optimize("HighOrderFastCurving")
gmsh.model.mesh.optimize("HighOrder")
gmsh.write("airfoil.msh")

# gmsh.fltk.run()
gmsh.finalize()
PROBLEM laplace 2D
static_steps = 20
READ_MESH airfoil.msh

# boundary conditions constant -> streamline
BC bottom  phi=0
BC top     phi=1

# initialize c = streamline constant for the wing
DEFAULT_ARGUMENT_VALUE 1 0.5
IF in_static_first
 c = $1
ENDIF
BC wing    phi=c

SOLVE_PROBLEM

PHYSICAL_GROUP kutta DIM 0
e = phi(kutta_cog[1], kutta_cog[2]) - c
PRINT TEXT "\# "step_static %.6f c %+.1e e HEADER

# check for convergence
done_static = done_static | (abs(e) < 1e-8)
IF done_static
  PHYSICAL_GROUP left DIM 1
  vx(x,y) = +dphidy(x,y) * left_length
  vy(x,y) = -dphidx(x,y) * left_length

  # write post-processing views
  WRITE_MESH $0-converged.msh   phi VECTOR vx vy 0
  WRITE_MESH $0-converged.vtk   phi VECTOR vx vy 0
  
  # circulation on profile
  INTEGRATE vx(x,y)*(+ny)+vy(x,y)*(-nx) OVER wing RESULT circ_profile
  PRINTF "circ_profile  = %g" -circ_profile

  # function unit tangent
  PHYSICAL_GROUP circle DIM 1
  cx = circle_cog[1]
  cy = circle_cog[2]
  tx(x,y) = -(y-cy)/sqrt((x-cx)*(x-cx)+(y-cy)*(y-cy))
  ty(x,y) = +(x-cx)/sqrt((x-cx)*(x-cx)+(y-cy)*(y-cy))
  
  # circulation on the circumference with tangent
  INTEGRATE vx(x,y)*tx(x,y)+vy(x,y)*ty(x,y) OVER circle RESULT circ_circle_t
  PRINTF "circ_circle_t = %g" -circ_circle_t

  # circulation on the outer box
  INTEGRATE vx(x,y)*(+ny)+vy(x,y)*(-nx) OVER circle RESULT circ_circle_n
  PRINTF "circ_circle_n = %g" -circ_circle_n
  
  INTEGRATE -vx(x,y) OVER top    GAUSS RESULT inttop
  INTEGRATE +vx(x,y) OVER bottom GAUSS RESULT intbottom
  INTEGRATE -vy(x,y) OVER left   GAUSS RESULT intleft
  INTEGRATE +vy(x,y) OVER right  GAUSS RESULT intright
  circ_box = inttop + intleft + intbottom + intright
  PRINTF "circ_box      = %g" -circ_box
  
  # pressure forces
  p(x,y) = 1.0 - vx(x,y)^2 + vy(x,y)^2
  INTEGRATE p(x,y)*(nx) OVER circle RESULT pFx
  PRINTF "pressure_drag = %g" pFx
  INTEGRATE p(x,y)*(ny) OVER circle RESULT pFy
  PRINTF "pressure_lift = %g" pFy
  
ELSE

 # update c
 IF in_static_first
  c_last = c
  e_last = e
  c = c_last - 0.1
 ELSE
  c_next = c_last - e_last * (c_last-c)/(e_last-e)
  c_last = c
  e_last = e
  c = c_next
 ENDIF
ENDIF
$ python airfoil.py
[...]
Info    : Writing 'airfoil.msh'...
Info    : Done writing 'airfoil.msh'
$ feenox airfoil.fee 
# #     step_static     c       e
#       1       0.500000        -3.4e-03
#       2       0.400000        +3.7e-03
#       3       0.452067        +2.0e-09
circ_profile  = 2.49918
circ_circle_t = 2.49975
circ_circle_n = 2.49907
circ_box      = 2.50379
pressure_drag = -0.00168068
pressure_lift = 2.60614
$
Potential and velocities on the whole domain
Potential and velocities zoomed over the airfoil

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