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11.2.4. Neumann Laplace Direct Method

keys: homogeneous Neumann bvp, hypersingular operator, unknown Dirichlet data

[1]:
from netgen.occ import *
from ngsolve import *
from ngsolve.webgui import Draw
from ngsolve.bem import *
from ngsolve.solvers import CG

Consider the Neumann boundary value problem

\[\begin{split}\left\{ \begin{array}{rlc l} \Delta u &=& 0, \quad &\Omega \subset \mathbb R^3\,,\\ \gamma_1 u&=& u_1, \quad &\Gamma = \partial \Omega\,.\end{array} \right.\end{split}\]

The unique solution \(u\in H^1(\Omega)\) can be written in the following form (representation forumula)

\[u(x) = \underbrace{ \int\limits_\Gamma \displaystyle{\frac{1}{4\,\pi}\, \frac{1}{\| x-y\|} } \, u_1(y)\, \mathrm{d}\sigma_y}_{\displaystyle{ \mathrm{SL}(u_1) }} + \underbrace{ \int\limits_\Gamma \displaystyle{\frac{1}{4\,\pi}\, \frac{\langle n(y), x-y \rangle}{\| x-y\|^3} } \, u_1( y)\, \mathrm{d}\sigma_y}_{\displaystyle{ \mathrm{DL}(u_1) }}\,.\]

Let’s carefully apply the Neumann trace \(\gamma_1\) to the representation formula and solve the resulting boundary integral equation for the Dirichlet trace \(u_0\) of \(u\) by discretisation of the following variational formulation:

\[\forall \, v\in H^{\frac12}(\Gamma): \left\langle v, \gamma_1 \left(\mathrm{DL}(u_0)\right) \right\rangle_{-\frac12} = \left\langle u_1, v\right\rangle_{-\frac12} - \left\langle v, \gamma_1 \left(\mathrm{SL}(u_1)\right) \right\rangle_{-\frac12}\,.\]

Define the geometry \(\Omega \subset \mathbb R^3\) and create a mesh:

[2]:
sp = Glue(Sphere( (0,0,0), 1).faces)
mesh = Mesh( OCCGeometry(sp).GenerateMesh(maxh=0.2)).Curve(2)

Define conforming finite element spaces for \(H^{\frac12}(\Gamma)\) and \(H^{-\frac12}(\Gamma)\) respectively:

[3]:
fesL2 = SurfaceL2(mesh, order=2, dual_mapping=False)
u,v = fesL2.TnT()
fesH1 = H1(mesh, order=2, definedon=mesh.Boundaries(".*"))
uH1,vH1 = fesH1.TnT()
print ("ndofL2 = ", fesL2.ndof, "ndof H1 = ", fesH1.ndof)
ndofL2 =  4332 ndof H1 =  1446

Project the given Neumann data \(u_1\) in finite element space and have a look at the boundary data:

[4]:
uexa = 1/ sqrt( (x-1)**2 + (y-1)**2 + (z-1)**2 )

graduexa = CF( (uexa.Diff(x), uexa.Diff(y), uexa.Diff(z)) )
n = specialcf.normal(3)
u1 = GridFunction(fesL2)
u1.Interpolate(graduexa*n, definedon=mesh.Boundaries(".*"))

Draw (u1, mesh, draw_vol=False, order=3);

The discretisation of the variational formulation leads to a system of linear equations, ie

\[\left( \mathrm{D} + \mathrm{S}\right) \, \mathrm{u}_0 = \left( \frac12 \,\mathrm{M} - \mathrm{K}' \right) \, \mathrm{u}_1\,,\]

where the linear operators are as follows

  • \(\mathrm{D}\) is the hypersingular operator and the stabilisation \((\mathrm D + \mathrm{S})\) is regular and symmetric.

  • \(\mathrm{M}\) is a mass matrix.

  • \(\mathrm{K}'\) is the adjoint double layer operator.

The stabilisation matrix \(S\) is needed to cope with the kernel of the hypersingular operator. The following stabilisation matrix is used here:

\[S \in \mathbb R^{n\times n}, \quad S_{ij} = \int\limits_{\Gamma} v_i(x) \,\mathrm{d} \sigma \cdot \int\limits_{\Gamma} v_j(x) \,\mathrm{d} \sigma\,,\]

where \(v_i, v_j\) being \(H^{\frac12}(\Gamma)\)-conforming shape functions.

[5]:
vH1m1 = LinearForm(vH1*1*ds(bonus_intorder=3)).Assemble()
S = (BaseMatrix(Matrix(vH1m1.vec.Reshape(1))))@(BaseMatrix(Matrix(vH1m1.vec.Reshape(fesH1.ndof))))

Now we assemble the stabilised system matrix and the right hand side and solve for the Dirichlet data \(u_0\) with an iterative solver:

[6]:
u0 = GridFunction(fesH1)
pre = BilinearForm(uH1*vH1*ds).Assemble().mat.Inverse(freedofs=fesH1.FreeDofs())
with TaskManager():
    D = LaplaceSL(curl(uH1)*ds)*curl(vH1)*ds

    M = BilinearForm( v.Trace() * uH1 * ds(bonus_intorder=3)).Assemble()
    Kt = LaplaceDL(u*ds)*vH1*ds
    rhs = ( (0.5 * M.mat.T - Kt.mat) * u1.vec ).Evaluate()

    CG(mat=D.mat+S, pre=pre, rhs=rhs, sol=u0.vec, tol=1e-8, maxsteps=200, printrates=True)
CG iteration 1, residual = 0.6734142037837124
CG iteration 2, residual = 0.35705284587486436
CG iteration 3, residual = 0.20848547590128977
CG iteration 4, residual = 0.1630691377997378
CG iteration 5, residual = 0.1285940923409068
CG iteration 6, residual = 0.09481213077257317
CG iteration 7, residual = 0.060941602084874216
CG iteration 8, residual = 0.048209014927422585
CG iteration 9, residual = 0.02978984442566913
CG iteration 10, residual = 0.021210625376812908
CG iteration 11, residual = 0.013531576526267783
CG iteration 12, residual = 0.008288866237733838
CG iteration 13, residual = 0.005443367814364068
CG iteration 14, residual = 0.0032123106862259982
CG iteration 15, residual = 0.0020387981973550105
CG iteration 16, residual = 0.001189555484176782
CG iteration 17, residual = 0.0006849500537641346
CG iteration 18, residual = 0.0004044845162947714
CG iteration 19, residual = 0.00022080861874748187
CG iteration 20, residual = 0.00012690272961924143
CG iteration 21, residual = 6.911746960365204e-05
CG iteration 22, residual = 3.737387421308229e-05
CG iteration 23, residual = 1.976201784581829e-05
CG iteration 24, residual = 9.964137900416136e-06
CG iteration 25, residual = 4.935087518301099e-06
CG iteration 26, residual = 2.4799107475607713e-06
CG iteration 27, residual = 1.19062290034754e-06
CG iteration 28, residual = 5.712439900390044e-07
CG iteration 29, residual = 2.623725373405627e-07
CG iteration 30, residual = 1.2096265541409491e-07
CG iteration 31, residual = 5.4962551046272146e-08
CG iteration 32, residual = 2.740344203250904e-08
CG iteration 33, residual = 1.8446422618841733e-08
CG iteration 34, residual = 1.8551212351087424e-08
CG iteration 35, residual = 2.1326134490020883e-08
CG iteration 36, residual = 1.5929329008365538e-08
CG iteration 37, residual = 7.846743037419083e-09
CG iteration 38, residual = 3.459786859356401e-09

Let’s have a look at the numerical and exact Dirichlet data and compare them quantitatively:

[7]:
Draw (u0, mesh, draw_vol=False, order=3);
u0exa = GridFunction(fesH1)
u0exa.Interpolate (uexa, definedon=mesh.Boundaries(".*"))
Draw (u0exa, mesh, draw_vol=False, order=3);
[8]:
print ("L2 error of surface gradients =", sqrt (Integrate ( (grad(u0exa)-grad(u0))**2, mesh, BND)))
L2 error of surface gradients = 0.004452375833735614
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