# laplace.py¶

In this script we solve the Laplace equation $$u_{,kk} = 0$$ on a unit square domain $$Ω$$ with boundary $$Γ$$, subject to boundary conditions:

\begin{align}\begin{aligned}u &= 0 && Γ_{\rm left}\\ ∂_n u &= 0 && Γ_{\rm bottom}\\ ∂_n u &= \cos(1) \cosh(x_1) && Γ_{\rm right}\\ u &= \cosh(1) \sin(x_0) && Γ_{\rm top} \end{aligned}\end{align}

This case is constructed to contain all combinations of homogenous and heterogeneous, Dirichlet and Neumann type boundary conditions, as well as to have a known exact solution:

$u_{\rm exact} = \sin(x_0) \cosh(x_1).$

We start by importing the necessary modules.

 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 from nutils import mesh, function, solver, export, cli, testing import treelog def main(nelems:int, etype:str, btype:str, degree:int): ''' Laplace problem on a unit square. .. arguments:: nelems  Number of elements along edge. etype [square] Type of elements (square/triangle/mixed). btype [std] Type of basis function (std/spline), availability depending on the selected element type. degree  Polynomial degree. ''' 

A unit square domain is created by calling the nutils.mesh.unitsquare() mesh generator, with the number of elements along an edge as the first argument, and the type of elements (“square”, “triangle”, or “mixed”) as the second. The result is a topology object domain and a vectored valued geometry function geom.

 49  domain, geom = mesh.unitsquare(nelems, etype) 

To be able to write index based tensor contractions, we need to bundle all relevant functions together in a namespace. Here we add the geometry x, a scalar basis, and the solution u. The latter is formed by contracting the basis with a to-be-determined solution vector ?lhs.

 56 57 58 59  ns = function.Namespace() ns.x = geom ns.basis = domain.basis(btype, degree=degree) ns.u = 'basis_n ?lhs_n' 

We are now ready to implement the Laplace equation. In weak form, the solution is a scalar field $$u$$ for which:

$∀ v: ∫_Ω v_{,k} u_{,k} - ∫_{Γ_n} v f = 0.$

By linearity the test function $$v$$ can be replaced by the basis that spans its space. The result is an integral res that evaluates to a vector matching the size of the function space.

 70 71  res = domain.integral('basis_n,i u_,i d:x' @ ns, degree=degree*2) res -= domain.boundary['right'].integral('basis_n cos(1) cosh(x_1) d:x' @ ns, degree=degree*2) 

The Dirichlet constraints are set by finding the coefficients that minimize the error:

$\min_u ∫_{\Gamma_d} (u - u_d)^2$

The resulting cons array holds numerical values for all the entries of ?lhs that contribute (up to droptol) to the minimization problem. All remaining entries are set to NaN, signifying that these degrees of freedom are unconstrained.

 83 84 85  sqr = domain.boundary['left'].integral('u^2 d:x' @ ns, degree=degree*2) sqr += domain.boundary['top'].integral('(u - cosh(1) sin(x_0))^2 d:x' @ ns, degree=degree*2) cons = solver.optimize('lhs', sqr, droptol=1e-15) 

The unconstrained entries of ?lhs are to be determined such that the residual vector evaluates to zero in the corresponding entries. This step involves a linearization of res, resulting in a jacobian matrix and right hand side vector that are subsequently assembled and solved. The resulting lhs array matches cons in the constrained entries.

 93  lhs = solver.solve_linear('lhs', res, constrain=cons) 

Once all entries of ?lhs are establised, the corresponding solution can be vizualised by sampling values of ns.u along with physical coordinates ns.x, with the solution vector provided via the arguments dictionary. The sample members tri and hull provide additional inter-point information required for drawing the mesh and element outlines.

 102 103 104  bezier = domain.sample('bezier', 9) x, u = bezier.eval(['x_i', 'u'] @ ns, lhs=lhs) export.triplot('solution.png', x, u, tri=bezier.tri, hull=bezier.hull) 

To confirm that our computation is correct, we use our knowledge of the analytical solution to evaluate the L2-error of the discrete result.

 109 110 111 112  err = domain.integral('(u - sin(x_0) cosh(x_1))^2 d:x' @ ns, degree=degree*2).eval(lhs=lhs)**.5 treelog.user('L2 error: {:.2e}'.format(err)) return cons, lhs, err 

If the script is executed (as opposed to imported), nutils.cli.run() calls the main function with arguments provided from the command line. For example, to keep with the default arguments simply run python3 laplace.py (view log). To select mixed elements and quadratic basis functions add python3 laplace.py etype=mixed degree=2 (view log).

 120 121 if __name__ == '__main__': cli.run(main) 

Once a simulation is developed and tested, it is good practice to save a few strategic return values for regression testing. The nutils.testing module, which builds on the standard unittest framework, facilitates this by providing nutils.testing.TestCase.assertAlmostEqual64() for the embedding of desired results as compressed base64 data.

 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 class test(testing.TestCase): @testing.requires('matplotlib') def test_default(self): cons, lhs, err = main(nelems=4, etype='square', btype='std', degree=1) with self.subTest('constraints'): self.assertAlmostEqual64(cons, ''' eNrbKPv1QZ3ip9sL1BgaILDYFMbaZwZj5ZnDWNfNAeWPESU=''') with self.subTest('left-hand side'): self.assertAlmostEqual64(lhs, ''' eNoBMgDN/7Ed9eB+IfLboCaXNKc01DQaNXM14jXyNR82ZTa+NpI2oTbPNhU3bjf7Ngo3ODd+N9c3SNEU 1g==''') with self.subTest('L2-error'): self.assertAlmostEqual(err, 1.63e-3, places=5) @testing.requires('matplotlib') def test_spline(self): cons, lhs, err = main(nelems=4, etype='square', btype='spline', degree=2) with self.subTest('constraints'): self.assertAlmostEqual64(cons, ''' eNqrkmN+sEfhzF0xleRbDA0wKGeCYFuaIdjK5gj2aiT2VXMAJB0VAQ==''') with self.subTest('left-hand side'): self.assertAlmostEqual64(lhs, ''' eNqrkmN+sEfhzF0xleRbrsauxsnGc43fGMuZJJgmmNaZ7jBlN7M08wLCDLNFZh/NlM0vmV0y+2CmZV5p vtr8j9kfMynzEPPF5lfNAcuhGvs=''') with self.subTest('L2-error'): self.assertAlmostEqual(err, 8.04e-5, places=7) @testing.requires('matplotlib') def test_mixed(self): cons, lhs, err = main(nelems=4, etype='mixed', btype='std', degree=2) with self.subTest('constraints'): self.assertAlmostEqual64(cons, ''' eNorfLZF2ucJQwMC3pR7+QDG9lCquAtj71Rlu8XQIGfC0FBoiqweE1qaMTTsNsOvRtmcoSHbHL+a1UD5 q+YAxhcu1g==''') with self.subTest('left-hand side'): self.assertAlmostEqual64(lhs, ''' eNorfLZF2ueJq7GrcYjxDJPpJstNbsq9fOBr3Gh8xWS7iYdSxd19xseMP5hImu5UZbv1xljOxM600DTW NN/0k2mC6SPTx6Z1pnNMGc3kzdaaPjRNMbMyEzWzNOsy223mBYRRZpPNJpktMks1azM7Z7bRbIXZabNX ZiLmH82UzS3Ns80vmj004za/ZPYHCD+Y8ZlLmVuYq5kHm9eahwDxavPF5lfNAWFyPdk=''') with self.subTest('L2-error'): self.assertAlmostEqual(err, 1.25e-4, places=6)