Non-orthogonal curvilinear coordinates

In this notebook we solve Poisson's equation on a 2D wavy domain, using non-orthogonal basis vectors.



In [1]:

    
import sympy as sp
import matplotlib.pyplot as plt
from IPython.display import Math
from shenfun import *
from shenfun.la import Solver2D

Define the non-orthogonal curvilinear coordinates. Here psi=(u, v) are the new coordinates and rv is the position vector

$$ \vec{r} = u \mathbf{i} + v\left(1- \frac{\sin 2u}{10} \right) \mathbf{j} $$

where $\mathbf{i}$ and $\mathbf{j}$ are the Cartesian unit vectors in $x$- and $y$-directions, respectively.



In [2]:

    
u = sp.Symbol('x', real=True, positive=True)
v = sp.Symbol('y', real=True)
psi = (u, v)
rv = (u, v*(1-sp.sin(2*u)/10))
rv = ()

Now choose basis functions and create tensor product space. Notice that one has to use complex Fourier space and not the real, because the integral measure is a function of u.



In [3]:

    
N = 20
#B0 = FunctionSpace(N, 'C', bc=(0, 0), domain=(0, 2*np.pi))
B0 = FunctionSpace(N, 'F', dtype='D', domain=(0, 2*np.pi))
B1 = FunctionSpace(N, 'L', bc=(0, 0), domain=(-1, 1))

T = TensorProductSpace(comm, (B0, B1), dtype='D', coordinates=(psi, rv, sp.Q.negative(sp.sin(2*u)-10) & sp.Q.negative(sp.sin(2*u)/10-1)))
p = TrialFunction(T)
q = TestFunction(T)
b = T.coors.get_covariant_basis()



In [4]:

    
Math(T.coors.latex_basis_vectors(covariant=True, symbol_names={u: 'u', v: 'v'}))









    Out[4]:





$\displaystyle  \mathbf{b}_{u} =\mathbf{i}- \frac{v \cos{\left(2 u \right)}}{5}\,\mathbf{j} \\ \mathbf{b}_{v} =\left(1 - \frac{\sin{\left(2 u \right)}}{10}\right)\,\mathbf{j} \\  $

Plot the mesh to see the domain.



In [5]:

    
mesh = T.local_curvilinear_mesh()
x, y = mesh
plt.figure(figsize=(10, 4))
for i, (xi, yi) in enumerate(zip(x, y)):
    plt.plot(xi, yi, 'b')
    plt.plot(x[:, i], y[:, i], 'r')

Print the Laplace operator in curvilinear coordinates. We use replace to simplify the expression.



In [6]:

    
dp = div(grad(p))
g = sp.Symbol('g', real=True, positive=True)
replace = [(1-sp.sin(2*u)/10, sp.sqrt(g)), (sp.sin(2*u)-10, -10*sp.sqrt(g)), (5*sp.sin(2*u)-50, -50*sp.sqrt(g))]
Math((dp).tolatex(funcname='p', symbol_names={u: 'u', v: 'v'}, replace=replace))









    Out[6]:





$\displaystyle \frac{\partial^2 p}{\partial u^2 }+\frac{2 v \cos{\left(2 u \right)}}{5 \sqrt{g}}\frac{\partial^2 p}{\partial u  \partial v  }+\frac{2 v \left(- 5 \sqrt{g} \sin{\left(2 u \right)} + \cos^{2}{\left(2 u \right)}\right)}{25 g}\frac{\partial  p}{\partial v  }+\frac{v^{2} \cos^{2}{\left(2 u \right)} + 25}{25 g}\frac{\partial^2 p}{\partial v^2 }$

Solve Poisson's equation. First define a manufactured solution and assemble the right hand side



In [7]:

    
ue = sp.sin(2*u)*(1-v**2)
f = (div(grad(p))).tosympy(basis=ue, psi=psi)
fj = Array(T, buffer=f)
f_hat = Function(T)
f_hat = inner(q, fj, output_array=f_hat)

Then assemble the left hand side and solve using a generic 2D solver



In [8]:

    
mats = inner(q, div(grad(p)))
#mats = inner(grad(q), -grad(p))
u_hat = Function(T)
Sol1 = Solver2D(mats)
u_hat = Sol1(f_hat, u_hat)
uj = u_hat.backward()
uq = Array(T, buffer=ue)
print('Error =', np.linalg.norm(uj-uq))









    



Error = 8.631598423869933e-15

Plot the solution in the wavy Cartesian domain



In [9]:

    
xx, yy = T.local_curvilinear_mesh()
plt.figure(figsize=(12, 4))
plt.contourf(xx, yy, uj.real)
plt.colorbar()









    Out[9]:





<matplotlib.colorbar.Colorbar at 0x7fb8d08031d0>

Inspect the sparsity pattern of the generated matrix on the left hand side



In [10]:

    
from matplotlib.pyplot import spy
plt.figure()
spy(Sol1.M, markersize=0.1)









    Out[10]:





<matplotlib.lines.Line2D at 0x7fb8c029e090>