Variable Coefficient Poisson

Derive the form of a test variable-coefficient elliptic equation with periodic boundary conditions for testing the variable-coefficient multigrid solver.

We want to solve an equation of the form $\nabla \cdot (\alpha \nabla \phi) = f$

Note: it is important for solvability that the RHS, f, integrate to zero over our domain. It seems sufficient to ensure that phi integrates to zero to have this condition met.



In [1]:

    
%pylab inline









    



Populating the interactive namespace from numpy and matplotlib



In [2]:

    
from sympy import init_session
init_session()









    



IPython console for SymPy 1.0 (Python 2.7.12-64-bit) (ground types: gmpy)

These commands were executed:
>>> from __future__ import division
>>> from sympy import *
>>> x, y, z, t = symbols('x y z t')
>>> k, m, n = symbols('k m n', integer=True)
>>> f, g, h = symbols('f g h', cls=Function)
>>> init_printing()

Documentation can be found at http://docs.sympy.org/1.0/

Variable coefficient elliptic problem, using periodic boundary conditions all around.



In [3]:

    
alpha = 2.0 + cos(2*pi*x)*cos(2*pi*y)



In [4]:

    
phi = sin(2*pi*x)*sin(2*pi*y)

we want to compute $\nabla \cdot (\alpha \nabla \phi)$



In [5]:

    
phi_x = diff(phi, x)
phi_y = diff(phi, y)



In [6]:

    
f = diff(alpha*phi_x, x) + diff(alpha*phi_y, y)



In [7]:

    
f = simplify(f)
f









    Out[7]:





$$- 16.0 \pi^{2} \left(\cos{\left (2 \pi x \right )} \cos{\left (2 \pi y \right )} + 1\right) \sin{\left (2 \pi x \right )} \sin{\left (2 \pi y \right )}$$



In [8]:

    
print(f)









    



-16.0*pi**2*(cos(2*pi*x)*cos(2*pi*y) + 1)*sin(2*pi*x)*sin(2*pi*y)

boundary conditions check



In [9]:

    
phi.subs(x, 0)









    Out[9]:





$$0$$



In [10]:

    
phi.subs(x, 1)









    Out[10]:





$$0$$



In [11]:

    
phi.subs(y, 0)









    Out[11]:





$$0$$



In [12]:

    
phi.subs(y, 1)









    Out[12]:





$$0$$

General Elliptic

Derive the form of a test variable-coefficient elliptic equation with periodic boundary conditions for testing the variable-coefficient multigrid solver.

We are solving

$$\alpha \phi + \nabla \cdot (\beta \nabla \phi ) + \gamma \cdot \nabla \phi = f$$

Note: it is important for solvability that the RHS, f, integrate to zero over our domain. It seems sufficient to ensure that phi integrates to zero to have this condition met.



In [13]:

    
phi = sin(2*pi*x)*sin(2*pi*y)



In [14]:

    
phi









    Out[14]:





$$\sin{\left (2 \pi x \right )} \sin{\left (2 \pi y \right )}$$



In [15]:

    
alpha = 1.0
beta = 2.0 + cos(2*pi*x)*cos(2*pi*y)
gamma_x = sin(2*pi*x)
gamma_y = sin(2*pi*y)



In [16]:

    
alpha









    Out[16]:





$$1.0$$



In [17]:

    
beta









    Out[17]:





$$\cos{\left (2 \pi x \right )} \cos{\left (2 \pi y \right )} + 2.0$$



In [18]:

    
gamma_x









    Out[18]:





$$\sin{\left (2 \pi x \right )}$$



In [19]:

    
gamma_y









    Out[19]:





$$\sin{\left (2 \pi y \right )}$$



In [20]:

    
phi_x = diff(phi, x)
phi_y = diff(phi, y)



In [21]:

    
f = alpha*phi + diff(beta*phi_x, x) + diff(beta*phi_y, y) + gamma_x*phi_x + gamma_y*phi_y



In [22]:

    
f = simplify(f)
f









    Out[22]:





$$\left(- 16.0 \pi^{2} \cos{\left (2 \pi x \right )} \cos{\left (2 \pi y \right )} + 2.0 \pi \cos{\left (2 \pi x \right )} + 2.0 \pi \cos{\left (2 \pi y \right )} - 16.0 \pi^{2} + 1.0\right) \sin{\left (2 \pi x \right )} \sin{\left (2 \pi y \right )}$$



In [23]:

    
print(f)









    



(-16.0*pi**2*cos(2*pi*x)*cos(2*pi*y) + 2.0*pi*cos(2*pi*x) + 2.0*pi*cos(2*pi*y) - 16.0*pi**2 + 1.0)*sin(2*pi*x)*sin(2*pi*y)



In [24]:

    
print(alpha)

1.0



In [25]:

    
print(beta)









    



cos(2*pi*x)*cos(2*pi*y) + 2.0



In [26]:

    
print(gamma_x)









    



sin(2*pi*x)



In [27]:

    
print (gamma_y)









    



sin(2*pi*y)



In [ ]:

Inhomogeneous BCs



In [28]:

    
phi = cos(Rational(1,2)*pi*x)*cos(Rational(1,2)*pi*y)



In [29]:

    
phi









    Out[29]:





$$\cos{\left (\frac{\pi x}{2} \right )} \cos{\left (\frac{\pi y}{2} \right )}$$



In [30]:

    
alpha = 10
gamma_x = 1
gamma_y = 1
beta = x*y + 1



In [31]:

    
phi_x = diff(phi, x)
phi_y = diff(phi, y)
f = alpha*phi + diff(beta*phi_x, x) + diff(beta*phi_y, y) + gamma_x*phi_x + gamma_y*phi_y



In [32]:

    
f









    Out[32]:





$$- \frac{\pi x}{2} \sin{\left (\frac{\pi y}{2} \right )} \cos{\left (\frac{\pi x}{2} \right )} - \frac{\pi y}{2} \sin{\left (\frac{\pi x}{2} \right )} \cos{\left (\frac{\pi y}{2} \right )} - \frac{\pi^{2}}{2} \left(x y + 1\right) \cos{\left (\frac{\pi x}{2} \right )} \cos{\left (\frac{\pi y}{2} \right )} - \frac{\pi}{2} \sin{\left (\frac{\pi x}{2} \right )} \cos{\left (\frac{\pi y}{2} \right )} - \frac{\pi}{2} \sin{\left (\frac{\pi y}{2} \right )} \cos{\left (\frac{\pi x}{2} \right )} + 10 \cos{\left (\frac{\pi x}{2} \right )} \cos{\left (\frac{\pi y}{2} \right )}$$



In [33]:

    
simplify(f)









    Out[33]:





$$- \frac{x y}{2} \pi^{2} \cos{\left (\frac{\pi x}{2} \right )} \cos{\left (\frac{\pi y}{2} \right )} - \frac{\pi x}{2} \sin{\left (\frac{\pi y}{2} \right )} \cos{\left (\frac{\pi x}{2} \right )} - \frac{\pi y}{2} \sin{\left (\frac{\pi x}{2} \right )} \cos{\left (\frac{\pi y}{2} \right )} - \frac{\pi}{2} \sin{\left (\pi \left(\frac{x}{2} + \frac{y}{2}\right) \right )} - \frac{\pi^{2}}{2} \cos{\left (\frac{\pi x}{2} \right )} \cos{\left (\frac{\pi y}{2} \right )} + 10 \cos{\left (\frac{\pi x}{2} \right )} \cos{\left (\frac{\pi y}{2} \right )}$$

boundary conditions



In [34]:

    
phi.subs(x,0)









    Out[34]:





$$\cos{\left (\frac{\pi y}{2} \right )}$$



In [35]:

    
phi.subs(x,1)









    Out[35]:





$$0$$



In [36]:

    
phi.subs(y,0)









    Out[36]:





$$\cos{\left (\frac{\pi x}{2} \right )}$$



In [37]:

    
phi.subs(y,1)









    Out[37]:





$$0$$



In [ ]: