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Fipy error:’’The Factor is exactly singular’’, when applying Neumann boundary conditions

We’re trying to solve a one-dimensional Coupled Continuity-Poisson problem in Fipy. When applying
Dirichlet’s conditions, it gives the correct results, but when we change the boundaries conditions to Neumann’s which is closer to our problem, it gives “The Factor is exactly singular’’ error.
Any help is highly appreciated. The code is as follows (0<x<2.5):
from fipy import *
from fipy import Grid1D, CellVariable, TransientTerm, DiffusionTerm, Viewer
import numpy as np
import math
import matplotlib.pyplot as plt
from matplotlib import cm
from cachetools import cached, TTLCache #caching to increase the speed of python
cache = TTLCache(maxsize=100, ttl=86400) #creating the cache object: the
#first argument= the number of objects we store in the cache.
#____________________________________________________
nx=50
dx=0.05
L=nx*dx
e=math.e
m = Grid1D(nx=nx, dx=dx)
print(np.log(e))
#____________________________________________________
phi = CellVariable(mesh=m, hasOld=True, value=0.)
ne = CellVariable(mesh=m, hasOld=True, value=0.)
phi_face = phi.faceValue
ne_face = ne.faceValue
x = m.cellCenters[0]
t0 = Variable()
phi.setValue((x-1)**3)
ne.setValue(-6*(x-1))
#____________________________________________________
#cached(cache)
def S(x,t):
f=6*(x-1)*e**(-t)+54*((x-1)**2)*e**(-2.*t)
return f
#____________________________________________________
#Boundary Condition:
valueleft_phi=3*e**(-t0)
valueright_phi=6.75*e**(-t0)
valueleft_ne=-6*e**(-t0)
valueright_ne=-6*e**(-t0)
phi.faceGrad.constrain([valueleft_phi], m.facesLeft)
phi.faceGrad.constrain([valueright_phi], m.facesRight)
ne.faceGrad.constrain([valueleft_ne], m.facesLeft)
ne.faceGrad.constrain([valueright_ne], m.facesRight)
#____________________________________________________
eqn0 = DiffusionTerm(1.,var=phi)==ImplicitSourceTerm(-1.,var=ne)
eqn1 = TransientTerm(1.,var=ne) ==
VanLeerConvectionTerm(phi.faceGrad,var=ne)+S(x,t0)
eqn = eqn0 & eqn1
#____________________________________________________
steps = 1.e4
dt=1.e-4
T=dt*steps
F=dt/(dx**2)
print('F=',F)
#____________________________________________________
vi = Viewer(phi)
with open('out2.txt', 'w') as output:
while t0()<T:
print(t0)
phi.updateOld()
ne.updateOld()
res=1.e30
#for sweep in range(steps):
while res > 1.e-4:
res = eqn.sweep(dt=dt)
t0.setValue(t0()+dt)
for m in range(nx):
output.write(str(phi[m])+' ') #+ os.linesep
output.write('\n')
if __name__ == '__main__':
vi.plot()
#____________________________________________________
data = np.loadtxt('out2.txt')
X, T = np.meshgrid(np.linspace(0, L, len(data[0,:])), np.linspace(0, T,
len(data[:,0])))
fig = plt.figure(3)
ax = fig.add_subplot(111,projection='3d')
ax.plot_surface(X, T, Z=data)
plt.show(block=True)

The issue with these equations, particularly eqn0, is that they admit an infinite number of solutions when Neumann boundary conditions are applied on both boundaries. You can fix this by pinning a value somewhere with an internal fixed value. E.g., based on the analytical solution given in the comments, phi = (x-1)**3 * exp(-t), we can pin phi = 0 at x = 1 with
mask = (m.x > 1-dx/2) & (m.x < 1+dx/2)
largeValue = 1e6
value = 0.
#____________________________________________________
eqn0 = (DiffusionTerm(1.,var=phi)==ImplicitSourceTerm(-1.,var=ne)
+ ImplicitSourceTerm(mask * largeValue, var=phi) - mask * largeValue * value)
At this point, the solutions still do not agree with the expected solutions. This is because, while you have called ne.faceGrad.constrain() for the left and right boundaries, does not appear in the discretized equations. You can see this if you plot ne; the gradient is zero at both boundaries despite the constraint because FiPy never "sees" the constraint.
What does appear is the flux . By applying fixed flux boundary conditions, I obtain the expected solutions:
ne_left = 6 * numerix.exp(-t0)
ne_right = -9 * numerix.exp(-t0)
eqn1 = (TransientTerm(1.,var=ne)
== VanLeerConvectionTerm(phi.faceGrad * m.interiorFaces,var=ne)
+ S(x,t0)
+ (m.facesLeft * ne_left * phi.faceGrad).divergence
+ (m.facesRight * ne_right * phi.faceGrad).divergence)
You can probably get better convergence properties with
eqn1 = (TransientTerm(1.,var=ne)
== DiffusionTerm(coeff=ne.faceValue * m.interiorFaces, var=phi)
+ S(x,t0)
+ (m.facesLeft * ne_left * phi.faceGrad).divergence
+ (m.facesRight * ne_right * phi.faceGrad).divergence)
but either formulation seems to work.
Note: phi.faceGrad.constrain() is fine, because the flux does appear in DiffusionTerm(coeff=1., var=phi).
Separately, it appears (based on "The Factor is exactly singular") that you are solving with the SciPy LinearLUSolver. The PETSc LinearLUSolver does better, but the baseline value of the solution wanders all over the place. Calling
res = eqn.sweep(dt=dt, solver=LinearGMRESSolver())
also seems to produce stable results (without pinning an internal value). This behavior probably shouldn't be relied on; pinning a value is the right thing to do.

FiPy evaluates different solutions on the same system using python2 or python3

I'm experimenting FiPy with a system of partial differential equations and I get different results running the same script in Python2 or Python3. More precisely, I get the result I expect running the script in Python 2.7, while I get a completely wrong result in Python 3.8.
Does somebody knows the reason? Am I missing something?
I know that some of the solvers FiPy uses in Python 2.7 don't support Python 3.x, but is this enough to explain a different behaviour?
Thank you in advance.
NOTE: I wanted to post images but I can't since I just subscribed to stack overflow.
Code
import fipy as fp
import fipy.tools as fpt
"""
Let's try to solve the reaction diffusion equation fith FiPy
A + B ---k---> C
"""
# define spatial domain (square)
nx = ny = 20
dx = dy = 1.
mesh = fp.Grid2D(dx=dx, dy=dy, nx=nx, ny=ny)
linear_dimension = (nx * ny,)
# define molecule as CellVariables
a = fp.CellVariable(name="molecule A",
mesh=mesh)
b = fp.CellVariable(name="molecule B",
mesh=mesh)
c = fp.CellVariable(name="molecule C",
mesh=mesh)
# define initial condition
def gauss_2d(mu_x, mu_y, sigma_x, sigma_y, x_1d, y_1d):
"""
Utility function to define initial conditions (see below). Provide a simil-gaussian
distribution (NOTICE: it's not an exact gaussian because I needed it to be 1 at the top)
"""
# initialize matrix
gauss_mat = fpt.numerix.empty((len(x_1d), len(y_1d)))
# define gaussian
gauss_x = fpt.numerix.exp((-1 / 2) * (((x_1d - mu_x) ** 2) / (sigma_x ** 2)))
gauss_y = fpt.numerix.exp((-1 / 2) * (((y_1d - mu_y) ** 2) / (sigma_y ** 2)))
# evaluate each point of the matrix
for i in range(0, len(x_1d)):
for j in range(0, len(y_1d)):
gauss_mat[i, j] = gauss_x[i] * gauss_y[j]
return gauss_mat
normal_distribution = gauss_2d(mu_x=nx / 2,
mu_y=ny / 2,
sigma_x=fpt.numerix.sqrt(10),
sigma_y=fpt.numerix.sqrt(10),
x_1d=fpt.numerix.arange(0, nx, dx),
y_1d=fpt.numerix.arange(0, ny, dy))
a_max = 100.
a0 = a_max * normal_distribution
a.setValue(a0.reshape(linear_dimension))
b_max = a_max / 2
b0 = b_max * normal_distribution
b.setValue(b0.reshape(linear_dimension))
c0 = 0.
c.setValue(c0)
# create viewer for the three molecules
vi = fp.Viewer(vars=(a, b, c), datamin=0., datamax=a_max)
vi.plot()
fp.input("Press enter to continue...")
# define the reaction term
k = 0.01
reaction_term = k * a * b
# define the equation for each molecule
D = 0.05
eq_a = fp.TransientTerm(var=a) == fp.DiffusionTerm(coeff=D, var=a) - reaction_term
eq_b = fp.TransientTerm(var=b) == fp.DiffusionTerm(coeff=D, var=b) - reaction_term
eq_c = fp.TransientTerm(var=c) == fp.DiffusionTerm(coeff=D, var=c) + reaction_term
# couple equations
sys = eq_a & eq_b & eq_c
# solve
dt = 0.25
steps = 400
for step in range(steps):
sys.solve(dt=dt)
vi.plot()
fp.input("Press enter to continue...")

tl;dr: You're probably seeing a difference in the behavior of the Matplotlib viewer and not in the solver.
When I adjust for bugs and regressions with Matplotlib, I see very similar evolution with Py27 and Py3k. Until we get that fixed, you can see if these changes help:
diff --git a/rxndiff.py b/rxndiff.py
index b41932a..ea30a95 100644
--- a/rxndiff.py
+++ b/rxndiff.py
## -1,5 +1,6 ##
import fipy as fp
import fipy.tools as fpt
+from matplotlib import pyplot as plt
"""
Let's try to solve the reaction diffusion equation fith FiPy
## -79,5 +80,8 ## steps = 400
for step in range(steps):
sys.solve(dt=dt)
vi.plot()
+ for vw in vi.viewers:
+ vw.fig.canvas.draw_idle()
+ plt.show(block=False)
The solver is different (I get PySparse LinearLUSolver with Py27 and PETSc LinearGMRESSolver with Py3k). This could matter, particularly since you don't sweep this nonlinear problem, but I don't see a huge difference.
The initial condition with Py27 is probably not what you want because of integer division. I recommend initializing with:
mu_x=nx / 2.
mu_y=ny / 2.
sigma_x=fpt.numerix.sqrt(10.)
sigma_y=fpt.numerix.sqrt(10.)
normal_distribution = (fpt.numerix.exp((-1./2) * (mesh.x - mu_x) ** 2 / sigma_x ** 2)
* fpt.numerix.exp((-1./2) * (mesh.y - mu_y) ** 2 / sigma_y ** 2))
a_max = 100.
a0 = a_max * normal_distribution
a.setValue(a0)
b_max = a_max / 2
b0 = b_max * normal_distribution
b.setValue(b0)
c0 = 0.
c.setValue(c0)

Solving differential equation with ODEINT in python

I have a coupled system of differential equations that I've already solved with Euler in Excel. Now I want to make it more precise with an ODE-solver in python.
However, there must be a mistake in my code because the curves look different than in Excel. I don't expect the curves to reach 1 and 0 in the end.
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
# define reactor
def reactor(x,z):
n_a = x[0]
n_b = x[1]
n_c = x[2]
dn_adz = A * (-1) * B * (n_a/(n_a + n_b + n_c)) / (1 + C * (n_c/(n_a + n_b + n_c)))
dn_bdz = A * (1) * B * (n_a/(n_a + n_b + n_c)) / (1 + C * (n_c/(n_a + n_b + n_c)))
dn_cdz = A * (1) * B * (n_a/(n_a + n_b + n_c)) / (1 + C * (n_c/(n_a + n_b + n_c)))
dxdz = [dn_adz,dn_bdz,dn_cdz]
return dxdz
# initial conditions
n_a0 = 0.5775
n_b0 = 0.0
n_c0 = 0.0
x0 = [n_a0, n_b0, n_c0]
# parameters
A = 0.12
B = 3.1e-9
C = 4.02e15
# number of steps
n = 100
# z step interval (m)
z = np.linspace(0,0.0274,n)
# solve ODEs
x = odeint(reactor,x0,z)
# Plot the results
plt.plot(z,x[:,0],'b-')
plt.plot(z,x[:,1],'r--')
plt.plot(z,x[:,2],'k:')
plt.show()
Is is a problem with the initial condition that stays constant and does not change from step to step?
Should it be like in Excel with Euler, where the next step uses the conditions/values of the precious step?

From the structure of the right sides you get constant combinations of the state variables, n_a+n_b=n_a0+n_b0 and n_a+n_c=n_a0+n_c0. This means that the dynamic reduces to the one-dimensional dynamic of n_a.
By the first equation, the derivative of n_a is negative for positive n_a, so that the solution is falling towards n_a=0. By the constants of the dynamics, n_b converges to n_a0+n_b0 and n_c converges to n_a0+n_c0.
It is unclear how you get convergence towards 1 in some components, as that is not supported by the initial conditions. Apart from that, the described odeint result fits this qualitative behavior.

How to solve a 9-equations system of non linear DE with python?

I'm desperately trying to solve (and display the graph) a system made of nine nonlinear differential equations which model the path of a boomerang. The system is the following:
All the letters on the left side are variables, the others are either constants or known functions depending on v_G and w_z
I have tried with scipy.odeint with no conclusive results (I had this issue but the workaround did not work.)
I begin to think that the problem is linked with the fact that these equations are nonlinear or that the function in denominator might cause a singularity that the scipy solver is simply unable to handle. However, I am not familiar with that sort of mathematical knowledge.
What possibilities python-wise do I have to solve this set of equations?
EDIT : Sorry if I was not clear enough. Since it models the path of a boomerang, my goal is not to solve analytically this system (ie I don't care about the mathematical expression of each function), but rather to get the values of each function for a specific time range (say, from t1 = 0s to t2 = 15s with an interval of 0.01s between each value) in order to display the graph of each function and the graph of the center of mass of the boomerang (X,Y,Z are its coordinates).
Here is the code I tried :
import scipy.integrate as spi
import numpy as np
#Constants
I3 = 10**-3
lamb = 1
L = 5*10**-1
mu = I3
m = 0.1
Cz = 0.5
rho = 1.2
S = 0.03*0.4
Kz = 1/2*rho*S*Cz
g = 9.81
#Initial conditions
omega0 = 20*np.pi
V0 = 25
Psi0 = 0
theta0 = np.pi/2
phi0 = 0
psi0 = -np.pi/9
X0 = 0
Y0 = 0
Z0 = 1.8
INPUT = (omega0, V0, Psi0, theta0, phi0, psi0, X0, Y0, Z0) #initial conditions
def diff_eqs(t, INP):
'''The main set of equations'''
Y=np.zeros((9))
Y[0] = (1/I3) * (Kz*L*(INP[1]**2+(L*INP[0])**2))
Y[1] = -(lamb/m)*INP[1]
Y[2] = -(1/(m * INP[1])) * ( Kz*L*(INP[1]**2+(L*INP[0])**2) + m*g) + (mu/I3)/INP[0]
Y[3] = (1/(I3*INP[0]))*(-mu*INP[0]*np.sin(INP[6]))
Y[4] = (1/(I3*INP[0]*np.sin(INP[3]))) * (mu*INP[0]*np.cos(INP[5]))
Y[5] = -np.cos(INP[3])*Y[4]
Y[6] = INP[1]*(-np.cos(INP[5])*np.cos(INP[4]) + np.sin(INP[5])*np.sin(INP[4])*np.cos(INP[3]))
Y[7] = INP[1]*(-np.cos(INP[5])*np.sin(INP[4]) - np.sin(INP[5])*np.cos(INP[4])*np.cos(INP[3]))
Y[8] = INP[1]*(-np.sin(INP[5])*np.sin(INP[3]))
return Y # For odeint
t_start = 0.0
t_end = 20
t_step = 0.01
t_range = np.arange(t_start, t_end, t_step)
RES = spi.odeint(diff_eqs, INPUT, t_range)
However, I keep getting the same problem as shown here and especially the error message :
Excess work done on this call (perhaps wrong Dfun type)
I am not quite sure what it means but it looks like the solver have troubles solving the system. In any case, when I try to display the 3D path thanks to the XYZ coordinates, I just get 3 or 4 points where there should be something like 2000.
So my questions are : - Am I doing something wrong in my code ?
- If not, is there an other maybe more sophisticated tool to solve this sytem ?
- If not, is it even possible to get what I want from this system of ODEs ?
Thanks in advance

There are several issues:
if I copy the code, it does not run
the workaround you mention does not work with odeint, the given
solution uses ode
The scipy reference for odeint says:"For new code, use
scipy.integrate.solve_ivp to solve a differential equation."
the call RES = spi.odeint(diff_eqs, INPUT, t_range) should be
consistent to the function head def diff_eqs(t, INP) . Mind the
order: RES = spi.odeint(diff_eqs,t_range, INPUT)
There are some issues about to mathematical formulas too:
have a look at the 3rd formula on your picture. It has no tendency term, it starts with a zero - what does that mean ?
it's hard to check wether you have translated the formula correctly into code since the code does not follow the formulas strictly.
Below I tried a solution with scipy solve_ivp. In case A I'm able to run a pendulum, but in case B no meaningful solution for the boomerang can be found. So check the maths, I guess some error in the mathematical expressions.
For the graphics use pandas to plot all variables together (see code below).
import scipy.integrate as spi
import numpy as np
import pandas as pd
def diff_eqs_boomerang(t,Y):
INP = Y
dY = np.zeros((9))
dY[0] = (1/I3) * (Kz*L*(INP[1]**2+(L*INP[0])**2))
dY[1] = -(lamb/m)*INP[1]
dY[2] = -(1/(m * INP[1])) * ( Kz*L*(INP[1]**2+(L*INP[0])**2) + m*g) + (mu/I3)/INP[0]
dY[3] = (1/(I3*INP[0]))*(-mu*INP[0]*np.sin(INP[6]))
dY[4] = (1/(I3*INP[0]*np.sin(INP[3]))) * (mu*INP[0]*np.cos(INP[5]))
dY[5] = -np.cos(INP[3])*INP[4]
dY[6] = INP[1]*(-np.cos(INP[5])*np.cos(INP[4]) + np.sin(INP[5])*np.sin(INP[4])*np.cos(INP[3]))
dY[7] = INP[1]*(-np.cos(INP[5])*np.sin(INP[4]) - np.sin(INP[5])*np.cos(INP[4])*np.cos(INP[3]))
dY[8] = INP[1]*(-np.sin(INP[5])*np.sin(INP[3]))
return dY
def diff_eqs_pendulum(t,Y):
dY = np.zeros((3))
dY[0] = Y[1]
dY[1] = -Y[0]
dY[2] = Y[0]*Y[1]
return dY
t_start, t_end = 0.0, 12.0
case = 'A'
if case == 'A': # pendulum
Y = np.array([0.1, 1.0, 0.0]);
Yres = spi.solve_ivp(diff_eqs_pendulum, [t_start, t_end], Y, method='RK45', max_step=0.01)
if case == 'B': # boomerang
Y = np.array([omega0, V0, Psi0, theta0, phi0, psi0, X0, Y0, Z0])
print('Y initial:'); print(Y); print()
Yres = spi.solve_ivp(diff_eqs_boomerang, [t_start, t_end], Y, method='RK45', max_step=0.01)
#---- graphics ---------------------
yy = pd.DataFrame(Yres.y).T
tt = np.linspace(t_start,t_end,yy.shape[0])
with plt.style.context('fivethirtyeight'):
plt.figure(1, figsize=(20,5))
plt.plot(tt,yy,lw=8, alpha=0.5);
plt.grid(axis='y')
for j in range(3):
plt.fill_between(tt,yy[j],0, alpha=0.2, label='y['+str(j)+']')
plt.legend(prop={'size':20})

Improving Numpy speed for Gauss-Seidel (Jacobi) Solver

This question is a follow-up to a recent question posted regarding MATLAB being twice as fast as Numpy.
I currently have a Gauss-Seidel solver implemented in both MATLAB and Numpy which acts on a 2D axisymmetric domain (cylindrical coordinates). The code was originally written in MATLAB and then transferred to Python. The Matlab code runs in ~20 s whereas the Numpy codes takes ~30 s. I would like to use Numpy, however, since this code is part of a larger program, the almost twice as long simulation time is a significant drawback.
The algorithm simply solves the discretized Laplace equation on a rectangular mesh (in cylindrical coordinates). It finishes when the maximum difference between updates on the mesh is less than the indicated tolerance.
The code in Numpy is:
import numpy as np
import time
T = np.transpose
# geometry
length = 0.008
width = 0.002
# mesh
nz = 256
nr = 64
# step sizes
dz = length/nz
dr = width/nr
# node position matrices
r = np.tile(np.linspace(0,width,nr+1), (nz+1, 1)).T
ri = r/dr
# equation coefficients
cr = dz**2 / (2*(dr**2 + dz**2))
cz = dr**2 / (2*(dr**2 + dz**2))
# initial/boundary conditions
v = np.zeros((nr+1,nz+1))
v[:,0] = 1100
v[:,-1] = 0
v[31:,29:40] = 1000
v[19:,54:65] = -200
# convergence parameters
tol = 1e-4
# Gauss-Seidel solver
tic = time.time()
max_v_diff = 1;
while (max_v_diff > tol):
v_old = v.copy()
# left boundary updates
v[0,1:nz] = cr*2*v[1,1:nz] + cz*(v[0,0:nz-1] + v[0,2:nz+2])
# internal updates
v[1:nr,1:nz] = cr*((1 - 1/(2*ri[1:nr,1:nz]))*v[0:nr-1,1:nz] + (1 + 1/(2*ri[1:nr,1:nz]))*v[2:nr+1,1:nz]) + cz*(v[1:nr,0:nz-1] + v[1:nr,2:nz+1])
# right boundary updates
v[nr,1:nz] = cr*2*v[nr-1,1:nz] + cz*(v[nr,0:nz-1] + v[nr,2:nz+1])
# reapply grid potentials
v[31:,29:40] = 1000
v[19:,54:65] = -200
# check for convergence
v_diff = v - v_old
max_v_diff = np.absolute(v_diff).max()
toc = time.time() - tic
print(toc)
This is actually not the full algorithm which I use. The full algorithm uses successive overrelaxation and a checkerboard iteration scheme to improve speed and remove solver directionality, but for purposes of simplicity I provided this easier to understand version. The speed drawbacks in Numpy are more pronounced for the full version (17s vs. 9s simulation times respectively in Numpy and MATLAB).
I tried the solution from the previous question, changing v to a column-major order array, but there was no performance increase.
Any suggestions?
Edit: The Matlab code for reference is:
% geometry
length = 0.008;
width = 0.002;
% mesh
nz = 256;
nr = 64;
% step sizes
dz = length/nz;
dr = width/nr;
% node position matrices
r = repmat(linspace(0,width,nr+1)', 1, nz+1);
ri = r./dr;
% equation coefficients
cr = dz^2/(2*(dr^2+dz^2));
cz = dr^2/(2*(dr^2+dz^2));
% initial/boundary conditions
v = zeros(nr+1,nz+1);
v(1:nr+1,1) = 1100;
v(1:nr+1,nz+1) = 0;
v(32:nr+1,30:40) = 1000;
v(20:nr+1,55:65) = -200;
% convergence parameters
tol = 1e-4;
max_v_diff = 1;
% Gauss-Seidel Solver
tic
while (max_v_diff > tol)
v_old = v;
% left boundary updates
v(1,2:nz) = cr.*2.*v(2,2:nz) + cz.*( v(1,1:nz-1) + v(1,3:nz+1) );
% internal updates
v(2:nr,2:nz) = cr.*( (1 - 1./(2.*ri(2:nr,2:nz))).*v(1:nr-1,2:nz) + (1 + 1./(2.*ri(2:nr,2:nz))).*v(3:nr+1,2:nz) ) + cz.*( v(2:nr,1:nz-1) + v(2:nr,3:nz+1) );
% right boundary updates
v(nr+1,2:nz) = cr.*2.*v(nr,2:nz) + cz.*( v(nr+1,1:nz-1) + v(nr+1,3:nz+1) );
% reapply grid potentials
v(32:nr+1,30:40) = 1000;
v(20:nr+1,55:65) = -200;
% check for convergence
max_v_diff = max(max(abs(v - v_old)));
end
toc

I've been able to reduce the running time in my laptop from 66 to 21 seconds by following this process:
Find the bottleneck. I profiled the code using line_profiler from the IPython console to find the lines that took most time. It turned out that over 80% of the time was spent in the line that does "internal updates".
Choose a way to optimise it. There are several tools to speed code up in numpy (Cython, numexpr, weave...). In particular, scipy.weave.blitz is well suited to compile numpy expressions, like the offending line, into fast code. In theory, that line could be wrapped inside "..." and executed as weave.blitz("...") but the array that's being updated is used in the computation, so as stated by point #4 in the docs a temporary array must be used to keep the same result:
expr = "temp = cr*((1 - 1/(2*ri[1:nr,1:nz]))*v[0:nr-1,1:nz] + (1 + 1/(2*ri[1:nr,1:nz]))*v[2:nr+1,1:nz]) + cz*(v[1:nr,0:nz-1] + v[1:nr,2:nz+1]); v[1:nr,1:nz] = temp"
temp = np.empty((nr-1, nz-1))
...
while ...
# internal updates
weave.blitz(expr)
After checking that the results are correct, runtime checks are disabled by using weave.blitz(expr, check_size=0). The code now runs in 34 seconds.
Building up on Jaime's work, precompute the constant factors A and B in the expression. The code runs in 21 seconds (with minimal changes but it now needs a compiler).
This is the core of the code:
from scipy import weave
# [...] Set up code till "# Gauss-Seidel solver"
tic = time.time()
max_v_diff = 1;
A = cr * (1 - 1/(2*ri[1:nr,1:nz]))
B = cr * (1 + 1/(2*ri[1:nr,1:nz]))
expr = "temp = A*v[0:nr-1,1:nz] + B*v[2:nr+1,1:nz] + cz*(v[1:nr,0:nz-1] + v[1:nr,2:nz+1]); v[1:nr,1:nz] = temp"
temp = np.empty((nr-1, nz-1))
while (max_v_diff > tol):
v_old = v.copy()
# left boundary updates
v[0,1:nz] = cr*2*v[1,1:nz] + cz*(v[0,0:nz-1] + v[0,2:nz+2])
# internal updates
weave.blitz(expr, check_size=0)
# right boundary updates
v[nr,1:nz] = cr*2*v[nr-1,1:nz] + cz*(v[nr,0:nz-1] + v[nr,2:nz+1])
# reapply grid potentials
v[31:,29:40] = 1000
v[19:,54:65] = -200
# check for convergence
v_diff = v - v_old
max_v_diff = np.absolute(v_diff).max()
toc = time.time() - tic

On my laptop your code runs in about 45 seconds. By trying to reduce creation of intermediate arrays to the bare minimum, including reuse of pre-allocated work arrays, I have managed to reduce that time to 27 seconds. That should put you back at the level of MATLAB, but your code would be less readable. Anyway, find below code to replace everything below your # Gauss-Seidel solver comment:
# work arrays
v_old = np.empty_like(v)
w1 = np.empty_like(v[0, 1:nz])
w2 = np.empty_like(v[1:nr,1:nz])
w3 = np.empty_like(v[nr, 1:nz])
# constants
A = cr * (1 - 1/(2*ri[1:nr,1:nz]))
B = cr * (1 + 1/(2*ri[1:nr,1:nz]))
# Gauss-Seidel solver
tic = time.time()
max_v_diff = 1;
while (max_v_diff > tol):
v_old[:] = v
# left boundary updates
np.add(v_old[0, 0:nz-1], v_old[0, 2:nz+2], out=v[0, 1:nz])
v[0, 1:nz] *= cz
np.multiply(2*cr, v_old[1, 1:nz], out=w1)
v[0, 1:nz] += w1
# internal updates
np.add(v_old[1:nr, 0:nz-1], v_old[1:nr, 2:nz+1], out=v[1:nr, 1:nz])
v[1:nr,1:nz] *= cz
np.multiply(A, v_old[0:nr-1, 1:nz], out=w2)
v[1:nr,1:nz] += w2
np.multiply(B, v_old[2:nr+1, 1:nz], out=w2)
v[1:nr,1:nz] += w2
# right boundary updates
np.add(v_old[nr, 0:nz-1], v_old[nr, 2:nz+1], out=v[nr, 1:nz])
v[nr, 1:nz] *= cz
np.multiply(2*cr, v_old[nr-1, 1:nz], out=w3)
v[nr,1:nz] += w3
# reapply grid potentials
v[31:,29:40] = 1000
v[19:,54:65] = -200
# check for convergence
v_old -= v
max_v_diff = np.absolute(v_old).max()
toc = time.time() - tic