# [Numpy-discussion] need help with simple conjugate gradient Laplace solver

rob rob at pythonemproject.com
Sat Mar 30 08:30:07 CST 2002

```I'm experimenting with electrostatics now.  I have an iterative Jacobian
Laplace solver working but it can be slow.  It creates a beautiful 3D
Animabob image.
So I decided to try out a conjugate-gradient solver, which should be an
order of mag better.  It runs but doesn't converge.  One thing I am
wondering, where is the conjugate?  In my FEM code, the solver realy
does use a conjugate, while this one here that I pieced together from
several other programs does not.  Why is it called conjugate gradient
without a conjugate ?  :)   Here is the code:

from math import *
from Numeric import *

#
#***  ENTER DATA
filename= "out"
#
bobfile=open(filename+".bob","w")

print "\n"*30

NX=30    # number of cells

NY=30

NZ=30

resmax=1e-3    # conj-grad tolerance

#allocate arrays

##Ex=zeros((NX+2,NY+2,NZ+2),Float)
##Ey=zeros((NX+2,NY+2,NZ+2),Float)
##Ez=zeros((NX+2,NY+2,NZ+2),Float)

p=zeros((NX+1,NY+1,NZ+1),Float)
q=zeros((NX+1,NY+1,NZ+1),Float)
r=zeros((NX+1,NY+1,NZ+1),Float)
u=zeros((NX+1,NY+1,NZ+1),Float)

u[0:30,0:30,0]=0    # box at 1V with one side at 0V
u[0:30,0,0:30]=1
u[0,0:30,0:30]=1
u[0:30,0:30,30]=1
u[0:30,30,0:30]=1
u[30,0:30,0:30]=1

r[1:NX-1,1:NY-1,1:NZ-1]=(u[1:NX-1,0:NY-2,1:NZ-1]+            #initialize
r matrix
u[1:NX-1,2:NY,1:NZ-1]+
u[0:NX-2,1:NY-1,1:NZ-1]+
u[2:NX,1:NY-1,1:NZ-1]+
u[1:NX-1,1:NY-1,0:NZ-2]+
u[1:NX-1,1:NY-1,2:NZ])

p[...]=r[...]   #initialize p matrix

#
#**** START ITERATIONS
#
N=(NX-2)*(NY-2)*(NZ-2)   # left over from Jacobi solution, not used
KK=0     # iteration counter

res=resmax=0.0;    #set  residuals=0

while(1):

q[1:NX-1,1:NY-1,1:NZ-1]=(p[1:NX-1,0:NY-2,1:NZ-1]+       # finite
difference eq
p[1:NX-1,2:NY,1:NZ-1]+
p[0:NX-2,1:NY-1,1:NZ-1]+
p[2:NX,1:NY-1,1:NZ-1]+
p[1:NX-1,1:NY-1,0:NZ-2]+
p[1:NX-1,1:NY-1,2:NZ])

#    Calculate r dot p and p dot q

rdotp = 0.0
pdotq = 0.0

rdotp = add.reduce(ravel( r[1:NX-1,1:NY-1,1:NZ-1] *
p[1:NX-1,1:NY-1,1:NZ-1]))
pdotq = add.reduce(ravel( p[1:NX-1,1:NY-1,1:NZ-1] *
q[1:NX-1,1:NY-1,1:NZ-1]))

#    Set alpha value

alpha = rdotp/pdotq

#   Update solution and residual

u[1:NX-1,1:NY-1,1:NZ-1] += alpha*p[1:NX-1,1:NY-1,1:NZ-1]
r[1:NX-1,1:NY-1,1:NZ-1] +=  - alpha*q[1:NX-1,1:NY-1,1:NZ-1]

#    calculate beta

rdotq = 0.0

rdotq =

beta = rdotq/pdotq

#   Set the new search direction

p[1:NX-1,1:NY-1,1:NZ-1] = r[1:NX-1,1:NY-1,1:NZ-1] -
beta*p[1:NX-1,1:NY-1,1:NZ-1]

res = sort(ravel(r[1:NX-1,1:NY-1,1:NZ-1]))[-1]    #find largest
residual
#    resmax = max(resmax,abs(res))

KK+=1
#
print "Iteration Number %d  Residual %1.2e" %(KK,abs(res))
if (abs(res)<=resmax): break     # if residual is small enough break
out

print "Number of Iterations ",KK

--
-----------------------------
The Numeric Python EM Project

www.pythonemproject.com

```