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Weird results obtained while solving a set of coupled differential equations (using a sparse array) in python

I have tried to no avail for a week while trying to solve a system of coupled differential equations and reproduce the results shown in the attached image. I seem to be getting weird results as shown also. I don't seem to know what I might be doing wrong.The set of coupled differential equations were solved using Newman's BAND. Here's a link to the python implementation: python solution using BAND . And another link to the original image of the problem in case the attached is not clear enough: here you find a clearer image of the problem. Now what I am trying to do is to solve the same problem by creating a sparse array directly from the discretized equations using a combination of sympy and numpy and then solving using scipy's spsolve. Here is my code below. I need some help to figure out what I am doing wrong.
I have represented the variables as c1 = cA, c2 = cB, c3 = cC, c4 = cD in my code. Equation 2 has been linearized and phi10 and phi20 are the trial values of the variables cC and cD.
# import modules
import numpy as np
import sympy
from sympy.core.function import _mexpand
import scipy as sp
import scipy.sparse as ss
import scipy.sparse.linalg as ssl
import matplotlib.pyplot as plt
# define functions
def flatten(t):
"""
function to flatten lists
"""
return [item for sublist in t for item in sublist]
def get_coeffs(coeff_dict, func_vars):
"""
function to extract coefficients from variables
and form the sparse symbolic array
"""
c = coeff_dict
for i in list(c.keys()):
b, _ = i.as_base_exp()
if b == i:
continue
if b in c:
c[i] = 0
if any(k.has(b) for k in c):
c[i] = 0
return [coeff_dict[val] for val in func_vars]
# Constants for the problem
I = 0.1 # A/cm2
L = 1.0 # distance (x) in cm
m = 100 # grid spacing
h = L / (m-1)
a = 23300 # 1/cm
io = 2e-7 # A/cm2
n = 1
F = 96500 # C/mol
R = 8.314 # J/mol-K
T = 298 # K
sigma = 20 # S/cm
kappa = 0.06 # S/cm
alpha = 0.5
beta = -(1-alpha)*n*F/R/T
phi10 , phi20 = 5, 0.5 # these are just guesses
P = a*io*np.exp(beta*(phi10-phi20))
j = sympy.symbols('j',integer = True)
cA = sympy.IndexedBase('cA')
cB = sympy.IndexedBase('cB')
cC = sympy.IndexedBase('cC')
cD = sympy.IndexedBase('cD')
# write the boundary conditions at x = 0
bc=[cA[1], cB[1],
(4/3) * cC[2] - (1/3)*cC[3], # use a three point approximation for cC_prime
cD[1]
]
# form a list of expressions from the boundary conditions and equations
expr=flatten([bc,flatten([[
-cA[j-1] - cB[j-1] + cA[j+1] + cB[j+1],
cB[j-1] - 2*h*P*beta*cC[j] + 2*h*P*beta*cD[j] - cB[j+1],
-sigma*cC[j-1] + 2*h*cA[j] + sigma * cC[j+1],
-kappa * cD[j-1] + 2*h * cB[j] + kappa * cD[j+1]] for j in range(2, m)])])
vars = [cA[j], cB[j], cC[j], cD[j]]
# flatten the list of variables
unknowns = flatten([[cA[j], cB[j], cC[j], cD[j]] for j in range(1,m)])
var_len = len(unknowns)
# # # substitute in the boundary conditions at x = L while getting the coefficients
A = sympy.SparseMatrix([get_coeffs(_mexpand(i.subs({cA[m]:I}))\
.as_coefficients_dict(), unknowns) for i in expr])
# convert to a numpy array
mat_temp = np.array(A).astype(np.float64)
# you can view the sparse array with this
fig = plt.figure(figsize=(6,6))
ax = fig.add_axes([0,0, 1,1])
cmap = plt.cm.binary
plt.spy(mat_temp, cmap = cmap, alpha = 0.8)
def solve_sparse(b0, error):
# create the b column vector
b = np.copy(b0)
b[0:4] = np.array([0.0, I, 0.0, 0.0])
b[var_len-4] = I
b[var_len-3] = 0
b[var_len-2] = 0
b[var_len-1] = 0
print(b.shape)
old = np.copy(b0)
mat = np.copy(mat_temp)
b_2 = np.copy(b)
resid = 10
lss = 0
while lss < 100:
mat_2 = np.copy(mat)
for j in range(3, var_len - 3, 4):
# update the forcing term of equation 2
b_2[j+2] = 2*h*(1-beta*old[j+3]+beta*old[j+4])*a*io*np.exp(beta*(old[j+3]-old[j+4]))
# update the sparse array at every iteration for variables cC and cD in equation2
mat_2[j+2, j+3] += 2*h*beta*a*io*np.exp(beta*(old[j+3]-old[j+4]))
mat_2[j+2, j+4] += 2*h*beta*a*io*np.exp(beta*(old[j+3]-old[j+4]))
# form the column sparse matrix
A_s = ss.csc_matrix(mat_2)
new = ssl.spsolve(A_s, b_2).flatten()
resid = np.sum((new - old)**2)/var_len
lss += 1
old = np.copy(new)
return new
val0 = np.array([[0.0, 0.0, 0.0, 0.0] for _ in range(m-1)]).flatten() # form an array of initial values
error = 1e-7
## Run the code
conc = solve_sparse(val0, error).reshape(m-1, len(vars))
conc.shape # gives (99, 4)
# Plot result for cA:
plt.plot(conc[:,0], marker = 'o', linestyle = '')

What happens seems pretty clear now, after having seen that the plotted solution indeed oscillates between the upper and lower values. You are using the central Euler method as discretization, for u'=F(u) this reads as
u[j+1]-u[j-1] = 2*h*F(u[j])
This method is only weakly stable and allows the sub-sequences of odd and even indices to evolve rather independently. As equation this would mean that the solution might approximate the system ue'=F(uo), uo'=F(ue) with independent functions ue, uo that follow the path of the even or odd sub-sequence.
These even and odd parts are only tied together by the treatment of the boundary points, two or three points deep. So to avoid or reduce the oscillation requires a very careful handling of boundary conditions and also the differential equations for the boundary points.
But one can avoid all this unpleasantness by using the trapezoidal method
u[j+1]-u[j] = 0.5*h*(F(u[j+1])+F(u[j]))
This also reduces the band-width of the system matrix.
To properly implement the implied Newton method correctly (linearizing via Taylor and solving the linearized equation is what the Newton-Kantorovich method does) you need to replace F(u[j]) with F(u_old[j])+F'(u_old[j])*(u[j]-u_old[j]). This then gives a linear system of equations in u for the iteration step.
For the trapezoidal method this gives
(I-0.5*h*F'(u_old[j+1]))*u[j+1] - (I+0.5*h*F'(u_old[j]))*u[j]
= 0.5*h*(F(u_old[j+1])-F'(u_old[j+1])*u_old[j+1] + F(u_old[j])-F'(u_old[j])*u_old[j])
In general, the derivatives values and thus the system matrix need not be updated every step, only the function value (else the iteration does not move forward).

Attempting to solve a system of linear equations, where one of my outputs is unknown but I do know it needs to be maximized

I have a set of linear equations that I am attempting to solve. I have five variables that are randomly assigned and sum to 1. I applied these variables row-wise to a matrix (i.e., Ax = B) However, the caveat is that one of my B variables needs to be maximized, subject to the constraint that the rest of my B variables are 0. My data is below:
import pandas as pd
A = [[0.031982, 0.02606, 0.055933, 0.004529, 0.064116],
[-0.000167, 0.181031, 0.145465, 0.120430, 0.114093],
[0.627812, 0.254377, 0.138211, 0.41156, -0.000801],
[-0.228139, 0.377169, 0.085415, 0.008888, -0.020791],
[-0.067697, -0.114113, 0.089583, 0.100222, -0.005291]]
B = [[maximized],
[0],
[0],
[0],
[0]]
x = [x1, x2, x3, x4, x5]
Note: 'maximized' is the value that I am attempting to maximize
What I've done so far:
import numpy as np
ABC=[]
A = DataFrame(A)
N=1000 #my attempt at maximizing by repeating the sequence N times and taking the largest value that results
for i in range(N):
x = np.random.rand(5) #creating random variables
x/= np.sum(x) #so that they sum to one
x=x*A.T[0] #I apply the variables to my data. I want to maximize the sum of the first column, so I transpose and take a slice that I sum below
x=x.sum()
ABC.append(x)
ABC = DataFrame(ABC)
A2=ABC.sort_values(by=0,ascending=False).head(1) #I sort by largest first and take the highest value and store in a new dataframe
maximized=np.array(A2) #I convert the dataframe back into an array
B = [[maximized],[0],[0],[0],[0]]
X = np.linalg.inv(A).dot(B)
X
Obviously this has a lot of error and isn't really achieving what I want. What I want to do is to run a maximization function that gives me the largest value and input that value into my matrix. I don't really know where to go from here, or what sort of maximization function applies in this case. If anyone has any ideas, that would be super appreciated!

If I understand you correctly, you have a given matrix A. Then you want to find a positive vector x with sum 1 such that if Ax = b then b is the all-zeroes vector except that the first entry is maximized, rather than zero.
Let A0 be the first row of matrix A and Ar be the rest. Then we can rephrase your problem as:
Find vector x such that its entries are non-negative, sum to 1, where Ar x = b, b = 0 and maximizing A0 x.
This is simply a standard linear programming instance (handling the 'sum-to-1' requirement by adding a single all-ones row to Ar and a 1 entry to b).

This isn't really a maximization problem I think. There's only one answer with the specified result. If our maximized number is c, then b = [[c,0,0,0,0]].T = c*[[1,0,0,0,0]].T Lets' set b_ as [[1,0,0,0,0]].T so b = c*b_
If we solve A # x_ = b_ by x_ = Ainv # b_, we get a single x_ matrix that still needs to be normalized to form x. If len_x_ = np.linalg.norm(x_), then:
A # (c * x_) = (c * b_) # multiplying `c` though
A # (c * x_) = b # substituting `b = c * b_`
c * x_ = x = x_ / len_x_ # therefore . . .
c = 1 / len_x_
c can only take one value if A is postive-definite, and thus x can only take one value:
b_ = np.array([[1,0,0,0,0]])
x_ = np.linalg.inv(A).dot(b_)
c = 1 / np.linalg.norm(x_)
b = b_ * c
x = x_ * c
Now, it'll get a lot more interesting if A isn't definite (i.e. it doesn't have an inverse). But I'm not sure that's within scope, since you're currently solving by inverting A.

generating random matrices in python

In the following code I have implemented Gaussian elimination with partial pivoting for a general square linear system Ax=b. I have tested my code and it produced the right output. However now I am trying to do the following but I am not quite sure how to code it, looking for some help with this!
I want to test my implementation by solving Ax=b where A is a random 100x100 matrix and b is a random 100x1 vector.
In my code I have put in the matrices
A = np.array([[3.,2.,-4.],[2.,3.,3.],[5.,-3.,1.]])
b = np.array([[3.],[15.],[14.]])
and gotten the following correct output:
[3. 1. 2.]
[3. 1. 2.]
but now how do I change it to generate the random matrices?
here is my code below:
import numpy as np
def GEPP(A, b, doPricing = True):
'''
Gaussian elimination with partial pivoting.
input: A is an n x n numpy matrix
b is an n x 1 numpy array
output: x is the solution of Ax=b
with the entries permuted in
accordance with the pivoting
done by the algorithm
post-condition: A and b have been modified.
'''
n = len(A)
if b.size != n:
raise ValueError("Invalid argument: incompatible sizes between"+
"A & b.", b.size, n)
# k represents the current pivot row. Since GE traverses the matrix in the
# upper right triangle, we also use k for indicating the k-th diagonal
# column index.
# Elimination
for k in range(n-1):
if doPricing:
# Pivot
maxindex = abs(A[k:,k]).argmax() + k
if A[maxindex, k] == 0:
raise ValueError("Matrix is singular.")
# Swap
if maxindex != k:
A[[k,maxindex]] = A[[maxindex, k]]
b[[k,maxindex]] = b[[maxindex, k]]
else:
if A[k, k] == 0:
raise ValueError("Pivot element is zero. Try setting doPricing to True.")
#Eliminate
for row in range(k+1, n):
multiplier = A[row,k]/A[k,k]
A[row, k:] = A[row, k:] - multiplier*A[k, k:]
b[row] = b[row] - multiplier*b[k]
# Back Substitution
x = np.zeros(n)
for k in range(n-1, -1, -1):
x[k] = (b[k] - np.dot(A[k,k+1:],x[k+1:]))/A[k,k]
return x
if __name__ == "__main__":
A = np.array([[3.,2.,-4.],[2.,3.,3.],[5.,-3.,1.]])
b = np.array([[3.],[15.],[14.]])
print (GEPP(np.copy(A), np.copy(b), doPricing = False))
print (GEPP(A,b))

You're already using numpy. Have you considered np.random.rand?
np.random.rand(m, n) will get you a random matrix with values in [0, 1). You can further process it by multiplying random values or rounding.
EDIT: Something like this
if __name__ == "__main__":
A = np.round(np.random.rand(100, 100)*10)
b = np.round(np.random.rand(100)*10)
print (GEPP(np.copy(A), np.copy(b), doPricing = False))
print (GEPP(A,b))

So I would use np.random.randint for this.
numpy.random.randint(low, high=None, size=None, dtype='l')
which outputs a size-shaped array of random integers from the appropriate distribution, or a single such random int if size not provided.
low is the lower bound of the ints you want in your range
high is one greater than the upper bound in your desired range
size is the dimensions of your output array
dtype is the dtype of the result
so if I was you I would write
A = np.random.randint(0, 11, (100, 100))
b = np.random.randint(0, 11, 100)

Basically you could create the desired matrices with ones and then iterate over them, setting each value to random.randint(0,100) for example.
Empty matrix with ones is:
one_array = np.ones((100, 100))
EDIT:
like:
for x in one_array.shape[0]:
for y in one_array.shape[1]:
one_array[x][y] = random.randint(0, 100)

A = np.random.normal(size=(100,100))
b = np.random.normal(size=(100,1))
x = np.linalg.solve(A,b)
assert max(abs(A#x - b)) < 1e-12
Clearly, you can use different distributions than normal, like uniform.

You can use numpy's native rand function:
np.random.rand()
In your code just define A and b as:
A = np.random.rand(100, 100)
b = np.random.rand(100)
This will generate 100x100 matrix and 100x1 vector (both numpy arrays) filled with random values between 0 and 1.
See the docs for this function to learn more.

How to create an array that can be accessed according to its indices in Numpy?

I am trying to solve the following problem via a Finite Difference Approximation in Python using NumPy:
$u_t = k \, u_{xx}$, on $0 < x < L$ and $t > 0$;
$u(0,t) = u(L,t) = 0$;
$u(x,0) = f(x)$.
I take $u(x,0) = f(x) = x^2$ for my problem.
Programming is not my forte so I need help with the implementation of my code. Here is my code (I'm sorry it is a bit messy, but not too bad I hope):
## This program is to implement a Finite Difference method approximation
## to solve the Heat Equation, u_t = k * u_xx,
## in 1D w/out sources & on a finite interval 0 < x < L. The PDE
## is subject to B.C: u(0,t) = u(L,t) = 0,
## and the I.C: u(x,0) = f(x).
import numpy as np
import matplotlib.pyplot as plt
# definition of initial condition function
def f(x):
return x^2
# parameters
L = 1
T = 10
N = 10
M = 100
s = 0.25
# uniform mesh
x_init = 0
x_end = L
dx = float(x_end - x_init) / N
#x = np.zeros(N+1)
x = np.arange(x_init, x_end, dx)
x[0] = x_init
# time discretization
t_init = 0
t_end = T
dt = float(t_end - t_init) / M
#t = np.zeros(M+1)
t = np.arange(t_init, t_end, dt)
t[0] = t_init
# Boundary Conditions
for m in xrange(0, M):
t[m] = m * dt
# Initial Conditions
for j in xrange(0, N):
x[j] = j * dx
# definition of solution to u_t = k * u_xx
u = np.zeros((N+1, M+1)) # NxM array to store values of the solution
# finite difference scheme
for j in xrange(0, N-1):
u[j][0] = x**2 #initial condition
for m in xrange(0, M):
for j in xrange(1, N-1):
if j == 1:
u[j-1][m] = 0 # Boundary condition
else:
u[j][m+1] = u[j][m] + s * ( u[j+1][m] - #FDM scheme
2 * u[j][m] + u[j-1][m] )
else:
if j == N-1:
u[j+1][m] = 0 # Boundary Condition
print u, t, x
#plt.plot(t, u)
#plt.show()
So the first issue I am having is I am trying to create an array/matrix to store values for the solution. I wanted it to be an NxM matrix, but in my code I made the matrix (N+1)x(M+1) because I kept getting an error that the index was going out of bounds. Anyways how can I make such a matrix using numpy.array so as not to needlessly take up memory by creating a (N+1)x(M+1) matrix filled with zeros?
Second, how can I "access" such an array? The real solution u(x,t) is approximated by u(x[j], t[m]) were j is the jth spatial value, and m is the mth time value. The finite difference scheme is given by:
u(x[j],t[m+1]) = u(x[j],t[m]) + s * ( u(x[j+1],t[m]) - 2 * u(x[j],t[m]) + u(x[j-1],t[m]) )
(See here for the formulation)
I want to be able to implement the Initial Condition u(x[j],t[0]) = x**2 for all values of j = 0,...,N-1. I also need to implement Boundary Conditions u(x[0],t[m]) = 0 = u(x[N],t[m]) for all values of t = 0,...,M. Is the nested loop I created the best way to do this? Originally I tried implementing the I.C. and B.C. under two different for loops which I used to calculate values of the matrices x and t (in my code I still have comments placed where I tried to do this)
I think I am just not using the right notation but I cannot find anywhere in the documentation for NumPy how to "call" such an array so at to iterate through each value in the proposed scheme. Can anyone shed some light on what I am doing wrong?
Any help is very greatly appreciated. This is not homework but rather to understand how to program FDM for Heat Equation because later I will use similar methods to solve the Black-Scholes PDE.
EDIT: So when I run my code on line 60 (the last "else" that I use) I get an error that says invalid syntax, and on line 51 (u[j][0] = x**2 #initial condition) I get an error that reads "setting an array element with a sequence." What does that mean?

Matrix vector multiplication where the vector has been interpolated - Python

I have used the finite element method to approximate the laplace equation and thus have turned it into a matrix system AU = F where A is the stiffness vector and solved for U (not massively important for my question).
I have now got my approximation U, which when i find AU i should get the vector F (or at least similar) where F is:
AU gives the following plot for x = 0 to x = 1 (say, for 20 nodes):
I then need to interpolate U to a longer vector and find AU (for a bigger A too, but not interpolating that). I interpolate U by the following:
U_inter = interp1d(x,U)
U_rich = U_inter(longer_x)
which seems to work okay until i multiply it with the longer A matrix:
It seems each spike is at a node of x (i.e. the nodes of the original U). Does anybody know what could be causing this? The following is my code to find A, U and F.
import numpy as np
import math
import scipy
from scipy.sparse import diags
import scipy.sparse.linalg
from scipy.interpolate import interp1d
import matplotlib
import matplotlib.pyplot as plt
def Poisson_Stiffness(x0):
"""Finds the Poisson equation stiffness matrix with any non uniform mesh x0"""
x0 = np.array(x0)
N = len(x0) - 1 # The amount of elements; x0, x1, ..., xN
h = x0[1:] - x0[:-1]
a = np.zeros(N+1)
a[0] = 1 #BOUNDARY CONDITIONS
a[1:-1] = 1/h[1:] + 1/h[:-1]
a[-1] = 1/h[-1]
a[N] = 1 #BOUNDARY CONDITIONS
b = -1/h
b[0] = 0 #BOUNDARY CONDITIONS
c = -1/h
c[N-1] = 0 #BOUNDARY CONDITIONS: DIRICHLET
data = [a.tolist(), b.tolist(), c.tolist()]
Positions = [0, 1, -1]
Stiffness_Matrix = diags(data, Positions, (N+1,N+1))
return Stiffness_Matrix
def NodalQuadrature(x0):
"""Finds the Nodal Quadrature Approximation of sin(pi x)"""
x0 = np.array(x0)
h = x0[1:] - x0[:-1]
N = len(x0) - 1
approx = np.zeros(len(x0))
approx[0] = 0 #BOUNDARY CONDITIONS
for i in range(1,N):
approx[i] = math.sin(math.pi*x0[i])
approx[i] = (approx[i]*h[i-1] + approx[i]*h[i])/2
approx[N] = 0 #BOUNDARY CONDITIONS
return approx
def Solver(x0):
Stiff_Matrix = Poisson_Stiffness(x0)
NodalApproximation = NodalQuadrature(x0)
NodalApproximation[0] = 0
U = scipy.sparse.linalg.spsolve(Stiff_Matrix, NodalApproximation)
return U
x = np.linspace(0,1,10)
rich_x = np.linspace(0,1,50)
U = Solver(x)
A_rich = Poisson_Stiffness(rich_x)
U_inter = interp1d(x,U)
U_rich = U_inter(rich_x)
AUrich = A_rich.dot(U_rich)
plt.plot(rich_x,AUrich)
plt.show()

comment 1:
I added a Stiffness_Matrix = Stiffness_Matrix.tocsr() statement to avoid an efficiency warning. FE calculations are complex enough that I'll have to print out some intermediate values before I can identify what is going on.
comment 2:
plt.plot(rich_x,A_rich.dot(Solver(rich_x))) plots nice. The noise you get is the result of the difference between the inperpolated U_rich and the true solution: U_rich-Solver(rich_x).
comment 3:
I don't think there's a problem with your code. The problem is with idea that you can test an interpolation this way. I'm rusty on FE theory, but I think you need to use the shape functions to interpolate, not a simple linear one.
comment 4:
Intuitively, with A_rich.dot(U_rich) you are asking, what kind of forcing F would produce U_rich. Compared to Solver(rich_x), U_rich has flat spots, regions where it's value is less than the true solution. What F would produce that? One that is spiky, with NodalQuadrature(x) at the x points, but near zero values in between. That's what your plot is showing.
A higher order interpolation will eliminate the flat spots, and produce a smoother back calculated F. But you really need to revisit the FE theory.
You might find it instructive to look at
plt.plot(x,NodalQuadrature(x))
plt.plot(rich_x, NodalQuadrature(rich_x))
The second plot is much smoother, but only about 1/5 as high.
Better yet look at:
plt.plot(rich_x,AUrich,'-*') # the spikes
plt.plot(x,NodalQuadrature(x),'o') # original forcing
plt.plot(rich_x, NodalQuadrature(rich_x),'+') # new forcing
In the model the forcing isn't continuous, it is a value at each node. With more nodes (rich_x) the magnitude at each node is less.