Solving matrix equations with some Boundary Conditions

Question

How can I solve a system of linear equations with some Boundary conditions, using Numpy?

Ax=B

Where x is a column vector with, let's say x1=0. For different iterations BCs are going to be different, so different variables of vector x going to be zero. [A] and [B] are known.

Here is an example from my FEM course:
{F} Is the column vector of known values
[k] is the stiffness matrix with the known values
{U} is the displacement column vector where U1 and U3 are known to be zero, but U2 and U4 need to be found.

Here is an example:

This would result in these values:

Naturally this would reduce to the 2X2 matrix equation, but I because for different elements the BC would be different, I'm looking for some numpy matrix equation solver where I can let it know that some of the unknowns must be this certain value and nothing else.
Is there something similar to np.linalg.solve() with conditions to it?

Thank you.

linear equations with boundary conditions? is your system of equations underdefined? (i.e. is there a span of solutins?). please give the most simple example of A and B that reproduces your problem. — hiro protagonist
please provide some code (even if it is only input)... a minimal, complete and verifiable example. why not just invert the matrix as usual? — hiro protagonist
ha! Why not indeed!!! Thank you! But then I'd need to make a logic loop to see if one of the Us equal to zero and eliminate that particular row and column. Certainly doable, but is there an easier way? a predefined numpy function? — Jek Denys

hiro protagonist hiro protagonist · Accepted Answer · 2018-10-29T20:45:55

the matrix k in your example is invertible. that means there is one and only one solution; you can not choose any of the Us. this is the solution:

import numpy as np

k = np.array(((1000, 0, -1000, 0),
              (0, 3000, 0, -3000),
              (-100, 0, 3000, -2000),
              (0, -3000, -2000, 5000)))

F = np.array((0, 0, 0, 5000))

U = np.linalg.solve(k, F)
print(U)
# # or:
# k_inv = np.linalg.inv(k)
# U = k_inv.dot(F)

# [ 5.55555556  8.05555556  5.55555556  8.05555556]

the same in sage:

k = matrix(((1000, 0, -1000, 0),
              (0, 3000, 0, -3000),
              (-100, 0, 3000, -2000),
              (0, -3000, -2000, 5000)))
F = vector((0, 0, 0, 5000))
U = k.inverse() * F

# (50/9, 145/18, 50/9, 145/18)

Solving matrix equations with some Boundary Conditions

1 Answers