Python finite difference method for differential equations

Question

I must solve the Euler Bernoulli differential beam equation which is:

u’’’’(x) = f(x) ; 

(x is the coordinate of the beam axis points)

and boundary conditions:

u(0)=0,    u’(0)=0,    u’’(1)=0,    u’’’(1)=a 

I have studied the theory of numerically finite differences which expresses the series of derivations as:

U’k = (1/2*h)(Uk+1 - Uk-1) 

U’’k = (1/h2)(Uk+1 - 2 Uk + Uk-1) 

U’’’k = (1/2h3)(Uk+2 - 2 Uk+1 + 2 Uk-1 + Uk-2)

U’’’’k = (1/h4)(Uk+2 - 4 Uk+1 + 6 Uk - 4 Uk-1 + Uk-2)

( k+1 , k+2 , etc. etc are subscripts)

and I found a script which expresses it as follows:

import numpy as np
from scipy.linalg import solveh_banded
import matplotlib.pyplot as plt

def beam4(n,ffun,a):
    x = np.linspace(0,1,n+1)

    h = 1.0/n

    stencil = np.array((1,-4,6))

    B = np.outer(stencil,np.ones(n))

    B[1,-1] = -2;    B[2,0] = 7;    B[2,-1] = 1;    B[2,-2] = 5

    f = ffun(x)

    f *= h** 4;     f[-1] *= 0.5;     f[-1] -= a*h**3

    u = np.zeros(n+1) 

    u[1:] = solveh_banded(B,f[1:],lower=False)

    return x,u

But I can't understand why the coefficient matrix is built this way:

stencil = np.array((1,-4,6))

B = np.outer(stencil,np.ones(n))

B[1,-1] = -2;    B[2,0] = 7;    B[2,-1] = 1;    B[2,-2] = 5    

f = ffun(x)

f *= h**4;    f[-1] *= 0.5;    f[-1] -= a*h**3  "

Thanks in advance!!!!

Answer 1

Hope this is helpful. (due to this is my first time to post an answer)

The coefficients of this Hermitian positive-definite banded matrix are due to applied of ghost node method. It is one of most efficient and popular method for treating the boundary conditions of FDM without lossing of accuracy (here these coefficients will give a second order converge rate in general). If you have trouble to visual the matrix please check the 'K' matrix in my code below:

from numpy import linspace,zeros,array,eye,dot
from numpy.linalg import solve
from pylab import plot,xlabel,ylabel,legend,show
a       = 0.2          ;b     = 0.0
LX,dx   = 1.0,0.05     ;nx    = int(LX/dx)+1
X       = linspace(0.0,LX,nx)
Fs      = X**2.0
""""""""""""""""""""""""""""""""""""""""""""""""""""""
def calcB(l,b):
    if   b==0:    return [0.0,0.0]
    elif b==1:    return 1.0/l/l/l/l*array([-4.0,7.0,-4.0,1.0])
    elif b==nx-2: return 1.0/l/l/l/l*array([1.0,-4.0,5.0,-2.0])
    elif b==nx-1: return 1.0/l/l/l/l*array([2.0,-4.0,2.0])
    else:         return 1.0/l/l/l/l*array([1.0,-4.0,6.0,-4.0,1.0])
U       = zeros(nx)     ;V       = zeros(nx)
M       = eye(nx)       ;K       = zeros((nx,nx))
F       = zeros(nx)     ;F[nx-2] = -b/dx/dx
F[nx-1] = 2.0*b/dx/dx-2.0*a/dx
for i in range(nx):
    if i == 0:      I   = [i,i+1]
    elif i == 1:    I   = [i-1,i,i+1,i+2]
    elif i == nx-2: I   = [i-2,i-1,i,i+1]
    elif i == nx-1: I   = [i-2,i-1,i]
    else:           I   = [i-2,i-1,i,i+1,i+2]
    for k,j in enumerate(I):
        K[i,j]  += calcB(dx,i)[k]
""""""""""""""""""""""""""""""""""""""""""""""""""""""
pn      = [0]        ;eq2     = pn
eq1     = [i for i in range(nx) if i not in pn]

MM1_    = K[eq1,:]; MM11,MM12 = MM1_[:,eq1],MM1_[:,eq2]
RR      = F+Fs

U[eq2]  = [0.0]
U[eq1]  = solve(MM11,(RR[eq1]-dot(MM12,U[eq2]))) 
######################Plotting#########################
Us = lambda x: x**6.0/360.0+x**3.0/6.0*(a-1.0/3.0)+x**2.0/2.0*(1.0/4.0-a)
plot(X,U,'bo',label='FDM') ;plot(X,Us(X),'g-',label='solution')
xlabel('X'); ylabel('U'); legend(loc='best')
show()

cheers