# Developing FEM in 1D¶

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In [1]:
import numpy as np
import numpy.linalg as la
import scipy as sp
import matplotlib.pyplot as plt
import scipy.sparse as sparse
import scipy.sparse.linalg as sla

## Boundary Value Problem¶

\begin{align*} - u'' &= f(x)\\ u(0) = u(1) &= 0 \end{align*}
In [2]:
if 1:
def f(x):
return 2+0*x

def uexact(x):
return x*(1-x)

elif 1:
wavenr = 5

def uexact(x):
return np.sin(wavenr * np.pi * x)
def f(x):
return (wavenr*np.pi)**2 * uexact(x)

else:
def f(x):
return 100*np.exp(-(x-0.5)**2 / 0.001)

uexact = None

## Grid Setup¶

V is a list of vertices. E is a list of elements (segments).

In [3]:
# number of points, crank me up
nx = 6

V = np.linspace(0,1,nx)
E = np.zeros((nx-1,2), dtype=int)
E[:,0] = np.arange(0,nx-1)
E[:,1] = np.arange(1,nx)
h = V[1] - V[0] # mesh spacing
In [4]:
if len(E) < 10:
print(E)
[[0 1]
[1 2]
[2 3]
[3 4]
[4 5]]

## COOrdinate Matrix Semantics¶

Note: What happened to the duplicated entry?

In [5]:
rows = [0,2,3,4,4]
cols = [4,1,2,4,4]
vals = [7,7,7,5,50]
sample_mat = sparse.coo_matrix((vals, (rows, cols))).toarray()
sample_mat
Out[5]:
array([[ 0,  0,  0,  0,  7],
[ 0,  0,  0,  0,  0],
[ 0,  7,  0,  0,  0],
[ 0,  0,  7,  0,  0],
[ 0,  0,  0,  0, 55]])

## Reference Matrix¶

Basis functions (on [0,1]): \begin{align*} \phi_1(x) &= 1-x,\\ \phi_2(x) &= x, \end{align*}

For both degrees of freedom in the element, figure: $$\hat A_{i,j} = \int \phi_i'(x) \phi_j'(x) dx$$

In [6]:
Aref = np.array([
[1, -1],
[-1, 1]
])

## Assembly Helper¶

In [7]:
class MatrixBuilder:
def __init__(self):
self.rows = []
self.cols = []
self.vals = []

for i, ri in enumerate(rows):
for j, cj in enumerate(cols):
self.rows.append(ri)
self.cols.append(cj)
self.vals.append(submat[i, j])

def coo_matrix(self):
return sparse.coo_matrix((self.vals, (self.rows, self.cols)))

## Assembly of the Linear System¶

Assemble $A$:

In [8]:
a_builder = MatrixBuilder()

for va, vb in E:
[va, vb], [va, vb],
h * 1/h * 1/h * Aref)

A = a_builder.coo_matrix()

For both degrees of freedom involved in each element, assemble the RHS vector: $$b_i=\int_E f(x) \phi_i(x) dx$$

In [9]:
b = np.zeros(nx)

for va, vb in E:
b[va] += f(V[va]) * h/2
b[vb] += f(V[vb]) * h/2

Examine the matrix.

In [10]:
print(A.toarray()*h)
[[ 1. -1.  0.  0.  0.  0.]
[-1.  2. -1.  0.  0.  0.]
[ 0. -1.  2. -1.  0.  0.]
[ 0.  0. -1.  2. -1.  0.]
[ 0.  0.  0. -1.  2. -1.]
[ 0.  0.  0.  0. -1.  1.]]

Notice anything?

In [11]:
if len(E) < 10:
u, s, vt = la.svd(A.toarray())
print(s)
print(vt[-1])
[1.86602540e+01 1.50000000e+01 1.00000000e+01 5.00000000e+00
1.33974596e+00 1.06418587e-15]
[-0.40824829 -0.40824829 -0.40824829 -0.40824829 -0.40824829 -0.40824829]

## Boundary Conditions¶

In [12]:
for i in range(A.nnz):
if A.row[i] in [0, nx-1]:
A.data[i] = 1 if A.row[i] == A.col[i] else 0

b[0] = 0
b[nx-1] = 0

A = A.tocsr()

Examine the matrix after applying BCs:

In [13]:
print(A.toarray()*h)
[[ 0.2  0.   0.   0.   0.   0. ]
[-1.   2.  -1.   0.   0.   0. ]
[ 0.  -1.   2.  -1.   0.   0. ]
[ 0.   0.  -1.   2.  -1.   0. ]
[ 0.   0.   0.  -1.   2.  -1. ]
[ 0.   0.   0.   0.   0.   0.2]]

## Computing the Solution¶

Plot the RHS $f$.

In [14]:
if len(E) < 10:
plotmode = "o-"
else:
plotmode = "-"

plt.plot(V, f(V), plotmode)
Out[14]:
[<matplotlib.lines.Line2D at 0x7efbdda97050>]

Solve and plot the solution.

In [15]:
u = sla.spsolve(A, b)
In [16]:
plt.plot(V, u, plotmode)

if uexact is not None:
plt.plot(V, uexact(V), plotmode)
In [17]:
if uexact is not None:
u_ex_h = uexact(V)
print(la.norm(u - u_ex_h)/la.norm(u_ex_h))
9.622514525208466e-17
In [ ]: