def dorr(n, theta=0.01, r_matrix=False, set_sparse=False):
"""
dorr dorr matrix - diagonally dominant, ill conditioned, tridiagonal.
c, d, e = dorr(n, theta) returns the vectors defining a row diagonally
dominant, tridiagonal m-matrix that is ill conditioned for small
values of the parameter theta >= 0.
If r_matrix is set to True, then c = full(tridiag(c,d,e)), i.e.,
the matrix iself is returned. The format for the matrix is set by
the boolean set_sparse. If set to true, return a matrix in sparse
matrix format. Default is False.
The columns of inv(c) vary greatly in norm. theta defaults to 0.01.
The amount of diagonal dominance is given by (ignoring rounding errors):
comp(c)*ones(n,1) = theta*(n+1)^2 * [1 0 0 ... 0 1]'.
Reference:
F.W. Dorr, An example of ill-conditioning in the numerical
solution of singular perturbation problems, Math. Comp., 25 (1971),
pp. 271-283.
"""
c = np.zeros(n)
e = np.zeros(n)
d = np.zeros(n)
# All length n for convenience. Make c, e of length n-1 later.
h = 1. / (n + 1)
m = np.floor((n + 1) / 2.)
term = theta / h ** 2
i = slice(0, m)
c[i] = -term * np.ones(m)
e[i] = c[i] - (0.5 - np.arange(1, m + 1) * h) / h
d[i] = -(c[i] + e[i])
i = slice(m, n)
e[i] = -term * np.ones(n - m)
c[i] = e[i] + (0.5 - np.arange(1, m + 1) * h) / h
d[i] = -(c[i] + e[i])
# shorten
c = c[1:n]
e = e[0:n - 1]
if r_matrix:
if set_sparse:
c = rogues.tridiag(c, d, e)
else:
c = np.diag(c, -1) + np.diag(d) + np.diag(e, 1)
# if we are to return the matrix, return _only_ the matrix
return c
# Return the diagonals of the matrix
return c, d, e
def dorr(n, theta=0.01, r_matrix=False, set_sparse=False):
"""
dorr dorr matrix - diagonally dominant, ill conditioned, tridiagonal.
c, d, e = dorr(n, theta) returns the vectors defining a row diagonally
dominant, tridiagonal m-matrix that is ill conditioned for small
values of the parameter theta >= 0.
If r_matrix is set to True, then c = full(tridiag(c,d,e)), i.e.,
the matrix iself is returned. The format for the matrix is set by
the boolean set_sparse. If set to true, return a matrix in sparse
matrix format. Default is False.
The columns of inv(c) vary greatly in norm. theta defaults to 0.01.
The amount of diagonal dominance is given by (ignoring rounding errors):
comp(c)*ones(n,1) = theta*(n+1)^2 * [1 0 0 ... 0 1]'.
Reference:
F.W. Dorr, An example of ill-conditioning in the numerical
solution of singular perturbation problems, Math. Comp., 25 (1971),
pp. 271-283.
"""
c = np.zeros(n)
e = np.zeros(n)
d = np.zeros(n)
# All length n for convenience. Make c, e of length n-1 later.
h = 1. / (n + 1)
m = np.floor((n + 1) / 2.)
term = theta / h**2
i = slice(0, m)
c[i] = -term * np.ones(m)
e[i] = c[i] - (0.5 - np.arange(1, m + 1) * h) / h
d[i] = -(c[i] + e[i])
i = slice(m, n)
e[i] = -term * np.ones(n - m)
c[i] = e[i] + (0.5 - np.arange(1, m + 1) * h) / h
d[i] = -(c[i] + e[i])
# shorten
c = c[1:n]
e = e[0:n - 1]
if r_matrix:
if set_sparse:
c = rogues.tridiag(c, d, e)
else:
c = np.diag(c, -1) + np.diag(d) + np.diag(e, 1)
# if we are to return the matrix, return _only_ the matrix
return c
# Return the diagonals of the matrix
return c, d, e
def test_tridiag_b():
"""Simple test of tridiag. Check the const diags from single arg"""
n = 10
a = rogues.tridiag(n)
b = a.todense()
x = []
x.append(-1 * np.ones(n - 1))
x.append(2 * np.ones(n))
x.append(-1 * np.ones(n - 1))
npt.assert_array_equal(x[0], np.diag(b, -1))
npt.assert_array_equal(x[1], np.diag(b, 0))
npt.assert_array_equal(x[2], np.diag(b, 1))
def test_tridiag_a():
"""Simple test of tridiag. Just recover the diagonals"""
n = 10 # size of array
x = []
x.append(np.ones(n - 1))
x.append(2 * np.ones(n))
x.append(3 * np.ones(n - 1))
a = rogues.tridiag(x[0], x[1], x[2])
b = a.todense()
npt.assert_array_equal(x[0], np.diag(b, -1))
npt.assert_array_equal(x[1], np.diag(b, 0))
npt.assert_array_equal(x[2], np.diag(b, 1))
def test_tridiag_c():
"""Simple test of tridiag. Check with 4 scalar args and known
eigenvalues"""
c = 1
d = 2
e = 3
n = 10
a = rogues.tridiag(n, c, d, e)
b = a.todense()
w, v = nl.eig(b)
w.sort()
tw = np.array([np.cos(i * np.pi / (n + 1)) for i in range(1, n + 1)])
tw = d + 2 * np.sqrt(c * e) * tw
tw.sort()
res = nl.norm(w - tw)
npt.assert_almost_equal(res, 0.0, 12)
