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/ integrate / _bvp.py
"""Boundary value problem solver."""
from __future__ import division, print_function, absolute_import
from warnings import warn
import numpy as np
from numpy.linalg import norm, pinv
from scipy.sparse import coo_matrix, csc_matrix
from scipy.sparse.linalg import splu
from scipy.optimize import OptimizeResult
EPS = np.finfo(float).eps
def estimate_fun_jac(fun, x, y, p, f0=None):
"""Estimate derivatives of an ODE system rhs with forward differences.
Returns
-------
df_dy : ndarray, shape (n, n, m)
Derivatives with respect to y. An element (i, j, q) corresponds to
d f_i(x_q, y_q) / d (y_q)_j.
df_dp : ndarray with shape (n, k, m) or None
Derivatives with respect to p. An element (i, j, q) corresponds to
d f_i(x_q, y_q, p) / d p_j. If `p` is empty, None is returned.
"""
n, m = y.shape
if f0 is None:
f0 = fun(x, y, p)
dtype = y.dtype
df_dy = np.empty((n, n, m), dtype=dtype)
h = EPS**0.5 * (1 + np.abs(y))
for i in range(n):
y_new = y.copy()
y_new[i] += h[i]
hi = y_new[i] - y[i]
f_new = fun(x, y_new, p)
df_dy[:, i, :] = (f_new - f0) / hi
k = p.shape[0]
if k == 0:
df_dp = None
else:
df_dp = np.empty((n, k, m), dtype=dtype)
h = EPS**0.5 * (1 + np.abs(p))
for i in range(k):
p_new = p.copy()
p_new[i] += h[i]
hi = p_new[i] - p[i]
f_new = fun(x, y, p_new)
df_dp[:, i, :] = (f_new - f0) / hi
return df_dy, df_dp
def estimate_bc_jac(bc, ya, yb, p, bc0=None):
"""Estimate derivatives of boundary conditions with forward differences.
Returns
-------
dbc_dya : ndarray, shape (n + k, n)
Derivatives with respect to ya. An element (i, j) corresponds to
d bc_i / d ya_j.
dbc_dyb : ndarray, shape (n + k, n)
Derivatives with respect to yb. An element (i, j) corresponds to
d bc_i / d ya_j.
dbc_dp : ndarray with shape (n + k, k) or None
Derivatives with respect to p. An element (i, j) corresponds to
d bc_i / d p_j. If `p` is empty, None is returned.
"""
n = ya.shape[0]
k = p.shape[0]
if bc0 is None:
bc0 = bc(ya, yb, p)
dtype = ya.dtype
dbc_dya = np.empty((n, n + k), dtype=dtype)
h = EPS**0.5 * (1 + np.abs(ya))
for i in range(n):
ya_new = ya.copy()
ya_new[i] += h[i]
hi = ya_new[i] - ya[i]
bc_new = bc(ya_new, yb, p)
dbc_dya[i] = (bc_new - bc0) / hi
dbc_dya = dbc_dya.T
h = EPS**0.5 * (1 + np.abs(yb))
dbc_dyb = np.empty((n, n + k), dtype=dtype)
for i in range(n):
yb_new = yb.copy()
yb_new[i] += h[i]
hi = yb_new[i] - yb[i]
bc_new = bc(ya, yb_new, p)
dbc_dyb[i] = (bc_new - bc0) / hi
dbc_dyb = dbc_dyb.T
if k == 0:
dbc_dp = None
else:
h = EPS**0.5 * (1 + np.abs(p))
dbc_dp = np.empty((k, n + k), dtype=dtype)
for i in range(k):
p_new = p.copy()
p_new[i] += h[i]
hi = p_new[i] - p[i]
bc_new = bc(ya, yb, p_new)
dbc_dp[i] = (bc_new - bc0) / hi
dbc_dp = dbc_dp.T
return dbc_dya, dbc_dyb, dbc_dp
def compute_jac_indices(n, m, k):
"""Compute indices for the collocation system Jacobian construction.
See `construct_global_jac` for the explanation.
"""
i_col = np.repeat(np.arange((m - 1) * n), n)
j_col = (np.tile(np.arange(n), n * (m - 1)) +
np.repeat(np.arange(m - 1) * n, n**2))
i_bc = np.repeat(np.arange((m - 1) * n, m * n + k), n)
j_bc = np.tile(np.arange(n), n + k)
i_p_col = np.repeat(np.arange((m - 1) * n), k)
j_p_col = np.tile(np.arange(m * n, m * n + k), (m - 1) * n)
i_p_bc = np.repeat(np.arange((m - 1) * n, m * n + k), k)
j_p_bc = np.tile(np.arange(m * n, m * n + k), n + k)
i = np.hstack((i_col, i_col, i_bc, i_bc, i_p_col, i_p_bc))
j = np.hstack((j_col, j_col + n,
j_bc, j_bc + (m - 1) * n,
j_p_col, j_p_bc))
return i, j
def stacked_matmul(a, b):
"""Stacked matrix multiply: out[i,:,:] = np.dot(a[i,:,:], b[i,:,:]).
In our case a[i, :, :] and b[i, :, :] are always square.
"""
# Empirical optimization. Use outer Python loop and BLAS for large
# matrices, otherwise use a single einsum call.
if a.shape[1] > 50:
out = np.empty_like(a)
for i in range(a.shape[0]):
out[i] = np.dot(a[i], b[i])
return out
else:
return np.einsum('...ij,...jk->...ik', a, b)
def construct_global_jac(n, m, k, i_jac, j_jac, h, df_dy, df_dy_middle, df_dp,
df_dp_middle, dbc_dya, dbc_dyb, dbc_dp):
"""Construct the Jacobian of the collocation system.
There are n * m + k functions: m - 1 collocations residuals, each
containing n components, followed by n + k boundary condition residuals.
There are n * m + k variables: m vectors of y, each containing n
components, followed by k values of vector p.
For example, let m = 4, n = 2 and k = 1, then the Jacobian will have
the following sparsity structure:
1 1 2 2 0 0 0 0 5
1 1 2 2 0 0 0 0 5
0 0 1 1 2 2 0 0 5
0 0 1 1 2 2 0 0 5
0 0 0 0 1 1 2 2 5
0 0 0 0 1 1 2 2 5
3 3 0 0 0 0 4 4 6
3 3 0 0 0 0 4 4 6
3 3 0 0 0 0 4 4 6
Zeros denote identically zero values, other values denote different kinds
of blocks in the matrix (see below). The blank row indicates the separation
of collocation residuals from boundary conditions. And the blank column
indicates the separation of y values from p values.
Refer to [1]_ (p. 306) for the formula of n x n blocks for derivatives
of collocation residuals with respect to y.
Parameters
----------
n : int
Number of equations in the ODE system.
m : int
Number of nodes in the mesh.
k : int
Number of the unknown parameters.
i_jac, j_jac : ndarray
Row and column indices returned by `compute_jac_indices`. They
represent different blocks in the Jacobian matrix in the following
order (see the scheme above):
* 1: m - 1 diagonal n x n blocks for the collocation residuals.
* 2: m - 1 off-diagonal n x n blocks for the collocation residuals.
* 3 : (n + k) x n block for the dependency of the boundary
conditions on ya.
* 4: (n + k) x n block for the dependency of the boundary
conditions on yb.
* 5: (m - 1) * n x k block for the dependency of the collocation
residuals on p.
* 6: (n + k) x k block for the dependency of the boundary
conditions on p.
df_dy : ndarray, shape (n, n, m)
Jacobian of f with respect to y computed at the mesh nodes.
df_dy_middle : ndarray, shape (n, n, m - 1)
Jacobian of f with respect to y computed at the middle between the
mesh nodes.
df_dp : ndarray with shape (n, k, m) or None
Jacobian of f with respect to p computed at the mesh nodes.
df_dp_middle: ndarray with shape (n, k, m - 1) or None
Jacobian of f with respect to p computed at the middle between the
mesh nodes.
dbc_dya, dbc_dyb : ndarray, shape (n, n)
Jacobian of bc with respect to ya and yb.
dbc_dp: ndarray with shape (n, k) or None
Jacobian of bc with respect to p.
Returns
-------
J : csc_matrix, shape (n * m + k, n * m + k)
Jacobian of the collocation system in a sparse form.
References
----------
.. [1] J. Kierzenka, L. F. Shampine, "A BVP Solver Based on Residual
Control and the Maltab PSE", ACM Trans. Math. Softw., Vol. 27,
Number 3, pp. 299-316, 2001.
"""
df_dy = np.transpose(df_dy, (2, 0, 1))
df_dy_middle = np.transpose(df_dy_middle, (2, 0, 1))
h = h[:, np.newaxis, np.newaxis]
dtype = df_dy.dtype
# Computing diagonal n x n blocks.
dPhi_dy_0 = np.empty((m - 1, n, n), dtype=dtype)
dPhi_dy_0[:] = -np.identity(n)
dPhi_dy_0 -= h / 6 * (df_dy[:-1] + 2 * df_dy_middle)
T = stacked_matmul(df_dy_middle, df_dy[:-1])
dPhi_dy_0 -= h**2 / 12 * T
# Computing off-diagonal n x n blocks.
dPhi_dy_1 = np.empty((m - 1, n, n), dtype=dtype)
dPhi_dy_1[:] = np.identity(n)
dPhi_dy_1 -= h / 6 * (df_dy[1:] + 2 * df_dy_middle)
T = stacked_matmul(df_dy_middle, df_dy[1:])
dPhi_dy_1 += h**2 / 12 * T
values = np.hstack((dPhi_dy_0.ravel(), dPhi_dy_1.ravel(), dbc_dya.ravel(),
dbc_dyb.ravel()))
if k > 0:
df_dp = np.transpose(df_dp, (2, 0, 1))
df_dp_middle = np.transpose(df_dp_middle, (2, 0, 1))
T = stacked_matmul(df_dy_middle, df_dp[:-1] - df_dp[1:])
df_dp_middle += 0.125 * h * T
dPhi_dp = -h/6 * (df_dp[:-1] + df_dp[1:] + 4 * df_dp_middle)
values = np.hstack((values, dPhi_dp.ravel(), dbc_dp.ravel()))
J = coo_matrix((values, (i_jac, j_jac)))
return csc_matrix(J)
def collocation_fun(fun, y, p, x, h):
"""Evaluate collocation residuals.
This function lies in the core of the method. The solution is sought
as a cubic C1 continuous spline with derivatives matching the ODE rhs
at given nodes `x`. Collocation conditions are formed from the equality
of the spline derivatives and rhs of the ODE system in the middle points
between nodes.
Such method is classified to Lobbato IIIA family in ODE literature.
Refer to [1]_ for the formula and some discussion.
Returns
-------
col_res : ndarray, shape (n, m - 1)
Collocation residuals at the middle points of the mesh intervals.
y_middle : ndarray, shape (n, m - 1)
Values of the cubic spline evaluated at the middle points of the mesh
intervals.
f : ndarray, shape (n, m)
RHS of the ODE system evaluated at the mesh nodes.
f_middle : ndarray, shape (n, m - 1)
RHS of the ODE system evaluated at the middle points of the mesh
intervals (and using `y_middle`).
References
----------
.. [1] J. Kierzenka, L. F. Shampine, "A BVP Solver Based on Residual
Control and the Maltab PSE", ACM Trans. Math. Softw., Vol. 27,
Number 3, pp. 299-316, 2001.
"""
f = fun(x, y, p)
y_middle = (0.5 * (y[:, 1:] + y[:, :-1]) -
0.125 * h * (f[:, 1:] - f[:, :-1]))
f_middle = fun(x[:-1] + 0.5 * h, y_middle, p)
col_res = y[:, 1:] - y[:, :-1] - h / 6 * (f[:, :-1] + f[:, 1:] +
4 * f_middle)
return col_res, y_middle, f, f_middle
def prepare_sys(n, m, k, fun, bc, fun_jac, bc_jac, x, h):
"""Create the function and the Jacobian for the collocation system."""
x_middle = x[:-1] + 0.5 * h
i_jac, j_jac = compute_jac_indices(n, m, k)
def col_fun(y, p):
return collocation_fun(fun, y, p, x, h)
def sys_jac(y, p, y_middle, f, f_middle, bc0):
if fun_jac is None:
df_dy, df_dp = estimate_fun_jac(fun, x, y, p, f)
df_dy_middle, df_dp_middle = estimate_fun_jac(
fun, x_middle, y_middle, p, f_middle)
else:
df_dy, df_dp = fun_jac(x, y, p)
df_dy_middle, df_dp_middle = fun_jac(x_middle, y_middle, p)
if bc_jac is None:
dbc_dya, dbc_dyb, dbc_dp = estimate_bc_jac(bc, y[:, 0], y[:, -1],
p, bc0)
else:
dbc_dya, dbc_dyb, dbc_dp = bc_jac(y[:, 0], y[:, -1], p)
return construct_global_jac(n, m, k, i_jac, j_jac, h, df_dy,
df_dy_middle, df_dp, df_dp_middle, dbc_dya,
dbc_dyb, dbc_dp)
return col_fun, sys_jac
def solve_newton(n, m, h, col_fun, bc, jac, y, p, B, bvp_tol, bc_tol):
"""Solve the nonlinear collocation system by a Newton method.
This is a simple Newton method with a backtracking line search. As
advised in [1]_, an affine-invariant criterion function F = ||J^-1 r||^2
is used, where J is the Jacobian matrix at the current iteration and r is
the vector or collocation residuals (values of the system lhs).
The method alters between full Newton iterations and the fixed-Jacobian
iterations based
There are other tricks proposed in [1]_, but they are not used as they
don't seem to improve anything significantly, and even break the
convergence on some test problems I tried.
All important parameters of the algorithm are defined inside the function.
Parameters
----------
n : int
Number of equations in the ODE system.
m : int
Number of nodes in the mesh.
h : ndarray, shape (m-1,)
Mesh intervals.
col_fun : callable
Function computing collocation residuals.
bc : callable
Function computing boundary condition residuals.
jac : callable
Function computing the Jacobian of the whole system (including
collocation and boundary condition residuals). It is supposed to
return csc_matrix.
y : ndarray, shape (n, m)
Initial guess for the function values at the mesh nodes.
p : ndarray, shape (k,)
Initial guess for the unknown parameters.
B : ndarray with shape (n, n) or None
Matrix to force the S y(a) = 0 condition for a problems with the
singular term. If None, the singular term is assumed to be absent.
bvp_tol : float
Tolerance to which we want to solve a BVP.
bc_tol : float
Tolerance to which we want to satisfy the boundary conditions.
Returns
-------
y : ndarray, shape (n, m)
Final iterate for the function values at the mesh nodes.
p : ndarray, shape (k,)
Final iterate for the unknown parameters.
singular : bool
True, if the LU decomposition failed because Jacobian turned out
to be singular.
References
----------
.. [1] U. Ascher, R. Mattheij and R. Russell "Numerical Solution of
Boundary Value Problems for Ordinary Differential Equations"
"""
# We know that the solution residuals at the middle points of the mesh
# are connected with collocation residuals r_middle = 1.5 * col_res / h.
# As our BVP solver tries to decrease relative residuals below a certain
# tolerance it seems reasonable to terminated Newton iterations by
# comparison of r_middle / (1 + np.abs(f_middle)) with a certain threshold,
# which we choose to be 1.5 orders lower than the BVP tolerance. We rewrite
# the condition as col_res < tol_r * (1 + np.abs(f_middle)), then tol_r
# should be computed as follows:
tol_r = 2/3 * h * 5e-2 * bvp_tol
# Maximum allowed number of Jacobian evaluation and factorization, in
# other words the maximum number of full Newton iterations. A small value
# is recommended in the literature.
max_njev = 4
# Maximum number of iterations, considering that some of them can be
# performed with the fixed Jacobian. In theory such iterations are cheap,
# but it's not that simple in Python.
max_iter = 8
# Minimum relative improvement of the criterion function to accept the
# step (Armijo constant).
sigma = 0.2
# Step size decrease factor for backtracking.
tau = 0.5
# Maximum number of backtracking steps, the minimum step is then
# tau ** n_trial.
n_trial = 4
col_res, y_middle, f, f_middle = col_fun(y, p)
bc_res = bc(y[:, 0], y[:, -1], p)
res = np.hstack((col_res.ravel(order='F'), bc_res))
njev = 0
singular = False
recompute_jac = True
for iteration in range(max_iter):
if recompute_jac:
J = jac(y, p, y_middle, f, f_middle, bc_res)
njev += 1
try:
LU = splu(J)
except RuntimeError:
singular = True
break
step = LU.solve(res)
cost = np.dot(step, step)
y_step = step[:m * n].reshape((n, m), order='F')
p_step = step[m * n:]
alpha = 1
for trial in range(n_trial + 1):
y_new = y - alpha * y_step
if B is not None:
y_new[:, 0] = np.dot(B, y_new[:, 0])
p_new = p - alpha * p_step
col_res, y_middle, f, f_middle = col_fun(y_new, p_new)
bc_res = bc(y_new[:, 0], y_new[:, -1], p_new)
res = np.hstack((col_res.ravel(order='F'), bc_res))
step_new = LU.solve(res)
cost_new = np.dot(step_new, step_new)
if cost_new < (1 - 2 * alpha * sigma) * cost:
break
if trial < n_trial:
alpha *= tau
y = y_new
p = p_new
if njev == max_njev:
break
if (np.all(np.abs(col_res) < tol_r * (1 + np.abs(f_middle))) and
np.all(np.abs(bc_res) < bc_tol)):
break
# If the full step was taken, then we are going to continue with
# the same Jacobian. This is the approach of BVP_SOLVER.
if alpha == 1:
step = step_new
cost = cost_new
recompute_jac = False
else:
recompute_jac = True
return y, p, singular
def print_iteration_header():
print("{:^15}{:^15}{:^15}{:^15}{:^15}".format(
"Iteration", "Max residual", "Max BC residual", "Total nodes",
"Nodes added"))
def print_iteration_progress(iteration, residual, bc_residual, total_nodes,
nodes_added):
print("{:^15}{:^15.2e}{:^15.2e}{:^15}{:^15}".format(
iteration, residual, bc_residual, total_nodes, nodes_added))
class BVPResult(OptimizeResult):
pass
TERMINATION_MESSAGES = {
0: "The algorithm converged to the desired accuracy.",
1: "The maximum number of mesh nodes is exceeded.",
2: "A singular Jacobian encountered when solving the collocation system.",
3: "The solver was unable to satisfy boundary conditions tolerance on iteration 10."
}
def estimate_rms_residuals(fun, sol, x, h, p, r_middle, f_middle):
"""Estimate rms values of collocation residuals using Lobatto quadrature.
The residuals are defined as the difference between the derivatives of
our solution and rhs of the ODE system. We use relative residuals, i.e.
normalized by 1 + np.abs(f). RMS values are computed as sqrt from the
normalized integrals of the squared relative residuals over each interval.
Integrals are estimated using 5-point Lobatto quadrature [1]_, we use the
fact that residuals at the mesh nodes are identically zero.
In [2] they don't normalize integrals by interval lengths, which gives
a higher rate of convergence of the residuals by the factor of h**0.5.
I chose to do such normalization for an ease of interpretation of return
values as RMS estimates.
Returns
-------
rms_res : ndarray, shape (m - 1,)
Estimated rms values of the relative residuals over each interval.
References
----------
.. [1] http://mathworld.wolfram.com/LobattoQuadrature.html
.. [2] J. Kierzenka, L. F. Shampine, "A BVP Solver Based on Residual
Control and the Maltab PSE", ACM Trans. Math. Softw., Vol. 27,
Number 3, pp. 299-316, 2001.
"""
x_middle = x[:-1] + 0.5 * h
s = 0.5 * h * (3/7)**0.5
x1 = x_middle + s
x2 = x_middle - s
y1 = sol(x1)
y2 = sol(x2)
y1_prime = sol(x1, 1)
y2_prime = sol(x2, 1)
f1 = fun(x1, y1, p)
f2 = fun(x2, y2, p)
r1 = y1_prime - f1
r2 = y2_prime - f2
r_middle /= 1 + np.abs(f_middle)
r1 /= 1 + np.abs(f1)
r2 /= 1 + np.abs(f2)
r1 = np.sum(np.real(r1 * np.conj(r1)), axis=0)
r2 = np.sum(np.real(r2 * np.conj(r2)), axis=0)
r_middle = np.sum(np.real(r_middle * np.conj(r_middle)), axis=0)
return (0.5 * (32 / 45 * r_middle + 49 / 90 * (r1 + r2))) ** 0.5
def create_spline(y, yp, x, h):
"""Create a cubic spline given values and derivatives.
Formulas for the coefficients are taken from interpolate.CubicSpline.
Returns
-------
sol : PPoly
Constructed spline as a PPoly instance.
"""
from scipy.interpolate import PPoly
n, m = y.shape
c = np.empty((4, n, m - 1), dtype=y.dtype)
slope = (y[:, 1:] - y[:, :-1]) / h
t = (yp[:, :-1] + yp[:, 1:] - 2 * slope) / h
c[0] = t / h
c[1] = (slope - yp[:, :-1]) / h - t
c[2] = yp[:, :-1]
c[3] = y[:, :-1]
c = np.rollaxis(c, 1)
return PPoly(c, x, extrapolate=True, axis=1)
def modify_mesh(x, insert_1, insert_2):
"""Insert nodes into a mesh.
Nodes removal logic is not established, its impact on the solver is
presumably negligible. So only insertion is done in this function.
Parameters
----------
x : ndarray, shape (m,)
Mesh nodes.
insert_1 : ndarray
Intervals to each insert 1 new node in the middle.
insert_2 : ndarray
Intervals to each insert 2 new nodes, such that divide an interval
into 3 equal parts.
Returns
-------
x_new : ndarray
New mesh nodes.
Notes
-----
`insert_1` and `insert_2` should not have common values.
"""
# Because np.insert implementation apparently varies with a version of
# numpy, we use a simple and reliable approach with sorting.
return np.sort(np.hstack((
x,
0.5 * (x[insert_1] + x[insert_1 + 1]),
(2 * x[insert_2] + x[insert_2 + 1]) / 3,
(x[insert_2] + 2 * x[insert_2 + 1]) / 3
)))
def wrap_functions(fun, bc, fun_jac, bc_jac, k, a, S, D, dtype):
"""Wrap functions for unified usage in the solver."""
if fun_jac is None:
fun_jac_wrapped = None
if bc_jac is None:
bc_jac_wrapped = None
if k == 0:
def fun_p(x, y, _):
return np.asarray(fun(x, y), dtype)
def bc_wrapped(ya, yb, _):
return np.asarray(bc(ya, yb), dtype)
if fun_jac is not None:
def fun_jac_p(x, y, _):
return np.asarray(fun_jac(x, y), dtype), None
if bc_jac is not None:
def bc_jac_wrapped(ya, yb, _):
dbc_dya, dbc_dyb = bc_jac(ya, yb)
return (np.asarray(dbc_dya, dtype),
np.asarray(dbc_dyb, dtype), None)
else:
def fun_p(x, y, p):
return np.asarray(fun(x, y, p), dtype)
def bc_wrapped(x, y, p):
return np.asarray(bc(x, y, p), dtype)
if fun_jac is not None:
def fun_jac_p(x, y, p):
df_dy, df_dp = fun_jac(x, y, p)
return np.asarray(df_dy, dtype), np.asarray(df_dp, dtype)
if bc_jac is not None:
def bc_jac_wrapped(ya, yb, p):
dbc_dya, dbc_dyb, dbc_dp = bc_jac(ya, yb, p)
return (np.asarray(dbc_dya, dtype), np.asarray(dbc_dyb, dtype),
np.asarray(dbc_dp, dtype))
if S is None:
fun_wrapped = fun_p
else:
def fun_wrapped(x, y, p):
f = fun_p(x, y, p)
if x[0] == a:
f[:, 0] = np.dot(D, f[:, 0])
f[:, 1:] += np.dot(S, y[:, 1:]) / (x[1:] - a)
else:
f += np.dot(S, y) / (x - a)
return f
if fun_jac is not None:
if S is None:
fun_jac_wrapped = fun_jac_p
else:
Sr = S[:, :, np.newaxis]
def fun_jac_wrapped(x, y, p):
df_dy, df_dp = fun_jac_p(x, y, p)
if x[0] == a:
df_dy[:, :, 0] = np.dot(D, df_dy[:, :, 0])
df_dy[:, :, 1:] += Sr / (x[1:] - a)
else:
df_dy += Sr / (x - a)
return df_dy, df_dp
return fun_wrapped, bc_wrapped, fun_jac_wrapped, bc_jac_wrapped
def solve_bvp(fun, bc, x, y, p=None, S=None, fun_jac=None, bc_jac=None,
tol=1e-3, max_nodes=1000, verbose=0, bc_tol=None):
"""Solve a boundary-value problem for a system of ODEs.
This function numerically solves a first order system of ODEs subject to
two-point boundary conditions::
dy / dx = f(x, y, p) + S * y / (x - a), a <= x <= b
bc(y(a), y(b), p) = 0
Here x is a 1-dimensional independent variable, y(x) is a n-dimensional
vector-valued function and p is a k-dimensional vector of unknown
parameters which is to be found along with y(x). For the problem to be
determined there must be n + k boundary conditions, i.e. bc must be
(n + k)-dimensional function.
The last singular term in the right-hand side of the system is optional.
It is defined by an n-by-n matrix S, such that the solution must satisfy
S y(a) = 0. This condition will be forced during iterations, so it must not
contradict boundary conditions. See [2]_ for the explanation how this term
is handled when solving BVPs numerically.
Problems in a complex domain can be solved as well. In this case y and p
are considered to be complex, and f and bc are assumed to be complex-valued
functions, but x stays real. Note that f and bc must be complex
differentiable (satisfy Cauchy-Riemann equations [4]_), otherwise you
should rewrite your problem for real and imaginary parts separately. To
solve a problem in a complex domain, pass an initial guess for y with a
complex data type (see below).
Parameters
----------
fun : callable
Right-hand side of the system. The calling signature is ``fun(x, y)``,
or ``fun(x, y, p)`` if parameters are present. All arguments are
ndarray: ``x`` with shape (m,), ``y`` with shape (n, m), meaning that
``y[:, i]`` corresponds to ``x[i]``, and ``p`` with shape (k,). The
return value must be an array with shape (n, m) and with the same
layout as ``y``.
bc : callable
Function evaluating residuals of the boundary conditions. The calling
signature is ``bc(ya, yb)``, or ``bc(ya, yb, p)`` if parameters are
present. All arguments are ndarray: ``ya`` and ``yb`` with shape (n,),
and ``p`` with shape (k,). The return value must be an array with
shape (n + k,).
x : array_like, shape (m,)
Initial mesh. Must be a strictly increasing sequence of real numbers
with ``x[0]=a`` and ``x[-1]=b``.
y : array_like, shape (n, m)
Initial guess for the function values at the mesh nodes, i-th column
corresponds to ``x[i]``. For problems in a complex domain pass `y`
with a complex data type (even if the initial guess is purely real).
p : array_like with shape (k,) or None, optional
Initial guess for the unknown parameters. If None (default), it is
assumed that the problem doesn't depend on any parameters.
S : array_like with shape (n, n) or None
Matrix defining the singular term. If None (default), the problem is
solved without the singular term.
fun_jac : callable or None, optional
Function computing derivatives of f with respect to y and p. The
calling signature is ``fun_jac(x, y)``, or ``fun_jac(x, y, p)`` if
parameters are present. The return must contain 1 or 2 elements in the
following order:
* df_dy : array_like with shape (n, n, m) where an element
(i, j, q) equals to d f_i(x_q, y_q, p) / d (y_q)_j.
* df_dp : array_like with shape (n, k, m) where an element
(i, j, q) equals to d f_i(x_q, y_q, p) / d p_j.
Here q numbers nodes at which x and y are defined, whereas i and j
number vector components. If the problem is solved without unknown
parameters df_dp should not be returned.
If `fun_jac` is None (default), the derivatives will be estimated
by the forward finite differences.
bc_jac : callable or None, optional
Function computing derivatives of bc with respect to ya, yb and p.
The calling signature is ``bc_jac(ya, yb)``, or ``bc_jac(ya, yb, p)``
if parameters are present. The return must contain 2 or 3 elements in
the following order:
* dbc_dya : array_like with shape (n, n) where an element (i, j)
equals to d bc_i(ya, yb, p) / d ya_j.
* dbc_dyb : array_like with shape (n, n) where an element (i, j)
equals to d bc_i(ya, yb, p) / d yb_j.
* dbc_dp : array_like with shape (n, k) where an element (i, j)
equals to d bc_i(ya, yb, p) / d p_j.
If the problem is solved without unknown parameters dbc_dp should not
be returned.
If `bc_jac` is None (default), the derivatives will be estimated by
the forward finite differences.
tol : float, optional
Desired tolerance of the solution. If we define ``r = y' - f(x, y)``
where y is the found solution, then the solver tries to achieve on each
mesh interval ``norm(r / (1 + abs(f)) < tol``, where ``norm`` is
estimated in a root mean squared sense (using a numerical quadrature
formula). Default is 1e-3.
max_nodes : int, optional
Maximum allowed number of the mesh nodes. If exceeded, the algorithm
terminates. Default is 1000.
verbose : {0, 1, 2}, optional
Level of algorithm's verbosity:
* 0 (default) : work silently.
* 1 : display a termination report.
* 2 : display progress during iterations.
bc_tol : float, optional
Desired absolute tolerance for the boundary condition residuals: `bc`
value should satisfy ``abs(bc) < bc_tol`` component-wise.
Equals to `tol` by default. Up to 10 iterations are allowed to achieve this
tolerance.
Returns
-------
Bunch object with the following fields defined:
sol : PPoly
Found solution for y as `scipy.interpolate.PPoly` instance, a C1
continuous cubic spline.
p : ndarray or None, shape (k,)
Found parameters. None, if the parameters were not present in the
problem.
x : ndarray, shape (m,)
Nodes of the final mesh.
y : ndarray, shape (n, m)
Solution values at the mesh nodes.
yp : ndarray, shape (n, m)
Solution derivatives at the mesh nodes.
rms_residuals : ndarray, shape (m - 1,)
RMS values of the relative residuals over each mesh interval (see the
description of `tol` parameter).
niter : int
Number of completed iterations.
status : int
Reason for algorithm termination:
* 0: The algorithm converged to the desired accuracy.
* 1: The maximum number of mesh nodes is exceeded.
* 2: A singular Jacobian encountered when solving the collocation
system.
message : string
Verbal description of the termination reason.
success : bool
True if the algorithm converged to the desired accuracy (``status=0``).
Notes
-----
This function implements a 4-th order collocation algorithm with the
control of residuals similar to [1]_. A collocation system is solved
by a damped Newton method with an affine-invariant criterion function as
described in [3]_.
Note that in [1]_ integral residuals are defined without normalization
by interval lengths. So their definition is different by a multiplier of
h**0.5 (h is an interval length) from the definition used here.
.. versionadded:: 0.18.0
References
----------
.. [1] J. Kierzenka, L. F. Shampine, "A BVP Solver Based on Residual
Control and the Maltab PSE", ACM Trans. Math. Softw., Vol. 27,
Number 3, pp. 299-316, 2001.
.. [2] L.F. Shampine, P. H. Muir and H. Xu, "A User-Friendly Fortran BVP
Solver".
.. [3] U. Ascher, R. Mattheij and R. Russell "Numerical Solution of
Boundary Value Problems for Ordinary Differential Equations".
.. [4] `Cauchy-Riemann equations
<https://en.wikipedia.org/wiki/Cauchy-Riemann_equations>`_ on
Wikipedia.
Examples
--------
In the first example we solve Bratu's problem::
y'' + k * exp(y) = 0
y(0) = y(1) = 0
for k = 1.
We rewrite the equation as a first order system and implement its
right-hand side evaluation::
y1' = y2
y2' = -exp(y1)
>>> def fun(x, y):
... return np.vstack((y[1], -np.exp(y[0])))
Implement evaluation of the boundary condition residuals:
>>> def bc(ya, yb):
... return np.array([ya[0], yb[0]])
Define the initial mesh with 5 nodes:
>>> x = np.linspace(0, 1, 5)
This problem is known to have two solutions. To obtain both of them we
use two different initial guesses for y. We denote them by subscripts
a and b.
>>> y_a = np.zeros((2, x.size))
>>> y_b = np.zeros((2, x.size))
>>> y_b[0] = 3
Now we are ready to run the solver.
>>> from scipy.integrate import solve_bvp
>>> res_a = solve_bvp(fun, bc, x, y_a)
>>> res_b = solve_bvp(fun, bc, x, y_b)
Let's plot the two found solutions. We take an advantage of having the
solution in a spline form to produce a smooth plot.
>>> x_plot = np.linspace(0, 1, 100)
>>> y_plot_a = res_a.sol(x_plot)[0]
>>> y_plot_b = res_b.sol(x_plot)[0]
>>> import matplotlib.pyplot as plt
>>> plt.plot(x_plot, y_plot_a, label='y_a')
>>> plt.plot(x_plot, y_plot_b, label='y_b')
>>> plt.legend()
>>> plt.xlabel("x")
>>> plt.ylabel("y")
>>> plt.show()
We see that the two solutions have similar shape, but differ in scale
significantly.
In the second example we solve a simple Sturm-Liouville problem::
y'' + k**2 * y = 0
y(0) = y(1) = 0
It is known that a non-trivial solution y = A * sin(k * x) is possible for
k = pi * n, where n is an integer. To establish the normalization constant
A = 1 we add a boundary condition::
y'(0) = k
Again we rewrite our equation as a first order system and implement its
right-hand side evaluation::
y1' = y2
y2' = -k**2 * y1
>>> def fun(x, y, p):
... k = p[0]
... return np.vstack((y[1], -k**2 * y[0]))
Note that parameters p are passed as a vector (with one element in our
case).
Implement the boundary conditions:
>>> def bc(ya, yb, p):
... k = p[0]
... return np.array([ya[0], yb[0], ya[1] - k])
Setup the initial mesh and guess for y. We aim to find the solution for
k = 2 * pi, to achieve that we set values of y to approximately follow
sin(2 * pi * x):
>>> x = np.linspace(0, 1, 5)
>>> y = np.zeros((2, x.size))
>>> y[0, 1] = 1
>>> y[0, 3] = -1
Run the solver with 6 as an initial guess for k.
>>> sol = solve_bvp(fun, bc, x, y, p=[6])
We see that the found k is approximately correct:
>>> sol.p[0]
6.28329460046
And finally plot the solution to see the anticipated sinusoid:
>>> x_plot = np.linspace(0, 1, 100)
>>> y_plot = sol.sol(x_plot)[0]
>>> plt.plot(x_plot, y_plot)
>>> plt.xlabel("x")
>>> plt.ylabel("y")
>>> plt.show()
"""
x = np.asarray(x, dtype=float)
if x.ndim != 1:
raise ValueError("`x` must be 1 dimensional.")
h = np.diff(x)
if np.any(h <= 0):
raise ValueError("`x` must be strictly increasing.")
a = x[0]
y = np.asarray(y)
if np.issubdtype(y.dtype, np.complexfloating):
dtype = complex
else:
dtype = float
y = y.astype(dtype, copy=False)
if y.ndim != 2:
raise ValueError("`y` must be 2 dimensional.")
if y.shape[1] != x.shape[0]:
raise ValueError("`y` is expected to have {} columns, but actually "
"has {}.".format(x.shape[0], y.shape[1]))
if p is None:
p = np.array([])
else:
p = np.asarray(p, dtype=dtype)
if p.ndim != 1:
raise ValueError("`p` must be 1 dimensional.")
if tol < 100 * EPS:
warn("`tol` is too low, setting to {:.2e}".format(100 * EPS))
tol = 100 * EPS
if verbose not in [0, 1, 2]:
raise ValueError("`verbose` must be in [0, 1, 2].")
n = y.shape[0]
k = p.shape[0]
if S is not None:
S = np.asarray(S, dtype=dtype)
if S.shape != (n, n):
raise ValueError("`S` is expected to have shape {}, "
"but actually has {}".format((n, n), S.shape))
# Compute I - S^+ S to impose necessary boundary conditions.
B = np.identity(n) - np.dot(pinv(S), S)
y[:, 0] = np.dot(B, y[:, 0])
# Compute (I - S)^+ to correct derivatives at x=a.
D = pinv(np.identity(n) - S)
else:
B = None
D = None
if bc_tol is None:
bc_tol = tol
# Maximum number of iterations
max_iteration = 10
fun_wrapped, bc_wrapped, fun_jac_wrapped, bc_jac_wrapped = wrap_functions(
fun, bc, fun_jac, bc_jac, k, a, S, D, dtype)
f = fun_wrapped(x, y, p)
if f.shape != y.shape:
raise ValueError("`fun` return is expected to have shape {}, "
"but actually has {}.".format(y.shape, f.shape))
bc_res = bc_wrapped(y[:, 0], y[:, -1], p)
if bc_res.shape != (n + k,):
raise ValueError("`bc` return is expected to have shape {}, "
"but actually has {}.".format((n + k,), bc_res.shape))
status = 0
iteration = 0
if verbose == 2:
print_iteration_header()
while True:
m = x.shape[0]
col_fun, jac_sys = prepare_sys(n, m, k, fun_wrapped, bc_wrapped,
fun_jac_wrapped, bc_jac_wrapped, x, h)
y, p, singular = solve_newton(n, m, h, col_fun, bc_wrapped, jac_sys,
y, p, B, tol, bc_tol)
iteration += 1
col_res, y_middle, f, f_middle = collocation_fun(fun_wrapped, y,
p, x, h)
bc_res = bc_wrapped(y[:, 0], y[:, -1], p)
max_bc_res = np.max(abs(bc_res))
# This relation is not trivial, but can be verified.
r_middle = 1.5 * col_res / h
sol = create_spline(y, f, x, h)
rms_res = estimate_rms_residuals(fun_wrapped, sol, x, h, p,
r_middle, f_middle)
max_rms_res = np.max(rms_res)
if singular:
status = 2
break
insert_1, = np.nonzero((rms_res > tol) & (rms_res < 100 * tol))
insert_2, = np.nonzero(rms_res >= 100 * tol)
nodes_added = insert_1.shape[0] + 2 * insert_2.shape[0]
if m + nodes_added > max_nodes:
status = 1
if verbose == 2:
nodes_added = "({})".format(nodes_added)
print_iteration_progress(iteration, max_rms_res, max_bc_res,
m, nodes_added)
break
if verbose == 2:
print_iteration_progress(iteration, max_rms_res, max_bc_res, m,
nodes_added)
if nodes_added > 0:
x = modify_mesh(x, insert_1, insert_2)
h = np.diff(x)
y = sol(x)
elif max_bc_res <= bc_tol:
status = 0
break
elif iteration >= max_iteration:
status = 3
break
if verbose > 0:
if status == 0:
print("Solved in {} iterations, number of nodes {}. \n"
"Maximum relative residual: {:.2e} \n"
"Maximum boundary residual: {:.2e}"
.format(iteration, x.shape[0], max_rms_res, max_bc_res))
elif status == 1:
print("Number of nodes is exceeded after iteration {}. \n"
"Maximum relative residual: {:.2e} \n"
"Maximum boundary residual: {:.2e}"
.format(iteration, max_rms_res, max_bc_res))
elif status == 2:
print("Singular Jacobian encountered when solving the collocation "
"system on iteration {}. \n"
"Maximum relative residual: {:.2e} \n"
"Maximum boundary residual: {:.2e}"
.format(iteration, max_rms_res, max_bc_res))
elif status == 3:
print("The solver was unable to satisfy boundary conditions "
"tolerance on iteration {}. \n"
"Maximum relative residual: {:.2e} \n"
"Maximum boundary residual: {:.2e}"
.format(iteration, max_rms_res, max_bc_res))
if p.size == 0:
p = None
return BVPResult(sol=sol, p=p, x=x, y=y, yp=f, rms_residuals=rms_res,
niter=iteration, status=status,
message=TERMINATION_MESSAGES[status], success=status == 0)