Coverage for /usr/lib/python3/dist-packages/scipy/integrate/_bvp.py : 7%

"""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): 

"""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. 

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 

# We also need to control residuals of the boundary conditions. But it 

# seems that they become very small eventually as the solver progresses, 

# i. e. the tolerance for BC are not very important. We set it 1.5 orders 

# lower than the BVP tolerance as well. 

tol_bc = 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(bc_res < tol_bc)): 

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}".format( 

"Iteration", "Max residual", "Total nodes", "Nodes added")) 

def print_iteration_progress(iteration, residual, total_nodes, nodes_added): 

print("{:^15}{:^15.2e}{:^15}{:^15}".format( 

iteration, 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." 

} 

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): 

"""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. 

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