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1159 lines
40 KiB
Python
1159 lines
40 KiB
Python
"""Boundary value problem solver."""
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from warnings import warn
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import numpy as np
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from numpy.linalg import pinv
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from scipy.sparse import coo_matrix, csc_matrix
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from scipy.sparse.linalg import splu
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from scipy.optimize import OptimizeResult
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EPS = np.finfo(float).eps
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def estimate_fun_jac(fun, x, y, p, f0=None):
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"""Estimate derivatives of an ODE system rhs with forward differences.
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Returns
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-------
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df_dy : ndarray, shape (n, n, m)
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Derivatives with respect to y. An element (i, j, q) corresponds to
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d f_i(x_q, y_q) / d (y_q)_j.
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df_dp : ndarray with shape (n, k, m) or None
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Derivatives with respect to p. An element (i, j, q) corresponds to
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d f_i(x_q, y_q, p) / d p_j. If `p` is empty, None is returned.
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"""
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n, m = y.shape
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if f0 is None:
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f0 = fun(x, y, p)
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dtype = y.dtype
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df_dy = np.empty((n, n, m), dtype=dtype)
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h = EPS**0.5 * (1 + np.abs(y))
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for i in range(n):
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y_new = y.copy()
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y_new[i] += h[i]
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hi = y_new[i] - y[i]
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f_new = fun(x, y_new, p)
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df_dy[:, i, :] = (f_new - f0) / hi
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k = p.shape[0]
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if k == 0:
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df_dp = None
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else:
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df_dp = np.empty((n, k, m), dtype=dtype)
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h = EPS**0.5 * (1 + np.abs(p))
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for i in range(k):
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p_new = p.copy()
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p_new[i] += h[i]
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hi = p_new[i] - p[i]
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f_new = fun(x, y, p_new)
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df_dp[:, i, :] = (f_new - f0) / hi
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return df_dy, df_dp
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def estimate_bc_jac(bc, ya, yb, p, bc0=None):
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"""Estimate derivatives of boundary conditions with forward differences.
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Returns
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-------
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dbc_dya : ndarray, shape (n + k, n)
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Derivatives with respect to ya. An element (i, j) corresponds to
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d bc_i / d ya_j.
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dbc_dyb : ndarray, shape (n + k, n)
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Derivatives with respect to yb. An element (i, j) corresponds to
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d bc_i / d ya_j.
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dbc_dp : ndarray with shape (n + k, k) or None
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Derivatives with respect to p. An element (i, j) corresponds to
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d bc_i / d p_j. If `p` is empty, None is returned.
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"""
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n = ya.shape[0]
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k = p.shape[0]
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if bc0 is None:
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bc0 = bc(ya, yb, p)
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dtype = ya.dtype
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dbc_dya = np.empty((n, n + k), dtype=dtype)
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h = EPS**0.5 * (1 + np.abs(ya))
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for i in range(n):
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ya_new = ya.copy()
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ya_new[i] += h[i]
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hi = ya_new[i] - ya[i]
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bc_new = bc(ya_new, yb, p)
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dbc_dya[i] = (bc_new - bc0) / hi
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dbc_dya = dbc_dya.T
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h = EPS**0.5 * (1 + np.abs(yb))
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dbc_dyb = np.empty((n, n + k), dtype=dtype)
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for i in range(n):
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yb_new = yb.copy()
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yb_new[i] += h[i]
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hi = yb_new[i] - yb[i]
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bc_new = bc(ya, yb_new, p)
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dbc_dyb[i] = (bc_new - bc0) / hi
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dbc_dyb = dbc_dyb.T
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if k == 0:
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dbc_dp = None
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else:
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h = EPS**0.5 * (1 + np.abs(p))
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dbc_dp = np.empty((k, n + k), dtype=dtype)
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for i in range(k):
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p_new = p.copy()
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p_new[i] += h[i]
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hi = p_new[i] - p[i]
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bc_new = bc(ya, yb, p_new)
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dbc_dp[i] = (bc_new - bc0) / hi
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dbc_dp = dbc_dp.T
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return dbc_dya, dbc_dyb, dbc_dp
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def compute_jac_indices(n, m, k):
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"""Compute indices for the collocation system Jacobian construction.
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See `construct_global_jac` for the explanation.
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"""
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i_col = np.repeat(np.arange((m - 1) * n), n)
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j_col = (np.tile(np.arange(n), n * (m - 1)) +
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np.repeat(np.arange(m - 1) * n, n**2))
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i_bc = np.repeat(np.arange((m - 1) * n, m * n + k), n)
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j_bc = np.tile(np.arange(n), n + k)
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i_p_col = np.repeat(np.arange((m - 1) * n), k)
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j_p_col = np.tile(np.arange(m * n, m * n + k), (m - 1) * n)
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i_p_bc = np.repeat(np.arange((m - 1) * n, m * n + k), k)
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j_p_bc = np.tile(np.arange(m * n, m * n + k), n + k)
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i = np.hstack((i_col, i_col, i_bc, i_bc, i_p_col, i_p_bc))
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j = np.hstack((j_col, j_col + n,
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j_bc, j_bc + (m - 1) * n,
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j_p_col, j_p_bc))
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return i, j
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def stacked_matmul(a, b):
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"""Stacked matrix multiply: out[i,:,:] = np.dot(a[i,:,:], b[i,:,:]).
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Empirical optimization. Use outer Python loop and BLAS for large
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matrices, otherwise use a single einsum call.
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"""
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if a.shape[1] > 50:
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out = np.empty((a.shape[0], a.shape[1], b.shape[2]))
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for i in range(a.shape[0]):
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out[i] = np.dot(a[i], b[i])
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return out
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else:
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return np.einsum('...ij,...jk->...ik', a, b)
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def construct_global_jac(n, m, k, i_jac, j_jac, h, df_dy, df_dy_middle, df_dp,
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df_dp_middle, dbc_dya, dbc_dyb, dbc_dp):
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"""Construct the Jacobian of the collocation system.
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There are n * m + k functions: m - 1 collocations residuals, each
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containing n components, followed by n + k boundary condition residuals.
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There are n * m + k variables: m vectors of y, each containing n
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components, followed by k values of vector p.
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For example, let m = 4, n = 2 and k = 1, then the Jacobian will have
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the following sparsity structure:
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1 1 2 2 0 0 0 0 5
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1 1 2 2 0 0 0 0 5
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0 0 1 1 2 2 0 0 5
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0 0 1 1 2 2 0 0 5
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0 0 0 0 1 1 2 2 5
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0 0 0 0 1 1 2 2 5
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3 3 0 0 0 0 4 4 6
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3 3 0 0 0 0 4 4 6
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3 3 0 0 0 0 4 4 6
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Zeros denote identically zero values, other values denote different kinds
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of blocks in the matrix (see below). The blank row indicates the separation
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of collocation residuals from boundary conditions. And the blank column
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indicates the separation of y values from p values.
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Refer to [1]_ (p. 306) for the formula of n x n blocks for derivatives
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of collocation residuals with respect to y.
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Parameters
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----------
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n : int
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Number of equations in the ODE system.
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m : int
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Number of nodes in the mesh.
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k : int
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Number of the unknown parameters.
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i_jac, j_jac : ndarray
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Row and column indices returned by `compute_jac_indices`. They
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represent different blocks in the Jacobian matrix in the following
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order (see the scheme above):
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* 1: m - 1 diagonal n x n blocks for the collocation residuals.
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* 2: m - 1 off-diagonal n x n blocks for the collocation residuals.
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* 3 : (n + k) x n block for the dependency of the boundary
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conditions on ya.
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* 4: (n + k) x n block for the dependency of the boundary
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conditions on yb.
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* 5: (m - 1) * n x k block for the dependency of the collocation
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residuals on p.
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* 6: (n + k) x k block for the dependency of the boundary
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conditions on p.
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df_dy : ndarray, shape (n, n, m)
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Jacobian of f with respect to y computed at the mesh nodes.
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df_dy_middle : ndarray, shape (n, n, m - 1)
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Jacobian of f with respect to y computed at the middle between the
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mesh nodes.
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df_dp : ndarray with shape (n, k, m) or None
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Jacobian of f with respect to p computed at the mesh nodes.
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df_dp_middle: ndarray with shape (n, k, m - 1) or None
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Jacobian of f with respect to p computed at the middle between the
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mesh nodes.
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dbc_dya, dbc_dyb : ndarray, shape (n, n)
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Jacobian of bc with respect to ya and yb.
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dbc_dp: ndarray with shape (n, k) or None
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Jacobian of bc with respect to p.
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Returns
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-------
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J : csc_matrix, shape (n * m + k, n * m + k)
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Jacobian of the collocation system in a sparse form.
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References
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----------
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.. [1] J. Kierzenka, L. F. Shampine, "A BVP Solver Based on Residual
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Control and the Maltab PSE", ACM Trans. Math. Softw., Vol. 27,
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Number 3, pp. 299-316, 2001.
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"""
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df_dy = np.transpose(df_dy, (2, 0, 1))
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df_dy_middle = np.transpose(df_dy_middle, (2, 0, 1))
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h = h[:, np.newaxis, np.newaxis]
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dtype = df_dy.dtype
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# Computing diagonal n x n blocks.
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dPhi_dy_0 = np.empty((m - 1, n, n), dtype=dtype)
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dPhi_dy_0[:] = -np.identity(n)
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dPhi_dy_0 -= h / 6 * (df_dy[:-1] + 2 * df_dy_middle)
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T = stacked_matmul(df_dy_middle, df_dy[:-1])
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dPhi_dy_0 -= h**2 / 12 * T
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# Computing off-diagonal n x n blocks.
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dPhi_dy_1 = np.empty((m - 1, n, n), dtype=dtype)
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dPhi_dy_1[:] = np.identity(n)
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dPhi_dy_1 -= h / 6 * (df_dy[1:] + 2 * df_dy_middle)
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T = stacked_matmul(df_dy_middle, df_dy[1:])
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dPhi_dy_1 += h**2 / 12 * T
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values = np.hstack((dPhi_dy_0.ravel(), dPhi_dy_1.ravel(), dbc_dya.ravel(),
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dbc_dyb.ravel()))
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if k > 0:
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df_dp = np.transpose(df_dp, (2, 0, 1))
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df_dp_middle = np.transpose(df_dp_middle, (2, 0, 1))
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T = stacked_matmul(df_dy_middle, df_dp[:-1] - df_dp[1:])
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df_dp_middle += 0.125 * h * T
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dPhi_dp = -h/6 * (df_dp[:-1] + df_dp[1:] + 4 * df_dp_middle)
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values = np.hstack((values, dPhi_dp.ravel(), dbc_dp.ravel()))
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J = coo_matrix((values, (i_jac, j_jac)))
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return csc_matrix(J)
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def collocation_fun(fun, y, p, x, h):
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"""Evaluate collocation residuals.
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This function lies in the core of the method. The solution is sought
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as a cubic C1 continuous spline with derivatives matching the ODE rhs
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at given nodes `x`. Collocation conditions are formed from the equality
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of the spline derivatives and rhs of the ODE system in the middle points
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between nodes.
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Such method is classified to Lobbato IIIA family in ODE literature.
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Refer to [1]_ for the formula and some discussion.
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Returns
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-------
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col_res : ndarray, shape (n, m - 1)
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Collocation residuals at the middle points of the mesh intervals.
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y_middle : ndarray, shape (n, m - 1)
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Values of the cubic spline evaluated at the middle points of the mesh
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intervals.
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f : ndarray, shape (n, m)
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RHS of the ODE system evaluated at the mesh nodes.
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f_middle : ndarray, shape (n, m - 1)
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RHS of the ODE system evaluated at the middle points of the mesh
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intervals (and using `y_middle`).
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References
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----------
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.. [1] J. Kierzenka, L. F. Shampine, "A BVP Solver Based on Residual
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Control and the Maltab PSE", ACM Trans. Math. Softw., Vol. 27,
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Number 3, pp. 299-316, 2001.
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"""
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f = fun(x, y, p)
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y_middle = (0.5 * (y[:, 1:] + y[:, :-1]) -
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0.125 * h * (f[:, 1:] - f[:, :-1]))
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f_middle = fun(x[:-1] + 0.5 * h, y_middle, p)
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col_res = y[:, 1:] - y[:, :-1] - h / 6 * (f[:, :-1] + f[:, 1:] +
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4 * f_middle)
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return col_res, y_middle, f, f_middle
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def prepare_sys(n, m, k, fun, bc, fun_jac, bc_jac, x, h):
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"""Create the function and the Jacobian for the collocation system."""
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x_middle = x[:-1] + 0.5 * h
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i_jac, j_jac = compute_jac_indices(n, m, k)
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def col_fun(y, p):
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return collocation_fun(fun, y, p, x, h)
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def sys_jac(y, p, y_middle, f, f_middle, bc0):
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if fun_jac is None:
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df_dy, df_dp = estimate_fun_jac(fun, x, y, p, f)
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df_dy_middle, df_dp_middle = estimate_fun_jac(
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fun, x_middle, y_middle, p, f_middle)
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else:
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df_dy, df_dp = fun_jac(x, y, p)
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df_dy_middle, df_dp_middle = fun_jac(x_middle, y_middle, p)
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if bc_jac is None:
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dbc_dya, dbc_dyb, dbc_dp = estimate_bc_jac(bc, y[:, 0], y[:, -1],
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p, bc0)
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else:
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dbc_dya, dbc_dyb, dbc_dp = bc_jac(y[:, 0], y[:, -1], p)
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return construct_global_jac(n, m, k, i_jac, j_jac, h, df_dy,
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df_dy_middle, df_dp, df_dp_middle, dbc_dya,
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dbc_dyb, dbc_dp)
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return col_fun, sys_jac
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def solve_newton(n, m, h, col_fun, bc, jac, y, p, B, bvp_tol, bc_tol):
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"""Solve the nonlinear collocation system by a Newton method.
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This is a simple Newton method with a backtracking line search. As
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advised in [1]_, an affine-invariant criterion function F = ||J^-1 r||^2
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is used, where J is the Jacobian matrix at the current iteration and r is
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the vector or collocation residuals (values of the system lhs).
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The method alters between full Newton iterations and the fixed-Jacobian
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iterations based
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There are other tricks proposed in [1]_, but they are not used as they
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don't seem to improve anything significantly, and even break the
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convergence on some test problems I tried.
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All important parameters of the algorithm are defined inside the function.
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Parameters
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----------
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n : int
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Number of equations in the ODE system.
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m : int
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Number of nodes in the mesh.
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h : ndarray, shape (m-1,)
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Mesh intervals.
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col_fun : callable
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Function computing collocation residuals.
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bc : callable
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Function computing boundary condition residuals.
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jac : callable
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Function computing the Jacobian of the whole system (including
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collocation and boundary condition residuals). It is supposed to
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return csc_matrix.
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y : ndarray, shape (n, m)
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Initial guess for the function values at the mesh nodes.
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p : ndarray, shape (k,)
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Initial guess for the unknown parameters.
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B : ndarray with shape (n, n) or None
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Matrix to force the S y(a) = 0 condition for a problems with the
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singular term. If None, the singular term is assumed to be absent.
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bvp_tol : float
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Tolerance to which we want to solve a BVP.
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bc_tol : float
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Tolerance to which we want to satisfy the boundary conditions.
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Returns
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-------
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y : ndarray, shape (n, m)
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Final iterate for the function values at the mesh nodes.
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p : ndarray, shape (k,)
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Final iterate for the unknown parameters.
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singular : bool
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True, if the LU decomposition failed because Jacobian turned out
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to be singular.
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References
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----------
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.. [1] U. Ascher, R. Mattheij and R. Russell "Numerical Solution of
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Boundary Value Problems for Ordinary Differential Equations"
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"""
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# We know that the solution residuals at the middle points of the mesh
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# are connected with collocation residuals r_middle = 1.5 * col_res / h.
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# As our BVP solver tries to decrease relative residuals below a certain
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# tolerance, it seems reasonable to terminated Newton iterations by
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# comparison of r_middle / (1 + np.abs(f_middle)) with a certain threshold,
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# which we choose to be 1.5 orders lower than the BVP tolerance. We rewrite
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# the condition as col_res < tol_r * (1 + np.abs(f_middle)), then tol_r
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# should be computed as follows:
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tol_r = 2/3 * h * 5e-2 * bvp_tol
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# Maximum allowed number of Jacobian evaluation and factorization, in
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# other words, the maximum number of full Newton iterations. A small value
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# is recommended in the literature.
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max_njev = 4
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# Maximum number of iterations, considering that some of them can be
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# performed with the fixed Jacobian. In theory, such iterations are cheap,
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# but it's not that simple in Python.
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max_iter = 8
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# Minimum relative improvement of the criterion function to accept the
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# step (Armijo constant).
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sigma = 0.2
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# Step size decrease factor for backtracking.
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tau = 0.5
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# Maximum number of backtracking steps, the minimum step is then
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# tau ** n_trial.
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n_trial = 4
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col_res, y_middle, f, f_middle = col_fun(y, p)
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bc_res = bc(y[:, 0], y[:, -1], p)
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res = np.hstack((col_res.ravel(order='F'), bc_res))
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njev = 0
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singular = False
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recompute_jac = True
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for iteration in range(max_iter):
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if recompute_jac:
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J = jac(y, p, y_middle, f, f_middle, bc_res)
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njev += 1
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try:
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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-D independent variable, y(x) is an N-D
|
|
vector-valued function and p is a k-D 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 an
|
|
(n + k)-D function.
|
|
|
|
The last singular term on 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, ith 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 4th 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
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>>> res_a = solve_bvp(fun, bc, x, y_a)
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>>> res_b = solve_bvp(fun, bc, x, y_b)
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Let's plot the two found solutions. We take an advantage of having the
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solution in a spline form to produce a smooth plot.
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>>> x_plot = np.linspace(0, 1, 100)
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>>> y_plot_a = res_a.sol(x_plot)[0]
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>>> y_plot_b = res_b.sol(x_plot)[0]
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>>> import matplotlib.pyplot as plt
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>>> plt.plot(x_plot, y_plot_a, label='y_a')
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>>> plt.plot(x_plot, y_plot_b, label='y_b')
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>>> plt.legend()
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>>> plt.xlabel("x")
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>>> plt.ylabel("y")
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>>> plt.show()
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We see that the two solutions have similar shape, but differ in scale
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significantly.
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In the second example, we solve a simple Sturm-Liouville problem::
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y'' + k**2 * y = 0
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y(0) = y(1) = 0
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It is known that a non-trivial solution y = A * sin(k * x) is possible for
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k = pi * n, where n is an integer. To establish the normalization constant
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A = 1 we add a boundary condition::
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y'(0) = k
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Again, we rewrite our equation as a first-order system and implement its
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right-hand side evaluation::
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y1' = y2
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y2' = -k**2 * y1
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>>> def fun(x, y, p):
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... k = p[0]
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... return np.vstack((y[1], -k**2 * y[0]))
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Note that parameters p are passed as a vector (with one element in our
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case).
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Implement the boundary conditions:
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>>> def bc(ya, yb, p):
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... k = p[0]
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... return np.array([ya[0], yb[0], ya[1] - k])
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Set up the initial mesh and guess for y. We aim to find the solution for
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k = 2 * pi, to achieve that we set values of y to approximately follow
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sin(2 * pi * x):
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>>> x = np.linspace(0, 1, 5)
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>>> y = np.zeros((2, x.size))
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>>> y[0, 1] = 1
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>>> y[0, 3] = -1
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Run the solver with 6 as an initial guess for k.
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>>> sol = solve_bvp(fun, bc, x, y, p=[6])
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We see that the found k is approximately correct:
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>>> sol.p[0]
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6.28329460046
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And, finally, plot the solution to see the anticipated sinusoid:
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>>> x_plot = np.linspace(0, 1, 100)
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>>> y_plot = sol.sol(x_plot)[0]
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>>> plt.plot(x_plot, y_plot)
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>>> plt.xlabel("x")
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>>> plt.ylabel("y")
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>>> plt.show()
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"""
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x = np.asarray(x, dtype=float)
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if x.ndim != 1:
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raise ValueError("`x` must be 1 dimensional.")
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h = np.diff(x)
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if np.any(h <= 0):
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raise ValueError("`x` must be strictly increasing.")
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a = x[0]
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y = np.asarray(y)
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if np.issubdtype(y.dtype, np.complexfloating):
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dtype = complex
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else:
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dtype = float
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y = y.astype(dtype, copy=False)
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if y.ndim != 2:
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raise ValueError("`y` must be 2 dimensional.")
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if y.shape[1] != x.shape[0]:
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raise ValueError("`y` is expected to have {} columns, but actually "
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"has {}.".format(x.shape[0], y.shape[1]))
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if p is None:
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p = np.array([])
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else:
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p = np.asarray(p, dtype=dtype)
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if p.ndim != 1:
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raise ValueError("`p` must be 1 dimensional.")
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if tol < 100 * EPS:
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warn("`tol` is too low, setting to {:.2e}".format(100 * EPS))
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tol = 100 * EPS
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if verbose not in [0, 1, 2]:
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raise ValueError("`verbose` must be in [0, 1, 2].")
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n = y.shape[0]
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k = p.shape[0]
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if S is not None:
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S = np.asarray(S, dtype=dtype)
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if S.shape != (n, n):
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raise ValueError("`S` is expected to have shape {}, "
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"but actually has {}".format((n, n), S.shape))
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# Compute I - S^+ S to impose necessary boundary conditions.
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B = np.identity(n) - np.dot(pinv(S), S)
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y[:, 0] = np.dot(B, y[:, 0])
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# Compute (I - S)^+ to correct derivatives at x=a.
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D = pinv(np.identity(n) - S)
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else:
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B = None
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D = None
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if bc_tol is None:
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bc_tol = tol
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# Maximum number of iterations
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max_iteration = 10
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fun_wrapped, bc_wrapped, fun_jac_wrapped, bc_jac_wrapped = wrap_functions(
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fun, bc, fun_jac, bc_jac, k, a, S, D, dtype)
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f = fun_wrapped(x, y, p)
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if f.shape != y.shape:
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raise ValueError("`fun` return is expected to have shape {}, "
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"but actually has {}.".format(y.shape, f.shape))
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bc_res = bc_wrapped(y[:, 0], y[:, -1], p)
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if bc_res.shape != (n + k,):
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raise ValueError("`bc` return is expected to have shape {}, "
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"but actually has {}.".format((n + k,), bc_res.shape))
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status = 0
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iteration = 0
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if verbose == 2:
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print_iteration_header()
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while True:
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m = x.shape[0]
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col_fun, jac_sys = prepare_sys(n, m, k, fun_wrapped, bc_wrapped,
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fun_jac_wrapped, bc_jac_wrapped, x, h)
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y, p, singular = solve_newton(n, m, h, col_fun, bc_wrapped, jac_sys,
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y, p, B, tol, bc_tol)
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iteration += 1
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col_res, y_middle, f, f_middle = collocation_fun(fun_wrapped, y,
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p, x, h)
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bc_res = bc_wrapped(y[:, 0], y[:, -1], p)
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max_bc_res = np.max(abs(bc_res))
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# This relation is not trivial, but can be verified.
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r_middle = 1.5 * col_res / h
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sol = create_spline(y, f, x, h)
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rms_res = estimate_rms_residuals(fun_wrapped, sol, x, h, p,
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r_middle, f_middle)
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max_rms_res = np.max(rms_res)
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if singular:
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status = 2
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break
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insert_1, = np.nonzero((rms_res > tol) & (rms_res < 100 * tol))
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insert_2, = np.nonzero(rms_res >= 100 * tol)
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nodes_added = insert_1.shape[0] + 2 * insert_2.shape[0]
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if m + nodes_added > max_nodes:
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status = 1
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if verbose == 2:
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nodes_added = "({})".format(nodes_added)
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print_iteration_progress(iteration, max_rms_res, max_bc_res,
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m, nodes_added)
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break
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if verbose == 2:
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print_iteration_progress(iteration, max_rms_res, max_bc_res, m,
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nodes_added)
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if nodes_added > 0:
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x = modify_mesh(x, insert_1, insert_2)
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h = np.diff(x)
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y = sol(x)
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elif max_bc_res <= bc_tol:
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status = 0
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break
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elif iteration >= max_iteration:
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status = 3
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break
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if verbose > 0:
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if status == 0:
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print("Solved in {} iterations, number of nodes {}. \n"
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"Maximum relative residual: {:.2e} \n"
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"Maximum boundary residual: {:.2e}"
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.format(iteration, x.shape[0], max_rms_res, max_bc_res))
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elif status == 1:
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print("Number of nodes is exceeded after iteration {}. \n"
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"Maximum relative residual: {:.2e} \n"
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"Maximum boundary residual: {:.2e}"
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.format(iteration, max_rms_res, max_bc_res))
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elif status == 2:
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print("Singular Jacobian encountered when solving the collocation "
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"system on iteration {}. \n"
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"Maximum relative residual: {:.2e} \n"
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"Maximum boundary residual: {:.2e}"
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.format(iteration, max_rms_res, max_bc_res))
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elif status == 3:
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print("The solver was unable to satisfy boundary conditions "
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"tolerance on iteration {}. \n"
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"Maximum relative residual: {:.2e} \n"
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"Maximum boundary residual: {:.2e}"
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.format(iteration, max_rms_res, max_bc_res))
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if p.size == 0:
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p = None
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return BVPResult(sol=sol, p=p, x=x, y=y, yp=f, rms_residuals=rms_res,
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niter=iteration, status=status,
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message=TERMINATION_MESSAGES[status], success=status == 0)
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