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//opt/alt/python35/lib64/python3.5/site-packages/scipy/integrate/_bvp.py
"""Boundary value problem solver.""" from __future__ import division, print_function, absolute_import from warnings import warn import numpy as np from numpy.linalg import norm, pinv from scipy.sparse import coo_matrix, csc_matrix from scipy.sparse.linalg import splu from scipy.optimize import OptimizeResult EPS = np.finfo(float).eps def estimate_fun_jac(fun, x, y, p, f0=None): """Estimate derivatives of an ODE system rhs with forward differences. Returns ------- df_dy : ndarray, shape (n, n, m) Derivatives with respect to y. An element (i, j, q) corresponds to d f_i(x_q, y_q) / d (y_q)_j. df_dp : ndarray with shape (n, k, m) or None Derivatives with respect to p. An element (i, j, q) corresponds to d f_i(x_q, y_q, p) / d p_j. If `p` is empty, None is returned. """ n, m = y.shape if f0 is None: f0 = fun(x, y, p) dtype = y.dtype df_dy = np.empty((n, n, m), dtype=dtype) h = EPS**0.5 * (1 + np.abs(y)) for i in range(n): y_new = y.copy() y_new[i] += h[i] hi = y_new[i] - y[i] f_new = fun(x, y_new, p) df_dy[:, i, :] = (f_new - f0) / hi k = p.shape[0] if k == 0: df_dp = None else: df_dp = np.empty((n, k, m), dtype=dtype) h = EPS**0.5 * (1 + np.abs(p)) for i in range(k): p_new = p.copy() p_new[i] += h[i] hi = p_new[i] - p[i] f_new = fun(x, y, p_new) df_dp[:, i, :] = (f_new - f0) / hi return df_dy, df_dp def estimate_bc_jac(bc, ya, yb, p, bc0=None): """Estimate derivatives of boundary conditions with forward differences. Returns ------- dbc_dya : ndarray, shape (n + k, n) Derivatives with respect to ya. An element (i, j) corresponds to d bc_i / d ya_j. dbc_dyb : ndarray, shape (n + k, n) Derivatives with respect to yb. An element (i, j) corresponds to d bc_i / d ya_j. dbc_dp : ndarray with shape (n + k, k) or None Derivatives with respect to p. An element (i, j) corresponds to d bc_i / d p_j. If `p` is empty, None is returned. """ n = ya.shape[0] k = p.shape[0] if bc0 is None: bc0 = bc(ya, yb, p) dtype = ya.dtype dbc_dya = np.empty((n, n + k), dtype=dtype) h = EPS**0.5 * (1 + np.abs(ya)) for i in range(n): ya_new = ya.copy() ya_new[i] += h[i] hi = ya_new[i] - ya[i] bc_new = bc(ya_new, yb, p) dbc_dya[i] = (bc_new - bc0) / hi dbc_dya = dbc_dya.T h = EPS**0.5 * (1 + np.abs(yb)) dbc_dyb = np.empty((n, n + k), dtype=dtype) for i in range(n): yb_new = yb.copy() yb_new[i] += h[i] hi = yb_new[i] - yb[i] bc_new = bc(ya, yb_new, p) dbc_dyb[i] = (bc_new - bc0) / hi dbc_dyb = dbc_dyb.T if k == 0: dbc_dp = None else: h = EPS**0.5 * (1 + np.abs(p)) dbc_dp = np.empty((k, n + k), dtype=dtype) for i in range(k): p_new = p.copy() p_new[i] += h[i] hi = p_new[i] - p[i] bc_new = bc(ya, yb, p_new) dbc_dp[i] = (bc_new - bc0) / hi dbc_dp = dbc_dp.T return dbc_dya, dbc_dyb, dbc_dp def compute_jac_indices(n, m, k): """Compute indices for the collocation system Jacobian construction. See `construct_global_jac` for the explanation. """ i_col = np.repeat(np.arange((m - 1) * n), n) j_col = (np.tile(np.arange(n), n * (m - 1)) + np.repeat(np.arange(m - 1) * n, n**2)) i_bc = np.repeat(np.arange((m - 1) * n, m * n + k), n) j_bc = np.tile(np.arange(n), n + k) i_p_col = np.repeat(np.arange((m - 1) * n), k) j_p_col = np.tile(np.arange(m * n, m * n + k), (m - 1) * n) i_p_bc = np.repeat(np.arange((m - 1) * n, m * n + k), k) j_p_bc = np.tile(np.arange(m * n, m * n + k), n + k) i = np.hstack((i_col, i_col, i_bc, i_bc, i_p_col, i_p_bc)) j = np.hstack((j_col, j_col + n, j_bc, j_bc + (m - 1) * n, j_p_col, j_p_bc)) return i, j def stacked_matmul(a, b): """Stacked matrix multiply: out[i,:,:] = np.dot(a[i,:,:], b[i,:,:]). In our case a[i, :, :] and b[i, :, :] are always square. """ # Empirical optimization. Use outer Python loop and BLAS for large # matrices, otherwise use a single einsum call. if a.shape[1] > 50: out = np.empty_like(a) for i in range(a.shape[0]): out[i] = np.dot(a[i], b[i]) return out else: return np.einsum('...ij,...jk->...ik', a, b) def construct_global_jac(n, m, k, i_jac, j_jac, h, df_dy, df_dy_middle, df_dp, df_dp_middle, dbc_dya, dbc_dyb, dbc_dp): """Construct the Jacobian of the collocation system. There are n * m + k functions: m - 1 collocations residuals, each containing n components, followed by n + k boundary condition residuals. There are n * m + k variables: m vectors of y, each containing n components, followed by k values of vector p. For example, let m = 4, n = 2 and k = 1, then the Jacobian will have the following sparsity structure: 1 1 2 2 0 0 0 0 5 1 1 2 2 0 0 0 0 5 0 0 1 1 2 2 0 0 5 0 0 1 1 2 2 0 0 5 0 0 0 0 1 1 2 2 5 0 0 0 0 1 1 2 2 5 3 3 0 0 0 0 4 4 6 3 3 0 0 0 0 4 4 6 3 3 0 0 0 0 4 4 6 Zeros denote identically zero values, other values denote different kinds of blocks in the matrix (see below). The blank row indicates the separation of collocation residuals from boundary conditions. And the blank column indicates the separation of y values from p values. Refer to [1]_ (p. 306) for the formula of n x n blocks for derivatives of collocation residuals with respect to y. Parameters ---------- n : int Number of equations in the ODE system. m : int Number of nodes in the mesh. k : int Number of the unknown parameters. i_jac, j_jac : ndarray Row and column indices returned by `compute_jac_indices`. They represent different blocks in the Jacobian matrix in the following order (see the scheme above): * 1: m - 1 diagonal n x n blocks for the collocation residuals. * 2: m - 1 off-diagonal n x n blocks for the collocation residuals. * 3 : (n + k) x n block for the dependency of the boundary conditions on ya. * 4: (n + k) x n block for the dependency of the boundary conditions on yb. * 5: (m - 1) * n x k block for the dependency of the collocation residuals on p. * 6: (n + k) x k block for the dependency of the boundary conditions on p. df_dy : ndarray, shape (n, n, m) Jacobian of f with respect to y computed at the mesh nodes. df_dy_middle : ndarray, shape (n, n, m - 1) Jacobian of f with respect to y computed at the middle between the mesh nodes. df_dp : ndarray with shape (n, k, m) or None Jacobian of f with respect to p computed at the mesh nodes. df_dp_middle: ndarray with shape (n, k, m - 1) or None Jacobian of f with respect to p computed at the middle between the mesh nodes. dbc_dya, dbc_dyb : ndarray, shape (n, n) Jacobian of bc with respect to ya and yb. dbc_dp: ndarray with shape (n, k) or None Jacobian of bc with respect to p. Returns ------- J : csc_matrix, shape (n * m + k, n * m + k) Jacobian of the collocation system in a sparse form. References ---------- .. [1] J. Kierzenka, L. F. Shampine, "A BVP Solver Based on Residual Control and the Maltab PSE", ACM Trans. Math. Softw., Vol. 27, Number 3, pp. 299-316, 2001. """ df_dy = np.transpose(df_dy, (2, 0, 1)) df_dy_middle = np.transpose(df_dy_middle, (2, 0, 1)) h = h[:, np.newaxis, np.newaxis] dtype = df_dy.dtype # Computing diagonal n x n blocks. dPhi_dy_0 = np.empty((m - 1, n, n), dtype=dtype) dPhi_dy_0[:] = -np.identity(n) dPhi_dy_0 -= h / 6 * (df_dy[:-1] + 2 * df_dy_middle) T = stacked_matmul(df_dy_middle, df_dy[:-1]) dPhi_dy_0 -= h**2 / 12 * T # Computing off-diagonal n x n blocks. dPhi_dy_1 = np.empty((m - 1, n, n), dtype=dtype) dPhi_dy_1[:] = np.identity(n) dPhi_dy_1 -= h / 6 * (df_dy[1:] + 2 * df_dy_middle) T = stacked_matmul(df_dy_middle, df_dy[1:]) dPhi_dy_1 += h**2 / 12 * T values = np.hstack((dPhi_dy_0.ravel(), dPhi_dy_1.ravel(), dbc_dya.ravel(), dbc_dyb.ravel())) if k > 0: df_dp = np.transpose(df_dp, (2, 0, 1)) df_dp_middle = np.transpose(df_dp_middle, (2, 0, 1)) T = stacked_matmul(df_dy_middle, df_dp[:-1] - df_dp[1:]) df_dp_middle += 0.125 * h * T dPhi_dp = -h/6 * (df_dp[:-1] + df_dp[1:] + 4 * df_dp_middle) values = np.hstack((values, dPhi_dp.ravel(), dbc_dp.ravel())) J = coo_matrix((values, (i_jac, j_jac))) return csc_matrix(J) def collocation_fun(fun, y, p, x, h): """Evaluate collocation residuals. This function lies in the core of the method. The solution is sought as a cubic C1 continuous spline with derivatives matching the ODE rhs at given nodes `x`. Collocation conditions are formed from the equality of the spline derivatives and rhs of the ODE system in the middle points between nodes. Such method is classified to Lobbato IIIA family in ODE literature. Refer to [1]_ for the formula and some discussion. Returns ------- col_res : ndarray, shape (n, m - 1) Collocation residuals at the middle points of the mesh intervals. y_middle : ndarray, shape (n, m - 1) Values of the cubic spline evaluated at the middle points of the mesh intervals. f : ndarray, shape (n, m) RHS of the ODE system evaluated at the mesh nodes. f_middle : ndarray, shape (n, m - 1) RHS of the ODE system evaluated at the middle points of the mesh intervals (and using `y_middle`). References ---------- .. [1] J. Kierzenka, L. F. Shampine, "A BVP Solver Based on Residual Control and the Maltab PSE", ACM Trans. Math. Softw., Vol. 27, Number 3, pp. 299-316, 2001. """ f = fun(x, y, p) y_middle = (0.5 * (y[:, 1:] + y[:, :-1]) - 0.125 * h * (f[:, 1:] - f[:, :-1])) f_middle = fun(x[:-1] + 0.5 * h, y_middle, p) col_res = y[:, 1:] - y[:, :-1] - h / 6 * (f[:, :-1] + f[:, 1:] + 4 * f_middle) return col_res, y_middle, f, f_middle def prepare_sys(n, m, k, fun, bc, fun_jac, bc_jac, x, h): """Create the function and the Jacobian for the collocation system.""" x_middle = x[:-1] + 0.5 * h i_jac, j_jac = compute_jac_indices(n, m, k) def col_fun(y, p): return collocation_fun(fun, y, p, x, h) def sys_jac(y, p, y_middle, f, f_middle, bc0): if fun_jac is None: df_dy, df_dp = estimate_fun_jac(fun, x, y, p, f) df_dy_middle, df_dp_middle = estimate_fun_jac( fun, x_middle, y_middle, p, f_middle) else: df_dy, df_dp = fun_jac(x, y, p) df_dy_middle, df_dp_middle = fun_jac(x_middle, y_middle, p) if bc_jac is None: dbc_dya, dbc_dyb, dbc_dp = estimate_bc_jac(bc, y[:, 0], y[:, -1], p, bc0) else: dbc_dya, dbc_dyb, dbc_dp = bc_jac(y[:, 0], y[:, -1], p) return construct_global_jac(n, m, k, i_jac, j_jac, h, df_dy, df_dy_middle, df_dp, df_dp_middle, dbc_dya, dbc_dyb, dbc_dp) return col_fun, sys_jac def solve_newton(n, m, h, col_fun, bc, jac, y, p, B, bvp_tol): """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 A = 1 we add a boundary condition:: y'(0) = k Again we rewrite our equation as a first order system and implement its right-hand side evaluation:: y1' = y2 y2' = -k**2 * y1 >>> def fun(x, y, p): ... k = p[0] ... return np.vstack((y[1], -k**2 * y[0])) Note that parameters p are passed as a vector (with one element in our case). Implement the boundary conditions: >>> def bc(ya, yb, p): ... k = p[0] ... return np.array([ya[0], yb[0], ya[1] - k]) Setup the initial mesh and guess for y. We aim to find the solution for k = 2 * pi, to achieve that we set values of y to approximately follow sin(2 * pi * x): >>> x = np.linspace(0, 1, 5) >>> y = np.zeros((2, x.size)) >>> y[0, 1] = 1 >>> y[0, 3] = -1 Run the solver with 6 as an initial guess for k. >>> sol = solve_bvp(fun, bc, x, y, p=[6]) We see that the found k is approximately correct: >>> sol.p[0] 6.28329460046 And finally plot the solution to see the anticipated sinusoid: >>> x_plot = np.linspace(0, 1, 100) >>> y_plot = sol.sol(x_plot)[0] >>> plt.plot(x_plot, y_plot) >>> plt.xlabel("x") >>> plt.ylabel("y") >>> plt.show() """ x = np.asarray(x, dtype=float) if x.ndim != 1: raise ValueError("`x` must be 1 dimensional.") h = np.diff(x) if np.any(h <= 0): raise ValueError("`x` must be strictly increasing.") a = x[0] y = np.asarray(y) if np.issubdtype(y.dtype, np.complexfloating): dtype = complex else: dtype = float y = y.astype(dtype, copy=False) if y.ndim != 2: raise ValueError("`y` must be 2 dimensional.") if y.shape[1] != x.shape[0]: raise ValueError("`y` is expected to have {} columns, but actually " "has {}.".format(x.shape[0], y.shape[1])) if p is None: p = np.array([]) else: p = np.asarray(p, dtype=dtype) if p.ndim != 1: raise ValueError("`p` must be 1 dimensional.") if tol < 100 * EPS: warn("`tol` is too low, setting to {:.2e}".format(100 * EPS)) tol = 100 * EPS if verbose not in [0, 1, 2]: raise ValueError("`verbose` must be in [0, 1, 2].") n = y.shape[0] k = p.shape[0] if S is not None: S = np.asarray(S, dtype=dtype) if S.shape != (n, n): raise ValueError("`S` is expected to have shape {}, " "but actually has {}".format((n, n), S.shape)) # Compute I - S^+ S to impose necessary boundary conditions. B = np.identity(n) - np.dot(pinv(S), S) y[:, 0] = np.dot(B, y[:, 0]) # Compute (I - S)^+ to correct derivatives at x=a. D = pinv(np.identity(n) - S) else: B = None D = None fun_wrapped, bc_wrapped, fun_jac_wrapped, bc_jac_wrapped = wrap_functions( fun, bc, fun_jac, bc_jac, k, a, S, D, dtype) f = fun_wrapped(x, y, p) if f.shape != y.shape: raise ValueError("`fun` return is expected to have shape {}, " "but actually has {}.".format(y.shape, f.shape)) bc_res = bc_wrapped(y[:, 0], y[:, -1], p) if bc_res.shape != (n + k,): raise ValueError("`bc` return is expected to have shape {}, " "but actually has {}.".format((n + k,), bc_res.shape)) status = 0 iteration = 0 if verbose == 2: print_iteration_header() while True: m = x.shape[0] col_fun, jac_sys = prepare_sys(n, m, k, fun_wrapped, bc_wrapped, fun_jac_wrapped, bc_jac_wrapped, x, h) y, p, singular = solve_newton(n, m, h, col_fun, bc_wrapped, jac_sys, y, p, B, tol) iteration += 1 col_res, y_middle, f, f_middle = collocation_fun(fun_wrapped, y, p, x, h) # This relation is not trivial, but can be verified. r_middle = 1.5 * col_res / h sol = create_spline(y, f, x, h) rms_res = estimate_rms_residuals(fun_wrapped, sol, x, h, p, r_middle, f_middle) max_rms_res = np.max(rms_res) if singular: status = 2 break insert_1, = np.nonzero((rms_res > tol) & (rms_res < 100 * tol)) insert_2, = np.nonzero(rms_res >= 100 * tol) nodes_added = insert_1.shape[0] + 2 * insert_2.shape[0] if m + nodes_added > max_nodes: status = 1 if verbose == 2: nodes_added = "({})".format(nodes_added) print_iteration_progress(iteration, max_rms_res, m, nodes_added) break if verbose == 2: print_iteration_progress(iteration, max_rms_res, m, nodes_added) if nodes_added > 0: x = modify_mesh(x, insert_1, insert_2) h = np.diff(x) y = sol(x) else: status = 0 break if verbose > 0: if status == 0: print("Solved in {} iterations, number of nodes {}, " "maximum relative residual {:.2e}." .format(iteration, x.shape[0], max_rms_res)) elif status == 1: print("Number of nodes is exceeded after iteration {}, " "maximum relative residual {:.2e}." .format(iteration, max_rms_res)) elif status == 2: print("Singular Jacobian encountered when solving the collocation " "system on iteration {}, maximum relative residual {:.2e}." .format(iteration, max_rms_res)) if p.size == 0: p = None return BVPResult(sol=sol, p=p, x=x, y=y, yp=f, rms_residuals=rms_res, niter=iteration, status=status, message=TERMINATION_MESSAGES[status], success=status == 0)
