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llmeval-env/lib/python3.10/site-packages/scipy/integrate/_bvp.py

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+"""Boundary value problem solver."""
+from warnings import warn
+
+import numpy as np
+from numpy.linalg import 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,:,:]).
+
+    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((a.shape[0], a.shape[1], b.shape[2]))
+        for i in range(a.shape[0]):
+            out[i] = np.dot(a[i], b[i])
+        return out
+    else:
+        return np.einsum('...ij,...jk->...ik', a, b)
+
+
+def construct_global_jac(n, m, k, i_jac, j_jac, h, df_dy, df_dy_middle, df_dp,
+                         df_dp_middle, dbc_dya, dbc_dyb, dbc_dp):
+    """Construct the Jacobian of the collocation system.
+
+    There are n * m + k functions: m - 1 collocations residuals, each
+    containing n components, followed by n + k boundary condition residuals.
+
+    There are n * m + k variables: m vectors of y, each containing n
+    components, followed by k values of vector p.
+
+    For example, let m = 4, n = 2 and k = 1, then the Jacobian will have
+    the following sparsity structure:
+
+        1 1 2 2 0 0 0 0  5
+        1 1 2 2 0 0 0 0  5
+        0 0 1 1 2 2 0 0  5
+        0 0 1 1 2 2 0 0  5
+        0 0 0 0 1 1 2 2  5
+        0 0 0 0 1 1 2 2  5
+
+        3 3 0 0 0 0 4 4  6
+        3 3 0 0 0 0 4 4  6
+        3 3 0 0 0 0 4 4  6
+
+    Zeros denote identically zero values, other values denote different kinds
+    of blocks in the matrix (see below). The blank row indicates the separation
+    of collocation residuals from boundary conditions. And the blank column
+    indicates the separation of y values from p values.
+
+    Refer to [1]_  (p. 306) for the formula of n x n blocks for derivatives
+    of collocation residuals with respect to y.
+
+    Parameters
+    ----------
+    n : int
+        Number of equations in the ODE system.
+    m : int
+        Number of nodes in the mesh.
+    k : int
+        Number of the unknown parameters.
+    i_jac, j_jac : ndarray
+        Row and column indices returned by `compute_jac_indices`. They
+        represent different blocks in the Jacobian matrix in the following
+        order (see the scheme above):
+
+            * 1: m - 1 diagonal n x n blocks for the collocation residuals.
+            * 2: m - 1 off-diagonal n x n blocks for the collocation residuals.
+            * 3 : (n + k) x n block for the dependency of the boundary
+              conditions on ya.
+            * 4: (n + k) x n block for the dependency of the boundary
+              conditions on yb.
+            * 5: (m - 1) * n x k block for the dependency of the collocation
+              residuals on p.
+            * 6: (n + k) x k block for the dependency of the boundary
+              conditions on p.
+
+    df_dy : ndarray, shape (n, n, m)
+        Jacobian of f with respect to y computed at the mesh nodes.
+    df_dy_middle : ndarray, shape (n, n, m - 1)
+        Jacobian of f with respect to y computed at the middle between the
+        mesh nodes.
+    df_dp : ndarray with shape (n, k, m) or None
+        Jacobian of f with respect to p computed at the mesh nodes.
+    df_dp_middle : ndarray with shape (n, k, m - 1) or None
+        Jacobian of f with respect to p computed at the middle between the
+        mesh nodes.
+    dbc_dya, dbc_dyb : ndarray, shape (n, n)
+        Jacobian of bc with respect to ya and yb.
+    dbc_dp : ndarray with shape (n, k) or None
+        Jacobian of bc with respect to p.
+
+    Returns
+    -------
+    J : csc_matrix, shape (n * m + k, n * m + k)
+        Jacobian of the collocation system in a sparse form.
+
+    References
+    ----------
+    .. [1] J. Kierzenka, L. F. Shampine, "A BVP Solver Based on Residual
+       Control and the Maltab PSE", ACM Trans. Math. Softw., Vol. 27,
+       Number 3, pp. 299-316, 2001.
+    """
+    df_dy = np.transpose(df_dy, (2, 0, 1))
+    df_dy_middle = np.transpose(df_dy_middle, (2, 0, 1))
+
+    h = h[:, np.newaxis, np.newaxis]
+
+    dtype = df_dy.dtype
+
+    # Computing diagonal n x n blocks.
+    dPhi_dy_0 = np.empty((m - 1, n, n), dtype=dtype)
+    dPhi_dy_0[:] = -np.identity(n)
+    dPhi_dy_0 -= h / 6 * (df_dy[:-1] + 2 * df_dy_middle)
+    T = stacked_matmul(df_dy_middle, df_dy[:-1])
+    dPhi_dy_0 -= h**2 / 12 * T
+
+    # Computing off-diagonal n x n blocks.
+    dPhi_dy_1 = np.empty((m - 1, n, n), dtype=dtype)
+    dPhi_dy_1[:] = np.identity(n)
+    dPhi_dy_1 -= h / 6 * (df_dy[1:] + 2 * df_dy_middle)
+    T = stacked_matmul(df_dy_middle, df_dy[1:])
+    dPhi_dy_1 += h**2 / 12 * T
+
+    values = np.hstack((dPhi_dy_0.ravel(), dPhi_dy_1.ravel(), dbc_dya.ravel(),
+                        dbc_dyb.ravel()))
+
+    if k > 0:
+        df_dp = np.transpose(df_dp, (2, 0, 1))
+        df_dp_middle = np.transpose(df_dp_middle, (2, 0, 1))
+        T = stacked_matmul(df_dy_middle, df_dp[:-1] - df_dp[1:])
+        df_dp_middle += 0.125 * h * T
+        dPhi_dp = -h/6 * (df_dp[:-1] + df_dp[1:] + 4 * df_dp_middle)
+        values = np.hstack((values, dPhi_dp.ravel(), dbc_dp.ravel()))
+
+    J = coo_matrix((values, (i_jac, j_jac)))
+    return csc_matrix(J)
+
+
+def collocation_fun(fun, y, p, x, h):
+    """Evaluate collocation residuals.
+
+    This function lies in the core of the method. The solution is sought
+    as a cubic C1 continuous spline with derivatives matching the ODE rhs
+    at given nodes `x`. Collocation conditions are formed from the equality
+    of the spline derivatives and rhs of the ODE system in the middle points
+    between nodes.
+
+    Such method is classified to Lobbato IIIA family in ODE literature.
+    Refer to [1]_ for the formula and some discussion.
+
+    Returns
+    -------
+    col_res : ndarray, shape (n, m - 1)
+        Collocation residuals at the middle points of the mesh intervals.
+    y_middle : ndarray, shape (n, m - 1)
+        Values of the cubic spline evaluated at the middle points of the mesh
+        intervals.
+    f : ndarray, shape (n, m)
+        RHS of the ODE system evaluated at the mesh nodes.
+    f_middle : ndarray, shape (n, m - 1)
+        RHS of the ODE system evaluated at the middle points of the mesh
+        intervals (and using `y_middle`).
+
+    References
+    ----------
+    .. [1] J. Kierzenka, L. F. Shampine, "A BVP Solver Based on Residual
+           Control and the Maltab PSE", ACM Trans. Math. Softw., Vol. 27,
+           Number 3, pp. 299-316, 2001.
+    """
+    f = fun(x, y, p)
+    y_middle = (0.5 * (y[:, 1:] + y[:, :-1]) -
+                0.125 * h * (f[:, 1:] - f[:, :-1]))
+    f_middle = fun(x[:-1] + 0.5 * h, y_middle, p)
+    col_res = y[:, 1:] - y[:, :-1] - h / 6 * (f[:, :-1] + f[:, 1:] +
+                                              4 * f_middle)
+
+    return col_res, y_middle, f, f_middle
+
+
+def prepare_sys(n, m, k, fun, bc, fun_jac, bc_jac, x, h):
+    """Create the function and the Jacobian for the collocation system."""
+    x_middle = x[:-1] + 0.5 * h
+    i_jac, j_jac = compute_jac_indices(n, m, k)
+
+    def col_fun(y, p):
+        return collocation_fun(fun, y, p, x, h)
+
+    def sys_jac(y, p, y_middle, f, f_middle, bc0):
+        if fun_jac is None:
+            df_dy, df_dp = estimate_fun_jac(fun, x, y, p, f)
+            df_dy_middle, df_dp_middle = estimate_fun_jac(
+                fun, x_middle, y_middle, p, f_middle)
+        else:
+            df_dy, df_dp = fun_jac(x, y, p)
+            df_dy_middle, df_dp_middle = fun_jac(x_middle, y_middle, p)
+
+        if bc_jac is None:
+            dbc_dya, dbc_dyb, dbc_dp = estimate_bc_jac(bc, y[:, 0], y[:, -1],
+                                                       p, bc0)
+        else:
+            dbc_dya, dbc_dyb, dbc_dp = bc_jac(y[:, 0], y[:, -1], p)
+
+        return construct_global_jac(n, m, k, i_jac, j_jac, h, df_dy,
+                                    df_dy_middle, df_dp, df_dp_middle, dbc_dya,
+                                    dbc_dyb, dbc_dp)
+
+    return col_fun, sys_jac
+
+
+def solve_newton(n, m, h, col_fun, bc, jac, y, p, B, bvp_tol, bc_tol):
+    """Solve the nonlinear collocation system by a Newton method.
+
+    This is a simple Newton method with a backtracking line search. As
+    advised in [1]_, an affine-invariant criterion function F = ||J^-1 r||^2
+    is used, where J is the Jacobian matrix at the current iteration and r is
+    the vector or collocation residuals (values of the system lhs).
+
+    The method alters between full Newton iterations and the fixed-Jacobian
+    iterations based
+
+    There are other tricks proposed in [1]_, but they are not used as they
+    don't seem to improve anything significantly, and even break the
+    convergence on some test problems I tried.
+
+    All important parameters of the algorithm are defined inside the function.
+
+    Parameters
+    ----------
+    n : int
+        Number of equations in the ODE system.
+    m : int
+        Number of nodes in the mesh.
+    h : ndarray, shape (m-1,)
+        Mesh intervals.
+    col_fun : callable
+        Function computing collocation residuals.
+    bc : callable
+        Function computing boundary condition residuals.
+    jac : callable
+        Function computing the Jacobian of the whole system (including
+        collocation and boundary condition residuals). It is supposed to
+        return csc_matrix.
+    y : ndarray, shape (n, m)
+        Initial guess for the function values at the mesh nodes.
+    p : ndarray, shape (k,)
+        Initial guess for the unknown parameters.
+    B : ndarray with shape (n, n) or None
+        Matrix to force the S y(a) = 0 condition for a problems with the
+        singular term. If None, the singular term is assumed to be absent.
+    bvp_tol : float
+        Tolerance to which we want to solve a BVP.
+    bc_tol : float
+        Tolerance to which we want to satisfy the boundary conditions.
+
+    Returns
+    -------
+    y : ndarray, shape (n, m)
+        Final iterate for the function values at the mesh nodes.
+    p : ndarray, shape (k,)
+        Final iterate for the unknown parameters.
+    singular : bool
+        True, if the LU decomposition failed because Jacobian turned out
+        to be singular.
+
+    References
+    ----------
+    .. [1]  U. Ascher, R. Mattheij and R. Russell "Numerical Solution of
+       Boundary Value Problems for Ordinary Differential Equations"
+    """
+    # We know that the solution residuals at the middle points of the mesh
+    # are connected with collocation residuals  r_middle = 1.5 * col_res / h.
+    # As our BVP solver tries to decrease relative residuals below a certain
+    # tolerance, it seems reasonable to terminated Newton iterations by
+    # comparison of r_middle / (1 + np.abs(f_middle)) with a certain threshold,
+    # which we choose to be 1.5 orders lower than the BVP tolerance. We rewrite
+    # the condition as col_res < tol_r * (1 + np.abs(f_middle)), then tol_r
+    # should be computed as follows:
+    tol_r = 2/3 * h * 5e-2 * bvp_tol
+
+    # Maximum allowed number of Jacobian evaluation and factorization, in
+    # other words, the maximum number of full Newton iterations. A small value
+    # is recommended in the literature.
+    max_njev = 4
+
+    # Maximum number of iterations, considering that some of them can be
+    # performed with the fixed Jacobian. In theory, such iterations are cheap,
+    # but it's not that simple in Python.
+    max_iter = 8
+
+    # Minimum relative improvement of the criterion function to accept the
+    # step (Armijo constant).
+    sigma = 0.2
+
+    # Step size decrease factor for backtracking.
+    tau = 0.5
+
+    # Maximum number of backtracking steps, the minimum step is then
+    # tau ** n_trial.
+    n_trial = 4
+
+    col_res, y_middle, f, f_middle = col_fun(y, p)
+    bc_res = bc(y[:, 0], y[:, -1], p)
+    res = np.hstack((col_res.ravel(order='F'), bc_res))
+
+    njev = 0
+    singular = False
+    recompute_jac = True
+    for iteration in range(max_iter):
+        if recompute_jac:
+            J = jac(y, p, y_middle, f, f_middle, bc_res)
+            njev += 1
+            try:
+                LU = splu(J)
+            except RuntimeError:
+                singular = True
+                break
+
+            step = LU.solve(res)
+            cost = np.dot(step, step)
+
+        y_step = step[:m * n].reshape((n, m), order='F')
+        p_step = step[m * n:]
+
+        alpha = 1
+        for trial in range(n_trial + 1):
+            y_new = y - alpha * y_step
+            if B is not None:
+                y_new[:, 0] = np.dot(B, y_new[:, 0])
+            p_new = p - alpha * p_step
+
+            col_res, y_middle, f, f_middle = col_fun(y_new, p_new)
+            bc_res = bc(y_new[:, 0], y_new[:, -1], p_new)
+            res = np.hstack((col_res.ravel(order='F'), bc_res))
+
+            step_new = LU.solve(res)
+            cost_new = np.dot(step_new, step_new)
+            if cost_new < (1 - 2 * alpha * sigma) * cost:
+                break
+
+            if trial < n_trial:
+                alpha *= tau
+
+        y = y_new
+        p = p_new
+
+        if njev == max_njev:
+            break
+
+        if (np.all(np.abs(col_res) < tol_r * (1 + np.abs(f_middle))) and
+                np.all(np.abs(bc_res) < bc_tol)):
+            break
+
+        # If the full step was taken, then we are going to continue with
+        # the same Jacobian. This is the approach of BVP_SOLVER.
+        if alpha == 1:
+            step = step_new
+            cost = cost_new
+            recompute_jac = False
+        else:
+            recompute_jac = True
+
+    return y, p, singular
+
+
+def print_iteration_header():
+    print("{:^15}{:^15}{:^15}{:^15}{:^15}".format(
+        "Iteration", "Max residual", "Max BC residual", "Total nodes",
+        "Nodes added"))
+
+
+def print_iteration_progress(iteration, residual, bc_residual, total_nodes,
+                             nodes_added):
+    print("{:^15}{:^15.2e}{:^15.2e}{:^15}{:^15}".format(
+        iteration, residual, bc_residual, total_nodes, nodes_added))
+
+
+class BVPResult(OptimizeResult):
+    pass
+
+
+TERMINATION_MESSAGES = {
+    0: "The algorithm converged to the desired accuracy.",
+    1: "The maximum number of mesh nodes is exceeded.",
+    2: "A singular Jacobian encountered when solving the collocation system.",
+    3: "The solver was unable to satisfy boundary conditions tolerance on iteration 10."
+}
+
+
+def estimate_rms_residuals(fun, sol, x, h, p, r_middle, f_middle):
+    """Estimate rms values of collocation residuals using Lobatto quadrature.
+
+    The residuals are defined as the difference between the derivatives of
+    our solution and rhs of the ODE system. We use relative residuals, i.e.,
+    normalized by 1 + np.abs(f). RMS values are computed as sqrt from the
+    normalized integrals of the squared relative residuals over each interval.
+    Integrals are estimated using 5-point Lobatto quadrature [1]_, we use the
+    fact that residuals at the mesh nodes are identically zero.
+
+    In [2] they don't normalize integrals by interval lengths, which gives
+    a higher rate of convergence of the residuals by the factor of h**0.5.
+    I chose to do such normalization for an ease of interpretation of return
+    values as RMS estimates.
+
+    Returns
+    -------
+    rms_res : ndarray, shape (m - 1,)
+        Estimated rms values of the relative residuals over each interval.
+
+    References
+    ----------
+    .. [1] http://mathworld.wolfram.com/LobattoQuadrature.html
+    .. [2] J. Kierzenka, L. F. Shampine, "A BVP Solver Based on Residual
+       Control and the Maltab PSE", ACM Trans. Math. Softw., Vol. 27,
+       Number 3, pp. 299-316, 2001.
+    """
+    x_middle = x[:-1] + 0.5 * h
+    s = 0.5 * h * (3/7)**0.5
+    x1 = x_middle + s
+    x2 = x_middle - s
+    y1 = sol(x1)
+    y2 = sol(x2)
+    y1_prime = sol(x1, 1)
+    y2_prime = sol(x2, 1)
+    f1 = fun(x1, y1, p)
+    f2 = fun(x2, y2, p)
+    r1 = y1_prime - f1
+    r2 = y2_prime - f2
+
+    r_middle /= 1 + np.abs(f_middle)
+    r1 /= 1 + np.abs(f1)
+    r2 /= 1 + np.abs(f2)
+
+    r1 = np.sum(np.real(r1 * np.conj(r1)), axis=0)
+    r2 = np.sum(np.real(r2 * np.conj(r2)), axis=0)
+    r_middle = np.sum(np.real(r_middle * np.conj(r_middle)), axis=0)
+
+    return (0.5 * (32 / 45 * r_middle + 49 / 90 * (r1 + r2))) ** 0.5
+
+
+def create_spline(y, yp, x, h):
+    """Create a cubic spline given values and derivatives.
+
+    Formulas for the coefficients are taken from interpolate.CubicSpline.
+
+    Returns
+    -------
+    sol : PPoly
+        Constructed spline as a PPoly instance.
+    """
+    from scipy.interpolate import PPoly
+
+    n, m = y.shape
+    c = np.empty((4, n, m - 1), dtype=y.dtype)
+    slope = (y[:, 1:] - y[:, :-1]) / h
+    t = (yp[:, :-1] + yp[:, 1:] - 2 * slope) / h
+    c[0] = t / h
+    c[1] = (slope - yp[:, :-1]) / h - t
+    c[2] = yp[:, :-1]
+    c[3] = y[:, :-1]
+    c = np.moveaxis(c, 1, 0)
+
+    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)
+
+    >>> import numpy as np
+    >>> 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])
+
+    Set up 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(f"`y` is expected to have {x.shape[0]} columns, but actually "
+                         f"has {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(f"`tol` is too low, setting to {100 * EPS:.2e}", stacklevel=2)
+        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(f"`S` is expected to have shape {(n, n)}, "
+                             f"but actually has {S.shape}")
+
+        # Compute I - S^+ S to impose necessary boundary conditions.
+        B = np.identity(n) - np.dot(pinv(S), S)
+
+        y[:, 0] = np.dot(B, y[:, 0])
+
+        # Compute (I - S)^+ to correct derivatives at x=a.
+        D = pinv(np.identity(n) - S)
+    else:
+        B = None
+        D = None
+
+    if bc_tol is None:
+        bc_tol = tol
+
+    # Maximum number of iterations
+    max_iteration = 10
+
+    fun_wrapped, bc_wrapped, fun_jac_wrapped, bc_jac_wrapped = wrap_functions(
+        fun, bc, fun_jac, bc_jac, k, a, S, D, dtype)
+
+    f = fun_wrapped(x, y, p)
+    if f.shape != y.shape:
+        raise ValueError(f"`fun` return is expected to have shape {y.shape}, "
+                         f"but actually has {f.shape}.")
+
+    bc_res = bc_wrapped(y[:, 0], y[:, -1], p)
+    if bc_res.shape != (n + k,):
+        raise ValueError(f"`bc` return is expected to have shape {(n + k,)}, "
+                         f"but actually has {bc_res.shape}.")
+
+    status = 0
+    iteration = 0
+    if verbose == 2:
+        print_iteration_header()
+
+    while True:
+        m = x.shape[0]
+
+        col_fun, jac_sys = prepare_sys(n, m, k, fun_wrapped, bc_wrapped,
+                                       fun_jac_wrapped, bc_jac_wrapped, x, h)
+        y, p, singular = solve_newton(n, m, h, col_fun, bc_wrapped, jac_sys,
+                                      y, p, B, tol, bc_tol)
+        iteration += 1
+
+        col_res, y_middle, f, f_middle = collocation_fun(fun_wrapped, y,
+                                                         p, x, h)
+        bc_res = bc_wrapped(y[:, 0], y[:, -1], p)
+        max_bc_res = np.max(abs(bc_res))
+
+        # This relation is not trivial, but can be verified.
+        r_middle = 1.5 * col_res / h
+        sol = create_spline(y, f, x, h)
+        rms_res = estimate_rms_residuals(fun_wrapped, sol, x, h, p,
+                                         r_middle, f_middle)
+        max_rms_res = np.max(rms_res)
+
+        if singular:
+            status = 2
+            break
+
+        insert_1, = np.nonzero((rms_res > tol) & (rms_res < 100 * tol))
+        insert_2, = np.nonzero(rms_res >= 100 * tol)
+        nodes_added = insert_1.shape[0] + 2 * insert_2.shape[0]
+
+        if m + nodes_added > max_nodes:
+            status = 1
+            if verbose == 2:
+                nodes_added = f"({nodes_added})"
+                print_iteration_progress(iteration, max_rms_res, max_bc_res,
+                                         m, nodes_added)
+            break
+
+        if verbose == 2:
+            print_iteration_progress(iteration, max_rms_res, max_bc_res, m,
+                                     nodes_added)
+
+        if nodes_added > 0:
+            x = modify_mesh(x, insert_1, insert_2)
+            h = np.diff(x)
+            y = sol(x)
+        elif max_bc_res <= bc_tol:
+            status = 0
+            break
+        elif iteration >= max_iteration:
+            status = 3
+            break
+
+    if verbose > 0:
+        if status == 0:
+            print(f"Solved in {iteration} iterations, number of nodes {x.shape[0]}. \n"
+                  f"Maximum relative residual: {max_rms_res:.2e} \n"
+                  f"Maximum boundary residual: {max_bc_res:.2e}")
+        elif status == 1:
+            print(f"Number of nodes is exceeded after iteration {iteration}. \n"
+                  f"Maximum relative residual: {max_rms_res:.2e} \n"
+                  f"Maximum boundary residual: {max_bc_res:.2e}")
+        elif status == 2:
+            print("Singular Jacobian encountered when solving the collocation "
+                  f"system on iteration {iteration}. \n"
+                  f"Maximum relative residual: {max_rms_res:.2e} \n"
+                  f"Maximum boundary residual: {max_bc_res:.2e}")
+        elif status == 3:
+            print("The solver was unable to satisfy boundary conditions "
+                  f"tolerance on iteration {iteration}. \n"
+                  f"Maximum relative residual: {max_rms_res:.2e} \n"
+                  f"Maximum boundary residual: {max_bc_res:.2e}")
+
+    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)

llmeval-env/lib/python3.10/site-packages/scipy/integrate/dop.py

@@ -0,0 +1,18 @@
+# This file is not meant for public use and will be removed in SciPy v2.0.0.
+
+from scipy._lib.deprecation import _sub_module_deprecation
+
+__all__ = [  # noqa: F822
+    'dopri5',
+    'dop853'
+]
+
+
+def __dir__():
+    return __all__
+
+
+def __getattr__(name):
+    return _sub_module_deprecation(sub_package="integrate", module="dop",
+                                   private_modules=["_dop"], all=__all__,
+                                   attribute=name)

llmeval-env/lib/python3.10/site-packages/scipy/integrate/lsoda.py

@@ -0,0 +1,15 @@
+# This file is not meant for public use and will be removed in SciPy v2.0.0.
+
+from scipy._lib.deprecation import _sub_module_deprecation
+
+__all__ = ['lsoda']  # noqa: F822
+
+
+def __dir__():
+    return __all__
+
+
+def __getattr__(name):
+    return _sub_module_deprecation(sub_package="integrate", module="lsoda",
+                                   private_modules=["_lsoda"], all=__all__,
+                                   attribute=name)

llmeval-env/lib/python3.10/site-packages/scipy/integrate/odepack.py

@@ -0,0 +1,17 @@
+# This file is not meant for public use and will be removed in SciPy v2.0.0.
+# Use the `scipy.integrate` namespace for importing the functions
+# included below.
+
+from scipy._lib.deprecation import _sub_module_deprecation
+
+__all__ = ['odeint', 'ODEintWarning']  # noqa: F822
+
+
+def __dir__():
+    return __all__
+
+
+def __getattr__(name):
+    return _sub_module_deprecation(sub_package="integrate", module="odepack",
+                                   private_modules=["_odepack_py"], all=__all__,
+                                   attribute=name)