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()
r