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def solve_bvp(fun, bc, x, y, p=None, S=None, fun_jac=None, bc_jac=None, tol=0.001, 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()
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