Skip to main content
`Ctrl`+`K`
scipy.integrate.

# solve_bvp

scipy.integrate.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.

  Added in version 0.18.0.

  References

  [1] (1,2)

  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.

  [2]

  L.F. Shampine, P. H. Muir and H. Xu, “A User-Friendly Fortran BVP Solver”, J. Numer. Anal., Ind. Appl. Math. (JNAIAM), Vol. 1, Number 2, pp. 201-217, 2006.

  [3]

  U. Ascher, R. Mattheij and R. Russell “Numerical Solution of Boundary Value Problems for Ordinary Differential Equations”, Philidelphia, PA: Society for Industrial and Applied Mathematics, 1995. DOI:10.1137/1.9781611971231

  [4]

  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()