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bundles / scipy 1.17.1 / scipy / optimize / _root / root

function

scipy.optimize._root:root

source: /scipy/optimize/_root.py :25

Signature

def   root ( fun ,  x0 ,  args = () ,  method = hybr ,  jac = None ,  tol = None ,  callback = None ,  options = None )

Summary

Find a root of a vector function.

Parameters

fun : callable

A vector function to find a root of.

Suppose the callable has signature f0(x, *my_args, **my_kwargs), where my_args and my_kwargs are required positional and keyword arguments. Rather than passing f0 as the callable, wrap it to accept only x; e.g., pass fun=lambda x: f0(x, *my_args, **my_kwargs) as the callable, where my_args (tuple) and my_kwargs (dict) have been gathered before invoking this function.

x0 : ndarray

Initial guess.

args : tuple, optional

Extra arguments passed to the objective function and its Jacobian.

method : str, optional

Type of solver. Should be one of

  • 'hybr' (see here) <optimize.root-hybr>

  • 'lm' (see here) <optimize.root-lm>

  • 'broyden1' (see here) <optimize.root-broyden1>

  • 'broyden2' (see here) <optimize.root-broyden2>

  • 'anderson' (see here) <optimize.root-anderson>

  • 'linearmixing' (see here) <optimize.root-linearmixing>

  • 'diagbroyden' (see here) <optimize.root-diagbroyden>

  • 'excitingmixing' (see here) <optimize.root-excitingmixing>

  • 'krylov' (see here) <optimize.root-krylov>

  • 'df-sane' (see here) <optimize.root-dfsane>

jac : bool or callable, optional

If jac is a Boolean and is True, fun is assumed to return the value of Jacobian along with the objective function. If False, the Jacobian will be estimated numerically. jac can also be a callable returning the Jacobian of fun. In this case, it must accept the same arguments as fun.

tol : float, optional

Tolerance for termination. For detailed control, use solver-specific options.

callback : function, optional

Optional callback function. It is called on every iteration as callback(x, f) where x is the current solution and f the corresponding residual. For all methods but 'hybr' and 'lm'.

options : dict, optional

A dictionary of solver options. E.g., xtol or maxiter, see show_options() for details.

Returns

sol : OptimizeResult

The solution represented as a OptimizeResult object. Important attributes are: x the solution array, success a Boolean flag indicating if the algorithm exited successfully and message which describes the cause of the termination. See OptimizeResult for a description of other attributes.

Notes

This section describes the available solvers that can be selected by the 'method' parameter. The default method is hybr.

Method hybr uses a modification of the Powell hybrid method as implemented in MINPACK [1].

Method lm solves the system of nonlinear equations in a least squares sense using a modification of the Levenberg-Marquardt algorithm as implemented in MINPACK [1].

Method df-sane is a derivative-free spectral method. [3]

Methods broyden1, broyden2, anderson, linearmixing, diagbroyden, excitingmixing, krylov are inexact Newton methods, with backtracking or full line searches [2]. Each method corresponds to a particular Jacobian approximations.

  • Method broyden1 uses Broyden's first Jacobian approximation, it is known as Broyden's good method.

  • Method broyden2 uses Broyden's second Jacobian approximation, it is known as Broyden's bad method.

  • Method anderson uses (extended) Anderson mixing.

  • Method Krylov uses Krylov approximation for inverse Jacobian. It is suitable for large-scale problem.

  • Method diagbroyden uses diagonal Broyden Jacobian approximation.

  • Method linearmixing uses a scalar Jacobian approximation.

  • Method excitingmixing uses a tuned diagonal Jacobian approximation.

Examples

The following functions define a system of nonlinear equations and its jacobian. 
import numpy as np
def fun(x):
    return [x[0]  + 0.5 * (x[0] - x[1])**3 - 1.0,
            0.5 * (x[1] - x[0])**3 + x[1]]

def jac(x):
    return np.array([[1 + 1.5 * (x[0] - x[1])**2,
                      -1.5 * (x[0] - x[1])**2],
                     [-1.5 * (x[1] - x[0])**2,
                      1 + 1.5 * (x[1] - x[0])**2]])
A solution can be obtained as follows. 
from scipy import optimize
sol = optimize.root(fun, [0, 0], jac=jac, method='hybr')
sol.x
**Large problem** Suppose that we needed to solve the following integrodifferential equation on the square :math:`[0,1]\times[0,1]`: .. math:: \nabla^2 P = 10 \left(\int_0^1\int_0^1\cosh(P)\,dx\,dy\right)^2 with :math:`P(x,1) = 1` and :math:`P=0` elsewhere on the boundary of the square. The solution can be found using the ``method='krylov'`` solver: 
from scipy import optimize
nx, ny = 75, 75
hx, hy = 1./(nx-1), 1./(ny-1)

P_left, P_right = 0, 0
P_top, P_bottom = 1, 0

def residual(P):
   d2x = np.zeros_like(P)
   d2y = np.zeros_like(P)

   d2x[1:-1] = (P[2:]   - 2*P[1:-1] + P[:-2]) / hx/hx
   d2x[0]    = (P[1]    - 2*P[0]    + P_left)/hx/hx
   d2x[-1]   = (P_right - 2*P[-1]   + P[-2])/hx/hx

   d2y[:,1:-1] = (P[:,2:] - 2*P[:,1:-1] + P[:,:-2])/hy/hy
   d2y[:,0]    = (P[:,1]  - 2*P[:,0]    + P_bottom)/hy/hy
   d2y[:,-1]   = (P_top   - 2*P[:,-1]   + P[:,-2])/hy/hy

   return d2x + d2y - 10*np.cosh(P).mean()**2

guess = np.zeros((nx, ny), float)
sol = optimize.root(residual, guess, method='krylov')
print('Residual: %g' % abs(residual(sol.x)).max())

import matplotlib.pyplot as plt
x, y = np.mgrid[0:1:(nx*1j), 0:1:(ny*1j)]
plt.pcolormesh(x, y, sol.x, shading='gouraud')
plt.colorbar()
plt.show()

See also

show_options

Additional options accepted by the solvers

Aliases

  • scipy.optimize.root

Referenced by

This package