### Table Of Contents

- Routines
- Examples
  - Small problem
  - Large problem

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scipy.optimize.minimize

This is a collection of general-purpose nonlinear multidimensional solvers. These solvers find x for which F(x) = 0. Both x and F can be multidimensional.

# Routines

Large-scale nonlinear solvers:

`newton_krylov`(F, xin[, iter, rdiff, method, ...]) | Find a root of a function, using Krylov approximation for inverse Jacobian.
`anderson`(F, xin[, iter, alpha, w0, M, ...]) | Find a root of a function, using (extended) Anderson mixing.

General nonlinear solvers:

`broyden1`(F, xin[, iter, alpha, ...]) | Find a root of a function, using Broyden’s first Jacobian approximation.
`broyden2`(F, xin[, iter, alpha, ...]) | Find a root of a function, using Broyden’s second Jacobian approximation.

Simple iterations:

`excitingmixing`(F, xin[, iter, alpha, ...]) | Find a root of a function, using a tuned diagonal Jacobian approximation.
`linearmixing`(F, xin[, iter, alpha, verbose, ...]) | Find a root of a function, using a scalar Jacobian approximation.
`diagbroyden`(F, xin[, iter, alpha, verbose, ...]) | Find a root of a function, using diagonal Broyden Jacobian approximation.

# Examples

## Small problem

    >>> def F(x):
    ...    return np.cos(x) + x[::-1] - [1, 2, 3, 4]
    >>> import scipy.optimize
    >>> x = scipy.optimize.broyden1(F, [1,1,1,1], f_tol=1e-14)
    >>> x
    array([ 4.04674914,  3.91158389,  2.71791677,  1.61756251])
    >>> np.cos(x) + x[::-1]
    array([ 1.,  2.,  3.,  4.])

## Large problem

Suppose that we needed to solve the following integrodifferential equation on the square \([0,1]\times[0,1]\):

\[\nabla^2 P = 10 \left(\int_0^1\int_0^1\cosh(P)\,dx\,dy\right)^2\]

with \(P(x,1) = 1\) and \(P=0\) elsewhere on the boundary of the square.

The solution can be found using the `newton_krylov` solver:

    import numpy as np
    from scipy.optimize import newton_krylov
    from numpy import cosh, zeros_like, mgrid, zeros

    # parameters
    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 = zeros_like(P)
        d2y = 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*cosh(P).mean()**2

    # solve
    guess = zeros((nx, ny), float)
    sol = newton_krylov(residual, guess, method='lgmres', verbose=1)
    print('Residual: %g' % abs(residual(sol)).max())

    # visualize
    import matplotlib.pyplot as plt
    x, y = mgrid[0:1:(nx*1j), 0:1:(ny*1j)]
    plt.pcolor(x, y, sol)
    plt.colorbar()
    plt.show()

(Source code)