The `scipy.optimize` package provides several commonly used
optimization algorithms. An detailed listing is available:
`scipy.optimize` (can also be found by `help(scipy.optimize)`).

The module contains:

- Unconstrained and constrained minimization and least-squares algorithms
(e.g.,
`fmin`: Nelder-Mead simplex,`fmin_bfgs`: BFGS,`fmin_ncg`: Newton Conjugate Gradient,`leastsq`: Levenberg-Marquardt,`fmin_cobyla`: COBYLA). - Global (brute-force) optimization routines (e.g.,
`anneal`) - Curve fitting (
`curve_fit`) - Scalar function minimizers and root finders (e.g., Brent’s method
`fminbound`, and`newton`) - Multivariate equation system solvers (
`fsolve`) - Large-scale multivariate equation system solvers (e.g.
`newton_krylov`)

Below, several examples demonstrate their basic usage.

The simplex algorithm is probably the simplest way to minimize a fairly well-behaved function. The simplex algorithm requires only function evaluations and is a good choice for simple minimization problems. However, because it does not use any gradient evaluations, it may take longer to find the minimum. To demonstrate the minimization function consider the problem of minimizing the Rosenbrock function of variables:

The minimum value of this function is 0 which is achieved when This minimum can be found using the `fmin` routine as shown in the example below:

```
>>> from scipy.optimize import fmin
>>> def rosen(x):
... """The Rosenbrock function"""
... return sum(100.0*(x[1:]-x[:-1]**2.0)**2.0 + (1-x[:-1])**2.0)
```

```
>>> x0 = [1.3, 0.7, 0.8, 1.9, 1.2]
>>> xopt = fmin(rosen, x0, xtol=1e-8)
Optimization terminated successfully.
Current function value: 0.000000
Iterations: 339
Function evaluations: 571
```

```
>>> print xopt
[ 1. 1. 1. 1. 1.]
```

Another optimization algorithm that needs only function calls to find
the minimum is Powell’s method available as `fmin_powell`.

In order to converge more quickly to the solution, this routine uses the gradient of the objective function. If the gradient is not given by the user, then it is estimated using first-differences. The Broyden-Fletcher-Goldfarb-Shanno (BFGS) method typically requires fewer function calls than the simplex algorithm even when the gradient must be estimated.

To demonstrate this algorithm, the Rosenbrock function is again used. The gradient of the Rosenbrock function is the vector:

This expression is valid for the interior derivatives. Special cases are

A Python function which computes this gradient is constructed by the code-segment:

```
>>> def rosen_der(x):
... xm = x[1:-1]
... xm_m1 = x[:-2]
... xm_p1 = x[2:]
... der = zeros_like(x)
... der[1:-1] = 200*(xm-xm_m1**2) - 400*(xm_p1 - xm**2)*xm - 2*(1-xm)
... der[0] = -400*x[0]*(x[1]-x[0]**2) - 2*(1-x[0])
... der[-1] = 200*(x[-1]-x[-2]**2)
... return der
```

The calling signature for the BFGS minimization algorithm is similar
to `fmin` with the addition of the *fprime* argument. An example
usage of `fmin_bfgs` is shown in the following example which
minimizes the Rosenbrock function.

```
>>> from scipy.optimize import fmin_bfgs
```

```
>>> x0 = [1.3, 0.7, 0.8, 1.9, 1.2]
>>> xopt = fmin_bfgs(rosen, x0, fprime=rosen_der)
Optimization terminated successfully.
Current function value: 0.000000
Iterations: 53
Function evaluations: 65
Gradient evaluations: 65
>>> print xopt
[ 1. 1. 1. 1. 1.]
```

The method which requires the fewest function calls and is therefore
often the fastest method to minimize functions of many variables is
`fmin_ncg`. This method is a modified Newton’s method and uses a
conjugate gradient algorithm to (approximately) invert the local
Hessian. Newton’s method is based on fitting the function locally to
a quadratic form:

where is a matrix of second-derivatives (the Hessian). If the Hessian is positive definite then the local minimum of this function can be found by setting the gradient of the quadratic form to zero, resulting in

The inverse of the Hessian is evaluted using the conjugate-gradient method. An example of employing this method to minimizing the Rosenbrock function is given below. To take full advantage of the NewtonCG method, a function which computes the Hessian must be provided. The Hessian matrix itself does not need to be constructed, only a vector which is the product of the Hessian with an arbitrary vector needs to be available to the minimization routine. As a result, the user can provide either a function to compute the Hessian matrix, or a function to compute the product of the Hessian with an arbitrary vector.

The Hessian of the Rosenbrock function is

if with defining the matrix. Other non-zero entries of the matrix are

For example, the Hessian when is

The code which computes this Hessian along with the code to minimize
the function using `fmin_ncg` is shown in the following example:

```
>>> from scipy.optimize import fmin_ncg
>>> def rosen_hess(x):
... x = asarray(x)
... H = diag(-400*x[:-1],1) - diag(400*x[:-1],-1)
... diagonal = zeros_like(x)
... diagonal[0] = 1200*x[0]-400*x[1]+2
... diagonal[-1] = 200
... diagonal[1:-1] = 202 + 1200*x[1:-1]**2 - 400*x[2:]
... H = H + diag(diagonal)
... return H
```

```
>>> x0 = [1.3, 0.7, 0.8, 1.9, 1.2]
>>> xopt = fmin_ncg(rosen, x0, rosen_der, fhess=rosen_hess, avextol=1e-8)
Optimization terminated successfully.
Current function value: 0.000000
Iterations: 23
Function evaluations: 26
Gradient evaluations: 23
Hessian evaluations: 23
>>> print xopt
[ 1. 1. 1. 1. 1.]
```

For larger minimization problems, storing the entire Hessian matrix
can consume considerable time and memory. The Newton-CG algorithm only
needs the product of the Hessian times an arbitrary vector. As a
result, the user can supply code to compute this product rather than
the full Hessian by setting the *fhess_p* keyword to the desired
function. The *fhess_p* function should take the minimization vector as
the first argument and the arbitrary vector as the second
argument. Any extra arguments passed to the function to be minimized
will also be passed to this function. If possible, using Newton-CG
with the hessian product option is probably the fastest way to
minimize the function.

In this case, the product of the Rosenbrock Hessian with an arbitrary vector is not difficult to compute. If is the arbitrary vector, then has elements:

Code which makes use of the *fhess_p* keyword to minimize the
Rosenbrock function using `fmin_ncg` follows:

```
>>> from scipy.optimize import fmin_ncg
>>> def rosen_hess_p(x,p):
... x = asarray(x)
... Hp = zeros_like(x)
... Hp[0] = (1200*x[0]**2 - 400*x[1] + 2)*p[0] - 400*x[0]*p[1]
... Hp[1:-1] = -400*x[:-2]*p[:-2]+(202+1200*x[1:-1]**2-400*x[2:])*p[1:-1] \
... -400*x[1:-1]*p[2:]
... Hp[-1] = -400*x[-2]*p[-2] + 200*p[-1]
... return Hp
```

```
>>> x0 = [1.3, 0.7, 0.8, 1.9, 1.2]
>>> xopt = fmin_ncg(rosen, x0, rosen_der, fhess_p=rosen_hess_p, avextol=1e-8)
Optimization terminated successfully.
Current function value: 0.000000
Iterations: 22
Function evaluations: 25
Gradient evaluations: 22
Hessian evaluations: 54
>>> print xopt
[ 1. 1. 1. 1. 1.]
```

All of the previously-explained minimization procedures can be used to solve a least-squares problem provided the appropriate objective function is constructed. For example, suppose it is desired to fit a set of data to a known model, where is a vector of parameters for the model that need to be found. A common method for determining which parameter vector gives the best fit to the data is to minimize the sum of squares of the residuals. The residual is usually defined for each observed data-point as

An objective function to pass to any of the previous minization algorithms to obtain a least-squares fit is.

The `leastsq` algorithm performs this squaring and summing of the
residuals automatically. It takes as an input argument the vector
function and returns the
value of which minimizes
directly. The user is also encouraged to provide the Jacobian matrix
of the function (with derivatives down the columns or across the
rows). If the Jacobian is not provided, it is estimated.

An example should clarify the usage. Suppose it is believed some measured data follow a sinusoidal pattern

where the parameters , and are unknown. The residual vector is

By defining a function to compute the residuals and (selecting an appropriate starting position), the least-squares fit routine can be used to find the best-fit parameters . This is shown in the following example:

```
>>> from numpy import *
>>> x = arange(0,6e-2,6e-2/30)
>>> A,k,theta = 10, 1.0/3e-2, pi/6
>>> y_true = A*sin(2*pi*k*x+theta)
>>> y_meas = y_true + 2*random.randn(len(x))
```

```
>>> def residuals(p, y, x):
... A,k,theta = p
... err = y-A*sin(2*pi*k*x+theta)
... return err
```

```
>>> def peval(x, p):
... return p[0]*sin(2*pi*p[1]*x+p[2])
```

```
>>> p0 = [8, 1/2.3e-2, pi/3]
>>> print array(p0)
[ 8. 43.4783 1.0472]
```

```
>>> from scipy.optimize import leastsq
>>> plsq = leastsq(residuals, p0, args=(y_meas, x))
>>> print plsq[0]
[ 10.9437 33.3605 0.5834]
```

```
>>> print array([A, k, theta])
[ 10. 33.3333 0.5236]
```

```
>>> import matplotlib.pyplot as plt
>>> plt.plot(x,peval(x,plsq[0]),x,y_meas,'o',x,y_true)
>>> plt.title('Least-squares fit to noisy data')
>>> plt.legend(['Fit', 'Noisy', 'True'])
>>> plt.show()
```

This module implements the Sequential Least SQuares Programming optimization algorithm (SLSQP).

The following script shows examples for how constraints can be specified.

```
"""
This script tests fmin_slsqp using Example 14.4 from Numerical Methods for
Engineers by Steven Chapra and Raymond Canale. This example maximizes the
function f(x) = 2*x*y + 2*x - x**2 - 2*y**2, which has a maximum at x=2,y=1.
"""
from scipy.optimize import fmin_slsqp
from numpy import array, asfarray, finfo,ones, sqrt, zeros
def testfunc(d,*args):
"""
Arguments:
d - A list of two elements, where d[0] represents x and
d[1] represents y in the following equation.
sign - A multiplier for f. Since we want to optimize it, and the scipy
optimizers can only minimize functions, we need to multiply it by
-1 to achieve the desired solution
Returns:
2*x*y + 2*x - x**2 - 2*y**2
"""
try:
sign = args[0]
except:
sign = 1.0
x = d[0]
y = d[1]
return sign*(2*x*y + 2*x - x**2 - 2*y**2)
def testfunc_deriv(d,*args):
""" This is the derivative of testfunc, returning a numpy array
representing df/dx and df/dy
"""
try:
sign = args[0]
except:
sign = 1.0
x = d[0]
y = d[1]
dfdx = sign*(-2*x + 2*y + 2)
dfdy = sign*(2*x - 4*y)
return array([ dfdx, dfdy ],float)
from time import time
print '\n\n'
print "Unbounded optimization. Derivatives approximated."
t0 = time()
x = fmin_slsqp(testfunc, [-1.0,1.0], args=(-1.0,), iprint=2, full_output=1)
print "Elapsed time:", 1000*(time()-t0), "ms"
print "Results",x
print "\n\n"
print "Unbounded optimization. Derivatives provided."
t0 = time()
x = fmin_slsqp(testfunc, [-1.0,1.0], args=(-1.0,), iprint=2, full_output=1)
print "Elapsed time:", 1000*(time()-t0), "ms"
print "Results",x
print "\n\n"
print "Bound optimization. Derivatives approximated."
t0 = time()
x = fmin_slsqp(testfunc, [-1.0,1.0], args=(-1.0,),
eqcons=[lambda x, y: x[0]-x[1] ], iprint=2, full_output=1)
print "Elapsed time:", 1000*(time()-t0), "ms"
print "Results",x
print "\n\n"
print "Bound optimization (equality constraints). Derivatives provided."
t0 = time()
x = fmin_slsqp(testfunc, [-1.0,1.0], fprime=testfunc_deriv, args=(-1.0,),
eqcons=[lambda x, y: x[0]-x[1] ], iprint=2, full_output=1)
print "Elapsed time:", 1000*(time()-t0), "ms"
print "Results",x
print "\n\n"
print "Bound optimization (equality and inequality constraints)."
print "Derivatives provided."
t0 = time()
x = fmin_slsqp(testfunc,[-1.0,1.0], fprime=testfunc_deriv, args=(-1.0,),
eqcons=[lambda x, y: x[0]-x[1] ],
ieqcons=[lambda x, y: x[0]-.5], iprint=2, full_output=1)
print "Elapsed time:", 1000*(time()-t0), "ms"
print "Results",x
print "\n\n"
def test_eqcons(d,*args):
try:
sign = args[0]
except:
sign = 1.0
x = d[0]
y = d[1]
return array([ x**3-y ])
def test_ieqcons(d,*args):
try:
sign = args[0]
except:
sign = 1.0
x = d[0]
y = d[1]
return array([ y-1 ])
print "Bound optimization (equality and inequality constraints)."
print "Derivatives provided via functions."
t0 = time()
x = fmin_slsqp(testfunc, [-1.0,1.0], fprime=testfunc_deriv, args=(-1.0,),
f_eqcons=test_eqcons, f_ieqcons=test_ieqcons,
iprint=2, full_output=1)
print "Elapsed time:", 1000*(time()-t0), "ms"
print "Results",x
print "\n\n"
def test_fprime_eqcons(d,*args):
try:
sign = args[0]
except:
sign = 1.0
x = d[0]
y = d[1]
return array([ 3.0*(x**2.0), -1.0 ])
def test_fprime_ieqcons(d,*args):
try:
sign = args[0]
except:
sign = 1.0
x = d[0]
y = d[1]
return array([ 0.0, 1.0 ])
print "Bound optimization (equality and inequality constraints)."
print "Derivatives provided via functions."
print "Constraint jacobians provided via functions"
t0 = time()
x = fmin_slsqp(testfunc,[-1.0,1.0], fprime=testfunc_deriv, args=(-1.0,),
f_eqcons=test_eqcons, f_ieqcons=test_ieqcons,
fprime_eqcons=test_fprime_eqcons,
fprime_ieqcons=test_fprime_ieqcons, iprint=2, full_output=1)
print "Elapsed time:", 1000*(time()-t0), "ms"
print "Results",x
print "\n\n"
```

Often only the minimum of a scalar function is needed (a scalar function is one that takes a scalar as input and returns a scalar output). In these circumstances, other optimization techniques have been developed that can work faster.

There are actually two methods that can be used to minimize a scalar
function (`brent` and `golden`), but `golden` is
included only for academic purposes and should rarely be used. The
brent method uses Brent’s algorithm for locating a minimum. Optimally
a bracket should be given which contains the minimum desired. A
bracket is a triple such that
and
. If this is not given, then alternatively two starting
points can be chosen and a bracket will be found from these points
using a simple marching algorithm. If these two starting points are
not provided 0 and 1 will be used (this may not be the right choice
for your function and result in an unexpected minimum being returned).

Thus far all of the minimization routines described have been
unconstrained minimization routines. Very often, however, there are
constraints that can be placed on the solution space before
minimization occurs. The `fminbound` function is an example of a
constrained minimization procedure that provides a rudimentary
interval constraint for scalar functions. The interval constraint
allows the minimization to occur only between two fixed endpoints.

For example, to find the minimum of near , `fminbound` can be called using the interval as a constraint. The result is :

```
>>> from scipy.special import j1
>>> from scipy.optimize import fminbound
>>> xmin = fminbound(j1, 4, 7)
>>> print xmin
5.33144184241
```

To find the roots of a polynomial, the command `roots` is useful. To find a root of a set of non-linear
equations, the command `fsolve` is needed. For example, the
following example finds the roots of the single-variable
transcendental equation

and the set of non-linear equations

The results are and .

```
>>> def func(x):
... return x + 2*cos(x)
```

```
>>> def func2(x):
... out = [x[0]*cos(x[1]) - 4]
... out.append(x[1]*x[0] - x[1] - 5)
... return out
```

```
>>> from scipy.optimize import fsolve
>>> x0 = fsolve(func, 0.3)
>>> print x0
-1.02986652932
```

```
>>> x02 = fsolve(func2, [1, 1])
>>> print x02
[ 6.50409711 0.90841421]
```

If one has a single-variable equation, there are four different root
finder algorithms that can be tried. Each of these root finding
algorithms requires the endpoints of an interval where a root is
suspected (because the function changes signs). In general
`brentq` is the best choice, but the other methods may be useful
in certain circumstances or for academic purposes.

A problem closely related to finding the zeros of a function is the
problem of finding a fixed-point of a function. A fixed point of a
function is the point at which evaluation of the function returns the
point: Clearly the fixed point of
is the root of
Equivalently, the root of is the fixed_point of
The routine
`fixed_point` provides a simple iterative method using Aitkens
sequence acceleration to estimate the fixed point of given a
starting point.

The `fsolve` function cannot deal with a very large number of
variables (*N*), as it needs to calculate and invert a dense *N x N*
Jacobian matrix on every Newton step. This becomes rather inefficent
when *N* grows.

Consider for instance the following problem: we need to solve the following integrodifferential equation on the square :

with the boundary condition on the upper edge and
elsewhere on the boundary of the square. This can be done
by approximating the continuous function *P* by its values on a grid,
, with a small grid spacing
*h*. The derivatives and integrals can then be approximated; for
instance . The problem is then equivalent to finding the root of
some function *residual(P)*, where *P* is a vector of length
.

Now, because can be large, `fsolve` will take a
long time to solve this problem. The solution can however be found
using one of the large-scale solvers in `scipy.optimize`, for
example `newton_krylov`, `broyden2`, or
`anderson`. These use what is known as the inexact Newton method,
which instead of computing the Jacobian matrix exactly, forms an
approximation for it.

The problem we have can now be solved as follows:

```
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 + 5*cosh(P).mean()**2
# solve
guess = zeros((nx, ny), float)
sol = newton_krylov(residual, guess, verbose=1)
#sol = broyden2(residual, guess, max_rank=50, verbose=1)
#sol = anderson(residual, guess, M=10, verbose=1)
print 'Residual', 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()
```

When looking for the zero of the functions ,
*i = 1, 2, ..., N*, the `newton_krylov` solver spends most of its
time inverting the Jacobian matrix,

If you have an approximation for the inverse matrix
, you can use it for *preconditioning* the
linear inversion problem. The idea is that instead of solving
one solves : since
matrix is “closer” to the identity matrix than
is, the equation should be easier for the Krylov method to deal with.

The matrix *M* can be passed to `newton_krylov` as the *inner_M*
parameter. It can be a (sparse) matrix or a
`scipy.sparse.linalg.LinearOperator` instance.

For the problem in the previous section, we note that the function to solve consists of two parts: the first one is application of the Laplace operator, , and the second is the integral. We can actually easily compute the Jacobian corresponding to the Laplace operator part: we know that in one dimension

so that the whole 2-D operator is represented by

The matrix of the Jacobian corresponding to the integral
is more difficult to calculate, and since *all* of it entries are
nonzero, it will be difficult to invert. on the other hand
is a relatively simple matrix, and can be inverted by
`scipy.sparse.linalg.splu` (or the inverse can be approximated by
`scipy.sparse.linalg.spilu`). So we are content to take
and hope for the best.

In the example below, we use the preconditioner .

```
import numpy as np
from scipy.optimize import newton_krylov
from scipy.sparse import spdiags, spkron
from scipy.sparse.linalg import spilu, LinearOperator
from numpy import cosh, zeros_like, mgrid, zeros, eye
# 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 get_preconditioner():
"""Compute the preconditioner M"""
diags_x = zeros((3, nx))
diags_x[0,:] = 1/hx/hx
diags_x[1,:] = -2/hx/hx
diags_x[2,:] = 1/hx/hx
Lx = spdiags(diags_x, [-1,0,1], nx, nx)
diags_y = zeros((3, ny))
diags_y[0,:] = 1/hy/hy
diags_y[1,:] = -2/hy/hy
diags_y[2,:] = 1/hy/hy
Ly = spdiags(diags_y, [-1,0,1], ny, ny)
J1 = spkron(Lx, eye(ny)) + spkron(eye(nx), Ly)
# Now we have the matrix `J_1`. We need to find its inverse `M` --
# however, since an approximate inverse is enough, we can use
# the *incomplete LU* decomposition
J1_ilu = spilu(J1)
# This returns an object with a method .solve() that evaluates
# the corresponding matrix-vector product. We need to wrap it into
# a LinearOperator before it can be passed to the Krylov methods:
M = LinearOperator(shape=(nx*ny, nx*ny), matvec=J1_ilu.solve)
return M
def solve(preconditioning=True):
"""Compute the solution"""
count = [0]
def residual(P):
count[0] += 1
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 + 5*cosh(P).mean()**2
# preconditioner
if preconditioning:
M = get_preconditioner()
else:
M = None
# solve
guess = zeros((nx, ny), float)
sol = newton_krylov(residual, guess, verbose=1, inner_M=M)
print 'Residual', abs(residual(sol)).max()
print 'Evaluations', count[0]
return sol
def main():
sol = solve(preconditioning=True)
# visualize
import matplotlib.pyplot as plt
x, y = mgrid[0:1:(nx*1j), 0:1:(ny*1j)]
plt.clf()
plt.pcolor(x, y, sol)
plt.clim(0, 1)
plt.colorbar()
plt.show()
if __name__ == "__main__":
main()
```

Resulting run, first without preconditioning:

```
0: |F(x)| = 803.614; step 1; tol 0.000257947
1: |F(x)| = 345.912; step 1; tol 0.166755
2: |F(x)| = 139.159; step 1; tol 0.145657
3: |F(x)| = 27.3682; step 1; tol 0.0348109
4: |F(x)| = 1.03303; step 1; tol 0.00128227
5: |F(x)| = 0.0406634; step 1; tol 0.00139451
6: |F(x)| = 0.00344341; step 1; tol 0.00645373
7: |F(x)| = 0.000153671; step 1; tol 0.00179246
8: |F(x)| = 6.7424e-06; step 1; tol 0.00173256
Residual 3.57078908664e-07
Evaluations 317
```

and then with preconditioning:

```
0: |F(x)| = 136.993; step 1; tol 7.49599e-06
1: |F(x)| = 4.80983; step 1; tol 0.00110945
2: |F(x)| = 0.195942; step 1; tol 0.00149362
3: |F(x)| = 0.000563597; step 1; tol 7.44604e-06
4: |F(x)| = 1.00698e-09; step 1; tol 2.87308e-12
Residual 9.29603061195e-11
Evaluations 77
```

Using a preconditioner reduced the number of evaluations of the
*residual* function by a factor of *4*. For problems where the
residual is expensive to compute, good preconditioning can be crucial
— it can even decide whether the problem is solvable in practice or
not.

Preconditioning is an art, science, and industry. Here, we were lucky in making a simple choice that worked reasonably well, but there is a lot more depth to this topic than is shown here.

References

Some further reading and related software:

[KK] | D.A. Knoll and D.E. Keyes, “Jacobian-free Newton-Krylov methods”, J. Comp. Phys. 193, 357 (2003). |

[PP] | PETSc http://www.mcs.anl.gov/petsc/ and its Python bindings http://code.google.com/p/petsc4py/ |

[AMG] | PyAMG (algebraic multigrid preconditioners/solvers) http://code.google.com/p/pyamg/ |