You can not select more than 25 topics
Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
289 lines
9.7 KiB
289 lines
9.7 KiB
5 months ago
|
from bisect import bisect
|
||
|
|
from ..libmp.backend import xrange
|
||
|
|
|
||
|
|
class ODEMethods(object):
|
||
|
|
pass
|
||
|
|
|
||
|
|
def ode_taylor(ctx, derivs, x0, y0, tol_prec, n):
|
||
|
|
h = tol = ctx.ldexp(1, -tol_prec)
|
||
|
|
dim = len(y0)
|
||
|
|
xs = [x0]
|
||
|
|
ys = [y0]
|
||
|
|
x = x0
|
||
|
|
y = y0
|
||
|
|
orig = ctx.prec
|
||
|
|
try:
|
||
|
|
ctx.prec = orig*(1+n)
|
||
|
|
# Use n steps with Euler's method to get
|
||
|
|
# evaluation points for derivatives
|
||
|
|
for i in range(n):
|
||
|
|
fxy = derivs(x, y)
|
||
|
|
y = [y[i]+h*fxy[i] for i in xrange(len(y))]
|
||
|
|
x += h
|
||
|
|
xs.append(x)
|
||
|
|
ys.append(y)
|
||
|
|
# Compute derivatives
|
||
|
|
ser = [[] for d in range(dim)]
|
||
|
|
for j in range(n+1):
|
||
|
|
s = [0]*dim
|
||
|
|
b = (-1) ** (j & 1)
|
||
|
|
k = 1
|
||
|
|
for i in range(j+1):
|
||
|
|
for d in range(dim):
|
||
|
|
s[d] += b * ys[i][d]
|
||
|
|
b = (b * (j-k+1)) // (-k)
|
||
|
|
k += 1
|
||
|
|
scale = h**(-j) / ctx.fac(j)
|
||
|
|
for d in range(dim):
|
||
|
|
s[d] = s[d] * scale
|
||
|
|
ser[d].append(s[d])
|
||
|
|
finally:
|
||
|
|
ctx.prec = orig
|
||
|
|
# Estimate radius for which we can get full accuracy.
|
||
|
|
# XXX: do this right for zeros
|
||
|
|
radius = ctx.one
|
||
|
|
for ts in ser:
|
||
|
|
if ts[-1]:
|
||
|
|
radius = min(radius, ctx.nthroot(tol/abs(ts[-1]), n))
|
||
|
|
radius /= 2 # XXX
|
||
|
|
return ser, x0+radius
|
||
|
|
|
||
|
|
def odefun(ctx, F, x0, y0, tol=None, degree=None, method='taylor', verbose=False):
|
||
|
|
r"""
|
||
|
|
Returns a function `y(x) = [y_0(x), y_1(x), \ldots, y_n(x)]`
|
||
|
|
that is a numerical solution of the `n+1`-dimensional first-order
|
||
|
|
ordinary differential equation (ODE) system
|
||
|
|
|
||
|
|
.. math ::
|
||
|
|
|
||
|
|
y_0'(x) = F_0(x, [y_0(x), y_1(x), \ldots, y_n(x)])
|
||
|
|
|
||
|
|
y_1'(x) = F_1(x, [y_0(x), y_1(x), \ldots, y_n(x)])
|
||
|
|
|
||
|
|
\vdots
|
||
|
|
|
||
|
|
y_n'(x) = F_n(x, [y_0(x), y_1(x), \ldots, y_n(x)])
|
||
|
|
|
||
|
|
The derivatives are specified by the vector-valued function
|
||
|
|
*F* that evaluates
|
||
|
|
`[y_0', \ldots, y_n'] = F(x, [y_0, \ldots, y_n])`.
|
||
|
|
The initial point `x_0` is specified by the scalar argument *x0*,
|
||
|
|
and the initial value `y(x_0) = [y_0(x_0), \ldots, y_n(x_0)]` is
|
||
|
|
specified by the vector argument *y0*.
|
||
|
|
|
||
|
|
For convenience, if the system is one-dimensional, you may optionally
|
||
|
|
provide just a scalar value for *y0*. In this case, *F* should accept
|
||
|
|
a scalar *y* argument and return a scalar. The solution function
|
||
|
|
*y* will return scalar values instead of length-1 vectors.
|
||
|
|
|
||
|
|
Evaluation of the solution function `y(x)` is permitted
|
||
|
|
for any `x \ge x_0`.
|
||
|
|
|
||
|
|
A high-order ODE can be solved by transforming it into first-order
|
||
|
|
vector form. This transformation is described in standard texts
|
||
|
|
on ODEs. Examples will also be given below.
|
||
|
|
|
||
|
|
**Options, speed and accuracy**
|
||
|
|
|
||
|
|
By default, :func:`~mpmath.odefun` uses a high-order Taylor series
|
||
|
|
method. For reasonably well-behaved problems, the solution will
|
||
|
|
be fully accurate to within the working precision. Note that
|
||
|
|
*F* must be possible to evaluate to very high precision
|
||
|
|
for the generation of Taylor series to work.
|
||
|
|
|
||
|
|
To get a faster but less accurate solution, you can set a large
|
||
|
|
value for *tol* (which defaults roughly to *eps*). If you just
|
||
|
|
want to plot the solution or perform a basic simulation,
|
||
|
|
*tol = 0.01* is likely sufficient.
|
||
|
|
|
||
|
|
The *degree* argument controls the degree of the solver (with
|
||
|
|
*method='taylor'*, this is the degree of the Taylor series
|
||
|
|
expansion). A higher degree means that a longer step can be taken
|
||
|
|
before a new local solution must be generated from *F*,
|
||
|
|
meaning that fewer steps are required to get from `x_0` to a given
|
||
|
|
`x_1`. On the other hand, a higher degree also means that each
|
||
|
|
local solution becomes more expensive (i.e., more evaluations of
|
||
|
|
*F* are required per step, and at higher precision).
|
||
|
|
|
||
|
|
The optimal setting therefore involves a tradeoff. Generally,
|
||
|
|
decreasing the *degree* for Taylor series is likely to give faster
|
||
|
|
solution at low precision, while increasing is likely to be better
|
||
|
|
at higher precision.
|
||
|
|
|
||
|
|
The function
|
||
|
|
object returned by :func:`~mpmath.odefun` caches the solutions at all step
|
||
|
|
points and uses polynomial interpolation between step points.
|
||
|
|
Therefore, once `y(x_1)` has been evaluated for some `x_1`,
|
||
|
|
`y(x)` can be evaluated very quickly for any `x_0 \le x \le x_1`.
|
||
|
|
and continuing the evaluation up to `x_2 > x_1` is also fast.
|
||
|
|
|
||
|
|
**Examples of first-order ODEs**
|
||
|
|
|
||
|
|
We will solve the standard test problem `y'(x) = y(x), y(0) = 1`
|
||
|
|
which has explicit solution `y(x) = \exp(x)`::
|
||
|
|
|
||
|
|
>>> from mpmath import *
|
||
|
|
>>> mp.dps = 15; mp.pretty = True
|
||
|
|
>>> f = odefun(lambda x, y: y, 0, 1)
|
||
|
|
>>> for x in [0, 1, 2.5]:
|
||
|
|
... print((f(x), exp(x)))
|
||
|
|
...
|
||
|
|
(1.0, 1.0)
|
||
|
|
(2.71828182845905, 2.71828182845905)
|
||
|
|
(12.1824939607035, 12.1824939607035)
|
||
|
|
|
||
|
|
The solution with high precision::
|
||
|
|
|
||
|
|
>>> mp.dps = 50
|
||
|
|
>>> f = odefun(lambda x, y: y, 0, 1)
|
||
|
|
>>> f(1)
|
||
|
|
2.7182818284590452353602874713526624977572470937
|
||
|
|
>>> exp(1)
|
||
|
|
2.7182818284590452353602874713526624977572470937
|
||
|
|
|
||
|
|
Using the more general vectorized form, the test problem
|
||
|
|
can be input as (note that *f* returns a 1-element vector)::
|
||
|
|
|
||
|
|
>>> mp.dps = 15
|
||
|
|
>>> f = odefun(lambda x, y: [y[0]], 0, [1])
|
||
|
|
>>> f(1)
|
||
|
|
[2.71828182845905]
|
||
|
|
|
||
|
|
:func:`~mpmath.odefun` can solve nonlinear ODEs, which are generally
|
||
|
|
impossible (and at best difficult) to solve analytically. As
|
||
|
|
an example of a nonlinear ODE, we will solve `y'(x) = x \sin(y(x))`
|
||
|
|
for `y(0) = \pi/2`. An exact solution happens to be known
|
||
|
|
for this problem, and is given by
|
||
|
|
`y(x) = 2 \tan^{-1}\left(\exp\left(x^2/2\right)\right)`::
|
||
|
|
|
||
|
|
>>> f = odefun(lambda x, y: x*sin(y), 0, pi/2)
|
||
|
|
>>> for x in [2, 5, 10]:
|
||
|
|
... print((f(x), 2*atan(exp(mpf(x)**2/2))))
|
||
|
|
...
|
||
|
|
(2.87255666284091, 2.87255666284091)
|
||
|
|
(3.14158520028345, 3.14158520028345)
|
||
|
|
(3.14159265358979, 3.14159265358979)
|
||
|
|
|
||
|
|
If `F` is independent of `y`, an ODE can be solved using direct
|
||
|
|
integration. We can therefore obtain a reference solution with
|
||
|
|
:func:`~mpmath.quad`::
|
||
|
|
|
||
|
|
>>> f = lambda x: (1+x**2)/(1+x**3)
|
||
|
|
>>> g = odefun(lambda x, y: f(x), pi, 0)
|
||
|
|
>>> g(2*pi)
|
||
|
|
0.72128263801696
|
||
|
|
>>> quad(f, [pi, 2*pi])
|
||
|
|
0.72128263801696
|
||
|
|
|
||
|
|
**Examples of second-order ODEs**
|
||
|
|
|
||
|
|
We will solve the harmonic oscillator equation `y''(x) + y(x) = 0`.
|
||
|
|
To do this, we introduce the helper functions `y_0 = y, y_1 = y_0'`
|
||
|
|
whereby the original equation can be written as `y_1' + y_0' = 0`. Put
|
||
|
|
together, we get the first-order, two-dimensional vector ODE
|
||
|
|
|
||
|
|
.. math ::
|
||
|
|
|
||
|
|
\begin{cases}
|
||
|
|
y_0' = y_1 \\
|
||
|
|
y_1' = -y_0
|
||
|
|
\end{cases}
|
||
|
|
|
||
|
|
To get a well-defined IVP, we need two initial values. With
|
||
|
|
`y(0) = y_0(0) = 1` and `-y'(0) = y_1(0) = 0`, the problem will of
|
||
|
|
course be solved by `y(x) = y_0(x) = \cos(x)` and
|
||
|
|
`-y'(x) = y_1(x) = \sin(x)`. We check this::
|
||
|
|
|
||
|
|
>>> f = odefun(lambda x, y: [-y[1], y[0]], 0, [1, 0])
|
||
|
|
>>> for x in [0, 1, 2.5, 10]:
|
||
|
|
... nprint(f(x), 15)
|
||
|
|
... nprint([cos(x), sin(x)], 15)
|
||
|
|
... print("---")
|
||
|
|
...
|
||
|
|
[1.0, 0.0]
|
||
|
|
[1.0, 0.0]
|
||
|
|
---
|
||
|
|
[0.54030230586814, 0.841470984807897]
|
||
|
|
[0.54030230586814, 0.841470984807897]
|
||
|
|
---
|
||
|
|
[-0.801143615546934, 0.598472144103957]
|
||
|
|
[-0.801143615546934, 0.598472144103957]
|
||
|
|
---
|
||
|
|
[-0.839071529076452, -0.54402111088937]
|
||
|
|
[-0.839071529076452, -0.54402111088937]
|
||
|
|
---
|
||
|
|
|
||
|
|
Note that we get both the sine and the cosine solutions
|
||
|
|
simultaneously.
|
||
|
|
|
||
|
|
**TODO**
|
||
|
|
|
||
|
|
* Better automatic choice of degree and step size
|
||
|
|
* Make determination of Taylor series convergence radius
|
||
|
|
more robust
|
||
|
|
* Allow solution for `x < x_0`
|
||
|
|
* Allow solution for complex `x`
|
||
|
|
* Test for difficult (ill-conditioned) problems
|
||
|
|
* Implement Runge-Kutta and other algorithms
|
||
|
|
|
||
|
|
"""
|
||
|
|
if tol:
|
||
|
|
tol_prec = int(-ctx.log(tol, 2))+10
|
||
|
|
else:
|
||
|
|
tol_prec = ctx.prec+10
|
||
|
|
degree = degree or (3 + int(3*ctx.dps/2.))
|
||
|
|
workprec = ctx.prec + 40
|
||
|
|
try:
|
||
|
|
len(y0)
|
||
|
|
return_vector = True
|
||
|
|
except TypeError:
|
||
|
|
F_ = F
|
||
|
|
F = lambda x, y: [F_(x, y[0])]
|
||
|
|
y0 = [y0]
|
||
|
|
return_vector = False
|
||
|
|
ser, xb = ode_taylor(ctx, F, x0, y0, tol_prec, degree)
|
||
|
|
series_boundaries = [x0, xb]
|
||
|
|
series_data = [(ser, x0, xb)]
|
||
|
|
# We will be working with vectors of Taylor series
|
||
|
|
def mpolyval(ser, a):
|
||
|
|
return [ctx.polyval(s[::-1], a) for s in ser]
|
||
|
|
# Find nearest expansion point; compute if necessary
|
||
|
|
def get_series(x):
|
||
|
|
if x < x0:
|
||
|
|
raise ValueError
|
||
|
|
n = bisect(series_boundaries, x)
|
||
|
|
if n < len(series_boundaries):
|
||
|
|
return series_data[n-1]
|
||
|
|
while 1:
|
||
|
|
ser, xa, xb = series_data[-1]
|
||
|
|
if verbose:
|
||
|
|
print("Computing Taylor series for [%f, %f]" % (xa, xb))
|
||
|
|
y = mpolyval(ser, xb-xa)
|
||
|
|
xa = xb
|
||
|
|
ser, xb = ode_taylor(ctx, F, xb, y, tol_prec, degree)
|
||
|
|
series_boundaries.append(xb)
|
||
|
|
series_data.append((ser, xa, xb))
|
||
|
|
if x <= xb:
|
||
|
|
return series_data[-1]
|
||
|
|
# Evaluation function
|
||
|
|
def interpolant(x):
|
||
|
|
x = ctx.convert(x)
|
||
|
|
orig = ctx.prec
|
||
|
|
try:
|
||
|
|
ctx.prec = workprec
|
||
|
|
ser, xa, xb = get_series(x)
|
||
|
|
y = mpolyval(ser, x-xa)
|
||
|
|
finally:
|
||
|
|
ctx.prec = orig
|
||
|
|
if return_vector:
|
||
|
|
return [+yk for yk in y]
|
||
|
|
else:
|
||
|
|
return +y[0]
|
||
|
|
return interpolant
|
||
|
|
|
||
|
|
ODEMethods.odefun = odefun
|
||
|
|
|
||
|
|
if __name__ == "__main__":
|
||
|
|
import doctest
|
||
|
|
doctest.testmod()
|