# -*- coding: iso-8859-1 -*-
# laplace.py with mpmath
# appropriate for high precision
# Talbot suggested that the Bromwich line be deformed into a contour that begins
# and ends in the left half plane, i.e., z \to \infty at both ends.
# Due to the exponential factor the integrand decays rapidly
# on such a contour. In such situations the trapezoidal rule converge
# extraordinarily rapidly.
# For example here we compute the inverse transform of F(s) = 1/(s+1) at t = 1
#
# >>> error = Talbot(1,24)-exp(-1)
# >>> error
# (3.3306690738754696e-015+0j)
#
# Talbot method is very powerful here we see an error of 3.3e-015
# with only 24 function evaluations
#
# Created by Fernando Damian Nieuwveldt
# email:fdnieuwveldt@gmail.com
# Date : 25 October 2009
#
# Adapted to mpmath and classes by Dieter Kadelka
# email: Dieter.Kadelka@kit.edu
# Date : 27 October 2009
#
# Reference
# L.N.Trefethen, J.A.C.Weideman, and T.Schmelzer. Talbot quadratures
# and rational approximations. BIT. Numerical Mathematics,
# 46(3):653 670, 2006.
from mpmath import mpf,mpc,pi,sin,tan,exp
# testfunction: Laplace-transform of exp(-t)
def F(s):
return 1.0/(s+1.0)
class Talbot(object):
def __init__(self,F=F,shift=0.0):
self.F = F
# test = Talbot() or test = Talbot(F) initializes with testfunction F
self.shift = shift
# Shift contour to the right in case there is a pole on the
# positive real axis :
# Note the contour will not be optimal since it was originally devoloped
# for function with singularities on the negative real axis For example
# take F(s) = 1/(s-1), it has a pole at s = 1, the contour needs to be
# shifted with one unit, i.e shift = 1.
# But in the test example no shifting is necessary
self.N = 24
# with double precision this constant N seems to best for the testfunction
# given. For N = 22 or N = 26 the error is larger (for this special
# testfunction).
# With laplace.py:
# >>> test.N = 500
# >>> print test(1) - exp(-1)
# >>> -2.10032517928e+21
# Huge (rounding?) error!
# with mp_laplace.py
# >>> mp.dps = 100
# >>> test.N = 500
# >>> print test(1) - exp(-1)
# >>> -5.098571435907316903360293189717305540117774982775731009465612344056911792735539092934425236391407436e-64
def __call__(self,t):
if t == 0:
print "ERROR: Inverse transform can not be calculated for t=0"
return ("Error");
# Initiate the stepsize
h = 2*pi/self.N
ans = 0.0
# parameters from
# T. Schmelzer, L.N. Trefethen, SIAM J. Numer. Anal. 45 (2007) 558-571
c1 = mpf('0.5017')
c2 = mpf('0.6407')
c3 = mpf('0.6122')
c4 = mpc('0','0.2645')
# The for loop is evaluating the Laplace inversion at each point theta i
# which is based on the trapezoidal rule
for k in range(self.N):
theta = -pi + (k+0.5)*h
z = self.shift + self.N/t*(c1*theta/tan(c2*theta) - c3 + c4*theta)
dz = self.N/t * (-c1*c2*theta/sin(c2*theta)**2 + c1/tan(c2*theta)+c4)
ans += exp(z*t)*self.F(z)*dz
return ((h/(2j*pi))*ans).real