具有复杂初始值的scipy odeint [英] scipy odeint with complex initial values
问题描述
我需要解决具有复杂初始值的复杂域定义的ODE系统。
scipy.integrate.odeint在复杂系统上不起作用。
我竭力削减系统的实部和虚部并分别求解，但是我的ODE系统的rhs涉及因变量本身及其复共轭之间的乘积。
aw，我这样做吗？这是我的代码，我尝试过破坏Re和Im零件中的RHS，但是我认为解决方案并不相同，因为复数之间的内部乘积使得我不会破坏它。
在我的脚本中，u1是一个（很长）的复数函数，比如说u1（Lm）= f_real（Lm）+ 1j * f_imag（Lm）。
from numpy import *
from scipy import积分
def cj（z）：return z.conjugate（）
def dydt（y，t = 0）：
＃符号
＃因变量
theta1 = y [0]
theta3 = y [1]
Lm = y [2]
u11 = u1（Lm）
u13 = u1（3 * Lm）
zeta1 = -2 * E * u11 * theta1
zeta3 = -2 * E * 3 * u13 * theta3
＃系数
A0 = theta1 * cj（zeta1）+ 3 * theta3 * cj（zeta3）
A2 = -zeta1 * theta1 + 3 * cj（ zeta1）* theta3 + zeta3 * cj（theta1）
A4 = -theta1 * zeta3-3 * zeta1 * theta3
A6 = -3 * theta3 * zeta3
A =-（A2 / 2 + A4 / 4 + A6 / 6）
＃RHS向量分量
dy1dt = Lm ** 2 *（theta1 *（A-cj（A））-cj（theta1）* A2 / 2
-3/2 * theta3 * cj（A2）
-3/4 * cj（theta3）* A4
-zeta1）
dy2dt = Lm ** 2 *（3 * theta3 *（ A-cj（A））-theta1 * A2 / 2
-cj（theta1）* A4 / 4
-1/2 * cj（theta3）* A6
-3 * zeta3）
dy3dt = Lm ** 3 *（A0 + cj（A0））
返回数组（[dy1dt，dy2dt，dy3dt]）
t = linspace（0 ，10000，100）＃积分时间步长
ry0 = array（[0.001，0，0.1]）＃Re（初始条件）
iy0 = array（[0.0，0.0，0.0]）＃Im （初始条件）
y0 = ry0 + 1j * iy0＃复数初始条件
def rdydt（y，t = 0）：＃Re（RHS）
return dydt（y ，t）.real
def idydt（y，t = 0）：＃Im（RHS）
return dydt（y，t）.imag
ry，rinfodict = integration .odeint（rdydt，y0，t，full_output = True）
iy，iinfodict = tegral.odeint（idydt，y0，t，full_output = True）
我得到的错误是此
TypeError：数组无法安全地强制转换为所需的类型
odepack.error：函数调用的结果不正确
浮点数的数组。
您已经发现，
I need to solve a complex-domain-defined ODE system, with complex initial values. scipy.integrate.odeint does not work on complex systems. I rod about cutting my system in real and imaginary part and solve separately, but my ODE system's rhs involves products between dependent variables themselves and their complex conjugates. Haw do I do that? Here is my code, I tried breaking RHS in Re and Im parts, but I don't think the solution is the same as if I wouldn't break it because of the internal products between complex numbers. In my script u1 is a (very)long complex function, say u1(Lm) = f_real(Lm) + 1j* f_imag(Lm).
from numpy import *
from scipy import integrate
def cj(z): return z.conjugate()
def dydt(y, t=0):
# Notation
# Dependent Variables
theta1 = y[0]
theta3 = y[1]
Lm = y[2]
u11 = u1(Lm)
u13 = u1(3*Lm)
zeta1 = -2*E*u11*theta1
zeta3 = -2*E*3*u13*theta3
# Coefficients
A0 = theta1*cj(zeta1) + 3*theta3*cj(zeta3)
A2 = -zeta1*theta1 + 3*cj(zeta1)*theta3 + zeta3*cj(theta1)
A4 = -theta1*zeta3 - 3*zeta1*theta3
A6 = -3*theta3*zeta3
A = - (A2/2 + A4/4 + A6/6)
# RHS vector components
dy1dt = Lm**2 * (theta1*(A - cj(A)) - cj(theta1)*A2/2
- 3/2*theta3*cj(A2)
- 3/4*cj(theta3)*A4
- zeta1)
dy2dt = Lm**2 * (3*theta3*(A - cj(A)) - theta1*A2/2
- cj(theta1)*A4/4
- 1/2*cj(theta3)*A6
- 3*zeta3)
dy3dt = Lm**3 * (A0 + cj(A0))
return array([dy1dt, dy2dt, dy3dt])
t = linspace(0, 10000, 100) # Integration time-step
ry0 = array([0.001, 0, 0.1]) # Re(initial condition)
iy0 = array([0.0, 0.0, 0.0]) # Im(initial condition)
y0 = ry0 + 1j*iy0 # Complex Initial Condition
def rdydt(y, t=0): # Re(RHS)
return dydt(y, t).real
def idydt(y, t=0): # Im(RHS)
return dydt(y, t).imag
ry, rinfodict = integrate.odeint(rdydt, y0, t, full_output=True)
iy, iinfodict = integrate.odeint(idydt, y0, t, full_output=True)
The error I get is this
TypeError: array cannot be safely cast to required type
odepack.error: Result from function call is not a proper array of
floats.
As you've discovered, odeint does not handle complex-valued differential equations, but there is scipy.integrate.complex_ode. complex_ode is a convenience function that takes care of converting the system of n complex equations into a system of 2*n real equations. (Note the discrepancy in the signatures of the functions used to define the equations for odeint and ode. odeint expects f(t, y, *args) while ode (and complex_ode) expect f(y, t, *args).)
A similar convenience function can be created for odeint. In the following code, odeintz is a function that handles the conversion of a complex system into a real system and solving it with odeint. The code includes an example of solving a complex system. It also shows how that system can be converted "by hand" to a real system and solved with odeint. But for a large system, that is a tedious and error prone process; using a complex solver is certainly a saner approach.
import numpy as np
from scipy.integrate import odeint
def odeintz(func, z0, t, **kwargs):
"""An odeint-like function for complex valued differential equations."""
# Disallow Jacobian-related arguments.
_unsupported_odeint_args = ['Dfun', 'col_deriv', 'ml', 'mu']
bad_args = [arg for arg in kwargs if arg in _unsupported_odeint_args]
if len(bad_args) > 0:
raise ValueError("The odeint argument %r is not supported by "
"odeintz." % (bad_args[0],))
# Make sure z0 is a numpy array of type np.complex128.
z0 = np.array(z0, dtype=np.complex128, ndmin=1)
def realfunc(x, t, *args):
z = x.view(np.complex128)
dzdt = func(z, t, *args)
# func might return a python list, so convert its return
# value to an array with type np.complex128, and then return
# a np.float64 view of that array.
return np.asarray(dzdt, dtype=np.complex128).view(np.float64)
result = odeint(realfunc, z0.view(np.float64), t, **kwargs)
if kwargs.get('full_output', False):
z = result[0].view(np.complex128)
infodict = result[1]
return z, infodict
else:
z = result.view(np.complex128)
return z
if __name__ == "__main__":
# Generate a solution to:
# dz1/dt = -z1 * (K - z2)
# dz2/dt = L - z2
# K and L are fixed parameters. z1(t) and z2(t) are complex-
# valued functions of t.
# Define the right-hand-side of the differential equation.
def zfunc(z, t, K, L):
z1, z2 = z
return [-z1 * (K - z2), L - z2]
# Set up the inputs and call odeintz to solve the system.
z0 = np.array([1+2j, 3+4j])
t = np.linspace(0, 4, 101)
K = 3
L = 1
z, infodict = odeintz(zfunc, z0, t, args=(K,L), full_output=True)
# For comparison, here is how the complex system can be converted
# to a real system. The real and imaginary parts are used to
# write a system of four coupled equations. The formulas for
# the complex right-hand-sides are
# -z1 * (K - z2) = -(x1 + i*y1) * (K - (x2 + i*y2))
# = (-x1 - i*y1) * (K - x2 + i(-y2))
# = -x1 * (K - x2) - y1*y2 + i*(-y1*(K - x2) + x1*y2)
# and
# L - z2 = L - (x2 + i*y2)
# = (L - x2) + i*(-y2)
def func(r, t, K, L):
x1, y1, x2, y2 = r
dx1dt = -x1 * (K - x2) - y1*y2
dy1dt = -y1 * (K - x2) + x1*y2
dx2dt = L - x2
dy2dt = -y2
return [dx1dt, dy1dt, dx2dt, dy2dt]
# Use regular odeint to solve the real system.
r, infodict = odeint(func, z0.view(np.float64), t, args=(K,L), full_output=True)
# Compare the two solutions. They should be the same. (As usual for
# floating point calculations, there could be a small difference.)
delta_max = np.abs(z.view(np.float64) - r).max()
print "Maximum difference between the complex and real versions is", delta_max
# Plot the real and imaginary parts of the complex solution.
import matplotlib.pyplot as plt
plt.clf()
plt.plot(t, z[:,0].real, label='z1.real')
plt.plot(t, z[:,0].imag, label='z1.imag')
plt.plot(t, z[:,1].real, label='z2.real')
plt.plot(t, z[:,1].imag, label='z2.imag')
plt.xlabel('t')
plt.grid(True)
plt.legend(loc='best')
plt.show()
Here's the plot generated by the script:
Update
This code has been significantly expanded into a function called odeintw that handles complex variables and matrix equations. The new function can be found on github: https://github.com/WarrenWeckesser/odeintw
这篇关于具有复杂初始值的scipy odeint的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
