python - scipy.integrate.solve_ivp 矢量化

Question

尝试使用solve_ivp的矢量化选项，奇怪的是它会抛出一个错误，即y0必须是一维的。MWE:

from scipy.integrate import solve_ivp
import numpy as np
import math

def f(t, y):
    theta = math.pi/4
    ham = np.array([[1,0],[1,np.exp(-1j*theta*t)]])
    return-1j * np.dot(ham,y)


def main():

    y0 = np.eye(2,dtype= np.complex128)
    t0 = 0
    tmax = 10**(-6)
    sol=solve_ivp( lambda t,y :f(t,y),(t0,tmax),y0,method='RK45',vectorized=True)
    print(sol.y)

if __name__ == '__main__':
    main()

The calling signature is fun(t, y). Here t is a scalar, and there are two options for the ndarray y: It can either have shape (n,); then fun must return array_like with shape (n,). Alternatively it can have shape (n, k); then fun must return an array_like with shape (n, k), i.e. each column corresponds to a single column in y. The choice between the two options is determined by vectorized argument (see below). The vectorized implementation allows a faster approximation of the Jacobian by finite differences (required for stiff solvers).

错误:

ValueError: y0 must be 1-dimensional.

Python 3.6.8

scipy.版本'1.2.1'

Answer 1

这里vectorize的含义有点令人困惑。这并不意味着 y0 可以是 2d，而是意味着传递给函数的 y 可以是 2d。换句话说，如果求解器需要的话，func 可以同时在多个点进行计算。多少分取决于求解器，而不是您。

更改 f 以在每次调用时显示形状 y:

def f(t, y):
    print(y.shape)
    theta = math.pi/4
    ham = np.array([[1,0],[1,np.exp(-1j*theta*t)]])
    return-1j * np.dot(ham,y)

示例调用:

In [47]: integrate.solve_ivp(f,(t0,tmax),[1j,0],method='RK45',vectorized=False) 
(2,)
(2,)
(2,)
(2,)
(2,)
(2,)
(2,)
(2,)
Out[47]: 
  message: 'The solver successfully reached the end of the integration interval.'
     nfev: 8
     njev: 0
      nlu: 0
      sol: None
   status: 0
  success: True
        t: array([0.e+00, 1.e-06])
 t_events: None
        y: array([[0.e+00+1.e+00j, 1.e-06+1.e+00j],
       [0.e+00+0.e+00j, 1.e-06-1.e-12j]])

相同的调用，但使用vectorize=True:

In [48]: integrate.solve_ivp(f,(t0,tmax),[1j,0],method='RK45',vectorized=True)  
(2, 1)
(2, 1)
(2, 1)
(2, 1)
(2, 1)
(2, 1)
(2, 1)
(2, 1)
Out[48]: 
  message: 'The solver successfully reached the end of the integration interval.'
     nfev: 8
     njev: 0
      nlu: 0
      sol: None
   status: 0
  success: True
        t: array([0.e+00, 1.e-06])
 t_events: None
        y: array([[0.e+00+1.e+00j, 1.e-06+1.e+00j],
       [0.e+00+0.e+00j, 1.e-06-1.e-12j]])

如果为 False，则传递给 f 的 y 为 (2,), 1d； True 则为 (2,1)。我猜如果求解器方法需要的话，它可能是 (2,2) 甚至 (2,3)。这可以加快执行速度，同时减少对 f 的调用。在这种情况下，这并不重要。

quadrature 有一个类似的 vec_func bool 参数:

Numerical Quadrature of scalar valued function with vector input using scipy

相关错误/问题讨论:

https://github.com/scipy/scipy/issues/8922