我正在尝试使用dopri5
中的scipy.integrate.ode
来解决一个ODE系统,该系统描述势能下二维空间中粒子在位置上运动的位置和速度:
我查找了解决此类ODE的示例,然后基本上复制了以下链接中的所有内容,只需切换变量和尺寸即可解决我的特殊问题:
How to use dorpi5 or dop853 in Python
代码如下:
from scipy.integrate import ode
from scipy.interpolate import RegularGridInterpolator
import numpy as np
import matplotlib.pyplot as plt
# Define V(x,y) and obtain the gradient function through interpolation:
def V(x, y):
return np.exp((x**2+y**2))-0.5
x = np.linspace(-10, 10, 1000)
y = np.linspace(-10, 10, 1000)
X, Y = np.meshgrid(x, y)
v = V(X, Y)
gradv_array = np.gradient(v, x, y)
gradv_x = RegularGridInterpolator((x, y), gradv_array[0])
gradv_y = RegularGridInterpolator((x, y), gradv_array[1])
def fun(t, z, m):
"""
Right hand side of the differential equations
dx/dt = v_x
dy/dt = v_y
dv_x/dt = -gradv_x(x, y)/m
dv_y/dt = -gradv_y(x, y)/m
"""
x, y, v_x, v_y = z
f = [v_x, v_y, -gradv_x(x, y)/m, -gradv_y(x, y)/m]
return f
# Create an `ode` instance to solve the system of differential
# equations defined by `fun`, and set the solver method to 'dopri5'.
solver = ode(fun)
solver.set_integrator('dopri5')
# Give the value of m to the solver. This is passed to
# `fun` when the solver calls it.
m = 1
solver.set_f_params(m)
# Set the initial value z(0) = z0.
t0 = 0.0
z0 = [0.5, 0.3, 0, 0]
solver.set_initial_value(z0, t0)
# Create the array `t` of time values at which to compute
# the solution, and create an array to hold the solution.
# Put the initial value in the solution array.
t1 = 2.5
N = 75
t = np.linspace(t0, t1, N)
sol = np.empty((N, 4))
sol[0] = z0
# Repeatedly call the `integrate` method to advance the
# solution to time t[k], and save the solution in sol[k].
k = 1
while solver.successful() and solver.t < t1:
solver.integrate(t[k])
sol[k] = solver.y
k += 1
# Plot the solution...
plt.plot(t, sol[:,0], label='v_x')
plt.plot(t, sol[:,1], label='v_y')
plt.xlabel('t')
plt.grid(True)
plt.legend()
plt.show()
plt.plot(t, sol[:,2], label='x')
plt.plot(t, sol[:,3], label='y')
plt.xlabel('t')
plt.grid(True)
plt.legend()
plt.show()
我尝试执行原始代码和修改后的代码,前者可行,而后者却无效,并显示以下错误消息:
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-39-2db0a4c50fa3> in <module>()
42 k = 1
43 while solver.successful() and solver.t < t1:
---> 44 solver.integrate(t[k])
45 sol[k] = solver.y
46 k += 1
~\Anaconda3\lib\site-packages\scipy\integrate\_ode.py in integrate(self, t, step, relax)
430 self._y, self.t = mth(self.f, self.jac or (lambda: None),
431 self._y, self.t, t,
--> 432 self.f_params, self.jac_params)
433 except SystemError:
434 # f2py issue with tuple returns, see ticket 1187.
~\Anaconda3\lib\site-packages\scipy\integrate\_ode.py in run(self, f, jac, y0, t0, t1, f_params, jac_params)
1170 def run(self, f, jac, y0, t0, t1, f_params, jac_params):
1171 x, y, iwork, istate = self.runner(*((f, t0, y0, t1) +
-> 1172 tuple(self.call_args) + (f_params,)))
1173 self.istate = istate
1174 if istate < 0:
<ipython-input-39-2db0a4c50fa3> in fun(t, z, m)
10 """
11 x, y, v_x, v_y = z
---> 12 f = [v_x, v_y, -gradv_x(x, y)/m, -gradv_y(x, y)/m]
13 return f
14
~\Anaconda3\lib\site-packages\scipy\interpolate\interpolate.py in __call__(self, xi, method)
2460 method = self.method if method is None else method
2461 if method not in ["linear", "nearest"]:
-> 2462 raise ValueError("Method '%s' is not defined" % method)
2463
2464 ndim = len(self.grid)
ValueError: Method '0.3' is not defined
显然,y
中的solver.y
被解释为该列表中的第二个元素:
z0 = [0.5, 0.3, 0, 0]
也许应该进行更改,因为有四个ODE而不是两个?但是如何?
编辑:发现错误。由scipy.interpolate.RegularGridInterpolator
生成的函数接受一个参数列表,但不接受多个参数,即gradv_x([x, y])
代替grad_x(x, y)
。