我想评估ODE的解决方案phi(+oo) 在哪里的价值phi(xi)
Eq(Derivative(phi(xi), (xi, 2)), (-K + xi**2)*phi(xi))
并且K是一个已知的实变量。通过dsolve,我得到了解决方案:
Eq(phi(xi), -K*xi**5*r(3)/20 + C2*(K**2*xi**4/24 - K*xi**2/2 + xi**4/12 + 1) + C1*xi*(xi**4/20 + 1) + O(xi**6))
r()在右侧的第一项中具有未知函数。这是我的代码:
import numpy as np
import matplotlib.pyplot as plt
import sympy
from sympy import I, pi, oo
sympy.init_printing()
def apply_ics(sol, ics, x, known_params):
"""
Apply the initial conditions (ics), given as a dictionary on
the form ics = {y(0): y0, y(x).diff(x).subs(x, 0): yp0, ...},
to the solution of the ODE with independent variable x.
The undetermined integration constants C1, C2, ... are extracted
from the free symbols of the ODE solution, excluding symbols in
the known_params list.
"""
free_params = sol.free_symbols - set(known_params)
eqs = [(sol.lhs.diff(x, n) - sol.rhs.diff(x, n)).subs(x, 0).subs(ics)
for n in range(len(ics))]
sol_params = sympy.solve(eqs, free_params)
return sol.subs(sol_params)
K = sympy.Symbol('K', positive = True)
xi = sympy.Symbol('xi',real = True)
phi = sympy.Function('phi')
ode = sympy.Eq( phi(xi).diff(xi, 2), (xi**2-K)*phi(xi))
ode_sol = sympy.dsolve(ode)
ics = { phi(0):1, phi(xi).diff(xi).subs(xi,0): 0}
phi_xi_sol = apply_ics(ode_sol, ics, xi, [K])
ode_sol解决方案在哪里,phi_xi_sol是应用初始条件后的解决方案。由于r()在 NumPy 中未定义,我无法通过以下方式评估结果
for g in [0.9, 0.95, 1, 1.05, 1.2]:
phi_xi = sympy.lambdify(xi, phi_xi_sol.rhs.subs({K:g}), 'numpy')
有谁知道这个功能是什么r()意思,我应该如何处理它?
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