python中ODE的求解系统；在此调用上完成了多余的工作

编程问答 2022-05-17

问题描述

我正在尝试在 python 中解决不同电位的耦合 ODE 系统。它适用于特定类型的势（指数），但是一旦势由幂律描述，python 生成的图就完全不连贯，它经常只为所有参数分配零值。我的编码适用于：

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      | Review Photos
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它不适用于（并打印 RuntimeWarning 消息）：

kr1 = 8*np.pi
#rho_m = a**(-3)
#V = np.e**(-kr1*x_1)
#dVdx = -kr1*np.e**(-kr1*x_1)

def quintessence (x,t):
    a = x[0]
    x_1 = x[1]
    x_2 = x[2]
    dadt = (kr1*a/np.sqrt(3))*np.sqrt(1/2 * x_2**2 + np.e**(-kr1*x_1) + a**(-3))
    dx_1dt = x_2
    dx_2dt = -np.sqrt(3)*kr1*np.sqrt(1/2 * x_2**2 + np.e**(-kr1*x_1) + a**(-3))*x_2 + kr1*np.e**(-kr1*x_1)

    return[dadt,dx_1dt,dx_2dt]

x0 = [0.0001,0]
t = np.linspace(0,80,1000)

x = odeint(quintessence,x0,t)

a = x[:,0]
x_1 = x[:,1]
x_2 = x[:,2]

plt.plot(t,a)
plt.show()

知道第二段代码有什么问题吗？我对 python 比较陌生，不知道它的局限性。

解决方法

精华暗能量模型（见physics.SE为任意参考）是

x'' + 3*H*x' + V'(x) = 0

H = a'/a

这里，根据所写，

H = kr1/sqrt(3) * sqrt(E) with

E = 0.5*x'^2 + V(x) + a^(-3)

(( which leads to
   E' = x'*(x''+V'(x)) - 3*a^(-4)*a' = - 3*H*x'^2 -3*a^(-3)*H 
      = -3*H*(x'^2+a^(-3))
))

我会在这些部分实施这个系统，因为

def V(x): return ...
def dV(x): return ...

def quintessence(u,t):
    a,x,dx = u
    E = 0.5*dx**2 + V(x) + a**-3
    H = kr1/3**0.5 * E**0.5
    da = a*H
    ddx = -3*H*dx -dV(x)
    return [da,dx,ddx]
odeodeintpython