Runge-Kutta fourth order method. Integrating backwards

Question

I am using a Runge-Kutta fourth order method to solve numerically the usual equation of motion of a background scalar field in curved spacetime with a quartic potential:

$\phi^{''}=-3\left(1+\frac{H^{'}}{3H}\right)\phi^{'}-\lambda\phi^3/H^2$,

$'$ denoting the derivative w.r.t. the e-folds number $\textrm{d}N=H\textrm{d}t$ and, from the Friedmann equation:

$H^2=\frac{\lambda \phi^4}{4}\frac{1}{3M_{Pl}^2-(1/2)\phi^{'2}}$;

$H^{'}=-\frac{1}{2M_{Pl}^2}H\phi^{'2}$.

The problem comes when integrating backwards using as the initial conditions the final values I got after integrating forward. The outcome blow up without matching the values obtained before, when integrating forward. I simply do not understand where the problem is as both the equation and the code are not unknown whatsoever. Firstly, I integrated from 0 to 64 e-folds. Then I simply reverse the integration direction.

I attach the code too:

def rk4trial(f,v0,t0,tf,n,V):  
    t=np.linspace(t0,tf,n)
    h=t[1]-t[0]
    v=np.array((n+1)*[v0])
    for j in range(n):  
        k1=f(v[j],t[j])*h
        k2=f(v[j]+(1/2)*k1,t[j]+(1/2)*h)*h
        k3=f(v[j]+(1/2)*k2,t[j]+(1/2)*h)*h
        k4=f(v[j]+k3,t[j]+h)*h
        v[j+1]=v[j]+(k1+2*k2+2*k3+k4)/6
    return v, t, h


def Fdet(v,t):
    phi, sigma = v
    H=(((lamb/4)*phi**4)/(3*mpl**2-(1/2)*sigma**2))**(1/2)
    HH=-((1/2)*(sigma/H)**2)*(1/mpl**2)
    return np.array([sigma,-3*(1+HH/3)*sigma-lamb*phi**3/(H**2)])

Answer 1

Based on my understanding from your question, I assume you are looking for to examine time reversibility of your ODE:

$$\phi^{''}(t) = - 3 (1+\frac{H^{'}(t)}{3 H(t)}) \phi^{'} (t) - \lambda \frac{\phi^{3}(t)}{H^{2}(t)} = 0$$

Where:

$$H(t) = \frac{\lambda \phi^{4}(t)}{4(3M_{Pl}^{2}-0.5\phi^{'}(t)^{2})}$$

and

$$H^{'}(t) = -\frac{1}{2M_{Pl}^{2}} \phi^{'}(t)^{2} H(t)$$

Your initial ODE is in the form of a Langevin equation with a viscous term of $3 (1+\frac{H^{'}(t)}{3 H(t)}) \phi^{'} (t)$. Your Langevin equation is not time-reversible due to the fact that the equation will not remain the same under T transformation. Simply because first-order derivative $\phi^{'}(t)$ change its sign to negative under T transformation: $$\phi^{'}(-t) = -\phi^{'}(t)$$