Runge-Kutta 4th order to solve 2nd order ODE system using Python

问题

I'm trying to solve system of two odes numerically by runge-kutta 4th order method. initial system: system to solve:

And I have very strange solution graph... I have:

Correct graph:

I can't find trouble in my runge-kutta. Please, help me.

My code is here:

dt = 0.04

#initial conditions
t.append(0)
zdot.append(0)
z.append(A)
thetadot.append(0)
theta.append(B)

#derrive functions
def zdotdot(z_cur, theta_cur):
   return -omega_z * z_cur - epsilon / 2 / m * theta_cur
def thetadotdot(z_cur, theta_cur):
   return -omega_theta * theta_cur - epsilon / 2 / I * z_cur 
i = 0
while True:
    # runge_kutta
    k1_zdot = zdotdot(z[i], theta[i])
    k1_thetadot = thetadotdot(z[i], theta[i])

    k2_zdot = zdotdot(z[i] + dt/2 * k1_zdot, theta[i])
    k2_thetadot = thetadotdot(z[i], theta[i]  + dt/2 * k1_thetadot)

    k3_zdot = zdotdot(z[i] + dt/2 * k2_zdot, theta[i])
    k3_thetadot = thetadotdot(z[i], theta[i]  + dt/2 * k2_thetadot)

    k4_zdot = zdotdot(z[i] + dt * k3_zdot, theta[i])
    k4_thetadot = thetadotdot(z[i], theta[i]  + dt * k3_thetadot)

    zdot.append (zdot[i] + (k1_zdot + 2*k2_zdot + 2*k3_zdot + k4_zdot) * dt / 6)
    thetadot.append (thetadot[i] + (k1_thetadot + 2*k2_thetadot + 2*k3_thetadot + k4_thetadot) * dt / 6)

    z.append (z[i] + zdot[i + 1] * dt)
    theta.append (theta[i] + thetadot[i + 1] * dt)
    i += 1

回答1:

Your state has 4 components, thus you need 4 slopes in each stage. It should be obvious that the slope/update for z can not come from k1_zdot, it has to be k1_z which is to be computed previously as

k1_z = zdot

and in the next stage

k2_z = zdot + dt/2*k1_zdot

etc.

But better is to use a vectorized interface

def derivs(t,u):
    z, theta, dz, dtheta = u
    ddz = -omega_z * z - epsilon / 2 / m * theta
    ddtheta = -omega_theta * theta - epsilon / 2 / I * z 
    return np.array([dz, dtheta, ddz, ddtheta]);

and then use the standard formulas for RK4

i = 0
while True:
    # runge_kutta
    k1 = derivs(t[i], u[i])
    k2 = derivs(t[i] + dt/2, u[i] + dt/2 * k1)
    k3 = derivs(t[i] + dt/2, u[i] + dt/2 * k2)
    k4 = derivs(t[i] + dt, u[i] + dt * k3)

    u.append (u[i] + (k1 + 2*k2 + 2*k3 + k4) * dt / 6)
    i += 1

and later unpack as

z, theta, dz, dtheta = np.asarray(u).T

来源：https://stackoverflow.com/questions/55989100/runge-kutta-4th-order-to-solve-2nd-order-ode-system-using-python

标签

python

numerical-methods

ode

runge-kutta