Runge-Kutta 4th order 使用 Python 解决二阶 ODE 系统 [英] Runge-Kutta 4th order to solve 2nd order ODE system using Python

问题描述

我正在尝试通过 runge-kutta 4 阶方法在数值上求解两个 ode 的系统.初始系统:系统解决:

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

而且我有非常奇怪的解决方案图...我有:

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

正确的图形:

我在我的 runge-kutta 中找不到问题.请帮帮我.

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

我的代码在这里:

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

推荐答案

您的状态有 4 个组件，因此您在每个阶段都需要 4 个斜坡.很明显，z 的斜率/更新不能来自 k1_zdot，它必须是 k1_z 之前要计算的

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

进入下一阶段

k2_z = zdot + dt/2*k1_zdot

等

但最好使用矢量化界面

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]);

然后使用 RK4 的标准公式

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

然后解压为

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

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