使用二阶 ODE 在 Python 中实现 3/8 Runge Kutta

Question

I want to solve the equation y'' + 5y' + 6y = cos(t). Since this is 2nd order, I first created a system of first order ODEs where dy/dt = z = f(t,y,z) and dz/dt = cos(t) - 5z - 6y = g(t,y,z). I'm a novice with python, and unsure how exactly to implement the program with a system of ODEs, but for my inputs I wrote func = [f,g] and ic = [y0,dy0].

Secondly, since Runge Kutta evaluates variable expression whose values are only available during runtime, I am using python's eval function, though I am receiving an error. Below is my code and the error message:

def f(t,y,z): 
    return z
def g(t,y,z): 
    return np.cos(t)-5*z-6*y
t = np.linspace(0,20,100)
y = np.zeros((1,len(t)))
z = np.zeros((1,len(t)))

fun = [f(t,y,z),g(t,y,z)]
ic = [1,0] # y0,dy0

def ruku(fun,h):
    t0=0
    tf=20
    
    t=t0
    y=ic
    fc=0

    while t < tf:
        if t+h > tf:
            exit()
        if h == tf-t:
            exit()
        k1=eval('fun',t,y)
        k2=eval('fun',t+h/3,y+h*k1/3)
        k3=eval('fun',t+2*h/3,y+h*(k2-k1/3))
        k4=eval('fun',t+h,y+h*(k3-k2+k1))
        
        y=y+h*(k1+3*(k2+k3)+k4)/8
        fc=fc+4
        t=t+h
    return y

ruku(fun,0.01)

-->TypeError: globals must be a dict

Thanks in advance for any suggestions; I'm really trying to make this work.

Answer 1

Turns out I was underestimating the simplicity of Python: Replaced my loop with:

    while t +h <= tf:
            
        k1=fun(t,y)
        k2=fun(t+h/3,y+h*k1/3)
        k3=fun(t+2*h/3,y+h*(k2-k1/3))
        k4=fun(t+h,y+h*(k3-k2+k1))
        
        y=y+h*(k1+3*(k2+k3)+k4)/8
        fc=fc+4
        t=t+h

and converted my inputs into np.arrays