CODEWITHSUNDEEP
pythonodedifferential-equationsrunge-kutta
newpythonuser
newpythonuser

Reputation: 133

Runge Kutta method in python

These are the functions I have written:

def rk4(f, t0, y0, h, N):
    t = t0 + arange(N+1)*h
    y = zeros((N+1, size(y0)))
    y[0] = y0
    for n in range(N):
        xi1 = y[n]
        f1 = f(t[n], xi1)
        xi2 = y[n] + (h/2.)*f1
        f2 = f(t[n+1], xi2)
        xi3 = y[n] + (h/2.)*f2
        f3 = f(t[n+1], xi3)
        xi4 = y[n] + h*f3
        f4 = f(t[n+1], xi4)
        y[n+1] = y[n] + (h/6.)*(f1 + 2*f2 + 2*f3 + f4)
    return y

def lorenzPlot():
    y = rk4(fLorenz, 0, array([0,2,20]), .01, 10000)
    fig = figure()
    ax = Axes3D(fig)
    ax.plot(*y.T)

def fLorenz(t,Y):   
    def Y(x,y,z):
        return array([10*(y-x), x*(28-z)-y, x*y-(8./3.)*z])

By implementing lorenzPlot, it's supposed to graph the numerical solution to fLorenz (the Lorenz system of equations) obtained using rk4 (4th order Runge Kutta method). I'm having a problem with the function fLorenz. When I define it as above and call lorenzPlot, I get an error that says 

File "C:/...", line 38, in rk4
    xi2 = y[n] + (h/2.)*f1
TypeError: unsupported operand type(s) for *: 'float' and 'NoneType'

I guessed that this had something to do with not being able to multiply the array correctly.
However, when I changed fLorenz to 

def fLorenz(t,x,y,z):   
    return array([10*(y-x), x*(28-z)-y, x*y-(8./3.)*z])

calling lorenzPlot gives me the error stating that fLorenz takes 4 arguments, but only 2 are given.

Additionally, rk4 and lorenzPlot both work correctly for functions composed of a singular equation.

How should I change fLorenz so that it can be used as f in rk4 and lorenzPlot?

Upvotes: 3

Views: 13282

Answers (3)

Lutz Lehmann
Lutz Lehmann

Reputation: 25992

Apart from the implementation errors, your understanding of the RK4 method is incomplete. The evaluation of the midpoint slopes has to happen at the midpoint of all components, which includes the time component. Thus it should be t[n]+h/2 instead of t[n+1] in the evaluation of f2 and f3.

Upvotes: 1

tzaman
tzaman

Reputation: 47780

Your first fLorenz function defines a sub-function, but doesn't actually return anything. Your second one does, except it's expecting four arguments (t, x, y, z), and you only give it t, Y.

From what I can understand, Y is a three-tuple; you can simply unpack it before you use the values:

def fLorenz(t, Y):
    x, y, z = Y
    return array([10*(y-x), x*(28-z)-y, x*y-(8./3.)*z])

Upvotes: 2

gboffi
gboffi

Reputation: 25023

Your fLorenz doesn't return anything, i.e., due to the well documented Python conventions, it returns None.

Your error message tells you

unsupported operand type(s) for *: 'float' and 'NoneType'

and that the error happens in the statement

    xi2 = y[n] + (h/2.)*f1

In the previous line you assigned to f1 what is returned by an invocation of fLorenz, i.e., None.

If you change the function fLorenz so that it returns a numerical value, either scalar or array, maybe you can proceed with your calculation.

Upvotes: 1

Related Questions