Python, RungeKutta and first-order differential equation system.

[MaxManus]

Make a function:

def rhs(u,t):

for returning the right-hand side of the first-order differential equation system
from Exercise 11.36. As usual, the u argument is an array or list with the two
solution components u[0] and u[1] at some time t. Inside rhs, assume that
you have access to three global Python functions friction(dudt), spring(u),
and external(t) for evaluating f(u˙ ), s(u), and F(t), respectively.

Homework Equations

The equation is, given to me by lurflurf,:
mv' + f(v ) + s(u) = F(t), t > 0, u(0) = U0, v (0) = V0
where v=u'
For the record the equation was:
mu'' + f(u' ) + s(u) = F(t), t > 0, u(0) = U0, u' (0) = V0 .

Test the rhs function in combination with the functions f(u' ) = 0, F(t) = 0, s(u) = u, and the choice m = 1. The differential equation then reads
u'' + u = 0. With initial conditions u(0) = 1 and u'(0) = 0, one can show
that the solution is given by u(t) = cos(t). Apply two numerical methods:
the 4th-order RungeKutta method and the Forward Euler method from the
ODESolver module developed in Chapter 11.4, using a time step dt = pi /20.

The Attempt at a Solution

 def rhs(u, t):
    return [u[1],
            (1./m)*(external(t) - friction(u[1]) - spring(u[0]))]

This is the solution, but I don't get how yuu go from the equation to rhs.
Can someone help me?

 

[MaxManus]

Here is the page with the exercises: http://www.ifi.uio.no/~inf1100/ODE_project.pdf"

The equation is given in 11.36 and it is 11.37 I need help with.
The solution is http://www.ifi.uio.no/~inf1100/live-programming/oscillator_v1.py"

ODESolver http://www.ifi.uio.no/~inf1100/src/oo/ODESolver.py"

Scitools contains array, plot etc

 

[quicknote]

I can understand your confusion about the process of converting the given first-order differential equation system into a function that returns the right-hand side (rhs) of the system. Let me explain it in a step-by-step manner:

1. The given equation is: mu'' + f(u') + s(u) = F(t), t > 0, u(0) = U0, u'(0) = V0

2. We can simplify this equation by substituting v = u', which gives us a new equation: mv' + f(v) + s(u) = F(t), t > 0, u(0) = U0, v(0) = V0

3. Now, the rhs function is essentially a way to represent the right-hand side of the given equation. In this case, the rhs function will take in two arguments: u and t, and will return the right-hand side of the equation at the given values of u and t.

4. To do this, we first need to define the functions f(v), s(u), and external(t). These functions are given to us in the problem and represent the friction, spring, and external forces acting on the system, respectively.

5. Next, we need to substitute the values of these functions in the given equation. This will give us the final form of the equation that we need to return in the rhs function.

6. Finally, we need to make sure that the rhs function takes into account the initial conditions of the system, which are u(0) = U0 and v(0) = V0.

I hope this helps to clarify the process of creating the rhs function for the given first-order differential equation system. Please feel free to ask any further questions if needed.