Need help solving a second order non-linear ODE in

Question

I don't really know where to start with this problem, as I haven't had much experience with this but it is required to solve this part of the project using a computer.

I have a 2nd order ODE which is:

m = 1220

k = 35600

g = 17.5

a = 450000

and b is between 1000 and 10000 with increments of 500.

x(0)= 0 

x'(0)= 5


m*x''(t) + b*x'(t) + k*x(t)+a*(x(t))^3 = -m*g

I need to find the smallest b such that the solution is never positive. I know what the graph should look like, but I just don't know how to use odeint to get a solution to the differential equation. This is the code I have so far:

from    numpy    import    *    
from    matplotlib.pylab    import    *    
from    scipy.integrate    import    odeint

m = 1220.0
k = 35600.0
g  = 17.5
a = 450000.0
x0= [0.0,5.0]

b = 1000

tmax = 10
dt = 0.01

def fun(x, t):
    return (b*x[1]-k*x[0]-a*(x[0]**3)-m*g)*(1.0/m)
t_rk = arange(0,tmax,dt)   
sol = odeint(fun, x0, t_rk)
plot(t_rk,sol)
show()

Which doesn't really produce much of anything.

Any thoughts? Thanks

Answer 1

To solve a second-order ODE using scipy.integrate.odeint, you should write it as a system of first-order ODEs:

I'll define z = [x', x], then z' = [x'', x'], and that's your system! Of course, you have to plug in your real relations:

x'' = -(b*x'(t) + k*x(t) + a*(x(t))^3 + m*g) / m

becomes:

z[0]' = -1/m * (b*z[0] + k*z[1] + a*z[1]**3 + m*g)
z[1]' = z[0]

Or, just call it d(z):

def d(z, t):
    return np.array((
                     -1/m * (b*z[0] + k*z[1] + a*z[1]**3 + m*g),  # this is z[0]'
                     z[0]                                         # this is z[1]'
                   ))

Now you can feed it to the odeint as such:

_, x = odeint(d, x0, t).T

(The _ is a blank placeholder for the x' variable we made)

In order to minimize b subject to the constraint that the maximum of x is always negative, you can use scipy.optimize.minimize. I'll implement it by actually maximizing the maximum of x, subject to the constraint that it remains negative, because I can't think of how to minimize a parameter without being able to invert the function.

from scipy.optimize import minimize
from scipy.integrate import odeint

m = 1220
k = 35600
g = 17.5
a = 450000
z0 = np.array([-.5, 0])

def d(z, t, m, k, g, a, b):
    return np.array([-1/m * (b*z[0] + k*z[1] + a*z[1]**3 + m*g), z[0]])

def func(b, z0, *args):
    _, x = odeint(d, z0, t, args=args+(b,)).T
    return -x.max()  # minimize negative max

cons = [{'type': 'ineq', 'fun': lambda b: b - 1000, 'jac': lambda b: 1},   # b > 1000
        {'type': 'ineq', 'fun': lambda b: 10000 - b, 'jac': lambda b: -1}, # b < 10000
        {'type': 'ineq', 'fun': lambda b: func(b, z0, m, k, g, a)}] # func(b) > 0 means x < 0

b0 = 10000
b_min = minimize(func, b0, args=(z0, m, k, g, a), constraints=cons)

Answer 2

I don't think you can solve your problem as stated: your initial conditions, with x = 0 and x' > 0 imply that the solution will be positive for some values very close to the starting point. So there is no b for which the solution is never positive...

Leaving that aside, to solve a second order differential equation, you first need to rewrite it as a system of two first order differential equations. Defining y = x' we can rewrite your single equation as:

x' = y
y' = -b/m*y - k/m*x - a/m*x**3 - g

x[0] = 0, y[0] = 5

So your function should look something like this:

def fun(z, t, m, k, g, a, b):
    x, y = z
    return np.array([y, -(b*y + (k + a*x*x)*x) / m - g])

And you can solve and plot your equations doing:

m, k, g, a = 1220, 35600, 17.5, 450000
tmax, dt = 10, 0.01
t = np.linspace(0, tmax, num=np.round(tmax/dt)+1)
for b in xrange(1000, 10500, 500):
    print 'Solving for b = {}'.format(b)
    sol = odeint(fun, [0, 5], t, args=(m, k, g, a, b))[..., 0]
    plt.plot(t, sol, label='b = {}'.format(b))
plt.legend()