Solve a nonlinear ODE system for the Frenet frame

Question

I have checked python non linear ODE with 2 variables , which is not my case. Maybe my case is not called as nonlinear ODE, correct me please.

The question isFrenet Frame actually, in which there are 3 vectors T(s), N(s) and B(s); the parameter s>=0. And there are 2 scalar with known math formula expression t(s) and k(s). I have the initial value T(0), N(0) and B(0).

diff(T(s), s) =             k(s)*N(s)
diff(N(s), s) = -k(s)*T(s)            + t(s)*B(s)
diff(B(s), s) =            -t(s)*N(s)

Then how can I get T(s), N(s) and B(s) numerically or symbolically?

I have checked scipy.integrate.ode but I don't know how to pass k(s)*N(s) into its first parameter at all

def model (z, tspan):
   T = z[0]
   N = z[1]
   B = z[2]

   dTds =            k(s) * N           # how to express function k(s)?      
   dNds = -k(s) * T          + t(s) * B
   dBds =           -t(s)* N

   return [dTds, dNds, dBds]


z = scipy.integrate.ode(model, [T0, N0, B0]

Answer 1

I found that the equation I listed in the first post does not work for my curve. So I read Gray A., Abbena E., Salamon S-Modern Differential Geometry of Curves and Surfaces with Mathematica. 2006 and found that for arbitrary curve, Frenet equation should be written as

diff(T(s), s) = ||r'||*           k(s)*N(s)
diff(N(s), s) = ||r'||*(-k(s)*T(s)            + t(s)*B(s))
diff(B(s), s) = ||r'||*          -t(s)*N(s)

where ||r'||(or ||r'(s)||) is diff([x(s), y(s), z(s)], s).norm()

now the problem has changed to be some different from that in the first post, because there is no r'(s) function or discrete data array. So I think this is suitable for a new reply other than comment.

I met 2 questions while trying to solve the new equation:

1. how can we program with r'(s) if scipy's solve_ivp is used?
2. I try to modify my gaussian solution, but the result is totally wrong.

thanks again

import numpy as np

from scipy.integrate import cumtrapz
import matplotlib.pylab as plt

# Define the parameters as regular Python function:
def k(s):
    return 1

def t(s):
    return 0

def dZ(s, Z, r_norm):
    return np.array([
        r_norm * k(s) * Z[1],
        r_norm*(-k(s) * Z[0] + t(s) * Z[2]),
        r_norm*(-t(s)* Z[1])
    ])

T0, N0, B0 = np.array([1, 0, 0]), np.array([0, 1, 0]), np.array([0, 0, 1])

deltaS = 0.1    # step to calculate dZ/ds
num = int(2*np.pi*1/deltaS) + 1 # how many points on the curve we have to calculate

T = np.zeros([num, ], dtype=object)
N = np.zeros([num, ], dtype=object)
B = np.zeros([num, ], dtype=object)
R0 = N0

T[0] = T0
N[0] = N0
B[0] = B0

for i in range(num-1):
    r_norm = np.linalg.norm(R0)
    temp_dZ = dZ(i*deltaS, np.array([T[i], N[i], B[i]]), r_norm)
    T[i+1] = T[i] + temp_dZ[0]*deltaS
    T[i+1] = T[i+1]/np.linalg.norm(T[i+1])

    N[i+1] = N[i] + temp_dZ[1]*deltaS
    N[i+1] = N[i+1]/np.linalg.norm(N[i+1])

    B[i+1] = B[i] + temp_dZ[2]*deltaS
    B[i+1] = B[i+1]/np.linalg.norm(B[i+1])

    R0 = R0 + T[i]*deltaS

coords = cumtrapz(
    [
        [i[0] for i in T], [i[1] for i in T], [i[2] for i in T]
    ]
    , x=np.arange(num)*deltaS
)


plt.figure()
plt.plot(coords[0, :], coords[1, :]);
plt.axis('equal'); plt.xlabel('x'); plt.xlabel('y');

plt.show()

Answer 2

thanks for your example. And I thought it again, found that since there is formula for dZ where Z is matrix(T, N, B), we can calculate Z[i] = Z[i-1] + dZ[i-1]*deltaS according to the concept of derivative. Then I code and find this idea can solve the circle example. So

1. is Z[i] = Z[i-1] + dZ[i-1]*deltaS suitable for other ODE? will it fail in some situation, or does scipy.integrate.solve_ivp/scipy.integrate.ode supply advantage over the direct usage of Z[i] = Z[i-1] + dZ[i-1]*deltaS?
2. in my code, I have to normalize Z[i] because ||Z[i]|| is not always 1. Why does it happen? A float numerical calculation error?

my answer to my question, at least it works for the circle

import numpy as np

from scipy.integrate import cumtrapz
import matplotlib.pylab as plt

# Define the parameters as regular Python function:
def k(s):
    return 1

def t(s):
    return 0

def dZ(s, Z):
    return np.array(
        [k(s) * Z[1], -k(s) * Z[0] + t(s) * Z[2], -t(s)* Z[1]]
    )

T0, N0, B0 = np.array([1, 0, 0]), np.array([0, 1, 0]), np.array([0, 0, 1])

deltaS = 0.1    # step to calculate dZ/ds
num = int(2*np.pi*1/deltaS) + 1 # how many points on the curve we have to calculate

T = np.zeros([num, ], dtype=object)
N = np.zeros([num, ], dtype=object)
B = np.zeros([num, ], dtype=object)

T[0] = T0
N[0] = N0
B[0] = B0

for i in range(num-1):
    temp_dZ = dZ(i*deltaS, np.array([T[i], N[i], B[i]]))
    T[i+1] = T[i] + temp_dZ[0]*deltaS
    T[i+1] = T[i+1]/np.linalg.norm(T[i+1])  # have to do this

    N[i+1] = N[i] + temp_dZ[1]*deltaS
    N[i+1] = N[i+1]/np.linalg.norm(N[i+1])

    B[i+1] = B[i] + temp_dZ[2]*deltaS
    B[i+1] = B[i+1]/np.linalg.norm(B[i+1])


coords = cumtrapz(
    [
        [i[0] for i in T], [i[1] for i in T], [i[2] for i in T]
    ]
    , x=np.arange(num)*deltaS
)

plt.figure()
plt.plot(coords[0, :], coords[1, :]);
plt.axis('equal'); plt.xlabel('x'); plt.xlabel('y');

plt.show()

Answer 3

Here is a code using solve_ivp interface from Scipy (instead of odeint) to obtain a numerical solution:

import numpy as np
from scipy.integrate import solve_ivp

from scipy.integrate import cumtrapz
import matplotlib.pylab as plt

# Define the parameters as regular Python function:
def k(s):
    return 1

def t(s):
    return 0

# The equations: dz/dt = model(s, z):
def model(s, z):
    T = z[:3]   # z is a (9, ) shaped array, the concatenation of T, N and B 
    N = z[3:6]
    B = z[6:]

    dTds =            k(s) * N  
    dNds = -k(s) * T          + t(s) * B
    dBds =           -t(s)* N

    return np.hstack([dTds, dNds, dBds])


T0, N0, B0 = [1, 0, 0], [0, 1, 0], [0, 0, 1] 

z0 = np.hstack([T0, N0, B0])

s_span = (0, 6) # start and final "time"
t_eval = np.linspace(*s_span, 100)  # define the number of point wanted in-between,
                                    # It is not necessary as the solver automatically
                                    # define the number of points.
                                    # It is used here to obtain a relatively correct 
                                    # integration of the coordinates, see the graph

# Solve:
sol = solve_ivp(model, s_span, z0, t_eval=t_eval, method='RK45')
print(sol.message)
# >> The solver successfully reached the end of the integration interval.

# Unpack the solution:
T, N, B = np.split(sol.y, 3)  # another way to unpack the z array
s = sol.t

# Bonus: integration of the normal vector in order to get the coordinates
#        to plot the curve  (there is certainly better way to do this)
coords = cumtrapz(T, x=s)

plt.plot(coords[0, :], coords[1, :]);
plt.axis('equal'); plt.xlabel('x'); plt.xlabel('y');

T, N and B are vectors. Therefore, there are 9 equations to solve: z is a (9,) array.

For constant curvature and no torsion, the result is a circle: