求解 Frenet 框架的非线性 ODE 系统

Question

我检查过 python non linear ODE with 2 variables ，这不是我的情况。也许我的情况不叫nonlinear ODE，请指正。

其实问题是Frenet Frame，其中有3个向量T(s)，N(s)和B(s)；参数 s>=0。并且有 2 个标量具有已知的数学公式表达式 t(s) 和 k(s)。我有初始值 T(0)、N(0) 和 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)

那我怎样才能得到T(s)、N(s)和B(s)的数值或符号呢？

我已经检查了 scipy.integrate.ode 但我根本不知道如何将 k(s)*N(s) 传递给它的第一个参数

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

这是使用 Scipy 中的 solve_ivp 接口（而不是 odeint）获得数值解的代码：

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 和 B 是向量。因此，有9个方程需要求解：z是一个(9,)数组。

对于恒定曲率且无扭转，结果是一个圆：

Answer 2

感谢您的示例。又想了想，发现既然Z为matrix(T, N, B)的dZ有公式，那么根据导数的概念就可以计算出Z[i] = Z[i-1] + dZ[i-1]*deltaS。然后我编码并发现这个想法可以解决圆圈的例子。所以

Z[i] = Z[i-1] + dZ[i-1]*deltaS是否适合其他 ODE？它会在某些情况下失败，还是 scipy.integrate.solve_ivp/scipy.integrate.ode 比直接使用 Z[i] = Z[i-1] + dZ[i-1]*deltaS?
在我的代码中，我必须规范化 Z[i] 因为 ||Z[i]||并不总是 1. 为什么会发生？一个float数值计算错误？

我对我的问题的回答，至少对圈子有用

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

我发现我在第一个 post 中列出的方程式不适用于我的曲线。所以我看了Gray A., Abbena E., Salamon S-Modern Differential Geometry of Curves and Surfaces with Mathematica. 2006，发现对于任意曲线，Frenet方程应该写成

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)

其中 ||r'||（或 ||r'(s)||）是 diff([x(s), y(s), z(s)], s).norm()

现在问题与第一个 post 中的问题有所不同，因为没有 r'(s) 函数或离散数据数组。 所以我认为这适合于评论以外的新回复。

我在尝试求解新方程式时遇到了 2 个问题：

如果使用 scipy 的 solve_ivp，我们如何使用 r'(s) 进行编程？
我尝试修改我的gaussian solution，但结果完全错误。

再次感谢

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()

求解 Frenet 框架的非线性 ODE 系统

Solve a nonlinear ODE system for the Frenet frame

python

scipy

differential-equations