contique 0.1.6

Numerical continuation of nonlinear equilibrium equations

contique

Numeric continuation of nonlinear equilibrium equations

Fig. 1 Archimedean spiral equation solved with contique

Example

A given set of equilibrium equations in terms of x and lpf (a.k.a. load-proportionality-factor) should be solved by numeric continuation of a given initial solution.

Function definition

def fun(x, lpf, a, b):
    return np.array([-a * np.sin(x[0]) + x[1]**2 + lpf, 
                     -b * np.cos(x[1]) * x[1]    + lpf])

with its initial solution

x0 = np.zeros(2)
lpf0 = 0.0

and function parameters

a = 1
b = 1

Run contique.solve and plot equilibrium states

Res = contique.solve(
    fun=fun,
    x0=x0,
    args=(a, b),
    lpf0=lpf0,
    dxmax=0.1,
    dlpfmax=0.1,
    maxsteps=75,
    maxcycles=4,
    maxiter=20,
    tol=1e-6,
)

For each step a summary is printed out per cylce. This contains needed Newton-Rhapson iterations per cycle if the step was successful and an information about the control component at the beginning and the end of a cycle. As an example the ouput of Steps 71, 72 and 73 are shown below.

| Step (Cycle) | Control Comp. | Equili. | Status        |
|--------------|---------------|---------|---------------|
|   71     (1) |   -1  =>   -1 | 1.1e-13 | Success ( 4#) |
|   72     (1) |   -1  =>   +2 | 1.4e-09 | Recycle       |
|          (2) |   +2  =>   +2 | 6.1e-11 | Success ( 3#) |
|   73     (1) |   +2  =>   +2 | 1.1e-10 | Success ( 3#) |

Next, we have to assemble the results

X = np.array([res.x for res in Res])

and plot the solution curve.

import matplotlib.pyplot as plt

plt.plot(X[:, 0], X[:, 1], "C0.-")
plt.xlabel('$x_1$')
plt.ylabel('$x_2$')
plt.plot([0],[0],'C0o',lw=3)
plt.arrow(X[-2,0],X[-2,1],X[-1,0]-X[-2,0],X[-1,1]-X[-2,1],
          head_width=0.075, head_length=0.15, fc='C0', ec='C0')
plt.gca().set_aspect('equal')

Fig. 2 Solution states of equilibrium equations solved with contique