contique 0.1.7

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-8,
    overshoot=1.05
)

For each step a summary is printed out per cylce. This contains an information about the control component at the beginning and the end of a cycle as well as the norm of the residuals along with needed Newton-Rhapson iterations per cycle. As an example the ouput of some interesting steps 31-33 and 38-40 are shown below. The last column contains messages about the solution. On the one hand, in step 32, cycle 1 the control component changed from +1 to -2, but the relative overshoot on the final control component -2 was inside the tolerated range of overshoot=1.05. Therefore the solver proceeds with step 33 without re-cycling step 32. On the other hand, in step 39, cycle 1 the control component changed from -2 to -1 and this time the overshoot on the final control component -1 was outside the tolerated range. A new cycle 2 is performed for step 39 with the new control component -1.

|Step,C.| Control Comp. | Norm (Iter.#) | Message     |
|-------|---------------|---------------|-------------|

(...)

|  31,1 |   +1  =>   +1 | 7.6e-10 ( 3#) |             |
|  32,1 |   +1  =>   -2 | 1.7e-14 ( 4#) |tol.Overshoot|
|  33,1 |   -2  =>   -2 | 4.8e-12 ( 3#) |             |

 (...)

|  38,1 |   -2  =>   -2 | 9.2e-12 ( 3#) |             |
|  39,1 |   -2  =>   -1 | 1.9e-13 ( 3#) | => re-Cycle |
|     2 |   -1  =>   -1 | 2.3e-13 ( 4#) |             |
|  40,1 |   -1  =>   -1 | 7.9e-09 ( 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