Numerical continuation of nonlinear equilibrium equations
Project description
contique
Numeric continuation of nonlinear equilibrium equations
Fig. 1 Archimedean spiral equation solved with contique
Theory of contique
's numeric continuation
A solution curve for (n)
equilibrium equations fun
in terms of (n)
unknowns x
and a load-proportionality-factor lpf
should be found by numeric continuation from an initial equilibrium state fun(x0, lpf0) = 0
. Contique's numeric continuation method is best classified as a
- component-based continuation with an adaptive
- magnitude-based control-component switching.
Extended equilibrium equations
The lpf
value is appended to the unknows x
which gives the so-called extended unknowns y = [x, lpf]
. One additional control equation is added to the equilibrium equations to ensure (n+1)
equations in terms of (n+1)
extended unknowns (see next section). This reduces the solution to a point on the initial solution curve.
Control Equation
The control equation is defined as follows: First, a needle-vector with dimension (n+1)
is created and filled with zeros needle = 0
. For a given initial signed control component j
the needle is positioned at needle[|j|] = 1
. The maximum allowed values per component are calculated as ymax = y0 + np.sign(j) dymax
. The control equation is finally formulated as f(y) = needle.T (y - ymax)
.
Solution technique
The numeric solution process is divided into three main parts:
- Step
- Cycle
- Iteration (...of a Newton-Rhapson root method)
- Cycle
As the name implies, a Step tries to find the extended unknowns for the next step forward of the equilibrium state. For each Cycle, the initial control component has to be evaluated first (see comment below). The additional control equation is evaluated with this initial control component. The generated extended equilibrium equations in terms of the extended unknows are now solved with the help of a root method (Newton-Rhapson Iterations). The solution of the root method dy
is further normalized as dy/dymax
and the final control component is evaluated as j = |j| sign((dy/dymax)[|j|])
with |j| = argmax(|dy/dymax|)
. If the control component changed, another Cycle is performed with the initial control component being now the final control component of the last cycle. This Cycle-loop is repeated until the control component does not change anymore.
A note on the pre-evaluation of the initial control component of a Step: This is performed by the linear solution of the extended equilibrium equations. It is equal to the result of the first Iteration of the Newton-Rhapson root method.
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 cycle
. 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
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