Stability of non-linear pendulum-like equation

Question

I am working with a differential equation of the from $$ \ddot{x} + a \dot{x} + b\sin{x} - c = 0, $$ where the initial condition is an equilibrium point: $\dot{x}(0) = 0, x(0) = \arcsin{\frac{c}{b}}$, and after some critical time $\tau > 0$, the constant $b$ changes to a different constant $b'$. The full equation would therefore look something like $$ \ddot{x} + a \dot{x} + b\sin{x} + \left(b'\Theta(t-\tau) - b\Theta(t-\tau)\right)\sin{x} - c = 0, $$ where $\Theta(\cdot)$ denotes the Heaviside step function. I am trying to determine for which values of the constants $a,b,c,b',\tau$ the solution will converge to a stable equilibrium, but have little clue where to start. Any input - either numerical, theoretical/analytical is of great interest.

Answer 1

Some ideas in case of near linearity.

Assuming $\sin x \approx x$ and transforming Laplace we have

$$ X(s) = \frac{(s^2+a s)x_0 + s x'_0 + c}{s(b-\delta b e^{s \tau}+a s+ s^2)} $$

Here the stability is associated with the root's location for $D(s)=s(b-\delta b e^{s \tau}+a s+ s^2)=0$. Making $s = x + i y$ we have

$$ \cases{ R(x,y,a,b,\delta b,\tau)=a x^2-a y^2+b x+\delta b e^{\tau x}( y \sin (\tau y)-x\cos (\tau y))+x^3-3 x y^2=0\\ S(x,y,a,b,\delta b,\tau)=2 a x y+b y-\delta b e^{\tau x}( x \sin (\tau y)+y \cos (\tau y))+3 x^2 y-y^3 = 0 } $$

now numerically we can calculate a set of parameters such that the $D(s)$ roots are located into a convenient stability region. Follows a MATHEMATICA script which implements a possible way.

sr = b x + a x^2 + x^3 - a y^2 - 3 x y^2 - db E^(tau x) x Cos[tau y] + db E^(tau x) y Sin[tau y];
si = b y + 2 a x y + 3 x^2 y - y^3 - db E^(tau x) y Cos[tau y] - db E^(tau x) x Sin[tau y];
sol = NMinimize[{sr^2 + si^2, x <= -1.5, y >= 1} /. {tau -> 0.8 }, {a,b, db, x, y}]
equ0 = X''[t] + a X'[t] + (b - db HeavisideTheta[t - tau]) Sin[X[t]] -  c /. sol[[2]] /. {tau -> 0.2 , c -> 1/2}
tmax = 10;
solX = NDSolve[{equ0 == 0, X[0] == 0.2, X'[0] == 0}, X, {t, 0, tmax}][[1]]
Plot[Evaluate[X[t] /. solX], {t, 0, tmax}, PlotRange -> All,  PlotStyle -> {Thick, Blue}]

NOTE

The results are tested against the nonlinear plant.

$$ X''(t)+a X'(t)+(b-\delta b \Theta (t-\tau ))\sin (X(t))-c=0 $$

Answer 2

The answer doesn't depend on either $b$ or $\tau$ , since you start on an equilibrium point you will remain there for all time $t<\tau$ so you only need to consider the second part which is tricky due to the $c$ term . It will be quite tedious to analyze the stability using lyapunov method so I will give an analysis by intuition and simple physics dealing with the system as inverted pendulum with damping and constant torque $c$ applied .

Notice that :

1 ) If $|c|>|b'|$ your system will not have an equilibrium point , in this case the damping coefficient $a$ determines the behavior :

1. If $a<0$ it will blow up or stabilize in a state where $\dot{x}$ is periodic .
2. If $a> 0$ it will stabilize in a state where $\dot{x}$ is periodic .

2 ) If $|c|<|b'|$ the system will have two sets of equilibrium points : $y=\{\arcsin(\frac{c}{b'})+2n \pi\}$ and $z=\{\arcsin(\frac{c}{b'})+(2n+1) \pi\}$ , where $n \in \mathbb{Z}$ ,then are two cases depending on $a$ :

1. If $a>0$ one set will be stable equilibrium and the other will be saddle equilibrium depending on the signs of $c$ and $b'$ , you might also get a semi-stable limit cycle .
2. If $a<0$ one set will be unstable of equilibrium and the other will be saddle equilibrium depending on the signs of $c$ and $b'$ , you might also get a semi-stable limit cycle .

3 ) If $|c|=|b'|$ the set of equilibrium points will be degenerate as result of bifurcation between the two sets $y$ and $z$ .