Finding the region of attraction using a Lyapunov function

basins-of-attractiondynamical systemslyapunov-functionsordinary differential equationsstability-theory

I'm trying to find an estimate for the region of attraction of an equilibrium point. The notes from Nonlinear Control by Khalil suggest that defining
$$
V(x) = x^TPx,
$$
where $P$ is the solution of
$$
PA+A^TP=-I,
$$
will yield the best results for an estimate. It also assures that the estimate can be found from these types of Lyapunov functions for exponentially stable equilibrium points.

The system I am studying is defined by
$$
\begin{gathered}
\dot{x}_1 = x_1 – x_1^3 + x_2 \\
\dot{x}_2 = x_1 – 3x_2.
\end{gathered}
$$
The linealization matrix around the equilibrium point $x^* = \{\frac{2}{\sqrt{3}},\frac{2}{3\sqrt{3}}\}$, is
$A =
\begin{bmatrix}
-3 & 1 \\
1 & -3\\
\end{bmatrix}
$,
and so,
$P =
\begin{bmatrix}
3/16 & 1/16 \\
1/16 & 3/16\\
\end{bmatrix}
$.

Given the previous values, I took the derivative of V(x) w.r.t. time and substituted the system in the equation (did it in Mathematica to try and simplify it as much as possible), giving
$$
\dot{V}(x) = \frac{1}{8}\left(4x_1^2-3x_1^4+4x_1x_2-x_1^3x_2-8x_2^2\right).
$$
Now I need to find a region where $\dot{V}(x)$ is negative definite, but when I try to use inequalities with the norm of $x$ I get only positive norms raised to a power, and so $\dot{V}(x)$ can never be negative definite.

Using
$$
|x_1|\leq||x||, \quad |x_1x_2|\leq\frac{1}{2}||x||^2,
$$
I arrived at
$$
\begin{gathered}
\dot{V}(x) \leq \frac{1}{8}\left(4||x||^2+3||x||^4+2||x||^2+\frac{1}{2}||x||^4+8||x||^2\right) \\
\leq \frac{1}{8}\left(14||x||^2+\frac{7}{2}||x||^4\right).
\end{gathered}
$$
As you can see, it never is negative.

Am I being overly aggresive with converting everything to norms? How can I find the region where $\dot{V}(x)$ is negative?

Thanks in advance.

Best Answer

Correcting an error done.

In my previous answer (this is the edition of changes) I considered

$$ \dot V = (x,y)P(x-x^3+y,x-3y) \ \ \text{translated to}\ \ (x^*, y^*)\ \ \text{which is wrong} $$

The correct procedure is to consider

$$ \dot V = (x-x^*,y-y^*)P(x-x^3+y,x-3y) $$

thus giving

$$ \dot V = -\frac{3 x^4}{8}-\frac{x^3 y}{8}+\frac{5 x^3}{6 \sqrt{3}}+\frac{x^2}{2}+\frac{x y}{2}-\frac{4 x}{3 \sqrt{3}}-y^2+\frac{2 y}{3 \sqrt{3}} $$

thus we have in light blue the region with $\dot V\le 0$ and in black the contour lines for $V$. As we can observe, for the choose Lyapounov function, the small attraction basin is shown by the tangent inner level curve.

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Basin of attraction of simple nonlinear coupled ODE

Hint.

A good stream plot can help a lot!

In red the equilibrium points for $\epsilon = 0.1$.

The corresponding MATHEMATICA script.

epsilon = 0.1;
str = StreamPlot[{x (x - 0.5) (x + 0.5) + epsilon y, 
    y (y - 0.5) (y + 0.5) + epsilon x}, {x, -1, 1}, {y, -1, 1}, 
   MeshFunctions -> Function[{x, y, vx, vy, n}, n], Mesh -> 55];
sols = Quiet[Solve[{x (x - 0.5) (x + 0.5) + epsilon y == 0, 
     y (y - 0.5) (y + 0.5) + epsilon x == 0}, {x, y}]];
equil = Table[Graphics[{Red, Disk[({x, y} /. sols[[k]]), 0.02]}], {k, 1, 
    Length[sols]}];
Show[str, equil]

Estimating the region of attraction of a non-linear system with a Lyapunov Function

You have the system

$$ \begin{align} \dot{x}_1&=\sin(x_2)\\ \dot{x}_2&=-x_1-\sin(x_2) \end{align} $$

which has the linearization matrix

$$ A(x)=\begin{pmatrix}0&\cos{x_2}\\-1&-\cos{x_2}\end{pmatrix} $$

which at the origin is

$$ A=A(0)=\begin{pmatrix}0&1\\-1&-1\end{pmatrix} $$

Now taking $Q=I$ you can solve

$$ PA+A^TP=-Q $$

for $P>0$ and get $P=\begin{pmatrix}3/2&1/2\\1/2&1\end{pmatrix}$ so that

$$ V(x)=x^TPx=3 x_1^2/2 + x_1 x_2 + x_2^2 $$

Now finding a region of attraction can be done by solving

$$ k=\min_x V(x) \text{ s.t. } \dot{V}(x)=0, x\neq0 $$

This is a constrained minimization problem:

$$ \begin{align} k=&\min_x\quad 3 x_1^2/2 + x_1 x_2 + x_2^2 \\ &\text{ s.t. } \quad 2 x_1\sin(x_2) - 2 x_1 x_2 - x_2 \sin(x_2) - x_1^2=0\\ &\phantom{\text{ s.t. }}\quad x\neq0 \end{align} $$

This is not easy to solve analytically but we can solve it numerically. We can solve $\dot{V}(x)=0$ for $x_2$ and get two different solutions which we can substitute in $V(x)$ to get:

$$ V_1(x_2)=\frac{\left(2\,x_{2}-3\,\sin\left(x_{2}\right)\right)\,\left(3\,x_{2}-2\,\sin\left(x_{2}\right)+2\,\sqrt{{x_{2}}^2-3\,x_{2}\,\sin\left(x_{2}\right)+{\sin\left(x_{2}\right)}^2}\right)}{2} $$

and

$$ V_2(x_2)=-\frac{\left(2\,x_{2}-3\,\sin\left(x_{2}\right)\right)\,\left(2\,\sin\left(x_{2}\right)-3\,x_{2}+2\,\sqrt{{x_{2}}^2-3\,x_{2}\,\sin\left(x_{2}\right)+{\sin\left(x_{2}\right)}^2}\right)}{2} $$

both assuming ${x_{2}}^2-3\,x_{2}\,\sin\left(x_{2}\right)+{\sin\left(x_{2}\right)}^2\geq 0.$ Now we can plot both functions:

In this image, only the parts of $V_1$ and $V_2$ are plotted where ${x_{2}}^2-3\,x_{2}\,\sin\left(x_{2}\right)+{\sin\left(x_{2}\right)}^2\geq 0.$ We can now see that we have two minima, which we can use to compute $x_1$. We end up with two solutions:

$$ \begin{align} (x_1,x_2)&=(0.9885, -2.2013)\\ (x_1,x_2)&=(-0.9885, 2.2013) \end{align} $$

The first corresponds to the minimum of the blue graph, the second to the minimum of the red graph. In both cases we have

$$ k=V(0.9885,-2.2013)=V(-0.9885,2.2013)=4.1355 $$

which is the level you are looking for. We can confirm this by checking another plot:

Teal: $\dot{V}(x)\leq 0$
Dark blue: $\dot{V}(x)> 0$
Yellow: $\dot{V}(x)\leq 0$ and $V(x)\leq k$
Black cross: origin
Blue dot: first solution $(x_1,x_2)=(0.9885, -2.2013)$
Red dot: second solution $(x_1,x_2)=(-0.9885, 2.2013)$

Note however that the yellow area in general is only a subset of the region of attraction.

Best Answer

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Basin of attraction of simple nonlinear coupled ODE

Estimating the region of attraction of a non-linear system with a Lyapunov Function

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