Exponential stability of a time varying system

Question

I want to show that the system below is exponentially stable and I want to estimate its region of attraction

\begin{align} \dot x_1 &= -x_1+x_2+(x_1^2+x_2^2)\sin(t) \\ \dot x_2 &= -x_1-x_2+(x_1^2+x_2^2)\cos(t) \end{align}

I tried using a Lyapunov function $V(x_1,x_2)=\frac 12 (x_1^2+x_2^2)$ and I have found

\begin{align} \dot V(x_1,x_2) &= -x_1^2+x_1x_2+x_1(x_1^2+x_2^2)\sin(t)-x_2x_1-x_2^2+x_2(x_1^2+x_2^2)\cos(t) \\ &\le -x_1^2-x_2^2+x_1(x_1^2+x_2^2)+x_2(x_1^2+x_2^2) \end{align}

And when $x_1^2+x_2^2 \le 1$ we have

\begin{align} \dot V &\le -x_1^2-x_2^2+x_1+x_2 \end{align}

so that $\dot V \le 0$ for $\lvert x_1 \rvert \ge 1$ and $\lvert x_2 \rvert \ge 1$. Hence, we can only have the 2 conditions when $\dot V = 0$ and $x_1^2 + x_2^2 = 1$.

I suppose the region of attraction should be the unit circle. I'm not sure if I need to find a Lyapunov function with a negative definite derivative to ensure exponential stability, as this condition is more closely related to asymptotic stability, isn't it?

EDIT:

The equilibrium is $(0,0)$, I linearised the system and got a Jacobian with a negative real part for the eigenvalues. Therefore, I can use the Lyapunov equation $A^TP+PA=-Q$ with A equal to the Jacobian and $Q=I$ and get a matrix $P$ so that $V=x^TPx$ and $\dot V = -x^TQx$.

However, I'm not sure if this gives the exponential stability?

Answer 1

Consider the same Lyapunov candidate, $$V(x) = \frac{1}{2} \left( x_1^2 + x_2^2 \right).$$

Differentiating along the vector field like you did,

$$ \begin{aligned} \dot{V}(x) &= -x_1^2+x_1\,x_2+x_1\,(x_1^2+x_2^2)\,\sin(t)-x_2\,x_1-x_2^2+x_2\,(x_1^2+x_2^2)\,\cos(t),\\ &= -x_1^2-x_2^2+x_1\,(x_1^2+x_2^2)\,\sin(t)+x_2\,(x_1^2+x_2^2)\,\cos(t)\end{aligned}$$

Now this is where I depart from your manipulation. Let me give you a hint to give you an idea of where I will take this.

Hint: $A \sin(t) + B \cos(t) = \sqrt{A^2 + B^2} \cos\left(t - \measuredangle A + j B\right)$ or, alternatively, use Cauchy-Scwharz as suggested by @Lutz (Sorry, I only saw your suggestion after I wrote this answer!).

Step 1:

Factor to find, $$\dot{V}(x) = -x_1^2 - x_2^2 + \left(x_1^2 + x_2^2\right)\left(x_1 \sin(t) + x_2 \cos(t)\right).$$ Use the trignometric identity, $$\dot{V}(x) = -x_1^2 - x_2^2 + \left(x_1^2 + x_2^2\right)^{3/2}\cos\left(t - \measuredangle x_1 + j x_2\right).$$ We now have a non-negative term multiplying the $\cos$ which can be bounded.

Step 2:

Now employ a bound on $\cos,$ $$\dot{V}(x) \leq -x_1^2 - x_2^2 + \left(x_1^2 + x_2^2\right)^{3/2},$$ and factor, $$\dot{V}(x) \leq 2\,\left(-1 + \sqrt{x_1^2 + x_2^2}\right)\,V(x)$$ Local exponential stability inside the unit circle is assured with an appropriate application of the exponential stability theorem with Lyapunov (I don't recall it having a name). The theorem requires (1) $V(x)$ to be bounded (above and below) by a multiple of $\|x\|$, (2) $\dot{V}(x)$ to be bounded above by a constant, negative multiple of $V(x)$ and (3) the norm of $\nabla V$ to be bounded above by a multiple of $\|x\|$. The Lyapunov candidate satisfies (1) and (3). We just showed (2) as long as you pick some open set contained within the unit circle: the exponential rate of convergence is determined by the initial condition's radius.

Answer 2

I'm unsure if the following analysis can be admitted. Perhaps there is a simpler approach.

From your original Lyapunov function candidate

$$ V(\mathbf{x}) = \frac{1}{2} x_{1}^{2} + \frac{1}{2} x_{2}^{2} .$$

The derivative of $V(\mathbf{x})$ along the trajectories of the system is given by

$$ \dot{V}(\mathbf{x}) = x_{1} \dot{x}_{1} + x_{2} \dot{x}_{2} ,$$

$$ \dot{V}(\mathbf{x}) = x_{1} \left[- x_{1} + x_{2} + \left(x_{1}^{2} + x_{2}^{2}\right) \sin(t)\right] + x_{2} \left[- x_{1} - x_{2} + \left(x_{1}^{2} + x_{2}^{2}\right) \cos(t)\right] ,$$

$$ \dot{V}(\mathbf{x}) = - x_{1}^{2} - x_{2}^{2} + x_{1} \left(x_{1}^{2} + x_{2}^{2}\right) \sin(t) + x_{2} \left(x_{1}^{2} + x_{2}^{2}\right) \cos(t) .$$

Let $\sin(t) = p$, and $\cos(t) = q$. Then

$$ \dot{V}(\mathbf{x}) = - x_{1}^{2} - x_{2}^{2} + p \left(x_{1}^{2} + x_{2}^{2}\right) x_{1} + q \left(x_{1}^{2} + x_{2}^{2}\right) x_{2} .$$

Since $p(t)$ and $q(t)$ are sinusoids, 8 points are chosen for $(p, q)$ to estimate the regions of attraction, and they are $(0, 1)$, $(0.7071, 0.7071)$, $(1, 0)$, $(0.7071, -0.7071)$, $(0, -1)$, $(-0.7071, -0.7071)$, $(-1, 0)$, $(-0.7071, 0.7071)$.

If we solve the 8 inequalities for $\dot{V} \leq 0$, then the 8 corresponding regions of attraction are found:

$$ x_{2} \leq 1 ,$$

$$ x_{2} \leq - \frac{\sqrt{2} x_{1} - 2}{\sqrt{2}} ,$$

$$ x_{1} \leq 1 ,$$

$$ x_{2} \geq \frac{\sqrt{2} x_{1} - 2}{\sqrt{2}} ,$$

$$ x_{2} \geq -1 ,$$

$$ x_{2} \geq - \frac{\sqrt{2} x_{1} + 2}{\sqrt{2}} ,$$

$$ x_{1} \geq -1 ,$$

$$ x_{2} \leq \frac{\sqrt{2} x_{1} + 2}{\sqrt{2}} .$$

The inequality plot is shown below. If there are more points, say 360 points, then the region of attraction should resemble a unit circle as you have predicted earlier.