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Problem 3.6 in Khalil's Nonlinear Control: Use given Lypunov candidate function to prove that the origin is exponentially stable. The system is $$\dot{x}=\begin{bmatrix}x_2\\-h(x_1)-2x_2\end{bmatrix},\hspace{0.5cm}h(x_1)=x_1\left(2+\frac{x_1^2}{1+x_1^2}\right)$$ and the given Lyapunov candidate function is $$V(x)=\int_0^{x_1}h(s) ds+\frac{1}{2}\left(x_1+x_2\right)^2$$

To prove exponential stability, we need to find four positive constants, $k_1,k_2,k_3,a$, such that $$k_1\lVert x \rVert^a \le V(x) \le k_2\lVert x \rVert^a$$ and $$\dot{V}(x)\le -k_3\lVert x \rVert^a$$ We can write the Lyapunov function more explicitly as $$V(x)=\frac{3x_1^2}{2} + \frac{1}{2}\left(x_1+x_2\right)^2 - \frac{1}{2}\ln(1+x_1^2)$$ and its time derivative as $$\dot{V}(x)=-\left(\frac{x_1}{\sqrt{2}} + \frac{x_2}{\sqrt{2}}\right)^2 - \frac{3x_1^2}{2}-\frac{x_2^2}{2}-\frac{x_1^4}{1+x_1^2}$$ It is relatively clear that we can use $a=2$, and that there exists some $k_1,k_2,k_3$ that will hold. But how do we find suitable values for the $k_i$'s ?

Rodrigo de Azevedo
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drC1Ron
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You don't necessarily have to find exact constants $k_1,k_2,k_3$, only need to show that there exists some positive constants. In your example above, I can say that

$$ \dot{V} \leq -\bigg(\frac{x_1 + x_2}{\sqrt{2}}\bigg)^2 = -\frac{1}{2}\|x\|^2, $$

so $k_3 = 1/2$ would work. For the function $V(x)$ itself, we know that it is positive definite. Note that the log term is always negative so

$$ V(x) \leq \frac{3}{2}x_1^2 + \frac{1}{2}\|x\|^2 = 2\|x||^2 - \frac{3}{2}x_2^2 \leq 2\|x\|^2, $$

so $k_2 = 2$ would work. For $k_1$, you can just pick an arbitrarily small positive constant that satisfies the lower bound.

Josh Pilipovsky
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