Derivative of a Lyapunov function for a nonlinear system

Question

Let $$\begin{aligned}\begin{cases}\dot{x}_{1}=-\left( 2x_{1}-x_{2}\right)^3+\left( x_{1}-x_{2}\right) \\ \dot{x}_{2}= -\left( 2x_{1}-x_{2}\right) ^{3}+2\left( x_{1}-x_{2}\right)\end{cases}\\ \end{aligned}$$ Given $$V\left( \mathbb{x}\right) =\mathbb{x}^{T}P\mathbb{x} \qquad P=\begin{bmatrix} 5 & -3 \\ -3 & 2 \end{bmatrix}$$ Answer: $\dot V\left( x_{1},x_{2}\right) =-2\left[ \left( 2x_{1}-x_{2}\right) ^{4}+\left( x_{1}-x_{2}\right) ^{2}\right]$

My attempt:

We can write $V(\mathbb{x})$ as $$\begin{aligned}V\left( x_1,x_2\right) &=5x_{1}^{2}-6x_{1}x_{2}+2x_{2}^{2}\\ &=\left( 2x_{1}-x_{2}\right) ^{2}+\left( x_{1}-x_{2}\right) ^{2}\end{aligned}$$

and defining the change of variables $z_{1}=2x_{1}-x_{2}$ and $z_{2}=x_{1}-x_{2}$ we obtain $$V(z_1,z_2)=z_1^2+z_2^2$$ and $$\begin{cases}\dot{z}_{1}=-z_{1}^{3}+z_{2}\\ \dot{z}_{2}=-z_{1}^{3}+2z_{2}\end{cases}$$

Now we can calculate $\dot V(z_1,z_2)$ as $$\begin{aligned}\dot V\left( z_{1},z_{2}\right) &=\dfrac{\partial V}{\partial z_{1}}\left( z_{1},z_{2}\right) \dot{z}_{1}+\dfrac{\partial V}{\partial z_{2}}\dot{z}_{2}\\ &=2z_{1}\left( -z_{1}^{3}+z_{2}\right) +2z_{2}\left( -z_{1}^{3}+2z_{2}\right) \end{aligned}$$

Answer 1

This is incorrect: $$\begin{cases}\dot{z}_{1}=-z_{1}^{3}+z_{2}\\ \dot{z}_{2}=-z_{1}^{3}+2z_{2}\end{cases}$$ It should be like this: $$ \dot z_1= 2\dot x_{1}-\dot x_{2}= 2(-\left( 2x_{1}-x_{2}\right)^3+\left( x_{1}-x_{2}\right)) -(-\left( 2x_{1}-x_{2}\right) ^{3}+2\left( x_{1}-x_{2}\right)) $$ $$ =2(-z_1^3+z_2)+z_1^3-2z_2= -z_1^3 $$ $$ \dot z_2= \dot x_1-\dot x_2= -\left( 2x_{1}-x_{2}\right)^3+\left( x_{1}-x_{2}\right) -(-\left( 2x_{1}-x_{2}\right) ^{3}+2\left( x_{1}-x_{2}\right)) $$ $$ =-z_1^3+z_2+z_1^3-2z_2= -z_2 $$

Answer 2

In addition to AVK's answer, I just wanted to point out how for this type of problem you can usually find $\dot{V}(x)$ by leveraging the symmetry of $P$. This sometimes saves a lot of busy work:

First, notice that $V(x)$ can be written as a scalar product: $$V(x)=\langle x,Px\rangle$$ Hence: $$\begin{aligned}\dot{V}(x)&=\langle \dot{x},Px\rangle +\langle x,P\dot{x}\rangle \\&= \langle P^T\dot{x},x\rangle +\langle x,P\dot{x}\rangle \\&= \langle x,P^T\dot{x}\rangle +\langle x,P\dot{x}\rangle\end{aligned}$$ And thanks to $P=P^T$ we obtain: $$\dot{V}(x)=2\langle x,P\dot{x}\rangle = 2x^TP\dot{x}$$

To demonstrate this technique on your case, using $A=-(2x_1-x_2)^3$ and $B=(x_1-x_2)$ as shorthands, we get:

$$\begin{aligned}\dot{V}(x)&=2x^T\left(\begin{array}{cc}5 & -3\\ -3 & 2\end{array}\right)\left(\begin{array}{c}A+B \\ A+2B\end{array}\right) \\ &=2x^T\left(\begin{array}{c}2A-B\\ -A+B\end{array}\right)\\&=2\left(x_1(2A-B)-x_2(A-B)\right) \\ &=2\left((2x_1-x_2)A-(x_1-x_2)B\right) \\&= -2\left( \left( 2x_{1}-x_{2}\right) ^{4}+\left( x_{1}-x_{2}\right) ^{2}\right)\end{aligned}$$

I just find the calculation of $\dot{V}(x)$ to be more straightforward this way in general !