construction Lyapunov function for global asymptotic stable systems

Question

We know that converse Lyapunov theorems for the conditions that a system is golobal asymptotic stable insure diferentiable Lyapunov functions. In the below I have a nonlinear system that is global asymptotic stable: \begin{gather}\label{equ.3} \begin{aligned} &{\dot{x\ }}_1=-x_1\\ &+h\left(x_1\right)({x_2}^2+{x_3}^2\sin^2\left(e^{x_1}\right)+{x_2}^4{{\mathrm{cos}}^2 (\mathrm{ln}\mathrm{}(1+x^2_1))\ }+{x_2}^2{x_3}^2), \\ &\dot{x_2}=-x_2+x^3_3+x^4_3+x_1x_3, \\ &\dot{x_3}=-x_3-x^2_3x_2-x^3_3x_2-x_1x_2, \end{aligned} \end{gather} where \begin{equation}\label{equ.4} h(x_1)= \begin{cases} 100\frac{\sin^2(x_1)}{x_1},\quad x_1\neq 0\\ 0, \quad x_1=0. \end{cases} \end{equation} To the best of my knowledge, using methods such as the gradient algorithm, sum of square, and Yoshizawa construction fail to compute the Lyapunov functions. I will be grateul if one propose me suitale Lyapunov function.

Answer 1

Consider the following Lyapunov function $V(x_1,x_2,x_3) = \frac{1}{2}x_1^{2} + \frac{100+b}{2}\left(x_3^{2} + x_3^{2}\right)$.

\begin{align} \dot{V} &= x_1\left(-x_1+h\left(x_1\right)({x_2}^2+{x_3}^2\sin^2\left(e^{x_1}\right)+{x_2}^4{{\mathrm{cos}}^2 (\mathrm{ln}\mathrm{}(1+x^2_1))\ }+{x_2}^2{x_3}^2)\right) \\ &\quad + (100+b)x_2\left(-x_2+x^3_3+x^4_3+x_1x_3\right) \\ &\quad + (100+b)x_3\left( -x_3-x^2_3x_2-x^3_3x_2-x_1x_2 \right) \\ &\leq -x^2_1 + 100\sin^2(x_1)({x_2}^2+{x_3}^2\sin^2\left(e^{x_1}\right)+{x_2}^4{{\mathrm{cos}}^2 (\mathrm{ln}\mathrm{}(1+x^2_1))\ }+{x_2}^2{x_3}^2) \\ &\quad -(100+b)x^2_2 -(100+b)x^2_3 \\ &\leq -x^2_1 + 100({x_2}^2+{x_3}^2 + {x_2}^4 +{x_2}^2{x_3}^2) -(100+b)x^2_2 -(100+b)x^2_3 \\ &\leq -x^2_1 + 100({x_2}^4 +{x_2}^2{x_3}^2) -bx^2_2 -bx^2_3 \end{align} Using the fact that $pq \leq \frac{a}{2}p + \frac{1}{2a}q$, then $x^2_2x^2_3 \leq \frac{\sqrt{2}-1}{2}x^4_2 + \frac{1}{2(\sqrt{2}-1)}x^4_3$ and, \begin{align} \dot{V} &\leq -x^2_1 + \frac{100}{2(\sqrt{2}-1)}({x_2}^4 +{x_3}^4) -bx^2_2 -bx^2_3 \\ &\leq -x^2_1 - x_2^2(b-Cx_2^2) - x_3^2(b-Cx_3^2) \end{align} where $C = \frac{100}{2(\sqrt{2}-1)}$. Therefore, if $x_2, x_3 \leq \sqrt{\frac{b}{C}}$ then $\dot{V} \leq 0$. Notice that for $W(x_2,x_3) = \frac{1}{2}(x_2^2+x_3^2)$ that, \begin{align} \dot{W} &= -x^2_2 - x_3^2 = -2W \end{align} Showing that the distance of $(x_2,x_3)$ to (0,0) is always decreasing. If $x^2_2 + x_3^2 \leq \frac{b}{C}$ then $x_2, x_3 \leq \sqrt{\frac{b}{C}}$ for all $t\geq 0$. This means that $\dot{V} \leq 0$ for all $t\geq 0$ if $x_2$ and $x_3$ are initialized such that, $x^2_2(0) + x_3^2(0) \leq \frac{b}{C}$. Lastly, notice that $b$ can be selected arbitrarily large. Therefore, for any finite initial condition $(x_1(0),x_2(0),x_3(0))$ we can select a $b$ such that $\dot{V}\leq 0$.

Answer 2

Let's attempt to find a suitable Lyapunov function in the form:

\begin{equation} V(x) = ax_1^2 + bx_2^2 + cx_3^2 \end{equation}

Now, calculate the time derivative of V(x) along the trajectories of the system:

\begin{equation} \begin{aligned} \dot{V}(x) &= 2ax_1\dot{x}_1 + 2bx_2\dot{x}_2 + 2cx_3\dot{x}_3 \\ &= 2ax_1\left(-x_1 + h(x_1)(x_2^2 + x_3^2\sin^2(e^{x_1}) + x_2^4\cos^2(\ln(1+x_1^2)) + x_2^2x_3^2)\right) \\ &\quad + 2bx_2\left(-x_2 + x_3^3 + x_3^4 + x_1x_3\right) \\ &\quad + 2cx_3\left(-x_3 - x_3^2x_2 - x_3^3x_2 - x_1x_2\right) \end{aligned} \end{equation}

Now, try to determine values for a, b, and c that make $\dot{V}(x)$ negative definite ($i.e., \dot{V}(x) < 0$ for all $x$ except at the equilibrium point). This may involve algebraic manipulations and solving inequalities. Numerical methods or computer algebra systems may be needed to find suitable values for a, b, c.

For example: Consider the simplified system:

\begin{align} \dot{x}_1 &= -x_1 + h(x_1)(x_2^2 + x_3^2) \\ \dot{x}_2 &= -x_2 + x_1x_3 \\ \dot{x}_3 &= -x_3 - x_2x_3 \end{align}

with

h(x_1) = \begin{cases} 100\frac{\sin^2(x_1)}{x_1},\quad x_1\neq 0\\ 0, \quad x_1=0. \end{cases}

We want to find a Lyapunov function of the form:

$V(x) = ax_1^2 + bx_2^2 + cx_3^2$

where a, b, c are positive constants.

Calculate the time derivative of V along the trajectories of the system:

\begin{align} \dot{V}(x) &= 2ax_1\dot{x}_1 + 2bx_2\dot{x}_2 + 2cx_3\dot{x}_3 \\ &= 2ax_1\left(-x_1 + h(x_1)(x_2^2 + x_3^2)\right) \nonumber \\ &\quad + 2bx_2\left(-x_2 + x_1x_3\right) \nonumber \\ &\quad + 2cx_3\left(-x_3 - x_2x_3\right) \end{align}

Now, let's choose suitable values for a, b, c to make $\dot{V}(x)$ negative definite. For simplicity, we'll choose:

$a = 1, b = 1, c = 1.$

Now, simplify $\dot{V}(x)$:

\begin{align} \dot{V}(x) &= -2x_1^2 + 2h(x_1)x_1(x_2^2 + x_3^2) - 2x_2^2 - 2x_3^2 - 2x_2x_3 \nonumber \end{align}

To analyze stability, we want to ensure that $\dot{V}(x) < 0$ for all x except at the equilibrium point (0, 0, 0). However, finding suitable values for a, b, c that guarantee this can be challenging.