Can you help me with next problems?
Find controls $u_2, u_2$ for system $$\left\{ \begin{array}{c} x_1'=ax_2x_3, \\ x_2'=bx_1x_3 +u_2, \\ x_3'=cx_1x_2 +u_3, \\ \end{array} \right.$$ using back-steping, for stabilization of the point $(0,0,0)$ (asymptotic stable), where $a>0, b>0, c>0$.
- Prove, that system $$\left\{ \begin{array}{c} x_1'=ax_2x_3, \\ x_2'=bx_1x_3, \\ x_3'=cx_1x_2 +u_3, \\ \end{array} \right.$$ are unstable (means point $(0,0,0)$) for any control $u_3$.
Thanks!
The second system is not stabilizable because it has a first integral. Consider $$ V(x_1,x_2,x_3)= bx_1^2-ax_2^2; $$ for any $u$ the derivative along the trajectories $$ \dot V= 2bx_1 \dot x_1-2a x_2\dot x_2= 2abx_1x_2x_3-2abx_1x_2x_3=0. $$ Update: As @Kwin van der Veen has suggested in his comment, you can substitute the variables $u_2=v_2-bx_1x_3$, $u_3=v_3-cx_1x_2$ and obtain $$ \left\{\begin{array}{lll} \dot x_1&=&ax_2x_3\\ \dot x_2&=&v_2\\ \dot x_3&=&v_3. \end{array}\right. $$ The subsystem $\dot x_1=ax_2x_3$ is stabilized by, among other, $x_2=-x_1$, $x_3=x_1^2$. Introduce the new variables $z_2= x_2+x_1$, $z_3=x_3-x_1^2$: $$ \left\{\begin{array}{lll} \dot x_1&=&a(z_2-x_1)(z_3+x_1^2)\\ \dot z_2&=&\dot x_1+\dot x_2=a(z_2-x_1)(z_3+x_1^2)+v_2\\ \dot z_3&=&\dot x_3-2x_1\dot x_1= v_3-2ax_1(z_2-x_1)(z_3+x_1^2) \end{array}\right. $$ and consider the Lyapunov function $V(x_1,z_2,z_3)=\frac12x_1^2+\frac12z_2^2+\frac12z_3^2$. The derivative $$ \dot V= x_1\dot x_1+z_2\dot z_2+z_3\dot z_3= ax_1(z_2-x_1)(z_3+x_1^2) $$ $$+ z_2 (a(z_2-x_1)(z_3+x_1^2)+v_2)+z_3(v_3-2ax_1(z_2-x_1)(z_3+x_1^2))=\ldots $$ Can you take it from here?