Stabilize system about origin using Backstepping Design

Question

I would like to use backstepping design to formulate a control $u(x_1,x_2)$ to globally stabilize the following system about the origin $(0,0)$. $$ \dot x_1 = -x_1+\bigg(\dfrac{1}{12}x_1-1\bigg)x_2$$ $$ \dot x_2 = u(x_1,x_2) $$ I am familiar with the process of backstepping design, but for this specific problem I can't seem to come up with a desired value for the virtual control $x_2$. At first I tried using a desired value $x_{2 \rm d}\triangleq 0$, but after the usual process (define error, change coords, evaluate lyapunov function derivative) I get the control $ u = 1 - \dfrac{1}{12}x_1 -x_2 $ which does stabilize the system, but not about the origin, since the control does not reduce to zero at the origin.

Answer 1

For $V_1=\frac{1}{2}x_1^2$ we have $$\dot{V}_1=x_1\dot{x}_1=-x_1^2+\left(\frac{x_1^2}{12}-x_1\right)x_2$$ If we now define the new error variable $z_2=x_2-\alpha_1$ with virtual control $\alpha_1$ we can write equivalently $$\dot{V}_1=-x_1^2+\left(\frac{x_1^2}{12}-x_1\right)\alpha_1+\left(\frac{x_1^2}{12}-x_1\right)z_2$$ The above equation guides us to select $$\alpha_1(x_1)=-\left(\frac{x_1^2}{12}-x_1\right)$$ so that $$\dot{V}_1=-x_1^2-\left(\frac{x_1^2}{12}-x_1\right)^2+\left(\frac{x_1^2}{12}-x_1\right)z_2$$ For the dynamics of $z_2$ we have $$\dot{z}_2=\dot{x}_2+\left(\frac{x_1}{6}-1\right)\dot{x}_1=u+\left(\frac{x_1}{6}-1\right)\left(-x_1-x_2+\frac{x_1x_2}{12}\right)$$ For $V_2:=V_1+\frac{1}{2}z_2^2$ we then have $$\dot{V}_2= -x_1^2-\left(\frac{x_1^2}{12}-x_1\right)^2+\left[\frac{x_1^2}{12}-x_1+u+\left(\frac{x_1}{6}-1\right)\left(-x_1-x_2+\frac{x_1x_2}{12}\right)\right]z_2$$ Choosing therefore $$u=x_1-\frac{x_1^2}{12}+\left(\frac{x_1}{6}-1\right)\left(x_1+x_2-\frac{x_1x_2}{12}\right)-z_2\\ =2x_1-x_2-\frac{x_1^2}{6}+\left(\frac{x_1}{6}-1\right)\left(x_1+x_2-\frac{x_1x_2}{12}\right)$$ we have that $$\dot{V}_2\leq -2V_2$$ and $V_2$ converges exponentially fast to zero. From this one easily deduces the exponential convergence of $x_1,x_2$ to zero.

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Edit: As pointed by Thomas $\alpha_1=0$ can also be used as a virtual control law. Then, $$\dot{V}_1=-x_1^2+\left(\frac{x_1^2}{12}-x_1\right)z_2.$$ Also, the dynamics of $z_2$ are $$\dot{z}_2=\dot{x}_2=u$$ and $$\dot{V}_2= -x_1^2+\left(\frac{x_1^2}{12}-x_1+u\right)z_2$$ Therefore choosing

$$u=x_1-\frac{x_1^2}{12}-z_2=x_1-\frac{x_1^2}{12}-x_2$$

we have that $$\dot{V}_2= -2V_2$$ and $V_2,x_1,x_2,u$ converge exponentially fast to zero.