Find the Lyapunov function for the nonlinear second order system.

Question

Consider the system $$ \begin{aligned} \dot{x}_1 &= -x_1+x_2,\\ \dot{x}_2 &= -x_2^3. \end{aligned} $$ The origin is obviously globally asymptotically stable. Indeed, $x_2\to 0$ and the dynamics of $x_1$ is linear and thus converges to zero for the converging input $x_2$.

However, I got stuck trying to find a Lyapunov function for this system. Could you please help?

Answer 1

Define $v=x_1-x_2$. Then $\dot{v} = -x_1+x_2+x_2^3 = -v+x_2^3$. Consider the system $$ \begin{aligned} \dot{v} &= -v + x_2^3, \\ \dot{x}_2 &= -x_2^3. \end{aligned} $$ Choose $V = v^2+\frac{1}{2}x_2^4$. Then $$\dot{V} = -2v^2 + 2vx_2^3-2x_2^6.$$ Since $2vx_2^3 \le v^2+x_2^6$ we have $$\dot{V}\le -v^2-x_2^6.$$ Thus the origin $(v,x_2)=(0,0)$ is globally asymptotically stable, and the same holds for $(x_1,x_2)$.

Answer 2

Best I could find is using SOSTOOLS (see also the manual, especially chapter 4.2), but its not a "nice" solution:

$$ V(x) = v(x)^TP v(x) $$

$$ v(x) = \begin{pmatrix} x_1 \\ x_2 \\ x_1^2 \\ x_1 x_2 \\ x_2^2 \end{pmatrix} $$

$$ P = \begin{pmatrix} 0.2270 & -0.2270 & 0.0000 & 0.0000 & -0.0000 \\ -0.2270 & 0.9522 & -0.0000 & -0.0000 &0.0000 \\ 0.0000 & -0.0000 & 0.3215 & -0.0000 & -0.0278 \\ 0.0000 & -0.0000 & -0.0000 & 0.5520 & 0.0293 \\ -0.0000 & 0.0000 & -0.0278 & 0.0293 & 0.3014 \end{pmatrix} $$

At least $P > 0$ and $\dot{V}(x) \leq 0$. So it should be a Lyapunov function, but unfortunatly not enough for asymptotic stability...

EDIT: If you include not the full monomials up to fourth degree but only

$$ m(x) = \begin{pmatrix} x_1^2 \\ x_2^2 \\ x_1 x_2 \\ x_2^4 \end{pmatrix} $$

for the optimization then SOSTOOLS computes

$$ \begin{align} V(x) &= (0.51201 x_1 - 0.21772 x_2)^2 + (0.21772 x_1 - 0.90981 x_2)^2 + 0.41052 x_2^4 \\ &= 0.30955 x_1^2 - 0.61911 x_1 x_2 + 0.41052 x_2^4 + 0.87515 x_2^2 \end{align} $$

and

$$ \begin{align} \dot{V}(x) = &-(-1.268 x_2^3 + 0.00000073 x_1 x_2^2 - 0.1309082 x_2 + 0.130908 x_1)^2 \\ &-(0.1309082 x_2^3 + 0.0000034 x_1 x_2^2 + 0.5486238 x_2 - 0.5486187 x_1)^2 \\ &-(0.130908 x_2^3 + 0.0000036 x_1 x_2^2 + 0.5486187 x_2 - 0.5486244 x_1)^2 \\ &- (1.063507 x_2^2 + 0.00006736922 x_1 x_2)^2 \\ &- (0.00000073 x_2^3 - 0.0000019 x_1 x_2^2 - 0.0000034 x_2 + 0.0000036 x_1)^2 \\ &- (0.00006736922 x_2^2 + 0.002748198 x_1 x_2)^2 \end{align} $$

If you scale $V$ this is just:

$$ \frac{V(x)}{0.30955} = x_1^2 - 2 x_1 x_2 + 1.326 x_2^4 + 2.827 x_2^2 $$

Your solution is:

$$ V(x) = x_1^2 - 2 x_1 x_2 + \frac{1}{2} x_2^4 + x_2^2 $$

So it finds a somewhat similar solution but the $\dot{V}$ is of course quite horrendous.

Answer 3

When $x_2$ gets close to the origin its derivative very quickly becomes very small, mean while the dynamics of $x_1$ drives $x_1$ closer to $x_2$. So $x_1$ should be driven to some manifold defined by $x_2$. In order to also utilize this behavior for the Lyapunov function one could define it as

$$ V(x) = \frac{1}{2} (x_1 - f(x_2))^2 + \frac{1}{2} x_2^2. \tag{1} $$

Taking the time time derivative of this yields

$$ \dot{V}(x) = (x_1 - f(x_2)) \left(-x_1 + x_2 + \frac{d\,f(x_2)}{d\,x_2} x_2^3\right) - x_2^4. \tag{2} $$

If there exists a function $f(x)$ such that

$$ x + \frac{d\,f(x)}{d\,x} x^3 = f(x), \tag{3} $$

then $(2)$ can also be written as

$$ \dot{V}(x) = (x_1 - f(x_2)) \left(-x_1 + f(x_2)\right) - x_2^4 = -(x_1 - f(x_2))^2 - x_2^4, \tag{4} $$

which would be negative definite.

The equation from $(3)$ is equivalent to the following differential equation in $f(x)$

$$ \frac{d\,f(x)}{d\,x} = \frac{f(x) - x}{x^3}, \tag{5} $$

which can be written into a form that is easier to solve by using an integrating factor of $e^{1/(2\,x^2)}$, yielding

\begin{align} \frac{d}{d\,x}\left(e^{\frac{1}{2\,x^2}}\,f(x)\right) &= -\frac{e^{\frac{1}{2\,x^2}}}{x^2}, \tag{6a} \\ e^{\frac{1}{2\,x^2}}\,f(x) &= -\int \frac{e^{\frac{1}{2\,x^2}}}{x^2} dx. \tag{6b} \end{align}

The integral from $(6b)$ can be shown to equal

$$ \int \frac{e^{\frac{1}{2\,x^2}}}{x^2} dx = -\sqrt{\frac{\pi}{2}}\, \text{erfi}\!\left(\frac{1}{\sqrt{2}\,x}\right) + c. \tag{7} $$

Therefore, the general solution for $f(x)$ can be expressed as

$$ f(x) = e^{\frac{-1}{2\,x^2}} \left(\sqrt{\frac{\pi}{2}}\, \text{erfi}\!\left(\frac{1}{\sqrt{2}\,x}\right) + c\right). \tag{8} $$

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Since $(8)$ contains expressions which are not very common (mainly the imaginary error function)/ Therefore, I also tried if I could approximate it with a "simpler" expression and I got a decent approximation for $c=0$ using

$$ f(x) \approx \frac{x + 5.506\,x^3 + 1.012\,x^5}{1 + 2.175\,x^2 + 6.107\,x^4 + x^6}. \tag{9} $$

The expression from $(8)$ is not well defined near $x=0$ (although the limit does exist) but the expression from $(9)$ does not suffer from this. However, I have not yet been able to show that the Lyapunov function is also negative definite using this expression for $f(x)$.