I have to determine the stability of the system:

$$\begin{cases}x' = xy^4 - 2x^3 - y \\ y' = 2x + 2x^2y^3 -y^7\end{cases}$$

How to fetermina what is Lyapunov function $V$ for this system?

I know that later I have to find:

$$\frac{dV}{dt} =\frac{\partial{V}}{\partial{x}}\frac{dx}{dt} + \frac{\partial{V}}{\partial{y}}\frac{dy}{dt}$$

but I can not realize how to find $V$ and where should I pay attention at.


2 Answers 2


Assume that a Lyapunov function $V$ exists then $$\dot{V} = \Big(x\dfrac{\partial V}{\partial x}-y^{3}\dfrac{\partial V}{\partial y}\Big)(y^4-2x^2)-y\dfrac{\partial V}{\partial x}+2x\dfrac{\partial V}{\partial y}$$ This suggests the function $V = x^2+\dfrac{y^2}{2}$ which is postive definite and radially unbounded. Further, we have $$\dot{V} = -(y^4-2x^2)^2$$ which is negative semi-definite. So we can conclude that the system is globally stable.

  • $\begingroup$ Nice, exactly the same example I found! $\endgroup$
    – SampleTime
    Jun 25, 2019 at 15:57
  • $\begingroup$ but how do we find $\dot{V}$? $\endgroup$
    – M.Mass
    Jun 25, 2019 at 16:27
  • $\begingroup$ @M.Mass You calculate it using the second formula from your post. $\endgroup$
    – SampleTime
    Jun 25, 2019 at 17:54

Using $x_1 = x$ and $x_2 = y$, you have the nonlinear planar system

$$ \begin{align} \dot{x}_1 &= x_1 x_2^4 - 2 x_1^3 - x_2 \\ \dot{x}_2 &= 2 x_1 + 2 x_1^2 x_2^3 - x_2^7 \end{align} \tag{1} $$

which has an equilibirum at $(x_1, x_2) = (0, 0)$. We can choose the Lyapunov function

$$ V(x_1, x_2) = x_1^2 + \frac{1}{2} x_2^2 $$

which is globally positive definite and radially unbounded. Its derivative is

$$ \dot{V}(x_1, x_2) = -(2 x_1^2 - x_2^4)^2 $$

so $\dot{V}(x_1, x_2) \leq 0$ and the system is Lyapunov stable. This was found with SOSTools, which can be applied here since the vector field is polynomial.

Edit: To account for comment by Hans Lundmark.

To show asymptotic stability with LaSalle, we need to show that no trajectory except the trivial solution can stay in the set $S = \{ (x_1, x_2) \in \mathbb{R}^2 : \dot{V}(x_1, x_2) = 0 \}$.

The condition $\dot{V}(x_1, x_2) = 0$ is true if $x_1 = \pm \frac{1}{\sqrt{2}}x_2^2$, so along the curve

$$ C(x_2) = \begin{bmatrix} \pm \frac{1}{\sqrt{2}}x_2^2 \\ x_2 \end{bmatrix} $$

The tangent vector to this curve is given as

$$ T(x_2) = \frac{d}{d x_2} C(x_2) = \begin{bmatrix} \pm \sqrt{2} x_2 \\ 1 \end{bmatrix} $$

Insert $x_1 = \pm \frac{1}{\sqrt{2}}x_2^2$ into the vector field in $(1)$ to get

$$ F(x_2) = \begin{bmatrix} -x_2 \\ \pm \sqrt{2} x_2^2 \end{bmatrix} $$

For $F$ and $T$ to be parallel, there must exist $k \in \mathbb{R}$ such that $T(x_2) = k F(x_2) \, \forall x_2 \in \mathbb{R}$. For the first row this is ensured by $k = \mp \sqrt{2}$. However, for the second row this choice of $k$ leads to $\mp 2 x_2^2 = 1$, which cannot be true for all $x_2 \in \mathbb{R}$.

So, the vector field is transversal to the curves $C$ and no solution (except the trivial) can stay in $S$ and so, the system is asymptotically stable.

  • 1
    $\begingroup$ Asymptotic stability follows from LaSalle's theorem, since (away from the origin) the vector field is transversal to the curves $x_1= \pm x_2^2/\sqrt2$ where $\dot{V}=0$, so that the only trajectory contained in the set where $\dot{V}=0$ is the equlilbrium point itself. $\endgroup$ Jun 25, 2019 at 19:53
  • $\begingroup$ @HansLundmark Is there an easy way to show that? (I mean not graphically) $\endgroup$
    – SampleTime
    Jun 25, 2019 at 20:04
  • 1
    $\begingroup$ Just substitute $x_1=\pm t^2/\sqrt2$ and $x_2=t$ into the vector field, and compare the result to the tangent vector to the curve at that point, and it should be algebraically obvious that they are non parallel. $\endgroup$ Jun 25, 2019 at 20:08
  • $\begingroup$ @HansLundmark Interesting, I have added that to the answer, hopefully done everything right. $\endgroup$
    – SampleTime
    Jun 25, 2019 at 21:30
  • 1
    $\begingroup$ It's looking fine! $\endgroup$ Jun 26, 2019 at 7:19

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