# How to find lyapunov function for the system?

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.

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.

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

• 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. Jun 25, 2019 at 19:53
• @HansLundmark Is there an easy way to show that? (I mean not graphically) Jun 25, 2019 at 20:04
• 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. Jun 25, 2019 at 20:08
• @HansLundmark Interesting, I have added that to the answer, hopefully done everything right. Jun 25, 2019 at 21:30
• It's looking fine! Jun 26, 2019 at 7:19