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For constants $a,b,c > 0 $

we're given the Lorenz-System:

$$\begin{cases}x'=a(y-x) \\ y'=bx -y - xz \\z'= xy -cz\end{cases}$$

I just want to find a Lyapunov function and don't know what I've done wrong:

Use $L(x,y,z)=x^2+ay^2+az^2$ We know that $$\dot L=\nabla L \cdot f= 2\begin{pmatrix}x \\ ay \\az\end{pmatrix}\cdot\begin{pmatrix}a(y-x) \\ -bx-y-xz \\xy-cz\end{pmatrix}$$

Edit:

As Robert mentioned :

$\dot L =-4ax^2+(2ab+4a)xy-2ay^2-2acz^2 $

So

$2x+2ay+2az=-4ax^2+(2ab+4a)xy-2ay^2-2acz^2$

$x+ay+az=-2ax^2+(ab+2a)xy-ay^2-acz^2$

  • 2
    It should be $\dot{L} = -4\,{x}^{2}a+ \left( 2\,ab+4\,a \right) yx-2\,a{y}^{2}-2\,ac{z}^{2}$ – 2017-01-27
  • 0
    @RobertIsrael: Thank you ! :D – 2017-01-27
  • 0
    @ Robert Israel: Would you check my edit, please ? – 2017-01-27

1 Answers 1

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For a global Lyapunov function $L$ you want $\dot{L} \le 0$ for all $x,y,z$. In this case, if $a > 0$ and $c > 0$ the quadratic form $−4ax^2+(2ab+4a)xy−2ay^2−2acz^2$ is negative semidefinite if and only if $ b^2 + 4 b - 4 \le 0$, i.e. $ -2 - 2 \sqrt{2} \le b \le -2 + 2 \sqrt{2}$.