# Lyapunov and Stability

$\lambda \in \mathbb{R}$. Find $a,b$ such that $V(x,y)=ax^2+by^2$ is Lyapunov with $$\begin{cases} x ' = \lambda x\\ y'= x+\lambda y \end{cases}$$ and find the stability of $(0,0).$

$V'=2ax(\lambda x)+2by(x+\lambda y)=2\lambda a x^2+2byx+2b\lambda y^2$. This quadratic form has the matrix $\left( \begin{array}{ccc} 2\lambda a & b \\ b & 2\lambda b \\ \end{array} \right)$ the characteristic polynomial is $(2\lambda a-z)(2\lambda b -z)-b^2 = 4\lambda^2 ab - z(2\lambda a+2\lambda b)+z^2 -b^2= z^2-z(2\lambda a + 2 \lambda b) + (4\lambda ^2 ab-b^2).$

The roots are $z=\frac{2\lambda a + 2\lambda b \pm \sqrt{4\lambda ^2 a^2 + 8 \lambda ^2 ab + 4 \lambda ^2 b^2 - (16 \lambda ^2 ab-8\lambda^2 a b^3+b^4)}}{2}$

But this seems hard to reduce in order to find the roots.

How can I find the stability of $(0,0)?$ I was thinking of finding the zeroes and then determine stability according to the sign of $V$.

• Why, oh why people forget about Vieta's formulas? Since you want this quadratic form to be positive-definite (all other options will lead to conclusions different from Lyapunov stability), you can easily use them here: all roots must be positive which will lead to positive determinant of quadratic form's matrix and negative trace. – Evgeny Nov 29 '16 at 10:01
• @Evgeny : The trace must be positive, as it is the sum of the roots. What you mean is that the linear coefficient must be negative. But as it is minus the trace, this comes out as the same. – LutzL Nov 29 '16 at 10:46
• @LutzL Yeah, thank you, I've confused signs in a hurry :) – Evgeny Nov 29 '16 at 11:32

## 1 Answer

Set $b=4|λ|$ to get $$\frac12\dot Vf=λax^2 + sign(λ)·((x+2λy)^2- x^2)\\ =sign(λ)·((|λ|a-1)x^2+(x+2λy)^2)$$ so you need only chose $a>\frac1{|λ|}$ to get a definite result.

• Thank you. So to determine stability I use values of $a$ higher than $1/\lambda$? Does it follow that it is lyapunov stable? – Cure Nov 29 '16 at 10:37
• Only for $λ>0$. For $λ=0$ you have to consider the exact solutions and for $λ<0$ the solutions grow away from the origin. – LutzL Nov 29 '16 at 10:42
• Excuse me. Can I ask how did you pick b? I see it works, but I'm not sure how I could have thought of doing that in the first place. – Cure Nov 29 '16 at 23:55
• The pair $(a,b)$ can be multiplied with a positive factor without changing the result. Thus you can fix one of them in a computationally convenient way. I chose to assign $b$ a value that allows a quotient-free form for the completed square of the terms $byx+bλy^2$. – LutzL Nov 30 '16 at 0:15