# Qualitative behavior of critical point at the origin

Determine the qualitative behavior of the critical point at the origin for the following system for all possible values of $a$:

$\dot{x} = -y + ax(x^2+y^2)$

$\dot{y} = x + ay(x^2+y^2)$

My question: I attempted to use the Local Center Manifold theorem to show that the center manifold: $x = h(y) = a_0 + a_1y+a_2y^2 +...$ for $a_0, a_1, ...$ are parameters to be determined and $h(0) = h'(0) = 0$, must be $0$. To do this, assume $h(y)\neq 0$ for $y\neq 0$. Now, we replace $x$ by $h(y)$ from the sysem above, and from the identity: $\dot{x} = \dot{y}\ h'(y)$, we get the following equation for all values of $a$:

$-y + a(a_0 + a_1y+ a_2y^2 + ...) (a_0^2+a_1^2y^2 + 2a_0a_1y+2a_0a_2y^2 +...+y^2) = (a_1 + 2a_2y + 3a_3y^2 + ...)[a_0 + a_1y+ a_2y^2 + ... + ay(a_0^2 + a_1^2y^2 + 2a_0a_1y + 2a_0a_2y^2 + ... +y^2)]$

Since $h(0) = h'(0) = 0$, we instantly get $a_0 = a_1 = 0$. But then the $-y$ term on the LHS of the equation above is never cancelled with anything, so the equation, after matching terms by terms, cannot be true for every $y\neq 0$. Thus $h(y)$ does not exist in this case.

Therefore, $h(y) = 0$ is the only choice, which implies $x = 0$. But if this is the case, then $0 = -y$, so $y = 0$ as well. Thus the critical point is a saddle point? Is this a correct conclusion?

• This vector field is a perfect candidate for Lyapunov stability. Are you familiar at all with Lyapunov functions and Lyapunov's second method for stability? – GaussTheBauss Dec 23 '15 at 6:52
• @GaussTheBauss: Thank you for your insight! I'm familiar with it, but it's quite hard to find explicitly a Lyapunov function. How do you decide to approach it by using Lyapunov function? Can you point out for me what went wrong with my method above? – user177196 Dec 23 '15 at 7:07
• Unfortunately, I am not familiar with your method. I did however elaborate a little bit on the Lyapunov method below. Find the correct function comes from practice. The example you have posted is a classic problem. When you vector field is polynomial (and symmetric in your case!), then the Lyapunov function will also be a polynomial, and will be symmetric. – GaussTheBauss Dec 23 '15 at 7:09
• To provide some context you could mention that the system does not satisfy the hypothesis of the en.wikipedia.org/wiki/Hartman-Grobman_theorem – Mikhail Katz Dec 23 '15 at 10:26
• First, you are constantly forgetting to mention people whom you are asking in comments. Second, I've told nothing about equivalence of nonlinear to linear and finding eigenspaces of Jacobi matrix doesn't require that. Third, as I've already said, you have two-dimensional system with two-dimensional center eigenspace. In this case your local center manifold is just any disk around origin. Because center manifold is tangent to center eigenspace and has the same dimension as the eigenspace. – Evgeny Dec 23 '15 at 20:37

I know this is not the method you were going for, but I could not resist offering another possible solution.

According to the Wikipedia article on Lyapunov Stability:

Lyapunov, in his original 1892 work, proposed two methods for demonstrating stability. The first method developed the solution in a series which was then proved convergent within limits. The second method, which is almost universally used nowadays, makes use of a Lyapunov function $V(x)$ which has an analogy to the potential function of classical dynamics. It is introduced as follows for a system $\dot{x} = f(x)$ having a point of equilibrium at x=0. Consider a function $V(x)$ : $\mathbb{R}^n \rightarrow \mathbb{R}$ such that:

• $V(x)=0$ if and only if $x=0$
• $V(x)>0$ if and only if $x \ne 0$
• $\dot{V}(x) = \frac{d}{dt}V(x) = \sum_{i=1}^n\frac{\partial V}{\partial x_i}f_i(x) \le 0$ for all values of $x$ (negative semidefinite). Note: for asymptotic stability, $\dot{V}(x)<0$ for $x > \ne 0$ is required (negative definite).

In your example, a candidate for a Lyapunov function is $V(x,y) = x^2 +y^2$.

Note: the method in the article, as it stands, will give you the behavior for $a=0$ and $a<0$. Here is the additional case:

Suppose $X$ is a $C^{1}$ vector field on an open set $\Omega \subset > \mathbb {R}^n$, $0 \in \Omega$ is a critical point of $X$, and $V : > \Omega \rightarrow \mathbb {R}$ is a continuous function such that

• $V (0) = 0$
• there exists $\Omega_{-} \subset \Omega$ such that $\Omega_{-} \cap B_{\delta}(0)= \emptyset$ for any $\delta >0$, $V (x) < 0 \ \forall > x\in \Omega_{-}$, $V (x) = 0 \ \forall x\in \partial \Omega_{-} \cap > B_{\epsilon} (0)$ for some $\epsilon >0$;
• $V$ is strictly decreasing on the part of orbits that stay in $\Omega$.

Then $0$ is unstable.

In the case of $a>0$, let $V(x,y) = -x^2 -y^2$ and use the theorem above.

• Thank you so much!! Can you try hard to find the result for the case $a > 0$? I still wonder why the Local Center Manifold doesn't work in this case:p – user177196 Dec 23 '15 at 7:09
• There you go! I have added the additional case for instability @user177196 – GaussTheBauss Dec 23 '15 at 7:24
• @GaussTheBauss Well, your method is not so offbeat. Because even when we reduced system to the center manifold, we have to prove the stability of equilibrium and (guess what) Lyapunov function is needed exactly in the same way :) – Evgeny Dec 23 '15 at 8:15
• @GaussTheBauss Also, proving instability can be done with Chetaev's theorem (en.wikipedia.org/wiki/Chetaev_instability_theorem). – Evgeny Dec 23 '15 at 8:24
• Aha! I did not know the name of the theorem. It is however exactly what I added at the end of my answer. Thanks for your input! @Evgeny – GaussTheBauss Dec 23 '15 at 9:06