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If I have a coupled nonlinear dynamical system, like $$\dot{x}=ax-bxy$$ $$\dot{y}=cxy-dy$$ by using jacobian matrix, I can find that the point ($\frac {d}{c}$,$\frac {a}{b}$) is a center. I think in a nonlinear dynamical system, we cannot ensure that a center obtained by jacobian matrix will be a true center, unless we can find some conserved quantity. But it is hard to guess a conserved quantity at a first glance.

I would like to know is there any analytical method to know that a center is really a center or not? Or given a pure imaginary eigenvalue obtained by jacobian matrix, can we know it is really marginal stable?

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+1. But $-dx$ is $-dy$. –  Did Apr 20 at 9:46
    
center manifold theory is what you want to look into. –  nonlinearism Apr 21 at 14:47

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Everything you say about centers of the linearized differential system vs centers of the true differential system holds. To show that a "linearized" center is a "true" center, I am afraid one is led to find an invariant of the dynamics, at least in a neighborhood of the center--and for that, I know no general approach.

In principle, some quantity $H(x,y)$ is invariant by the differential system $$\dot x=f(x,y),\qquad\dot y=g(x,y),$$ if and only if $H$ solves the PDE $$ \partial_xH(x,y)\cdot f(x,y)+\partial_yH(x,y)\cdot g(x,y)=0. $$ hence to find an invariant of the dynamics is equivalent to solving this PDE.

In the present case however, finding an invariant of the dynamics is relatively direct, once one notes that some quantity $$ H(x,y)=F(x)+G(y) $$ is invariant by the differential system $$ \dot x=f(x,y),\qquad\dot y=g(x,y), $$ if and only if $$ F'(x)\cdot f(x,y)+G'(y)\cdot g(x,y)=0. $$ Here, one gets $$ \frac{x}{d-cx}F'(x)=\frac{y}{a-by}G'(y), $$ hence both sides are constant, say without loss of generality that they are both equal to $1$. Thus, $$ F'(x)=\frac{d}x-c,\qquad G'(y)=\frac{a}y-b, $$ which shows that an invariant of the dynamics on $xy\ne0$ is $$ H(x,y)=d\log|x|+a\log|y|-cx-by. $$ To get an invariant of the dynamics on the whole $(x,y)$-plane, consider $$ H(x,y)=|x|^d\,|y|^a\,\mathrm e^{-cx-by}. $$

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It is very clear, thanks –  John Brown Apr 20 at 21:21

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