Talk:Andronov-Hopf bifurcation

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    In section "Two-dimensional Case" the sytem of ODEs would imply that \(\dot{x}_1 = \dot{x}_2\) for an initial value \(x_1(0) = x_2(0)\). Shouldn't this be two different functions?

    Yu.A.: Of course, \(x_1(0) = x_2(0)=a\) does not imply \(\dot{x}_1 = \dot{x}_2\), since

    \[ \dot{x}_1=\beta a - a + \sigma a(a^2+a^2)=(\beta-1)a+2\sigma a^3 \]


    \[ \dot{x}_2= a + \beta a + \sigma a(a^2+a^2)=(\beta+1)a+2\sigma a^3 \]

    So, function \(f\) isn't unique? It is used for both equations in the ODE system.

    Yu.A. The "function \(f\)" assigns to each \((x_1,x_2)\) two numbers \((f_1(x_1,x_2),f_2(x_1,x_2))\). Please, read some basic ODE textbook.

    So this isn't unique. I already read ODE books but I am used to identifying indexes etc.. I am sorry for misunderstanding. This article is altogether more than excellent. But I have to admit that in my eyes the dimensionality of "the" function f is not correct. As you wrote there is a tuple \((f_1(x_1,x_2),f_2(x_1,x_2))\) which components should be used in these two equations instead of f.

    Yu.A. Both. f is one vector-function of the vector-argument.

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