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I haven't been able to find an answer to this question anywhere, so I thought I should post it here. In doing a fixed-points/stability analysis, one is required to find the fixed points of a dynamical system that finding the eigenvalues of the corresponding Jacobian, etc... It is understood what to do when the eigenvalues have complex parts, negative, positive real parts, etc...

However, my question is what if the fixed points themselves are complex? All the examples I have seen involve fixed points of dynamical systems that are real. Does the stability analysis remain the same if the fixed points are complex?

Thank you.

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  • $\begingroup$ If the dynamical system looks for real valued functions $t\mapsto(x(t),y(t))$ defined on some interval $I$ of $\mathbb R$ with values in $\mathbb R^2$ such that $x'(t)=f(x(t),y(t))$, $y'(t)=g(x(t),y(t))$ for every $t$ in $I$, then non real solutions $(z,w)$ in $\mathbb C^2\setminus\mathbb R^2$ of the system $f(z,w)=g(z,w)=0$ are simply irrelevant to the study of the dynamical system. $\endgroup$ – Did Apr 3 '16 at 9:08
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Nothing changes on $\mathbb C^n$, provided that we understand that the conditions (linearization, eigenvalues, etc) are computed for the corresponding dynamics on $\mathbb R^{2n}$. Otherwise one would be tempted to compute the derivatives in $\mathbb C$, while one may even consider dynamics that are not holomorphic.

On the other hand, for an holomorphic dynamics on $\mathbb C^n$, you can simply replace the usual conditions for stability on $\mathbb R^n$ for analogous conditions (really the same) on $\mathbb C^n$, although using the derivative in $\mathbb C$.

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  • $\begingroup$ Hi Jonas. That is a very informative answer, thanks! So, in this case, can one simply take the complex fixed point and evaluate the Jacobian at that point and analyze the eigenvalues as usual? $\endgroup$ – Thomas Moore Apr 2 '16 at 22:33
  • $\begingroup$ Indeed, the same as usual. $\endgroup$ – John B Apr 2 '16 at 22:35
  • $\begingroup$ Much appreciated, thanks!! $\endgroup$ – Thomas Moore Apr 2 '16 at 22:38
  • $\begingroup$ Hi Jonas. A further question to your comment. If we are computing are conditions for the corresponding dynamics on R^{2n}, what is the physical meaning or rather, what is the interpretation for a complex fixed point to be a local sink or source in the real plane? One could not even perceivably draw the phase plot? $\endgroup$ – Thomas Moore Apr 2 '16 at 23:17
  • $\begingroup$ The same as usual: for example $z'=z$ is equivalent to $x'=x$, $y'=y$, since $z=x+iy\equiv(x,y)$. $\endgroup$ – John B Apr 2 '16 at 23:26

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