Almost linear ODE: how node becomes a spiral - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T04:28:47Z http://mathoverflow.net/feeds/question/45253 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/45253/almost-linear-ode-how-node-becomes-a-spiral Almost linear ODE: how node becomes a spiral Igor Belegradek 2010-11-08T02:26:15Z 2011-10-17T16:37:41Z <p>Most introductory ODE books contain a discussion of almost linear systems, and there are two cases when the behavior of an almost linear system near an equilbrium point can differ from the behaviour of a linearlized system. One of these two cases is when the linearlized system has a repeated nonzero real eigenvalue; then the equilibrium solution of the linearlized system is a node, while the original almost linear system could be a node or a spiral, and <b>I wish to see an explicit example of when the spiral occurs.</b> </p> <p>To give you an idea of what I need, let's discuss the other case when the linearized system is a center, but the almost linear system could be a center or a spiral, and the example when the spiral occurs is $x^\prime=y+x(x^2+y^2)$ and $y^\prime=-x+y(x^2+y^2)$. </p> <p>UPDATE: I now feel much better about the question because it turned out not as silly as I initially feared. Apparently, given a $2\times 2$ almost linear system $X^\prime=F(X)$ if the solution $X=0$ of the linearized system $X^\prime=AX$ is a node, then it is also a node for $X^\prime=F(X)$, as long as $F$ is $C^2$. If $F$ is merely $C^1$ there is a counterexample indicated in the comments. In cartezian coordinates the counterexample is $x^\prime=-x-\frac{2y}{\ln(x^2+y^2)}$ and $y^\prime=-y+\frac{2x}{\ln(x^2+y^2)}$; the right hand side is not $C^2$.</p>