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I've been hearing about using the sign of the real part of eigenvalues of the multivariate derivative of the change vector field of an ODE at an equilibrium point to determine whether that equilibrium is stable or unstable. Unfortunately, the source is not very good at things like getting the statement of theorems correct, so I was wondering:

Is there a statement and proof of such a theorem I could access online?

I'm particularly interested in if anything can be said about the non-diagonalizable case where the eigenvalues have zero real part.

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says when the equilibrium is stable in the linear system, but not when it is stable in the original system. –  Ricky Demer Aug 21 '10 at 19:12
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Ricky, easy examples such as $y'=\pm y^n, n=2,3$ show that if linearization has purely imaginary eigenvalues, the long term behavior of the system is not determined by its linearization (i.e. you need to know more). –  Victor Protsak Aug 21 '10 at 19:30
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The linear case is of course completely solved via Jordan form. The nonlinear hyperbolic case (=no eigenvalues on the imaginary axis) reduces to the linear via the Grobman-Hartman theorem. The case of purely imaginary eigenvalues is more delicate as Victor says. –  Pietro Majer Aug 21 '10 at 21:23
    
Victor, what are the easy examples for the non-diagonalizable case with purely imaginary eigenvalues? –  Ricky Demer Aug 22 '10 at 6:25
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2 Answers 2

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I got curious about "the non-diagonalizable case" with purely imaginary eigenvalues and came up with the system $$ y' = A y $$ where $$ A \; = \; \left( \begin{array}{rrrr} 0 & -1 & 1 & 0 \\\ 1 & 0 & 0 & 1 \\\ 0 & 0 & 0 & -1 \\\ 0 & 0 & 1 & 0 \end{array} \right) . $$ Taking the matrix of eigenvectors and "generalized eigenvectors" (see http://en.wikipedia.org/wiki/Generalized_eigenvector ) $$ B \; = \; \left( \begin{array}{rrrr} 1 & i & 0 & 0 \\\ i & 1 & 0 & 0 \\\ 0 & 0 & 1 & i \\\ 0 & 0 & i & 1 \end{array} \right) . $$ we get a permuted Jordan form in the shape I prefer for this case, $$ B^{-1} A B \; = \; \left( \begin{array}{rrrr} -i & 0 & 1 & 0 \\\ 0 & i & 0 & 1 \\\ 0 & 0 & -i & 0 \\\ 0 & 0 & 0 & i \end{array} \right) . $$ This is something I made up years ago and forgot, with an original real matrix and a repeated characteristic value $\alpha$ with non-diagonal Jordan block, we get the same block for the complex conjugate $\bar{\alpha}.$ For both the original real matrix $A$ and $B^{-1} A B $ we can rearrange everything into convenient 2 by 2 blocks, in particular the off-diagonal 1's become little 2 by 2 identity matrices. Try it for a six by six example where there is a 3 by 3 block for $\alpha$ with two off-diagonal ones, then a 3 by 3 block for $\bar{\alpha}$ with two off-diagonal ones. Permute the diagonal elements to $ \alpha, \; \bar{\alpha}, \; \alpha, \; \bar{\alpha}, \; \alpha, \; \bar{\alpha} $ and see what happens. By the way, these "forms" probably have standard names.

After a bunch of work I found that the "fundamental matrix" for the system is $$ e^{A t} \; = \; \left( \begin{array}{rrrr} \cos t & - \sin t & t \cos t & - t \sin t \\\ \sin t & \cos t & t \sin t & t \cos t \\\ 0 & 0 & \cos t & - \sin t \\\ 0 & 0 & \sin t & \cos t \end{array} \right) . $$ which means that any solution of the system is $ y = e^{A t} y_0.$ So, if the third and fourth entries in $y_0$ are $0,$ the orbit is a circle. If not, the orbit leaves the origin as $t$ increases. The trick from my first answer, making a not quite linear system, can force the periodic orbits to switch to attracting.

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As Victor says:

I am confident that the system $$ x' = -y + ( x^2 + y^2) x, $$ $$ y' = x + ( x^2 + y^2) y $$ has the origin repelling nearby trajectories, while $$ x' = -y - ( x^2 + y^2) x, $$ $$ y' = x - ( x^2 + y^2) y $$ has the origin attracting nearby trajectories, and $$ x' = -y $$ $$ y' = x $$ has just periodic orbits near the origin. But all three linearize to the same thing at the origin, $$ \left( \begin{array}{rr} 0 & -1 \\\ 1 & 0 \end{array} \right) . $$ with eigenvalues $\pm i.$

EDIT: Indeed, given a constant real number $\lambda$ and system $$ x' = -y + \lambda ( x^2 + y^2) x, $$ $$ y' = x + \lambda ( x^2 + y^2) y , $$ we find that $$ \frac{d}{dt} \; (x^2 + y^2) = 4 \lambda (x^2 + y^2)^2. $$

EDIT some more: so, for the nonconstant paths, if we set time to $0$ when the trajectory crosses the unit circle, we get $$ x^2 + y^2 = \frac{1}{1 - 4 \lambda t} $$ showing that when $\lambda > 0$ the path reaches infinite radius in finite time, while with $\lambda < 0$ the path spirals in to the origin, as expected. Then, if we set $$ x = r \cos \theta, \; y = r \sin \theta $$ as usual, the rate of change of $ \theta $ does not depend on $ \lambda $ and $ \forall \lambda,t$ we have $$ \frac{d \theta}{d t} = 1. $$

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