Sternberg’s theorem and smooth conjugacy

Suppose we have a dynamical system $\dot x=f(x)$ and a steady state $x_0$ . Let $\phi$ be a homeomorphism defined on a neighbourhood of $x_0$ with $\phi (x_0)=0$ . If we transform the system to coordinates $y=\phi(x)$ under what circumstances can $\phi$ be chosen so that the transformed system is equal to the linearization of the original system about $x_0$ ? The answer depends on the regularity required for $f$ and $\phi$ . If $f$ is $C^1$ and $x_0$ is hyperbolic (i.e. all eigenvalues of the linearization at $x_0$ have non-vanishing real part) then there always exists a continuous mapping $\phi$ with this property. (In that case we cannot necessarily transform the vector field but we can transform its integral curves.) This is the Grobman-Hartman theorem. Even if $f$ is $C^\infty$ it is not in general possible to choose $\phi$ to be $C^2$ .

Sternberg’s theorem shows that if $f$ is $C^\infty$ and there are no resonances then $\phi$ can be chosen to be $C^\infty$ . The definition of a resonance in this context is as follows. Let $\lambda_i$ be the eigenvalues of the linearization. A resonance is a relation of the form $\lambda_j=\sum_i m_i\lambda_i$ with positive integers $m_i$ satisfying $\sum_i m_i\ge 2$ . For example, if the eigenvalues are $\lambda_1=1$ and $\lambda_2=2$ we have the resonance $\lambda_2=2\lambda_1$ . Consider the two-dimensional case. If there the eigenvalues are real with opposite signs then there can be no resonance and Sternberg’s theorem applies to all saddles in two dimensions. If both eigenvalues are positive then it is known that $\phi$ can be chosen to be $C^1$ but resonances can occur and there are cases where although $f$ is analytic $\phi$ cannot be chosen to be $C^2$ . Even when there are no resonances it is in general not the case that if $f$ is analytic the mapping $\phi$ can also be chosen analytic. It is necessary to make a further assumption (no small divisors). This says roughly speaking that the expressions $\lambda_j-\sum_i m_i\lambda_i$ should not only be non-zero but that in addition they should not be able to approach zero too fast as the number of summands increases. This is the content of a theorem of Siegel.

All the results discussed up to now concern the hyperbolic case. What happens in the presence of purely imaginary eigenvalues? The Grobman-Hartman theorem has a generalization, the theorem of Shoshitaishvili, also known as the reduction theorem. In words it says that there is a continuous mapping $\phi$ which reduces the system to the product of the flow on any centre manifold and a standard saddle. A standard saddle is linear and so if the centre manifold is trivial this reduces to Grobman-Hartman. The next question is what can be achieved with mappings $\phi$ of higher regularity. A fact which is implicit in the reduction theorem is that linear systems where the origin is a hyperbolic fixed point whose stable and unstable manifolds have the same dimension are topologically equivalent. Evidently they are in general not smoothly equivalent since any smooth mapping preserving the origin will not change the eigenvalues of the linearization. However once this has been taken into account the theorem does generalise under the assumptions that there are no resonances among the eigenvalues with non-zero real parts.More precisly the eigenvalues $\lambda_i$ in the definition of a resonance must have non-zero real parts. By contrast quantity $\lambda_j$ should either be an eigenvalue with non-zero real part or zero. The result being referred to here is Takens’ theorem.

Suppose that for a steady state we introduce coordinates $x$ , $y$ and $z$ corresponding to the stable, unstable and centre manifolds. Then the form achieved by the reduction in Takens theorem is that we get the equations $\dot x=A(z)x$ , $\dot y=A(z)y$ and $\dot z=g(z)$ for matrices $A$ and $B$ and a function $g$ depending on the central variables $z$ . One technical point is that while for a smooth system the reduction can be done by a mapping which is $C^k$ for any finite $k$ it cannot in general be done by a mapping which is itself infinitely differentiable. As an example, consider a two-dimensional system and a fixed point where the linearization has one zero and one non-zero eigenvalue. In this case there can be no resonances and the theorem says that there is a $C^k$ coordinate transformation which reduces the system to the form $\dot x=a(y)x$ , $\dot y=b(y)$ with $a(0)\ne 0$ , $b(0)=0$ and $b'(0)=0$ .

This entry was posted on August 23, 2017 at 3:40 pm and is filed under dynamical systems. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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