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Consider the two one-dimensional linear odes $$\dot x=\lambda_1x\qquad\dot x=\lambda_2x$$ Here $\lambda_1\not=\lambda_2$ and they have the same sign.

Now the solutions to those equations are $x_ie^{\lambda_i t}$, where $x_i$ are some initial conditions. $(i=1,2)$

The flows are $\phi_i^t(x)=xe^{\lambda_i t}$. By Hartman-Grobman's theorem, a homeomorphism that conjugates these flows exists. I want to find one in explicit form. Additionally, can I find a diffeomorphism that conjugates the flows?

Any help would be appreciated.

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Try it out yourself, using a map of the form $x\mapsto x^\gamma$ for some $\gamma$. – Harald Hanche-Olsen May 9 '13 at 14:42
Is that correct, @HaraldHanche-Olsen? – Student May 10 '13 at 4:46
up vote 0 down vote accepted

Thank you, Harald. A desired map is $$x\to x^{\frac{\lambda_1}{\lambda_2}}$$ Another is $$x\to x^{\frac{\lambda_2}{\lambda_1}}$$ Since $\lambda_1\not=\lambda_2$, and they have the same sign, the above maps are diffeomorphisms on the domain of positive reals.

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Actually, $x\mapsto x^{1/3}$ is not differentiable at the origin. It is a homemorphism, however. – Harald Hanche-Olsen May 10 '13 at 7:02
just to add a little bit to the comment, for a diffeomorphism you would require $\lambda_1=\lambda_2$ which does not lead to a nice classification. There the importance to ask for a homeomorphism instead of a diffeomorphism. Naturally the same applies for conjugation between vectorfields of higher dimension – PepeToro May 15 '13 at 7:15
But if we restrict ourselves to the positive reals, both maps are diffeomorphisms, aren't they. Furthermore, if the quotient happens to be a whole number, we still get a diffeomorphism, don't we? @user58533 – Student May 20 '13 at 17:22

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