Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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

1 Answer 1

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.

share|improve this answer
2  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.