This is problem 1.53 from Ordinary Differential Equations by Chicone (2nd Ed).

Prove that there are open intervals $U, V\subset \mathbb{R}$ both containing the origin and a differentiable map $H: U \to V$ with a differentiable inverse such that the flow $\phi_t$ of $\dot x=-x$ is conjugate to the flow $\psi_t$ of $\dot y = y^2 -y$. That is $$ H(\phi_t(H^{-1}(y)))=\psi_t(y)$$ whenever $y \in V$.

So far, I have that $$\phi_t(x)=xe^{-t} \quad \mbox{and}\quad\psi_t(y) = \frac{1}{1-e^t\left(1-\frac1y\right)}$$

Which are the two flows induced by the above equations. Now I'm trying to find the open intervals $U$ and $V$ and function $H$ such that $$H(H^{-1}(y)e^{-t}) = \frac{1}{1-e^t\left(1-\frac1y\right)}$$

Judging by the phase portrait of these systems, I'm pretty sure that I can take both $U$ and $V$ to be $(-\infty,\epsilon)$, where $\epsilon$ is some small number. I am not sure how to find the conjugacy $H$.


You almost have it. You just need to put the transformation $H$ in the correct way. If you define $y=H(x)$, clearly $H$ must satisfy some ODE.

$$ (-x)\frac{dH}{dx}=H^2-H. $$

The integration of the ODE would give you the transformation $H$. By doing the computation you'd get

$$ H(x)=\frac{y_0x}{y_0x-y_0x_0+x_0} $$

where $y_0=H(x_0)$. It is now a matter of checking that $H(x_0e^{-t})=y(t)$.

  • $\begingroup$ Thank you for the help! I was trying to find $H$ by starting with the assumption that $H(xe^{-t}) = y(t)$ but I was having trouble getting the solution using this method for some reason. Your way works. $\endgroup$ – Bob_Bobingston Mar 18 '14 at 15:55

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