Consider the system 1
$$ \dot x = f(x) = 1 − x^2$$
Consider this above equation on the domain $x > 1$ (something very similar should be true for the case x < 1).
Find explicitly the solutions of the system 2 given below,
$$\dot y = \frac{f(y)}{(1 + |f(y)|)}$$
Where f is defined above. We need to show that for the second dynamical system given above
i) solutions exist for all times τ (maximal interval of existence is (−∞, ∞)),
and
ii) find the mapping of time and/or phase space which shows that these two equations are topological conjugate to each other.
=== My Attempt
I noticed the typo above, I got an imaginary number when I put -1 limit in $\sqrt{y^2-2}$. I should have written $\sqrt{y^2-2}$ in the final equation. But clearly, it seems wrong so I did not bother correcting it.
So I thought of breaking the second integral into 3 cases in order to handle the modulus function. But this assumes the initial condition is less than $-1$. I am pretty sure my answer is wrong since when I differentiate my solution I don't get back my question.
Please help me integrate this second system and find an explicit solution for the flow of the differential equation. After that, I can try to find a homeomorphism that shows these two systems are topologically conjugate.
Basically, help me do this integral -
$$\int \frac{|1-y^2|}{1-y^2} = \int \frac{|z|dz}{z \times -2y}$$
okay, rearranging we get: $$\frac{1+|f(y)|}{f(y)}\dot{y}=1$$ so we can express it as: $$\int\frac{1+|f(y)|}{f(y)}dy=\int dt\tag{1}$$ now we also know that: $$\frac{1+|f(y)|}{f(y)}=\frac{1+|1-y^2|}{1-y^2}=\frac{1}{1-y^2}+U_1(y)$$ where we will define: $$U_1(y)=\begin{cases}1&-1<y<1\\-1&\text{otherwise}\end{cases}$$
now the first part of the integral is easy but lets look at the second part: $$\int_0^YU_1(y)dy\tag{2}$$ it is clear that if $-1\le Y\le1$ then we have: $$\int_0^YU_1(y)dy=Y$$ Now just do the same for the other regions: $$\int_0^YU_1(y)dy=\begin{cases}-(Y+2) & Y<-1\\Y & -1\le Y\le1\\-(Y-2)& Y>1\end{cases}$$ Hopefully this is what you were looking for
To visualise what this looks like, try doing the following in a graphing calculator: $$f(y)=1-y^2$$ $$g(y)=\frac{1+|f(y)|}{f(y)}$$ Plot: $$x=\int_0^yg(z)dz$$ this will give you a plot for $t$ vs $y$