Exploring a Dimensionless ODE and its Double Root

[Dustinsfl]

dimensionless ODE

$\displaystyle\frac{du}{dt}=ru\left(1-\frac{u}{q}\right)-\frac{u^2}{1+u^2}$

$\displaystyle U=r-\frac{ru}{q}$ and $\displaystyle V=\frac{u}{1+u^2}$

Show using conditions of a double root that the steady state is given parametrically by

$\displaystyle r=\frac{2a^3}{(1+a^2)^2}$, $\displaystyle q=\frac{2a^3}{a^2-1}$

$\displaystyle U'=-\frac{r}{q} = \frac{1-u^2}{(1+u^2)^2}=V'$

Solving for r

$\displaystyle r=\frac{(u^2-1)q}{(1+u^2)^2}$

Now substitution

$\displaystyle\frac{(u^2-1)q}{(1+u^2)^2}-\frac{(u^2-1)u}{q}=\frac{u}{1+u^2}$

If I solve for q, I will have a quadratic. Is there a mistake or something I am not seeing?

---------- Post added at 01:27 PM ---------- Previous post was at 12:06 PM ----------

I solved via Mathematica. Everything was correct.

 

[Aaron Bex]

The steady state is given parametrically by

$\displaystyle r=\frac{2a^3}{(1+a^2)^2}$, $\displaystyle q=\frac{2a^3}{a^2-1}$