- #1
Dustinsfl
- 2,281
- 5
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.
$\displaystyle\frac{du}{dt}=ru\left(1-\frac{u}{q}\right)-\frac{u^2}{1+u^2}$
$\displaystyle U=r-\frac{ru}{q}$ 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.