Let $u'(t)=Au^2-Bu$. Find conditions on A, B to guarantee global solution.

Question

I'm taking a real variable course and we have just covered the Banach Contraction Principle. Our professor sometimes makes problems up on the spot for us to try and figure out together. This is one for which we were unable to come up with a solution.

Given $u'(t)=Au^2 - Bu$, $t>0$, $u(0)=u_0$, $u(x): \Bbb R \to \Bbb R$. Prove $u$ has a unique solution, and find conditions on $A$, $B$, $u_0$ to guarantee global solution for $A$, $B >0$.

We tried to consider the Banach Contraction Principle but failed to force our c value to be <1. From there we attempted various substitutions (like $u'=ve^{Bt}$) but nothing seemed to work.

He said that this problem highlights difficulties that occur with nonlinear problems and that he would work on it but has yet to give us a solution. I am curious if anyone here has any hints for solutions that may work?

Answer 1

Let us transform $u'(t)=Au^2 - Bu$ via dividing by $u^2$ to $$u^{-2}u'=-B u^{-1}+ A$$. Then subtitute $z:= u^{-1} $, hence $z' = -u^{-2}u' $ and then the upper equation becomes $$z' = Bz -A$$ where you now can apply your Banach Contraction Principle

Answer 2

$$u'=Au^2-Bu=u(Au-B)$$ has solutions that grow for $u<0$, fall for $0<u<B/A$ and grow again superlinearly for $u>B/A$. Thus you get overall bounded solutions in the middle interval, solutions bounded for $t\ge0$ for $u_0<0$ and unbounded solutions (with a pole at finite time) for $u_0>B/A$.