This problem is taken from here. Consider the initial value problem: $$ u'=f(u,t)=u^2-u^3$$ $$ u(0)= 2/a>0 $$ $a$ is a small constant.
How can I determine wheter a unique solution tom $t=0$ to $t=a$?
My own attempt/thoughts:
I know Picard-Lindelöf Theorem is the way to go here, and in order to make a conclusion one has to find Lipschitz constant (and thereby show that $f$ is lipschitz continuous). So I compute $f_u=2u-3u^2$. I observe that $|f_u|\rightarrow \infty$ as $u\rightarrow\infty$ as well as $u\rightarrow -\infty.$ This means that we need u as negative as possible or positive as possible to select $\max_{u,t}|f_u|$. But which one?
Our domain in which we need the Lipschitz constant is $D=\{ (u,t): |u-a|<c;0 \leq t \leq a\}$ for some positive constant c. $f_u$ is a polymial as must be bounded on a closed ($D$).
I don't know what to make of all this? Please help me reach a conclusion
As I remarked on the previous incarnation of this question, if you reproduce the source correctly it should be $y(0)=a\gtrapprox 0$ for the time interval $[0,2/a]$ with a near vertical jump from zero to one expected at $t=1/a$. (Or to stay in that geometric setup, $y(0)=1/a$, $a\gg1$, on the interval $[0,2a]$.)
Your question has an easy answer:
The right side is polynomial, thus locally Lipschitz, thus all solutions are unique on their domain.
The right side $u^2-u^3=u^2(1-u)$ has roots at $u=0$ and $u=1$ which give constant equilibrium solutions. As the initial value is inside $(0,1)$, the solution under consideration is bounded by these constant solutions, thus can not diverge and so exists on the whole of $\Bbb R$.