Global existence without solving a differential equation

I have the following cauchy problem:

$$y(t)' = (y(t) - t^3 - 1)^3$$ $$y(0) = \alpha$$ Discuss the global existence when

$1)\,\, \alpha < 1$

$2) \,\,\alpha = 1$

I tried the following:

If you let be $F(y,t) = (y - t^3 - 1)^3$, the function $F(y,t)$ exist for every $(y,t) \in \Re \times \Re$, being continuous and then a lipschitz function. That means the equation has a unique solution given the initial conditions for every $[a,b] \subseteq \Re$ with $a < b$. I know the function isn't sublinear, but this is a sufficent condition. If I would be able of solve the problem I also would find the domain of the solution, but I can't solve it.

I had attemp making $z = y -t^3 - 1$, so I have

$$z(t)' = z^3 - 3t^2$$ $$y(0) = \alpha - 1$$

but I can't solve neither this. That's why I believe that I need to do this without solving it.

• $F$ is not globally Lipschitz with respect to $y$. Although being merely locally Lipschitz with respect to $y$ and continuous with respect to $t$ does guarantee a unique local solution, you can still have blowup in finite time if you do not have a global Lipschitz property. For example, $y'=y^2,y(0)=1$ has a solution on $(-1,1)$ but blows up at $1$. Looking at your equation involving $z$, it should be clear that $z$ is a decreasing function if $z(0) \leq 0$, and so by comparison to my example you should expect blowup in finite time whenever $\alpha \leq 1$. – Ian Aug 2 '14 at 19:24
• That's what I mean, I can see that $y′=y2,y(0)=1$ blow up at t = 1, but I need to solve it. Neither I can't see how ( or why ) the function blow up, only for being a decreasing function. Maybe I'm forgotten a theorem? – Leonhard Aug 2 '14 at 20:18
• Suppose $z'=z^3-3t^2$, $y'=y^3$, and $z(0)=y(0)<0$. Then it should not be too hard to see that $z(t) \leq y(t)$ for all $t \geq 0$ (where both are defined). So because $y$ blows up in finite time, $z$ also blows up in finite time. – Ian Aug 2 '14 at 20:58
• Unfortunately I do not know of a nice theorem for guaranteeing blowup in finite time. – Ian Aug 2 '14 at 21:04
• Oh, that's make sense. Thanks you very much. – Leonhard Aug 2 '14 at 21:05