# Prove that solution blows up in finite time [duplicate]

I want to show that solution of this Cauchy problem

$$\begin{cases} u'(t)=u(t)^2 + t \\ u(0)=0 \end{cases}$$

is defined for $$t \in [0,\alpha]$$, with $$\alpha <3$$

I tried to integrate, but it's not possible because of the $$t$$. I can see that $$u'>0$$ for all $$t \geq 0$$ , and that's all. I don't know how to argue honestly

• These equations are known as Riccatti equation – Surb Aug 23 '20 at 14:47
• I was trying to use the substitution $u^2 = w$, as I didn't know these equations @Surb – andereBen Aug 23 '20 at 14:50

Note that $$u(t) \geq \int_0^t sds = t^2/2$$ so that $$u(\sqrt{2})\ge 1$$.
Since $$u'(t)/u(t)^2 \ge 1$$, we can integrate both sides from $$\sqrt{2}$$ to $$t$$ and we see that $$-\frac{1}{u(t)}+\frac1{u(\sqrt{2})}\ge t-\sqrt{2},$$ whenever $$t>\sqrt{2}$$.
Thus $$\frac{1}{u(t)} \le \sqrt{2}-t+\frac1{u(\sqrt{2})}$$. We know $$u(\sqrt{2})\geq 1$$, so that $$\frac{1}{u(\sqrt{2})}\leq 1$$.
Thus, if $$t \ge \sqrt{2}$$ then $$u(t)>\frac{1}{\sqrt{2}+1-t}$$. Therefore the blow-up time is at most $$\sqrt{2}+1<3$$.