Do we have uniqueness of solutions to the IVP $\dot{x}(t) = x(t)^2, x(0) = 0?$

Question

Consider the following nonlinear initial value problem:

$$\dot{x}(t) = x(t)^2, \qquad x(0) = 0.$$

Clearly, one solution to the above equation, which holds for all time $t \in (-\infty, \infty)$, is simply $x \equiv 0$.

Is $x \equiv 0$ the only $C^1$ solution to the above IVP that can be defined in an interval about $t =0$? Or do there exists other solutions?

I do not have much experience with studying nonlinear equations, and I am wondering if there is a standard technique for investigating uniqueness of solutions for this type of IVP.

One naive attempt I have made is to try separation of variables:

$$\dot{x}(t) = x(t)^2, x(0) =0 \implies \int_0^{u(t)} \frac{du}{u^2} = \int_0^t dt.$$ But of course the first integral doesn't make any sense because $u^{-2}$ is not integrable near $u =0$.

Hints or solutions are greatly appreciated!

Answer 1

By Picard-Lindelöf's existence and uniqueness theorem, $x \equiv 0$ is indeed the only solution on an interval containing $0$.

Answer 2

As has been mentioned, there exists a general theorem that handles this. Here's an ad hoc proof of uniqueness:

There exists $d\in(0,1)$ so that if $I=[-d,d]$ then $$|x(t)|\le\frac12\quad(t\in I).$$

Since $x'=x^2$ this shows that $|x'|\le1/4$ on $I$. Hence, since $x(0)=0$, $$|x(t)|\le\frac14d<\frac14\quad(t\in I).$$

Hence $|x'|\le1/16$ on $I$, so that $|x|\le1/16$ on $I$. Etc. Officially, an induction shows that $|x|<???$ on $I$, hence $x=0$ on $I$.

(And now since $x(d/2)=0$, the same argument shows that $x=0$ on $[-d/2,3d/2]$. Etc, so $x=0$ on $\Bbb R$.)