So I have the problem

$$u'(x) = u^2(x), (x\in [0,\infty), u(0) = 0$$

Now if I define

$$f:[0,a] \times B_R(0)\rightarrow\mathbb{R}$$

then $|f| \leq R^2$ on $x \in [0,a]$ $u B_R(0) $

also $\partial_uu^2 = 2u \leq 2R \implies f$ is Lipschitz.

So by the Picard Lindelof theorem there exists a unique solution on

$$[0, \min(a,\frac{1}{R}) =\frac{1}{R} ]$$ since we can make $a$ as large as we like.

Now, I have to answer the question

"Is the maximally extended solution unique?"

Well, I am not sure, I know we have local uniqueness, but I am unsure about uniqueness of the maximally extended solution. I would like to say that we can send $a \rightarrow \infty$ and $R \rightarrow 0$ and take upper bound on the interval of uniqueness to be $\min(\infty,\infty) = \infty$

I have another problem to answer which is almost identical, the only difference is the initial condition $u(0) = 1$ this time. A similar process checks that $|f| \leq (R+1)^2$ if

$$f:[0,a] \times B_R(1)\rightarrow\mathbb{R}$$ So this time we have uniqueness on $[0,\min(a,\frac{R}{(R+1)^2})]$

so my question is

$$\text{How do I correctly interpret these results?}$$

As a side note, the first problem has a maximally extended solution $v(x) = 0, \forall x \in [0,\infty)$ and the second has $u = \frac{1}{1-x}, x \in [0,1]$


The estimate $\alpha=\min(a,\frac{R}{M})$ for the existence interval is always only a lower bound, it means that any curve of maximal slope/velocity $M$ can reach a distance of $R$ from its origin only after time $R/M$, there is no necessity that it actually goes to this distance. Thus any manipulations in the second part are unsatisfactory.

Take two different solutions. Then there is a minimal time where they start to be different. Consider the IVP from that point on and what uniqueness says about the curves actually splitting.

  • $\begingroup$ So regarding the question. "Is the maximally extended solution unique?" The answer would be. We cant say for sure.? $\endgroup$ – user197848 May 13 '16 at 13:16
  • $\begingroup$ No, the conclusion should be that a branching point is not possible. $\endgroup$ – LutzL May 13 '16 at 13:43

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