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I'll show my doubt with the Cauchy problem $\begin{cases} y'=1+y^2 \\ y(0)=0 \end{cases}$. I know the solution is $y(t)=tg(t)$, but let's say I don't know the explicit solution. If I look at $1+y^2$, then I have all the hypothesis for the application of Cauchy-Lipshitz Theorem in $[0, 2\pi]$ for instance, because it is lipshitz there (or not?). So I have a unique solution in $[0,2\pi]$, but now if I wonder if I can prolong the solution, how can I understand that in fact I cannot prolong the solution to $2\pi$ but at best to $\pi/2$ without solving the differential equation? I mean, where do I commit a mistake in using the Cauchy-Lipshitz Theorem? If you have some other (not simple like mine) examples that have the same problem, I would like to see how to solve them in the right way. Thanks for the help.

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  • $\begingroup$ The right hand side $1+y^2$ is indeed Lipschitz continuous for $y \in [-M,M]$ for any $M$. (note that right hand side does not depend on $t$). But the Lipschitz constant goes to infinity if $M \to \infty$. Therefore the solution is indeed unique, but it could go to infinity in finite time, just as $y = \tan t$ does. $\endgroup$ – Hans Engler Mar 24 '17 at 18:40
  • $\begingroup$ Sorry, could you explain it? I don't understand why it can go to infinity finite times. I mean, the theorem states that I have a unique solution defined on all the interval (which is $\mathbb{R}$ in this case), but $tg(t)$ is not always defined. What am I missing? $\endgroup$ – tommy1996q Mar 24 '17 at 20:26
  • $\begingroup$ @tommy1996q, false. You have only local existence: en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 24 '17 at 21:23
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Here is the statement of Cauchy-Lipschitz:

Suppose we are given an ODE, $$y'(t) = f(y(t)), \ \ \ \ \ \ \ y(0) = y_0,$$ where $f$ is a real function defined for $y \in [y_0 - b, y_0 + b]$, obeying the boundedness condition $$ |f(y)| \leq m,$$ and obeying the Lipschitz condition $$|f(y_1) - f(y_2)| \leq k |y_1 - y_2 |.$$

Then for any $$a < \min \left(\tfrac b m, \tfrac 1 k \right),$$ there exists a unique solution $y(t)$ to the ODE, valid for $t \in [-a, a]$, with $y(t)$ taking values in the range $[y_0 - b, y_0 + b]$.

Note that Cauchy-Lipschitz does NOT give us a solution for all values of $t$! It only gives us a solution for $t$ in the range $[-a, a]$, where $a$ is restricted by (i) the range of $y$ on which $f(y)$ is defined compared to how big $f(y)$ gets on this range of $y$, and by (ii) how "contracting" the function $f(y)$ is.

Let's now apply this theorem to our ODE. In our ODE, $f(y) = 1 + y^2$ and $y_0 = 0$. For the time being, let us fix a value of $b$ (so we are looking for solutions such that $y(t)$ always stays within the interval $[-b,b]$ at all values of $t$.)

On the interval $[-b,b]$, we have $$ |f(y)| \leq 1 + b^2,$$ so $|f(y)|$ is bounded by $m = 1 + b^2$. Furthermore, $$ |f'(y)| = 2|y| \leq 2b,$$ so, by the mean value theorem, we learn that $f$ is $k$-Lipschitz with $ k = 2b.$

Taking $$a < \min \left( \tfrac{b}{1 + b^2}, \tfrac 1 {2b} \right) = \begin{cases} \tfrac{b}{1 + b^2} & b \in (0, 1); \\ \tfrac 1 {2b} & b \in (1,\infty),\end{cases}$$ Cauchy-Lipschitz tells us that there exists a solution, valid for $t \in [-a, a]$, which takes values within the range $y(t) \in [-b,b]$.

The largest possible bound on $a$ is obtained when we take $b = 1$: in this case, we learn that there exists a solution valid for $t \in (-\tfrac 1 2, \tfrac 1 2)$ (and this solution takes values in the range $y(t) \in [-1,1]$, though this is less interesting).

In fact, the solution is $$ y(t) = \tan t,$$ which is valid for $t \in (- \tfrac \pi 2, \tfrac \pi 2)$. So our method of applying Cauchy-Lipschitz has given us an under-estimate for the range of $t$ on which the solution is valid.

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  • $\begingroup$ I think I have understood now. Tell me if I'm wrong. The problem is that I have local existence, so once fixed $b$, that is the maximum variability of $y$ allowed, I am bound to take $a$ like that, and the interval I described couldn't be used for the theorem. Just one thing, I knew you needed local lipshitz for $y$ and uniform continuity for $t$, never heard about finiteness of $f$, is this hypothesis equivalent to the uniform continuity in this case? $\endgroup$ – tommy1996q Mar 25 '17 at 14:18
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    $\begingroup$ I have looked at the wiki page of the theorem and I think I got it now. Long story short, the proof is based on a contraction, and the values $b/m$ and $1/k$ are needed because the former makes you fall into the interval you chose for $y$, while the latter gives you a contraction. Thank you very much, you helped a lot. $\endgroup$ – tommy1996q Mar 25 '17 at 15:25
  • $\begingroup$ @tommy1996q Yes, that's exactly right! :-) $\endgroup$ – Kenny Wong Mar 25 '17 at 19:45
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So I have a unique solution in $[0,2\pi]$...

False. We have only local existence:

$$\cdots$$ Then, for some value $\epsilon > 0$, there exists a unique solution $y(t)$ to the initial value problem on the interval $[t_0-\epsilon ,t_0+\epsilon]$.
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