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From: Rob Johnson on 26 Jun 2010 06:24
In article <4c254961$0$31370$4fafbaef(a)reader1.news.tin.it>,
Joubert <trappedinthecloset9985(a)yahoo.com> wrote:
>On 26/06/2010 1.27, Robert Israel wrote:
>
>> No, I mean for |y(x)| to go to infinity as x approaches a finite value.
>
>Then take for instance
>
>y'=(y^2 - 1)(y^2 + x^2)
>y(0)=y*
>
>I know for sure that if y* > 1 there is not a globally defined solution
>(as opposed to when |y*| < 1) but why?
>It doesn't look like the right hand member ever goes to infinity (except
>when y does).

However, |y(x)| goes to infinity as x approaches a finite value, just
as Robert Israel specified. If y(0) > 1, then for x > 0, we have

y' > (y(0) - 1) y^3

Separating variables and integrating from 0 to x, we get

|\y 1
(y(0) - 1) x < | --- dt
\| y(0) t^3

1 1 1
= - ( ------ - --- )
2 y(0)^2 y^2

Thus, y goes to infinity for some x no greater than

1
-------------------
2 (y(0) - 1) y(0)^2

Rob Johnson <rob(a)trash.whim.org>
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From: Joubert on 26 Jun 2010 09:42
On 06/26/2010 12:24 PM, Rob Johnson wrote:

> However, |y(x)| goes to infinity as x approaches a finite value, just
> as Robert Israel specified. If y(0)> 1, then for x> 0, we have
>
> y'> (y(0) - 1) y^3
>
> Separating variables and integrating from 0 to x, we get
>
> |\y 1
> (y(0) - 1) x< | --- dt
> \| y(0) t^3
>
> 1 1 1
> = - ( ------ - --- )
> 2 y(0)^2 y^2
>
> Thus, y goes to infinity for some x no greater than
>
> 1
> -------------------
> 2 (y(0) - 1) y(0)^2
>

Thanks. I was trying to convince myself that it cannot be globally
defined "by hand", I mean trying to "draw" the graph of a possible
solution but I was going nowhere.
Partially changing topic: do you have knowledge of any type of inversion
for the existence and uniqueness theorem? I mean what can one say about
f if y'=f(x,y) has a unique solution?
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