# Globally defined and unique solutions.

1. Jun 25, 2010

### Malmstrom

Take a Cauchy problem like:

$$y'=(y^2-1)(y^2+x^2)$$
$$y(0)=y_0$$

Show that the problem has a unique maximal solution.
Show that if $$|y_0| < 1$$ the solution is globally defined on R whereas if $$y_0 > 1$$ it is not.
I'm having trouble with this type of questions: how does one prove global uniqueness? Only via the usual theorems? How does one show a solution is "globally" defined on R?

Another example is
$$y'=1/y - 1/x$$
$$y(1)=1$$
with a solution that should be globally defined on (0,+infinity).
Any insight is helpful.

2. Jul 1, 2010

### Eynstone

Let's try to make a qualitative sketch of y. First, note that when |y|<1 ,y' will be negative (x^2+y^2 is always nonnegative). If we start with |y_0|<1 , y'(0) will be negative and y will keep decreasing. When y approaches -1,y' approaches 0. Thus, the rate of decrese of y becomes negligible. The same happens with decreasing x : y becomes +1 asymptotically. Hence, we have a global solution (i.e. , over (-inf. ,inf)).
This is not the case when y_0>1 and x-> -inf. ; hence the solution is not global. The second problem can be solved by a similar reasoning.