Globally defined and unique solutions.

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SUMMARY

The discussion focuses on the uniqueness and global definition of solutions to specific Cauchy problems in differential equations. The first problem, defined by the equation y'=(y^2-1)(y^2+x^2) with initial condition y(0)=y_0, demonstrates that if |y_0| < 1, the solution is globally defined on R, while if y_0 > 1, the solution is not globally defined. The second problem, y'=1/y - 1/x with initial condition y(1)=1, also exhibits a globally defined solution on the interval (0, +infinity). The discussion emphasizes the importance of analyzing the behavior of y and its derivative to establish global uniqueness.

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Malmstrom
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Take a Cauchy problem like:

[tex]y'=(y^2-1)(y^2+x^2)[/tex]
[tex]y(0)=y_0[/tex]

Show that the problem has a unique maximal solution.
Show that if [tex]|y_0| < 1[/tex] the solution is globally defined on R whereas if [tex]y_0 > 1[/tex] 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
[tex]y'=1/y - 1/x[/tex]
[tex] y(1)=1[/tex]
with a solution that should be globally defined on (0,+infinity).
Any insight is helpful.
 
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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.
 

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