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Globally defined and unique solutions.

  1. Jun 25, 2010 #1
    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.
     
  2. jcsd
  3. Jul 1, 2010 #2
    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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