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Problem with Nonunique Solutions

  1. Sep 24, 2005 #1
    Hello all,

    I am just having a problem understanding a problem in my textbook on nonunique solutions. Let me get to the problem:

    So, consider the initial value problem:

    [tex] y' = y^{1/3}\mbox{, } y(0) = y_0 = 0[/tex]
    [tex]\mbox{for t}\geq 0[/tex]

    So, solving for the differential equation, I get:

    [tex]y = [\frac{2}{3}(t + C)]^{3/2}[/tex]

    So, satisfying initial condition, [tex]0 = [\frac{2}{3}(0 + C)]^{3/2}[/tex]

    So, [tex] C = 0[/tex]

    [tex]y = [\frac{2}{3}(t)]^{3/2}[/tex]
    , for [tex]t\geq 0[/tex]

    So, that's all understandable to me.
    But the answer in the book goes on to say that:

    [tex]y = -[\frac{2}{3}(t)]^{3/2}[/tex]
    , for [tex]t\geq 0[/tex]

    is also a solution. And:

    [tex] y = 0[/tex]
    , for [tex]t\geq 0[/tex]

    is also a solution. Finally, the answer says you can generalize the solultion to:

    [tex]y = \chi (t) =\left\{\begin{array}{cc}0,&\mbox{ if }
    0\leq t< t_0\\ \pm [\frac{2}{3}(t - t_0)]^{3/2}, & \mbox{ if } t\geq 0\end{array}\right[/tex]

    This last part is very confusing for me. If someone could explain it, it would be very helpful. For example, if the value of [tex]t_0[/tex] was given, and it followed the generalization above, wouldn't values for [tex]0 \leq t < t_0[/tex] not be 0, but instead be undefined? Since, you can't do [tex](negative number)^{3/2}[/tex]
    Right?

    Thanks in advance for all the help.
    -mk
     
  2. jcsd
  3. Sep 24, 2005 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, it is.
    [tex](-[\frac{2}{3}t]^{3/2})= -\frac{2}{3}[\frac{3}{2}t^{\frac{1}{2}[/tex]
    while [tex](-[\frac{2}{3}t]^\frac{3}{2})^\frac{1}{3}][/tex] is exactly the same thing. Also, of course, y(0)= 0.

    Absolutely: (0)'= 0 and (0)1/3= 0.

    Well, yes. Perhaps that is why they specifically said [tex]t\geq 0[/tex]?

    You'er welcome.
     
    Last edited: Sep 24, 2005
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