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Differential Euqation (infinite solutions)

  1. Oct 4, 2006 #1
    Hi, I have a question which asks: Prove that there are infinitly many different solutions to the differential equation x' = x^(1/3) with x(0) =0

    Well, I solved the differential equation with the five initial value.

    I got:

    x = [ (2/3)(t) ]^(3/2)

    And I know that x = 0 is also a solution...

    Can somebody show me how to show that this differential equation has infinitly many solutions?
  2. jcsd
  3. Oct 4, 2006 #2
    I think it only has infinitely many solutions without applying the initial condition. Once you apply this, I don't see where infinitely many more solutions will come from...
  4. Oct 5, 2006 #3


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    No, there are an infinite number of solutions satisfying the boundary conditions. You get others by "patching" the two you have together. For example, take x=[ (2/3)(t-a) ]3/2 for x>= a, 0 for x< a where a can be any real number.
  5. Oct 5, 2006 #4
    Do you mean [itex]t<a[/itex] and [itex]t\geq a[/itex] ? (As opposed to x)

    Anyway I don't think the above solution would work because [itex]x[/itex] as you have given it is a continuous function of [itex]t[/itex], while it has a finite discontinuity in the derivative at [itex]t=a[/itex]. If I look at the original ODE,

    x^{\prime} = x^{1/3}

    If [itex]x[/itex] is continuous, then so is [itex]x^{\prime}[/itex]. Probably you would have to choose the range of [itex]t[/itex] for these functions so that you have a finite discontinuity in [itex]x[/itex], to match the discontinuity in the derivative... ? I'm not so sure about this though. Would a derivative even be defined then?
    Last edited: Oct 5, 2006
  6. Oct 5, 2006 #5

    matt grime

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    A finite discontinuity? What on earth is that? The deriviative of the function given by Halls is not discontinuous, unless you think the square root of 0 is not 0.
  7. Oct 5, 2006 #6
    Yes, you're right, I was picturing the wrong function as it was going to a - the derivative is in fact continuous everywhere. It's a nice solution, although you would require that [itex]a\geq 0[/itex].

    Matt, maybe you know a finite discontinuity as a jump discontinuity.
  8. Oct 5, 2006 #7

    matt grime

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    No, I just know that as a discontinuity. Why would anyone make up that definition? Presumably you have that, say 1/x for x>0 and 0 otherwise as an infinite discontinuity (and that isn't a 'jump'). Seems a strange thing to bother to name.
  9. Oct 5, 2006 #8
    Well i am confused out of my mind.

    Our text book SUCKS and has no solutions to it's practice problems.

    My prof. posted her soultions, but it makes no sense...

    ill post back here soon..
  10. Oct 5, 2006 #9
    Yes, it does seem pointless in naming them, I don't see any benefit from their definitions. And yes 1/x for x>0, 0 otherwise, would be an infinite discontinuity.

    rad: What are you confused with?
    Last edited: Oct 5, 2006
  11. Oct 5, 2006 #10
    I'm confuzed with the question "Prove that this differential equation had infinite solutions"

    How on earth do you prove that?

    I'll have to read through the above posts and take sometime to think things through...it's late now and i'm off to sleep..
  12. Oct 5, 2006 #11
    Well Halls' solution shows you that this equation has infinitely many solutions. For every [itex]a[/itex] that you pick, you will have a solution which satisfies your initial condition. Since [itex]a[/itex] is an arbitrary constant, you have inifinitely many choices of [itex]a[/itex], and so infinitely many solutions.
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