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I Uniqueness and Existence Theorm

  1. Apr 29, 2017 #1
    Consider y' = 1/sqrt(y)

    I seem to be able to find a unique solution given the initial condition of the form y(c) = 0, but the theorem says I won't be able to do so, so I am kind of confused.

    I just want some clarifications. Does the uniqueness and existence theorem say anything about the region outside the rectangle where y' and dy'/dy is continuous? Say in this example the theorem guarantees uniqueness when y > 0, but does it necessarily mean that when y = 0, there is no unique solution (i.e. you always find multiple solutions)?
     
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  3. Apr 29, 2017 #2

    mfb

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    You cannot plug in y(c) = 0 in the formula - the function cannot satisfy the differential equation there. You cannot apply a theorem using a differential equation in a region where this equation is not valid.
     
  4. Apr 29, 2017 #3

    Charles Link

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    In this case, it appears the uniqueness theorem doesn't guarantee anything. There may be a unique solution, but from the uniqueness theorem you can not determine whether this is the case or not.
     
  5. Apr 29, 2017 #4
    If I understand correctly, uniqueness and existence theorem only guarantees solutions in the region where y' and dy'/dy are continuous, so initial conditions outside this region may or may not result in multiple solutions?
     
  6. Apr 29, 2017 #5
    I see that the only reason we cannot substitute in y(c) = 0 as the solution is that y' goes to infinity as y approaches 0? But that only means the tangent at y = 0 tends to a vertical line. If I directly solve it and find the general solution, they still give me unique solutions that don't cross each other...
     
  7. Apr 29, 2017 #6

    mfb

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    You need an applicable initial condition for a unique solution. y(c)=d for d>0 for example.
     
  8. Apr 30, 2017 #7

    Charles Link

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    This reminds me of a similar problem that results when you do a Taylor expansion of ## ln(1+x) ## about ## x=0 ##. Because ## x=-1 ## clearly will diverge, the circle of convergence of this series in the complex plane has ## r=1 ##. However, if you use ## x=1 ## you do get a converging series for ## ln(2)=1-1/2+1/3-1/4... ##.
     
  9. May 6, 2017 #8
    Actually it is easy to give sense to the original question. Just reformulate it for the system
    $$\dot x=\sqrt y,\quad \dot y= 1$$
     
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