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Existence and Uniqueness Theorem

  1. Mar 29, 2013 #1
    Suppose you have an ODE [itex] y' = F(x,y) [/itex] that is undefined at x=c but defined and continuous everywhere else. Now suppose you have an IVP at the point (c,y(c)). Then is it impossible for there to be a solution to this IVP on any interval containing c, given that the derivative of the function, i.e. y', does not even exist at that point?

    So say you found a solution P(x) which does go through (c,y(c)) and satisfies the ODE [itex] y' = F(x,y) [/itex] everywhere where F(x,y) exists. Would P(x) be considered a solution of the IVP?

    This is a rather technical issue. I would appreciate if anyone addressed it.
    Thanks!

    BiP
     
  2. jcsd
  3. Mar 30, 2013 #2

    HallsofIvy

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    The "existance and uniqueness" theorems are all of the form "if **** is true, then there exists a unique solution". They do not say anything about what happens if **** is not true.
     
  4. Mar 30, 2013 #3
    I see. I guess my question is not so much about the E-U theorem, but really about what it means to "solve an IVP". To solve an IVP, certainly the function must pass through the initial condition. It must also certainly solve the ODE at points where both sides of the ODE are defined (i.e. exist). What about at points in the ODE for which y' = F(x,y) is not defined?

    So if you have y' = F(x,y), with some initial condition and you claim that P(x) is a solution to the IVP, then first P(x) should certainly pass through the initial condition. But suppose that y'=F(x,y) is undefined at the initial condition. If P(x) satisfies the ODE everywhere, except at the initial condition where the ODE is not even defined, then can P(x) be considered a solution to the IVP?

    BiP
     
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