Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Initial Value Problem

  1. Oct 3, 2009 #1
    Hi all!

    I'm having some trouble with the Existence and Uniqueness (E&U) thms. for ODEs.

    Consider the IVP:

    [tex]x'(t)=f(t,x(t)), x(\xi)=\eta[/tex]

    In order to prove Existence, we need f to be Lipschitz or at least locally Lipschitz. For the Uniqueness we use the Banach's fixed point thm. for the operator:

    [tex](Tx)(t):=\eta+\int_{\xi}^t f(s,x(s))ds[/tex] which is equavalent to the initial ODE.

    The problem lies in proving the operator T has only one fixed point: To do this, we need f to be bounded on a certain domain (which also corresponds to the Lipschitz condition), so we actually need the domain of f.

    But defining the domain of f is in special cases what causes the problem.

    Consider the IVP:

    [tex]x'=\frac{t}{1-x}, f(0)=2[/tex]

    From the ODE we obtain the domain of f:

    f: R x R\{1} -> R
    (t,x) -> f(t,x)

    So f does not appear to be bounded at all ?! We could conclude that f is continuously differentiable in x, so f is also locally Lipschitz continuous and therefore must be bounded by some const. at least in some interval of R, but we cannot dfind neither this interval, nor the constant from the initial data. (can we?)

    So the boundness is what we seek but cannot find.

    I hope someone to bring more light into this topic for me :)


  2. jcsd
  3. Oct 3, 2009 #2


    User Avatar
    Science Advisor

    You mean Lipschitz in the variable x. f only has to be continuous in t.

    But why are you looking for "boundedness"? There is nothing in the "existence and uniqueness" theorem that says anything about f being bounded- except, of course, locally, since f must be continuous in both variables. And t/(1- x) certainly is bounded in some neighborhood of (0, 2).
  4. Oct 4, 2009 #3
    *which neighbourhood do you mean - this one of x or of t ?

    So you cut out the interval containing the point(s) that make(s) problems (i.e. x=1) and solve for the rest of R (i.e. R\(0,2)?

    But the solution to the problem [tex]x(t)=1+\sqrt{1-t^2}[/tex] is defined only for (-1,1), for x could not be 1. And (-1,1) does not fully lie in the neighbourhood of (0,2), since they intersect?

    I'd assume we could chose a smaller interval to change the considered neighbourhood. But how do we see this from the equations?!

    And how to proceed here since this interval has to be open and extended up to t= 1, where f blows up?!

    What do we do if we cannot expicitly solve?

    (sorry, it's lots of questions I posted, but it's still kind of vague to me)
  5. Oct 8, 2009 #4
    The existence and uniqueness theorem is a local theorem. It guarantees you a unique solution in a neighbourhood of the initial point, (0,2). And as you have shown, the solution exists and is unique in a neighbourhood of t=0 . . . in fact, it exists in (-1,1) . . . but that's something you deduce from the solution itself, the theorem cannot determine how big the neighbourhood in which the unique solution it guarantees is.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook