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Picard existence theorem and IVP

  1. Apr 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the initial value problem
    \displaystyle \frac{dy}{dx} = \exp(xy) \\
    y(0) = 1

    1. Verify that this IVP has a unique solution in a neighborhood
    of [tex]x = 0[/tex].

    2. Following the notation of the lectures, find the values of [tex]K[/tex],
    [tex]M[/tex], and [tex]\delta[/tex] that will work for this case.

    2. Relevant equations

    3. The attempt at a solution

    1. Let [tex]\displaystyle f(x,y) = \frac{dy}{dx} = \exp(xy)[/tex]. Then
    [tex]\displaystyle \frac{df}{dy} = x \exp(xy)[/tex] which is continuous and
    hence has upper bound [tex]K[/tex]. Hence according to Picard's theorem the
    IVP has a unique solution in [tex]\abs{x - x_0} \leq \delta[/tex].

    2. How can I find the values of [tex]M[/tex], [tex]K[/tex] and [tex]\delta[/tex]? Does [tex]M[/tex] mean the upper bound of function [tex]f(x,y)[/tex]?
  2. jcsd
  3. Apr 15, 2010 #2
    We cannot possibly help you with this unless you explain "the notation of the lectures".
  4. Apr 16, 2010 #3
    I think [tex]\delta[/tex] relates to the neighborhood [tex]|x - x_0| < \delta[/tex] where the differential equation has a unique solution. [tex]K[/tex] is the constant in Lipschitz condition. [tex]M[/tex] is the upper bound of function [tex]f(x,y)[/tex].
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