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Homework Help: Initial-value problem

  1. May 7, 2006 #1
    Hello, I have trouble showing that the following initial-value problem has a unique solution. I also need to find this unique solution.

    y' = e^(t-y), where 0 <= t <= 1, and y(0) = 1.

    How can I test the Lipschitz condition on this?

    Thanks in advance.
  2. jcsd
  3. May 8, 2006 #2


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    Well what have you done?

    [tex] \frac{dy}{dt} = e^{t} e^{-y} [/tex]
  4. May 8, 2006 #3


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    What does your theorem on uniqueness say? My book has one where y' = F(y) for some function F of y. Here, your function is F(y,t) = et-y, it's a function of 2 variables. So maybe you have to show that for each t in [0,1], the function Ft(y) = et-y is Lipschitz as a function of y. Also, I think you only need to satisfy the Lipschitz condition for y=1, since that's the initial condition you're solving for. Again, just look at the precise statement of the theorem that I would assume is given to you.
  5. May 9, 2006 #4
    Yes, that's what I'm trying to show. For each pair of points (t,y1) and (t,y2) where t is in [0,1], we have |F(t,y1) - F(t,y2)| <= L |y1 - y2| where L is a Lipschitz constant for F. And y1 and y2 can be anything between positive and negative infinity. But it seems that it doesn't satisfy the Lipschitz condition if you write it out -- |F(t,y1) - F(t,y2)| isn't bounded. I'm not sure what other ways there are to show uniqueness.

    I found the IVP's solution using separable variables. I got y(t) = ln[(e^t) + e - 1].
  6. May 9, 2006 #5
    Oh sorry, did you say that it only needs to satisfy the Lipschitz condition for y=1? How do you know that?

    My theorem for uniqueness says that f needs to satisfy a Lipschitz condition on D, where D is the convex set {(t,y)}, where 0 <= t <= 1 and -infinity < y < +infinity.

  7. May 10, 2006 #6


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    Because the existence-uniqueness theorem for ODEs is a local result. It needs to be Lipschitz near 1 for there to be a unique solution with initial value 1. There just needs to be a unique trajectory going through 1, not a unique trajectory everywhere. Of course, if its not Lipschitz everywhere, then you may not have a unique solution everywhere, but that's not really important.

    Actually, looking at the following graph (black = |1-e1-y|, green = |1-y|), even if you steepen the green graph by increasing L, it will never dominate the black curve, so I don't think it is Lipschitz)

    Attached Files:

    Last edited: May 10, 2006
  8. May 10, 2006 #7
    Ok I see. Thanks.
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