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Math Help - Simple question

  1. #1
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    Simple question

    Why does  f'(x)=0 \; \forall \; x\in\mathbb{R}\implies f(x)=c ?
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  2. #2
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    Integrate both sides. What do you get?
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  3. #3
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    I understand that, but does that imply that's the unique solution?
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  4. #4
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    Well, you have an infinite family of solutions depending on your initial condition. However, the fact that the RHS of your DE, namely 0, is Lipschitz and continuous, is enough to guarantee the uniqueness of the solution to an initial value problem (an initial value problem is your original DE plus an initial condition to determine the constant of integration). Does that make sense?
    Last edited by Ackbeet; June 22nd 2010 at 12:09 PM. Reason: Saved early.
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  5. #5
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    Why does Lipschitz continuity guarantee uniqueness of the solution (up to a constant)?
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  6. #6
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    Because of the Picard-Lindelof theorem. If you're curious, by all means investigate the proof of that theorem.
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  7. #7
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    Quote Originally Posted by mathman88 View Post
    Why does  f'(x)=0 \; \forall \; x\in\mathbb{R}\implies f(x)=c ?
    For this specific question, you don't need the full "Picard-Lindelof", just the mean value theorem.

    Let x_0 and x_1 be two distinct values of x: x_0\ne x_1.

    By the mean value theorem, \frac{f(x_1)- f(x_0)}{x_1- x_0}= f'(c) where c is some number between x_0 and x_1. If f'(x)= 0 for all x, then f'(x)= 0 so f(x_1)- f(x_0)= 0 and f(x_1)= f(x_0).
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