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Thread: Absolute max of differentiable function

  1. #1
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    Absolute max of differentiable function

    Let f(x) be a differentiable function \mathbb{R} \to \mathbb{R} which satisfies:

    f(0) = a
    f'(0) = b

    f''(x) = -f(x)^n

    where a,b are real, n is a natural number.

    Find the absolute max of f(x) in terms of a,b, and n.
    Last edited by SworD; Jun 21st 2015 at 06:44 PM.
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  2. #2
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    Re: Absolute max of differentiable function

    For such an f, get that:

    f''(x) + f(x)^n = 0 \Rightarrow f'(x)f''(x) + f'(x)f(x)^n = 0

    \Rightarrow \frac{d}{dx}\left( \frac{1}{2}f'(x)^2 + \frac{1}{n+1}f(x)^{n+1} \right) = 0

    \Rightarrow \frac{d}{dx}\left( \frac{n+1}{2}f'(x)^2 + f(x)^{n+1} \right) = 0

    \Rightarrow \frac{n+1}{2}f'(x)^2 + f(x)^{n+1} = c,

    and from f(0) = a, f'(0) = b, get that c = \frac{(n+1)}{2}f'(0)^2 + f(0)^{n+1} = \frac{(n+1)}{2}b^2 + a^{n+1}.

    Note that, if n is odd, then c is non-negative, so that the real (n+1)-root of c always exists for any n. Call it K = c^\frac{1}{n+1} (taking the positive root as usual if n is odd).

    From \frac{(n+1)}{2}f'(x)^2 + f(x)^{n+1} = c, get that f(x)^{n+1} = c - \frac{(n+1)}{2}f'(x)^2 \le c = K^{n+1}, so f(x)^{n+1} \le K^{n+1}.

    Thus f(x) \le K for n even, and |f(x)| \le K for n odd. So K is an upper bound for f regardless of the parity of n. (Also, f is additionally bounded below if n is odd).

    ##############################

    If there exists x_0 such that f'(x_0) = 0, then from \frac{(n+1)}{2}f'(x)^2 + f(x)^{n+1} = c = K^{n+1} get that f(x_0)^{n+1} = K^{n+1},

    in which case f(x_0) = K if n odd, and f(x_0) \in \{-K, K\} if n even.

    Note that, since K is an upper bound, any conclusion that f(x_0) = K proves that K is an attained global maximum for the function over all the reals.

    That means that, if there exists x_0 such that f'(x_0) = 0, then K is either the maximum of f over all the reals, or: f(x_0) = -K and n is even.

    ##############################

    If the derivative is never 0, then by Rolle's Theorem f itself can have at most one zero on the entire real line, and by continuity of the derivative (since the second derivative exists), the derivative is either always positive or always negative.

    Going through what's possible case by, using that f''(x)  = - f(x)^n, and that f is bounded above always, and below too for n odd, get:

    If the derivative is never 0, then either:
    1) n is even and f is always concave down & decreasing.
    2) n is even and f is always concave down & increasing.
    3) n is odd, and f has exactly one inflection point x_1, which is also its only zero, and f is always increasing, and f is concave up and negative on (-\infty, x_1), and concave down and positive on (x_1, \infty).
    4) n is odd, and f has exactly one inflection point x_1, which is also its only zero, and f is always decreasing, and f is concave down and positive on (-\infty, x_1), and concave up and negative on (x_1, \infty).

    Note that f being bounded above everywhere, and differentiable everywhere, and having such fixed concavity on intervals out to infinity and minus infinity, means that either \lim_{x \to \infty} f'(x) = 0 and/or \lim_{x \to -\infty} f'(x) = 0,

    and so from \frac{(n+1)}{2}f'(x)^2 + f(x)^{n+1} = c = K^{n+1} get that \lim_{x \to \infty} f(x)^{n+1} = K^{n+1} and/or \lim_{x \to -\infty} f(x)^{n+1} = K^{n+1}.

    So in the case the derivative is never 0 and n is even (so n+1 odd), that guarantees that K isn't just an upper bound, but is the least upper bound of f over the reals.

    It probably guarantees that when n is odd too, but I don't feel like futzing around with all the special cases. I'll finish with one final section, showing that f can be explicitly found (in terms of the inverse of a function defined as an integral):

    ##############################

    Now from \frac{(n+1)}{2}f'(x)^2 + f(x)^{n+1} = K^{n+1}, get that \frac{f'(x)^2}{K^{n+1} - f(x)^{n+1}} = \frac{2}{n+1} whenever f(x) \ne K, so whenever f(x) < K.

    Define G(t) = \int_{a}^{t} \frac{du}{\sqrt{K^{n+1} - u^{n+1} }} (on the largest possible domain... and will need to look back and rexamine this to see if it makes sense in the case K = a (i.e. b = 0).

    Then G(a) = 0, and G'(t) = \frac{1}{\sqrt{K^{n+1} - t^{n+1} }}. Note that G is increasing (G' > 0) on every interval where it's defined.

    By the chain rule \frac{d}{dx} G(f(x)) = G'(f(x)) f'(x) = \frac{f'(x)}{\sqrt{K^{n+1} - (f(x))^{n+1} }}, so that \left( \frac{d}{dx} G(f(x)) \right)^2 = \frac{f'(x)^2}{|K^{n+1} - (f(x))^{n+1}|} = \frac{f'(x)^2}{K^{n+1} - (f(x))^{n+1}}.

    Thus, for f(x) \ne K (so whenever f(x) < K), get that \left( \frac{d}{dx} G(f(x)) \right)^2 = \frac{f'(x)^2}{K_{1}^{n+1} - (f(x))^{n+1}} = \frac{2}{n+1}.

    Thus, for f(x) \ne K, have \frac{d}{dx} G(f(x)) = \epsilon(x) \sqrt{\frac{2}{n+1}}, where \epsilon(x) \in \{ \pm 1 \}.

    To keep it simple at first, treat epsilon as a constant (as it surely will be on various intervals, especially the one containing x=0). Specifically, all the continuity in sight guarantees epsilon is constant and can only potentially flip signs (-1 to 1, or 1 to -1) at points making \frac{d}{dx} G(f(x)) = 0, or when going over a point not in the domain.

    Then get \int_0^x \frac{d}{dt} G(f(t)) dt = \int_0^x \epsilon \sqrt{\frac{2}{n+1}} dt = \epsilon \sqrt{\frac{2}{n+1}} \int_0^x dt = \left( \epsilon \sqrt{\frac{2}{n+1} \right) x,

    and \int_0^x \frac{d}{dt} G(f(t)) dt = G(f(x)) - G(f(0)) = G(f(x)) - G(a) = G(f(x)), and so G(f(x)) = \left( \epsilon \sqrt{\frac{2}{n+1} \right) x.

    Therefore f(x) = G^{-1}(c_1 x) where c_1 = \epsilon \sqrt{\frac{2}{n+1}. Not that G increasing wherever defined, so on any interval such an inverse exists.

    That's enough. There are now a host of special cases and little complexities to take into account - pesky details that I don't feel like dealing with.
    Last edited by johnsomeone; Jul 2nd 2015 at 01:53 PM.
    Thanks from sakonpure6 and SworD
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