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Thread: difficult stationary points question

  1. #1
    Junior Member
    Mar 2014
    United Kingdom

    difficult stationary points question

    the interval is x and y between 0.25pi and 0.5pi (the square thing kind of confused me!)

    I got

    ∂u/∂x = siny sin(2x+y) =0

    siny = 0

    y = 0 ± npi (not in interval so doesn't count ; npi means integer number of pi)


    2x +y =0


    ∂u/∂y = sinx sin(2y+x) =0

    Again sinx =0 ; x = 0 ±npi is outside the given interval

    2y + x =0

    if you try to find x:

    2x -0.5x = 0 ±npi

    x = n * 2/3 * pi

    even with n = 1 this is outside the interval!!

    Please help!?
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  2. #2
    MHF Contributor
    Nov 2010

    Re: difficult stationary points question

    First, you got $2x+y = n_1\pi$. Then you got $2y+x = n_2\pi$. So, solving for $y$ in the first equation gives: $y = n_1\pi - 2x$. Plugging into the second gives:

    $2(n_1\pi - 2x)+x = n_2\pi$

    $(2n_1 - n_2)\pi = 3x$

    Note that the map $\mathbb{Z}^2 \mapsto \mathbb{Z}$ that takes $(n_1,n_2) \mapsto (2n_1-n_2)$ is a surjection, so we can replace $(2n_1-n_2)$ with $n$:

    So, $3x = n\pi$ where $n$ is any integer.

    $x = \dfrac{n\pi}{3}$, and $\dfrac{\pi}{3} \in \left[\dfrac{\pi}{4},\dfrac{\pi}{2}\right]$

    Plugging in: $y = n_1\pi - 2\dfrac{\pi}{3} = \dfrac{(3n_1-2)\pi}{3}$

    So, if $y = \dfrac{\pi}{3}$, it too is in the interval. Those are the only values for $x$ and $y$ that work.
    Thanks from Applestrudle
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