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Thread: Stationary points / two variables (help)

  1. #1
    Member billym's Avatar
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    Stationary points / two variables (help)

    I have this function and I am looking for the stationary points:

    $\displaystyle
    f(x,y) = x^4 + 2x^2y^2 + y^4 + 4x^2 - 4y^2 + 6
    $

    I started by trying to solve the two partial derivatives:

    $\displaystyle
    4x(x^2 + y^2 + 1) = 0, (E.1)
    $
    $\displaystyle
    4y(y^2 + x^2 - 1) = 0, (E.2)
    $

    If $\displaystyle x = 0$, then $\displaystyle (E.2)$ becomes:

    $\displaystyle
    4y(y^2 - 1) = 0
    $

    Thus $\displaystyle y = 0$ , $\displaystyle y = 1$ , or $\displaystyle y = -1$, so there are three stationary points at $\displaystyle (0,0)$ , $\displaystyle (0,1)$ , and $\displaystyle (0,-1)$ , correct?

    Next, if:

    $\displaystyle
    x^2 + y^2 + 1 = 0,
    $

    i.e.

    $\displaystyle
    y^2 = -x^2 - 1
    $

    then $\displaystyle (E.2)$ becomes... uh...

    $\displaystyle
    -8y = 0
    $
    this is where I run into trouble.

    Is $\displaystyle x \ge 0$ ?

    How do I go about finding other points? Are there other stationary points?
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  2. #2
    Senior Member
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    Anchorage, AK
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    Note that for real x and y, $\displaystyle x^2\ge0$ and $\displaystyle y^2\ge0$, thus $\displaystyle x^2+y^2\ge0$, and so $\displaystyle x^2+y^2+1\ge1$, and thus $\displaystyle x^2+y^2+1$ is never zero.

    --Kevin C.
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  3. #3
    Super Member

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    Lexington, MA (USA)
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    Hello, billym!

    A small error in your derivatives, but your intentions were correct.


    $\displaystyle f(x,y) \:=\: x^4 + 2x^2y^2 + y^4 + 4x^2 - 4y^2 + 6$
    $\displaystyle f_x\;=\;4x^3 + 4xy^2 + {\color{red}8}x \;=\;4x(x^2+y^2+2)$

    $\displaystyle f_y\;=\;4x^2y + 4y^2 - {\color{red}8}y \;=\;4y(x^2+y^2-2)$


    We have: .$\displaystyle \begin{array}{cccc}x(x^2 + y^2+2) &=&0 & {\color{blue}[1]} \\
    y(x^2+y^2-2) &=&0 & {\color{blue}[2]} \end{array}$


    From [1], we have: .$\displaystyle \begin{array}{cc}x \:= \:0 \\ x^2+y^2 \:=\:\text{-}2& \leftarrow\text{ impossible}\end{array}$

    If $\displaystyle x = 0$, [2] becomes: .$\displaystyle y(y^2-2) \:=\:0\quad\Rightarrow\quad y \:=\:0,\:\pm\sqrt{2}$

    . . So far, we have: .$\displaystyle (0,\,0),\;\;(0,\,\sqrt{2}),\;\;(0,\,\text{-}\sqrt{2}) $


    From [2], we have: .$\displaystyle \begin{array}{c}y \:=\:0 \\ x^2+y^2 \:=\:2 \end{array}$

    If $\displaystyle y = 0$, [1] becomes: .$\displaystyle x(x^2+2) \:=\:0\quad\Rightarrow\quad x \:=\:0$

    If $\displaystyle x^2+y^2\:=\:2$, [1] becomes: .$\displaystyle x(2+2) \:=\:0\quad\Rightarrow\quad x \:=\:0 $
    . . and we've already encountered these solutions.


    Therefore, the stationary points are: .$\displaystyle \boxed{(0,\,0),\;\;(0,\,\sqrt{2}),\;\;(0,\,\text{-}\sqrt{2})} $

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  4. #4
    Member billym's Avatar
    Joined
    Feb 2008
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    whoops

    Sorry i made a mistake in the original equation. It actually works out to:

    $\displaystyle
    f(x,y)= x^4 + 2x^2y^2 + y^4 + 2x^2 - 2y^2 + 6
    $

    So I have a saddle point at (0,0), and local minimums at (0,1) and (0,-1). (I hope).

    Thanks for the help guys.
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