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Volume preserving mapping in R^3

  1. Jan 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Find all [tex]\mathcal{C}^1[/tex] functions [tex]f(\mathbf{x})[/tex] in [tex]\mathbb{R}^3[/tex] such that the mapping [tex]\psi : \mathbb{R}^3 \to \mathbb{R}^3[/tex] also preserves volumes, where

    [tex]
    \begin{equation*}
    \psi(\mathbf{x}) = \left(
    \begin{array}{c}
    x_1 \\
    x_1^2 + x_2 \\
    f(\mathbf{x})
    \end{array} \right).
    \end{equation*}
    [/tex]

    Here, the mapping preserves volumes in the sense that for any Jordan domain [tex]D \in \mathbb{R}^3[/tex] the sets [tex]D[/tex] and [tex]\phi(D)[/tex] have the same volume.


    2. Relevant equations

    The obvious relevant equation is the Change of Variable theorem. Another one that I used was the fact that the determinant of a triangular matrix is the product of its diagonal entries.


    3. The attempt at a solution

    Using those two facts, I got the solution [tex]f(\mathbf{x})[/tex] has to be a continuously differentiable function of [tex]x_1, x_2[/tex] plus or minus [tex]x_3[/tex], i.e. it must be of the form

    [tex]
    \begin{equation*}
    f(\mathbf{x}) = g(x_1, x_2) \pm x_3,
    \end{equation*}
    [/tex]

    where [tex]g(x_1, x_2)[/tex] is a continuously differentiable function of two variables. Am I right? Can someone please help?

    Thanks very much for your time.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 7, 2010 #2

    Dick

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    Science Advisor
    Homework Helper

    I think you've got it right. Setting the Jacobian=1 gives |df/dx3|=1. That's what you did, correct?
     
  4. Jan 9, 2010 #3
    yes, but can i take the jacobean for a nonlinear mapping like this? is the change of coordinates theorem applicable?

    thanks.
     
  5. Jan 9, 2010 #4

    Dick

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    Sure it applies. A change of coordinates doesn't have to be linear.
     
  6. Jan 9, 2010 #5
    it has to be bijective though, right? but now it seems to me that this *is* going to be bijective, because the first two coordinates fix x_1 and x_2, and the last one therefore fixes x_3.

    am i right?

    thanks fr ur help, very mcuh.
     
  7. Jan 9, 2010 #6

    Dick

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    Science Advisor
    Homework Helper

    Sure. If you write psi(x1,x2,x3)=(a,b,c) it's pretty easy to solve for x1, x2 and x3 in terms of a, b and c.
     
  8. Jan 9, 2010 #7
    thanks much.
     
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