Volume preserving mapping in R^3

[shoplifter]

1. Homework Statement

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

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

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

Homework 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.

The Attempt at a Solution

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

<br /> \begin{equation*}<br /> f(\mathbf{x}) = g(x_1, x_2) \pm x_3,<br /> \end{equation*}<br />

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

Thanks very much for your time.

 

[Dick]

I think you've got it right. Setting the Jacobian=1 gives |df/dx3|=1. That's what you did, correct?

 

[shoplifter]

yes, but can i take the jacobean for a nonlinear mapping like this? is the change of coordinates theorem applicable?

thanks.

 

[Dick]

Sure it applies. A change of coordinates doesn't have to be linear.

 

[shoplifter]

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.

 

[Dick]

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.

 

[shoplifter]

thanks much.