# Volume preserving mapping in R^3

1. Jan 7, 2010

### shoplifter

1. The problem statement, all variables and given/known data

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

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

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.

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 $$f(\mathbf{x})$$ 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

$$\begin{equation*} f(\mathbf{x}) = g(x_1, x_2) \pm x_3, \end{equation*}$$

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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 7, 2010

### Dick

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

3. Jan 9, 2010

### shoplifter

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

thanks.

4. Jan 9, 2010

### Dick

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

5. Jan 9, 2010

### 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.

6. Jan 9, 2010

### 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.

7. Jan 9, 2010

thanks much.