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Given a vector valued function $\bf{f}:\mathbb{R}^n\rightarrow\mathbb{R}^n$, what is a scalar function $g:\mathbb{R}^n\rightarrow\mathbb{R}$ that satisfies the following $$g({\bf x} +\mathbf{f}(\mathbf{x}))=g(\mathbf{x})\det(I+\mathbf{f}'(\mathbf{x})),$$ where $\bf{f}'(\bf{x})$ is the Jacobian matrix of $\bf{f}(\bf{x})$ and $I$ is the $n\times n$ identity matrix.

If a solution can't be found for arbitrary $\bf{f}$, what structure can one impose on $\bf{f}$ for there to exist a particular function $g$ that satisfies the above condition.

One particular example of a function $g$ that satisfies the above is all I'm after (i.e., I don't need the most general solution).

Alternatively: Can one prove that there does not exist a function $g$ that satisfies the above?

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    $\begingroup$ $g\equiv 0$ works. $\endgroup$ – zhw. Sep 24 '19 at 19:51
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A couple of observations in 1D that may be helpful (but I have not tried to generalize),

Let, $$ g(x) = \exp{\int_0^x h(y) dy} $$

Then, $$ g(x+f(x))=\exp{\int_0^{x+f(x)} h(y) dy} = g(x) \exp{\int_x^{f(x)}h(y)dy} $$

If we specialize to $h(y) = 1/y$, and $f(x) = -\int\log(x)$, the last exponential becomes

$$ \exp{\log(\frac{-\int \log(x)}{x})} = 1 - \log(x) = 1 + f'(x) $$

So in 1D we can write, $$ g(x+f(x)) = g(x)(1+f'(x)) $$

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  • $\begingroup$ That is very nice solution to the problem in 1D! Looks like the solution relies on $f(x)$ having a particular form so I guess it is quite hard to do solve the problem 1D while keeping $f$ arbitrary, which probably means that in higher dimensions it's even more difficult as we have to deal with the determinant. Perhaps a general $n$-D solution is a bit ambitious... $\endgroup$ – Ben Tapley Sep 25 '19 at 13:34

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