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Consider the transformation $x=G(u)$ where $x=(u_{1},u_{2},u_{3})$ given by

$x_{1} = u_{1} +(u_{3})^{2}$

$x_{2} = u_{3} -(u_{1})^{2}$

$x_{3} = u_{1} +u_{2} + u_{3}$

1) compute the derivative of this transformation. i am having a bit of trouble understanding what this all means i know that i am supposed to be Computing Df from the other example in my textbook but the way $u_{1}$ and $x_{1}$ are defined has me completely baffled.

2) Use the inverse function theorem to show that the transformation is locally invertable if $u_{1}u_{2} > 0$

Could someone perhaps layout the inverse function theorem in a more palatable way to this question? my textbook gives only a proof of the theorem and im not sure how to apply it.

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1 Answer 1

up vote 1 down vote accepted

I suspect that you have an error in the definition of $G$ or the invertibility condition.

Let $G_1(u) = u_{1} +(u_{3})^{2}$, $G_{2}(u) = u_{3} -(u_{1})^{2}$, $G_{3}(u) = u_{1} +u_{2} + u_{3}$.

Compute $DG(u) = \begin{bmatrix} \frac{\partial G_1(u)}{\partial u_1} & \frac{\partial G_1(u)}{\partial u_2} & \frac{\partial G_1(u)}{\partial u_3} \\ \frac{\partial G_2(u)}{\partial u_1} & \frac{\partial G_2(u)}{\partial u_2} & \frac{\partial G_2(u)}{\partial u_3} \\ \frac{\partial G_3(u)}{\partial u_1} & \frac{\partial G_3(u)}{\partial u_2} & \frac{\partial G_3(u)}{\partial u_3} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 2 u_3 \\ -2 u_1 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix} $. Computing the determinant gives $\det( DG(u)) = -1 - 4 u_1 u_3$, hence $DG(u)$ is invertible iff $u_1 u_3 \neq - \frac{1}{4}$.

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