# Why $F(u_\gamma) = \int_{\Omega} | \nabla u (D \tau_{\gamma})^{-1} |^2 \det (D \tau_\gamma) ?$ Is this by change of variables?

I'd like to understand the following: Let $$F(u) = \int_{\Omega } |\nabla u|^2$$ where $\Omega$ is adomain in $\mathbb{R}^{n} \cdots$ For $| \gamma |$ small enough define $\tau_\gamma \Omega \rightarrow \Omega$ by $\tau_\gamma (x)= x +\gamma \psi(x)$ where $\psi \in C^\infty_0 (\Omega,\mathbb{R}^n)$. Setting $u_\gamma = u \circ \tau_{\gamma}^{-1}$ Why $$F(u_\gamma) = \int_\Omega | \nabla u (D \tau_{\gamma})^{-1} |^2 \det (D \tau_\gamma) ?$$

I believe that this follows by the change of variables. If $G: \Omega \rightarrow \mathbb{R}^n$ is a $C^1$ diffeomorphism $$\int_{G(\Omega)} f = \int_\Omega f \circ G | \det D_x G|$$ but I don't get the calculations.

The motivation for this is to undertand the theorem 5.2 in this article here. By the way if you explain the details in the proof of the thorem above I will be very grateful.

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What is $G$ in the last displayed equation? –  Davide Giraudo Nov 5 '12 at 20:56
Shouldn't it read $u_\gamma = u \circ \tau_\gamma^{-1}$ instead of $\tau_{\gamma^{-1}}$? –  martini Nov 5 '12 at 21:09
Sorry David Giraudo I forgot to say What is $G$? You can ask another thing you want. –  user29999 Nov 5 '12 at 21:12
Yes Martini, I did a typo. –  user29999 Nov 5 '12 at 21:13

$\def\abs#1{\left|#1\right|}$I don't have access to the linked article, but we have \begin{align*} F(u_\gamma) &= F(u \circ \tau_\gamma^{-1})\\ &= \int_\Omega \abs{\nabla(u \circ \tau_\gamma^{-1})}^2\\ &= \int_\Omega \abs{(\nabla u) \circ \tau_\gamma^{-1} \cdot D\tau_\gamma^{-1}}^2\\ &= \int_{\tau_\gamma^{-1}\Omega} \abs{(\nabla u) \circ \tau_\gamma^{-1}\circ \tau_\gamma\cdot D\tau_\gamma^{-1}\circ \tau_\gamma}^2\abs{\det(D\tau_\gamma)}\\ &= \int_\Omega \abs{\nabla u\cdot (D\tau_\gamma)^{-1}}^2\abs{\det(D\tau_\gamma)} \end{align*} If $\gamma$ is small, $\det(D\tau_\gamma)$ will be positive, as $\det$ is continuous and $D\tau_0 = \mathrm{id}$, so $F(u_\gamma) = \int_\Omega \abs{\nabla u\cdot (D\tau_\gamma)^{-1}}^2\det(D\tau_\gamma)$
Very good. Could you give me a referece or to say How can I obtain $\nabla (u \circ \tau_{\gamma}^{-1}) = (\nabla u) \circ \tau_{\gamma}^{-1} \cdot D \tau_{\gamma}^{-1}$. I will be happy in accept your answer. –  user29999 Nov 5 '12 at 21:25
It's the chain rule ... componentwise we have (writing $\tau$ for $\tau_\gamma^{-1}$) $\partial_i(u \circ \tau) = \sum_j (\partial_j u)\circ \tau \cdot \partial_i\tau_j$ right? Combine these equations to get $\nabla(u \circ \tau) = (\nabla u)\circ \tau \cdot D\tau$. –  martini Nov 5 '12 at 22:16