I'd like to understand the following: Let F(u)=∫Ω|∇u|2 where Ω is adomain in Rn⋯ For |γ| small enough define τγΩ→Ω by τγ(x)=x+γψ(x) where ψ∈C∞0(Ω,Rn). Setting uγ=u∘τ−1γ Why F(uγ)=∫Ω|∇u(Dτγ)−1|2det
I believe that this follows by the change of variables. If G: \Omega \rightarrow \mathbb{R}^n is a C^1 diffeomorphism \begin{equation} \int_{G(\Omega)} f = \int_\Omega f \circ G | \det D_x G| \end{equation} 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.
Answer
\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)
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