Let $B\subset\mathbb{R}^2$ be the unit ball centred at the origin, and let $u\in W^{1, 2}\cap L^{\infty}(B;\mathbb{R}^N)$. For each $m\in\mathbb{N}$, suppose that $v_m\in [u+W^{1, 2}\cap L^{\infty}(B;\mathbb{R}^N)]\cap C^{0, \gamma}(\overline B;\mathbb{R}^N)$ (with $\gamma$ independent of $m$) is a minimiser of the functional \begin{equation}\tag{1} u+W^{1, 2}_0(B;\mathbb{R}^N)\ni w\mapsto\mathcal{G}_m[w]\equiv \int_{B}|Dw|^2\ \mathrm{d}x +m\int_{B}\frac{|Dv_m-Dw|}{1+|Dv_m-Dw|}\ \mathrm{d}x \end{equation}Furthermore, assume that we also have \begin{equation*}\tag{2} \int_B\frac{|Dv_m-Du|}{1+|Dv_m-Du|}\ \mathrm{d}x\leq m^{-1}\quad (m\in\mathbb{N}). \end{equation*}I want to derive the Euler-Lagrange system for $\mathcal{G}_m$ at $v_m$. I was thinking that $v_m$ should satisfy Laplace's equation. Below is my work, but I am not sure if it's correct.
Let $\varphi\in C_c^{\infty}(B;\mathbb{R}^N)$ and let $\varepsilon>0$. In computing the first variation of $\mathcal{G}_m$ at $v_m$ we have \begin{align*} \lim_{\varepsilon\rightarrow 0}\frac{\mathcal{G}_m[v_m+\varepsilon\varphi]-\mathcal{G}_m[v_m]}{\varepsilon}&=\lim_{\varepsilon\rightarrow 0}\int_{B}\frac{|Dv_m+\varepsilon D\varphi|^2-|Dv_m|^2}{\varepsilon}\ \mathrm{d}x+m\lim_{\varepsilon\rightarrow 0}\int_{B}\frac{|D\varphi|}{1+\varepsilon|D\varphi|}\ \mathrm{d}x\\ &=\lim_{\varepsilon\rightarrow 0}\int_{B}2Dv_m\cdot D\varphi+\varepsilon|D\varphi|^2\ \mathrm{d}x+m\int_{B}|D\varphi|\ \mathrm{d}x\\ &=\int_{B}2Dv_m\cdot D\varphi\ \mathrm{d}x+m\int_{B}|D\varphi|\ \mathrm{d}x. \end{align*}Since $v_m$ is a minimiser of $\mathcal{G}_m$, it follows that the first variation must vanish. Consequently, \begin{equation*}\tag{3} -\int_{B}2Dv_m\cdot D\varphi\ \mathrm{d}x=m\int_{\Omega}|D\varphi|\ \mathrm{d}x \end{equation*}for each $(m, \varphi)\in \mathbb{N}\times C_c^{\infty}(B;\mathbb{R}^N)$. Now we fix an arbitrary $\varphi\in C_c^{\infty}(B;\mathbb{R}^N)$.
Note that the minimisation property of $v_m$, (1) and (2) imply that \begin{equation*}\tag{4} \int_B|Dv_m|^2\ \mathrm{d}x=\mathcal{G}_m[v_m]\leq \mathcal{G}_m[u]\leq \|Du\|_2^2+1\quad (m\in\mathbb{N}). \end{equation*}
Thus, applying Young's inequality to the left-hand side of (3) and then using (4), for example, we have that \begin{equation*} m\int_B|D\varphi|\ \mathrm{d}x\leq C\quad(m\in\mathbb{N}) \end{equation*}for some $C$ independent of $m$. But this must mean that \begin{equation*} \int_B |D\varphi|\ \mathrm{d}x=0 \end{equation*}so that in (3), we have \begin{equation*} \int_BDv_m\cdot D\varphi\ \mathrm{d}x=0\quad \big(\varphi\in C_c^{\infty}(B;\mathbb{R}^N)\big). \end{equation*}Is this correct?
The first variation of $\mathcal{G}_m$ at $v_{m}$ vanishes, if it exists. This is the problem. In the derivation of the first variation, if the limit exists, it should be independent of how $\varepsilon$ approaches zero. The calculation above holds for $\varepsilon\rightarrow 0^+$. If, however, $\varepsilon\rightarrow 0^-$, then \begin{equation} \lim_{\varepsilon\rightarrow 0^-}\frac{1}{\varepsilon}\int_B\frac{|\varepsilon|| D\varphi|}{1+|\varepsilon||D\varphi|}\ \mathrm{d}x=\lim_{\varepsilon\rightarrow 0^-}-\int_{B}\frac{|D\varphi|}{1+|\varepsilon||D\varphi|}\ \mathrm{d}x=-\int_{B}|D\varphi|\ \mathrm{d}x\neq \int_{B}|D\varphi|\ \mathrm{d}x \end{equation}unless $\varphi\equiv\text{const}$.