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In Giaquinta's and Giusti's 1982 paper entitled "On the regularity of the minima of variational integrals", they look at the following quadratic functional: \begin{equation} F(u)=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)D_{\alpha}u^iD_{\beta}u^i\ \mathrm{d}x \end{equation}where the coefficients are differentiable, uniformly bounded and $D_{\alpha}\equiv\frac{\partial}{\partial x_{\alpha}}$. Here, $\Omega\subset\mathbb{R}^n$ and the integrand is the mapping \begin{equation} \Omega\times \mathbb{R}^N\times\mathbb{R}^{nN}\ni(x, u(x), Du(x))\mapsto \sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)D_{\alpha}u^iD_{\beta}u^i \end{equation} They say that the corresponding system of Euler equations (satisfied by every bounded local minimum $u$) is: \begin{equation} \int_{\Omega} \sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)D_{\alpha}u^iD_{\beta}\varphi^i\ \mathrm{d}x=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}\sum_{h=1}^{N}-\frac{1}{2}A^{\alpha\beta}_{u^i}(x, u)\frac{\partial u^h}{\partial x_{\alpha}}\frac{\partial u^h}{\partial x_{\beta}}\varphi^i\ \mathrm{d}x \end{equation} for all $\varphi\in L^{\infty}(\Omega, \mathbb{R}^N)\cap H_{0}^{1}(\Omega, \mathbb{R}^N)$. I am trying to deduce the above equation for $\varphi\in C_{c}^{\infty}(\Omega, \mathbb{R}^N)$ since I suppose it follows by approximation that it is true for $\varphi\in L^{\infty}(\Omega, \mathbb{R}^N)\cap H_{0}^{1}(\Omega, \mathbb{R}^N)$. However, I'm having difficulty in showing this. Below is my working.

Let $\varphi\in C_{c}^{\infty}(\Omega)$. Define, \begin{equation} i(\tau)\equiv F(u+\tau\varphi)\quad\tau\in\mathbb{R}. \end{equation}For $\tau\neq 0$ we have \begin{align} i'(\tau)&=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u+\tau\varphi)\Big[D_{\alpha}\varphi^iD_{\beta}[u^i+\tau\varphi^i]+D_{\beta}\varphi^iD_{\alpha}[u^i+\tau\varphi^i]\Big]\\ &+\sum_{h=1}^{N}A^{\alpha\beta}_{u^h}(x, u+\tau\varphi)\varphi^h\Big[D_{\alpha}[u^i+\tau\varphi^i]D_{\beta}[u^i+\tau\varphi^i]\Big]. \end{align}Consequently, \begin{equation} i'(0)=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)\Big[D_{\alpha}\varphi^iD_{\beta}u^i+D_{\beta}\varphi^iD_{\alpha}u^i\Big]+\sum_{h=1}^{N}A^{\alpha\beta}_{u^h}(x, u)\varphi^h\Big[D_{\alpha}u^iD_{\beta}u^i\Big]. \end{equation}After integrating by parts and equating to zero (since $i'(0)=0$), we obtain: \begin{equation*} \int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}D_{\alpha}\Big[A^{\alpha\beta}(x, u)D_{\beta}u^i\Big]\varphi^i+D_{\beta}\Big[A^{\alpha\beta}(x, u)D_{\alpha}u^i\Big]\varphi^i\ \mathrm{d}x=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}\sum_{h=1}^{N}A^{\alpha\beta}_{u^h}(x, u)\varphi^hD_{\alpha}u^iD_{\beta}u^i\ \mathrm{d}x. \end{equation*} So this is not the system I was supposed to end up with. What am I doing wrong?

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Why have you integrated by parts? The author of your paper didn't do that, starting from your line \begin{equation} i'(0)=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)\Big[D_{\alpha}\varphi^iD_{\beta}u^i+D_{\beta}\varphi^iD_{\alpha}u^i\Big]+\sum_{h=1}^{N}A^{\alpha\beta}_{u^h}(x, u)\varphi^h\Big[D_{\alpha}u^iD_{\beta}u^i\Big]\tag{$*$} \end{equation} we note that the first sum is symmetric in $\alpha$, $\beta$, hence $$\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)\Big[D_{\alpha}\varphi^iD_{\beta}u^i+D_{\beta}\varphi^iD_{\alpha}u^i\Big] = 2 \sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)D_{\alpha}u^iD_{\beta}\varphi^i $$ Now divide by 2, rename some of the indices (exchange $h$ and $i$), and we arrive at \begin{equation} \int_{\Omega} \sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)D_{\alpha}u^iD_{\beta}\varphi^i\ \mathrm{d}x=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}\sum_{h=1}^{N}-\frac{1}{2}A^{\alpha\beta}_{u^i}(x, u)\frac{\partial u^h}{\partial x_{\alpha}}\frac{\partial u^h}{\partial x_{\beta}}\varphi^i\ \mathrm{d}x \end{equation}

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Oh yeah! Thanks...I can't believe I overlooked that...too many symbols I guess! –  Nirav Oct 8 '13 at 9:50

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