# Derivation of weak form of Euler Lagrange Equation

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: $$F(u)=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)D_{\alpha}u^iD_{\beta}u^i\ \mathrm{d}x$$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 $$\Omega\times \mathbb{R}^N\times\mathbb{R}^{nN}\ni(x, u(x), Du(x))\mapsto \sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)D_{\alpha}u^iD_{\beta}u^i$$ They say that the corresponding system of Euler equations (satisfied by every bounded local minimum $u$) is: $$\int_{\Omega} \sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)D_{\alpha}u^iD_{\beta}\varphi^i\ \mathrm{d}x=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}\sum_{h=1}^{N}-\frac{1}{2}A^{\alpha\beta}_{u^i}(x, u)\frac{\partial u^h}{\partial x_{\alpha}}\frac{\partial u^h}{\partial x_{\beta}}\varphi^i\ \mathrm{d}x$$ 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, $$i(\tau)\equiv F(u+\tau\varphi)\quad\tau\in\mathbb{R}.$$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, $$i'(0)=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)\Big[D_{\alpha}\varphi^iD_{\beta}u^i+D_{\beta}\varphi^iD_{\alpha}u^i\Big]+\sum_{h=1}^{N}A^{\alpha\beta}_{u^h}(x, u)\varphi^h\Big[D_{\alpha}u^iD_{\beta}u^i\Big].$$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 $$i'(0)=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)\Big[D_{\alpha}\varphi^iD_{\beta}u^i+D_{\beta}\varphi^iD_{\alpha}u^i\Big]+\sum_{h=1}^{N}A^{\alpha\beta}_{u^h}(x, u)\varphi^h\Big[D_{\alpha}u^iD_{\beta}u^i\Big]\tag{*}$$ 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 $$\int_{\Omega} \sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)D_{\alpha}u^iD_{\beta}\varphi^i\ \mathrm{d}x=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}\sum_{h=1}^{N}-\frac{1}{2}A^{\alpha\beta}_{u^i}(x, u)\frac{\partial u^h}{\partial x_{\alpha}}\frac{\partial u^h}{\partial x_{\beta}}\varphi^i\ \mathrm{d}x$$