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I'd appreciate some help with the following problem:

Let $F = F(x, \{p_\alpha\}_{|\alpha|\le m})$ be a smooth function of the variables $x\in \overline \Omega$, and $p_\alpha \in \mathbb R, |\alpha|\le m$. Find the Euler-Lagrange operator $\mathcal M$ of the functional $$\mathcal F(u) = \int_\Omega F(x, \{D^\alpha u\}) \, dx$$ and show that its linearization at a smooth funciton $u_0$ is the operator $D^\beta(a_{\alpha \beta} D^\alpha u)$, where $a_{\alpha \beta}(x) = F_{p_\alpha p_\beta}(x,\{D^\delta u_0(x_0)\}_{|\delta |\le m}$. (The subscripts $p_\alpha, p_\beta$ here denote partial derivatives; also, recall that the linearization of $\mathcal M(u)$ at $u_0$ is given by $\mathcal L(v) = \frac{d}{ds}\Big|_{s=0} \mathcal M(u_0+sv)$.)

$\Omega$ presumably is some bounded domain in $\mathbb R^n$ (but you may assume any niceness conditions necessary).

I must admit, I'm not quite sure what is meant by the "Euler-Lagrange operator" of a functional $\mathcal F$. Google couldn't explain it to me, either. So I'm asking here.

I know that one normally looks for a stationary part of the functional in hope of finding a minimizer/maximizer. So I simply tried differenting under the integral sign to see what comes out: For an extremal point $u$ and any function $\varphi$ we have

\begin{align} 0 &= \frac{d}{ds}\Big|_{s=0} \mathcal F(u+s\varphi) \\ &= \int_{\Omega} \sum_{\alpha} F_{p_\alpha}(x,\{D^\delta u\}) D^\alpha \varphi \, dx \\ &= \int_{\Omega}\left[ \sum_{\alpha} (-1)^{|\alpha|} D^\alpha \left(F_{p_\alpha}(x,\{D^\delta u\})\right) \right] \varphi \, dx \end{align}

which implies $\sum_{\alpha} (-1)^{|\alpha|} D^\alpha \left(F_{p_\alpha}(x,\{D^\delta u\})\right) = 0$. Should I try to go on from here? I this what I'm supposed to do at all? If yes, is there a way to avoid a messy calculation in trying to evaluate $D^\alpha \left(F_{p_\alpha}(x,\{D^\delta u\})\right)$? And what is this $\mathcal M$ in the exercise statement?

I think the exercise is largely to motivate why we might be interested in the properties of equations of the form $$\sum_{|\alpha|, |\beta| \le m} D^\beta (a_{\alpha \beta} D^\alpha u) = \sum_{|\beta|\le m} D^\beta f_\beta$$ which were the content of the last chapter. So I would like to understand this in some detail. Also, I could not find a definition of the Euler-Lagrange operator $\mathcal M$ in any earlier notes, either.

Thanks for your help!

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I assume the EL operator for this problem is $$\mathcal{M}:u\mapsto\sum_\alpha(-1)^{|\alpha|}D^\alpha\left(F_{p_\alpha}(x,\{D‌​^\sigma u\})\right),$$ and you're allowed to use the $a_{\alpha\beta}$ abbreviation to verify the linearization. – anon Jun 30 '12 at 2:20
@anon: Thanks for the commment! I think you're probably right. – Sam Jun 30 '12 at 10:35
up vote 0 down vote accepted

We have $\mathcal M(u) = \sum_\beta (-1)^{|\beta|}D^\alpha(F_{p_\beta}(x,\{D^\sigma u\})$ as was pointed out above. From the definition of $\mathcal L$, we derive \begin{align} \mathcal L(v) &= \frac{d}{dt}\Big|_{t=0} \mathcal M(u+tv) \\ &= \sum_\beta (-1)^{|\beta|}D^\beta(F_{p_\alpha p_\beta}(x,\{D^\sigma u\})D^\alpha v) \\ &= \sum_\beta (-1)^{|\beta|}D^\beta(a_{\alpha \beta}D^\alpha v). \end{align}

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