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Let $\Omega \subset \mathbb{R}^n$ and $F=F(z,p):\Omega \times \mathbb{R} \times \mathbb{R}^n$ be smooth and independent of $x\in\Omega$. Let $u$ be a solution of the Euler-Lagrange equation of $\mathcal{F}(u)= \int_\Omega F(u,Du)dx$. I'd like to show that $\sum_i (\partial_{p_i}F \partial_{x_j}u-F \delta_{i,j})_{x_i}=0$.

Since $u$ is a solution of the Euler-Lagrange equation and we've defined the Euler-Lagrange operator $L_F(u)=-\sum_i \partial_i (\partial_{p_i}F + \partial_zF)$ it follows that $L_F(u)=0$. I've been told that one should simply transform and compute the left side of the equation to prove that it equals $0$. I tried to transform the equation in order to use the fact that $u$ is a solution of the ELE of $\mathcal{F}(u)$. We have

$\sum_i (\partial_{p_i}F \partial_{x_j}u-F \delta_{i,j})_{x_i} \\ = \sum_i \partial_{x_i}(\partial_{p_i}F \partial_{x_j}u-F \delta_{i,j}) \\ = \sum_i \partial_{x_i}(\partial_{p_i}F \partial_{x_j}u) - \sum_i \partial_{x_i}F \delta_{i,j} \\ = \sum_i \partial_{x_i}(\partial_{p_i}F) \partial_{x_j}u + \sum_i \partial_{p_i}F \partial_{x_i}( \partial_{x_j}u) -\partial_{x_j}F \; \mathrm{(product \, rule)} \\ = \sum_i div(\partial_{p_i}F \partial_{x_j}u) - \sum_i \partial_{p_i}F \partial_{x_i} (\partial_{x_j}u) + \sum_i \partial_{p_i}F \partial_{x_i}( \partial_{x_j}u) -\partial_{x_j}F \; \mathrm{(chain \, rule \, for \, divergence)}\\ = \sum_i div(\partial_{p_i}F \partial_{x_j}u) -\partial_{x_j}F.$

But I don't see how I can transform this expression any further or how I could use that $u$ is a solution of the Euler-Lagrange equation. I would appreciate if someone could give me tips so that I will be able to prove this assumption.

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Hint: One has to distinguish between explicit $x$-derivatives $\partial_{x^i}$ and total $x$-derivatives $$d_{x^i}~=~\partial_{x^i} +~ d_{x^j}u ~\partial_{z}+d_{x^j}\partial_{x^i}u~\partial_{p_i}\tag{0}.$$ No explicit $x$-dependence $$ \partial_{x^i}F~=~0 \tag{1}$$ of the Lagrangian $F$ leads (via Noether's theorem) to the on-shell continuum equations for the energy-momentum tensor: $$\begin{align} (\partial_{p_i}F ~\partial_{x^j}u - F \delta_i^j)_{x^i} &~~~=~ d_{x^i}(\partial_{p_i}F ~\partial_{x^j}u - F \delta_i^j) \cr &~~~=~ d_{x^i}\partial_{p_i}F~ \partial_{x^j}u + \partial_{p_i}F~d_{x^i}\partial_{x^j}u - d_{x^j}F \cr &~~~\stackrel{(0)}{=}~ d_{x^i}\partial_{p_i}F~ \partial_{x^j}u + \partial_{p_i}F~\partial_{x^i}\partial_{x^j}u - (\partial_{x^i}F +\partial_{z}F~ d_{x^j}u +\partial_{p_i}F~d_{x^j}\partial_{x^i}u)~\cr &\stackrel{(1)+(3)}{=}~0. \end{align}\tag{2}$$ The last equality follows from eq. (1) and the Euler-Lagrange (EL) equation $$ d_{x^i}\partial_{p_i}F~=~\partial_{z}F.\tag{3}$$ The word on-shell here means that the EL eq. (3) is satisfied. It is implicitly understood in eq. (2) that the Lagrangian $F$ is applied $F(u,\partial u)$ to a solution $u$ of the EL eq. (3). In contrast, the Lagrangian $F$ in eq. (1) is not applied to a solution $u$.

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  • $\begingroup$ Thanks for your help! I've got another question regarding your answer: Do we get in the last equation $d_{x^j}F=\partial_zF d_{x^j}u+\partial_{p_i}F d_{x^j} \partial_{x^i}u$ because the Lagrangian $F$ is independent of $x$ and therefore $\partial_{x^j}F=0$? I thought the Lagrangian $F(u,Du)$ was still dependent, even if not explicitly, on $x$, because $u$ and $Du$ are dependent on $x$ and then the partial derivative of the Lagrangian with respect to $x^j$ wouldn't be $0$ or am I mistaken? $\endgroup$ – mathstu Jan 14 at 10:16
  • $\begingroup$ $\uparrow$ You are right, not mistaken. $\endgroup$ – Qmechanic Jan 14 at 10:26
  • $\begingroup$ But how do we get this equation? With the Noether Theorem? Then I still have to check that $F$ is infinitesimal invariant. Is this correct? $\endgroup$ – mathstu Jan 14 at 11:57
  • $\begingroup$ So x-independence is equivalent to $\partial_{x^i}F=0$? I thought this wasn't true. I'm sorry but I'd like to understand every detail and I still don't completely understand this. $\endgroup$ – mathstu Jan 14 at 13:53
  • $\begingroup$ Explicit $x$-independence (1) is an assumption. $\endgroup$ – Qmechanic Jan 14 at 14:49

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