PhyslibAlpha.ClassicalFieldTheory.Local.EulerLagrangeEquation

Given a local Lagrangian $L$ of order $k$ and a field $f: \mathbb{R}^d \to \mathbb{R}^m$, the field satisfies the local Euler-Lagrange equations if the Euler-Lagrange operator $\mathcal{E}(L)$ applied to $f$ vanishes identically. That is, $\mathcal{E}(L)[f] = 0$, or more explicitly, for every component $a \in \{1, \dots, m\}$ and at every point $x \in \mathbb{R}^d$: $$\sum_{|I| \le k} (-1)^{|I|} D_I \left( \frac{\partial L}{\partial u^a_I} \right) [f](x) = 0$$ where the sum is over multi-indices $I$ of total order $|I| \le k$, $\frac{\partial L}{\partial u^a_I}$ is the partial derivative of the Lagrangian with respect to the jet coordinate $u^a_I$, and $D_I$ is the iterated total derivative.

For a local Lagrangian $L$ of order $k$ and a field $f: \mathbb{R}^d \to \mathbb{R}^m$, the field satisfies the local Euler-Lagrange equations if and only if the local Euler-Lagrange operator $\mathcal{E}(L)$ applied to $f$ vanishes identically, i.e., $\mathcal{E}(L)[f] = 0$.

Let $L$ be a local Lagrangian of order $k$ and $f: \mathbb{R}^d \to \mathbb{R}^m$ be a field. If $f$ satisfies the local Euler-Lagrange equations (denoted as `SatisfiesEulerLagrange L f`), then the local Euler-Lagrange operator $\mathcal{E}(L)$ applied to $f$ vanishes identically: $$\mathcal{E}(L)[f] = 0$$ where the $a$-th component of the operator is defined by $\sum_{|I| \le k} (-1)^{|I|} D_I \left( \frac{\partial L}{\partial u^a_I} \right) [f]$.

Let $L$ be a local Lagrangian of order $k$ and $f: \mathbb{R}^d \to \mathbb{R}^m$ be a field. If the local Euler-Lagrange operator $\mathcal{E}(L)$ applied to $f$ vanishes identically (i.e., $\mathcal{E}(L)[f] = 0$), then the field $f$ satisfies the local Euler-Lagrange equations.

Let $L$ be a local Lagrangian of order $k$ and $f: \mathbb{R}^d \to \mathbb{R}^m$ be a smooth ($C^\infty$) field. Suppose $L$ is smooth in its jet coordinates $(x, u^a_I)$ and the action of $f$ is finite, meaning the action density $x \mapsto L(j^k f(x))$ is integrable over $\mathbb{R}^d$. Then, $f$ is a critical point for the action functional (i.e., the first variation of the action vanishes for all admissible variations) if and only if $f$ satisfies the local Euler-Lagrange equations: $$\sum_{|I| \le k} (-1)^{|I|} D_I \left( \frac{\partial L}{\partial u^a_I} \right) [f](x) = 0$$ for every component $a \in \{1, \dots, m\}$ and at every point $x \in \mathbb{R}^d$.

Let $L$ be a local Lagrangian of order $k$ and $f: \mathbb{R}^d \to \mathbb{R}^m$ be a field. Suppose that the pair $(L, f)$ is admissible for the action (meaning $f$ is $C^\infty$ and the action density $x \mapsto L(j^k f(x))$ is integrable over $\mathbb{R}^d$) and that $L$ is smooth in its jet coordinates (meaning its partial derivatives $\frac{\partial L}{\partial u^a_I}$ are $C^\infty$). Then, the field $f$ is critical for the action functional $S[f] = \int_{\mathbb{R}^d} L(j^k f(x)) \, dx$ if and only if it satisfies the local Euler-Lagrange equations $\mathcal{E}(L)[f] = 0$.