PhyslibAlpha.ClassicalFieldTheory.Local.FirstVariation.Criterion

Let $L$ be a local Lagrangian of order $k$ and $f: \text{Space } d \to \mathbb{R}^m$ be a field. The property `AllVariationsHaveFiniteAction` states that for every admissible variation $\eta$, the action density of the varied field $f + s\eta$ is integrable over $\text{Space } d$ for all scalar parameters $s \in \mathbb{R}$. Mathematically, this means that for any admissible variation $\eta$ and any $s \in \mathbb{R}$, the function $x \mapsto L(j^k(f + s\eta)(x))$ is integrable.

Let $L$ be a local Lagrangian of order $k$ and $f: \mathbb{R}^d \to \mathbb{R}^m$ be a field. The property `HasFirstVariationFormula L f` holds if, for every admissible variation $\eta: \mathbb{R}^d \to \mathbb{R}^m$, whenever the action density of the varied field $f + s\eta$ is integrable for all $s \in \mathbb{R}$, the derivative of the action variation $s \mapsto S(L, f + s\eta)$ at $s = 0$ is equal to the first-variation value. Mathematically, this is expressed as: $$\left. \frac{d}{ds} \right|_{s=0} \int_{\mathbb{R}^d} L(j^k(f + s\eta)(x)) \, dx = \int_{\mathbb{R}^d} \langle \mathcal{E}(L)[f](x), \eta(x) \rangle \, dx$$ where $\mathcal{E}(L)[f]$ denotes the Euler-Lagrange operator applied to $f$ under $L$, $j^k$ denotes the $k$-jet of the field, and $\langle \cdot, \cdot \rangle$ is the standard Euclidean inner product on $\mathbb{R}^m$.

Let $L$ be a local Lagrangian of order $k$ and $f: \mathbb{R}^d \to \mathbb{R}^m$ be a field. Suppose that the first-variation formula holds for $L$ at $f$, which states that for any admissible variation $\eta$, the derivative of the action $S$ satisfies: $$\left. \frac{d}{ds} \right|_{s=0} S(L, f + s\eta) = \int_{\mathbb{R}^d} \langle \mathcal{E}(L)[f](x), \eta(x) \rangle \, dx$$ where $\mathcal{E}(L)[f]$ is the Euler-Lagrange operator. If the Euler-Lagrange operator is identically zero at $f$ ($\mathcal{E}(L)[f] = 0$), then $f$ is a critical field for the action functional.

Let $L$ be a local Lagrangian of order $k$ and $f: \mathbb{R}^d \to \mathbb{R}^m$ be a field. Suppose that the following conditions hold: 1. The first-variation formula holds for $f$ under $L$, such that for every admissible variation $\eta$, the derivative of the action $S$ satisfies: $$\left. \frac{d}{ds} \right|_{s=0} S(L, f + s\eta) = \int_{\mathbb{R}^d} \langle \mathcal{E}(L)[f](x), \eta(x) \rangle \, dx$$ where $\mathcal{E}(L)[f]$ is the local Euler-Lagrange operator. 2. For every admissible variation $\eta$, the action density $x \mapsto L(j^k(f + s\eta)(x))$ is integrable for all $s \in \mathbb{R}$ (all variations have finite action). 3. The function $x \mapsto \mathcal{E}(L)[f](x)$ is continuous. 4. The field $f$ is a critical point of the action functional, meaning $\left. \frac{d}{ds} \right|_{s=0} S(L, f + s\eta) = 0$ for all admissible variations $\eta$. Then, the Euler-Lagrange equations are satisfied everywhere: $\mathcal{E}(L)[f] = 0$.

Let $L$ be a local Lagrangian of order $k$ and $f: \mathbb{R}^d \to \mathbb{R}^m$ be a field. Suppose that the first-variation formula holds for $L$ and $f$, all admissible variations of $f$ have finite action under $L$, and the Euler-Lagrange operator $\mathcal{E}(L)[f]$ is continuous. Then, the field $f$ is critical for the action functional if and only if the Euler-Lagrange operator applied to $f$ vanishes identically: $$\mathcal{E}(L)[f] = 0$$ Here, $f$ being critical means that for every admissible variation $\eta$, the derivative of the action variation vanishes at zero: $$\left. \frac{d}{ds} \right|_{s=0} S(L, f + s\eta) = 0$$ where $S$ is the action functional defined by the integral of the Lagrangian density.

Let $L$ be a local Lagrangian of order $k$ for fields $f: \mathbb{R}^d \to \mathbb{R}^m$, and let $f$ be a smooth ($C^\infty$) field. Suppose that: 1. The partial derivatives of the Lagrangian with respect to the jet coordinates, $\frac{\partial L}{\partial u^a_I}$, are infinitely differentiable ($C^\infty$) functions of the coordinates. 2. The Lagrangian $L$ is continuous in its local jet coordinates $(x, u^a_I)$. 3. The field $f$ has finite action, meaning the action density $x \mapsto L(j^k f(x))$ is integrable over $\mathbb{R}^d$. Then the field $f$ is critical for the action functional (i.e., its first variation vanishes) if and only if it satisfies the Euler-Lagrange equations $\mathcal{E}(L)[f] = 0$ everywhere, where the components of the Euler-Lagrange operator are given by: $$(\mathcal{E}(L)[f](x))_a = \sum_{|I| \le k} (-1)^{|I|} D_I \left( \frac{\partial L}{\partial u^a_I} \right) [f](x) = 0$$ for each $a \in \{0, \dots, m-1\}$.