Spacetime $\mathbb{R}^{1+n}$ as the function type $\{0, 1, \ldots, n\} \to \mathbb{R}$. The index $0$ is the time coordinate; indices $1, \ldots, n$ are spatial.
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A scalar field on $\mathbb{R}^{1+n}$, i.e. a real-valued function $\phi : \mathbb{R}^{1+n} \to \mathbb{R}$.
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A Lagrangian density $L(\phi, \partial \phi, x)$: a real-valued function of the field value, the spacetime gradient (an $(n+1)$-tuple of partial derivatives), and the spacetime point.
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The spacetime gradient $(\partial_{\mu} \phi)_{\mu = 0, \ldots, n}$ of a scalar field $\phi$ at the point $x$, i.e. the $(n+1)$-tuple of partial derivatives.
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The action functional $\mathcal{S}_K[\phi] = \int_K L(\phi(x), \partial \phi(x), x)\, dx$ of a scalar field $\phi$ on a region $K \subset \mathbb{R}^{1+n}$.
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A variation $\psi$ supported in $K$: a smooth ($C^{\infty}$) scalar field whose support is contained in the region $K$. This is the class of admissible test perturbations of $\phi$.
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The one-parameter perturbation $\phi_{\varepsilon}(x) = \phi(x) + \varepsilon \psi(x)$ used to define the first variation of the action.
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A scalar field $\phi$ is a stationary point of the action: for every compact region $K$ and every variation $\psi$ supported in $K$, the first variation of the action vanishes, $\left. \frac{d}{d\varepsilon} \right|_{\varepsilon = 0} \mathcal{S}_K[\phi + \varepsilon \psi] = 0$.
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Partial derivative of the Lagrangian with respect to the field value $\phi$: $\partial L / \partial \phi$ evaluated at $(\phi(x), \partial \phi(x), x)$.
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Partial derivative of the Lagrangian with respect to the $\alpha$-th component of the gradient: $\partial L / \partial(\partial_{\alpha} \phi)$ evaluated at $(\phi(x), \partial \phi(x), x)$.
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The Euler-Lagrange operator applied to $\phi$ at $x$: $\mathrm{EL}\phi = \dfrac{\partial L}{\partial \phi}
- \sum_{\alpha} \partial_{\alpha} !\left( \dfrac{\partial L}{\partial(\partial_{\alpha} \phi)} \right)$. A scalar field is stationary iff this operator vanishes identically.
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Fundamental lemma of the calculus of variations: if $f$ is continuous and $\int_K f \psi \, dx = 0$ for every compact $K$ and every smooth variation $\psi$ supported in $K$, then $f \equiv 0$.
Smooth differentiation under the integral sign on a compact domain: if at $\varepsilon_0$ the parameter derivative $\partial_{\varepsilon} F(\varepsilon, x)$ exists pointwise with limit $F'(x)$ and both $F(\varepsilon, \cdot)$ and $F'$ are continuous on $K$, then the parameterized integral $\varepsilon \mapsto \int_K F(\varepsilon, x) \, dx$ has derivative $\int_K F'(x) \, dx$ at $\varepsilon_0$.
Chain rule for the Lagrangian along a perturbation: at $\varepsilon = 0$, the derivative of $\varepsilon \mapsto L(\phi + \varepsilon \psi, \partial(\phi + \varepsilon \psi), x)$ equals $\partial_{\phi} L \cdot \psi + \sum_{\alpha} \partial_{(\partial_{\alpha} \phi)} L \cdot \partial_{\alpha} \psi$.
Continuity hypotheses needed to apply differentiation under the integral: both the parameter-dependent Lagrangian integrand $L(\phi_{\varepsilon}, \partial \phi_{\varepsilon}, x)$ and the limit integrand $\partial_{\phi} L \cdot \psi + \sum_{\alpha} \partial_{(\partial_{\alpha} \phi)} L \cdot \partial_{\alpha} \psi$ are continuous on $K$.
Integration by parts in a single coordinate direction: for $\psi$ compactly supported in $K$ and $f$ of class $C^1$, $\int_K f \, \partial_{\alpha} \psi \, dx = -\int_K (\partial_{\alpha} f) \, \psi \, dx$. The boundary term vanishes because $\psi$ is supported in $K$.
Integrability of all terms appearing in the Euler-Lagrange expansion: each piece of $\partial_{\phi} L \cdot \psi$, $\sum_{\alpha} \partial_{(\partial_{\alpha} \phi)} L \cdot \partial_{\alpha} \psi$, the integration-by-parts substitute $\sum_{\alpha} (-\partial_{\alpha} \partial_{(\partial_{\alpha} \phi)} L) \cdot \psi$, and their per-coordinate components are integrable on $K$.
Smoothness of the gradient-derivative of the Lagrangian: if $L$ is $C^2$ in its arguments and $\phi$ is $C^2$, then $x \mapsto \partial_{(\partial_{\alpha} \phi)} L (\phi(x), \partial \phi(x), x)$ is $C^1$.
Per-coordinate integration by parts applied to the gradient term: the integral $\int_K \sum_{\alpha} \partial_{(\partial_{\alpha} \phi)} L \cdot \partial_{\alpha} \psi \, dx$ equals $\int_K \sum_{\alpha} \bigl(-\partial_{\alpha} \partial_{(\partial_{\alpha} \phi)} L\bigr) \cdot \psi \, dx$, together with the integrability of all involved terms.
Combining the pointwise chain rule with differentiation under the integral, the first variation of the action equals the integral of the linearized integrand: $\left. \frac{d}{d\varepsilon} \right|_{\varepsilon = 0} \mathcal{S}_K[\phi + \varepsilon \psi] = \int_K \left( \partial_{\phi} L \cdot \psi + \sum_{\alpha} \partial_{(\partial_{\alpha} \phi)} L \cdot \partial_{\alpha} \psi \right) dx$.
Integrating by parts reorganizes the linearized integrand as the Euler-Lagrange operator times $\psi$: $\int_K \left( \partial_{\phi} L \cdot \psi + \sum_{\alpha} \partial_{(\partial_{\alpha} \phi)} L \cdot \partial_{\alpha} \psi \right) dx = \int_K \mathrm{EL}[\phi] \cdot \psi \, dx$.
First-variation formula: combining the Leibniz/chain rule and the integration by parts, the first variation of the action equals the $L^2$-pairing of the Euler-Lagrange operator with the test variation, $\left. \frac{d}{d\varepsilon} \right|_{\varepsilon = 0} \mathcal{S}_K[\phi + \varepsilon \psi] = \int_K \mathrm{EL}[\phi](x) \cdot \psi(x) \, dx$.
Euler-Lagrange equations: a sufficiently regular field $\phi$ is a stationary point of the action if and only if it satisfies the Euler-Lagrange equation $\mathrm{EL}\phi = \dfrac{\partial L}{\partial \phi}
- \sum_{\alpha} \partial_{\alpha} !\left( \dfrac{\partial L}{\partial(\partial_{\alpha} \phi)} \right) = 0$ for all $x$.