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I started to study classical field theory using the book "Field Quantization" of Greiner and Reinhardt, and I have some doubts. First, the book write the Lagrangian $L(t)$ as a functional of a field $\phi(\textbf{x},t)$: $$L(t) = L[\phi(\textbf{x},t), \dot{\phi}(\textbf{x},t)].$$ After, it defines the Lagrange density $\mathcal{L}$ by $$L(t) = \int d^{3}x \mathcal{L} ( \phi(\textbf{x},t), \nabla \phi(\textbf{x},t), \dot{\phi}(\textbf{x},t)),$$ and the book compute the variation in $L$: $$\delta L (t) = \int d^{3}x\ \delta \mathcal{L} = \int d^{3}x\ \left( \frac{\partial \mathcal{L}}{\partial \phi}\delta \phi + \frac{\partial \mathcal{L}}{\partial (\nabla \phi)}\delta \nabla \phi + \frac{\partial \mathcal{L}}{\partial \dot{\phi}}\delta \dot{\phi} \right),$$ Using that $\delta \nabla \phi = \nabla \delta \phi $ and integrating by parts the second term, it gives $$\int d^{3}x \nabla \frac{\partial \mathcal{L}}{\partial (\nabla \phi)}\nabla \delta \phi = \int d^{3}x \nabla \left( \frac{\partial \mathcal{L}}{\partial (\nabla \phi)}\delta \phi \right) - \int d^{3}x \left(\nabla \frac{\partial \mathcal{L}}{\partial (\nabla \phi)} \right)\delta \phi.$$ But the book consider only the second term of the second member of the expression above. The first integral should be zero, but why?

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Presumably, Greiner & Reinhardt assume Dirichlet boundary conditions (BC) so that $\delta \phi=0$ vanishes at the boundary.

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