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Mathematics
Calculus
Constrained variational problem
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[QUOTE="jambaugh, post: 6015112, member: 76054"] One more point. As this functional differential is just a conventional differential on a function space it obeys the same rules. It commutes with linear operators: [itex]\delta L[\phi] = L[\delta\phi][/itex] and more generally obeys the Leibniz rule for multilinear forms: [tex]\delta F[\phi_1,\phi_2,\phi_3,\ldots] = F[\delta \phi_1,\phi_2,\phi_3,\ldots] + F[\phi_1,\delta \phi_2,\phi_3,\ldots] + \cdots[/tex] This is how you evaluated the differential of your inner product. Note that since an definite integral is a linear functional on its argument you get likewise: [tex] \delta \int_{\Omega}Fdx = \int_{\Omega} \delta F dx[/tex] Oh and since [itex]d[/itex] is linear [itex]\delta [/itex] and [itex]d[/itex] commute: [tex]\delta dF = d\delta F[/tex] That's a crucial step in deriving Euler-Lagrange equations as you may recall. [/QUOTE]
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Constrained variational problem
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