First variation of a functional F(a, u(a), u(n)d /6( F(a, u(a), u()) Su tSu'dx du Extremum of a functional "ulo" is the minumum of a functional if: m)2/(o A necessary condition for a functional to attain an extremum atuo 6I(0)=0 (uo +av, uo+ai 0 Note analogy with differential calculus. Also difference since here we require 出=0ata=0. OF aF Su+ dx Integrate by parts the second term to get rid of du OF d/aF d/OF aF d/aF OF。中b Require du to satisfy homogeneous displacement boundary conditions 6u(b)=6u(a) 6Ⅰ= af d/aF\lOud.r=0�� � � � � � � � First variation of a functional: δI = δ F(x, u(x), u� (x))dx = δ F(x, u(x), u� (x)) dx � �∂F ∂F � � δI = δu + dx ∂u ∂u� δu Extremum of a functional “u0” is the minumum of a functional if: I(u) ≥ I(u0)∀u A necessary condition for a functional to attain an extremum at “u0” is: dI δI(u0) = 0, or dα(u0 + αv, u� 0 + αv� ) � � = 0 α=0 Note analogy with differential calculus. Also difference since here we require dF = 0 at α = 0. dα b�∂F ∂F � � δI = δu + ∂u� δu dx a ∂u Integrate by parts the second term to get rid of δu� . b�∂F d �∂F � d �∂F �� δI = δu + ∂u� δu − δu dx ∂u� dx a ∂u dx � b�∂F d �∂F �� ∂F � �b = − δudx + � ∂u dx ∂u� ∂u� δu a a Require δu to satisfy homogeneous displacement boundary conditions: δu(b) = δu(a) = 0 Then: � b�∂F d �∂F �� δI = − δudx = 0, ∂u dx ∂u� a 4