first property of the first variation so we can identify at with the first variation of the derivative of u But d=dr (6) Consider a function of the following form F=F(x,(x),(x) It depends on an independent variable x, another function of x(u(a)and derivative(u()). Consider the change in F, when u(therefore u) chang △F=F(x,u+6,u2+6u)-F(x,u,) F(a,u+ )-F(x expanding in Taylor series OF OF 1 aF 1 aF 2!2(a) 2 dudu (av)(a')+ OF OF +~0t+h.o.t First total variation of F. SF lim =alim aut aute OF a OF OF SF� As a first property of the first variation: du¯ du dv = + α dx dx ���� dx so we can identify αdv with the first variation of the derivative of u: dx �du dx = α dv dx δ But: dv dx αdv dx d dx α = = (δu) We conclude that: �du� d δ = dx dx(δu) Consider a function of the following form: F = F(x, u(x), u� (x)) It depends on an independent variable x, another function of x (u(x)) and its derivative (u� (x)). Consider the change in F, when u (therefore u� ) changes: ΔF = F(x, u + δu, u� + δu� ) − F(x, u, u� ) = F(x, u + αv, u� + αv� ) − F(x, u, u� ) expanding in Taylor series: ∂F ∂F 1 ∂2F 1 ∂2F ΔF = F + αv + ∂u� αv� + ∂u2 (αv) 2 + ∂u∂u� (αv)(αv� ) + · · · − F ∂u 2! 2! ∂F ∂F = αv + ∂u� αv� + h.o.t. ∂u First total variation of F: � ΔF � δF = α limα→0 α �F(x, u + αv, u� + αv� ) − F(x, u, u� )� = α limα→0 α � ∂F αv + ∂F = α lim ∂u ∂u�αv� � = ∂F αv + ∂F � ∂u� αv α→0 α ∂u δF = ∂F ∂u δu + ∂F ∂u� δu� 2