6 Variational Methods

In this case findF (x, y, u, u x ,u y )=u 2 x + u 2 yand the Euler equation isu xx + u yy =0.This equation is called Laplace’s equation.Problems with Contraints: Lagrange Multipliers:In many problems we not only need to minimize a functional but this must be carried outsubject to some additional constraints. An important method for dealing with problems of thistype is the Method of Lagrange Multipliers. Let us suppose that we want to minimize a functionalJ[y] =∫ 10F (x, y, y ′ ) dx (6.1.21)where y is prescribed at the endpoints by y(0) = y 0 and y(1) = y 1 and y is subject to the extraconstraint thatK =∫ 10G(x, y, y ′ ) dx = l. (6.1.22)We consider a variation of y in the form y + α 1 η 1 + α 2 η 2 where η 1 and η 2 are continuouslydifferentiable and vanish at the endpoints. This we considerJ(α 1 ,α 2 ) ≡ ∫ 1F (x, y + α ⎫0 1η 1 + α 2 η 2 ,y ′ + α 1 η 1 ′ + α 2 η 2) ′ dx ⎬K(α 1 ,α 2 ) ≡ ∫ 1G(x, y + α 0 1η 1 + α 2 η 2 ,y ′ + α 1 η 1 ′ + α 2 η 2) ′ ⎭ ,dxand it must be true that J(α 1 ,α 2 ) takes on a minimum value subject to the constraint K(α 1 ,α 2 )=0when α 1 = α 2 =0. Thus as in multidimensional calculus we want to find a minimum of a functionof (α 1 ,α 2 ) subject to a constraint. One of the most useful tools for solving problems of this typeis the method of Lagrange Multipliers.We consider∂[J(α 1 ,α 2 )+λK(α 1 ,α 2 )] ∣ ∂α α1 =α 2=0, (6.1.23)=0 1∂[J(α 1 ,α 2 )+λK(α 1 ,α 2 )] ∣ ∂α α1 =α 2=0. (6.1.24)=0 2If we carry out the necessary steps and do integration by parts in each of the approriate terms, weend up with the two equations∫ 1[( ∂F∂y − ddx0∫ 10( ∂F∂y ′ ))+ λ( ∂G∂y − d ( ))] ∂Gηdx ∂y ′ 1 dx =0, (6.1.25)[( ∂F∂y − d ( )) ( ∂F ∂G+ λdx ∂y ′ ∂y − d ( ))] ∂Gηdx ∂y ′ 2 dx =0. (6.1.26)18