Minimizing function using Calc of Variations and LaGrange Equations

[shank8]

My instructor likes to explain his topics at light speed and I could barely understand how to use Calculus of variations and the La Grange equations to solve this so I need some help please.

This is the problem:

Consider the functional for W = w(x,y) prescribed on partial(D),

I(W) = ∫∫√1+(Wx)^2+(Wy)^2)dxdy

subject to the integral constraint that

J(W) = ∫∫Wdxdy = C, where C is the constant.

Use a Calculus of Variations approach in conjunction with the intro of a Lagrange multiplier, λ, to show that:

λ = (Wxx(1+(Wy)^2) - 2WxWyWxy + Wyy(1+(Wx)^2)) / (1+(Wx)^2+(Wy)^2)^(3/2)

where Wx, Wy, Wxx, Wyy, Wxy are all partials.

Now demonstrate that:

λ = -2/a = -2(2∏/3C)^(1/3)

Any help at all would be great. I don't even know where to start.

 

[gtoguy97]

Solution: We can use the calculus of variations and the introduction of a Lagrange multiplier, λ, to solve this problem. The Euler-Lagrange equation is a necessary condition for the extremum of a functional and is given by: F(x,y,u,u',u'') - λg(x,y,u)=0 Where F represents the functional, g represents the constraint, and λ is the Lagrange multiplier. For our problem, the functional is I(W) and the constraint is J(W); therefore, we can rewrite the Euler-Lagrange equation as: I'(W) - λJ'(W) = 0 The functional derivative of I(W) is given by: I'(W) = Wxx(1+(Wy)^2) - 2WxWyWxy + Wyy(1+(Wx)^2)) / (1+(Wx)^2+(Wy)^2)^(3/2) The functional derivative of J(W) is given by: J'(W) = 1 Substituting these into the Euler-Lagrange equation, we get:Wxx(1+(Wy)^2) - 2WxWyWxy + Wyy(1+(Wx)^2)) / (1+(Wx)^2+(Wy)^2)^(3/2) - λ = 0 Rearranging, we get: λ = (Wxx(1+(Wy)^2) - 2WxWyWxy + Wyy(1+(Wx)^2)) / (1+(Wx)^2+(Wy)^2)^(3/2) Now, for the second part of the question, we have that the integral constraint is J(W) = ∫∫Wdxdy = C, where C is the constant. We can rewrite this as: ∫∫Wdxdy =