Euler-Lagrange Equation for Functional S

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Homework Statement

Let P be a rectangle , f_{0} : \partial P \rightarrow R) continuous and Lipschitz, C_{0} = \{ f \in C^{2}(P) : f=f_{0} \ on \ \partial P \}. and finally S : C_{0} \rightarrow R a functional:

S(f) = \int^b_a (\int^d_c (\frac{\partial f}{\partial x})^{2}\,dy)\,dx + \int^d_c (\int^a_b (\frac{\partial f}{\partial y})^{2}\,dx)\,dy.

Write Euler-Lagrange equation for S.

Homework Equations

The Attempt at a Solution

I tried writing: S(f) = \int^b_a (\int^d_c (\frac{\partial f}{\partial x})^{2} + (\frac{\partial f}{\partial y})^{2}\,dy)\,dx, so the proper Lagrangian would be L(x) = \int^d_c (\frac{\partial f}{\partial x})^{2} + (\frac{\partial f}{\partial y})^{2}\,dy.

Then the Euler-Lagrange equation should be \frac{d}{dx}\frac{\partial L}{\partial f^{'}_{x}} = 0 \leftrightarrow \frac{d}{dx}\frac{\partial }{\partial f^{'}_{x}}\int^d_c (\frac{\partial f}{\partial x})^{2}\,dy = 0, (L^{'} = \int^d_c (\frac{\partial f}{\partial x})^{2}\,dy) and now since \frac{dL^{'}}{dx} = \frac{\partial L^{'}}{\partial f_{x}^{'}}\frac{\partial f_{x}^{'}}{\partial x}, we can rewrite that as \frac{d}{dx}\frac{\frac{d}{dx}\int^d_c (\frac{\partial f}{\partial x})^{2}\,dy}{f_{xx}^{''}} = 0 \leftrightarrow \frac{d}{dx}\frac{\int^d_c \frac{\partial}{\partial x}(\frac{\partial f}{\partial x})^{2}\,dy}{f_{xx}^{''}} = 0 \leftrightarrow \frac{d}{dx}\frac{\int^d_c 2f_{x}^{'}f_{xx}^{''}\,dy}{f_{xx}^{''}} = 0. but what then?

 

[Dick]

You are making this too hard. Let's call \frac{\partial f}{\partial x}=f_x and \frac{\partial f}{\partial y}=f_x. Then the form of the Euler-Lagrange equations for two independent variables is \frac{\partial L}{\partial f}-\frac{\partial}{\partial x} \frac{\partial L}{\partial f_x}-\frac{\partial}{\partial y}\frac{\partial L}{\partial f_y}=0 where L=(f_x)^2+(f_y)^2.