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neworder1
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Homework Statement
Let [tex]P[/tex] be a rectangle , [tex]f_{0} : \partial P \rightarrow R)[/tex] continuous and Lipschitz, [tex]C_{0} = \{ f \in C^{2}(P) : f=f_{0} \ on \ \partial P \}[/tex]. and finally [tex]S : C_{0} \rightarrow R[/tex] a functional:
[tex]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[/tex].
Write Euler-Lagrange equation for S.
Homework Equations
The Attempt at a Solution
I tried writing: [tex]S(f) = \int^b_a (\int^d_c (\frac{\partial f}{\partial x})^{2} + (\frac{\partial f}{\partial y})^{2}\,dy)\,dx [/tex], so the proper Lagrangian would be [tex]L(x) = \int^d_c (\frac{\partial f}{\partial x})^{2} + (\frac{\partial f}{\partial y})^{2}\,dy[/tex].
Then the Euler-Lagrange equation should be [tex]\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[/tex], ([tex]L^{'} = \int^d_c (\frac{\partial f}{\partial x})^{2}\,dy[/tex]) and now since [tex]\frac{dL^{'}}{dx} = \frac{\partial L^{'}}{\partial f_{x}^{'}}\frac{\partial f_{x}^{'}}{\partial x}[/tex], we can rewrite that as [tex]\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[/tex]. but what then?
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