- #1
bugatti79
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Homework Statement
The functional ##I(u,v)=\int_\Omega F(x,y,u,v,u_x,u_y, v_x,v_y)dxdy##
The partial variation of a functional is given as
## \displaystyle \delta I =\int_\Omega (\frac{\partial F}{\partial u} \delta u+\frac{\partial F}{\partial u_x} \delta u_x+\frac{\partial F}{\partial u_y} \delta u_y+\frac{\partial F}{\partial v} \delta v+\frac{\partial F}{\partial v_x} \delta v_x+\frac{\partial F}{\partial v_y} \delta v_y)dxdy##
Homework Equations
The Attempt at a Solution
The next step in developing this equation is to integrate by parts the 2nd, 3rd, 5th and 6th terms.
1) Why do we not integrate the 1st term since F and ##\delta u## are both functions of x and y, ie a product..? Similarly for the 4th term?
2) For the second term I let ##U= \frac{\partial F}{\partial u_x}## and ## \displaystyle dV=\frac{\partial \delta u}{\partial x}## Therefore
## \displaystyle \int_\Omega (\frac{\partial F}{\partial u_x} \frac{\partial \delta u_x}{\partial x} )dxdy=\frac{ \partial F}{\partial u_x} \delta u - \int_\Omega [\frac{\partial }{\partial x}(\frac{\partial F}{\partial u_x}) \delta u]dxdy=\int_\Omega [\frac{\partial}{\partial x}(\frac{ \partial F}{\partial u_x} \delta u) - \frac{\partial }{\partial x}(\frac{\partial F}{\partial u_x}) \delta u]dxdy##
We seem to be integrating wrt x, how do we justify not doing anything with dy...? Becasue F and u are also functions of y..? Thanks