I have a question related to these two posts:
(1) Euler-Lagrange, Gradient Descent, Heat Equation and Image Denoising
and
(2) When the Euler Lagrange equation simplifies to zero
Background
Suppose we have a functional (similar to (1)):
$E(u,u_x,u_y)=\int_\Omega(u_x^2 + u_y^2)dxdy$,
where $\Omega$ is the image domain, and $u(x,y)$ is a grayscale $256\times256$ image.
As I understand it, one can calculate the first variation of $E$:
$\delta E(u,u_x,u_y)(h(x,y))$ according to http://en.wikipedia.org/wiki/First_variation,
where $h$ is basiacly some unknown image (i.e. scalar valued 2D function).
In doing so, one eventually reaches a point where integration by parts is used, and it assumed that the end points of $u$, i.e. the bounds of the image, are fixed. The equation then simplifies, and becomes:
$\delta E(u,u_x,u_y)(h(x,y))=\int_\Omega(2u_{xx} + 2u_{yy})h(x,y)dxdy$
Okay. Now, if we knew $h$ then calculating $\delta E(u,u_x,u_y)(h(x,y))$ would be trivial. However, in general we don't know $h$ (though methods exist for solving for it, see http://tau.ac.il/~barleah/papers/Siam09.pdf).
However, if we can force $2u_{xx}+2u_{yy}=0$ for all $x,y\in\Omega$ then by definition $\delta E(u,u_x,u_y)(h(x,y))=0$. These equations are the Euler Lagrange equations.
Now one typically minimizes $E$ by using gradient descent (see Euler-Lagrange, Gradient Descent, Heat Equation and Image Denoising ).
Introducing a time step, $t$, and given an intial image, $u(x,y,0)$, then performing updates as:
$u(x,y,t+1) = u(x,y,t) + u_{xx}(x,y,t) + u_{yy}(x,y,t)$
Question
In order to arrive at:
$\delta E(u,u_x,u_y)(h(x,y))=\int_\Omega(2u_{xx} + 2u_{yy})h(x,y)dxdy$
we assumed that $h(x,y)=0$ for the boundaries of the image (see When the Euler Lagrange equation simplifies to zero).
What consequnces does this assumption of $h(x,y)=0$ for the boundaries of the image, have on the solution? (Source, "the Euler-Lagrange equation is derived using integration by parts on the assumption that all candidate functions take the same values at the endpoints" --When the Euler Lagrange equation simplifies to zero)
Are only variations that don't modify the boundaries of the image explored? i.e. is the solution we obtain the minimum of $E$ assuming that $u(x,y,\infty)$ = $u(x,y,0)$ at the boundaries of $\Omega$?