Minimisation problem in 2D with Dubois-Reymond, deriving the Euler-Lagrange equation and natural boundary conditions. Confusion over the latter.

I find I am confused about the `natural' boundary conditions of this problem.

I will first formulate the problem, the method in which I find my final result so far, and then I will state my confusion more clearly. In short, the confusion is that the essential and natural boundary conditions I am supposed to find seem to imply that Dubois-Reymond (2D) cannot be used to find the Euler-Lagrange equation.

Given a domain $\Omega \in \mathbb{R}^2$ with boundary $\Gamma = \Gamma_1 \cup \Gamma_2 \cup \Gamma_3$, try to find a solution $u \in \Sigma$, $\Sigma \equiv \left\{ u: \: u(\underline{x}) = g(\underline{x}) \quad \forall \underline{x} \in \Gamma_1\right\}$ of the minimisation problem: $\text{min}_{\:u\in \Sigma}\left[ \int_\Omega \: F\left(x,y,u, u_x, u_y\right) \:d\Omega + \int_{\Gamma_2}\: f(x,y,u) \:d\Gamma_2 \right]$

So, following the methods in the book I define $\hat{u}$ as the smooth solution to the minimisation problem with the given boundary. I then write $u=\hat{u} + \epsilon \eta(x,y)$, where the latter is an arbitrary function. I then substitute this into the integral given, derivate to $\epsilon$ and so forth; I'm sure you are familiar with this method. After applying the divergence theorem, writing $\frac{\partial F}{\partial \nabla u}$ as the vector $\begin{pmatrix} \frac{\partial F}{\partial u_x} & \frac{\partial F}{\partial u_y} \end{pmatrix}$ for convenience:

$\int_\Omega \eta \left(\frac{\partial F}{\partial u} - \nabla \cdot \frac{\partial F}{\partial \nabla u}\right) d\Omega + \oint_\Gamma \eta \frac{\partial F}{\partial \nabla u} \cdot \underline{n} d\Gamma + \int_{\Gamma_2} \eta \frac{\partial f}{\partial u} d\Gamma_2 = 0 \quad \quad (\text{my final result})$.

Technically, I would now take $\epsilon = 0$, because $\hat{u}$ is the solution; it's simpler to just say that now, $u=\hat{u}$ and be done with it.

IF we were to apply Dubois-Reymond's Lemma on the first term, we'd find the differential equation $\frac{\partial F}{\partial u} - \frac{\partial}{\partial x} \frac{\partial F}{\partial u_x} - \frac{\partial}{\partial y} \frac{\partial F}{\partial u_y} = 0$, which is exactly the requested Euler-Lagrange equation.

However, to do that we must find some expression $\int_\Omega M(x,y) \eta(x,y) d\Omega = 0$. To me, this implies that the sum of the second and third term of my final result must be zero.

Since we have $u=g$ on $\Gamma_1$, it follows that $\int_{\Gamma / \Gamma_1} \eta \frac{\partial F}{\partial \nabla u} \cdot \underline{n} d\left\{\Gamma / \Gamma_1\right\} + \int_{\Gamma_2} \eta \frac{\partial f}{\partial u} d\Gamma_2 = - \int_{\Gamma_1} \eta \frac{\partial F}{\partial \nabla u} \cdot \underline{n} d\Gamma_1$.

This would then be the natural boundaries opposing the essential boundary.

The solution provided in my text book - which only gives the Theorem, not its derivation - has $u=g$ on $\Gamma_1$, $\frac{\partial F}{\partial \nabla u} \cdot \underline{n} + \frac{\partial f}{\partial u} = 0$ on $\Gamma_2$ and finally $\frac{\partial F}{\partial \nabla u} \cdot \underline{n} = 0$ on $\Gamma_3$.

To me, it seems that this is larger than the opposing condition I just proposed. This is exactly my confusion. They have a simpler example which they refer back to (same method), but this example has $g=0$, and thus requires exactly that the second/third term sum to zero.

Can anyone provide me with an answer as to why the natural boundary conditions are as defined in the book?

NB: Book is Numerical Methods in Scientific Computing, J. van Kan, A. Segal, F. Vermolen, where the latter is my actual teacher (whom is currently unavailable).

• I just realised the problem might be that they never formulated dubois-raymond on 2D. I'm guessing that, since $\eta$ is arbitrary, and the boundaries are lines, we can basically parametrise it and apply 1D dubois-raymond. If so, then the natural conditions are as given. I do not, however, have much confidence in this explanation. – Daimonie Nov 22 '14 at 11:09
Because $u = \hat{u} + \epsilon \eta$ is the variation around the minimal solution, we require that $u, \hat{u} \in \Sigma$. As this requres $u=g$ on $\Gamma_1$ it follows that $\eta = 0$ on $\Gamma 1$. This solves the problem I had.