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).

Thanks in advance for your help.

PS: Sadly, this isn't even the homework, but me getting stuck on a theorem.

  • $\begingroup$ 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. $\endgroup$ – Daimonie Nov 22 '14 at 11:09
  • $\begingroup$ Dubois-reymond * $\endgroup$ – Daimonie Nov 22 '14 at 22:11

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.


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