How to interpret the vector fields $F_p(x,u,Du)$ in a Lagrangian optimization problem

Question

Let $\Omega$ be an open bounded subset of $\mathbb{R}^n$ with $C^1$ boundary. Let $$ \begin{matrix} F: \mathbb{R}^n \times \mathbb{R}^N \times \mathbb{R}^{nN} \to \mathbb{R},& \\ (x,z,p) \mapsto F(x,z,p)& \end{matrix} $$ be a $C^2$ Lagrangian corresponding to the functional $$ \mathcal{F}(u) = \int_{\Omega} F(x,u(x),Du(x)) dx $$ on the class $C^2(\bar{\Omega}, \mathbb{R}^N)$. By applying integration by parts and the divergence theorem (see Giaquinta and Hildebrandt, Chapter 1.2.4), we can show that if $u$ is an extrema, then $$ F_{p^i_\alpha}(x,u(x),Du(x)) \nu_\alpha = 0 \quad\text{ for }i = 1, \ldots, N, $$ where $\nu$ is the outward unit normal of $\partial \Omega$. Here, for fixed $1 \leq i \leq N$, $F_{p^i_\alpha}$ is a vector in $\mathbb{R}^n$ indexed by $\alpha$, and so $F_{p^i_\alpha} \nu_\alpha$ is a sum of $n$ vectors scaled by the components of $\nu$.

For example, when $N = 1$ and $$ F(x,z,p) = \sqrt{1 + |p|^2}, $$ then $\mathcal{F}(u)$ is the area of the graph of $u$, and the stationary points are functions whose graphs are minimal surfaces. The single vector field $F_p(x,u,Du)$ is $$ \frac{Du}{\sqrt{1 + |Du|^2}}, $$ which is the unit surface normal of the graph. The natural boundary condition $F_p \nu = 0$ translates into the statement that $F_p$ and $\nu$ are orthogonal. However, on the graph of $u$ defined as $$ \Gamma(u) = \{ (x,u(x)) : x \in \bar{\Omega} \} \subseteq \mathbb{R}^n \times \mathbb{R}, $$ the unit surface normal is given by $$ n = \frac{1}{\sqrt{1 + |Du|^2}}\left( Du , -1 \right), $$ and the normal to the boundary cylinder $\partial \Omega \times \mathbb{R}$ is $\bar{\nu} = (\nu, 0)$. Hence the boundary condition implies $n$ and $\bar{\nu}$ are also orthogonal, which means $\Gamma(u)$ and $\partial \Omega \times \mathbb{R}$ intersect perpendicularly.

In general Lagrangian problems, how are we supposed to interpret the vector fields $F_p$, particularly with general candidate solutions $u$? In principle, I wouldn't expect these vector fields to have a geometric relationship with the graph of $u$, yet the natural boundary conditions provide a geometric condition to check whether a candidate function is an extrema. Here, I'm ignoring questions regarding existence of solutions and regularity. I would presume that whatever interpretation is given has an appropriate weak analogue.

Answer 1

There is the following interpretation coming from physics and continuum mechanics, which is a bit too long for a comment but might be helpful:

If you think of $\mathcal{F}$ as an energy that you want to minimize, then its variation (with the opposite sign) abstractly corresponds to a force (Change of energy = Work = Force*distance), so critical points are those where the force is zero and (at first order) no work can be done.

If the energy is given by an integral, like in your example, one can then localize this into a "force density" on $\Omega$ in the form of $$ -F_z+\nabla \cdot F_p$$ as in the Euler-Lagrange equation (I will play a bit loose with notation). Now the $-F_z$-part acts more like a bulk force and has no impact on the boundary. On the other hand $F_p$ is what is called a Stress tensor in continuum mechanics.

The standard interpretation of a Stress tensor is that, if at a point you would cut $\Omega$ in half by a plane with the normal $\nu$, then the force (or rather its surface density) of one of the halves at that point is $F_p\cdot \nu$. In the interior, the forces of the two halves more or less cancel, which is why we are left with only the divergence. At the boundary on the other hand, there is only one side.

So at the boundary, $F_p \cdot \nu$ is the surface force density that is coming from the energy of $u$. For any minimizer that has to be zero, otherwise you could locally move into that direction and decrease $\mathcal{F}$. Thus $F_p \cdot \nu = 0$.

Keep in mind however that $F_p$ is a tensor, a thing that takes surface normals in $\mathbb{R}^n$ and maps them to vectors in $\mathbb{R}^N$. So even for $N=1$, one could at most interpret it as a vector in $\mathbb{R}^n$, but only its extent in normal direction actually matters. I am not entirely sure what you did in your graph-example, but there the dimensions do not add up. If you want to have a vector in $\mathbb{R}^{n+N}$, then you can rewrite the problem as a minimization (in your example of the area-functional) of maps $v:\Omega \to \mathbb{R}^{n+N}$ with the constraints that $v_i(x) =x_i$ for $i\in \{1,\dots,n\}$. Then $F_p \cdot \nu \in \mathbb{R}^{n+N}$, but the constraints will give you some additional Lagrange-multipliers, which make this a bit superfluous.