what to make out of it when Euler-Lagrange Equation is constantly zero

Question

I'm new to Calculus of Variations, and I'm trying to apply it to a simple vector calculus problem. Let's consider finding a curve $C$ along which the work $W$ done by a given vector field $\textbf{F}$ is maximized. Obviously $W$ can be expressed by

$$W = \int_C \textbf{F}\cdot d\textbf{r}$$

Since a line integral in essence is a single-variable integral, we can rewrite the above integral as

$$W = \int_C \textbf{F}\cdot d\textbf{r} = \int_C \textbf{F}(\textbf{r}(t))\cdot \textbf{r'}(t)dt=\int_C L(t,\textbf{r},\textbf{r'})dt$$

where $\textbf{r}(t)$ is the displacement

I used the one-dimensional Euler-Lagrange Equation,

$$\frac{d}{dx}(\frac{\partial L}{\partial \textbf{r'}}) - \frac{\partial L}{\partial \bf{r}}=0$$ which actualy yields $$\textbf{F'}(\textbf{r}) \cdot \textbf{r'} - \textbf{F'}(\textbf{r}) \cdot \textbf{r'} = 0$$

My question is :

1. Is there any mistakes in the steps?
2. what can we say about this constantly zero result? I noticed that the original derivation in 1-D Euler-Lagrange equation is to find the functional $y$ (in my case is $\textbf{r}(t)$) in terms of $t$ or $x$ that maximize a certain integral, and E-L equation is a necessary condition but not a sufficient for the extremal (similar to setting the derivative/gradient to zero). Does it mean this methods(E-L equation) fails in this case(we can't tell what curve maximize the line integral) or am I missing some underlying constraints here?
3. I've found a similar thread When the Euler Lagrange equation simplifies to zero. I'm not sure if I get this right but it seems the answer here implies a constantly zero E-L equation only tells us that the integral is of the same form regardless of the candidate function $\textbf{r}(t)$ here.
4. If we consider a special case, say, $C$ is a closed curve ($\textbf{r}(t)$ takes the same value at the both endpoints), then the original line integral (now turns into a contour integral) can be rewrite into a double integral of the curl using Stokes' theorem:

$$W = \oint_C \textbf{F}\cdot d\textbf{r} = \iint_D (\nabla \times \textbf{F}) \cdot d\textbf(A)$$

where $D$ is the region enclosed by the closed curve $C$, then clearly the extrema of the integral is also dependent on the vector field $\textbf{F}$, but how can I include this info in the E-L equation?

Answer 1

There was a mistake in your deviation. First we note that $L = \sum_j F_j \dot r_j$ $$ \frac{\partial L}{ \partial r_i} = \sum_j \frac{\partial F_j}{ \partial r_i} \dot r_j$$

$$ \frac{\partial L}{ \partial \dot r_i} = F_i$$ $$\frac{d}{dt} \frac{\partial L}{ \partial \dot r_i} = \sum_j \frac{\partial F_i}{ \partial r_j} \dot r_j$$

Hence the Euler lagrange equation reads

$$ \sum_j\left (\frac{\partial F_i}{ \partial r_j} - \frac{\partial F_j}{ \partial r_i} \right) \dot r_j = 0$$

This it clearly trivial iff $\nabla \times F = 0.$ In this case the field in conservative and the integral constant if the endpoints are fixed.

The consistently $0$ result is what happens when your integral is constant. It is basically analogous to how constant functions have derivative $0$.