I'm studying calculus of variations and currently am trying to understand problems with variable endpoints in three dimensions.
In two dimensions, my understanding is:
For a functional $ I(y) = \int_{a}^{b}L(x,y(x),y'(x))dx $, where endpoint $A[a,\alpha]$ is an arbitrary point on a curve $g(x)$ and endpoint $B[b,\beta]$ is an arbitrary point on a curve $h(x)$ we obtain stationary curves (not sure about the translation from my language of stationary curves, by that I mean solutions that are "suspect" of being an extrema, fulfilling the necessary condition) by solving Euler-Lagrange equation
$ \frac{\partial L}{\partial y} - \frac{d}{dx}(\frac{\partial L}{\partial y'})=0, $
while satisfying transversality conditions
$(L(x,y,y')+\frac{\partial L}{\partial y'}(g'-y'))|_{x=a} = 0$
$(L(x,y,y')+\frac{\partial L}{\partial y'}(h'-y'))|_{x=b} = 0$.
My confusion begins in three dimensional case:
There are two options for variable ends, either space curve or a surface, let's consider space curve for now. In my textbook, there is a very brief description of this topic. It uses this denotation $ I(y^{\sigma}) = \int_{a}^{b}L(x,y^{\sigma},y^{\sigma}{'})dx $, where $y^{\sigma}$ means $(y^1(x),y^2(x),...,y^n(x))$, superscripts are now upper indices not powers. Here for a case $\sigma = 1,2$ I'm going to use $L(x,y,z,y',z')$.
For a functional $I(y,z) = \int_{a}^{b}L(x,y(x),z(x),y'(x),z'(x))dx$ where endpoint $A[a,\alpha_1,\alpha_2]$ is an arbitrary point on a space curve $c_A$ given by $g_1(x), g_2(x)$ and endpoint $B[b,\beta_1,\beta_2]$ is an arbitrary point on a space curve $c_B$ given by $h_1(x), h_2(x)$ we obtain stationary curves by solving
$ \frac{\partial L}{\partial y} - \frac{d}{dx}(\frac{\partial L}{\partial y'})=0 $
$ \frac{\partial L}{\partial z} - \frac{d}{dx}(\frac{\partial L}{\partial z'})=0, $
while satisfying transversality conditions
$(L+\frac{\partial L}{\partial y'}(g_1'-y')+\frac{\partial L}{\partial z'}(g_2'-z'))|_{x=a} = 0$
$(L+\frac{\partial L}{\partial y'}(h_1'-y')+\frac{\partial L}{\partial z'}(h_2'-z'))|_{x=b} = 0$.
Now for the questions and confusions I have about this. I somehow thought that when we're dealing with three dimensions, the Lagrangian would be a function of two independent variables and one dependent $(L(x,y,z(x,y),z_x,z_y), $ but that would require use of double integral and now notion of endpoints wouldn't make sense, because we would integrate over some area. So how does use of Lagrangian $L(x,y(x),z(x),y'(x),z'(x))$ work here?
Usually I like to calculate few examples to get a better grasp, unfortunately my textbook doesn't offer any for this case so I'd like to come up with my own, the only thing I can think of is extending catenary problem from 2d to 3d and constraining it by two curves. Let's consider this problem:
Let's say we want to erect two poles and connect them with wire (what I'm aiming at here is getting the solution as catenary, where we minimize potential energy of the wire) and we're constrained with two contour lines (curves on a map where the height for every point on the curve is equal, could be an ellipse or something simple in this case).
Now, since it is an example that I'm trying to make up and I don't understand the three dimensional case, I don't know how to work out the assignment and computational details. If anyone could help me here, I would be grateful. Or perhaps offer another example.