Euler-Lagrange equations for functions with multiple dependent arguments

Question

Consider the following functional equation:

$$ g[f] = \int_{t_1}^{t_2} f(x_1, x_2, x_1',x_2',u ) du.\tag{1}$$

Now, suppose there is some dependency between $x_2$ and $x_1$ that is there exists $x_2(x_1)$, without substituting directly the dependency, can I use euler-lagrange equation? I tried deriving it as follows by petrubing $x_1 \to x_1 + \epsilon \eta(u)$

$$ \delta g = \int_{t_1}^{t_2} \frac{\partial f}{\partial x_1} \eta + \frac{\partial f}{\partial x_2} \frac{ dx_2}{dx_1} \eta + \frac{\partial f}{\partial x_1' } \eta' + \frac{ \partial f}{\partial x_2' } \frac{ \partial x_2'}{\partial x_1' } \eta' du.\tag{2}$$

Or by algebra and integration by parts

$$ \delta g = \int_{t_1}^{t_2} \eta \left[ \partial_{x_1} f + (\partial_{x_2} f) \frac{dx_2}{dx_1} \right] - \frac{d}{du} \left[ \partial_{x_1'} f + (\partial_{x_2'} f )\frac{dx_2'}{dx_1'} \right] \eta du.\tag{3}$$

Hence, for the above expression to be true for all $ \eta$, it must be that: $$ \partial_{x_1} f + (\partial_{x_2} f) \frac{dx_2}{dx_1}=\frac{d}{du} \left[ \partial_{x_1'} f + (\partial_{x_2'} f) \frac{dx_2'}{dx_1'} \right].\tag{4}$$

Is the condition for optimum under optimizing with $x_1$?

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Questions:

1. Is my result correct?
2. Suppose $x_2 \to x_1$ is not a bijective map, then there would be different forms of $f$ depending on which branch we substitute into the functional equation, so seems to me there would be multipile optimums but doing it without substituting lead me to only one answer. So, what's the buisness going on there? eg: say $x_1 = (x_2)^2 $ then $x_1 = \pm x_2$

I considered an analogy with implicit equations, say that of a circle

$$ y = \pm \sqrt{1-x^2}$$

If we isolate for $y$ and differentiate, it turns out that we get two branches for slope value for same $x$ but if we directly different $x^2+y^2=1$ then we would get a derivative which is function of $(x,y)$ , so the branch is directly accounted for , so is that a similar idea here?

Note:

$$ \partial_z f = \frac{\partial f}{\partial z}$$

Answer 1

Comments:

1. The dependence between $x_1$ and $x_2$ can usually be implemented via a constraint $$\Phi(x_1,x_2)~=~0,$$ and a Lagrange multiplier $\lambda$ with extended Lagrangian $$\widetilde{f}=f+\lambda \Phi .$$ This is e.g. the case with OP's circle example.

2. Often the constraint function $\Phi$ is of graph type $$\Phi(x_1,x_2)~=~x_2-\phi(x_1),$$ i.e. the constraint is a graph $$x_2~=~\phi(x_1).$$ It is not difficult to see (by eliminating $\lambda$) that this is equivalent to OP's approach (4).