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I have a Lagrangian $L(q,\dot q, t)$ that defines a variational problem, along with a constraint that takes the form $$ df = 0 \tag{1}$$ with $f(q,t)$. I take it is what is called a "semi-holonomic" constraint (see here and here), which here is integrable. It implies that solutions of the constraint obey $$ f(q(t),t) = constant_0 \tag{2}$$

Then when I want to solve the variational problem over all paths obeying to the constraint, I need to consider variations preserving the constraint. A part of these variation are nonlocal. I can consider solutions obeying to say $f=0$ and then essentially do as is my configuration space was that $f=0$ subspace. But I also need to consider these variations with $$\partial_q f \delta q + \partial_t f = \delta constant_0 = constant_1\tag{3}$$ that would change that $constant_0$ to another constant. For $\partial_t f\neq 0$, these can have to be nonlocal, so they do not enter the local Euler-Lagrange equation frame, but still have the global form

$$ \int (\partial_t (\partial_{\dot q}L \cdot\delta q) + \delta L\cdot\delta q)dt = 0\tag{4}$$ where $\delta L$ is the so-called variational derivative of $L$.

Now, this global integral doesn't even have to exist. As far as I understand, the action defined by a Lagrangian is essentially a tool to justify the Euler-Lagrange equations, which we care about as local equations. One could argue that these $constant_0$ define different "topological sectors" of the solutions, but in my case I want to be able to make that constant vary continuously.

Question: Is my analysis right, and if so how do I deal with this extra nonlocal equation? Ideally I want to express this constraint in the Hamiltonian formalism.

Note: My actual problem involves field theory and some kind of more complicated higher degrees constraints, but I think I need to understand this simpler case first.

jpdm
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  • FWIW, eq. (1) is an integrable semi-hol. constr, equivalent to the hol. constr, (2). Not all semi-hol. constr. are integrable/hol. Is that what you're asking? – Qmechanic May 04 '20 at 19:22
  • Thanks for the edit. My point is that all paths obeying (1) have a constant for which (2) holds, but the constant might differ. Considering the variational problem across all such field needs to take in account the variation between paths with different constants. In other words, "df=0" and "f=0" don't define the same kind of variational problem. Does that make sense? – jpdm May 04 '20 at 19:35

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