# Functional extrema and the Euler-Lagrange equation

For a functional of the form:

$$S(q)=\int_{t_{1}}^{t_{2}}L(q,\dot{q})dt,\tag{1}$$ where $\dot{q}=\frac{\partial q}{\partial t}$, one finds that extrema are reached (to first order) for the condition:

$$0=S(q+\delta q)-S(q)=\int_{t_{1}}^{t_{2}}dt(\frac{dL(q,\dot{q})}{dq}-\frac{\partial}{\partial t}\frac{\partial L(q,\dot{q})}{\partial\dot{q}})\delta q(t).\tag{2}$$

Then it's simply stated that:

$$0=\frac{dL(q,\dot{q})}{dq}-\frac{\partial}{\partial t}\frac{\partial L(q,\dot{q})}{\partial\dot{q}}.\tag{3}$$

I could be wrong, but it seems like the integral form yields a more general set of solutions than the lattermost equation? In the particular problem I'm working on, the domain of integration is a compact space such that $t_{1}=t_{2}$, lets just say it's over a circle of some radius $R$. Then especially here, won't these two equations diverge in their solutions? Maybe I'm missing something.

1. Euler-Lagrange (EL) equation (3) is a necessary condition for a stationary path for the functional $S$. This follows from localization, i.e. the fundamental lemma of calculus of variations. In particular, eq. (2) does not lead to more stationary solutions than eq. (3), as OP seems to ponder.
2. OP does not mention any boundary conditions (BCs). BCs are usually needed to have a well-posed variational problem. E.g. typically for a problem where the independent variable $t$ is 1D, and where the Lagrangian $L$ depends on first-order derivatives, the EL eq. (3) becomes a 2nd-order ODE, and two BCs are often needed.