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.

  3. For the periodic case that OP mentions, the solution should be periodic. The periodicity condition corresponds effectively to one BC, i.e. we are one BC short compared to the above counting.

  • $\begingroup$ The actual problem I'm working on has a Lagrangian density. The issue I'm having is that for a compact space, I can't correctly claim that there exists some boundary far away for which the variation vanishes. This seems to make me unable to drop the integral $\endgroup$ – R. Rankin Aug 6 '16 at 3:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.