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In Lagrangian formalism, given two points $(x_1,t_1)$ and $(x_2,t_2)$, we ask the question which paths $x(t)$ make the action $S=\displaystyle \int_{t_1}^{t_2}L\ \mathrm dt$ stationary and satisfy the boundary condition $x(t_1)=x_1,\ x(t_2)=x_2$. This question is equivalent to solving the Euler-Lagrange equation $$\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial \dot{x}}=\frac{\partial L}{\partial q}$$ with boundary condtion $x(t_1)=x_1,\ x(t_2)=x_2$.

My question is why we are authorized to use the Euler-Lagrange equation to solve the initial condition problem $x(t_1)=x_1,\ \dot{x}(t_1)=v_1$.

It seems that these are two different problems. One problem is to find a path satisfying the boundary condition $x(t_1)=x_1,\ x(t_2)=x_2$ and make the action stationary.

The other problem is to find a path with initial condition $x(t_1)=x_1,\ \dot{x}(t_1)=v_1$ and I even don't know how to put other requirements such that its equation of motion is the Euler-Lagrange equation.

How can you prove these two problems are equivalent if you can make the second problem clear? Or maybe it is an axiom that we require the initial condition problem is solved by Euler-Lagrange equation. I'm confused about the logic of Lagrangian formalism.

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    $\begingroup$ I answered a similar question here, explaining how this is not a problem in practice. I also asked a question here about how bad the loophole can get (pretty bad!). $\endgroup$ – knzhou Apr 29 '16 at 7:17
  • $\begingroup$ Why we are authorized to use this equation? You don't need authorization to use an equation. $\endgroup$ – immibis Apr 30 '16 at 0:54
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Indeed the problem with boundary conditions, generally speaking, is not well-posed.

There are boundary conditions admitting no curves or admitting many curves, satisfying both these conditions and Euler-Lagrange equations.

Examples.

(1) Think of a particle constrained to stay on a smooth sphere where it can freely move. If you assign the North and the South pole of the sphere as boundary conditions, you get infinitely many solutions as the motion always describes a geodesic.

(2) Similarly, if you remove an open ball in $\mathbb R^3$ you do not have solutions when assigning boundary conditions on the opposite sides of the ball for a free particle in $\mathbb R^3$.

If $L$ is quadratic with respect to the $\dot{q}$ variables and this quadratic form is strictly positively defined, as is the case for systems of classical particles (also satisfying ideal holonomic constraints), the problem with initial conditions is always well-posed provided $L$ is sufficiently regular. There is exactly one maximal solution satisfying both Euler-Lagrange equations and initial conditions.

With these hypotheses also the problem with boundary conditions is well-posed with the additional condition that the two boundary conditions are sufficiently close to each other (this is evident from the two examples I presented above).

For these reasons a safer (mathematically minded) viewpoint is assuming that the variational principle determines the equation of motion, but not the solutions themselves.

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    $\begingroup$ Thanks. So your meaning is that E-L equation with boundary condition sometimes is ill-defined. But E-L equation with intial condition is mostly well-defined. Given this motivation, we require that equation of motion should be E-L equation. And this is kind of axiom in Lagrangian formalism. $\endgroup$ – 346699 Apr 29 '16 at 7:35
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    $\begingroup$ I don't think that your "safer viewpoint" is justified. This whole negative "spirit" in the interpretation of these things is largely irrational. When some laws of physics are formulated in terms of the action - and they often are in practice - and the solution to the least action principle says that there are no solutions or many solutions, then it is a right and important insight that means something and we should take it very seriously, regardless of the (naive!) prejudice that there always has to be a unique solution. $\endgroup$ – Luboš Motl Apr 29 '16 at 8:00
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    $\begingroup$ For example, in the Euclidean path integral in QFT (using the action principle), one can find numerous solutions even with given boundary conditions, the instantons. That doesn't mean that something is wrong and we should discard these configurations or the theory. These instantons (even though they are just "local minima" of the action) are genuine and they give contributions to the probability amplitudes - and allow some otherwise forbidden processes. $\endgroup$ – Luboš Motl Apr 29 '16 at 8:03
  • $\begingroup$ Dear Lubos, my viewpoint is completely mathematical (regarded the well-posedness of the problem), I completely agree with you concerning physical interpretation and the use of variational principle by physicists... $\endgroup$ – Valter Moretti Apr 29 '16 at 8:17
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    $\begingroup$ I want to ask a question: what's the definition of boundary conditions are sufficiently close to each other? And what's the definition of L is regular. You can directly tell me the reference if you feel it's not conventient. Thanks. $\endgroup$ – 346699 Apr 29 '16 at 8:47
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The issue is that the underlying classical physics is determined by equations of motion (EOMs) (i.e. Newton's 2nd law), which are common for initial value problems (IVPs) and boundary value problems (BVPs). For BVPs , the EOMs can often alternatively be formulated as Euler-Lagrange (EL) equations of a stationary action principle. The latter approach does never work for IVPs, because one does not have the correct boundary conditions to deduce EL equations via integration by parts. See also this related Phys.SE post and links therein.

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