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When we derive Euler-Lagrange equations in classical mechanics following the Lagrangian approach we introduce Boundary conditions at the starting- and end-points of the path in the configuration space. Usually,though not necessarily, one requires that $q(t_{initial})=q(t_{final})=0$. But the standard problem in classical mechanics is to instead assume Initial (not Boundary) conditions, in practice it is assumed that the initial position $q(t_{initial})$ and initial velocity $q'(t_{initial})$ are known.

Is there any way to modify the Lagrangian such that the extremisation of the Action would yield the Euler-Lagrange equations together with the above mentioned Initial conditions?

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  • $\begingroup$ The Euler Lagrange equations don't depend on initial conditions? Do you mean $\delta q(t_{initial}=\delta q(t_{final})=0$? The variation of $q$ has to be zero because any varied path should still match the boundary conditions but there are no restrictions on the B.C. of $q$. $\endgroup$ – AccidentalTaylorExpansion Nov 21 '19 at 17:03
  • $\begingroup$ Related: physics.stackexchange.com/q/38348/2451 and links therein. $\endgroup$ – Qmechanic Nov 21 '19 at 17:22
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Well that is a problem, I've faced once as well, you see the Euler Lagrange equations are basically used to give you the equation of motion, they for classical mechanics are as good as F=ma.

If you have used The second law to get equations of motion, you might have definitely used constraints or initial conditions to get the equation of motion, for example when you solve for the equations of the simple harmonic oscillator, you get the general equation to be

$$x=Asin(\omega t)$$

But you don't know what $A$ and $\omega$ are without substituting certain conditions.

That is also the case with the Euler Lagrange Equations, they provide you with the general equation of motion, you will have to substitute the conditions to find the exact equations

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