Suppose we're given two action integrals:

$$ S_1 = \int_{t_0}^{t_1} L_1(t,y,\dot{y}) dt \\ S_2 = \int_{t_1}^{t_2} L_2(t,y,\dot{y}) dt$$

The minimization of action integral $S = S_1 + S_2$ with respect to $y(t)\in C^2$ entails

$$ \dfrac{d}{d t}\dfrac{\partial L_1}{\partial \dot{y}} - \dfrac{\partial L_1}{\partial y} = 0 \; \forall \;t_0\leq t\leq t_1\\ \dfrac{d}{d t}\dfrac{\partial L_2}{\partial \dot{y}} - \dfrac{\partial L_2}{\partial y} = 0 \; \forall \;t_1\leq t\leq t_2$$

Now suppose I want to find $y(t)$ given boundary conditions $y(t_0) = y_0$ and $y(t_2) = y_2$. Here $y(t_1)$ is unconstrained. This can be done by solving each Euler Lagrange equation. However, I am stuck at enforcing continuity of $y(t)$ and $\dot{y}(t)$ at $t=t_1$. How should one incorporate them?


1 Answer 1


TL;DR: Assuming that the variation of $y$ at $t=t_1$ is unconstrained, the 2 pertinent conditions at $t=t_1$ are $$\lim_{t\to t_1^-}\frac{\partial L}{\partial \dot{y}}~=~\lim_{t\to t_1^+}\frac{\partial L}{\partial \dot{y}}\qquad\text{and}\qquad \lim_{t\to t_1^-}y(t)~=~\lim_{t\to t_1^+}y(t).\tag{A}$$


  1. Let us fix the framework by assuming a priori that the Lagrangian $L$ and the paths $y$ are (not necessarily continuous) piecewise $C^2$ with one-sided limits and one-sided 1st and 2nd derivatives everywhere.

  2. The conditions (A) follow by adjusting the standard proof of the Euler-Lagrange (EL) equation to OP's situation. Concretely, the task is to analyze possible boundary terms $\int_{t_0}^{t_2}\! dt\frac{d}{dt}\left[\frac{\partial L}{\partial \dot{y}}\delta y \right]$.

  3. Physically, the conditions (A) mean that momentum $p=\frac{\partial L}{\partial \dot{y}}$ and position are continuous functions of time $t$, respectively. Note that the continuity of velocity $\dot{y}$ and the force $\dot{p}$ might not be achievable at $t=t_1$.

  4. The 2 conditions (A) together with the 2 boundary conditions $y(t_0) = y_0$ and $y(t_2) = y_2$ are in total 4 conditions.

  5. Since each of the 2 interval solutions come with 2 integration constants, a total of 4 conditions are just the right number in order to have a unique solution.

  • $\begingroup$ Aha, got you now. we don't need $\dot{y}$ to be continuous. All we need is to make sure that the variational term gotten outside the integral after applying integration parts to be zero. That makes perfect sense. However, I sense that we're faced with 6 conditions now, which are : $y(t_0)$, $y(t_2)$ and yours 4. I understand that $y(t_0)$ and $y(t_2)$ cannot be enforced. Am I correct? $\endgroup$ Commented Jul 8, 2020 at 14:43
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Jul 8, 2020 at 15:31
  • $\begingroup$ It has been a while. Could I ask you a short relevant question? how should one interpret enforcing $\dot{y}(t_1)=\dot{y}(t_2)$? In particular, one can enforce that by modifying the variation applied to the system with a variation vanishing at the intermediary point in addition to its usual requirements (i.e., $\delta y(t_0)=\delta y(t_1) = \delta y(t_2) = 0$). However, I have no intuition whatsoever how to interpret consideration of a more limited variation like this? $\endgroup$ Commented Jul 25, 2020 at 1:44
  • $\begingroup$ FWIW, a non-local BC is typically unphysical. $\endgroup$
    – Qmechanic
    Commented Jul 25, 2020 at 8:25
  • $\begingroup$ I made a typo. It should have been $\dot{y}(t_1^-) = \dot{y}(t_1^+)$. I'm sorry. Basically, continuity of intermediate point joining components of piecewise Lagrangian. Suppose, this is what the physics of my problem dictates. I am understanding that this corresponds to a variations with $\delta y(t_0) = \delta y(t_1) = \delta y(t_2) =0$ rather than variations vanishing only at the initial and end points. Can one consider any variation viable depending on details of their problem? What would be a general interpretation of variations in this restricted form outside the specifics of a problem? $\endgroup$ Commented Jul 25, 2020 at 16:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .