# Principle of Stationary Action and Euler-Lagrange Equation

Principle of Stationary Action:

Given a mechanical system, there exists an action $$S$$ such that it is extremitized, or $$\delta S=0$$, for the actual motion of the system.

$$S = \int_{t_1}^{t_2}L(q, \dot{q}, t)dt$$

where $$L$$ is the Lagrangian of the system.

Euler-Lagrangian Equation: $$\frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot{q}}\bigg) = \frac{\partial L}{\partial q}$$

My understanding that the extremum of S implies that the E-L Equation is satisfied.

My question is: Does it work the other way? i.e. Given a mechanical system, is demanding $$\delta S = 0$$ for its action equivalent to demanding $$\frac{d}{dt}\big(\frac{\partial L}{\partial \dot{q}}\big) = \frac{\partial L}{\partial q}$$ for its Lagrangian?

• The Euler- lagranges equations are only a necessary condition for action to be extremum -not sufficient condition - For other sufficient conditions, see Gelfand & Fomin 2000. Chapter 5: see Gelfand and Fomin (2000)Silverman, Richard A., ed. Calculus of variations (Unabridged repr. ed.). Mineola, New York: Dover Publications. p. 3. ISBN 978-0486414485."The Second Variation.Chapter 6: "Fields. Sufficient Conditions for a Strong Extremum". Sufficient conditions for a strong minimum are given by the theorem on p. 148. Mar 6, 2016 at 17:16

The functional derivative of a functional $S[q]$ with respect to the function $q(t)$ is defined as $$\frac{\delta S[q]}{\delta q(t)}\equiv \lim_{\alpha\to 0}\frac{S[q+\alpha\delta_t]-S[q]}{\alpha}$$ where $\delta_t$ is the Dirac delta function centered at $t$.
Your professor/book probably proved that the functional derivative coincides with the Euler-Lagrange derivative, $$\frac{\delta S[q]}{\delta q(t)}=\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial L}{\partial\dot q}\right)-\frac{\partial L}{\partial q}$$ which means $\delta S=0$ iff E-L is satisfied. This means: as the functional derivative equals the E-L derivative, both are zero or neither is.