As mentioned on the Wikipedia page en.wikipedia.org/w/index.php?title=Stationary_Action_Principle&oldid=1020413171, "principle of least action" is not accurate since it could not necessarily be a minima, we could just be in a saddle-point.

Calculus of variations is the field that searches for maxima and minima of Functionals, rather than the more elementary case of functions from $R$ to $R$.

A function that takes input function and outputs a real number.

Let's start with the one dimensional case. Let the $q:R→R$ and a Functional $I$ defined by a function of three variables $F:R_{3}→R$:

$I(q)=∫_{t_{0}}F(t,q(t),q˙ (t))dt$

Then, the Euler-Lagrange equation gives the maxima and minima of the that type of functional. Note that this type of functional is just one very specific type of functional amongst all possible functionals that one might come up with. However, it turns out to be enough to do most of physics, so we are happy with with it.

Given $F$, the Euler-Lagrange equations are a system of ordinary differential equations constructed from that $F$ such that the solutions to that system are the maxima/minima.

In the one dimensional case, the system has a single ordinary differential equation:

$∂q∂F(t,q(t),q˙ (t)) −dtd ∂q˙ ∂F(t,q(t),q˙ (t)) =0$

By $∂q∂F $ and $∂q˙ ∂F $ we simply mean "the partial derivative of $F$ with respect to its second and third arguments". The notation is a bit confusing at first, but that's all it means.

Therefore, that expression ends up being at most a second order ordinary differential equation where $q$ is the unknown, since:

- the term $∂q∂F(t,q(t),q˙ (t)) $ is a function of $t,q(t),q˙ (t)$
- the term $∂q˙ ∂F(t,q(t),q˙ (t)) $ is a function of $t,q(t),q˙ (t)$. And so it's derivative with respect to time will contain only up to $q¨ $

Now let's think about the multi-dimensional case. Instead of having $q:R→R$, we now have $q:R→R_{n}$. Think about the Lagrangian mechanics motivation of a double pendulum where for a given time we have two angles.

Let's do the 2-dimensional case then. In that case, $F$ is going to be a function of 5 variables $F:R_{5}→R$ rather than 3 as in the one dimensional case, and the functional looks like:

$I(q)=∫_{t_{0}}F(t,q_{1}(t),q_{2}(t),q_{1}˙ (t),q_{2}˙ )(t)dt$

This time, the Euler-Lagrange equations are going to be a system of two ordinary differential equations on two unknown functions $q_{1}$ and $q_{2}$ of order up to 2 in both variables:
At this point, notation is getting a bit clunky, so people will often condense the $q_{i}$ vector
or just omit the arguments of $F$ entirely:

$∂q_{1}∂F(t,q_{1}(t),q_{2}(t),q_{1}˙ (t),q_{2}˙ (t)) −dtd ∂q_{1}˙ ∂F(t,q_{1}(t),q_{2}(t),q_{1}˙ (t),q_{2}˙ (t) =0∂q_{2}∂F(t,q_{1}(t),q_{2}(t),q_{1}˙ (t),q_{2}˙ (t)) −dtd ∂q_{2}˙ ∂F(t,q_{1}(t),q_{2}(t),q_{1}˙ (t),q_{2}˙ (t) =0$

$∂q_{1}∂F(t,q(t),q˙ (t)) −dtd ∂q_{1}˙ ∂F(t,q(t),q˙ (t)) =0∂q_{2}∂F(t,q(t),q˙ (t)) −dtd ∂q_{2}˙ ∂F(t,q(t),q˙ (t)) =0$

$∂q_{i}∂F −dtd ∂q_{i}˙ ∂F =0$

These are the final equations that you derive from the Lagrangian via the Euler-Lagrange equation which specify how the system evolves with time.