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# Euler-Lagrange equation

nosplit "Calculus of variations" words: 451
Let's start with the one dimensional case. Let the and a Functional defined by a function of three variables : $$I(q)=∫t0​t​F(t,q(t),q˙​(t))dt (6)$$
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 , the Euler-Lagrange equations are a system of ordinary differential equations constructed from that such that the solutions to that system are the maxima/minima.
In the one dimensional case, the system has a single ordinary differential equation: $$∂q∂F(t,q(t),q˙​(t))​−dtd​∂q˙​∂F(t,q(t),q˙​(t))​=0 (7)$$
By and we simply mean "the partial derivative of 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 is the unknown, since:
• the term is a function of
• the term is a function of . And so it's derivative with respect to time will contain only up to
Now let's think about the multi-dimensional case. Instead of having , we now have . 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, is going to be a function of 5 variables rather than 3 as in the one dimensional case, and the functional looks like: $$I(q)=∫t0​t​F(t,q1​(t),q2​(t),q1​˙​(t),q2​˙​)(t)dt (8)$$
This time, the Euler-Lagrange equations are going to be a system of two ordinary differential equations on two unknown functions and of order up to 2 in both variables: $$∂q1​∂F(t,q1​(t),q2​(t),q1​˙​(t),q2​˙​(t))​−dtd​∂q1​˙​∂F(t,q1​(t),q2​(t),q1​˙​(t),q2​˙​(t)​=0∂q2​∂F(t,q1​(t),q2​(t),q1​˙​(t),q2​˙​(t))​−dtd​∂q2​˙​∂F(t,q1​(t),q2​(t),q1​˙​(t),q2​˙​(t)​=0 (9)$$ At this point, notation is getting a bit clunky, so people will often condense the vector $$∂q1​∂F(t,q(t),q˙​(t))​−dtd​∂q1​˙​∂F(t,q(t),q˙​(t))​=0∂q2​∂F(t,q(t),q˙​(t))​−dtd​∂q2​˙​∂F(t,q(t),q˙​(t))​=0 (10)$$ or just omit the arguments of entirely: $$∂qi​∂F​−dtd​∂qi​˙​∂F​=0 (11)$$