$$
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Euler-Lagrange Equations
- Lagrangian
- Calculus of Variations
- TODO Change of Coordinates
- TODO Hamilton-Pontryagin Principle
- Notes
Lagrangian
\[\LL(t, q, v) : \RR \times \RR^n \times \RR^n \to \RR\]
Let us consider a smooth curve \(q: \RR \to \RR^n\), the Lagrangian
\(\LL\) is evaluated along \(q\) by composing it with:
\[t \mapsto \block{t, q(t), \dot{q}(t)}\]
The action \(S\) of \(\LL\) along \(q\) is a functional that
integrates \(\LL\) along the path \(q\) between end-points \(q_0 =
q(0)\) and \(q_1 = q(1)\):
\[S: \mathcal{C}^\infty\block{[0, 1], \RR^n} \to \RR, \quad S(q) = \int_{t=0}^{t=1}\LL\block{t, q(t), \dot{q}(t)}.\dd t\]
The action \(S\) computes the cost associated with a given path
\(q\). Assuming the end-points are kept fixed, it is natural to ask
which curve minimizes the cost. As usual, a necessary condition for a
minimizer is to be a critical point of the action:
\[\delta S(q) = 0\]
Calculus of Variations
Let us now consider the space of smooth functions
\(\mathcal{C}^\infty\block{[0, 1], \RR^n}\) with fixed end-points
\(q_0, q_1\) in more details: it is a vector space of infinite
dimension, and it is not too difficult to see that its tangent space
is the space of smooth functions with zero end-points. Let \(\delta
q\) be such a tangent vector at \(q\), called a variation of \(q\).
By definition, \(\delta S(q).\delta q\) is equal to the directional
derivative of \(S\) in direction \(\delta q\), which can be obtained
as:
\[\delta S(q).\delta q = \left.\frac{\dd}{\dd \epsilon}S\block{q + \epsilon \delta q} \right|_{\epsilon = 0}\]
Expanding the right-hand side a little bit:
\[\begin{align}
\frac{\dd}{\dd \epsilon}S(q + \epsilon \delta q) &=\frac{\dd}{\dd \epsilon} \int_0^1\LL\block{t, q(t) + \epsilon \delta q(t), \dot{q}(t) + \epsilon \delta \dot{q}(t)}.\dd t \\
&= \int_0^1 \frac{\dd}{\dd \epsilon} \LL\block{t, q(t) + \epsilon \delta q(t), \dot{q}(t) + \epsilon \delta \dot{q}(t)}.\dd t \\
&= \int_0^1 \block{\ddd{\LL}{q}.\delta q(t) + \ddd{\LL}{v}.\delta \dot{q}(t)}.\dd t \\
&= \int_0^1 \ddd{\LL}{q}.\delta q(t).\dd t + \int_0^1\ddd{\LL}{v}.\delta \dot{q}(t).\dd t \\
\end{align}\]
Let us now integrate the second term by parts:
\[\begin{align}
\int_0^1\ddd{\LL}{v}.\delta \dot{q}(t).\dd t &= \underbrace{\left[\ddd{\LL}{v}.\delta q(t) \right]_0^1}_0 - \int_0^1\frac{\dd}{\dd t}\ddd{\LL}{v}.\delta q(t).\dd t \\
\end{align}\]
So that we finally obtain:
\[\delta S(q).\delta q = \int_0^1\block{\ddd{\LL}{q} -\frac{\dd}{\dd t}\ddd{\LL}{v} } \delta q(t).\dd t\]
But since we look for a critical point, we have \(\delta S(q).\delta q = 0\)
for all variations \(\delta q\), which implies:
\[\ddd{\LL}{q} -\frac{\dd}{\dd t}\ddd{\LL}{v} = 0\]
This equation is known as the
Euler-Lagrange
equation, rewritten below with more details:
\[\frac{\dd}{\dd t}\ddd{\LL}{v}\block{t, q(t), \dot{q}(t)} = \ddd{\LL}{q}\block{t, q(t), \dot{q}(t)}\]
TODO Change of Coordinates
TODO Hamilton-Pontryagin Principle
\[S(q, v, \lambda) = \int_{t=0}^{t=1}\LL\block{t, q(t), v(t)} - \lambda(t)^T\block{\dot{q}(t) - v(t)}.\dd t\]
Notes
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