$$ \htmlClass{fragment}{ S(q) = \int_{t_0}^{t_1} L_t(q_t,\dot{q}_t) \d t } \qquad \htmlClass{fragment}{ \delta S(q) = 0 } $$

Letting $q^\eps = q + \eps v$, write $$ \htmlClass{fragment}{ \delta S(q) } \htmlClass{fragment}{ = \left.\frac{\d}{\d\eps} \int_{t_0}^{t_1} L(q_t^\eps, \dot{q}_t^\eps) \d t\right|_{\eps=0} } \htmlClass{fragment}{ = \left.\int_{t_0}^{t_1} \frac{\d}{\d\eps}L(q_t^\eps, \dot{q}_t^\eps) \right|_{\eps=0} \d t } $$ and, letting $L^\eps = L(q_t^\eps, \dot{q}_t^\eps)$, the total derivative is $$ \htmlClass{fragment}{ \frac{\d L^\eps}{\d\eps} } \htmlClass{fragment}{ = \frac{\d t}{\d\eps} \frac{\pd L^\eps}{\pd t} + \frac{\d q^\eps}{\d\eps} \frac{\pd L^\eps}{\pd q^\eps} + \frac{\d \dot{q}^\eps}{\d\eps} \frac{\pd L^\eps}{\pd \dot{q}^\eps} } \htmlClass{fragment}{ = v \frac{\pd L^\eps}{\pd q^\eps} + \dot{v} \frac{\pd L^\eps}{\pd \dot{q}^\eps} . } $$

Hence, choosing variations which vanish at the boundary, $$ \htmlClass{fragment}{ \delta S(q) } \htmlClass{fragment}{ = \int_{t_0}^{t_1} v \frac{\pd L}{\pd q} + \dot{v} \frac{\pd L}{\pd \dot{q}} \d t } \htmlClass{fragment}{ = \int_{t_0}^{t_1} v \del{\frac{\pd L}{\pd q} - \frac{\d}{\d t} \frac{\pd L}{\pd \dot{q}} }\d t } \htmlClass{fragment}{ = 0 } $$ using integration by parts, we obtain the differential equation $$ \htmlClass{fragment}{ \frac{\d}{\d t} \frac{\pd L}{\pd \dot{q}} - \frac{\pd L}{\pd q} = 0 } $$ by applying the fundamental lemma of variational calculus.

Consider the Lagrangian $$ \htmlClass{fragment}{ L(q,\dot{q}) } \htmlClass{fragment}{ = \frac{1}{2}m\dot{q}^2 - V(q) } $$ and write $\htmlClass{fragment}{\frac{\pd L}{\pd q}}\htmlClass{fragment}{ = - \frac{\pd V}{\pd q}}\htmlClass{fragment}{ = F}$ along with $\htmlClass{fragment}{\frac{\d}{\d t} \frac{\pd L}{\pd \dot{q}}}\htmlClass{fragment}{ = \frac{\d}{\d t}m\dot{q}}\htmlClass{fragment}{ = ma}$. Then $$ \htmlClass{fragment}{ \frac{\d}{\d t} \frac{\pd L}{\pd \dot{q}} - \frac{\pd L}{\pd q} = 0 } \qquad \htmlClass{fragment}{ \implies } \qquad \htmlClass{fragment}{ F = ma . } $$

$$ \frac{\d}{\d t} \frac{\pd L}{\pd \dot{q}} - \frac{\pd L}{\pd q} = 0 $$

(a) Find the contact time $\small t_c$

(b) Calculate true trajectory

Collisions lead to discontinuous jumps in velocity

$$ \htmlClass{fragment}{ \delta S(q,\dot{q}) = 0 } \qquad \htmlClass{fragment}{ S(q,\dot{q}) = \int_{t_0}^{t_1} L(q_t,\dot{q}_t) \d t } $$

Use neural ODEs to design networks that can model contacts

Principled and accurate handling of discontinuity

Principled and accurate handling of discontinuity