0
$\begingroup$

Considering an ordinary differential equation of first order in the implicit form

$$ F(q(t),\dot q (t))=\alpha,\,\,\, q(0)=q_0 $$

with $\alpha,\, q_0$ constants, what is the relation of the solution $q(t)$ to solutions of the variational problem for $S[y]$, obtained by taking $F$ (or a suitable function thereof, e.g. $F^2$) as a Lagrange function

$$ S[y] = \int_0^T F(y(t),\dot y (t))\, dt, \\ y(0)=q_0, \, \, \, y(T)=q(T) $$

where $T$ is a conveniently chosen positive constant.

$\bf{Corrected:}$

The solutions of the variational problem for $F(y(t),\dot y (t))$ imply another implicit equation

$$F(y(t),\dot y (t))-\dot y(t) \frac{\partial F}{\partial {\dot y(t)}} = \it{const.}$$

With $\dot y(t) = z$, the correct function $G(y,z)$ to take in the variational problem should therefore satify

$$ \frac{\partial}{\partial z} \left( \frac{G}{z} \right) =\frac{F(y,z)}{z^2} $$

which relates $G$ to $F$ as a Lagrangian relates to its corresponding Hamilton function.

It seems therefore that a variational problem can be associated to any implicit, first order ODE.

$\endgroup$
  • 1
    $\begingroup$ It is often the opposite: your equation may come from a variational problem, but it would be tortuous to vary what was already varied. Simpler answer: in general there is no relation, unless you take some integral of the equation. $\endgroup$ – John B Jan 29 '16 at 17:50
  • $\begingroup$ @Jonas by integral of the equation you mean an integral of motion, like the Hamilton function (energy, conserved for a time-independent ODE). You are right, and I have now included this case. $\endgroup$ – Lupercus Jan 31 '16 at 18:45
  • $\begingroup$ Yes, that's what I meant. For example $F$ may have been obtained from some variational problem, but you are talking about using it for kind of a "second" variational problem. $\endgroup$ – John B Feb 1 '16 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.