The Hamilton-Jacobi equation is: \begin{equation} H\left(q,\frac{\partial S}{\partial q},t\right)+\frac{\partial S}{\partial t}~=~0, \end{equation} where $S$ is Hamilton's principal function.

If we take a second-order $L$ such that $$L=L(\{q_j,\dot{q}_j,\ddot{q}_j\},t),$$ does the Hamilton-Jacobi equation change, or we could always consider that $\displaystyle\frac{\partial S}{\partial t}=-H$?

  • 2
    $\begingroup$ Which definition of the "Hamiltonian" are you using for higher-order Lagrangians? There are several (e.g. that by Ostrogradsky) that are similar, but not the same. $\endgroup$
    – ACuriousMind
    Sep 30, 2016 at 1:26
  • $\begingroup$ umm if it's true to consider that ∂S/∂t=−H, I would define my Hamiltonian from \delta S (wich is equal to L \delta t), so H will be =∑j p\dot{q}+∑j ∂S/∂\dot{q} \ddot{q}−L (all will j index) $\endgroup$
    – Milou
    Sep 30, 2016 at 1:39
  • $\begingroup$ @ACuriousMind am I allowed to write this? I've never worked with a higher-order L. $\endgroup$
    – Milou
    Sep 30, 2016 at 17:47
  • $\begingroup$ The point is - what space is that new $H$ living on? In the first-order setting, we switch from the Lagrangian tangent bundle with coordinates $(q,\dot{q})$ to the Hamiltonian phase space/cotangent bundle with coordinates $(q,p)$. But here you can't just Legendre transform the $\dot{q}$, you need to somehow handle the $\ddot{q}$ dependence. There are difference schemes for that, leading to different notions of the final Hamiltonian, so it's not exactly clear what you want to do here. $\endgroup$
    – ACuriousMind
    Sep 30, 2016 at 17:55
  • $\begingroup$ What I've written in my comment is absolutely wrong :) I'm sorry! So I've decided to use the Ostogradsky Hamiltonian. But I still don't know if the Hamilton-jacobi equation is still true anyway. $\endgroup$
    – Milou
    Oct 2, 2016 at 16:40

1 Answer 1


I) Let us suppress explicit time dependence $t$ from the notation in the following. Let there be given a second-order Lagrangian

$$L(q,v,a); \tag{1}$$

where $q^i$ are positions, $v^i$ are velocities, $a^i$ are accelerations, and where $i\in\{1,\ldots,n\}$.

II) We would like to find the corresponding Ostrogradsky Hamiltonian formulation. Let us for simplicity assume that the Hessian

$$ H_{ij}~=~\frac{\partial^2L}{\partial a^i\partial a^j} \tag{2}$$

is invertible.$^1$ Then the Ostrogradsky Hamiltonian is defined as

$$ H(Q,P)~:=~ p_iv^i + \sup_a\left(\pi_i a^i-L(q,v,a)\right) ,\tag{3}$$

where we have introduced the collective notation

$$Q^I~=~\{q^i;v^i\},\qquad P_I~=~\{p_i;\pi_i\},\qquad I~\in~\{1,\ldots,2n\}.\tag{4} $$

III) In the spirit of my Phys.SE answer here, we introduce an extended Lagrangian$^2$

$$ L_E(Q,\dot{Q},P,a)~:=~p_i(\dot{q}^i-v^i)+\pi_i(\dot{v}^i-a^i)+L(q,v,a)\tag{5}$$

If we integrate out $P_I$, $v^i$ and $a^i$ in the extended Lagrangian (5), we get back the Lagrangian itself

$$ L(q,\dot{q},\ddot{q})\tag{6} .$$

If we only integrate out $a^i$ in the extended Lagrangian (5), we get the Ostrogradsky Hamiltonian Lagrangian

$$ L_H(Q,\dot{Q},P)~:=~P_I\dot{Q}^I - H(Q,P).\tag{7}$$

This implies that the higher-order Euler-Lagrange (EL) equations of (5) is equivalent to a standard Hamilton's equations in $Q^I$ and $P_I$! In other words, in the non-singular case (2), we can re-use the standard Hamilton-Jacobi (HJ) theory for this case! The only difference is that the phase space (4) is twice as big.


$^1$ If the Hessian matrix is singular, there will appear constraints, and the Hamiltonian formulation and the Hamilton-Jacobi theory become modified as a result.

$^2$ If we vary the extended Lagrangian (5) wrt. to $a^i$ and $v^i$, we get the Ostrogradsky momenta $$ \pi_i~\approx~\frac{\partial L}{\partial a^i} ,\tag{8} $$ and $$ p_i ~\approx~\frac{\partial L}{\partial v^i}- \dot{\pi}_i~\approx~\frac{\partial L}{\partial v^i}- \frac{d}{dt}\frac{\partial L}{\partial a^i} ,\tag{9} $$ respectively. [The $\approx$ sign means equality modulo equations of motion.]


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