# Hamilton-Jacobi equation with a second-order Lagrangian

The Hamilton-Jacobi equation is: $$$$H\left(q,\frac{\partial S}{\partial q},t\right)+\frac{\partial S}{\partial t}~=~0,$$$$ where $$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 $$\displaystyle\frac{\partial S}{\partial t}=-H~?$$

• 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. Commented Sep 30, 2016 at 1:26
• 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) Commented Sep 30, 2016 at 1:39
• @ACuriousMind am I allowed to write this? I've never worked with a higher-order L. Commented Sep 30, 2016 at 17:47
• 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. Commented Sep 30, 2016 at 17:55
• 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. Commented Oct 2, 2016 at 16:40

1. 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 $$i\in\{1,\ldots,n\}$$.
2. 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 $$Q^I~=~\{q^i;v^i\},\qquad P_I~=~\{p_i;\pi_i\},\qquad I~\in~\{1,\ldots,2n\}.\tag{4}$$
3. 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.]