Derivation of Hamilton-Jacobi equation

I am trying my own way of deriving the Hamilton Jacobi equation

$$\frac{\partial S}{\partial t} = -H \tag{1}$$

through direct variation. I think the difficulty of doing this is that the upper limit of integral:

$$S = \int_0^t L dt$$

is actually varying. So I try to rewrite the integral in an alternative form:

$$S = \int_0^1 L(q(z),\dot{q}(z)) \frac{\partial t}{\partial z} dz$$.

Here $$z$$ parameterize the 'progress' of the motion from the start to the end, which is alway from 0 to 1. Then:

$$\delta S = \int_0^1 \{(\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial \dot{q}}\delta \dot{q}) \frac{\partial t}{\partial z}+L\frac{\partial \delta t}{\partial z}\} dz$$.

Consider the first two terms first:

$$\int_0^1 (\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial \dot{q}}\delta \dot{q}) \frac{\partial t}{\partial z}dz$$

By the euler lagrange equation (it holds because we are considering a real trajectory) , also replace time derivative by $$\frac{dz}{dt}\frac{d}{dz}$$ the first term could be written as:

$$=\int_0^1 (\frac{dz}{dt}\frac{d}{dz}\frac{\partial L}{\partial \dot{q}}\delta q+\frac{\partial L}{\partial \dot{q}}\frac{dz}{dt}\frac{d}{dz}\delta q) \frac{\partial t}{\partial z}dz$$

$$=\int_0^1 (\frac{d}{dz}\frac{\partial L}{\partial \dot{q}}\delta q+\frac{\partial L}{\partial \dot{q}}\frac{d}{dz}\delta q) dz$$

Integrate the second term by parts:

$$=\int_0^1 (\frac{d}{dz}\frac{\partial L}{\partial \dot{q}}\delta q-\frac{d}{dz}\frac{\partial L}{\partial \dot{q}}\delta q) dz=0$$

The boundary term vanishes because $$\delta q$$ vanish at the start and the end (we are only varying the arrival time). Now consider the second part:

$$\int_0^1 L\frac{\partial \delta t}{\partial z} dz$$.

integration by part again:

$$=L \delta t|^{z=1}_{z=0} - \int_0^1 \frac{\partial }{\partial z} L \delta tdz$$.

$$=L \delta t|^{z=1}_{z=0} - \int_0^1 (\frac{\partial L}{\partial q}\frac{dq}{dz}+\frac{\partial L}{\partial \dot{q}}\frac{d\dot{q}}{dz})\delta t dz$$.

$$=L \delta t|^{z=1}_{z=0} - \int_0^1 (\frac{dz}{dt}\frac{d}{dz}\frac{\partial L}{\partial \dot{q}}\frac{dq}{dz}+\frac{\partial L}{\partial \dot{q}}\frac{d\dot{q}}{dz})\delta t dz$$.

$$=L \delta t|^{z=1}_{z=0} - \int_0^1 (\frac{d}{dz}\frac{\partial L}{\partial \dot{q}}\dot{q}+\frac{\partial L}{\partial \dot{q}}\frac{d\dot{q}}{dz})\delta t dz$$.

$$=L \delta t|^{z=1}_{z=0} - \int_0^1 (\frac{d}{dz}(\frac{\partial L}{\partial \dot{q}}\dot{q})\delta t dz$$.

$$=L \delta t|^{z=1}_{z=0} - (\frac{\partial L}{\partial \dot{q}}\dot{q})\delta t |_{z=0}^{z=1}+\int_0^1 \frac{\partial L}{\partial \dot{q}}\dot{q}\frac{d}{dz}\delta t dz$$.

The first two terms is just what I need ($$-H\delta t|_{z=0}^{z=1}$$). However, the last term also shows up, which does not seem to be zero.

Is this approach to derive H-J equation viable? If not, where did I make the mistake?

2 Answers

The place that I made mistake is here:

$$\delta \dot{q} = \delta (\frac{d}{dt}q)$$

$$= \delta (\frac{dz}{dt}\frac{d}{dz}q)$$

Here what I did is to simply took $$\frac{dz}{dt}$$ outside $$\delta$$. What need to be done is:

$$\delta \dot{q} = \delta (\frac{dz}{dt}\frac{d}{dz}q)$$

$$= \delta (\frac{dz}{dt})\frac{d}{dz}q+\frac{dz}{dt}\frac{d\delta q}{dz}$$

Here I give the correct version of the derivation:

$$S = \int_0^1 L(q(z),\dot{q}(z)) \frac{\partial t}{\partial z} dz$$.

$$\delta S = \int_0^1 \{(\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial \dot{q}}\delta \dot{q}) \frac{\partial t}{\partial z}+L\frac{\partial \delta t}{\partial z}\} dz$$.

Consider the first two terms first:

$$\int_0^1 (\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial \dot{q}}\delta \dot{q}) \frac{\partial t}{\partial z}dz$$

By the euler lagrange equation (it holds because we are considering a real trajectory) , also replace time derivative by $$\frac{dz}{dt}\frac{d}{dz}$$ the first term could be written as:

$$=\int_0^1 (\frac{dz}{dt}\frac{d}{dz}\frac{\partial L}{\partial \dot{q}}\delta q+\frac{\partial L}{\partial \dot{q}}\delta (\frac{dz}{dt})\frac{d}{dz}q+\frac{\partial L}{\partial \dot{q}}\frac{dz}{dt}\frac{d\delta q}{dz}) \frac{\partial t}{\partial z}dz$$

$$=\int_0^1 (\frac{dz}{dt}\frac{d}{dz}\frac{\partial L}{\partial \dot{q}}\delta q-\frac{\partial L}{\partial \dot{q}}(\frac{dz}{dt})^2\frac{\partial \delta t}{\partial z}\frac{d}{dz}q+\frac{\partial L}{\partial \dot{q}}\frac{dz}{dt}\frac{d\delta q}{dz}) \frac{\partial t}{\partial z}dz$$

$$=\int_0^1 (\frac{d}{dz}\frac{\partial L}{\partial \dot{q}}\delta q+\frac{\partial L}{\partial \dot{q}}\frac{d}{dz}\delta q-\frac{\partial L}{\partial \dot{q}}\dot{q}\frac{\partial \delta t}{\partial z}) dz$$

Integrate the second term by parts:

$$=\int_0^1 (\frac{d}{dz}\frac{\partial L}{\partial \dot{q}}\delta q-\frac{d}{dz}\frac{\partial L}{\partial \dot{q}}\delta q)-\frac{\partial L}{\partial \dot{q}}\dot{q}\frac{\partial \delta t}{\partial z}) dz$$

$$=-\int_0^1 \frac{\partial L}{\partial \dot{q}}\dot{q}\frac{\partial \delta t}{\partial z} dz$$

The boundary term vanishes because $$\delta q$$ vanish at the start and the end (we are only varying the arrival time). Now consider the second part:

$$\int_0^1 L\frac{\partial \delta t}{\partial z} dz$$.

integration by part again:

$$=L \delta t|^{z=1}_{z=0} - \int_0^1 \frac{\partial }{\partial z} L \delta tdz$$.

$$=L \delta t|^{z=1}_{z=0} - \int_0^1 (\frac{\partial L}{\partial q}\frac{dq}{dz}+\frac{\partial L}{\partial \dot{q}}\frac{d\dot{q}}{dz})\delta t dz$$.

$$=L \delta t|^{z=1}_{z=0} - \int_0^1 (\frac{dz}{dt}\frac{d}{dz}\frac{\partial L}{\partial \dot{q}}\frac{dq}{dz}+\frac{\partial L}{\partial \dot{q}}\frac{d\dot{q}}{dz})\delta t dz$$.

$$=L \delta t|^{z=1}_{z=0} - \int_0^1 (\frac{d}{dz}\frac{\partial L}{\partial \dot{q}}\dot{q}+\frac{\partial L}{\partial \dot{q}}\frac{d\dot{q}}{dz})\delta t dz$$.

$$=L \delta t|^{z=1}_{z=0} - \int_0^1 (\frac{d}{dz}(\frac{\partial L}{\partial \dot{q}}\dot{q})\delta t dz$$.

$$=L \delta t|^{z=1}_{z=0} - (\frac{\partial L}{\partial \dot{q}}\dot{q})\delta t |_{z=0}^{z=1}+\int_0^1 \frac{\partial L}{\partial \dot{q}}\dot{q}\frac{d}{dz}\delta t dz$$.

Add the up first and the second part:

$$\delta S = (L - (\frac{\partial L}{\partial \dot{q}}\dot{q}))\delta t |_{z=0}^{z=1}$$

$$\delta S = -H\delta t |_{z=0}^{z=1}$$

The R.H.S only depend on the arrival time difference. Therefore

$$\frac{\partial S}{\partial t} = -H$$

OP's eq. (1) superficially looks like the Hamilton-Jacobi (HJ) equation, but the devil is in the detail. While the HJ equation is a non-linear first-order PDE for Hamilton's principal function, it seems OP is actually talking about a property $$\frac{\partial S(q_f,t_f;q_i,t_i)}{\partial t_f}~=~-h_f \tag{12}$$ of the (Dirichlet) on-shell action $$S(q_f,t_f;q_i,t_i)$$. See eq. (12) in my Phys.SE answer here, where a proof is provided.

See also this related Phys.SE post.