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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?

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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.

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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

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

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