Geometry of Hamilton-Jacobi Equation

Question

I'm trying to understand the geometry of the Hamilton-Jacobi equation (working from Gelfand + Fomin), but I'm stuck. I know that:

If we define the function $S(t,y;t_0, y_0)$ as:

$$S(t,y;t_0,y_0) = \int_{t_0}^{t_1} L(t,y_{cl},y_{cl}') dt$$

where $y_{cl}$ is the unique (by assumption) extremal joining $(t_0, y_0)$ and $(t,y)$ then, holding $(t_0, y_0)$ fixed, $S(t,y)$ will satisfy the Hamilton-Jacobi equation:

$$\frac{\partial S}{\partial t} + H(t,y,\nabla_y S)=0.$$

I'm happy working in this direction: define $S(t,y,t_0, y_0)$, then observe it obeys a PDE.

However, I'm a little confused going the other way - from the PDE to the complete solution.

We know from above that all of the functions

$$S(t,y;t_0,y_0) + S_0$$

for any fixed $t_0,y_0$ and $S_0$ will be solutions of the Hamilton-Jacobi equation (we can add $S_0$ because the Hamilton-Jacobi equation doesn't depend on $S$ explicitly.) Thus, if we imagine $y = (y_1,y_2,\ldots,y_n)$, the possible solutions will be a family with at least $n+2$ independent parameters: $$S(t,y;t_0,y_0,S_0) = S(t,y;\alpha_1,\alpha_2,\ldots,\alpha_{n+2})$$

However, the H-J equation is a First Order PDE in $n+1$ variables, and so it's complete integral depends on only $n+1$ parameters:

$$S = S(t,y;\alpha_1,\alpha_2,\ldots,\alpha_{n+1})$$

It seems the complete integral doesn't contain enough parameters to cover the solutions I know about, let alone any others!

My questions are:

1. What am I misunderstanding about the H-J equations/complete integrals?

2. Are all solutions of the H-J equations of the form: $$S(t,y;t_0,y_0) + S_0$$ i.e., does each solution correspond to some point $(t_0,y_0)$ in phase space, with $S$ giving us the minimal action between that point and every other in phase space (at least up to an additive constant)?

Answer 1

Spot the error in this argument:

Consider this simple ODE:

$$\frac{dy}{dt} = 2$$

Because it's a first order equation in one independent variable, the families of solutions will be completely parameterised by one parameter, which we'll call $t_0$. Thus we can say the complete solution is:

$$y = y(t;t_0)$$

However, this ODE doesn't explicitly depend on $y$, so any solution $y(t)$ will give a solution $y(t) + y_0$. Thus it seems the most general solution would be:

$$y(t;t_0) + y_0 = y(t;t_0,y_0)$$

But this depends on 2 parameters, and we started with a 1st order ODE. What's happening?

It's obvious in this case - the general solution is:

$$y = 2(t-t_0)$$

and so adding the constant:

$$y = 2(t-t_0) + y_0$$

is superfluous. While $t_0$ and $y_0$ are independent, each pair $(t_0,y_0)$ doesn't give a unique function $y(t)$, so the solution doesn't intrinsically depend on two parameters, only one.

But this is just the error made above. Any complete solution of the Hamilton-Jacobi PDE will depend on $n+1$ parameters, and we know, by the form of the PDE, that any solution $S(t,y)$ will give a family of solutions $S(t,y) + S_0$. But this doesn't introduce a new independent parameter, it just tells us that one of the $n+1$ parameters we know exist must be this simple, additive one.

Regarding the second question, the answer is yes (correcting for the misleading extra parameter) - if $y_{cl}$ is an extremal between two points $(t_0,y_0)$ and $(t_1,y_1)$ then any solution $S$ of the Hamilton-Jacobi equation will satisfy

$$S(t_1,y_1) - S(t_0,y_0) = \int_{t_0}^{t_1}L(t,y_{cl}, y'_{cl}) dt$$

and in particular, if the $n+1$ parameters are chosen so that

$$S(t_0,y_0) = 0$$

then $S$ simply measures the minimal action from $(t_0,y_0)$ to the rest of space. Specifying the $n+1$ parameters effectively specifies the 'base point' in phase space, which we're measuring the minimal action from (though for some choices of the parameters it may be the case that $S(t,y)\neq 0$ everywhere on the domain - ultimately this discussion is only accurate up to an additive constant.)

The formula above is derived at the end of this answer:

Hamilton-Jacobi Equation