# Solving functions for S in a q-q* Hamilton-Jacobi diffeq

• dynamicskillingme
In summary, the student is trying to solve for the Hamiltonian and Hamiltonian jacobi, but is having difficulty because of a mistake he made early on. He is hoping to use the usual type of reasoning to separate variables and then solve for the functions. However, he needs to simplify the equation first.

## The Attempt at a Solution

So far I have a solution for a) as

For b) I formulate the equation as

and so far for c) I have

My main idea at the moment is that as the Lagrangian was not time dependent, the Hamiltonian will not be. Following on maybe I can say that when separating variables T is constant so I can separate it so
but this is as far as I could get, as I couldn't think of how to separate them.

Any ideas, or are my problems due to a mistake in working?

dynamicskillingme said:

## The Attempt at a Solution

So far I have a solution for a) as
The last three terms do not have the correct dimesions. I believe you have dropped some factors of ##2## and ##m##. So, check your derivation.

Alternately, you might see if you can first take the Lagrangian as given in terms of a general vector potential ##\bf{A}## and show that the Hamiltonian can be written nicely as ##H = \frac{1}{2m} (\mathbf{p}-e \mathbf{A}) \cdot (\mathbf{p}-e \mathbf{A})##. Then substitute the given expression for ##\mathbf{A}##.
and so far for c) I have
You probably didn't mean to have an equal sign after the ##T(t, \overline{x}, \overline{y})##

My main idea at the moment is that as the Lagrangian was not time dependent, the Hamiltonian will not be. Following on maybe I can say that when separating variables T is constant so I can separate it so
but this is as far as I could get, as I couldn't think of how to separate them.
You cannot conclude that ##T## is a constant. But you should be able to argue that ##\frac{\partial T}{\partial t}## is a constant.
Also, you should hope that ##X## and ##Y## are not identically equal to zero. If they were, there would be no hope of finding the correct equations for ##x(t)## and ##y(t)##.

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My Hamilton was a mess from a mistake early on. After redoing it I get
.

I can then formulate the hamiltonian jacobi again and get

Then (and I'm mostly going from an example in my lecture notes and the question hints) I can separate the variables out as

That can be substituted as

In my notes the example only contains two separate variables e.g. X and T. He then makes every non time dependent term equal the time dependent terms and says they are both equal to a constant. He then solves X and T wrt this constant.

I'm not sure how to take this question further and solve for these functions.

dynamicskillingme said:
My Hamilton was a mess from a mistake early on. After redoing it I get
.
The numerical factor of -3 is not correct for the ##(eBx)^2## term. Also, I believe the last term has the wrong sign.
Then (and I'm mostly going from an example in my lecture notes and the question hints) I can separate the variables out as
Yes
That can be substituted as

In my notes the example only contains two separate variables e.g. X and T. He then makes every non time dependent term equal the time dependent terms and says they are both equal to a constant. He then solves X and T wrt this constant.

I'm not sure how to take this question further and solve for these functions.

Here you want to use the usual type of reasoning when separating variables.
Suppose you have an equation of the form ##f(x) + g(y)## = 0 that must hold for all values of ##x## and ##y## within some domain. Here, ##f(x)## is a function only of ##x## and ##g(y)## is a function only of ##y##. Then you can argue that ##f(x)## must be a constant: ##f(x) = C## for all ##x##. Moreover, ##g(y)## must also be a constant which is the negative of ##C##; i.e., ##g(y) = -C##. Try to construct such an argument.

For your H-J equation, you have something of the form ##f(x,y) + \frac{dT(t)}{dt} = 0## that must hold for a range of values of ##x, y##, and ##t##. Use the type of argument mentioned above to deduce something about ##\frac{dT}{dt}##.

Okay, so I must have assumed something wrong early on or have made a pretty bad mistake in my work (not unheard of)

dynamicskillingme said:
Okay, so I must have assumed something wrong early on or have made a pretty bad mistake in my work (not unheard of)
So far so good. Just a matter of simplifying.

The Hamiltonian is the Legendre transformation of the Lagrangian:
## H = \sum_{i} \dot q_i \frac {\partial L} {\partial \dot q_i} - L = m(\dot x^2 + \dot y^2) + eBx\dot y - \frac {m} {2}(\dot x^2 + \dot y^2)- eBx\dot y##
## = \frac {1} {2m} (p_x^2 + p_y^2)##
In this problem you need to find constants of motion. If you solve for the Lagrangian eqns. of motion, you will find:
## m\ddot x -eB\dot y = 0 →\frac {d} {dt}(m\dot x -eBy) = 0## and
## m\ddot y +eB\dot x = 0→\frac {d} {dt}(m\dot y +eBx) = 0##
Thus we have the constants of motion:
##\tilde p_x =(p_x -eBy) ##,
##\tilde p_y =(p_y +eBx)##.
We write the HJ equation with ## T(t,\tilde x,\tilde y) = \tilde xt## and substitute ## p_y = \tilde p_y - eBx## to get:
## (\frac {dX} {dx})^2 + ( \tilde p_y - eBx)^2 = 2m \tilde x##,
## X= \int dx\sqrt { 2m \tilde x- ( \tilde p_y - eBx)^2}##
## S= \int dx\sqrt { 2m \tilde x- ( \tilde p_y - eBx)^2} - \tilde xt##
##p_x = \frac {\partial S} {\partial x} = \sqrt { 2m \tilde x- ( \tilde p_y - eBx)^2} ##
##\tilde p_x = -\frac {\partial S} {\partial \tilde x} = -m \int dx\frac {1} { \sqrt { 2m \tilde x- ( \tilde p_y - eBx)^2}} +t##
We now make the substitution in the integral:
## x' =\tilde p_y - eBx ##,
## dx' = -eBdx ##
##\tilde p_x = -\frac {\partial S} {\partial \tilde x} = \frac {m} {eB} \int \frac {dx'} { \sqrt { 2m \tilde x- x'^2}} +t = \frac {m} {eB}sin^{-1}(\frac{\tilde p_y - eBx} { 2m \tilde x}) + t##
You can do the algebra to solve for x in terms of ##\tilde p_x ,\tilde p_y## and ##\tilde x##.
To get y just use the constant of motion equation ## y = \frac {m\dot x - \tilde p_x} {eB}##

## 1. What is a q-q* Hamilton-Jacobi diffeq?

A q-q* Hamilton-Jacobi diffeq is a type of differential equation used to solve for the function S. It is commonly used in physics and mathematics to describe the evolution of a system over time.

## 2. How do you solve for S in a q-q* Hamilton-Jacobi diffeq?

To solve for S in a q-q* Hamilton-Jacobi diffeq, you can use the Hamilton-Jacobi equation, which is a partial differential equation that describes the dynamics of the system. This equation can be solved using methods such as separation of variables, characteristic curves, or the method of characteristics.

## 3. What are the applications of solving functions for S in a q-q* Hamilton-Jacobi diffeq?

Solving functions for S in a q-q* Hamilton-Jacobi diffeq has many applications, especially in physics and engineering. It can be used to describe the motion of particles in a system, the behavior of waves, and the evolution of quantum systems. It is also used in optimal control theory, which is used to find the most efficient way to control a system.

## 4. What are the challenges in solving functions for S in a q-q* Hamilton-Jacobi diffeq?

One of the main challenges in solving functions for S in a q-q* Hamilton-Jacobi diffeq is that it is a nonlinear partial differential equation, which can be difficult to solve analytically. It often requires advanced mathematical techniques and computer simulations to find solutions. Additionally, the specific form of the Hamilton-Jacobi equation will vary depending on the system being studied, which adds to the complexity of solving for S.

## 5. What are some resources for learning how to solve functions for S in a q-q* Hamilton-Jacobi diffeq?

There are many resources available for learning how to solve functions for S in a q-q* Hamilton-Jacobi diffeq. Some helpful resources include textbooks, online courses, and academic papers. Additionally, consulting with experts in the field or joining a study group can also be beneficial. It is important to have a strong foundation in calculus and differential equations before attempting to solve a q-q* Hamilton-Jacobi diffeq.