# Precursor to brachistochrone problem

1. Jan 31, 2006

### stunner5000pt

Another long question but not that hard. Most of the writing is my work/questions
According to my prof if i cna solve this... the resulting relation can be used to solve Bernoulli's problem

For $$\delta \int_{x_{1}}^{x_{2}} F(x,y(x),\dot{y}(x)) dx = 0$$
where $$\dot{y} = \frac{dy}{dx}$$
Show that if $F = F(y,\dot{y})$ only (not dependant of x explicitly) then the quantity
$$H = \dot{y} \frac{\partial F}{\partial \dot{y}} - F$$
is a constant. That is $\frac{dH}{dx} = 0$.

is this the total derivative (like it asks) of H wrt x? Since F does depend on y and y dot... which depend on x... then would hteir derivatives wrt x be included in the rightmost expression below?
$$\frac{dH}{dx} = \frac{d \dot{y}}{dx} \frac{\partial F}{\partial \dot{y}} + \dot{y} \frac{\partial F}{\partial \dot{y} dx} - \frac{dF}{dx} [/tex is this true >> [tex] \frac{d \dot{y}}{\partial \dot{y}} =1$$ ??
in that case all we're left with (if the dH/dx is correct) is
$$\frac{dH}{dx} = \dot{y} \frac{\partial F}{\partial \dot{y} dx}$$
if i were to differentiate the expression for F
$$\delta \int_{x_{1}}^{x_{2}} F(x,y(x),\dot{y}(x)) dx = 0$$
wrt to y dot and x i would get zero, yes?
SO dH/dx = 0?

Determine the expression for H if F = 1/2 mv^2 - V(x) and explain the physical significance of this quantity.
is this correct? chain rule application everywhere
$$\frac{\partial F}{\partial \dot{y}} = m \dot{x} \frac{d \dot{x}}{d \dot{y}} - \frac{\partial V}{\partial \dot{y}} \frac{\partial x}{\partial \dot{y}}$$

then $$H = \dot{y} m \dot{x} \frac{d \dot{x}}{d \dot{y}} - \frac{\partial V}{\partial \dot{y}} \frac{\partial x}{\partial \dot{y}} - \frac{1}{2} m \dot{x}^2 + V(x)$$

the physical significance... hmm ...
H is a consnat quantity wrt x. But im not sure how to interpret this past that part.

2. Jan 31, 2006

### Gokul43201

Staff Emeritus
1. You are asked for the total derivative and you can show that it will reduce to the partial derivative (the two extra terms : $\dot{y}\partial H/\partial y} + \ddot{y}\partial H/\partial{\dot{y}$ will cancel off) which is clearly = 0.

2. You've started off on the wrong foot...

First notice that F = F(x, dx/dt), since v=dx/dt

So, $$H = \dot{x} \frac{\partial F}{\partial \dot{x}} - F$$

When you do it correctly, the pysical significance will pop right out !

3. Jan 31, 2006

### stunner5000pt

where did you get $\dot{y}\partial H/\partial y} + \ddot{y}\partial H/\partial{\dot{y}$ from?

also domnt u mena F(y,dy/dt)? Or do you actually mean it for x?
The former is in the question...

4. Feb 1, 2006

### Gokul43201

Staff Emeritus
$$dH/dx = \partial H/\partial x + (\partial H/\partial y)dy/dx + (\partial H/\partial\dot{y}) d\dot{y}/dx$$

No, it's the latter which is in the question. What you are given in part 1 is the general form in terms of some general parameters. In part 2, you are given :

$$F(x,\dot{x}) = \frac{1}{2}m \dot{x}^2 + V(x)$$

where $\dot{x} = dx/dt$

Last edited: Feb 1, 2006
5. Feb 2, 2006

### dextercioby

That H is a constant ON the equations of motion !

That is

$$\frac{dH}{dx}= - \frac{dy}{dx}\frac{\delta F}{\delta y}$$

,where $\frac{\delta F}{\delta y}$ is the Euler-Lagrange (variational) derivative of the lagrangian.

Daniel.

6. Feb 2, 2006

### stunner5000pt

also in this question... what is the partial derivative of F wrt x zero? Is it because the action is zero?

for hte second one since $$F = \frac{1}{2} m \dot{x}^2 + V(x)$$
then
$$\frac{\partial F}{\partial \dot{x}} = m \dot{x} + \frac{\partial V}{\partail \dot{x}} \frac{\partial x}{\partial \dot{x}}$$ ...(1 )

so far so good?
im just wondering if the partial of V wrt to x dot is zero becuase V would not depend on x dot because of its definition?
$$F(x,\dot{x}) = - \nabla V(x)$$ ?
so then when i sub 1 into the expression for H
$$H = \frac{1}{2} m \dot{x}^2 - V(x)$$ this is given that the partial of V wrt x dot is zero. THis means that H is the force exerted on the particle? Doesnt this mean that H is invariant of F, for a potential that does not depend on the time deriavitve of x?

7. Feb 2, 2006

### Gokul43201

Staff Emeritus
Correct. But there's a sign mistake in the solution. The sign of V in F(x,v) has been changed (looks like I made that mistake in my previous post - sorry !)

PS : No time to look at the rest now.

Last edited: Feb 2, 2006
8. Feb 2, 2006

### stunner5000pt

Correction to the mistake

so the expression for H (the hamiltonian, apparently)
$$H = \frac{1}{2} m \dot{x}^2 + V(x) = T + V = E$$

Thus H represents the total energy of the system.