Precursor to brachistochrone problem

[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 dependent 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?
[tex]\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<br /> is this true >> $$\frac{d \dot{y}}{\partial \dot{y}} =1$$ ??<br /> in that case all we're left with (if the dH/dx is correct) is <br /> $$\frac{dH}{dx} = \dot{y} \frac{\partial F}{\partial \dot{y} dx}$$<br /> if i were to differentiate the expression for F <br /> $$\delta \int_{x_{1}}^{x_{2}} F(x,y(x),\dot{y}(x)) dx = 0$$<br /> wrt to y dot and x i would get zero, yes? <br /> SO dH/dx = 0?<br /> <br /> <br /> <b>Determine the expression for H if F = 1/2 mv^2 - V(x) and explain the physical significance of this quantity.</b><br /> is this correct? chain rule application everywhere<br /> $$\frac{\partial F}{\partial \dot{y}} = m \dot{x} \frac{d \dot{x}}{d \dot{y}} - \frac{\partial V}{\partial \dot{y}} \frac{\partial x}{\partial \dot{y}}$$<br /> <br /> then $$H = \dot{y} m \dot{x} \frac{d \dot{x}}{d \dot{y}} - \frac{\partial V}{\partial \dot{y}} \frac{\partial x}{\partial \dot{y}} - \frac{1}{2} m \dot{x}^2 + V(x)$$<br /> <br /> the physical significance... hmm ...<br /> H is a consnat quantity wrt x. But I am not sure how to interpret this past that part. <br /> <br /> Please help! Thank you![/tex]

 

[Gokul43201]

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 physical significance will pop right out !

 

[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...

 

[Gokul43201]

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

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

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

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 $\dot{x} = dx/dt$

 

[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.

 

[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 because 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?

 

[Gokul43201]

im just wondering if the partial of V wrt to x dot is zero because V would not depend on x dot because of its definition?

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.

 

[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.