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Precursor to brachistochrone problem

  1. Jan 31, 2006 #1
    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 [tex] \delta \int_{x_{1}}^{x_{2}} F(x,y(x),\dot{y}(x)) dx = 0 [/tex]
    where [tex] \dot{y} = \frac{dy}{dx} [/tex]
    Show that if [itex] F = F(y,\dot{y}) [/itex] only (not dependant of x explicitly) then the quantity
    [tex] H = \dot{y} \frac{\partial F}{\partial \dot{y}} - F [/tex]
    is a constant. That is [itex] \frac{dH}{dx} = 0 [/itex].

    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
    is this true >> [tex] \frac{d \dot{y}}{\partial \dot{y}} =1 [/tex] ??
    in that case all we're left with (if the dH/dx is correct) is
    [tex] \frac{dH}{dx} = \dot{y} \frac{\partial F}{\partial \dot{y} dx} [/tex]
    if i were to differentiate the expression for F
    [tex] \delta \int_{x_{1}}^{x_{2}} F(x,y(x),\dot{y}(x)) dx = 0 [/tex]
    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
    [tex] \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}} [/tex]

    then [tex] 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) [/tex]

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

    Please help! Thank you!
  2. jcsd
  3. Jan 31, 2006 #2


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    1. You are asked for the total derivative and you can show that it will reduce to the partial derivative (the two extra terms : [itex]\dot{y}\partial H/\partial y} + \ddot{y}\partial H/\partial{\dot{y} [/itex] 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, [tex] H = \dot{x} \frac{\partial F}{\partial \dot{x}} - F [/tex]

    When you do it correctly, the pysical significance will pop right out !
  4. Jan 31, 2006 #3
    where did you get [itex]\dot{y}\partial H/\partial y} + \ddot{y}\partial H/\partial{\dot{y} [/itex] from?

    also domnt u mena F(y,dy/dt)? Or do you actually mean it for x?
    The former is in the question...
  5. Feb 1, 2006 #4


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

    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 :

    [tex]F(x,\dot{x}) = \frac{1}{2}m \dot{x}^2 + V(x) [/tex]

    where [itex]\dot{x} = dx/dt[/itex]
    Last edited: Feb 1, 2006
  6. Feb 2, 2006 #5


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    That H is a constant ON the equations of motion !

    That is

    [tex] \frac{dH}{dx}= - \frac{dy}{dx}\frac{\delta F}{\delta y} [/tex]

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

  7. Feb 2, 2006 #6
    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 [tex] F = \frac{1}{2} m \dot{x}^2 + V(x) [/tex]
    [tex] \frac{\partial F}{\partial \dot{x}} = m \dot{x} + \frac{\partial V}{\partail \dot{x}} \frac{\partial x}{\partial \dot{x}} [/tex] ...(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?
    [tex] F(x,\dot{x}) = - \nabla V(x) [/tex] ?
    so then when i sub 1 into the expression for H
    [tex] H = \frac{1}{2} m \dot{x}^2 - V(x) [/tex] 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?
  8. Feb 2, 2006 #7


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    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
  9. Feb 2, 2006 #8
    Correction to the mistake

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

    Thus H represents the total energy of the system.
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