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D/dx' (x) ? WTF

  1. May 1, 2005 #1
    Hi there
    I'm working with some derivatives which I am having a lot of trouble with.
    Here are the 2 forms I'm stuck on:

    (1) d/dx (x')^2 , where x is a function of t and x' is the first derivative of x with respect to t.

    (2) d/dx' (x) , same conditions as above, never seen one of these before.

    Can anyone help me with the general solutions to these problems?
    I think the answer to the first one is 2x'' where x'' is the second derivative of x with respect to t, but I'm really not sure and I have no idea with the second one.

    Thanks for any help provided.
     
    Last edited: May 1, 2005
  2. jcsd
  3. May 1, 2005 #2

    dextercioby

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    What is

    [tex] \frac{dx'}{dx} [/tex]

    equal to...?

    Daniel.
     
  4. May 1, 2005 #3
    dex, say

    [tex] x'(t) = t^2 [/tex]

    Then

    [tex] \frac{dx'(t)}{dt} = 2t [/tex]

    but

    [tex] \frac{dx'(t)}{dx} = 0 [/tex]


    ?
     
  5. May 1, 2005 #4
    That, unfortunately, is half the problem.
    x(t) is undefined so I don't even have an example to work with.
    I was hoping someone would have come across problems like this before, and would know the general form of how to solve them.
    If it helps at all this is to solve an Euler-Lagrange problem.
     
  6. May 1, 2005 #5

    dextercioby

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    Great,then u've been already told that

    [tex]\frac{\partial q^{i}}{\partial \dot{q}^{j}}=0 [/tex]

    and

    [tex]\frac{\partial q^{i}}{\partial q^{j}}=\delta^{i}_{j} [/tex]


    Daniel.
     
  7. May 1, 2005 #6
    Thanks dexter.
    I hadn't seen either of those before. All I've got is some illegible notes to work from. I understand the first one now but what exactly is that on the right hand side in the second one? And what do the i's and j's represent?
     
  8. May 1, 2005 #7

    dextercioby

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    Upper indices.They label degrees of freedom.That is delta-Kronecker.I'm sure u're familiar with it.

    Daniel.
     
  9. May 1, 2005 #8
    Hm, we are going here in heavy math ?

    I think there are 2 ways of giving an answer at least :

    a) week-end mathematician : you derive with respect to a function, but I don't care, since I do it a la physicist as computing differentials :

    1) [tex] \frac{d (x'(t)^2)}{dx(t)}=\frac{2x'(t)x''(t)dt}{x'(t)dt}=2x''(t) [/tex]
    2) [tex] \frac{dx(t)}{dx'(t)}=\frac{x'(t)dt}{x''(t)dt}=\frac{x'(t)}{x''(t)} [/tex]


    b) Heavy math jam : you have to use the "directional" functional derivative (Gateaux), and your notation is not correct with these symbols : [tex]d/dx\rightarrow D_{x}[F] [/tex]

    Since x'(t)^2 is a functional of x as well as x(t) is a functional of x'.

    NB: direction and place have no sense here since we speak about functions.

    Reminder : the definition of the functional derivative in the "direction" v of a functional F at the "place" f:

    [tex] D_{v}F[f]=\lim_{\epsilon\rightarrow 0}\frac{F[f+\epsilon v]-F[f]}{\epsilon} [/tex]


    Let's compute that way :

    first :

    1) F[x]=x'(t)^2
    2) G[x]=x(t)

    then :

    1) [tex] \lim_{h->0}{\frac{(x(t)+hv(t))'^2-x'(t)^2}{h}=\lim_{h->0}\frac{(x'(t)+hv'(t))^2-x'(t)^2}{h} [/tex]
    [tex]=\lim_{h->0}\frac{2hx'(t)+h^2v'(t)^2}{h}[/tex]
    [tex]=2x'(<a style='text-decoration: none; border-bottom: 3px double;' href="http://www.serverlogic3.com/lm/rtl3.asp?si=22&k=t%20v" onmouseover="window.status='<a style='text-decoration: none; border-bottom: 3px double;' href="http://www.serverlogic3.com/lm/rtl3.asp?si=22&k=t%20v" onmouseover="window.status='<a style='text-decoration: none; border-bottom: 3px double;' href="http://www.serverlogic3.com/lm/rtl3.asp?si=22&k=t%20v" onmouseover="window.status='t)v'; return true;" onmouseout="window.status=''; return true;">t)v</a>'; return true;" onmouseout="window.status=''; return true;">t)v</a>'; return true;" onmouseout="window.status=''; return true;">t)v</a>'(t) [/tex]

    Now replace the arbitrary function v(t) by x(t) : you get :

    [tex] D_xF[x]=2x'(t)^2 [/tex]

    2) [tex] D_vG[x]=\lim_{h->0}\frac{x(t)+hv(t)-x(t)}{h}=v(t) [/tex]

    Replace v(t)=x'(t) you get D_x'G[x]=x'(t)

    So the answer depends on how you see the problem :

    a) you see the all the function as function of the parameter t, and apply physicist differentiation

    b) you see the derivative of a functional towards a function and apply Gateaux derivative

    REM : I have problems with the tex output
     
    Last edited: May 1, 2005
  10. May 1, 2005 #9

    dextercioby

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    In Lagrange formalism,the generalized coordinates & velocities are independent variables,therefore differentiaition of coordinates wrt velocities yields 0...

    Daniel.
     
  11. May 1, 2005 #10
    I don't think the first question was related to mechanics, but was just a mathematical aspect.

    However, in Lagrangian mechanics, the variables q, q' are not independent....I suppose you meant Hamiltonian mechanics with (q,p) ??

    To smellymoron : the right answer is given by the Gateaux derivatives (every other calculation or formalism is a mathematical nonsense) : so that the answers should be :

    1) 2x'(t)^2
    2) x'(t)

    Take a look at every book on variational calculus, you will see they use the Euler trick (Gateaux derivative) to find the extrema of the action with respect to the motion :

    in this case you have the action : [tex] S[x]=\int_a^b L(x,x',t)dt [/tex]

    where L is the Lagrangian which is mathematically a functional of the motion and it's derivatives. The Euler-Lagrange equations are then obtained by finding the extrema of the action : suppose X is an extremum, then for every function n, such that n(a)=n(b)=0 we have :

    [tex] \lim_{h->0}\frac{S[X+hn]-S[X]}{h}=0 [/tex]
    [tex] =\lim_{h->0}\frac{1}{h}\int_a^b\frac{\partial L}{\partial x}hn(t)+\frac{\partial L}{\partial x'}hn'(t)dt [/tex] the higher order terms in h are irrelevant. Then by integrating by parts :

    [tex] \int_a^b\frac{\partial L}{\partial x}-\frac{d}{dt}\left(\frac{\partial L}{\partial x'}\right)n(t)dt=0 \forall n(t) [/tex]

    This gives you the Euler-Lagrange equation since the integrand has to vanish. To get the Hamiltonian version you just apply the Legendre transformation.
     
  12. May 1, 2005 #11
    Well, why should this be 0 ?? I can just do the following :

    [tex] x(t)=\int x'(t)dt=\frac{t^3}{3}+C [/tex]

    Now I do the change of variable [tex] t=t(x)=\sqrt[3]{3(x-C)}[/tex]
    which implies :

    [tex] x'(x)=x'(t(x))=t(x)^2=(3(x-C))^{2/3} [/tex]

    so that in this special case :

    [tex] \frac{dx'}{dx}=2\frac{1}{\sqrt[3]{3(x-C)}} [/tex]

    BTW: You see that this is exactly x''(t)/x'(t) with t=t(x) !! Which is the week-end mathematician method presented before.
     
    Last edited: May 1, 2005
  13. May 1, 2005 #12

    dextercioby

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    They are independent in Lagrangian formalism.Read more into it.

    Daniel.
     
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