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Thread: Hamiltonian mechanics, finding extremals through solving S

  1. #1
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    Hamiltonian mechanics, finding extremals through solving S

    given : Consider $\displaystyle L(x,\dot x) = \dot x ^2 -x^2$
    Show that the extremals are given by $\displaystyle x=\sqrt{\alpha}sin(t+\beta)$



    Now i did a previous example as follows:Given $\displaystyle L(x,\dot x) = \dot x ^2$

    $\displaystyle p=\frac{\partial L}{\partial \dot x} = 2\dot x $ Long story short, hamiltonian is
    $\displaystyle H = \frac{p^2}{4}$ Which yields the Hamilton Jacobi $\displaystyle \frac{1}{4}( \frac{\partial S}{\partial x})^2 + \frac{\partial S}{\partial t}=0$

    Now given the hint to use S of Form S=u(t) + v(x), hence $\displaystyle \frac{\partial S}{\partial t} = \frac{du}{dt} = -\frac{1}{4}( \frac{dv}{dx})^2 = K$
    K a constant,i assume since no t in expression. From here, we let k= $\displaystyle -\alpha ^2$ and solve for t by direct integration, i.e $\displaystyle \frac{du}{dt} = - \alpha ^2$ and we do the same for $\displaystyle \frac{dv}{dx}$ and finally we get $\displaystyle S=-\alpha ^2 t+2\alpha x + \beta$ Where $\displaystyle \beta$ constant of integration. $\displaystyle \frac{\partial S}{\partial \alpha} $ then provides me with an expression in x for which i solve to get $\displaystyle x(t) = \alpha t + \frac{E}{2}$ Where E is the expression found from $\displaystyle \frac{\partial S}{\partial \alpha}$ ,which proves the extremal to be a straight line.

    Now my problem is that for this example,$\displaystyle L(x,\dot x) = \dot x ^2 -x^2$
    i get $\displaystyle p=\frac{\partial L}{\partial \dot x}\ \ so \ \dot x = \frac{p}{2} \ \ therefore \ \ H = -\frac{p^2}{4} + x^2 $
    Which yields the Hamilton jacobi $\displaystyle -\frac{1}{4}(\frac{\partial S}{\partial x})^2 + x^2 +\frac{\partial S}{\partial t}}=0$
    I have so little time, not enough to complete this question properly, but i believe i'm on the wrong track with this example, should x be here? I am asked to use the exact same hints and form for S as the first example but it just doesn't work out that way ! Because of x,I can no longer use total derivatives in place of partials for S! I get $\displaystyle S=-\alpha ^2t-\frac{x^3}{3} +4\alpha^2x + \beta$ Which doesn't solve the HJ! Any words of advice? Sorry for sloppy presentation of the question! I write in a week, have so much to learn!
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  2. #2
    A Plied Mathematician
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    Are you required to use the HJ equation? If you just do Euler-Lagrange, you get $\displaystyle \ddot{x}=-x.$ The function in question, $\displaystyle x(t)=\sqrt{\alpha}\sin(t+\beta)$, is a solution of this DE.
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  3. #3
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    It seems so unfortunately, they say " Use the previous problem as an example and solve this problem" But i have too many other fish to fry ! Thanx!
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  4. #4
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    Are you working with the following definition of the Hamiltonian:

    $\displaystyle \displaystyle H=\sum_{j}p_{j}\dot{q}_{j}-L?$

    If so, then in your case, I get

    $\displaystyle \displaystyle H=p\dot{x}-L.$

    Now, I agree with the equation

    $\displaystyle \displaystyle p=\frac{\partial L}{\partial\dot{x}}=2\dot{x},$ and hence $\displaystyle \dot{x}=p/2.$

    It follows that

    $\displaystyle \displaystyle H=\frac{p^{2}}{2}-\left(\left(\frac{p}{2}\right)^{2}-x^{2}\right)=\frac{p^{2}}{4}+x^{2},$

    which differs from your Hamiltonian by a minus sign.

    Now then, the HJ equation, for your setup here, is

    $\displaystyle \displaystyle\frac{\partial S}{\partial t}+H(x,p,t)=0,$ where we replace the $\displaystyle p$'s by $\displaystyle S_{x}$ thus:

    $\displaystyle \displaystyle\frac{\partial S}{\partial t}+H(x,\frac{\partial S}{\partial x},t)=0,$ which leads to the pde

    $\displaystyle \displaystyle\frac{\partial S}{\partial t}+\frac{1}{4}\left(\frac{\partial S}{\partial x}\right)^{\!\!2}+x^{2}=0.$

    Following Landau and Lifschitz's Mechanics, page 149, let us seek an additively separated solution in the form of

    $\displaystyle S=T(t)+X(x).$ Then $\displaystyle S_{t}=\dot{T}(t),$ and $\displaystyle S_{x}=X'(x).$ Dots are time derivatives, primes are spatial derivatives. If we substitute these into the pde, we obtain

    $\displaystyle \displaystyle\dot{T}(t)+\frac{1}{4}(X'(x))^{2}+x^{ 2}=0.$

    Separating out yields

    $\displaystyle \displaystyle\dot{T}(t)=-\frac{1}{4}(X'(x))^{2}-x^{2}=-\lambda^{2}.$

    Therefore, $\displaystyle T(t)=-\lambda^{2}t+C.$ For the $\displaystyle x$ equation, we obtain

    $\displaystyle (X'(x))^{2}+4x^{2}=4\lambda^{2}.$

    The solutions are

    $\displaystyle X(x)=\pm\left(x\sqrt{\lambda^{2}-x^{2}}+\lambda^{2}\tan^{-1}\left(\frac{x}{\sqrt{\lambda^{2}-x^{2}}}\right)\right).$

    Unfortunately, I have not really studied the HJ equation in depth. There appears to be more to do here, such as on pages 150-1 of Landau and Lifschitz, but I'm not really qualified to do more. Do you know what to do next?
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  5. #5
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    Thanx Adrian,it definitely makes allot more sense stripped down like that! Would send you a beer if i could!
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  6. #6
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    You're welcome. Have a good one!
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