# Thread: Hamiltonian mechanics, finding extremals through solving S

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

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

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

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

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

Now my problem is that for this example, $L(x,\dot x) = \dot x ^2 -x^2$
i get $p=\frac{\partial L}{\partial \dot x}\ \ so \ \dot x = \frac{p}{2} \ \ therefore \ \ H = -\frac{p^2}{4} + x^2$
Which yields the Hamilton jacobi $-\frac{1}{4}(\frac{\partial S}{\partial x})^2 + x^2 +\frac{\partial S}{\partial t}}=0$
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 $S=-\alpha ^2t-\frac{x^3}{3} +4\alpha^2x + \beta$ 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!

2. Are you required to use the HJ equation? If you just do Euler-Lagrange, you get $\ddot{x}=-x.$ The function in question, $x(t)=\sqrt{\alpha}\sin(t+\beta)$, is a solution of this DE.

3. 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!

4. Are you working with the following definition of the Hamiltonian:

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

If so, then in your case, I get

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

Now, I agree with the equation

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

It follows that

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

which differs from your Hamiltonian by a minus sign.

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

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

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

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

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

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

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

Separating out yields

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

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

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

The solutions are

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

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?

5. Thanx Adrian,it definitely makes allot more sense stripped down like that! Would send you a beer if i could!

6. You're welcome. Have a good one!