Hamiltonian mechanics, finding extremals through solving S

**punkstart** · October 19th 2010, 10:30 AM

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!

**Ackbeet** · October 22nd 2010, 09:48 AM

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.

**punkstart** · October 23rd 2010, 12:09 AM

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!

**Ackbeet** · October 23rd 2010, 04:13 AM

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?

**punkstart** · October 26th 2010, 05:10 AM

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

**Ackbeet** · October 26th 2010, 05:49 AM

You're welcome. Have a good one!

Thread: Hamiltonian mechanics, finding extremals through solving S

LinkBack

Thread Tools

Display

Hamiltonian mechanics, finding extremals through solving S

Similar Math Help Forum Discussions

[SOLVED] Mechanics, solving λ throught j and i

Hamiltonian mechanics- show that the following are extremals...

extremals no. 2

Mechanics - Finding tension

Extremals

Search Tags