2
$\begingroup$

I'm asked to consider the 1-param. group of transformations generated by $V = \dfrac{\partial}{\partial u} + \alpha t \dfrac{\partial}{\partial x}$, which easily enough yields $g^{\epsilon}(x,t,u) = (x+\alpha t \epsilon, t, u+\epsilon)$.

Then I'm asked to find $\alpha$ s.t. $V$ generates a Lie point symmetry of $\Delta[x,t,u] = u_t + u_{xxx} - 6uu_x$, which is where I come unstuck.

I wrote $g^{\epsilon} = (\tilde{x}, \tilde{t}, \tilde{u})$ and attempted to compute $\Delta[\tilde{x}, \tilde{t}, \tilde{u}]$, but I continually obtain results such as $\tilde{u}_{\tilde{t}} = -\alpha \epsilon u_x + u_t$, which do not seem correct (given the presence of the parameter epsilon).

Any ideas on the best way to find $\alpha$?

Thanks in advance!

$\endgroup$
  • $\begingroup$ You may glean a lot of information about this problem in m-hikari.com/imf-password2009/1-4-2009/… $\endgroup$ – Autolatry Jan 14 '15 at 15:24
  • $\begingroup$ @Autolatry, thanks for the response. I have taken a read of that, but I was hoping to avoid any such computation. I'm looking for a direct route from this general generator to the generator I want. $\endgroup$ – FH93 Jan 14 '15 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.