Solving a PDE to Yield Determining Equations

Question

I'm going through an example in Peter Hydon's book "Symmetry Methods for Differential Equations" which finds the basis for the Lie Algebra of the point symmetry generators for Burgers' equations. Burgers' equation is given in this case as:

$u_t+uu_x=u_{xx}$

So x and t are the independent variables. One part of the linearised symmetry condition is given as:

$2\xi_x+2\xi_uu_x-\tau_t=0$

The book continues to state $\xi_u=0$ - how does one show this exactly? Using information given later on I can reason this but for the life of my I can't figure out how to take the above line and show that $\xi_u=0$. I know from earlier working that $\tau=\tau(t)$ but as far as I can tell I shouldn't know at this point what $\xi$ is a function of so that doesn't really help me much to show $\xi_u=0$ right? Also I understand that from Burger's equation I know that $u_x\neq0$ or else the solution would be trivial, so just need to figure out how to show that term is equal to 0.

Thank you!

Answer 1

Ok, the answer is quite simple, I simplified the question too much. The book actually states the equations as:

$(\eta_u-\tau_t)u_t-\xi_u u_x u_t = (\eta_u - 2\xi_x - 3\xi_u u_x )u_t $

As can be seen the results they got were simply obtained by equating terms with the same coefficients $u$ of a particular derivative. By removing them I was unable to solve it. Had a feeling was something silly. :P

Answer 2

Do not waste your time solving these determining equations, determination of symmetries for PDE with constant coefficients(even variable coefficients) is fully automated using in PDEtools package of Maple.

For relevant details related to package PDEtools you can see posts of Ecterreb and book by Bluman. This book by Bluman is much better than Hydon book. Both authors are quite responsive to queries of readers through e-mails.

And related to your query that why $\xi_{u}=0$, you must see that in expression

$2\xi_{x}+2\xi_{u}u_{x}-\tau_{t}=0$

$\xi$ and $\tau $ are functions of $(x,t,u)$, therefore from independence of functions in equation the coefficient of $u_{x}$ must vanish.