Register to reply

Notes on symmetries of the KdV equation

by MarkovMarakov
Tags: kdv, lie algebra, symmetry
Share this thread:
MarkovMarakov
#1
Nov25-12, 07:10 PM
P: 34
I am having trouble understanding a section in these notes: . It is on page 3. Section 3 -- Discretization of the Korteweg-de Vries equation. I don't understand why [tex]V_4=x∂_x+3t∂_t-2u∂_u[/tex] generates a symmetry group of the KdV. I see that it generates the transformation
[tex](x',t',u')= (x\exp(\epsilon), 3t\exp(\epsilon), -2u\exp(\epsilon))[/tex]
So [tex]u'_{t'}-6u'u'_{x'}+u'_{x'x'x'}=-{2\over 3}u_t-24\exp(\epsilon)uu_x-2\exp(-2\epsilon)u_{xxx}[/tex] How does this vanish (so that we get symmetry) given that [itex]u[/itex] satisfies the KdV?
Phys.Org News Partner Science news on Phys.org
An interesting glimpse into how future state-of-the-art electronics might work
Tissue regeneration using anti-inflammatory nanomolecules
C2D2 fighting corrosion
Vargo
#2
Nov26-12, 12:24 PM
P: 350
Shouldn't the transformation be:

(x exp(e), t exp(3e), u exp(-2e)) ?
MarkovMarakov
#3
Dec10-12, 05:31 PM
P: 34
Indeed!

mathnerd15
#4
May16-13, 11:58 PM
P: 110
Notes on symmetries of the KdV equation

can these equations be used to model black holes for instance in analogy to water waves?


Register to reply

Related Discussions
Internal (gauge) symmetries and spacetime symmetries High Energy, Nuclear, Particle Physics 54
Symmetries in QM Quantum Physics 12
Symmetries in qft Quantum Physics 4
Mathematical justification for a crystallographic point group General Physics 0
Obtaining Dirac equation from symmetries High Energy, Nuclear, Particle Physics 2