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Notes on symmetries of the KdV equation

  1. Nov 25, 2012 #1
    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?
     
  2. jcsd
  3. Nov 26, 2012 #2
    Shouldn't the transformation be:

    (x exp(e), t exp(3e), u exp(-2e)) ?
     
  4. Dec 10, 2012 #3
    Indeed!
     
  5. May 16, 2013 #4
    can these equations be used to model black holes for instance in analogy to water waves?
     
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