Notes on symmetries of the KdV equation

  • #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?
 
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Answers and Replies

  • #2
349
1
Shouldn't the transformation be:

(x exp(e), t exp(3e), u exp(-2e)) ?
 
  • #4
109
0
can these equations be used to model black holes for instance in analogy to water waves?
 

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