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Notes on symmetries of the KdV equation 
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#1
Nov2512, 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 Kortewegde Vries equation. I don't understand why [tex]V_4=x∂_x+3t∂_t2u∂_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_t24\exp(\epsilon)uu_x2\exp(2\epsilon)u_{xxx}[/tex] How does this vanish (so that we get symmetry) given that [itex]u[/itex] satisfies the KdV? 


#2
Nov2612, 12:24 PM

P: 350

Shouldn't the transformation be:
(x exp(e), t exp(3e), u exp(2e)) ? 


#3
Dec1012, 05:31 PM

P: 34

Indeed!



#4
May1613, 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?



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