Notes on symmetries of the KdV equation

[MarkovMarakov]

I am having trouble understanding a section in http://www.mathstat.dal.ca/~francisv/publications/XXV-ICGTMP-Proceeding/dkdv.pdf: . It is on page 3. Section 3 -- Discretization of the Korteweg-de Vries equation. I don't understand why $$V_4=x∂_x+3t∂_t-2u∂_u$$ generates a symmetry group of the KdV. I see that it generates the transformation
$$(x',t',u')= (x\exp(\epsilon), 3t\exp(\epsilon), -2u\exp(\epsilon))$$
So $$u'_{t'}-6u'u'_{x'}+u'_{x'x'x'}=-{2\over 3}u_t-24\exp(\epsilon)uu_x-2\exp(-2\epsilon)u_{xxx}$$ How does this vanish (so that we get symmetry) given that $u$ satisfies the KdV?

 

[Vargo]

Shouldn't the transformation be:

(x exp(e), t exp(3e), u exp(-2e)) ?

 

[MarkovMarakov]

Indeed!

 

[mathnerd15]

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