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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 $$V_4=x∂_x+3t∂_t-2u∂_u$$ 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?

Is it possible that I have misunderstood something, such that from "I see that..." onwards I have been barking up the wrong tree?

Help would be very much appreciated!

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This might be helpful: – Lukas Geyer Nov 26 '12 at 1:38
up vote 0 down vote accepted

The transformation generated by $V_4$ is not the one you've written, but $$ (x',t',u')=(x \exp(\epsilon), t \exp(3 \epsilon) , u \exp(-2\epsilon)). $$ (You can tell right away that something is wrong with your transformation, since it doesn't reduce to the identity map when $\epsilon=0$.)

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Ahhh!! Thank you, Hans! – jude Nov 26 '12 at 11:19
You're welcome. – Hans Lundmark Nov 26 '12 at 12:20

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