- #1
MarkovMarakov
- 32
- 1
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?
(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?