
#1
Nov2512, 07:07 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? 


Register to reply 
Related Discussions  
Interest in Linear algebra and Differential Equations?  Academic Guidance  4  
Differential Equations or Linear Algebra?  Academic Guidance  7  
Linear Algebra or Differential equations?  Academic Guidance  6  
linear algebra ordinary differential equations  Calculus & Beyond Homework  11  
Differential Equations vs Linear Algebra  General Math  6 