What is
a symmetry of the KDV $$\frac{\partial u}{\partial t}=6u\frac{\partial u}{\partial x}-\frac{\partial^3 u}{\partial x^3}$$ generated by $$V=A(t,x,u)\frac{\partial }{\partial t}+B(t,x,u)\frac{\partial }{\partial x}+C(t,x,u)\frac{\partial }{\partial u}$$?
I don't understand what that means. From class notes, I know that $$V=A(t,x,u)\frac{\partial }{\partial t}+B(t,x,u)\frac{\partial }{\partial x}+C(t,x,u)\frac{\partial }{\partial u}$$ generates a symmetry given by the set of ODEs $$\frac{d t}{d \epsilon}=A\\\frac{d x}{d \epsilon}=B\\\frac{d u}{d \epsilon}=C $$ But I don't understand what
"a symmetry of ... generated by V" means.
Could someone please explain?
Yes, this statement means that the continuous symmetry generated by the vector field $V$ preserves the PDE.