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This nonlinear dispersive equation replaces the quadratic KdV nonlinearity by a cubic one. In the displayed sign convention, the Miura transformation maps its solutions to $u_t-6u{u}_x+u_{xxx}=0$. Its scaling symmetry is $(x,t,v)\mapsto (e^{s}x,e^{3s}t,e^{-s}v)$.

Integrating the vector-field equations gives the one-parameter transformationsUnder this scaling, $v_t,v^2{v}_x,v_{xxx}$ have weights $\gamma -\beta ,3\gamma -\alpha ,\gamma -3\alpha$. Equality of all three weights givesThus the nontrivial Lie point symmetry generator can be normalized to $x{\partial }_x+3t{\partial }_t-v{\partial }_v$.

For the Miura transformation $u=v^2+v_x$, direct differentiation gives the operator identityFor example $u_x=2v{v}_x+v_{xx}$ and $u_{xxx}=6v_x{v}_{xx}+2v{v}_{xxx}+v_{xxxx}$; inserting them makes the $v_x{v}_{xx}$ terms cancel and leaves the right side. Hence $u$ solves $u_t-6u{u}_x+u_{xxx}=0$ whenever $v$ solves the mKdV equation. Both $v^2$ and $v_x$ have weight $-2\alpha$, so the corresponding Korteweg-De Vries equation symmetry is