 Solution

For $k,n>0$, substitute $x^{\prime }=\alpha x$, $t^{\prime }=\beta t$, $u^{\prime }=\gamma u$ and compare the two distinct right-hand terms. Invariance requires $\beta ={\alpha }^{2k+1}$ and ${\gamma }^n={\alpha }^{-2k}$. Taking $\alpha =e^s$ gives the Lie point symmetryTranslations in $x$ and $t$ are also Lie point symmetries, generated by $X={\partial }_x$ and $T={\partial }_t$. Direct differentiation givesso their span is a three-dimensional Lie algebra.

The printed nonnegative-integer range also includes cases where the generic scaling would divide by zero. If $k>0,n=0$, the equation is linear, $u_t=u_x+{\partial }_x^{2k+1}u$. Amplitude scaling $(x,t,u)\mapsto (x,t,e^{s}u)$ is a symmetry, with $D=u{\partial }_u$. Along with $X,T$ all brackets are zero, giving an abelian three-dimensional Lie algebra. A nontrivial pure rescaling of space and time cannot preserve both derivative orders in this case.

If $k=0$, the equation becomes $u_t=(u^n+1)u_x$, including $u_t=2u_x$ when $n=0$. In either case choose $(x,t,u)\mapsto (e^{s}x,e^{s}t,u)$, with $D=x{\partial }_x+t{\partial }_t$. Then $[D,X]=-X$, $[D,T]=-T$, and $[X,T]=0$. This covers every allowed pair $(k,n)$; when both vanish, there is also an independent amplitude scaling.