So I am working through Peter Olver's book Applications of Lie Groups to Differential equations and there is this example on page 329 that I cannot seem to figure out.
In the first highlighted part, from the formula provided I do not see how there is that extra $-u_x$ term since the formula does not require you to apply $D_x$ to $u$.
The last two highlighted parts make sense to me based on the formula that the adjoint of $D_x^2 +u $ is $(-D_x)^2 + u$ but then for the second one if I apply the same logic as the very first highlighted example I get $(-D_x)^3 - 2(D_x)\cdot u +u_x = -D_x^3 -2u_xD_x - 2u_x +u_x = -D_x^3 -2uD_x - u_x$ but again I don't understand where the extra term is coming from when the formula is simply a multiplication, so I must be interpreting this wrong could somebody please clear this up?
Note that the differential operator $(-D_x)\cdot u$ acts on a function $f$ as follows
$$((-D_x)\cdot u)(f) = -D_x(uf) = -u_xf - uf_x = (-u_x - uD_x)(f).$$
That is, as differential operators $(-D_x)\cdot u = -u_x - uD_x$ which is why the two expressions for $\mathscr{D}^*$ are equal.