Consider
I believe that there is something wrong with this text. In particular, how is $$\Delta=0 \quad \Longrightarrow \quad V(\Delta)=0$$ completely non trivial by linearity of operators?
Moreover, I do not understand how this is used to find lie point symmetries.
More context For context, the text also mentions
so I assume that we plug such $V$ in $$ V(\Delta)=0$$ and then try to solve for the coefficients. But again, I do not understand why all coefficient dont simply work.
Question: Why is $$\Delta=0 \quad \Longrightarrow \quad V(\Delta)=0$$ not true for any $V$? Moreover, can someone give me a simple example of how this technique is used to generate lie point symmetries?
The implication $\Delta = 0 \Longrightarrow V(\Delta) = 0$ is not trivially true, despite the linearity of $V$, because the object $\Delta$ has to be understood as a formal expression representing the differential equation for $u$ itself (when set to zero). As an element of the tangent space, the vector field $V$ is defined by its action on the space of smooth functions of $(x,u,u_x,\ldots)$ but not on the space where $\Delta$ lives, in such a way that the linearity of $V$ doesn't have to be guaranteed.
Here is a (counter-)example with the differential equation $\Delta[x,u,u'] = u' + xu = 0$, which is solved by $u(x) = Ce^{-\frac{1}{2}x^2}$, with $C$ a constant. For example, the translation $x \to x + \varepsilon$, represented by $g^\varepsilon = e^{\varepsilon V}$ with $V = \partial_x$, is not a symmetry of this equation of motion because of the $x$ prefactor. Then, one has $V(\Delta) = \partial_x\Delta = u' + u$, which doesn't vanish for the same solution $u$.