For a nonlinear operator $\mathcal{N}$, $\mathcal{N}(u)=0$ can be solved through Newton iteration, where many steps is usually needed.
If the operator is linear, and we still use the Netwen iteration to solve. Can we solve $\mathcal{L}(u)=0$ in a single step?
For example, if the exact solution is $\bar{u}$, and the initial guess is $\hat{u}$, $\frac{\partial \mathcal{L}(\hat{u})}{\partial u} \Delta u = -\mathcal{L}(\hat{u})$, and $\hat{u}+\Delta u = \bar{u}$?
For a scaler problem it is quite trivial, but I'm not sure this result can be generalized to all the linear operaor, i.e. multi-dimensional, $\mathcal{L}(u)=\mathcal{L}(u,u')$ etc.