In one of my textooks I found Laplace's equation listed as
$$\nabla^2f=0$$
together with the following quote: Laplace’s equation is linear, and hence any number of particular solutions to [equation above] can be added together to obtain another solution.
I am not entirely certain if I understood this part correctly, so I would appreciate it if you could check my understanding and correct me if I misunderstand something.
As far as I understand, the critical thing is not that $\nabla^2f$ is linear, but that $\nabla^2f=f''=0$ implies linearity of $f$ (all linear functions have zero second derivative). And linear functions can be added together (due to additivity) to obtain another linear function. This isn't quite what the textbook wrote (but maybe what they meant), which is why I'd like to double-check my interpretation of it first.
This works for two reasons:
1: $\nabla^2$ is a linear operator
2: The PDE is homogeneous
I'll state without proof that for scalar fields $\Phi$ and $\Psi$, and scalars $a$ and $b$, $$\nabla^2(a\Phi + b\Psi)= a\nabla^2(\Phi)+b\nabla^2(\Psi)$$ Therefore, if $\Phi$ and $\Psi$ are solutions of Laplace's equation, then $\nabla^2(\Phi + \Psi)=\nabla^2(\Phi) + \nabla^2(\Psi)=0+0=0$ thus $\Phi + \Psi$ is also a solution of Laplace's equation. Note that this does not work for Poisson's equation: say $\Phi$ and $\Psi$ satisfy a special case of Poisson's equation, say $\nabla^2(U)=C$. Then $$\nabla^2(\Phi + \Psi) = \nabla^2(\Phi)+\nabla^2(\Psi) = C+C=2C \neq C$$ Thus $\Phi + \Psi$ is not a solution.