If $B$ is magnetic field and $E$ electric Field, then
$$B=\nabla\times A,$$
$$E= -\nabla V+\frac{\partial A}{\partial t}.$$
There is Gauge invariance for the trnasformation
$$A'\rightarrow A+\frac{\nabla L}{dt},$$
$$V'\rightarrow V-\frac{dL}{dt}.$$
Now, we can write:
Coulomb Gauge (CG): the choice of a $L$ that implies $\nabla\cdot A=0$.
Lorenz Gauge (LG): the choice of a $L$ that implies $\nabla \cdot A+\frac{1}{c^2} \frac{\partial V}{\partial t}$.
Now, I'm trying to demonstrate that MATHEMATICALLY it's always possible to find a $L$ that satisfies $CG$ or $LG$.
Answer
The correct Gauge transformation formula should be $$\begin{aligned} \mathbf A &\mapsto \mathbf A + \nabla \lambda \\ \mathbf V &\mapsto V - \frac{\partial\lambda}{\partial t}, \end{aligned} $$ not something with "gradL/dt". The Coulomb gauge requires $\nabla\cdot\mathbf A=0$, not "rotA = 0". The Lorenz gauge requires $\nabla\cdot\mathbf A + \frac1{c^2}\frac{\partial V}{\partial t}=0$, not "gradA+1/c^2 dV/dt".
The Coulomb gauge can be chosen by solving the Poisson equation $$ \nabla^2 \lambda = -\nabla\cdot\mathbf A$$
The Lorenz gauge can be chosen by solving the inhomogeneous wave equation $$ \nabla^2 \lambda - \frac1{c^2}\frac{\partial^2\lambda}{\partial t^2} = -\nabla\cdot\mathbf{A} - \frac1{c^2}\frac{\partial{V}}{\partial{t}}$$
(Substitute the transformed potentials into the conditions to get the PDEs)
Existence of solutions of these PDEs are guaranteed as long as the source terms (stuff on the RHS) are "well-behaved" (e.g. $\nabla\cdot\mathbf A$ should grow slower than $1/r$ in the Poisson equation)
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