Existence of $(\tilde{V},\mathbf{\tilde{A}})$ for each $(V,\mathbf{A})$ that gives rise to the same EM-fields The electric field $\mathbf{E}$ and the magnetic induction $\mathbf{B}$ can be parameterized in terms of potentials $V$ and $\mathbf{A}$: $$ \mathbf{E}=-\nabla V-\frac{\partial \mathbf{A}}{\partial t},\quad \mathbf{B}=\nabla\times \mathbf{A}.$$ This parameterization is not unique, as we can find a scalar function $\theta$ and define a couple $(\tilde{V},\mathbf{\tilde{A}})$ via $\tilde{V} = V-\partial \theta/\partial t$ and $\mathbf{\tilde{A}}=A+\nabla \theta$ $(*)$. Then both $(V,\mathbf{A})$ and $(\tilde{V},\mathbf{\tilde{A}})$ will give rise to the same $(\mathbf{E},\mathbf{B})$.
Via Maxwell's equations we can find a coupled system of differential equations for $V$ and $\mathbf{A}$: $$ \begin{cases} \square \mathbf{A} = -\mu \mathbf{J}+\nabla\left( \nabla\cdot \mathbf{A}+\varepsilon\mu \frac{\partial V}{\partial t}\right) \\ \square V = -\frac{\rho}{\varepsilon} -\frac{\partial}{\partial t}\left( \nabla\cdot \mathbf{A}+\varepsilon\mu \frac{\partial V}{\partial t}\right)\end{cases},\quad \square=\nabla^2-\frac{\partial^2}{\partial t^2}.$$
These can be made independent by considering the Lorenz-gauge, in which we set $\nabla\cdot \mathbf{A}+\varepsilon\mu \frac{\partial V}{\partial t} = 0$. How can one explicitly show that for each $(V,\mathbf{A})$ there is $(\tilde{V},\mathbf{\tilde{A}})$ (i.e. give rise to the same fields) such that this couple will satisfy the Lorenz-gauge condition. Is it enough to consider the expressions $(*)$ and deduce that both couples of potentials need to satisfy the Lorenz-gauge condition, resulting in the condition $\square \theta = 0$, i.e. we can always choose a scalar function $\theta$ for which $\square \theta=0$ and consider a new potentials via $(*)$?
Thanks in advance.
 A: Let's say that you have a pair of potentials $(V,\mathbf{A})$ that do not satisfy the Lorenz Gauge. i.e. $$\mathbf{\nabla \cdot A} + \frac{1}{c^2}\frac{\partial V}{\partial t} = f(x,t) \neq 0.$$
Let's now perform a gauge change to some new $(V',\mathbf{A'})$ using the function $\theta(x,t)$ as you mentioned.
\begin{equation*}
\begin{aligned}
\mathbf{A'} &= \mathbf{A} + \nabla\theta\\
V' &= V - \frac{\partial\theta}{\partial t}
\end{aligned}
\end{equation*}
Of course, these new potentials have been constructed such that they will also produce the same fields as $(V,\mathbf{A})$, by Gauge Invariance. Substituting these relations into the equation above, we get
$$\mathbf{\nabla \cdot \left(A' - \nabla\theta\right)} + \frac{1}{c^2}\frac{\partial}{\partial t}\left(V' + \frac{\partial\theta}{\partial t}\right) = f(x,t),$$
and rearranging, we have
$$\mathbf{\nabla \cdot A'} + \frac{1}{c^2}\frac{\partial V'}{\partial t}  - \nabla^2\theta + \frac{1}{c^2}\frac{\partial^2\theta}{\partial t^2} = f(x,t).$$
From this it should be clear that if the function $\theta(x,t)$ that we chose satisfies the condition
$$\boxed{-\nabla^2\theta + \frac{1}{c^2}\frac{\partial^2\theta}{\partial t^2} = f(x,t),}$$
then
$$\mathbf{\nabla \cdot A'} + \frac{1}{c^2}\frac{\partial V'}{\partial t}=0.$$
The problem now reduces to finding a function $\theta(x,t)$ that solves the above wave equation with a source $f(x,y)$, and from the properties of the wave equation, we can always find such a $\theta(x,t)$, provided it's "source" $f(x,t)$ isn't too wacky.

EDIT: As @hyportnex points out in the comments and in this answer, in order to solve a wave equation like the one above, one needs to completely specify the boundary conditions on $\theta$ which could make solving the equation non-trivial, though I still feel that a solution should always exist. However, I'd appreciate it if anyone could correct me.
