I was trying to show that a coupled system of PDEs has a solution in two different ways. One of them works, but for the other way, I am not able to conclude. Therefore, I would like to know where my mistke is or how I can continue.
The system I am considering is for a smooth function $f:\mathbb{R}^{3}\to\mathbb{R}$ and a smooth vector-valued and time-dependent function $\vec{V}:I\times \mathbb{R}^{3}\to\mathbb{R}^{3}$ and has the form
$$\begin{cases}\vec{\nabla}f=-\partial_{t}\vec{V}\\\Delta\vec{V}=\vec{S}\end{cases}$$
where the (smooth) source is a vector-valued function such that $\partial_{t}^{2}\vec{S}=0$ and such that $\partial_{t}\vec{S}=\vec{\nabla}g$ for some scalar-valued function $g$, which by $\partial_{t}^{2}\vec{S}=0$ is time-independent, i.e. $g:\mathbb{R}^{3}\to\mathbb{R}$.
The first approach, which works (if I am not doing any error) is the following:
Take a solution of the equation $\Delta f=-g$, which under appropriate boundary/initial conditions exists. By assumption, $g$ is time-independent (since $\partial_{t}^{2}S=0$) and hence, we can take the solution $f$ to be time-independent (which means the first equation is satisfied). Next, we use the first equation and define $$\vec{V}(\vec{r},t):=-[\vec{\nabla}f(\vec{r})]\cdot t+\vec{C}(\vec{r})$$ for some time-independent function $\vec{C}$ to be determined later. Now, we have fixed $f$ and $\vec{V}$ and we have to check that the second equation is satisfied. For this, we do the following steps:
$$\Delta f=-g\\\vec{\nabla}\Delta f=-\vec{\nabla}g\\-\partial_{t}\Delta\vec{V}=-\partial_{t}\vec{S}\\-\Delta\vec{V}+\Delta\vec{C}=-\vec{S}+\vec{S}\vert_{t=0}$$
where we integrated in the last step. Hence, we set the constants $\vec{C}$ to be solutions of the time-independent Poisson equation $$\Delta\vec{C}=\vec{S}\vert_{t=0}$$ which again exists under suitable assumptions. This concludes the proof.
For the second approach, I do the following:
I start by taking a solution of equation 2, which in general has the form
$$\vec{V}(\vec{r},t)=\int_{\mathbb{R}^{3}}\frac{\vec{S}(\vec{r}^{\prime},t)}{\Vert\vec{r}-\vec{r}^{\prime}\Vert}\mathrm{d}^{3}\vec{r}^{\prime}$$ Now, the second equation states that $\vec{\nabla}f=-\partial_{t}\vec{V}$, or in other words, that the vector field $\partial_{t}\vec{V}$ is the gradient of a funcion. By Helmholtz decomposition, this can only be the case if $\partial_{t}\vec{V}$ is rotation free, which I don't see why it should be the case:
$$\vec{\nabla}\times \partial_{t}\vec{V}(\vec{r},t)= \int_{\mathbb{R}^{3}}\vec{\nabla}\times\frac{\partial_{t}\vec{S}(\vec{r}^{\prime},t)}{\Vert\vec{r}-\vec{r}^{\prime}\Vert}\mathrm{d}^{3}\vec{r}^{\prime}=\int_{\mathbb{R}^{3}}\vec{\nabla}\times\frac{\vec{\nabla}^{\prime}g(\vec{r}^{\prime})}{\Vert\vec{r}-\vec{r}^{\prime}\Vert}\mathrm{d}^{3}\vec{r}^{\prime}$$
I don't see how this should vanish and hence how to solve the first equation for $f$. Does anyone see how to continue? Maybe it also involves using the freedom of adding to $\vec{V}$ a harmonic vector-valued function?
If we can assume $g$ is bounded (and I believe we can), we can easily show $$ \int_{\mathbb R^3}\nabla' \frac{g(\vec{r}')}{||\vec{r}-\vec{r}'||}d^3\vec{r}' = 0. $$ Then we note that for any function $p$, $\nabla'p(\vec{r}-\vec{r}') = -\nabla p(\vec{r}-\vec{r}')$ by the chain rule. This lets us integrate by parts to convert the $\nabla'$ into a $\nabla$: \begin{multline} \partial_t \vec{V} = \int_{\mathbb R^3}\frac{\nabla'g(\vec{r}')}{||\vec{r}-\vec{r}'||} d^3\vec{r}'=\int_{\mathbb R^3}\nabla' \frac{g(\vec{r}')}{||\vec{r}-\vec{r}'||}d^3\vec{r}'-\int_{\mathbb R^3}g(\vec{r}')\nabla'\frac{1}{||\vec{r}-\vec{r}'||}d^3\vec{r}' \\ = \int_{\mathbb R^3}g(\vec{r}')\nabla\frac{1}{||\vec{r}-\vec{r}'||}d^3\vec{r}' = \nabla \int_{\mathbb R^3}\frac{g(\vec{r}')}{||\vec{r}-\vec{r}'||} d^3\vec{r}'. \end{multline} So $\partial_t\vec{V}$ is indeed a gradient field.