How to find $\vec{v}$ if I know $\vec{\nabla}\times\vec{v}$ and $\vec{\nabla}\cdot\vec{v}$?

Question

So let's assume we know that $\vec{\nabla}\times\vec{v}=5\hat{\imath} + 3\hat{k}$ and $\vec{\nabla}\cdot{\vec{v}}=0$ ($\hat{\imath}$, $\hat{\jmath}$, $\hat{k}$ being the unit vectors in the $x$, $y$ and $z$ direction). What whould be a general/good/efficient/clever process of finding $\vec{v}$?

Edit: Trying to clarify things a bit: the reason for my question is the fact, that Maxwell's equations define the curl of the magnetic and the electric field. So I'm wondering how to find a vector if all I know is it's curl and it's div (I assume it's constant component to be zero).

In contrast to that, finding $\vec{\nabla}\times{\vec{v}}$ and $\vec{\nabla}\cdot{\vec{v}}$ is simple if $\vec{v}$ is known. But what kind of approach is advisable if I want to go the oppesit direction?

Answer 1

Hint (for finding one solution): a rigid rotation about some fixed axis has the same curl everywhere and zero divergence.

Answer 2

Suprisingly, this question was never fully answered, although the solution is straightforward. As @Winther already mentioned, there is a theorem readily describing what you want: the Helmholtz decomposition. It states that given a vector field that decomposes into a divergence- and a curl-free component $$\vec{F}=-\vec{\nabla}\Phi+\vec{\nabla}\times\vec{A},$$ we can reconstruct the original potentials $\Phi$ and $\vec{A}$ using \begin{align} \Phi(\vec{r})&=\frac{1}{4\pi}\int_V \frac{\vec{\nabla} \cdot \vec{F} (\vec{r}{}^{\hspace{0.13ex}\prime})}{\lVert\vec{r}-\vec{r}{}^{\hspace{0.13ex}\prime}\rVert} \, \mathrm{d}^3r{}^{\hspace{0.13ex}\prime}\\\vec{A}(\vec{r})&=\frac{1}{4\pi}\int_V \frac{\vec{\nabla} \times \vec{F} (\vec{r}{}^{\hspace{0.13ex}\prime})}{\lVert\vec{r}-\vec{r}{}^{\hspace{0.1ex}\prime}\rVert} \, \mathrm{d}^3r{}^{\hspace{0.13ex}\prime}. \end{align} Conversely, given the curl $\vec{\nabla}\times\vec{v}=\vec{C}$ and divergence $\vec{\nabla}\cdot\vec{v}=D$ of a vector field (and given it vanishes at infinity) you can construct a unique vector field $\vec{v}$ that has these properties, namely: $$\boxed{\vec{v}=\frac{1}{4\pi}\vec{\nabla}\biggl(\int_V \frac{D(\vec{r}{}^{\hspace{0.13ex}\prime})}{\lVert\vec{r}-\vec{r}{}^{\hspace{0.13ex}\prime}\rVert} \, \mathrm{d}^3r{}^{\hspace{0.13ex}\prime}\biggr)+\frac{1}{4\pi}\vec{\nabla}\times\biggl(\int_V \frac{\vec{C} (\vec{r}{}^{\hspace{0.13ex}\prime})}{\lVert\vec{r}-\vec{r}{}^{\hspace{0.1ex}\prime}\rVert} \, \mathrm{d}^3r{}^{\hspace{0.13ex}\prime}\biggr)}$$