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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?

Caesar.tcl
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K. Nick
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    There is not enough information to uniquely determine $\vec{v}$ (adding a constant $\vec{c}$ to $\vec{v}$ gives the same curl and divergence) – Winther Mar 09 '18 at 17:11
  • Related: finding a vector field when given curl(F) – Winther Mar 09 '18 at 17:17
  • @Winther if this is the problem that op actually was given (on which I have doubts), I don't see them as particularly related. – operatorerror Mar 09 '18 at 17:28
  • @Winther Does setting $\vec{c} = \vec{0}$ is anything needed to determine $\vec{v}$ uniquely? I'm interested in the parts of $\vec{v}$ which do contain curl and divergence. – K. Nick Mar 09 '18 at 18:12
  • @Winther I'm an engineering student (not maths) so please excuse if I failed to express my problem properly. My lecturer (doing his job not particularly well) mentioned the concept that $\vec{v} = \vec{v}_1 + \vec{v}_2 + \vec{v}_3$ with $\operatorname{curl} \vec{v}_1 = \vec{0}, \operatorname {div} \vec{v}_2 =0 \ \text{and} \ \vec{v}_3=const$. Does this always apply? Has this concept a name? What are its limitations? What are the consequences. Hints to places, where I can find more information on that would be great. – K. Nick Mar 09 '18 at 18:22
  • Ok, then I'm sure all they want is to find one solution, i.e. find a $\vec{v}$ such that when you take the curl and divergence you get what you are given. So you should just ignore the comment above. – Winther Mar 09 '18 at 18:25
  • Sounds like you are talking about this: https://en.wikipedia.org/wiki/Helmholtz_decomposition – Winther Mar 09 '18 at 18:25

2 Answers2

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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: $$\boxed{\vec{v}=\frac{1}{4\pi}\vec{\nabla}\biggl(\int_V \frac{D(\vec{r}{}^{\hspace{0.13ex}\prime})}{\lVert\vec{r}-\vec{r}{}^{\hspace{0.13ex}\prime}\rVert} \, \mathrm{d}^3r{}^{\hspace{0.13ex}\prime}\biggr)+\frac{1}{4\pi}\vec{\nabla}\times\biggl(\int_V \frac{\vec{C} (\vec{r}{}^{\hspace{0.13ex}\prime})}{\lVert\vec{r}-\vec{r}{}^{\hspace{0.1ex}\prime}\rVert} \, \mathrm{d}^3r{}^{\hspace{0.13ex}\prime}\biggr)}$$

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

Arthur
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