A system of linear PDEs in cylindrical coordinates. I have the following system of PDEs:
$$\partial_r b_2 - \partial_{\theta}b_1 = 0, \\
\partial_{\theta} b_3 - \partial_{z}b_2 = 0, \\
~~~~~~~\partial_rb_3 - \partial_{z}b_1 = \xi(r, z), $$
where $\xi(r, z)$ is a given function depending only on $(r, z)$ and $b_1, b_2, b_3$ are unknown functions of $(r, \theta, z)$. This system is written in cylindrical coordinates in $\mathbb{R}^3$.
Fix, once and for all, a function $\xi(r, z)$ such that the following holds:
Assumption: The above system has a unique solution ($b_1$, $b_2$, $b_3$).
Question: For this solution, can we claim that $b_2 = 0$?
My thoughts on this: I wanna say that, knowing a solution exists, one could in particular solve the third equation for $b_1, b_3$ depending on $r$ and $z$ alone. Then, since they don't depend on $\theta$, one could take $b_2 = 0$ and find a solution. Since it's unique, that has to be the solution. Is this reasoning correct?
Thank you very much!
 A: First, from the third equation it follows that we can write $b_1(r,\theta,z) = \beta_1(r,z) + z f(\theta)$ and $b_3(r,\theta,z) = \beta_3(r,z) + r f(\theta)$, with $\frac{\partial \beta_3}{\partial r} - \frac{\partial \beta_1}{\partial z} = \xi(r,z)$. From the first equation, we see that
\begin{equation}
b_2(r,\theta,z) = r z f'(\theta) + g(r), \tag{1}
\end{equation}
and from the second equation, we see that
\begin{equation}
b_2(r,\theta,z) = z r f'(\theta) + h(z). \tag{2}
\end{equation}
Comparing $(1)$ and $(2)$, it's clear that $g(r)$ and $h(z)$ must be zero. However, it's also clear that any function $b_2$ of the form $b_2(r,\theta,z) = r z f'(\theta)$ obeys the set of partial differential equations. To conclude: the system does not force $b_2$ to be zero. Moreover, as you see, it does not follow that $b_1$ and $b_3$ are independent of $\theta$. 
Generally speaking, you have a system of the form
\begin{equation}
 \nabla \times \mathbf{b} = \mathbf{f}. \tag{3}
\end{equation}
Because the divergence of the curl vanishes, taking the divergence on both sides of $(3)$ yields
\begin{equation}
 0 = \nabla \cdot \mathbf{f} = \frac{\partial \xi}{\partial z}.
\end{equation}
Therefore, we can conclude that $(3)$ has a solution only if $\frac{\partial \xi}{\partial z} = 0$.
Any remarks on the uniqueness of a solution of $(3)$ depend on boundary conditions. If you don't specify boundary conditions, any solution to $(3)$ is not unique. In this particular case, there is an additional source of non-uniqueness from the equation itself: you can add any gradient of a function to $\mathbf{b}$, i.e. replace $\mathbf{b} \to \mathbf{b} + \nabla \phi$, and it will still obey $(3)$. For more information, see here.
