Solve the system of partial differential equations 
How to solve the system of partial differential equations $$u_{1x_1}=u_{2x_2}=u_{3x_3=0}$$$$\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}=0$$  for $i=1,2,3$

I tried so hard and it becomes so complicated. 
I first conclude $u_1=u_1(x_2,x_3), u_2=u_2(x_1,x_3),u_3=u_3(x_1,x_2)$.
Then by $\frac{\partial u_1}{\partial x_2}+\frac{\partial u_2}{\partial x_1}=0$, I get $u_1(x_2,x_3)=c_1(x_3)x_2+b_1(x_3)$, and $u_2(x_1,x_3)=-c_1(x_3)x_1+b_2(x_3)$. I did these for three of the system of equations and in the end it becomes very complicated. I cannot solve for the coefficient $c_1,b_1,c_2,b_2$.etc. at all. 
Could someone kindely provide some help ? Thanks!
 A: I prefer to change the notations in order to make it clearer :

NOTE :
in the wording of the question, it is written : $\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}=0$  for $i=1,2,3$. But nothing is specified about $j$. In my answer above I supposed $j=1,2,3$ with $j\neq i$ because if the equations with $i=j$ are allowed, the problem becomes trivial (too simple).
A: In THIS ANSWER, I showed that each component $u_i$ is a linear function of $x_j$ and $x_k$, with $i$, $j$ and $k$ distinct.  In that development, we first used that since the diagonal terms add to zero, we obtain
$$\frac{\partial u_i}{\partial x_i}=0 \tag 1$$
which implies that the $i$th component of $u$ is independent of the $i$th coordinate variable.  We can write, therefore
$$\begin{align}
u_1=u_1(x_2,x_3)\\\\
u_2=u_2(x_1,x_3)\\\\
u_3=u_3(x_1,x_2)\\\\
\end{align}$$
In this problem, we are given that
$$\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}=0 \tag 2$$
Observe $(2)$ for the case for which $i=1$ and $j=2$.  Then, we have
$$\frac{\partial u_1(x_2,x_3)}{\partial x_2}=-\frac{\partial u_2(x_1,x_3)}{\partial x_1} \tag 3$$
The left-hand side of $(3)$ is independent of $x_1$ while the right-hand side of $(3)$ is independent of $x_2$.  But this implies that $u_1(x_2,x_3)$ must be a linear function of $x_2$; seen by straightforward integration, $u_1(x_2,x_3)=A_1(x_3)x_2+B_1(x_3)$.  And similarly, $u_2$ must be a linear function of $x_1$ with $u_2(x_1,x_3)=A_2(x_3)x_1+B_2(x_3)$.  Continuing to use $(2)$ with various combinations of $i$ and $j$ gives us the result
$$\begin{align}
u_1&=A_{12}x_2+A_{13}x_3+Bx_2x_3+C_1\\\\
u_2&=A_{12}x_1+A_{23}x_3+Bx_1x_3+C_2\\\\
u_3&=A_{13}x_1+A_{23}x_2+Bx_1x_2+C_3
\end{align}$$
Without additional information on $\vec u$, the constants are arbitrary.
