Linear vector field

Question

Let $u: R^3 \rightarrow R^3$ be $C^1$ and assume $\nabla u + (\nabla u)^T =0$ then we can deduce $$u(x_1,x_2,x_3)= \langle f_1(x_2,x_3), f_2(x_1,x_3),f_3(x_1,x_2) \rangle$$ for some functions $f_j$. How can I continue to prove that $f_j$'s are in fact linear?

Answer 1

Since the diagonal terms add to zero, we obtain

$$\frac{\partial u_i}{\partial x_i}=0 \tag 1$$

which indeed implies that the $i$th component of $u$ is independent of the $i$th coordinate variable.

Going further, we also see that

$$\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}=0 \tag 2$$

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If we are not given any information on the existence of the second partials, then we proceed as follows. 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.