Show that if $\nabla { {\bf{u}}}+(\nabla { {\bf{u}})^T}=0$ then $\bf{u}(x)=\bf{a}\times \bf{x}+\bf{b}$

I am working on this problem:

Prove that if a smooth($C^1$) function $\bf{u}$ satisfies $\nabla { {\bf{u}}}+(\nabla { {\bf{u}})^T}=0$ then there exist constant vectors ${\bf{a},\bf{b}}\subset \Bbb R^3$, such that $\bf{u}(x)=\bf{a}\times \bf{x}+\bf{b}$, where $\bf{x} =<x_1,x_2,x_3>$.

I really have no clue about this problem and I am not familiar with matrix calculations. So I just tried to verify that $\bf{u}(x)=\bf{a}\times \bf{x}+\bf{b}$ actually satisfy $\nabla { {\bf{u}}}+(\nabla { {\bf{u}})^T}=0$ .

I tried to calculate $\nabla ({ {\bf{\bf{a}\times \bf{x}+\bf{b}}}}）$ and get $\bf{a}\times \nabla {\bf{x}}$.

Then if I calculate $(\nabla ({ {\bf{\bf{a}\times \bf{x}+\bf{b}}}})^T$, is the result $(\bf{a}\times \nabla {\bf{x}})^T$? It doesn't seem right to me, since I cannot conclude $\bf{a}\times \nabla {\bf{x}}+(\bf{a}\times \nabla {\bf{x}})^T=0$.

Could anyone kindly provide some help? For example, some reference or theorem can be used here? Thanks very much！

• Is $u$ a function from $\mathbb{R}^3 \rightarrow \mathbb{R}^3$? – NoseKnowsAll Sep 12 '15 at 20:26
• You are calculating a derivative with respect to $x$. We have $$\nabla \mathbf x = I$$ where $I$ is the identity matrix. Also, you'll need to write the cross product using matrix notation – Omnomnomnom Sep 12 '15 at 20:34

You are confusing divergence ($\nabla \cdot$) with gradient ($\nabla$).

The gradient can be used on vector functions as well, but this gives you are matrix-valued instead of an scalar-valued function.

Try to compute the first line of

$${ {\bf{\bf{a}\times \bf{x}+\bf{b}}}}.$$ which is $$a_2x_3-a_3x_2+b_1$$

Then use the gradient on it. This gives you $$\nabla (a_2x_3-a_3x_2+b_1) = \begin{pmatrix}0\\-a_3 \\a_2\end{pmatrix}$$

This gives you the first column (or row, not 100% certain, see bottom) of your $\nabla { {\bf{\bf{a}\times \bf{x}+\bf{b}}}}$. Do so for the rest.
$$\nabla ( {\bf{\bf{a}\times \bf{x}+\bf{b}}}) =\begin{pmatrix}0 &a_3&-a_2\\-a_3 & 0 & a_1\\a_2 &-a_1&0\end{pmatrix} (=:A)$$ The matrix you get, should be antisymmetric ($A^T = -A$), which it is.
This is exactly what you need to show.

IMPORTANT: I'm not certain if $\nabla (a_2x_3-a_3x_2+b_1)$ should be $\begin{pmatrix}0\\-a_3 \\a_2\end{pmatrix}$ or $\begin{pmatrix}0 &-a_3 &a_2\end{pmatrix}$. Depending on the convention you need to transpose everthing above.

Hint:

Once you figure out how to take the derivative of $a \times x$ properly, you'll have no trouble showing that if $u(x) = a \times x + b$, then it will satisfy $\nabla u + \nabla u^T = 0$.

Next: suppose that $$u = (u_1(x_1,x_2,x_3),u_2(x_1,x_2,x_3),u_3(x_1,x_2,x_3))$$ then $\nabla u + \nabla u^T = 0$ gives us the equations $$\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} = 0 \qquad i,j = 1,2,3$$