3
$\begingroup$

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!

$\endgroup$
  • 3
    $\begingroup$ Is $u$ a function from $\mathbb{R}^3 \rightarrow \mathbb{R}^3$? $\endgroup$ – NoseKnowsAll Sep 12 '15 at 20:26
  • 2
    $\begingroup$ 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 $\endgroup$ – Omnomnomnom Sep 12 '15 at 20:34
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $$ \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} = 0 \qquad i,j = 1,2,3 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.