3
$\begingroup$

I want to prove the following:

If $\nabla \cdot \mathbf{v}= 0$ and $\mathbf{v}$ is parallel to $\partial D$ then

$$\int_D \nabla f \cdot (\mathbf{v} \cdot \nabla \mathbf{y}) dx =0 $$

where $f$ is a scalar potential and $\mathbf{v},\ \mathbf{y}$ are vector fields.

It must be something very straightforward like applying the divergence theorem or Stokes's but I cannot see it. I'd appreciate any help.

Thanks

  • 2
    what are $f,y,v$? functions/vector fields? What does $v//\partial D$ mean?2012-03-20
  • 0
    $f$ is a scalar field, $v$ and $y$ are vector fields. $v // \partial D$ means that $v$ is parallel to the boundary of the domain $D$.2012-03-20
  • 0
    (Off topic, but it's Stokes's theorem, not Stoke's.)2012-03-20
  • 0
    How can $v||\partial D$ and $\nabla v=0$ simultaneously; doesn't the latter mean $v$ is constant, or do you mean e.g. $\nabla\cdot v=0$?2012-03-20
  • 0
    What does $\nabla v =0$ mean? That the matrix is the zero matrix? If so, $v$ is constant and can't possibly be parallel everywhere to the boundary....2012-03-20
  • 0
    Sorry, you are right. I meant to write $\nabla \cdot v = 0$ (ie $v$ is divergence free).2012-03-20

1 Answers 1

1

The statement is false.

Consider D the unit circle, $\textbf{v}=(y,-x)$, $\textbf{y}=(0,x)$, and $f=y^2$.

We have \begin{align*}\int_D \nabla f \cdot(\textbf{v}\cdot\nabla\textbf{y})\,dA &= \int_D (0,2y)\cdot \left[\begin{array}{cc}0 & 0 \\1 & 0\end{array}\right]\left[\begin{array}{c}y\\-x\end{array}\right]\,dA\\ &= \int_D 2y^2\,dA \end{align*} which clearly can't vanish.

  • 0
    Your counterexample seems to be ok. Thank you!2012-03-26