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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

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what are $f,y,v$? functions/vector fields? What does $v//\partial D$ mean? –  Ilya Mar 20 '12 at 14:47
    
$f$ is a scalar field, $v$ and $y$ are vector fields. $v // \partial D$ means that $v$ is parallel to the boundary of the domain $D$. –  chango Mar 20 '12 at 15:12
    
(Off topic, but it's Stokes's theorem, not Stoke's.) –  Hans Lundmark Mar 20 '12 at 15:33
    
How can $v||\partial D$ and $\nabla v=0$ simultaneously; doesn't the latter mean $v$ is constant, or do you mean e.g. $\nabla\cdot v=0$? –  anon Mar 20 '12 at 15:50
    
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.... –  user7530 Mar 20 '12 at 15:53
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1 Answer

up vote 1 down vote accepted

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.

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Your counterexample seems to be ok. Thank you! –  chango Mar 26 '12 at 10:56
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