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\begin{align} \Delta\overrightarrow{v}&=\nabla(\nabla\cdot\overrightarrow{v})-\nabla\times(\nabla\times\overrightarrow{v})\\ \nabla\cdot(\overrightarrow{v}\times\overrightarrow{w})&=\overrightarrow{w}\cdot\nabla\times\overrightarrow{v}-\overrightarrow{v}\cdot\nabla\times\overrightarrow{w}\\ \nabla\cdot(f\overrightarrow{v})&=(\nabla f)\cdot\overrightarrow{v}+f\nabla\cdot\overrightarrow{v} \end{align} Suppose $\overrightarrow{v}$ satisfies $\Delta\overrightarrow{v}=0$ and that $\overrightarrow{v}$ vanishes outside some bounded region $V \subset \Bbb R^3$. Show that $\nabla\times\overrightarrow{v}=0$ and $\nabla\cdot\overrightarrow{v}=0$. (Hint: Integrate $\overrightarrow{v}\Delta \overrightarrow{v}$ over a suitable region. You will need to make use of the divergence theorem.)

I started by integrating the hint, then using the first identity for $\Delta\overrightarrow{v}$, then separated the integral. I tried to get it into the form for Gauss theorem, and know that the dot product is commutative so I tried that, but neither of the integrals that I separated the original one from can be solved with Gauss theorem.

I feel that I am very far off the solution, any hints as to the next step would be greatly appreciated. The original document is linked here in case that would be easier to read. Thanks in advance for the help. I am new to math.stackexchange, please let me know if I make any mistakes with inputting my question.

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    $\begingroup$ You might need to edit that question. $\endgroup$ – Thomas Andrews Aug 17 '14 at 23:52
  • $\begingroup$ I've adjusted your question's Mathjax a bit, and I've moved the picture to a link so that it all reads better. On a mathematical level, it's worth noting that $\Delta$ in the above question is necessarily the vector Laplacian rather than the scalar Laplacian. $\endgroup$ – Semiclassical Aug 18 '14 at 0:45
  • $\begingroup$ $\Delta \vec v =\nabla^2 \vec v$, right? $\endgroup$ – Robert Lewis Aug 18 '14 at 2:55
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    $\begingroup$ I'm a bit confused on that equality. How can the divergence be equal to the curl? One is a scalar and one is vector... $\endgroup$ – ClassicStyle Aug 18 '14 at 3:42
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I assume the statement $\nabla \cdot \vec v = \nabla \times \vec v = 0$ is a minor abuse of notation, since as pointed out in the comments, $\nabla \cdot \vec v$ is a scalar but $\nabla \times \vec v$ is a vector. Furthermore, I assume that $\vec v$ is a $C^2$ vector field on $\Bbb R^3$, so that all the derivatives make sense.

These things being said, here is a proof which, though not directly employing the given hint, goes a little further than requested and shows that $\vec v = 0$ identically, from which the required results immediately and trivially follow.

By $\Delta$ we mean $\nabla^2$, and so $\Delta \vec v = \nabla^2 \vec v$, where by $\nabla^2 \vec v$ we mean, in a global Cartesian coordinate system, that $\nabla^2$ is to be applied to each component of $\vec v$ separately, so that taking

$\vec v = (v_1, v_2, v_3) \tag{1}$

we have

$\nabla^2 \vec v = (\nabla^2 v_1, \nabla^2 v_2, \nabla^2 v_3). \tag{2}$

(2) may also be directly validated by working through our OP Dan Smith's given equation

$\nabla^2 \vec v = \Delta \vec v = \nabla (\nabla \cdot \vec v) - \nabla \times (\nabla \times \vec v) \tag{3}$

in Cartesian coordinates, a straightforward task requiring mainly careful bookkeeping of indices:

$\nabla \times \vec v = (v_{3, 2} - v_{2, 3}, v_{1, 3} - v_{3, 1}, v_{2, 1} - v_{1, 2}) \tag{4}$

where we have used the notation

$v_{1, 2} = \dfrac{\partial v_1}{\partial x_2} \tag{5}$

and so forth, where $x_1$, $x_2$, $x_3$ are the coordinates on $\Bbb R^3$. From (4) it follows that the $1$-component of $\nabla \times (\nabla \times \vec v)$ is

$(\nabla \times (\nabla \times \vec v))_1 = v_{2, 12} - v_{1, 22} - v_{1, 33} + v_{3, 13}, \tag{6}$

and the $1$-component of $\nabla(\nabla \cdot \vec v)$ is

$(\nabla(\nabla \cdot \vec v))_1 = v_{1, 11} + v_{2, 21} + v_{3, 31}; \tag{7}$

we see from (6) and (7) that

$(\nabla(\nabla \cdot \vec v) - \nabla \times (\nabla \times \vec v))_1 = v_{1, 11} + v_{1, 22} + v_{1, 33}, \tag{8}$

and the corresponding results hold for the $2$ and $3$ components, so

$\nabla^2 \vec v = \Delta \vec v = (\nabla^2 v_1, \nabla^2 v_2, \nabla^2 v_3) \tag{9}$

as claimed. It follows then from $\Delta \vec v = 0$ that $\nabla^2 v_i = 0$ for $1 \le i \le 3$; each component of $\vec v$ is a harmonic function on $\Bbb R^3$. But a harmonic function which vanishes outside of a bounded set such as $V$ must be identically zero; this follows from the mean value property of harmonic functions $u$, which in the present circumstances states that for $p \in \Bbb R^3$

$u(p) = \dfrac{1}{4 \pi R^2}\int_{\partial B(p, R)} u dS, \tag{10}$

where $B(p, R)$ is the ball of radius $R$ centered at $p$, and $dS$ is the surface area measure on the sphere bounding said ball. By taking $R$ sufficiently large for any particular $p$, we see that the integral in (10) must vanish if $u$ vanishes outside of some bounded set $V$; $u = 0$ identically on a sufficiently large sphere. Thus we see that $\vec v = 0$ everywhere, whence in particular $\nabla \cdot \vec v$ and $\nabla \times \vec v = 0$ as well. QED.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

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Here are some signposts towards the solution. First, here are two applications of the two 'product rule' vector identities given above, specifically for the cases $\vec{w}=\nabla\times\vec{v}$ and $f=\nabla\cdot\vec{v}$:

\begin{align} \vec{v}\cdot(\nabla\times (\nabla\times\vec{v})) &=-\nabla\cdot(\vec{v}\times(\nabla\times\vec{v}))+(\nabla\times \vec{v})^2\\ \\ \vec{v}\cdot\nabla(\nabla\cdot\vec{v}) &=\nabla\cdot(\vec{v}\;(\nabla\cdot\vec{v}))-(\nabla\cdot\vec{v})^2 \end{align} That, along with the first identity for the vector Laplacian of $\vec{v}$, is enough to get to the application of the Divergence Theorem. What remains is the choice of region $R$; for that you should think carefully about the bounded region $V$.

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