Prove vector integral identity $ \int_{{ \Bbb R}^3} (\nabla\times\nabla\times({\bf u} f))\cdot(\nabla\times({\bf u} g)){\rm d}x{\rm d}y{\rm d}z = 0 $

Question

For sufficiently well behaved ($L^2$ and two times differentiable) scalar functions $f$ and $g$ from $\mathbb{R}^3\to\mathbb{R}$ and a constant vector $\mathbf{u}\in\mathbb{R}^3$

$$ \int_{\mathbb{R}^3} (\nabla\times\nabla\times(\mathbf{u} f))\cdot (\nabla\times(\mathbf{u} g)) \mathrm{d}x\mathrm{d}y\mathrm{d}z = 0 $$

I stumbled across this problem in a text about vector field decompostion, where this was stated but left to be proven by the reader.

I have tried simplifying the integrand using standard vector calculus identities, but it seems that the integrand in the general case becomes not equal $0$ and that also the intergation has to be carried out in order to prove it.

Any ideas?

Answer 1

Using Einstein summation & index notation, \begin{align*} &\quad \ (\nabla\times\nabla\times(\mathbf{u} f))\cdot (\nabla\times(\mathbf{u} g))\\ &=[\epsilon_{ijk} \partial_i (\epsilon_{jlm} \partial_l (u_m f))] [\epsilon_{kpq} \partial_p (u_q g) ]\\ &= [(\delta_{kl} \delta_{im} - \delta_{km} \delta_{il} )\partial_i \partial_l (u_m f) ] [\epsilon_{kpq} \partial_p (u_q g) ] \\ &= [(\partial_i \partial_k (u_i f) - \partial_i \partial_i (u_k f)) ] [\epsilon_{kpq} \partial_p (u_q g)] \\ &\sim - \epsilon_{kpq}\partial_p [(\partial_i \partial_k (u_i f) - \partial_i \partial_i (u_k f)) ] u_q g \\ &= \epsilon_{kpq} \{ \partial_p \partial_k \partial_i (u_i f) u_q g - u_k u_q \partial_p (\partial_i \partial_i f) g \} \\ &= 0 \text{ by antisymmetry of } \epsilon \end{align*}

In the step marked "$\sim$" we have discarded a total derivative: $$\partial_p \{ [\partial_i \partial_k (u_i f) - \partial_i \partial_i (u_k f)) ] \epsilon_{kpq} u_q g \}. $$ This will lead to a boundary term when integrated, and assuming that all fields fall off quickly enough as $|\mathbf{r}| \to \infty$, it will vanish when integrated over all of $\mathbb{R}^3$.

Answer 2

From Proof for the curl of a curl of a vector field we have $\nabla\times \nabla \times v= \nabla(\nabla\cdot v) - \Delta v $. Using this formula, the (integral of the) first term can be dropped as we can use that the adjoint of the gradient is the negative divergence, i.e. integration by parts, i.e.

$$ \int \nabla A \cdot B = -\int A \nabla\cdot B$$ with $A=\nabla\cdot (u f), \ B=\nabla\times (u g)$ and then use that $\nabla\cdot\nabla\times \equiv 0$.

(The integration by parts formula is precisely due to the divergence theorem and the vanishing of our functions at infinity $\int \nabla\cdot (A B) = 0$ and the product rule $\nabla\cdot (A B) = \nabla A \cdot B + A\nabla \cdot B$ when $A$ is a scalar field and $B$ is a vector field. $\nabla\cdot\nabla\times \equiv 0$ is an explicit calculation that you can do from high school. Or use antisymmetry.)

So we now have (minus one times) $$\int \Delta (uf) \cdot \nabla\times (ug) $$ Since $\Delta$ acts component-wise this is $$\int (\Delta f) (u \cdot \nabla\times (ug)) $$ Now we expand with indices $\let\del\partial$ $$u \cdot \nabla\times (ug) = u_i \epsilon_{ijk} \del_j (u_k g) = u_i \epsilon_{ijk} (\del_j g) u_k = u\cdot (\nabla g\times u) $$ whence the result, as $v\times w$ is orthogonal to $w$.

If you follow the above on paper, you should obtain the following calculations.

\begin{align} &\int (\nabla\times\nabla\times (u f)) \cdot \nabla\times (ug) \\ &=\int (\nabla(\nabla\cdot (u f)) - \Delta(u f)) \cdot \nabla\times (ug) \\ &=\int \nabla(\nabla\cdot (u f))\cdot \nabla\times (ug) - \int \Delta(u f) \cdot \nabla\times (ug) \\ &=-\int \nabla\cdot (u f)\nabla \cdot \nabla\times (ug) - \int \Delta f (u \cdot \nabla g \times u \\ &=0. \end{align} Note - at no point is it claimed that the original integrand is identically zero.

Answer 3

For notational simplicity, let

$$ \mathbf{F} = \nabla \times (f \mathbf{u}) \qquad\text{and}\qquad \mathbf{G} = \nabla \times (g \mathbf{u}) $$

Also, let $\partial_{\mathbf{u}} = \mathbf{u} \cdot \nabla$ denote the directional derivative in the direction of $\mathbf{u}$. Then utilizing the vector cross-product identity $\nabla \times (\nabla \times \mathbf{A}) = \nabla (\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A} $ and noting that $\mathbf{u}$ is a constant vector, we get

$$ \begin{align*} \mathbf{G} &= (\nabla g) \times \mathbf{u}, & \nabla \times \mathbf{F} &= \partial_{\mathbf{u}} \nabla f - (\nabla^2 f) \mathbf{u}. \end{align*} $$

This shows that $\mathbf{G}$ is perpendicular to the term $(\nabla^2 f) \mathbf{u}$, hence

$$ \begin{align*} (\nabla \times \mathbf{F}) \cdot \mathbf{G} &= (\partial_{\mathbf{u}} \nabla f) \cdot \mathbf{G}. \end{align*} $$

However, since $\partial_{\mathbf{u}}$ behaves just like other partial derivative operators (indeed, partial derivatives themselves are also directional derivatives!), it commutes with other differential operators, yielding

$$ \begin{align*} (\nabla \times \mathbf{F}) \cdot \mathbf{G} &= (\nabla \partial_{\mathbf{u}} f) \cdot \mathbf{G} \\ &= \nabla \cdot ((\partial_{\mathbf{u}} f) \mathbf{G}) - (\partial_{\mathbf{u}} f) \underbrace{ (\nabla \cdot \mathbf{G}) }_{=0}. \end{align*} $$

In the last step, we utilized the identity $\nabla \cdot (\nabla \times \mathbf{A}) = \mathbf{0} $. Therefore, for any nice bounded region $\Omega \subseteq \mathbb{R}^3$,

$$ \begin{align*} \int_{\Omega} (\nabla \times \mathbf{F}) \cdot \mathbf{G} \, \mathrm{d}x\mathrm{d}y\mathrm{d}z &= \int_{\Omega} \nabla \cdot ((\partial_{\mathbf{u}} f) \mathbf{G}) \, \mathrm{d}x\mathrm{d}y\mathrm{d}z \\ &= \int_{\partial \Omega} ((\partial_{\mathbf{u}} f) \mathbf{G}) \cdot \mathrm{d}\mathbf{A} \\ &\to 0 \qquad \text{as $\Omega \uparrow \mathbb{R}^3$}, \end{align*} $$

assuming $\nabla f$ and $\nabla g$ decay sufficiently fast as $\|\mathbf{x}\| \to \infty$.