Index function gradient in 3D

I have written up a proof of a lemma, but it is very heuristic, and it would be cool if you could comment on whether it is cheating or not

There is a domain $\Omega \in \mathbb{R}^3$ consisting of two disjoint subdomains $\Omega_1$ and $\Omega_2$.There is a function defined on $\mathbb{R}^3$, namely

$F(\vec{x}) = Ind(\vec{x} \in \Omega_1)$

where the index function is defined as

$Ind(a) = \begin{cases} 1, \mathrm{if\ a\ is\ True} \\ 0, \mathrm{if\ a\ is\ False} \end{cases}$

I want to compute the integral

$\iiint_{\Omega} \vec{G}(\vec{x}) \cdot \nabla F(\vec{x}) d^3 x$

where $\vec{G}(\vec{x})$ is some well-behaved function on $\Omega$.

My first guess is that

$\nabla F(\vec{x}) = -\vec{n}_{\Omega_1}(\vec{x}) Ind(\vec{x} \in \partial \Omega _1)$

where $\vec{n}_{\Omega_1}$ is the unit outer normal of $\Omega_1$ and $\partial \Omega_1$ is the boundary of the subdomain. My second guess is that I can plug the gradient into the integral and rewrite it as

$\iiint_{\Omega} \vec{G}(\vec{x}) \cdot \nabla F(\vec{x}) d^3 x = -\iint_{\partial \Omega_1} \vec{G}(\vec{x}) \cdot \vec{n}_{\Omega_1}(\vec{x}) d^2 x$

Are the two above steps justified?

• To take gradients (and derivatives in general) you need an open subset of, say, $\mathbb{R}^n$. Thus, $\nabla F$ is defined only in the interior of $\Omega$, where it is clearly zero. – lisyarus Dec 2 '16 at 14:28
• The gradient is indeed defined on $\mathbb{R}^n$, but it is only relevant within $\Omega$, because that is the domain I am integrating over. The integral is not zero, since there is a step between the two subdomains – Aleksejs Fomins Dec 2 '16 at 15:00
• Ok, I edited the definitions a bit so that it is more clearly written – Aleksejs Fomins Dec 2 '16 at 15:04
• The gradient at the "step" is like the derivative at jump discontinuity - that is, it does not exist at the boundary, at least in the usual sence. You'll have to work with distributions, as Joce says in his answer, to make this mathematically meaningfull. – lisyarus Dec 2 '16 at 15:43

$F$ is discontinuous over $\Omega$, thus its gradient is defined only in the sense of distributions, so in some sense what you want is $\nabla F = -n \delta_{\partial\Omega_1}$. This will give you the equality of the integrals provided that $G$ is smooth enough.
• I'm not much into electromagnetism, but you also find this in mechanics across interfaces. Usually, integrals such as $\int G\cdot \nabla F$ arise from applying Green's formula (integration by parts) to $\int \nabla G \cdot F$. So, if you want to avoid regularity difficulties and distributions, you can always split this initial integral over the 2 subdomains and apply interface conditions on $G$ and $\nabla G \cdot n$. – Joce Dec 2 '16 at 15:50