I need help in clarifying a few concepts related to integrals involving the derivatives of indicator functions when they appear inside integrals.
Specifically, I am trying to solve a problem where I have three different sufficiently well-behaved functions:
- A scalar function $f(x):\Omega \subset \mathbb{R}^3 \to \mathbb{R}$.
- A vector-valued function $\mathbf{u}(x):\Omega \subset \mathbb{R}^3 \to \mathbb{R}^3$.
- A second order tensor function $\mathbf{L}(x):\Omega \subset \mathbb{R}^3 \to \mathbb{R}^3\times \mathbb{R}^3$
In this $\Omega$ space, there is a certain bounded domain $V \subset \mathbb{R}^3$, with a smooth, closed surface $D = \partial V$. To all effects, and as far as I can understand it, the surface $D$ is a Caccioppoli set, although I don't know whether this is relevant. In the most idealised case, the domain $V$ may be a sphere, and $D$ its surface.
Associated with the surface $D$, I am given indicator function $\mathbb{I}_D(x)$, defined in the usual way $$ \mathbb{I}_D(x) = \begin{cases} 1 & x \in D \\ 0 & x \notin D \end{cases} $$
Now, I need to consider the following integrals, defined over all points in $\mathbb{R}^3$:
- $$ I_1 = \int_{\mathbb{R}^3} f(x) \nabla \mathbb{I}_D(x) \text{d}x $$
- $$ I_2 = \int_{\mathbb{R}^3} \mathbf{u}(x) \nabla \mathbb{I}_D(x) \text{d}x $$
- $$ I_3 = \int_{\mathbb{R}^3} \mathbf{L}(x) \nabla \mathbb{I}_D(x) \text{d}x $$
where $ \nabla \mathbb{I}_D(x)$ is the gradient of the indicator function (understood in the sense of distributions).
What are the values of these integrals? Can they be reduced to surface integrals on $D$, using a divergence theorem like approach? How does one operate with gradients (or other derivatives) of indicator functions when inside an integral?
My approach: I believe I understand that $ \int_{\mathbb{R}^3} f(x) \mathbb{I}_D(x) \text{d}x $ implies that the integral is really $ \int_{D} f(x) \text{d}x $, that is, the surface integral of $f(x)$.
My reasoning then is (say, for the vector function $\mathbf{u}(x)$), that I can integrate all these integrals by parts, using $\mathbf{u} \nabla \mathbb{I}_D = \nabla \cdot (\mathbb{I}_D \mathbf{u}) - \mathbb{I}_D \nabla \cdot \mathbf{v}$. This leads to: $$ I_2 = \int_{\mathbb{R}^3} \mathbf{u}(x) \nabla \mathbb{I}_D(x) \text{d}x = \int_{\mathbb{R}^3} \nabla \cdot (\mathbb{I}_D(x) \mathbf{u}(x)) \text{d}x - \int_{D} \nabla \mathbf{u} (x) \text{d}x $$ The second integral on the right hand side, $\int_{D} \nabla \mathbf{u} (x) \text{d}x$, is performed on the surface. It is, as far as I understand these things, a surface integral, of the sort some people would write as $$ \int_{D} \nabla \mathbf{u} (x) \text{d}x = \int_D \mathbf{n} \cdot \nabla \mathbf{u} \text{d} D $$ for $\mathbf{n}(x)$ surface $D$'s normal vector.
The first integral on the right hand side, in turn, is the integral of the divergence of a function that is supported over a much smaller region than $\mathbb{R}^3$, so my reasoning is that this integral vanishes, and we get: $$ I_2 = - \int_{D} \nabla \mathbf{u} (x) \text{d}x $$ Presumably, then $$ I_1 = - \int_{D} \nabla f (x) \text{d}x $$ and $$ I_3 = - \int_{D} \nabla \mathbf{L} (x) \text{d}x $$ Am I right? I do feel I am missing some important detail. In particular, I am not confident that the cancellation of $\int_{\mathbb{R}^3} \nabla \cdot (\mathbb{I}_D(x) \mathbf{u}(x)) \text{d}x$ is valid. I would really appreciate it if someone could comment on whether I went wrong, and how to generally operate with the derivatives of indicator functions.