I have a question concerning this paper, page 8.
On page 8, it is said that $$ \int_{K_l}1\delta(x-\hat{x})\, dx=\begin{cases}1, & \hat{x}\in K_l\\ 0, & \hat{x}\notin K_l\end{cases}.\tag{1} $$ (Here, $K_l$ denotes a cell or control volume and $\bar{\Omega}=\bigcup_{l=1}^M\bar{K_l}$.)
It is also said that the Dirac delta function is defined by requiring that
- $\delta(x-\hat{x})=0$ for all $x\neq \hat{x}$,
- $\int_{\mathbb{R}^3}\psi(x)\delta(x-\hat{x})\, dx=\psi(\hat{x})$ for any function $\psi\in C^0(\Omega)$.
How do these two conditions imply (1)? Do not see it.
Of course, the first condition implies that $\int_{K_l}1\delta(x-\hat{x})\, dx=0$ if $\hat{x}\notin K_l$. But I do not see that $\int_{K_l}1\delta(x-\hat{x})\, dx=1$ if $\hat{x}\in K_l$.
EDIT:
Is this just splitting the integral, i.e., $$ 1=\int_{\mathbb{R}^3}1\delta(x-\hat{x})\, dx=\underbrace{\int_{\bar{\Omega}}1\delta(x-\hat{x})\, dx}_{=\int_{K_l}1\delta(x-\hat{x})\, dx}+\underbrace{\int_{\bar{\Omega}^C}1\delta(x-\hat{x})\, dx}_{=0} $$ assuming that $\hat{x}\in K_l$ and $\delta(x-\hat{x})=0$ for all $x\neq\hat{x}$?
Firstly, $$ \int_{K_{l}} 1 \delta(x-\hat{x})~dx = \int_{\mathbb{R}^{3}} \mathbf{1}_{x \in K_{l}} \cdot \delta(x-\hat{x})~dx $$
Then using your second property with $\psi = \mathbf{1}_{x \in K_{l}}$ gives
$$ \int_{\mathbb{R}^{3}} \mathbf{1}_{x \in K_{l}} \cdot \delta(x-\hat{x})~dx = \mathbf{1}_{x \in K_{l}} =\begin{cases}1, & \hat{x}\in K_l\\ 0, & \hat{x}\notin K_l\end{cases}$$ as required. However we cannot take $\psi = \mathbf{1}_{x \in K_{l}}$ directly since your property only holds for $\psi$ continuous, but nonetheless we can take a sequence of continuous functions $\{\psi_{n}\}_{n}$ which converge uniformly to $\mathbf{1}_{x \in K_{l}}$ and apply property 2 for this sequence:
$$ \int_{\mathbb{R}^{3}} \psi_{n} \cdot \delta(x-\hat{x})~dx = \psi_{n}(\hat{x}) $$ Passing to the limit $n \to \infty$ leads to the result.
** this is the formal idea, I avoided full rigour because even the definition of the dirac they give is not mathematically rigorous. To fully justify this entire computation we need some distribution theory which I tried to avoid since you don't seem to be familiar with it **