Question in Helmholtz Theorem

In page 555, Appendix B of Intro to electrodynamics by D Griffiths:
[tex]\nabla\cdot\vec F=D,\;\nabla\times\vec F=\vec C\;\Rightarrow\;\nabla\cdot \vec C=0[/tex]
Griffiths let ##\vec F=-\nabla U+\nabla\times \vec W##
[tex]\nabla\cdot \vec F=-\nabla^2U=-\frac{1}{4\pi}\int D\nabla^2\left(\frac{1}{\vec{\vartheta}}\right)d\tau'=\int D(\vec r')\delta^3(\vec r-\vec r')d\tau'=D(\vec r) \;\hbox { where }\;\;\vec{\vartheta}=\vec r-\vec r'[/tex]

My questions are:
(1)Why the theorem use ##D(\vec r),\; and \;\vec C(\vec r)## where ##\vec r## is a position vector pointing at the OBSERVATION point ##P##?
In electrostatic, ##\nabla\cdot\vec E=\frac{\rho(\vec r')}{\epsilon}## where ##\rho(\vec r')## represented the charge density at the SOURCE point pointed by position vector ##\vec r'##. We measure the field at the OBSERVATION point ##P(\vec r)## where position vector ##\vec r## pointing at the point ##P##. Why in Helmholtz Theorem using ##D(\vec r)##?

(2) Why ##D(\vec r)## and ##\vec C(\vec r)## goes to zero when ##|\vec r| \rightarrow \;\infty##? Which, pretty much back to question (1)......why use ##D(\vec r)##?

Thanks