My task is to derive $\frac{1}{2} T(p)= \int_{\infty} [G(p,p')\frac{\partial T(p')}{\partial n} - T(p')\frac{\partial G(p,p')}{\partial n}] dS , p \in \partial \Omega$(smooth)
from
$T(p)= \int_{\partial\Omega} [G(p,p')\frac{\partial T(p')}{\partial n} - T(p')\frac{\partial G(p,p')}{\partial n}] dS , p \in \Omega $
Note p is the vector (x,y) in two dimensions.
I am struggling to understand the concepts behind this therefore do not have a starting point for this problem. I have undertaken some research online but can not find anything to help me derive this equation.
The first equation is called Green's third identity, and it comes about by integrating the relation $G(p,p') \nabla^2 T(p')=0$ with respect to $p'$, over $\Omega$ with a disk centered at $p$ removed. You integrate by parts twice and then contract the disk to a point, and the separate $T(p)$ term comes from the integral over the tiny circle, which does not go to zero even as the disk is contracted.
You can imitate this derivation with a point located on the boundary, where you exclude a disk centered at the point on the boundary. But now the disk isn't really a disk, it is the intersection of the disk with the domain, which is always a proper subset of the disk no matter how small the disk is.
In general this intersection is some funny shape, but for small radius and smooth boundary, the intersection is approximately a half-disk, essentially because a small enough piece of a smooth boundary is approximately straight. (You can visualize a piece of the boundary playing the role of the diameter of the half-disk.)
Now the $1/2 T(p)$ comes about from the integral over the tiny semicircle, with the 1/2 appearing because this semicircle is half as long as the circle was in the off-boundary case.
Note that there are some subtleties going on here: for example the value of $T(p)$ that is really appearing on the right hand side is the limit of $T(p')$ as $p'$ approaches $p$ from inside the domain.