Green's third identity - unit inconsistency

Question

I have been trying to understand Green's functions and using them to solve differential equations. I have hit a road block in terms of dimensionality and units.

My understanding is that the Green's function has dimensionality of $L^{-n}$ where $L$ represents length and $n$ is the dimensionality of the problem. So for a 2D problem the Green's function would have dimensionality of -2.

From the definitions I have seen, we have Green's second theorem in 2D

$$\iint(u\nabla^2G-G\nabla^2u)dA = \int(u{\nabla}G-G{\nabla}u).\hat{n}\,dS$$

where $A$ is the area of the problem and $S$ is the boundary. $u$ is some function we wish to find, knowing $\nabla^2u$ and $G$ is the Green's function for the problem.

To get Green's third theorem we use the fact that

$$\nabla^2G(x, x') = \delta(x-x')$$

and

$${\int}f(x)\delta(x-a)\,dx = f(a).$$

We split the area integral of Green's second theorem

$$\iint u\nabla^2G\,dA-\iint G\nabla^2u\,dA = \int(u{\nabla}G-G{\nabla}u).\hat{n}\,dS$$

then apply the two equations above to remove the integral over $u\nabla^2G\,dA$

$$u-\iint G\nabla^2u\,dA = \int(u{\nabla}G-G{\nabla}u).\hat{n}\,dS$$

However, at this point the equation stops being dimensionally consistent. We have lost the dimensionality from multiplying by $G$.

How do we reconcile this?

Answer 1

The dimension $[G]$ of $G$ depends on the problem (the order of the differential operator) and on the dimension of space.

First, since $\int \delta(x) \, d^nx = 1$ we have $[\delta]=L^{-n}.$ Then, from $\nabla^2 G = \delta$ and $[\nabla^2] = L^{-2}$ we get $[G] = L^{2-n}.$ Thus, when $n=2$ we have $[G] = 1.$