# Green's third identity - unit inconsistency

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?

• "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." What makes you think this? Commented Nov 3, 2023 at 20:11
• You are correct to ask this as I was wrong to think this Commented Nov 8, 2023 at 12:53

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 $$[G] = 1.$$
• Thank-you, this explains it. However, I understand that a 2D laplacian Green's function is $G(x,s)=1/(2\pi)/\sqrt((x-s_x)^2+(y-s_y)^2)$, which implies units of $L^-1$. Does the normalisation constant $1/(2\pi)$ have dimension of $L$? Commented Nov 8, 2023 at 14:40
• @PhilRosenberg. Green's function for the Laplacian in 2 dimensions is $(2\pi)^{-1} \ln\sqrt{x^2+y^2}.$ Commented Nov 8, 2023 at 17:10
• @PhilRosenberg. Consider $$(2\pi)^{-1} \ln \sqrt{\frac{x^2+y^2}{l_0^2}}.$$ Commented Nov 9, 2023 at 20:14