Green’s First and Second Identities

Question

I was doing some research on potential theory and Green's identities and noticed that most of the literature I find on the subject tends to define the vector field

$$\tag{1}\vec F=\phi\,\text{grad}(\psi)$$

where $\phi$ and $\psi$ are $C^2$ scalar fields.

Question 1:
If $\vec F$ is supposed to be a conservative field (i.e, curl-free), then wouldn't the $\phi$ term make the curl of the gradient non-zero and hence make $\vec F$ non-conservative? If potential theory is supposed to be built on conservative fields, then what is the purpose of the $\phi$ term? Are we just assuming some general form of $\vec F$?

Question 2: By plugging in $(1)$ into the Divergence theorem in $\mathbb R^3$

$$\iint_{\partial V}\vec F\cdot \hat n\;dS=\iiint_V\text{div}(\vec F)\;dV,$$

we can obtain

$$\tag{G1}\iint_{\partial V}\phi\frac{\partial \psi}{\partial \hat n}\;dS=\iiint_V(\vec\nabla\phi\cdot\vec\nabla\psi+\phi\nabla^2\psi)\;dV,$$
which is known as Green's first identity.

By interchanging $\phi$ and $\psi$ in $(1)$ (i.e, letting $\vec F=\psi\,\text{grad}(\phi)$), we can obtain

$$\tag{G1'}\iint_{\partial V}\psi\frac{\partial \phi}{\partial \hat n}\;dS=\iiint_V(\vec\nabla\psi\cdot\vec\nabla\phi+\psi\nabla^2\phi)\;dV.$$

Subtracting equation $(G1')$ from $(G1)$ will then yield the famous Green's second identity

$$\tag{G2}\iint_{\partial V}\bigg(\phi\frac{\partial \psi}{\partial \hat n}-\psi\frac{\partial \phi}{\partial \hat n}\bigg)\;dS=\iiint_V\big(\phi\nabla^2\psi-\psi\nabla^2\phi\big)\;dV.$$

My question: Why are we allowed to interchange $\phi$ and $\psi$ in $(1)$? Doesn't switching these terms results in a completely new vector field?

I appreciate any and all clarification! 🙂

Answer 1

1. I’m not really sure what you’re asking here. Yes, in general if $\phi,\psi$ are non-trivial functions (e.g. $\phi(x,y,z)=x$, and $\psi(x,y,z)=y$), then the curl of $\phi \,\text{grad}(\psi)$ is non-zero. But I don’t see how this causes any issues for potential theory. This vector field usually only appears as part of an intermediate step for much larger calculations.
2. Green’s first identity tells us that for ALL smooth functions $f,g$, we have \begin{align} \int_{\partial V}f\frac{\partial g}{\partial n}\,dS&=\int_{V}\left(\nabla f\cdot \nabla g+ f\nabla^2 g\right)\,dV\tag{$G1, (f,g)$} \end{align} Now, let us fix two nice functions $\phi,\psi$. The above identity holds for all pairs of smooth functions $(f,g)$. So, in particular it holds for the pair $(\phi,\psi)$. But also, $(\psi,\phi)$ is just as good a pair of functions. Thus, we get two identities \begin{align} \int_{\partial V}\phi\frac{\partial \psi}{\partial n}\,dS&=\int_{V}\left(\nabla \phi\cdot \nabla \psi+ \phi\nabla^2 \psi\right)\,dV\tag{$G1, (\phi,\psi)$}\\ \int_{\partial V}\psi\frac{\partial \phi}{\partial n}\,dS&=\int_{V}\left(\nabla \psi\cdot \nabla \phi+ \psi\nabla^2 \phi\right)\,dV.\tag{$G1, (\psi,\phi)$} \end{align} Now, we can subtract these equations (and note that on the RHS, the first term cancels out due to symmetry of the dot product) to get Green’s second identity for the pair of functions $(\phi,\psi)$ \begin{align} \int_{\partial V}\left(\phi\frac{\partial \psi}{\partial n}-\psi\frac{\partial \phi}{\partial n}\right)\,dS&=\int_{V}\left( \phi\nabla^2 \psi-\psi\nabla^2 \phi \right)\,dV.\tag{$G2, (\phi,\psi)$} \end{align} This is simply the fact that you can interchange two universal quantifiers. Another way of saying this is that you first apply the divergence-theorem reasoning to the vector field $\phi\text{ grad}(\phi)$, and then next you apply that exact same calculation to the vector field $\psi\,\text{grad}(\phi)$ (but really this is redundant… you should convince yourself that the interchange at the end is justified).

If this is not clear, think of the following simpler example: the inequality \begin{align} xy\leq x^2+3y^2 \end{align} holds for all $x,y\in\Bbb{R}$. Therefore, the inequality $xy\leq y^2+3x^2$ holds for all $x,y\in\Bbb{R}$ (i.e I apply the previous inequality with $x$ and $y$ interchanged). Side remark: the inequality above is not sharp, it is just an illuestration (the sharp inequality is $xy\leq \frac{x^2+y^2}{2}$).