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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! :)

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    $\begingroup$ You are too preoccupied with thinking the name “potential theory” requires that one deal only with conservative fields. Think of Green’s identities as multivariable versions of the product rule, which set you up to do integration by parts. $\endgroup$ Commented Jan 1, 2023 at 0:33
  • $\begingroup$ @Ted Shifrin Yes! Guilty as charged! Thank you for this mathematical point of view. I am trying to find some physical intuition behind this as well, but from what I've read, most of this is just mathematical machinery used to solve PDEs. Potential theory that is. $\endgroup$
    – whitenoise
    Commented Jan 1, 2023 at 0:39

1 Answer 1

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  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 $xy\leq \frac{x^2+y^2}{2}$).

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  • $\begingroup$ Thank you for your reply! With regards to Question 1, my confusion is mostly conceptual. My assumption is that $\vec F$ takes the form given in $(1)$ because is it somehow related to conservative (irrotational) fields. We commonly see such fields written as $\vec F=\vec\nabla \psi$, for $\psi\in C^2$. The curl of this grad field is guaranteed to vanish (i.e, plugging in $\vec F$ back into Stokes' theorem should return 0). However, with the form given by $(1)$, this isn't the case. The quantity may be non-zero and thus make $\vec F$ non-conservative. I hope I made myself clear this time. $\endgroup$
    – whitenoise
    Commented Dec 31, 2022 at 20:00
  • $\begingroup$ @whitenoise no the vector field takes that form merely so you can apply thr divergence theorem and derive Green's identities (or maybe some other fancier identities). As I mentioned in the example, this vector field is not conservative in general (and I doubt any textbook claimed it is) but that's not an issue because it's only part of an intermediate calculation. $\endgroup$
    – peek-a-boo
    Commented Dec 31, 2022 at 20:19
  • $\begingroup$ Interesting. Green's identities apply to any nice-enough field (conservative or not), but if we wanted it to apply to a conservative field, we could set $\vec F =\vec\nabla\psi$. In this case, Green's first identity would be $$\int_S\frac{\partial \psi}{\partial \hat n}\;dS=\int_V\nabla^2\psi \;dV.$$ Beautiful! $\endgroup$
    – whitenoise
    Commented Dec 31, 2022 at 20:41

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