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Below is an excerpt from the book "Partial Differential Equations" by Evans. The underlined equation confuses me. Clearly it is an application of Stokes theorem, and the implication seems to be that if $f$ is any compactly supported smooth function (for simplicity say on all of $\mathbb{R}^n$) then $$\int_{\mathbb{R}^n-B_\epsilon(0)} f_{x^i} d{x}=\int_{\partial B_\epsilon(0)} f\cdot \frac{(-x^i)}{\epsilon} d{S}$$ (here i am just replacing $u\phi$ with $f$ and $\nu^i$ with $ \frac{(-x^i)}{\epsilon} $). But I can't get this equation to come out of stokes theorem. E.g. assume for simplicity $n=2$ (set $(x^1,x^2)=(x,y)$), and $\epsilon =1$. Let $d\theta$ be the 1-form gotten by pulling back (via $\mathbb{R}^2-0\rightarrow S^1, v\mapsto v/|v|$) the volume form on $S^1$. Then stokes gives $$\int_{S^1}f\cdot(-x) d\theta=\int_{\mathbb{R}^2-B_1(0)}d(f\cdot(-x) d\theta)=\int_{\mathbb{R}^2-B_1(0)}\frac{\partial f\cdot(-x)}{\partial x}dx\wedge d\theta+\int_{\mathbb{R}^2-B_1(0)}\frac{\partial f\cdot(-x)}{\partial y}dy\wedge d\theta.$$

Now it seems $dx\wedge d\theta = \frac{x}{x^2+y^2}~~ dx\wedge dy$ and $dy\wedge d\theta = \frac{y}{x^2+y^2}~~ dx\wedge dy$ so this gives

$$\int_{\mathbb{R}^2-B_1(0)}\frac{\partial f\cdot(-x)}{\partial x} \cdot \frac{x}{x^2+y^2} ~~dx dy+\int_{\mathbb{R}^2-B_1(0)}\frac{\partial f\cdot(-x)}{\partial y} \cdot \frac{y}{x^2+y^2}~~dx dy=$$ $$\int_{\mathbb{R}^2-B_1(0)}- \frac{x}{x^2+y^2}\cdot f~~dx dy+\int_{\mathbb{R}^2-B_1(0)}- \frac{x^2}{x^2+y^2}\cdot f_x ~~dx dy+\int_{\mathbb{R}^2-B_1(0)}- \frac{xy}{x^2+y^2}\cdot f_y~~dx dy.$$

I see no cancellations here or any way to make this look like $$\int_{\mathbb{R}^2-B_1(0)}f_x~~dx dy.$$

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    $\begingroup$ Are you sure the underlined equation in the text isn't just integration by parts? $\endgroup$ Mar 22, 2020 at 15:27
  • $\begingroup$ @EricTowers inasmuch as IBP is just FTC which is just Stokes Thm, i agree. Do you mean it might follow trivially from IBP on $\mathbb{R}$? $\endgroup$
    – bart
    Mar 22, 2020 at 15:44

2 Answers 2

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It follows from the divergence theorem: $$ \int_\Omega\nabla\cdot F = \int_{\partial\Omega}n\cdot F, $$ where $n$ is the outward pointing unit normal to $\partial\Omega$.

Apply this with $F=u\phi e_i$ and $\Omega=U-B(0,\varepsilon)$, where $e_i$ is the $i$-th standard vector.

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    $\begingroup$ Ok, this is much simpler than what I wrote - oh well. $\endgroup$
    – user17945
    Mar 27, 2020 at 23:18
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I think it's easier to go the other direction: $$ \begin{split} \int_{\mathbb{R^2}-B_1(0)}f_x\,dx\wedge dy &= \int_{\mathbb{R}^2-B_1(0)}d(f\,dy) = \int_{S^1}f\,dy \\ &= -\int_0^{2\pi} f\,d(\sin\theta) = -\int_0^{2\pi}f\cos\theta\,d\theta = -\int_{\partial B_1(0)} fx\, dS. \end{split} $$ The additional minus sign after the third equality is because $S^1$ has the orientation induced by the outward pointing normal, which points towards the origin.

The reason you're not seeing this in your original calculation is that you need to add an additional term that vanishes when pulled back to $S^1$. In $\mathbb{R}^2-B_1(0)$, $$ \begin{split} &f(-x)\,d\theta = f\left(-\frac{x^2\,dy + xy\,dx}{x^2+y^2}\right) = -f\,dy + \frac{1}{2}y\,d(\ln(x^2+y^2))\\ \implies &f\,dy = f x\,d\theta + \frac{1}{2}y\,d(\ln(x^2+y^2)). \end{split} $$ The second term vanishes when pulled back to $S^1$.

Edit: to address the more general question: taking $i=1$ for simplicity $$ \int_{\mathbb{R}^n-B_\epsilon(0)} f_{x^1} dx^1\wedge\ldots\wedge dx^n = \int_{\mathbb{R}^n-B_\epsilon(0)} d(fdx^2\wedge\ldots\wedge dx^n) = -\int_{\partial B_\epsilon(0)} f\,dx^2\wedge\ldots\wedge dx^n. $$ However $dS = i_{-\nu}dV$ (since $\nu$ is the inward pointing normal), so $$ \begin{split} f\nu^1 dS &= -f\frac{x^1}{\epsilon^2}(x^1 dx^2\wedge\ldots\wedge dx^n - x^2 dx^1\wedge dx^3\wedge\ldots\wedge dx^n + x^3 dx^1\wedge dx^2\wedge\ldots\wedge dx^n\ldots)\\ &= -\frac{f}{\epsilon^2}\big\lbrace(x^1)^2 dx^2\wedge\ldots\wedge dx^n \\ &\qquad\quad+ \frac{1}{2}d(x^1)^2\wedge(-x^2\widehat{dx^2}\wedge dx^3\wedge\ldots\wedge dx^n+x^3dx^2\wedge\widehat{dx^3}\wedge\ldots\wedge dx^n-\ldots)\big\rbrace\\ &= -\frac{f}{\epsilon^2}\big\lbrace ((x^1)^2+(x^2)^2+\ldots+(x^n)^2)\,dx^2\wedge\ldots\wedge dx^n\big\rbrace \\ & \qquad+ \frac{1}{2}d((x^1)^2+(x^2)^2+\ldots+(x^n)^2)\wedge(-x^2\widehat{dx^2}\wedge dx^3\wedge\ldots\wedge dx^n+x^3dx^2\wedge\widehat{dx^3}\wedge\ldots\wedge dx^n-\ldots)\big\rbrace \end{split} $$ Pulling back to the surface of the sphere, the first term simplifies to $-f\,dx^2\wedge\ldots\wedge dx^n$, while the second term vanishes. So $$ \int_{\partial B_\epsilon(0)}f\nu^1 dS = - \int_{\partial B_\epsilon(0)} f\,dx^2\wedge\ldots\wedge dx^n = \int_{\mathbb{R}^n-B_\epsilon(0)}f_{x^1}dx^1\wedge\ldots\wedge dx^n $$

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  • $\begingroup$ thanks so much (+1). what about the general case though? for $n>2$, $dS$ doesnt have a nice expression in terms of all the $dx^1\wedge dx^2 \wedge ... \wedge \widehat{dx^i}\wedge ... \wedge dx^n$, does it? $\endgroup$
    – bart
    Mar 25, 2020 at 7:47
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    $\begingroup$ @bart, if you are undecided about which answer to accept, I would encourage you to accept timur's, since it's clearly the better one. $\endgroup$
    – user17945
    Mar 31, 2020 at 13:05

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