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Let $p_1,...p_n$ be points in $\mathbb{R}^n$. Let $P(x,y), Q(x,y)$ be functions with continuous derivatives in $ D=\mathbb{R}^2\setminus\{p_1,...p_n\}$ such that $Q_x-P_y=1$ for all $(x,y)\in D$. For all $i$ let $C_i$ be a circle of radius $r_i$ around $p_i$ such that $C_i$ doesn't contain $p_j$ for all $j\neq i$. Let $C$ be a circle of radius $R$ around $(0,0)$, such that $\forall i=1,...n, p_i \in C$. Let $I=\int_C (Pdx+Qdy), I_i =\int_C(Pdx+Qdy)$. Express $I$ as a function of $I_1,...I_n, r_1,...r_n, R$.

I was able to prove this considering that $C_1,...C_n$ are all disjoint and $\cup_{i=1}^nC_i \subseteq C$, but wasn't able to prove this when they were not disjoint. Here's the proof for disjoint $C_1,...C_n$ and $\cup_{i=1}^nC_i \subseteq C$:

Let $A$ be $C\setminus \cup_{i=1}^nC_i$. $A$ is a Green region, since it's boundary is $C\cup \cup_{i=1}^nC_i$, and therefore:

$I+\sum_{i=1}^nI_i=\int_C(Pdx+Qdy)+\sum_{i=1}^n\int_{C_i}(Pdx+Qdy)=\int_{\partial A}(Pdx+Qdy)=\iint_A(Q_x-P_y)dxdy=\iint_A1dxdy=\mu (A)$

And therefore, $I=\mu(A)-\sum_{i=1}^nI_i$, but since $C_1,..C_n$ are all disjoint and if we write $B$ to be the interior of $C$ and $B_i$ to be the interior of $C_i$, then, we get that $\mu(A)-\mu(B)-\sum_{i=1}^n\mu(B_i)=\pi(R-\sum_{i=1}^nr_i^2)$, such that $I=\pi(R-\sum_{i=1}^nr_i^2)-\sum_{i=1}^nI_i$.

In the case that $C,C_1,...C_n$ are not disjoint I looked at the case of only two that intersect. Let's assume that $B_1\cap B_2=E\neq \emptyset$. Since $B_1$ only contains $p_1$ of the set $\{p_1,...p_n\}$ and $B_2$ only contains $p_2$ of $\{p_1,...p_n\}$, then $E$ doesn't contain any of the points $\{p_1,...p_n\}$ and therefore, $P,Q$ and their derivatives are continuous on $E$. Also, if $E$ contains more than one point, then it can be proven that $E$'s boundary is a Jordan curve, and therefore, Green's theorem applies on $E$. Therefore, I figured out that $\int_{C_1\cup C_2}(Pdx+Qdy)=\int_{C_1}(Pdx+Qdy)+\int_{C_2}(Pdx+Qdy)-\iint_E(Q_x-P_y)dxdy=I_1+I_2-\mu(E)$.

From here on, I couldn't find a direction that worked.

AnonymouseCat
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1 Answers1

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Write $Pdx+Qdy=:\omega$ for short. Denote by $B_{i,\epsilon}$ the disk of radius $\epsilon>0$ around the point $p_i$. Choose $\epsilon<\min_i r_i$ so small that the $B_{i,\epsilon}$ do not intersect the large circle $C$.

Green's theorem, applied to the annular regions bounded by $C_i$ and $\partial B_{i,\epsilon}$ gives $$\int_{C_i}\omega\ -\int_{\partial B_{i,\epsilon}}\!\!\omega\ =\ {\rm area}\bigl(B_{i, r_i}\setminus B_{i,\epsilon}\bigr)=\pi(r_i^2-\epsilon^2)\qquad(1\leq i\leq n)\ .\tag{1}$$ We now apply Green's theorem to the "Emmentaler cheese" region generated by removing the $n$ small disks $B_{i,\epsilon}$ from the large disk $B$ bounded by $C$: $$\int_C\omega\ -\sum_{i=1}^n \int_{\partial B_{i,\epsilon}}\!\!\omega\ =\pi(r^2-n\epsilon^2)\ .\tag{2}$$ Subtracting the sum of the equations $(1)$ from $(2)$ we obtain $$I -\sum_{i=1}^n I_i=\pi\left(r^2-\sum_{i=1}^n r_i^2\right)\ ,$$ since all other contributions cancel. It follows that $$I=\pi r^2+\sum_{i=1}^n (I_i-\pi r_i^2)\ ,$$ whether or not some of the $C_i$ intersect $C$.

Up to this point there was no limit $\epsilon\to0$ involved. But note that you can consider $$I_i-\pi r_i^2=\lim_{\epsilon\to0}\int_{\partial B_{i,\epsilon}}\omega$$ as a kind of residue of $\omega$ at the singularity $p_i$.

Christian Blatter
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