Multiply Connected Domains

The vortex field puzzle, picked back up

We've met the vortex before, at the end of §5.6:

$$\mathbf{u}_{\text{vortex}} \;=\; \left\langle \frac{-y}{x^2+y^2},\; \frac{x}{x^2+y^2}\right\rangle, \qquad D = \mathbb{R}^2 \setminus \{(0,0)\}.$$

A direct calculation gives $\partial P/\partial y = \partial Q/\partial x$ at every point of $D$ - the curl condition holds everywhere the field is defined. By Theorem III's converse we'd love to conclude path independence. But Theorem IV demanded one more thing - the domain has to be simply connected - and the punctured plane is not. It has a hole at the origin. And the unit-circle calculation,

$$\oint_C \frac{-y\,dx + x\,dy}{x^2+y^2} \;=\; \int_0^{2\pi} 1\,dt \;=\; 2\pi \;\neq\; 0,$$

certifies that the integral really does depend on the path on this domain. So Theorem IV failed - exactly where its hypothesis failed.

Here's the question that §5.7 sets out to answer. If the integral isn't a single function of endpoints anymore, what does it depend on? Looking at the hook: A and B both gave $2\pi$ even though one is a tiny circle and the other is some lopsided medium loop. C gave $0$. D - the loop wrapping the origin twice - gave $4\pi$. The shape doesn't matter; the holes you wrap and the number of wraps do. That's the whole story, and we're about to make it precise.

Domains with holes - the vocabulary

First, language. A planar domain $D$ (open and connected) is called $n$-tuply connected if its boundary consists of exactly $n$ disjoint simple closed curves. Equivalently: $D$ has $n - 1$ holes punched out of a single outer boundary.

• $n = 1$ - simply connected. Just an outer boundary. Disk, half-plane, square, all of $\mathbb{R}^2$.
• $n = 2$ - doubly connected. Outer boundary plus one hole. The annulus. The punctured plane $\mathbb{R}^2\setminus\{0\}$ (think of the "outer boundary" as the circle at infinity).
• $n = 3$ - triply connected. Outer boundary plus two holes. A disk with two pebbles removed.

Each shaded region is the domain $D$. Its boundary is the union of the highlighted closed curves.

Why does this matter? Theorem IV (§5.6) needed simply connected so that for any closed curve $C$ in $D$, the region inside $C$ also lies in $D$ and Green's theorem can fire. As soon as $D$ has a hole, you can draw a loop around the hole - and the inside is no longer entirely in $D$. The proof breaks. We need a replacement.

The deformation theorem

Here is the central statement of §5.7. Kaplan calls it just "the theorem." We'll call it the deformation theorem because of how we'll use it: it lets us deform the curve we integrate over, as long as we don't cross any holes.

Read the second equation again - it's the headline. Outer-curve integral equals the sum of inner-curve integrals, whenever the region between them is curl-free. The integrals over loops that all enclose the same set of holes are tied together: they don't have to be equal pointwise, but their sum-relations are forced.

The geometric punch line is even better. Take any two simple closed CCW curves $C$ and $C'$ in $D$ that enclose the same holes in the same order, so that we can find a curl-free annular region between them. Apply the theorem with $C_0 = C$ and $C_1 = C'$: we get $\oint_C = \oint_{C'}$. We can deform one curve into the other without changing the integral. The integral depends only on which holes $C$ wraps - not on the shape of $C$.

That's why loops A and B in the hook gave the same answer: both wrap the origin once, CCW. They don't have to look alike.

Why it's true - the cut trick

The proof is a beautiful little construction. We can't apply Green's theorem directly to the annular region $R$ because it's not simply connected. So we make it simply connected by hand.

For each inner curve $C_k$ we draw an auxiliary arc - a "cut" - from $C_0$ to $C_k$, picking the cuts so they don't cross each other. With one cut per hole, the original ring-with-holes region $R$ splits into a finite collection of simply connected sub-regions $R_1, \ldots, R_m$. (For a single hole, one cut already does the job - see the figure.)

Now Green's theorem applies on every simply connected piece. Adding the closed-loop integrals around each piece's boundary,

$$\sum_{j} \oint_{\partial R_j} P\,dx + Q\,dy \;=\; \sum_{j} \iint_{R_j} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA \;=\; \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA.$$

And here's the cancellation. Each cut-arc is shared by two adjacent sub-regions, and the sub-region boundaries traverse it in opposite directions (we always keep the sub-region on our left). So the contributions of every cut-arc cancel pairwise. What's left is the integral around the outer curve $C_0$ in the CCW direction minus the integrals around the inner curves $C_1, \ldots, C_n$ in the CCW direction (since to keep $R$ on our left, we traverse the inner curves clockwise).

Putting it together,

$$\oint_{C_0} P\,dx + Q\,dy \;-\; \sum_{k=1}^n \oint_{C_k} P\,dx + Q\,dy \;=\; \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA.$$

Exactly the theorem. The cuts were a piece of accounting; once they cancelled, the curl-free hypothesis did the rest.

The vortex revisited - winding number

Let's apply the deformation principle to the vortex and see exactly why the hook integers came out as they did.

We're working on $D = \mathbb{R}^2 \setminus \{0\}$, with $\partial Q/\partial x = \partial P/\partial y$ on all of $D$. There's one hole, at the origin. Take any small CCW circle $C_*$ around the origin: we computed

$$\oint_{C_*} \frac{-y\,dx + x\,dy}{x^2+y^2} \;=\; 2\pi.$$

Call this number $\alpha$ - the fundamental period of the form around the origin. It's a property of the singularity, not of the loop we used: by the deformation theorem, every CCW simple closed curve around the origin in $D$ gives the same $2\pi$. Loops A and B in the hook both wrap the origin once CCW, so both equal $\alpha = 2\pi$. Loop C avoids the origin, so it lies in a simply connected sub-domain (e.g. a half-plane not containing $0$); Theorem IV gives $\oint = 0$ there.

What about loop D, which wraps the origin twice CCW? It's not a simple closed curve - it self-intersects, or doubles back on itself. Decompose it into two simple loops, each wrapping the origin once CCW; each contributes $\alpha = 2\pi$, total $4\pi$. More generally:

Three remarks about $n(C, \mathbf{0})$ that are worth pausing on. First, it's an integer, always. The integral comes out a multiple of $2\pi$ because polar angle is well defined modulo $2\pi$ - going around the loop, $\theta$ has to return to a value differing from its start by a multiple of $2\pi$. Second, it depends only on the homotopy class of $C$ in $D$: any two loops you can deform into each other through $D$ have the same winding number. Third, you can read it off a picture by counting net CCW crossings of a ray from $0$. No integral needed.

So the hook table reads as a winding-number census: $n_A = 1, n_B = 1, n_C = 0, n_D = 2$. Multiply by $\alpha = 2\pi$. Done.

How to evaluate $\oint$ over a multiply connected domain

Putting it all into a recipe. You're handed a closed curve $C$ in some domain $D$, and an integral $\oint_C P\,dx + Q\,dy$ to compute. Here's the decision tree.

1. Compute $Q_x - P_y$. If it's nonzero on $C$ or on the region enclosed, your best move is usually plain Green's theorem - reduce $\oint$ to a 2D integral $\iint(Q_x - P_y)\,dA$. Done.
2. If $Q_x = P_y$ everywhere on $D$, look at the holes (singularities of $P, Q$).
   • Zero holes (i.e. $D$ already simply connected): Theorem IV gives $\oint_C = 0$ for free.
   • One or more holes $A_1, \ldots, A_k$: for each hole compute the fundamental period $\alpha_j = \oint$ of a small CCW loop encircling $A_j$ alone. Then for any CCW closed curve $C$ in $D$, $$\oint_C P\,dx + Q\,dy \;=\; \sum_{j=1}^k m_j(C)\,\alpha_j,$$ where $m_j(C)$ is the winding number of $C$ about $A_j$.

This is the deformation theorem in working dress. Steps: catalog the singularities, compute one period per singularity (these are the constants of the field, computed once), then read the integral off any new loop just by counting wraps.

If we'd had to integrate $\int_0^{2\pi} \cos^2 t / (1 + 2t \sin t \cos t + t^2 \sin^2 t)\,dt$ directly - which is what Step 3 would have given without the deformation theorem - we'd still be sweating. The whole point of §5.7 is that the topology does the integration for us.