Singular Fields on Punctured Planar Regions

When a Singularity Changes the Problem

A vector field may be smooth at every point of a region except one, yet that one exception can prevent direct application of Green’s theorem. The issue is not simply whether a formula for the curl exists almost everywhere. Green’s theorem requires the field to have the required differentiability throughout a neighborhood of the region on which the theorem is applied.

When a singular point lies inside the region, one standard approach is to remove a small disk around it. The remaining punctured region no longer contains the singular point, so the theorem can apply there if the field is sufficiently smooth on that region and its boundary. The new inner boundary is essential: it contributes to the boundary integral, and its orientation is opposite to that of an outer boundary.

Key idea: If a field is singular inside a region, do not apply Green’s theorem across the singularity. Remove the singular point, apply the theorem to the punctured region when its hypotheses hold, and include the integral around the new inner boundary.

The Hypothesis That Fails

Write a planar vector field as \(\mathbf{F}=(P,Q)\). In its circulation form, Green’s theorem states that if \(D\) is a suitable bounded region and \(P,Q\) have continuous first partial derivatives on an open neighborhood of \(D\), then

\oint_{\partial D} P\,dx+Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA.

The boundary \(\partial D\) is given the positive orientation: as it is traversed, the region lies to the left. For a region with a hole, this means that the outer boundary is counterclockwise and each inner boundary is clockwise.

A singularity is a point where the field is undefined or fails to have the required regularity. If such a point lies in \(D\), then \(P,Q\) are not continuously differentiable on an open neighborhood of \(D\), so the stated hypotheses fail. The fact that a single point has zero area does not repair that failure. Nor is it enough to calculate the curl away from the singular point and then ignore the point in the area integral.

To handle an isolated singularity at \(a\), remove a small disk centered at \(a\). Denote the punctured region by \(D_\varepsilon\), where the disk has radius \(\varepsilon\). If the field is continuously differentiable on an open neighborhood of the closure of \(D_\varepsilon\), Green’s theorem applies to \(D_\varepsilon\). Its positively oriented boundary consists of the original outer boundary together with the new inner boundary, traversed clockwise.

\oint_{\partial D} P\,dx+Q\,dy + \oint_{C_\varepsilon^{\,\mathrm{clockwise}}} P\,dx+Q\,dy = \iint_{D_\varepsilon} \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA.

This identity is the relevant replacement for a direct application across the singularity. In particular, the inner-boundary integral cannot be discarded unless a separate argument shows what it does. It may approach a nonzero value as \(\varepsilon\) shrinks.

Why the Inner Boundary Has the Opposite Orientation

Positive orientation keeps the region on the left of the direction of travel. On an outer boundary, that condition gives counterclockwise travel. On the boundary of a hole, the region is outside the hole, so clockwise travel keeps the remaining region on the left.

The circulation identity above follows directly by applying Green’s theorem to the punctured region, not to the original region containing the singularity. The boundary of that punctured region has two parts: the original outer curve and the clockwise inner curve. Green’s theorem sums the integrals over both parts. Therefore, if the inner integral is nonzero, the outer integral need not equal the area integral of the curl alone.

Locate the singularities.
Check whether any point where the field is undefined or nonsmooth lies in the region or on its boundary.

Check the hypotheses on the proposed region.
Green’s theorem requires the field’s first partial derivatives to be continuous on a neighborhood of the region being used.

Puncture if needed.
Remove a small disk around an interior singularity, making sure the field is smooth on a neighborhood of the remaining region.

Use the full oriented boundary.
Include the inner boundary with clockwise orientation, then evaluate the curl integral on the punctured region.

Worked Example: A Singular Field with Zero Curl Away from the Origin

Worked Example: Puncturing a Disk Around an Angular Field

Consider \(\mathbf{F}=(P,Q)\), where

P(x,y)=\frac{-y}{x^2+y^2}, \qquad Q(x,y)=\frac{x}{x^2+y^2}.

The field is undefined at \((0,0)\). Let \(D\) be the disk \(x^2+y^2\leq 4\). Explain why Green’s theorem cannot be applied directly to \(D\), then compute the circulation around its outer boundary by applying the theorem to the disk with the smaller disk \(x^2+y^2<1\) removed.

Step 1: Check the original region. The singularity \((0,0)\) lies inside \(D\), since \(0^2+0^2\leq4\). The field is not defined there, so the differentiability hypotheses fail on the original disk. Direct application of Green’s theorem is not justified.

Step 2: Compute the curl away from the origin. For \(x^2+y^2>0\),

\frac{\partial Q}{\partial x} = \frac{(x^2+y^2)-2x^2}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}, \qquad \frac{\partial P}{\partial y} = \frac{-(x^2+y^2)+2y^2}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}.

Thus \(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0\) wherever the field is defined.

Step 3: Apply Green’s theorem to the annulus. The punctured region is \(1\leq x^2+y^2\leq4\). The field is continuously differentiable on a neighborhood of this closed annulus. Its outer boundary \(C_2\) is counterclockwise, and its inner boundary \(C_1\) is clockwise. The area integral is

\iint_{1\leq x^2+y^2\leq4} 0\,dA=0.

On a counterclockwise circle of radius \(r\), use \(x=r\cos t\), \(y=r\sin t\), for \(0\leq t\leq2\pi\). Then \(dx=-r\sin t\,dt\), \(dy=r\cos t\,dt\), and

P\,dx+Q\,dy = \frac{-r\sin t}{r^2}(-r\sin t\,dt) + \frac{r\cos t}{r^2}(r\cos t\,dt) = (\sin^2t+\cos^2t)\,dt = dt.

Therefore the counterclockwise circulation on either circle is \(\int_0^{2\pi}1\,dt=2\pi\). The inner boundary of the annulus is clockwise, so its integral is \(-2\pi\). Green’s theorem gives

\oint_{C_2}P\,dx+Q\,dy + \oint_{C_1^{\,\mathrm{clockwise}}}P\,dx+Q\,dy = 2\pi-2\pi=0.

Hence the circulation around the outer boundary is \(2\pi\), since its integral plus \(-2\pi\) equals zero.

Check: Parametrizing the outer circle directly gives the integrand \(dt\) as calculated above, so its circulation is \(\int_0^{2\pi}dt=2\pi\). The result agrees. The zero curl on the punctured annulus does not make the outer circulation zero; the inner boundary contributes \(-2\pi\).

Worked Example: A Singularity Outside the Region

Worked Example: A Disk That Avoids the Singular Point

Let \(D\) be the unit disk \(x^2+y^2\leq1\), and consider

P(x,y)=\frac{x-3}{(x-3)^2+y^2}, \qquad Q(x,y)=\frac{y}{(x-3)^2+y^2}.

The field is singular at \((3,0)\). Determine whether Green’s theorem applies on \(D\), and find the circulation around its counterclockwise boundary.

Step 1: Locate the singularity relative to the disk. For every \((x,y)\in D\), the distance from \((x,y)\) to \((3,0)\) is at least \(3-\sqrt{x^2+y^2}\geq2\). Thus the disk, and some open neighborhood of it, avoids the singularity. The field is continuously differentiable there.

Step 2: Compute the curl. Set \(R=(x-3)^2+y^2\). Then

\frac{\partial Q}{\partial x} = \frac{-2y(x-3)}{R^2}, \qquad \frac{\partial P}{\partial y} = \frac{-2y(x-3)}{R^2}.

Therefore the curl density is zero throughout \(D\). Green’s theorem gives a circulation of \(\iint_D0\,dA=0\).

Check by parametrizing the boundary. Set \(x=\cos t\), \(y=\sin t\), \(0\leq t\leq2\pi\). Then \(dx=-\sin t\,dt\), \(dy=\cos t\,dt\), and the denominator is \(10-6\cos t\). Hence

P\,dx+Q\,dy = \frac{-(\cos t-3)\sin t+\sin t\cos t}{10-6\cos t}\,dt = \frac{3\sin t}{10-6\cos t}\,dt.

The integral over a full period is zero: the denominator is unchanged when \(t\) is replaced by \(2\pi-t\), while \(\sin t\) changes sign. The boundary calculation therefore agrees with Green’s theorem.

Conclusion: A field’s singularity elsewhere in the plane does not prevent Green’s theorem from applying. What matters is whether the required smoothness holds on a neighborhood of the region in question.

Worked Example: A Punctured Annulus with Nonzero Curl

Worked Example: Separating the Inner-Boundary Contribution

On the annulus \(1\leq x^2+y^2\leq4\), consider

P(x,y)=\frac{-y}{x^2+y^2}, \qquad Q(x,y)=\frac{x}{x^2+y^2}+x.

Find the circulation around the complete positively oriented boundary and verify Green’s theorem, including the inner boundary.

Step 1: Find the curl density. The angular terms have equal partial derivatives, as computed above. The added \(x\) in \(Q\) contributes \(\frac{\partial x}{\partial x}=1\), so

\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1.

The area of this annulus is \(\pi(2^2)-\pi(1^2)=3\pi\). Thus the double integral is \(\iint_D1\,dA=3\pi\).

Step 2: Compute the outer-boundary integral. For a counterclockwise circle of radius \(r\), the angular part contributes \(\int_0^{2\pi}1\,dt=2\pi\). The added term contributes \(x\,dy=(r\cos t)(r\cos t\,dt)=r^2\cos^2t\,dt\), whose integral is \(r^2\pi\), since \(\int_0^{2\pi}\cos^2t\,dt=\pi\). At \(r=2\), the outer integral is \(2\pi+4\pi=6\pi\).

Step 3: Compute the inner-boundary integral. Traversed counterclockwise, the radius-one circle would contribute \(2\pi+\pi=3\pi\). The annulus’s inner boundary is clockwise, so its actual contribution is \(-3\pi\).

Step 4: Compare both sides. The complete boundary integral is \(6\pi-3\pi=3\pi\), equal to the area integral \(3\pi\).

Check: Green’s theorem on the annulus gives the same value on both sides. In this example the inner-boundary contribution includes both the singular angular field and the regular added term; its orientation reverses the entire counterclockwise integral.

What the Examples Show

These examples distinguish three situations. A singularity inside the proposed region blocks direct application, even when the curl is zero wherever it is defined. A singularity outside the region is harmless if the field is smooth on a neighborhood of the region. After an interior singularity is removed, Green’s theorem applies to the punctured region under its hypotheses, but the resulting identity includes the inner-boundary integral.

For the angular field \(\left(-y/(x^2+y^2),x/(x^2+y^2)\right)\), the curl is zero away from the origin, yet a counterclockwise circle around the origin has circulation \(2\pi\), regardless of its radius. Consequently, shrinking the inner circle does not make its circulation disappear. In general, one must calculate or otherwise justify the behavior of that boundary term; its limit is not determined merely by the fact that the circle becomes small.

Common pitfall: It is incorrect to say, “The curl is zero except at one point, so Green’s theorem says the circulation is zero.” The theorem cannot be applied across that point. Apply it to a region that excludes the singularity, and retain every boundary component with its correct orientation.

Using the Method Carefully

Before applying this method, identify the singular set and check whether the region being used stays a positive distance from it. When you remove a disk, verify that the field is continuously differentiable on a neighborhood of the resulting closed region, including its boundaries. Then write the complete oriented boundary integral before simplifying. This makes the inner-boundary term visible and reduces the chance of confusing the circulation around a hole with circulation in the opposite direction.

The method does not automatically justify taking the puncture radius to zero. If a limiting statement is needed, the inner-boundary integral and the area integral must have appropriate limits, and the limiting argument must be established. The examples above avoid any such assumption: Green’s theorem is applied to a fixed punctured region with a fixed inner boundary.

Check Your Understanding

Use the hypotheses and orientation rules discussed above to answer these questions.

Why can the zero curl of a field away from one point fail to determine the circulation around a curve enclosing that point?
What regularity must the field have on the punctured region for Green’s theorem to apply there?
Why is the inner boundary of a punctured region traversed clockwise?
For the field \(\left(-y/(x^2+y^2),x/(x^2+y^2)\right)\), what is the circulation on a counterclockwise circle of radius \(r>0\)?
What additional fact is needed before one can discard an inner-boundary integral in a limit as the puncture shrinks?