Curl-Free Fields on Simply Connected Domains

From Zero Curl to a Potential

A conservative field has zero scalar curl under the usual differentiability assumptions. The converse is more delicate: zero curl alone does not guarantee a potential on every domain. The shape of the domain matters.

The planar curl test gives a useful sufficient condition. It says that if a continuously differentiable field has zero scalar curl throughout an open simply connected domain, then the field is conservative there. The key idea is that zero curl first gives potentials locally, on small disks. Simple connectivity lets us combine that local information without ambiguity from paths that wind around holes.

Definition: A domain $D\subseteq\mathbb R^2$ is an open, connected set. It is simply connected if every closed curve in $D$ can be continuously contracted to a point while remaining in $D$.

Write a planar vector field as $\mathbf{F}(x,y)=(P(x,y),Q(x,y))$. Its scalar curl is defined by

\operatorname{curl}\mathbf{F}(x,y) = \frac{\partial Q}{\partial x}(x,y) - \frac{\partial P}{\partial y}(x,y).

We use this order of subtraction throughout. A potential for $\mathbf{F}$ is a scalar function $f$ such that $\nabla f=\mathbf{F}$, or equivalently $f_x=P$ and $f_y=Q$. A field with a potential is also called conservative.

Theorem (planar curl test): Let $D$ be an open simply connected domain in $\mathbb R^2$, and let $P,Q$ have continuous first partial derivatives on $D$. If $$ \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y} \quad\text{throughout }D, $$ then there exists a differentiable function $f:D\to\mathbb R$ such that $f_x=P$ and $f_y=Q$. Thus $\mathbf{F}=(P,Q)$ is conservative on $D$.

The hypotheses do separate jobs. Continuous first partial derivatives make the local calculations below valid. Zero curl gives local potentials. Simple connectivity is the global condition that allows those local potentials to fit together into a potential on all of $D$. The conclusion also gives path independence: the line integral of $\mathbf{F}$ between two fixed endpoints does not depend on the chosen path in $D$.

Why Zero Curl Gives a Local Potential

First consider an open disk $B$ contained in $D$. A disk is convex, so the straight segment from its center $a=(a_1,a_2)$ to any point $(x,y)\in B$ stays inside the disk. Define

g(x,y) = \int_0^1 \left[ P\bigl(a_1+t(x-a_1),a_2+t(y-a_2)\bigr)(x-a_1) + Q\bigl(a_1+t(x-a_1),a_2+t(y-a_2)\bigr)(y-a_2) \right]dt.

This is the line integral from the center to $(x,y)$ along that straight segment. We can verify directly that $g_x=P$ and $g_y=Q$, provided $Q_x=P_y$ on $B$. To see the first identity, put $u=x-a_1$, $v=y-a_2$, and abbreviate the point $(a_1+tu,a_2+tv)$ by $z_t$. Differentiating under the integral gives

\begin{aligned} g_x(x,y) &=\int_0^1 \left[P(z_t)+tuP_x(z_t)+tvQ_x(z_t)\right]dt\\ &=\int_0^1 \left[P(z_t)+t\bigl(uP_x(z_t)+vP_y(z_t)\bigr)\right]dt\\ &=\int_0^1\frac{d}{dt}\left[tP(z_t)\right]dt\\ &=P(x,y). \end{aligned}

The second line uses $Q_x=P_y$; the third follows from the chain rule. For the other partial derivative, the calculation is

\begin{aligned} g_y(x,y) &=\int_0^1 \left[tuP_y(z_t)+Q(z_t)+tvQ_y(z_t)\right]dt\\ &=\int_0^1 \left[Q(z_t)+t\bigl(uQ_x(z_t)+vQ_y(z_t)\bigr)\right]dt\\ &=\int_0^1\frac{d}{dt}\left[tQ(z_t)\right]dt\\ &=Q(x,y). \end{aligned}

Here $P_y=Q_x$, and the final equalities in both calculations use the fundamental theorem of calculus. Thus zero curl guarantees a potential on every disk contained in the domain. This local result alone does not yet guarantee one potential on all of $D$.

Why Simple Connectivity Makes the Result Global

We now explain how to pass from local potentials to a potential on the whole domain. The key is to show that the integral around every closed piecewise smooth curve $\gamma$ in $D$ is zero.

Because $D$ is simply connected, $\gamma$ can be continuously deformed to a constant curve within $D$. The deformation can be taken piecewise smooth after subdivision. Its image is compact and lies inside the open set $D$, so it can be covered by small disks contained in $D$. Subdivide the deformation finely enough that each small cell of its parameter square maps into one such disk.

On each disk, the local construction above supplies a potential. Therefore the line integral around the image of each small cell is zero: it is the integral of a gradient around a closed curve. Add these cell integrals. Integrals along shared edges cancel because each shared edge is traversed once in each direction. What remains is the integral around the original curve, together with the integral around the final constant curve. The latter is zero. Hence the original closed-curve integral is zero.

\oint_\gamma \mathbf{F}\cdot d\mathbf{r}=0.

Now fix a base point $p\in D$. Since an open connected subset of the plane is path-connected, every point $z\in D$ can be joined to $p$ by a piecewise smooth path in $D$. Define $f(z)$ to be the line integral of $\mathbf{F}$ along any such path from $p$ to $z$. This definition is independent of the path: following one path forward and the other backward makes a closed curve, whose integral is zero by the argument above.

It remains to check that $f$ is a potential. Near any point $z$, choose a small disk contained in $D$. Within that disk, changing the endpoint by a short horizontal or vertical segment changes the path integral by the integral along that segment. The one-variable fundamental theorem of calculus then gives $f_x(z)=P(z)$ and $f_y(z)=Q(z)$. Thus $\nabla f=\mathbf{F}$ throughout $D$, proving the theorem.

Applying the Curl Test

Worked Example: A Polynomial Field on the Plane

Let $\mathbf{F}(x,y)=(2xy+3,\ x^2+4y)$ on $D=\mathbb R^2$. First identify $P=2xy+3$ and $Q=x^2+4y$. Then

P_y=2x,\qquad Q_x=2x,\qquad \operatorname{curl}\mathbf{F}=Q_x-P_y=2x-2x=0.

The plane is open and simply connected: any closed curve can be contracted to a point by straight-line contraction. The components have continuous first partial derivatives everywhere. All the hypotheses hold, so the theorem guarantees that $\mathbf{F}$ has a potential on $\mathbb R^2$. Consequently its line integrals depend only on the endpoints.

Check: the calculated curl is identically zero for every $(x,y)$, not merely at a selected point, and the domain contains the entire field.

Worked Example: A Field on a Nonconvex Domain

Consider $D=((0,2)\times(0,3))\cup((0,3)\times(0,2))$ and $\mathbf{F}(x,y)=(x^2y+2y,\ x^3/3+2x+3y^2)$. This L-shaped open region is not convex, but it is star-shaped with respect to $(1,1)$: for every point in $D$, the segment joining it to $(1,1)$ remains in at least one of the two rectangles. Therefore $D$ is simply connected.

Set $P=x^2y+2y$ and $Q=x^3/3+2x+3y^2$. The relevant derivatives are

P_y=x^2+2,\qquad Q_x=\frac{3x^2}{3}+2=x^2+2.

Thus $Q_x-P_y=(x^2+2)-(x^2+2)=0$ at every point of $D$. Both components have continuous first partial derivatives on $D$. The curl test applies, so the field has a potential throughout this nonconvex domain.

Check: simple connectivity, rather than convexity, is the domain condition in the theorem. This example satisfies that condition even though some segments between points of $D$ leave $D$.

Worked Example: A Field That Fails the Curl Condition

Let $\mathbf{F}(x,y)=(2y,x)$ on $\mathbb R^2$. Here $P=2y$ and $Q=x$, so

P_y=2,\qquad Q_x=1,\qquad \operatorname{curl}\mathbf{F}=Q_x-P_y=1-2=-1.

The curl is not zero, so the curl test does not establish that this field has a potential. To check the consequence directly, traverse the unit square counterclockwise. On the bottom edge, $y=0$, so $P\,dx+Q\,dy=0$. On the right edge, $x=1$ and $dy$ runs from $0$ to $1$, giving integral $\int_0^1 1\,dy=1$. On the top edge, $y=1$, $P=2$, and $x$ runs from $1$ to $0$, giving $\int_1^0 2\,dx=-2$. On the left edge, $x=0$, so the integral is zero. The total is $0+1-2+0=-1$, not zero. A conservative field would have zero integral around every closed curve, so this field is not conservative.

Check: the nonzero curl and the nonzero closed-curve integral agree; the theorem's zero-curl hypothesis is essential.

A Common Sign Error

A frequent mistake is to compute a derivative combination other than the defined scalar curl. For $\mathbf{F}=(2y,x)$, one might incorrectly calculate $P_x-Q_y=0-0=0$ and claim that the curl test guarantees a potential. But the scalar curl is $Q_x-P_y=1-2=-1$, not $P_x-Q_y$. The square calculation above confirms that the field is not conservative. Always identify $P$ and $Q$ first, then use $Q_x-P_y$ in that order.

Takeaway: Zero scalar curl is a local compatibility condition. On an open simply connected domain, it also guarantees a global potential. If a hypothesis is missing, the theorem gives no conclusion; in particular, zero curl on a domain that is not simply connected cannot be treated as sufficient without further analysis.

Check Your Understanding

For each question, identify the relevant hypotheses before deciding what the curl test implies.

For $\mathbf{F}=(P,Q)$, which partial derivatives are compared in the planar scalar curl, and in what order?
Why does zero curl give a potential on a disk contained in the domain?
In the global argument, what role does simple connectivity play in proving that closed-curve integrals vanish?
Does the curl test guarantee a potential for a field whose curl is nonzero at some point? Explain.
Why can the curl test apply on the nonconvex L-shaped region in the second worked example?