Vector fields, gradient, divergence and curl

Learning goals and definitions

Compute vector differential operators, relate them to local expansion and circulation, and decide when a potential exists. Let $U\subseteq\mathbb R^3$ be open. A vector field is a map $F=(P,Q,R):U\to\mathbb R^3$. For a scalar $C^1$ function $\phi$, its gradient is $\nabla\phi=(\phi_x,\phi_y,\phi_z)$. For a $C^1$ vector field define

$$\operatorname{div}F=P_x+Q_y+R_z, \quad \operatorname{curl}F=(R_y-Q_z,\ P_z-R_x,\ Q_x-P_y).$$

Divergence is a scalar, curl a vector, and the gradient requires a scalar input. These types prevent meaningless substitutions. A potential for $F$ is a differentiable scalar $\phi$ with $F=\nabla\phi$. Adding a constant does not change its gradient; on a connected domain any two potentials differ by a constant.

If $\phi$ is $C^2$, equality of continuous mixed partials gives $\operatorname{curl}(\nabla\phi)=0$. If $F$ is $C^2$, it also gives $\operatorname{div}(\operatorname{curl}F)=0$. Expanding the displayed definitions shows cancellation term by term; the second identity involves second derivatives of the components, not merely first derivatives.

Theorem: a potential on a star-shaped domain

Suppose $U$ is open and star-shaped with respect to zero: $tx\in U$ for every $x\in U$, $0\le t\le1$. If $F\in C^1(U)$ and $\operatorname{curl}F=0$, then

$$\phi(x)=\int_0^1F(tx)\cdot x\,dt$$

is a potential. For a different star center, translate coordinates first.

Proof. Fix $x$. The compact segment $\{tx:0\le t\le1\}$ has a small surrounding neighborhood contained in $U$, permitting differentiation under this finite integral; continuous first derivatives give a uniform difference-quotient bound and convergence on that compact set. For coordinate $j$,

$$\partial_j\phi(x)=\int_0^1\left(F_j(tx)+t\sum_i x_i\partial_jF_i(tx)\right)dt.$$

Zero curl is precisely the symmetry $\partial_jF_i=\partial_iF_j$. The integrand therefore equals $d[tF_j(tx)]/dt$. The one-variable fundamental theorem gives $\partial_j\phi(x)=F_j(x)$, since the value at $t=0$ is zero. This proves existence with explicit domain and regularity assumptions, not just a candidate formula.

For $F=\nabla\phi$ and a piecewise $C^1$ path $r:[a,b]\to U$, the chain rule on each smooth piece gives $\int_rF\cdot dr=\phi(r(b))-\phi(r(a))$. Summing cancels internal endpoints. Thus every closed-path integral is zero. Conversely, let $F$ be continuous on a nonempty open path-connected domain. If every closed piecewise $C^1$ path has zero integral, define $\phi(x)$ by integration from a base point. Concatenation shows path independence; adding a short coordinate segment proves each partial derivative equals $F_j$. This is another potential criterion, with no star-shaped assumption but a stronger integral hypothesis.

Proof

Constructing the potential from path independence

Let $\Omega\subset\mathbb R^m$ be nonempty, open and path connected, and suppose $F:\Omega\to\mathbb R^m$ is continuous. The points reachable from a base point by polygonal paths form a relatively open subset of $\Omega$: a small ball inside $\Omega$ lets us append one line segment. Its complement is relatively open by the same argument; otherwise a reachable point in such a ball would make its center reachable. Connectedness therefore makes every point reachable. Polygonal paths are piecewise $C^1$.

Fix $x_0\in\Omega$ and define $\phi(x)$ as the integral of $F$ along any such path from $x_0$ to $x$. Path independence makes this a single number. For $x\in\Omega$, choose a ball about $x$ contained in $\Omega$. If $e_i$ is the $i$th coordinate vector and $h$ is sufficiently small, append the coordinate segment to obtain

$$\phi(x+he_i)-\phi(x)=\int_0^h F_i(x+te_i)\,dt.$$

After division by $h\ne0$, continuity gives $\partial_i\phi(x)=F_i(x)$ as $h\to0$ from either side. These partial derivatives are continuous because $F$ is, so $\phi$ is $C^1$ and $\nabla\phi=F$. This proves the converse without assuming differentiability of $F$. If $F$ is $C^1$, the potential is $C^2$, and equality of mixed partials gives $P_y=Q_x$ in the plane.

Worked examples and domain warnings

Construct a potential. For $F=(2xy+z,x^2, x)$, the mixed partials match. Integrating $\phi_x=2xy+z$ gives $\phi=x^2y+xz+h(y,z)$. Matching the second and third components forces $h_y=h_z=0$, so $\phi=x^2y+xz+C$ on $\mathbb R^3$.

Zero curl without a global potential. On $U=\{(x,y,z):x^2+y^2>0\}$, set $F=(-y/(x^2+y^2),x/(x^2+y^2),0)$. Direct differentiation gives zero curl. But on $r(t)=(\cos t,\sin t,0)$, $0\le t\le2\pi$, the dot product $F(r)\cdot r'$ equals one, so circulation is $2\pi$. A potential would force zero circulation, so none exists globally. The domain is not star-shaped and has a hole; local differential conditions alone miss that obstruction.

Graded practice

1. Foundation. Compute divergence and curl of $F=(x^2,y^2,z^2)$.
2. Core. Find a potential for $F=(yz,xz,xy)$ and integrate it from $(0,0,0)$ to $(1,2,3)$ along any piecewise smooth path.
3. Core. For $F=(-y,x,0)$, compute the unit-circle circulation and compare with its curl.
4. Proof. Prove that potentials on a connected open set differ by a constant. State why connectedness is needed.

Full solutions

1. Divergence is $2x+2y+2z$, and all curl components are zero. A potential is $(x^3+y^3+z^3)/3$.
2. The potential is $xyz$, verified by differentiation. The path integral is its endpoint difference $6-0=6$.
3. Curl is $(0,0,2)$. The standard counterclockwise unit-circle parametrization gives $F(r)\cdot r'=1$, hence circulation $2\pi$, matching curl flux through the upward disk.
4. Their difference $h$ has zero gradient. On each small ball in the domain, the mean value argument along a segment makes $h$ constant. Thus $h$ is locally constant. The set where $h$ equals its value at one chosen point and its complement are both relatively open; connectedness forces the complement empty. On a disconnected set, different constants can be chosen on different components.

Summary and term definitions

A vector field assigns a vector to each point. Divergence is infinitesimal net outward flux density, while curl measures oriented circulation density. A potential turns line integrals into endpoint differences. Star-shaped means every segment to one center remains in the domain. Check regularity, domain geometry and operator types before inferring a global potential from zero curl.