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Green’s Theorem in the Plane · Tutorial 812 of 999

Divergence as Outward Flux per Unit Area

Learn how divergence measures local net outward flux per unit area and how its sign and magnitude describe flux across a planar boundary.

Advanced 9 min read

What You'll Learn

  • Define divergence for a planar vector field and state the conditions for its use.
  • Derive divergence as the limiting outward flux per unit area near a point.
  • Relate total boundary flux to the area integral of divergence using Green’s theorem.
  • Interpret positive, negative, and zero divergence without overclaiming what happens along each boundary segment.
  • Distinguish divergence at a point from the average divergence over a region.

From Boundary Flux to Local Density

A vector field can carry flow across a boundary. The outward flux through a closed curve records the net amount crossing that curve in the outward direction: outward contributions count positively, and inward contributions count negatively.

For a planar field \(\mathbf{F}(x,y)=(P(x,y),Q(x,y))\), Green’s theorem expresses this boundary flux as an area integral. The integrand is the field’s divergence. The important interpretation is local: divergence measures net outward flux per unit area near a point. It does not measure how much flux crosses the entire boundary of a large region; that total depends on the divergence throughout the region.

Definition: If \(P\) and \(Q\) have continuous first partial derivatives, the divergence of \(\mathbf{F}=(P,Q)\) is \[ \nabla\cdot\mathbf{F}=P_x+Q_y. \] At a point, positive divergence indicates local net outward flow, negative divergence indicates local net inward flow, and zero divergence indicates no local net production of outward flux per unit area.

Green’s Theorem and the Flux Convention

Let \(D\) be a bounded planar region with a positively oriented, piecewise-smooth simple boundary \(C\). Positive orientation means that \(C\) is traversed counterclockwise, so the region stays to the left as the boundary is followed. Suppose \(P\) and \(Q\) have continuous first partial derivatives on an open set containing \(D\). The flux form of Green’s theorem states that the outward flux across \(C\) is the area integral of the divergence:

$$ \oint_C P\,dy-Q\,dx = \iint_D (P_x+Q_y)\,dA. $$

The boundary expression has this form because, along a positively oriented curve, the outward normal multiplied by arc length is \((dy,-dx)\). Thus the field’s outward component multiplied by arc length is \(P\,dy-Q\,dx\). The integral adds those signed contributions around the boundary.

The theorem makes a local-to-global connection. The divergence describes the flux density at each point of the region; integrating that density over the area gives the net outward flux through the whole boundary.

Why Divergence Is Flux per Unit Area

Consider a small rectangle centered at \((a,b)\), with horizontal side length \(h\) and vertical side length \(k\). Assume \(P\) and \(Q\) have continuous first partial derivatives near the rectangle. The outward flux through its two vertical sides is the value of \(P\) on the right side integrated along that side, minus the corresponding value on the left. The horizontal sides contribute the top value of \(Q\) minus the bottom value, integrated across the width.

Dividing the total flux by the rectangle’s area \(hk\) gives an average of the two relevant partial derivatives over the rectangle. To see this precisely, the vertical-side contribution divided by \(hk\) is

$$ \frac{1}{hk}\int_{b-k/2}^{b+k/2} \bigl(P(a+h/2,y)-P(a-h/2,y)\bigr)\,dy = \frac{1}{hk}\iint_{R_{h,k}}P_x(x,y)\,dA. $$

The equality follows from the Fundamental Theorem of Calculus applied in the \(x\)-direction: for each fixed \(y\), the difference of the two boundary values is the integral of \(P_x\) between them. Applying the same reasoning in the \(y\)-direction shows that the horizontal-side contribution divided by \(hk\) is \(\frac{1}{hk}\iint_{R_{h,k}}Q_y\,dA\). Therefore the flux per unit area of the rectangle is

$$ \frac{\text{outward flux across }\partial R_{h,k}}{hk} = \frac{1}{hk}\iint_{R_{h,k}}(P_x+Q_y)\,dA. $$

As \(h\) and \(k\) shrink to zero, continuity of \(P_x+Q_y\) implies that its average over the rectangle tends to its value at the center. Hence

$$ \lim_{\substack{h\to 0\\k\to 0}} \frac{\text{outward flux across }\partial R_{h,k}}{hk} = P_x(a,b)+Q_y(a,b). $$
Interpretation: Divergence at a point is the limiting outward flux per unit area around that point. For a small region, its flux-to-area ratio is approximately the divergence near the region’s center; the approximation becomes exact in the limit as the region shrinks.

Worked Example: Constant Positive Divergence

Worked Example: Constant Positive Divergence

Let \(\mathbf{F}(x,y)=(2x,3y)\), and let \(D\) be the disk \(x^2+y^2\leq 4\). First compute the divergence:

$$ P_x=2,\qquad Q_y=3,\qquad \nabla\cdot\mathbf{F}=2+3=5. $$

The disk has area \(4\pi\). Green’s theorem therefore gives total outward flux \(5(4\pi)=20\pi\). Since the divergence is constantly \(5\), the outward flux per unit area is \(5\) throughout the disk.

Check the boundary flux directly. Parametrize the counterclockwise circle by \(x=2\cos t\), \(y=2\sin t\), for \(0\leq t\leq 2\pi\). Then \(dx=-2\sin t\,dt\), \(dy=2\cos t\,dt\), \(P=4\cos t\), and \(Q=6\sin t\). Thus

$$ \begin{aligned} \oint_C P\,dy-Q\,dx &=\int_0^{2\pi}\left(4\cos t\cdot 2\cos t -6\sin t\cdot(-2\sin t)\right)\,dt\\ &=\int_0^{2\pi}(8\cos^2 t+12\sin^2 t)\,dt\\ &=8\pi+12\pi\\ &=20\pi. \end{aligned} $$

This agrees with the area integral. The positive divergence describes net outward flow locally, and its constant value times the disk’s area gives the total outward flux.

Sign and Magnitude over a Region

For a region \(D\) of positive area, Green’s theorem also gives the average divergence over the region:

$$ \frac{1}{\operatorname{Area}(D)} \oint_C P\,dy-Q\,dx = \frac{1}{\operatorname{Area}(D)} \iint_D \nabla\cdot\mathbf{F}\,dA. $$

This is an average, not necessarily the divergence at any particular point. If divergence is positive everywhere in the region, the total outward flux is positive. If it is negative everywhere, the total outward flux is negative. If it changes sign, positive and negative contributions can partly or completely cancel. A large positive value in a small part of the region may also outweigh smaller negative values elsewhere.

Worked Example: Positive Net Flux with Nonconstant Divergence

Let \(\mathbf{F}(x,y)=(x^2,y^2)\) on the rectangle \(D=[0,2]\times[0,1]\). Its divergence is

$$ \nabla\cdot\mathbf{F}=2x+2y. $$

Integrating over \(D\),

$$ \begin{aligned} \iint_D(2x+2y)\,dA &=\int_0^2\int_0^1(2x+2y)\,dy\,dx\\ &=\int_0^2\left(2x+1\right)\,dx\\ &=\left[x^2+x\right]_0^2\\ &=6. \end{aligned} $$

The rectangle has area \(2\), so its average divergence is \(6/2=3\). The divergence is not \(3\) at every point: at \((1,0)\) it is \(2\), while at \((1,1)\) it is \(4\).

Check the total flux along the four sides. On the right side \(x=2\), the outward flux is \(\int_0^1 4\,dy=4\). On the left side \(x=0\), it is \(0\). On the top side \(y=1\), it is \(\int_0^2 1\,dx=2\). On the bottom side \(y=0\), it is \(0\). The total is \(4+0+2+0=6\), as Green’s theorem requires.

Negative Divergence and Net Inward Flow

Worked Example: Constant Negative Divergence

Let \(\mathbf{F}(x,y)=(-2x,-y)\), and take the square \(D=[-1,1]\times[-1,1]\). The divergence is

$$ P_x=-2,\qquad Q_y=-1,\qquad \nabla\cdot\mathbf{F}=-3. $$

The square has area \(4\). Therefore the total outward flux is \((-3)(4)=-12\). The negative sign means that the net flux is inward, and the average outward flux per unit area is \(-3\).

Check by adding the flux on each side. On the right side \(x=1\), \(P=-2\), so the flux is \(-2\cdot 2=-4\). On the left side \(x=-1\), \(P=2\) but the outward normal points left, so the outward flux is again \(-2\cdot 2=-4\). On the top side \(y=1\), \(Q=-1\), giving \(-1\cdot 2=-2\). On the bottom side \(y=-1\), \(Q=1\) but the outward normal points downward, giving \(-1\cdot 2=-2\). Their sum is \(-4-4-2-2=-12\), which matches the area integral.

A Common Misinterpretation: Zero Divergence

Zero divergence means zero net outward flux per unit area locally. It does not mean that the vector field is zero, nor does it mean that there is no flux through any part of the boundary. Flux entering one part of a closed boundary can be balanced by flux leaving another part.

For example, take the constant field \(\mathbf{F}(x,y)=(1,0)\) on the square \([-1,1]\times[-1,1]\). A mistaken conclusion would be: “The divergence is zero, so there is no boundary flux.” In fact, the right side contributes \(1\cdot 2=2\), while the left side contributes \(-1\cdot 2=-2\). The top and bottom contribute zero. The total is \(2-2+0+0=0\), consistent with \(\nabla\cdot\mathbf{F}=0\). The correction is that divergence describes the net balance, not each boundary contribution separately.

What Divergence Tells You—and What It Does Not

  • Positive divergence: a small neighborhood has net outward flux per unit area.
  • Negative divergence: a small neighborhood has net inward flux per unit area.
  • Zero divergence: outward and inward contributions balance locally in the limiting flux-to-area sense.
  • Magnitude: the absolute value measures the strength of that local net flux density.

These statements concern the net flow near a point. They do not say that every vector on the boundary points outward when divergence is positive, or inward when it is negative. Nor does divergence alone give the total flux across a particular boundary without accounting for the area over which the divergence acts. Green’s theorem supplies that relationship by integrating divergence across the region.

Check Your Understanding

Use the definition of divergence and the flux interpretation to answer the following.

  1. For \(\mathbf{F}(x,y)=(3x,-2y)\), find the divergence and interpret its sign.
  2. If divergence is constantly \(4\) on a region of area \(5\), what is the total outward flux across its positively oriented boundary?
  3. Why is the average divergence over a region not necessarily equal to the divergence at its center?
  4. Can a field with zero divergence have nonzero flux across individual sides of a closed boundary? Explain using the constant field \((1,0)\).
  5. State the regularity and boundary-orientation conditions needed to use the flux form of Green’s theorem as presented here.

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