Interpretation of the divergence of a two-dimensional vector field as the rate of change
of small areas flowing along the vector field. Laurent Demanet, Math 52, Winter 07.
Let F (x, y) = (F1 (x, y), F2 (x, y)) be a C 1 vector field in R2 . We will sometimes write (x, y) = x
in what follows. The vector field can be interpreted as the rate of change of position of particles
flowing along the flow lines, because the flow line equation
x0 (t) = F (x(t))
(1)
precisely states that the velocity vector of the particle x(t) is the vector field F evaluated at x(t).
Equation (1) is called a differential equation, and its solution x(t) (which depends on the starting
point x(0)) is called a flow line, or trajectory. In this note, we will see that ∇ · F , the divergence
of F , measures something quite different but equally physical, namely the rate of change of areas
as we follow the same trajectories.
For illustration, we consider a small rectangle defined by the 4 points (0, 0), (∆x, 0), (0, ∆y),
and (∆x, ∆y). We denote this rectangle by R(0), because we take it as the “initial” rectangle at
time t = 0. Its sidelengths are ∆x and ∆y, so its area is
Area(R(0)) = ∆x∆y.
As time goes by, each point inside the rectangle follows the flow (1), so the region R(t) comprising
these points moves with the flow and becomes somewhat distorted. By “rate of change of area”,
we mean the time derivative of the area of R(t). It will be too difficult to compute it exactly, so
instead we will approximate the derivative at t = 0 by
Area(R(∆t)) − Area(R(0))
dArea(R)
(0) '
,
dt
∆t
(2)
where ∆t is vanishingly small. The negligible remainder is of order (∆t)2 . It turns out that
considering a small time step ∆t also allows us to approximate R(t) by a parallelogram, up to an
error that goes to zero as ∆t → 0.
So which parallelogram does the rectangle turn into after a small time ∆t, in good approximation? We can follow each corner point independently. Since ∆t is small, we can also discretize the
derivative in (1) and approximate the flow locally by the tangent line:
x(∆t) ' x(0) + ∆t F (x(0)).
We can apply this evolution rule to the each of the three corner points x1 (0) = (0, 0), x2 (0) =
(∆x, 0), and x3 (0) = (0, ∆y), each taken as the initial position:
(0, 0) 7→ x1 (∆t) ' (0, 0) + ∆t F (0, 0),
(∆x, 0) 7→ x2 (∆t) ' (∆x, 0) + ∆t F (∆x, 0),
(0, ∆y) 7→ x3 (∆t) ' (0, ∆y) + ∆t F (0, ∆y).
Since ∆x and ∆y are also small, we can Taylor expand the vector field around the origin to obtain,
again in good approximation,
(0, 0) 7→ x1 (∆t) ' (0, 0) + ∆t F (0, 0),
∂F
(0, 0),
∂x
∂F
(0, 0).
(0, ∆y) 7→ x3 (∆t) ' (0, ∆y) + ∆t F (0, 0) + ∆t∆y
∂y
(∆x, 0) 7→ x2 (∆t) ' (∆x, 0) + ∆t F (0, 0) + ∆t∆x
(The notation ∂F
∂x means that every component of F is differentiated with respect to x.)
The rectangle R(0) is defined by the two perpendicular vectors x2 (0) − x1 (0) and x3 (0) −
x1 (0). Similarly, the new parallelogram approximating R(t) is defined by the two vectors x2 (∆t) −
x1 (∆t) and x3 (∆t) − x1 (∆t) (which are not in general perpendicular anymore). From the previous
equations, we have
∂F
(0, 0),
∂x
∂F
(0, 0).
x3 (∆t) − x1 (∆t) ' (0, ∆y) + ∆t∆y
∂y
x2 (∆t) − x1 (∆t) ' (∆x, 0) + ∆t∆x
The are of the parallelogram formed by two vectors is the determinant of the matrix whose
columns are the two vectors, so we have
"
!#
∂F1
1
∆x ∂F
(0,
0)
∆y
(0,
0)
∆x 0
∂x
∂y
+ ∆t
Area(R(∆t)) ' det
,
∂F2
2
0 ∆y
(0,
0)
∆y
∆x ∂F
∂x
∂y (0, 0)
∂F2
∂F2
∂F1
∂F1
(0, 0)
∆y + ∆t∆y
(0, 0) − (∆t)2 ∆x∆y
(0, 0)
(0, 0),
= ∆x + ∆t∆x
∂x
∂y
∂x
∂y
∂F1
∂F2
' ∆x∆y + ∆t∆x∆y
(0, 0) +
(0, 0) .
∂x
∂y
(Contributions of order (∆t)2 are considered negligible.) The quantity in parentheses in the last
line is precisely ∇ · F .
If we substitute this result in equation (2), we obtain
dArea(R)
(0) ' ∆x∆y (∇ · F ).
dt
This result shows that the rate of change of area (of a small rectangle) is proportional to the
divergence of F , and also of course to the original area Area(R(0)) = ∆x∆y.
All the approximation signs ' become equality signs as ∆t, ∆x and ∆y tend to zero, because
all the Taylor remainders go to zero. (In a rigorous proof this point would have be be addressed
more carefully.)