Some Applications of Vector Calculus

Introduction and Notation

For many standard applications of partial differential equations, a key ingredient is Green’s
identity, which is useful to view as a version if integration by parts. In what follows Ω is a
bounded region in Rn with smooth boundary ∂Ω. The element of “Volume” in Ω is dV
while the element of “Area” on ∂Ω is dA. The outer normal directional derivative of a
dϕ
function ϕ on ∂Ω is . In the usual cartesian coordinates in Rn , the Laplacian is
dN
∆ϕ = ∇ · ∇ϕ = ∇2 ϕ = ϕx1 x1 + · · · + ϕxn xn .

Green’s First Identity for smooth functions u and v is:

ZZ Z ZZ
∂v
u∆v dV = u dA − ∇u · ∇v dV. (1)
Ω ∂Ω ∂N Ω

In fact, this identity is the simplest way to define the Laplacian as the differential operator
that satisfies (1) for all smooth functions u, v that are zero near the boundary ∂Ω (to kill
the boundary integral). For instance, to find the Laplacian in spherical coordinates, first
compute ∇u · ∇v and dV and then integrate by parts. This avoids the messy computation
of the second derivatives using the chain rule.
To prove (1), use the divergence theorem
ZZ Z
∇ · F dV = F · N dA
Ω ∂Ω

with F = u ∇v , so ∇F = ∇u · ∇v + u∆v .

Problems

1. Maxwell’s equations in a vacuum are

Et = c ∇ × B, Bt = −c ∇ × E, ∇ · B = 0, ∇ · E = 0,

where the constant c is the speed of light.

a) If div B = 0 at t = 0, show this remains true for all t > 0.
b) If div E = 0 at t = 0, show this remains true for all t > 0.
c) Show that E satisfies the wave equation Ett = c2 ∆E (here ∆E means the Lapla-
cian of each component of E. [The identity ∇ × (∇ × F) = ∇(∇ · F) − ∆F for any
vector field F is essential for this.]
d) Show that B satisfies the wave equation Btt = c2 ∆B.

2. Let C denote the unit circle centered at the origin of the plane, and D denote the
circle of radius 5 centered at (2, 1), both oriented counterclockwise. Let Q denote the

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 annular region between
R these curves.
R If a vector field V satisfies div V = 0, show that
the line integral C V · N ds = D V · N ds [This extends immediately to the situation
where C and D are more general curves and Q is the region between them. For fluid
flow it is an expression of conservation of mass, since div V = 0 means there are no
sources or sinks in the region Q.]

3. Let Ω be a bounded region in the plane with smooth boundary ∂Ω. Show that
Z
1
Area(Ω) = 2 x dy − y dx.
∂Ω

Use this to find the area inside the ellipse (x, y) = (a cos θ, b sin θ) for 0 ≤ θ ≤ 2π .

4. Given the vector field V(x, y, z) = (4y, x, 2z) in 3-space.

a) Find the value of the integral
ZZ
(curl V)·N dA
H

where H is the hemisphere x2 + y 2 + z 2 = a2 , z ≥ 0, n is the unit outward

normal, and dA is the element of area.
b) What changes if in part a) you replace H by the disk x2 + y 2 ≤ a2 in the plane
z = 0?
c) What changes if in part a) you replace H by a more general surface in R3 whose
boundary is the circle x2 + y 2 = a2 in the plane z = 0? [One specific – but far too
restrictive – example is if H is the ellipsoid x2 + y 2 + 4z 2 = a2 for z > 0.]

5. Let Ω be a bounded region in the plane, and let ∂Ω be its boundary.

a) Use the divergence theorem (or any related formula you know) to show that for
any smooth function v(x, y)
ZZ Z
∂v
∆v dx dy = ds
Ω ∂Ω ∂N

where ∂v/∂N := ∇v · N is the outer normal directional derivative on ∂Ω.

b) Let u(x, y, t) be a solution of the heat equation ut = ∆u for (x, y) in Ω. Assume
that the boundary, ∂Ω, is insulated, so the outer normal derivative there is zero:
∂u/∂N = 0 on ∂Ω.
ZZ
Show that Q(t) := u(x, y, t) dx dy is a constant. Thus, the average tem-
Ω
perature is a constant for all time.

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6. Let u(x, y, t) be a solution of the heat equation ut = ∆u for (x, y) in Ω. Assume that
∂Ω, u(x, y, t) = 0 for all points (x, y) on the boundary ∂Ω, and t ≥ 0.
a) Show that the function K(t) := 21 Ω u2 (x, y, t) dx dy has the property dK/dt ≤ 0
RR

(this is improved in Problem 11).

b) Use this to show that with these zero boundary conditions, if the initial temperature
is zero, u(x, y, 0) = 0, then u(x, y, t) = 0 throughout Ω for all t ≥ 0.

7. Let u(x, y, t) describe the motion of a vibrating drumhead Ω. A reasonable mathe-

matical model shows that u satisfies the wave equation utt = ∆u in Ω with boundary
condition u(x, y, t) = 0 for all (x, y) on the boundary ∂Ω.
Physical reasoning leads one to define the energy as
ZZ
1
E(t) := (u2 + |∇u|2 ) dA.
2 Ω t

a) Show that energy is conserved: E(t) = E(0). [Hint: Show dE/dt = 0.]
b) If in addition one knows that the initial position u(x, y, 0) = 0 and that the initial
velocity ut (x, y, 0) = 0, show that E(t) = 0 for all t and deduce that u(x, y, t) ≡ 0.
[This is hardly a surprise on physical grounds, but it should be interpreted as
reassuring us that this mathematical model is indeed more plausible.]

8. Let ψ(t) be a scalar-valued function with a continuous derivative for 0 < t < ∞ and
let X = (x, y, z) ∈ R3 . Define the vector field F(X) := ψ(∥X∥)X for all X ̸= 0. Show
that this vector field is conservative by finding a scalar-valued function φ(r) with the
property that F(X) := ∇φ(∥X∥). In particular, this shows that every central force
field is conservative.

9. Let A be a symmetric matrix with eigenvalue λ and corresponding eigenvector v , so

Av = λv .
a) Show that
⟨v, Av⟩
λ= .
∥v∥2
b) If A is positive definite with positive definite square root P , so A = P 2 , where P
is a symmetric matrix, show that
∥P v∥2
λ= .
∥v∥2
c) A standard ingredient in many problems involves the eigenvalues λ and correspond-
ing eigenfunctions u of the Laplacian,
−∆u = λu in Ω with u=0 on ∂Ω,

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 As usual, to be useful one wants numbers λ so that there is a solution u other than
the trivial solution u ≡ 0. Show that
ZZ
|∇u|2 dA
Ω
λ = ZZ .
2
u dA
Ω

In particular, deduce that λ > 0 (so you must somehow exclude λ = 0).
d) If in the previous part one changes the boundary condition to ∂u/∂N = 0, show
that λ ≥ 0. What can you conclude about u if λ = 0?

10. a) Let c(x) be a given smooth function and u(x) ̸≡ 0 satisfy the ordinary differential
equation −u′′ + c(x)u = λu on the bounded interval Ω = {a < x < b} with u = 0
on ∂Ω. Show that
(u′ 2 + cu2 ) dx
R
λ= Ω R 2 .
Ω u dx

b) Let c(x, y) be a given smooth function and u(x, y) ̸≡ 0 satisfy the differential
equation −(uxx + uyy ) + cu = λu on a bounded set Ω ⊂ R2 with u = 0 on the
boundary of Ω. Show that

(|∇u|2 + cu2 ) dx dy
RR
λ = Ω RR 2 . (2)
Ω u dx dy

c) If c(x, y) ≥ 0, and −(uxx + uyy ) + cu = 0 with u = 0 on ∂Ω, show that u(x, y) ≡ 0

throughout Ω. Thus, in (2), λ > 0 unless u(x, y) ≡ 0. The example u′′ + u = 0
on 0 ≤ x ≤ π which has the solution u(x) = sin x shows that the condition c ≥ 0
cannot be removed entirely.
d) How do the result of parts b)-c) change if the boundary condition is changed to
∂u/∂N = 0? Note there are two cases: when c(x, y) > 0 somewhere and when
c(x, y) = 0 throughout Ω.
e) Say the region Ω has a hole – such as the annular region between two circles. Let
Γ be the boundary of the hole and B be the outer boundary of Ω. If the boundary
conditions are now u = 0 on B while ∂u/∂N = 0 on Γ, how does the result of
parts b)-c) change?

11. Let u(x, y, t) be a solution of the heat equation ut = ∆u for (x, y) in Ω with boundary
condition u(x, y, t) = 0 for (x, y) on ∂Ω and t ≥ 0. Let
ZZ
K(t) := u2 (x, y, t) dx dy t ≥ 0.
Ω

In Problem 6 we found that dK/dt ≤ 0. Here we improve this estimate.

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 a) By Problem 14, there is a constant c > 0 depending only on Ω so that for any
smooth function ϕ with ϕ = 0 on ∂Ω the inequality
ZZ ZZ
2
ϕ dx dy ≤ c |∇ϕ|2 dx dy
Ω Ω
′
Use this to show that dK/dt ≤ −2cK(t) and thus e2ct K(t) ≤ 0.


b) Show that K(t) ≤ e−2ct K(0) for all t ≥ 0 and thus K(t) → 0 at t → ∞. Therefore,
in this sense u(x, y, t) → 0 as t → ∞. [One can also show this using separation of
variables].

12. In applying the divergence theorem where the region is all of R3 , the integral over the
boundary is not well defined. Instead, one works on the ball of radius R and then lets
R → ∞.
Suppose V(x, y, z) is a vector-valued function defined everywhere in 3-dimensional
space. Further, suppose that V is differentiable and that for some constant c
c
∥V(x, y, z)∥ ≤
1 + (x2 + y 2 + z 2 )3/2

for all (x, y, z). Show that

ZZZ
∇ · V(x, y, z) dx dy dz = 0. (3)
R3

In other words, if B(0, R) is the ball of radius R centered at the origin, then (3) means
that ZZZ
lim ∇ · V(x, y, z) dx dy dz = 0.
R→∞ B(0,R)

13. Let Ψ(x, y, z, t) be a (complex) solution of the Schrödinger equation

iΨt = −∆Ψ + V (x, y, z)Ψ

where V is a continuous real-valued function. Assume Ψ is sufficiently small at infinity

[in the space variables (x, y, z)] so that the boundary terms when integrating by parts
are zero. Let ZZZ
1
Q(t) := |Ψ(x, y, z, t)|2 dx dy dz.
2 R3

Show that Q(t) = constant. In particular, if Q(0) = 1, then Q(t) = 1 for all t ≥ 0.
[This observation was an important ingredient in Max Born’s successfully interpretation
of Ψ as a probability amplitude whose absolute square is equal to probability density.]

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14. [Zaremba’s Inequality] This problem outlines two proof’s of an important inequal-
ity. Given a bounded domain Ω ⊂ Rn , for any smooth real-valued function ϕ that
vanishes on the boundary: ϕ = 0 on ∂Ω,
Z Z
2 2
ϕ(x) dx ≤ γ |∇ϕ|2 dx, (4)
Ω Ω

where γ(Ω) is the width of Ω and dx is the element of volume. We write x =

(x1 , . . . , xn ). Say Ω is contained in the strip a ≤ x1 ≤ b where γ; = b − a and
extend f (x) to be zero outside Ω.
Proof 1 . Write ϕ1 (x) = ∂ϕ(x)/∂x1 ). Given x = (x1 , . . . , xn ) ∈ Ω, let x1b be the x1
coordinate of a point on the boundary of Ω with the same x2 , . . . , xn coordinate
as x. Since ϕ(x) is zero on the boundary, for any x ∈ Ω, show that
Z x1 Z b
ϕ(x) = ϕ1 (t, x2 , . . . , xn ) dt ≤ |ϕ1 (t, x2 , . . . , xn )| dt.
x1b a

Use the Schwarz inequality to deduce that

Z b
2
ϕ (x) ≤ γ |ϕ1 (t, x2 , . . . , xn )|2 dt.
a

Finally integrate both sides over Ω to find (4).

Proof 2 . This is a variant of Proof 1. Show that for any smooth vector field V
Z Z
ϕ2 ∇ · V dx = −2 ϕ∇ϕ · V dx.
Ω Ω

Let V = (x1 −(a+b)/2, . . . , 0) and use the Schwarz inequality to obtain Zaremba’s
inequality.
[Remark: The best constant c is c = 1/λ1 where λ1 > 0 is the lowest eigenvalue
of −∆u = λu in Ω with u = 0 on ∂Ω].

15. [Marsden-Tromba p. 438–9 # 29-34]

Notation: Let u(x, y) be a smooth function on the plane (actually, we will only use
that the second derivatives are continuous) and D ⊂ R2 be an open region. Given a
point P ∈ D , let Br (P ) be the closed disk of radius r centered at P and contained in
D for 0 < r ≤ R (so just pick R sufficiently small). Define I(r) by
Z
1
I(r) := u ds.
2πr ∂Br (P )

This is just the average of u on this circle. Note that in polar coordinates centered at
P we have ds = r dθ and u(r, θ).

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 a) Show that limr→0 I(r) = u(P ).
b) Let N denote the unit outer normal to ∂Br (P ) and define ∂u/∂N := ∇u · N (this
is the directional derivative of u in the direction of the outer normal). Show that
Z ZZ
∂u
ds = ∆u dA.
∂Br (P ) ∂N Br (P )

ZZ
1
c) Use this to show that I ′ (r) = ∆u dA.
2πr Br (P )
d) Suppose that u is a harmonic function, that is, ∆u = 0 in D . Use the above to
deduce the mean value property of harmonic functions
Z
1
u(P ) = u ds.
2πr ∂Br (P )

This states the the value of u at the center of a disk is the average of its values on
the circumference.
e) From the previous part, deduce the “solid mean value property”
ZZ
1
u(P ) = u dA.
πR2 BR (P )

f) If u is harmonic in D and has a local maximum at some point p in D , show that

u must be a constant in some small disk centered at P .
g) Assuming that D is connected, show that if u is harmonic in D and has its absolute
maximum at some point P in D (so u(P ) ≥ u(Q) for all points Q ∈ D ), then u
must be a constant D . [This is called the strong maximum principle for harmonic
functions.]
Similarly, if u has its absolute minimum at some point P in D , then u must be a
constant in D .

There is also a nice application of line integrals to prove the isoperimetric inequality for
curves in the plane in the article
Chern, S.S., “Curves and Surfaces in Euclidean Space,” pp.16-56 in the book Studies in
Global Geometry and Analysis, MAA Studies in Mathematics, Vol. 4, 1967.
[See also http://www.math.upenn.edu/~kazdan/260S12/notes/isoperimetric-ineq.pdf]

[Last revised: February 27, 2023]