Powers D.L. Boundary Value Problems and Partial Differential Equations
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328 Chapter 5 Higher Dimensions and Other Coordinates
Figure 9 Nodal curves: The curves in these graphs represent solutions of
φ
mn
(r ,θ) = 0. Adjacent regions bulge up or down, according to the sign. Only
those φ’s containing the factor cos(mθ) have been used. (See the cover photo-
graph.)
3.
List the five lowest frequencies of vibration of a circular membrane.
4. Sketch the function J
0
(λ
n
r) for n = 1, 2, 3.
5. What boundary conditions must the function φ of Eq. (18) satisfy?
6. Justify the derivation of Eqs. (19) and (20) from Eqs. (12)–(14) and (18).
7. Show that
a
0
J
m
(λ
mn
r)J
m
(λ
mq
r)rdr=0, n = q,
if
J
m
(λ
ms
a) =0, s = 1, 2,....
8. Sketch the nodal curves of the eigenfunctions Eq. (21) corresponding to
λ
31
, λ
32
,andλ
33
.
9. In the simple case of symmetric vibrations, we found the eigenfunctions
φ
0n
(r,θ) = J
0
(λ
0n
r),whereJ
0
(λ
0n
a) = 0 for n = 1, 2, 3 ....Thenodal
curves of φ
03
are concentric circles. What are their radii (as multiples
5.8 Some Applications of Bessel Functions 329
Figure 10 Exercise 10.
of a)? What are the radii of the circles that are the nodal curves of φ
0n
(r,θ)
for general n?
10. The nodal curves of φ
mn
(r,θ) are shown in Fig. 10.
a. By examining the figure, determine what values m and n have.
b. What is the numerical value of the eigenvalue λ
mn
(as a multiple of a)
for this eigenfunction?
c. What is the formula for the function φ
mn
(r,θ) whose nodal curves are
shown?
d. What is the frequency of vibration for the drumhead when it is vibrat-
ing in this mode? (“In this mode” means “so that the displacement u
equals a product solution in which this eigenfunction is a factor.”)
5.8 Some Applications of Bessel Functions
After the elementary functions, the Bessel functions are among the most useful
in engineering and physics. One reason for their usefulness is they solve a fairly
general differential equation. The general solution of
φ
+
1 −2α
x
φ
+
λγ x
γ −1
2
−
p
2
γ
2
−α
2
x
2
φ = 0(1)
is given by
φ(x) = x
α
AJ
p
(λx
γ
) +BY
p
(λx
γ
)
.
330 Chapter 5 Higher Dimensions and Other Coordinates
Several problems in which the Bessel functions play an important role follow.
The details of separation of variables, which should now be routine, are kept
to a minimum.
A. Potential Equation in a Cylinder
The steady-state temperature distribution in a circular cylinder with insulated
surface is determined by the problem
1
r
∂
∂r
r
∂u
∂r
+
∂
2
u
∂z
2
=0, 0 < r < a, 0 < z < b, (2)
∂u
∂r
(a, z) = 0, 0 < z < b, (3)
u(r, 0) = f (r), 0 < r < a, (4)
u(r, b) = g(r), 0 < r < a. (5)
Here we are considering the boundary conditions to be independent of θ,sou
is independent of θ also.
Assuming that u = R(r)Z(z) we find that
rR
+λ
2
rR =0, 0 < r < a, (6)
R
(a) =0, (7)
R(0)
bounded, (8)
Z
−λ
2
Z = 0. (9)
Condition(8)hasbeenaddedbecauser = 0 is a singular point. The solution
of Eqs. (6)–(8) is
R
n
(r) = J
0
(λ
n
r), (10)
where the eigenvalues λ
2
n
are defined by the solutions of
R
(a) =λ J
0
(λa) =0. (11)
Because J
0
=−J
1
,theλ’s are related to the zeros of J
1
. The first three eigen-
values are 0, (3.832/a)
2
,and(7.016/a)
2
.NotethatR(0) = J
0
(0) =1.
The solution of the problem Eqs. (2)–(5) may be put in the form
u(r, z) = a
0
+b
0
z +
∞
n=1
J
0
(λ
n
r)
a
n
sinh(λ
n
z)
sinh(λ
n
b)
+b
n
sinh(λ
n
(b −z))
sinh(λ
n
b)
. (12)
5.8 Some Applications of Bessel Functions 331
The a’s and b’s are determined from Eqs. (4) and (5) by using the orthogonality
relation
a
0
J
0
(λ
n
r)J
0
(λ
m
r)rdr=0, n = m.
B. Spherical Waves
In spherical (ρ,θ,φ)coordinates (see Section 5.9), the Laplacian operator ∇
2
becomes
∇
2
u =
1
ρ
2
∂
∂ρ
ρ
2
∂u
∂ρ
+
1
ρ
2
sin(φ)
∂
∂φ
sin(φ)
∂u
∂φ
+
1
ρ
2
sin
2
(φ)
∂
2
u
∂θ
2
.
Consider a wave problem in a sphere when the initial conditions depend only
on the radial coordinate ρ:
1
ρ
2
∂
∂ρ
ρ
2
∂u
∂ρ
=
1
c
2
∂
2
u
∂t
2
, 0 <ρ<a, 0 < t,(13)
u(a, t) = 0, 0 < t,(14)
u(ρ, 0) = f (ρ), 0 <ρ<a,(15)
∂u
∂t
(ρ, 0) = g(ρ), 0 <ρ<a.(16)
Assuming u(ρ, t) = R(ρ)T(t), we separate variables and find
T
+λ
2
c
2
T =0, (17)
ρ
2
R
+λ
2
ρ
2
R =0, 0 <ρ<a, (18)
R(a) =0, (19)
R(0)
bounded. (20)
Again, the condition (20) has been added because ρ = 0 is a singular point.
Equation (18) may be put into the form
R
+
2
ρ
R
+λ
2
R =0,
and comparison with Eq. (1) shows that α =−1/2, γ = 1, and ρ =1/2; thus
the general solution of Eq. (18) is
R(ρ) = ρ
−1/2
AJ
1/2
(λρ) + BY
1/2
(λρ)
.
We know that near ρ =0,
J
1/2
(λρ) ∼ const ×ρ
1/2
,
Y
1/2
(λρ) ∼ const ×ρ
−1/2
.
332 Chapter 5 Higher Dimensions and Other Coordinates
Thus in order to satisfy Eq. (20), we must have B = 0. It is possible to show
that
J
1/2
(λρ) =
2
π
sin(λρ)
√
λρ
, Y
1/2
(λρ) =−
2
π
cos(λρ)
√
λρ
.
Our solution to Eqs. (18) and (20) is, therefore,
R(ρ) =
sin(λρ)
ρ
, (21)
and Eq. (19) is satisfied if λ
2
n
= (nπ/a)
2
. The solution of the problem of
Eqs. (13)–(16) can be written in the form
u(ρ, t) =
∞
n=1
sin(λ
n
ρ)
ρ
a
n
cos(λ
n
ct) + b
n
sin(λ
n
ct)
. (22)
The a’s and b’s are, as usual, chosen so that the initial conditions Eqs. (15)
and (16) are satisfied.
C. Pressure in a Bearing
The pressure in the lubricant inside a plane-pad bearing satisfies the problem
∂
∂x
x
3
∂p
∂x
+x
3
∂
2
p
∂y
2
=−1, a < x < b, −c < y < c,(23)
p(a, y) = 0, p(b, y) = 0, −c < y < c ,(24)
p(x, −c) = 0, p(x, c) = 0, a < x < b.(25)
(Here a and c are positive constants and b = a + 1.) Equation (23) is ellip-
tic and nonhomogeneous. To reduce this equation to a more familiar one, let
p(x, y) = v(x) +u(x, y),wherev(x) satisfies the problem
x
3
v
=−1, a < x < b, (26)
v(a) = 0,v(b) = 0. (27)
Then, when v is found, u must be the solution of the problem
∂
∂x
x
3
∂u
∂x
+x
3
∂
2
u
∂y
2
=0, a < x < b, −c < y < c,(28)
u(a, y) = 0, u(b, y) = 0, −c < y < c,(29)
u(x, ±c) =−v(x), a < x < b.(30)
If we now assume that u(x, y) = X(x)Y(y), the variables can be separated:
5.8 Some Applications of Bessel Functions 333
x
3
X
+λ
2
x
3
X = 0, a < x < b, (31)
X(a) = 0, X(b) =0, (32)
Y
−λ
2
Y = 0, −c < y < c. (33)
Equation (31) may be put in the form
X
+
3
x
X
+λ
2
X = 0, a < x < b.
By comparing to Eq. (1) we find that α =−1, γ = 1, and p = 1 and that the
generalsolutionofEq.(31)is
X(x) =
1
x
AJ
1
(λx) +BY
1
(λx)
.
Because the point x = 0 is not included in the interval a < x < b,thereisno
problem with boundedness. Instead we must satisfy the boundary conditions
Eq. (32), which after some algebra have the form
AJ
1
(λa) +BY
1
(λa) = 0,
AJ
1
(λb) +BY
1
(λb) = 0.
Not both A and B may be zero, so the determinant of these simultaneous
equations must be zero:
J
1
(λa)Y
1
(λb) −J
1
(λb)Y
1
(λa) =0.
Some solutions of the equation are tabulated for various values of b/a.For
instance, if b/a =2.5, the first three eigenvalues λ
2
are
2.156
a
2
,
4.223
a
2
,
6.307
a
2
.
We n ow c an t ake X
n
to be
X
n
(x) =
1
x
Y
1
(λ
n
a)J
1
(λ
n
x) − J
1
(λ
n
a)Y
1
(λ
n
x)
, (34)
and the solution of Eqs. (28)–(30) has the form
u(x, y) =
∞
n=1
a
n
X
n
(x)
cosh(λ
n
y)
cosh(λ
n
c)
. (35)
The a’s are chosen to satisfy the boundary conditions Eq. (30), using the or-
thogonality principle
b
a
X
n
(x)X
m
(x)x
3
dx = 0, n = m.
Notice that Eqs. (31) and (32) make up a regular Sturm–Liouville problem.
334 Chapter 5 Higher Dimensions and Other Coordinates
EXERCISES
1. Find the general solution of the differential equation
x
n
φ
+λ
2
x
n
φ = 0,
where n = 0, 1, 2,....
2. Find the solution of the equation in Exercise 1 that is bounded at x =0.
3. Find the solutions of Eq. (9), including the case λ
2
= 0, and prove that
Eq. (12) is a solution of Eqs. (2)–(4).
4. Show that any function of the form
u(ρ, t) =
1
ρ
φ(ρ +ct) +ψ(ρ −ct)
is a solution of Eq. (13) if φ and ψ have at least two derivatives.
5. Find functions φ and ψ such that u(ρ, t) asgiveninExercise4satisfies
Eqs. (14)–(16).
6. Give the formula for the a’s and b ’s in Eq. (12).
7. What is the orthogonality relation for the eigenfunctions of Eqs. (18)–
(20)? Use it to find the a’s and b’s in Eq. (22).
8. Sketch the first few eigenfunctions of Eqs. (18)–(20).
9. Find the function v(x) that is the solution of Eqs. (26) and (27).
10. Use the technique of Example C to change the following problem into a
potential problem:
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=−f (x ), 0 < x < a, 0 < y < b,
u =0 on all boundaries.
11. In Exercise 10, will the same technique work if f (x) is replaced by f (x, y)?
12. Verify that Eqs. (31) and (32) form a regular Sturm–Liouville problem.
Show the eigenfunctions’ orthogonality by using the orthogonality of the
Bessel functions.
13. Find a formula for the a
n
of Eq. (35).
14. Verify that Eq. (34) is a solution of Eqs. (28)–(30).
5.9 Spherical Coordinates; Legendre Polynomials 335
5.9 Spherical Coordinates;
Legendre Polynomials
After the Cartesian and cylindrical coordinate systems, the one most fre-
quently encountered is the spherical system (Fig. 11), in which
x = ρ sin(φ) cos(θ),
y = ρ sin(φ) sin(θ),
z =ρ cos(φ).
The variables are restricted by 0 ≤ ρ,0≤ θ<2π ,0≤ φ ≤ π .Inthiscoordi-
nate system the Laplacian operator is
∇
2
u =
1
ρ
2
∂
∂ρ
ρ
2
∂u
∂ρ
+
1
sin(φ)
∂
∂φ
sin(φ)
∂u
∂φ
+
1
sin
2
(φ)
∂
2
u
∂θ
2
.
From what we have seen in other cases, we expect solvable problems in
spherical coordinates to reduce to one of the following.
Problem 1. ∇
2
u =−λ
2
u in R, plus homogeneous boundary conditions.
Problem 2. ∇
2
u = 0inR , plus homogeneous boundary conditions on facing
sides (where
R is a generalized rectangle in spherical coordinates).
Problem 1 would come from a heat or wave equation after separating out
the time variable. Problem 2 is a part of the potential problem.
The complete solution of either of these problems is very complicated, but a
number of special cases are simple, important, and not uncommon. We have
already seen Problem 1 solved (Section 5.8) when u is a function of ρ only.
A second important case is Problem 2, when u is independent of the variable θ .
We shall state a complete boundary value problem and solve it by separation
Figure 11 Spherical coordinates.
336 Chapter 5 Higher Dimensions and Other Coordinates
of variables:
1
ρ
2
∂
∂ρ
ρ
2
∂u
∂ρ
+
1
sin(φ)
∂
∂φ
sin(φ)
∂u
∂φ
=0,
0 <ρ<c, 0 <φ<π, (1)
u(c,φ)=f (φ), 0 <φ<π. (2)
From the assumption u(ρ, φ) = R(ρ)(φ), it follows that
(ρ
2
R
(ρ))
R(ρ)
+
(sin(φ)
(φ))
sin(φ)(φ)
=0.
Both terms are constant, and the second is negative, −µ
2
, because the bound-
ary condition at ρ =c will have to be satisfied by a linear combination of func-
tions of φ. The separated equations are
ρ
2
R
−µ
2
R =0, 0 <ρ<c,(3)
sin(φ)
+µ
2
sin(φ) = 0, 0 <φ<π. (4)
Neither equation has a boundary condition. However, ρ =0 is a singular point
of the first equation, and both φ =0andρ = π are singular points of the sec-
ond equation. (At these points, the coefficient of the highest-order derivative
is zero, while some other coefficient is nonzero.) At each of the singular points,
we impose a boundedness condition:
R(0) bounded,(0) and (π) bounded.
Equation (4) can be simplified by the change of variables x = cos(φ),
(φ) =y(x).(Ofcourse,x is not the Cartesian coordinate.) By the chain rule,
the relevant derivatives are
d
dφ
=−sin(φ)
dy
dx
,
d
dφ
sin(φ)
d
dφ
= sin
3
(φ)
d
2
y
dx
2
−2sin(φ) cos(φ)
dy
dx
.
The differential equation becomes
sin
2
(φ)
d
2
y
dx
2
−2cos(φ)
dy
dx
+µ
2
y = 0,
or, in terms of x alone,
1 −x
2
y
−2xy
+µ
2
y = 0, −1 < x < 1. (5)
In addition, we require that y(x) be bounded at x =±1.
5.9 Spherical Coordinates; Legendre Polynomials 337
Solutions of the differential equation are usually found by the power series
method. Assume that y(x) = a
0
+ a
1
x +···+a
k
x
k
+···.Thetermsofthe
differential equations are then
y
= 2a
2
+3 ·2a
3
x +4 ·3a
4
x
2
+··· +(k +2)(k +1)a
k+2
x
k
+···
−x
2
y
=−2a
2
x
2
−··· −k(k −1)a
k
x
k
+···
−2xy
=−2a
1
x −2a
2
x
2
−··· −2ka
k
x
k
−···
µ
2
y = µ
2
a
0
+µ
2
a
1
x +µ
2
a
2
x
2
+··· +µ
2
a
k
x
k
+···.
When this tableau is added vertically, the left-hand side is zero, according to
the differential equation. The right-hand side adds up to a power series, each
of whose coefficients must be zero. We therefore obtain the following relations:
2a
2
+µ
2
a
0
= 0,
6a
3
+
µ
2
−2
a
1
= 0,
(k +2)(k +1)a
k+2
+
µ
2
−k(k +1)
a
k
= 0.
The last equation actually includes the first two, apparently special, cases. We
may write the general relation as
a
k+2
=
k(k + 1) −µ
2
(k +2)(k +1)
a
k
,
valid for k = 0, 1, 2,....
Suppose for the moment that µ
2
is given. A short calculation gives the first
few coefficients:
a
2
=
−µ
2
2
a
0
, a
3
=
2 −µ
2
6
a
1
,
a
4
=
6 −µ
2
12
a
2
, a
5
=
12 −µ
2
20
a
3
,
=
6 −µ
2
12
·
−µ
2
2
a
0
, =
12 −µ
2
20
·
2 −µ
2
6
a
1
.
It is clear that all the a’s with even index will be multiples of a
0
and those with
odd index will be multiples of a
1
.Thusy(x ) equals a
0
times an even function
plus a
1
times an odd function, with both a
0
and a
1
arbitrary.
It is not difficult to prove that odd and even series produced by this process
diverge at both x =±1, for general µ
2
.However,whenµ
2
has one of the spe-
cial values
µ
2
=µ
2
n
=n(n +1), n =0, 1, 2,...,
one of the two series turns out to have all zero coefficients after a
n
.Forin-
stance, if µ
2
= 3 · 4, then a
5
= 0, and all subsequent coefficients with odd