Problems
1. Show that if u, v both satisfy the boundary conditions (6.1.9)-(6.1.10) then
p (uv
− vu
)
|
b
a
= 0.
2.
Show that the right hand side of (6.3.4) is zero even if u, v satisfy periodic boundary
conditions, i.e.
u(a) = u(b)
p(a)u
(a) = p(b)u
(b),
and similarly for v.
3.
What can be proved about eigenvalues and eigenfunctions of the circularly symmetric
heat flow problem.
Give details of the proof.
Note: This is a singular Sturm-Liouville problem.
4. Consider the heat flow with convection
u
t
= ku
xx
+ V
0
u
x
,
0 < x < L,
t > 0.
a.
Show that the spatial ordinary differential equation obtained by separation of
variables is not in Sturm-Liouville form.
b.
How can it be reduced to S-L form?
c.
Solve the initial boundary value problem
u(0, t) = 0,
t > 0,
u(L, t) = 0,
t > 0,
u(x, 0) = f (x),
0 < x < L.
136
6.4
Linearized Shallow Water Equations
In this section, we give an example of an eigenproblem where the eigenvalues appear also
in the boundary conditions. We show how to find all eigenvalues in such a case. The
eigenfunctions relate to the confluent hypergeometric functions.
The shallow water equations are frequently used in simplified dynamical studies of at-
mospheric and oceanographic phenomena. When the equations are linearized, the thickness
of the fluid is often assumed to be a linear function of one of the spatial variables, see Stan-
iforth, Williams and Neta [1993]. The basic equations are derived in Pedlosky [1979]. The
thickness of the fluid layer is given by
H(x, y, t) = H
0
(y) + η(x, y, t)
(6.4.1)
where
|η| H.
If u, v are small velocity perturbations, the equations of motion become
∂u
∂t
− fv = −g
∂η
∂x
(6.4.2)
∂v
∂t
+ f u =
−g
∂η
∂y
(6.4.3)
∂η
∂t
+ H
0
∂u
∂x
+
∂
∂y
(vH
0
) = 0
(6.4.4)
where f is the Coriolis parameter and g is the acceleration of gravity. We assume periodic
boundary conditions in x and wall conditions in y where the walls are at
±L/2. We also let
H
0
= D
0
1
−
s
L
y
with D
0
the value at y = 0 and s is a parameter not necessarily small as long as H
0
is
positive in the domain.
It was shown by Staniforth et al [1993] that the eigenproblem is given by
−
d
dy
1
−
s
L
y
dφ
dy
+ k
2
1
−
s
L
y
φ
− λ(c)φ = 0
(6.4.5)
dφ
dy
+
1
c
f φ = 0
on
y =
±
L
2
(6.4.6)
where
λ(c) =
k
2
c
2
− f
2
gD
0
−
f s
Lc
(6.4.7)
and k is the x-wave number.
Using the transformation
z = 2k
L
s
− y
,
(6.4.8)
137
the eigenvalue problem becomes
d
dz
z
dφ
dz
−
z
4
−
λ(c)L
2sk
φ = 0,
z
−
< z < z
+
(6.4.9)
dφ
dz
−
1
c
f
2k
φ = 0,
z = z
±
,
(6.4.10)
where
z
±
=
2kL
s
1
∓
s
2
.
(6.4.11)
Notice that the eigenvalues λ(c) appear nonlinearly in the equation and in the boundary
conditions.
Another transformation is necessary to get a familiar ODE, namely
φ = e
−z/2
ψ.
(6.4.12)
Thus we get Kummer’s equation (see e.g. Abramowitz and Stegun, 1965)
zψ
+ (1
− z)ψ
− a(c)ψ = 0
(6.4.13)
ψ
(z
±
)
−
1
2
1 +
1
c
f
k
ψ (z
±
) = 0
(6.4.14)
where
a(c) =
1
2
−
λ(c)L
2sk
.
(6.4.15)
The general solution is a combination of the confluent hypergeometric functions M (a, 1; z)
and U (a, 1; z) if a(c) is not a negative integer. For a negative integer, a(c) =
−n, the solution
is L
n
(z), the Laguerre polynomial of degree n. We leave it as an exercise for the reader to
find the second solution in the case a(c) =
−n.
138
Problems
1. Find the second solution of (6.4.13) for a(c) =
−n.
Hint: Use the power series solution method.
2.
a.
Find a relationship between M (a, b; z) and its derivative
dM
dz
.
b.
Same for U .
3.
Find in the literature a stable recurrence relation to compute the confluent hypergeo-
metric functions.
139
6.5
Eigenvalues of Perturbed Problems
In this section, we show how to solve some problems which are slightly perturbed. The first
example is the solution of Laplace’s equation outside a near sphere, i.e. the boundary is
perturbed slightly.
Example Find the potential outside the domain
r = 1 + P
2
(cos θ)
(6.5.1)
where P
2
is a Legendre polynomial of degree 2 and is a small parameter. Clearly when
= 0 the domain is a sphere of radius 1.
The statement of the problem is
∇
2
φ = 0
in
r
≥ 1 + P
2
(cos θ)
(6.5.2)
subject to the boundary condition
φ = 1
on
r = 1 + P
2
(cos θ)
(6.5.3)
and the boundedness condition
φ
→ 0
as
r
→ ∞.
(6.5.4)
Suppose we expand the potential φ in powers of ,
φ(r, θ, ) = φ
0
(r, θ) + φ
1
(r, θ) +
2
φ
2
(r, θ) +
· · ·
(6.5.5)
then we expect φ
0
to be the solution of the unperturbed problem, i.e. φ
0
=
1
r
. This will be
shown in this example. Substituting the approximation (6.5.5) into (6.5.2) and (6.5.4), and
then comparing the coefficients of
n
, we find that
∇
2
φ
n
= 0
(6.5.6)
φ
n
→ 0
as
r
→ ∞.
(6.5.7)
The last condition (6.5.3) can be checked by using Taylor series
1 = φ
|
r=1+ P
2
(cos θ)
=
∞
n=0
(P
2
)
n
n!
∂
n
φ
∂r
n
|
r=1
.
(6.5.8)
Now substituting (6.5.5) into (6.5.8) and collect terms of the same order to have
1 = φ
0
(1, θ) +
φ
1
(1, θ) + P
2
(cos θ)
∂φ
0
(1,θ)
∂r
+
2
φ
2
(1, θ)) + P
2
(cos θ)
∂φ
1
(1,θ)
∂r
+
1
2
P
2
2
(cos θ)
∂
2
φ
0
(1,θ)
∂r
2
+
· · ·
140
Thus the boundary conditions are
φ
0
(1, θ) = 1
(6.5.9)
φ
1
(1, θ) =
−P
2
(cos θ)
∂φ
0
(1, θ)
∂r
(6.5.10)
φ
2
(1, θ) =
−P
2
(cos θ)
∂φ
1
(1, θ)
∂r
−
1
2
P
2
2
(cos θ)
∂
2
φ
0
(1, θ)
∂r
2
(6.5.11)
The solution of (6.5.6)-(6.5.7) for n = 0 subject to the boundary condition (6.5.9) is then
φ
0
(r, θ) =
1
r
(6.5.12)
as mentioned earlier. Now substitute the solution (6.5.12) in (6.5.10) to get
φ
1
(1, θ) = P
2
(cos θ)
1
r
2
|
r=1
= P
2
(cos θ)
(6.5.13)
Now solve (6.5.6)-(6.5.7) for n = 1 subject to the boundary condition (6.5.13) to get
φ
1
(r, θ) =
P
2
(cos θ)
r
3
.
(6.5.14)
Using these φ
0
, φ
1
in (6.5.11), we get the boundary condition for φ
2
φ
2
(1, θ) = 2P
2
2
(cos θ) =
36
35
P
4
(cos θ) +
4
7
P
2
(cos θ) +
2
5
P
0
(cos θ)
(6.5.15)
and one can show that the solution of (6.5.6)-(6.5.7) for n = 2 subject to the boundary
condition (6.5.15) is
φ
2
(r, θ) =
36
35
P
4
(cos θ)
r
5
+
4
7
P
2
(cos θ)
r
3
+
2
5
1
r
.
(6.5.16)
Thus
φ(r, θ) =
1
r
+
P
2
(cos θ)
r
3
+
2
=
36
35
P
4
(cos θ)
r
5
+
4
7
P
2
(cos θ)
r
3
+
2
5
1
r
+
· · ·
(6.5.17)
The next example is a perturbed equation but no perturbation in the boundary.
Example Consider a near uniform flow with a parabolic perturbation, i.e.
u = 1 + y
2
at infinity.
(6.5.18)
In steady, inertially dominated inviscid flow the vorticity ζ is constant along a streamline.
Thus the streamfunction ψ(x, y, ) satisfies
∇
2
ψ =
−ζ(ψ, )
in r > 1,
(6.5.19)
141
subject to the boundary conditions
ψ = 0,
on r = 1,
(6.5.20)
and
ψ
→ y +
1
3
y
3
as r
→ ∞.
(6.5.21)
To find ζ, we note that in the far field
ψ = y +
1
3
y
3
(6.5.22)
and thus
ζ =
−∇
2
ψ =
−2y,
(6.5.23)
or in terms of ψ
ζ =
−2ψ +
2
3
2
ψ
3
+
· · ·
(6.5.24)
Now we suppose the streamfunction is given by the Taylor series
ψ = ψ
0
(r, θ) + ψ
1
(r, θ) +
· · ·
(6.5.25)
Substitute (6.5.25) and (6.5.24) in (6.5.19)-(6.5.20) we have upon comparing terms with no
:
∇
2
ψ
0
= 0
in r > 1,
ψ
0
= 0
on r = 1,
ψ
0
→ r sin θ
as r
→ ∞
which has a solution
ψ
0
= sin θ
r
−
1
r
.
(6.5.26)
Using (6.5.26) in the terms with
1
we have
∇
2
ψ
1
= 2 sin θ
r
−
1
r
in r > 1,
ψ
1
= 0
on r = 1,
ψ
1
→
1
3
r
3
sin
3
θ
as r
→ ∞
The solution is (see Hinch [1991])
ψ
1
=
1
3
r
3
sin
3
θ
− r ln r sin θ −
1
4
sin θ
r
+
1
12
sin 3θ
r
3
.
(6.5.27)
The last example is of finding the eigenvalues and eigenfunctions of a perturbed second
order ODE.
142
Example Find the eigenvalues and eigenfunctions of the perturbed Sturm Liouville prob-
lem
X
(x) + (λ + Λ(x)) X(x) = 0,
0 < x < 1
(6.5.28)
subject to the boundary conditions
X(0) = X(1) = 0.
(6.5.29)
Assume a perturbation for the n
th
eigenpair
λ
n
= λ
(0)
n
+ λ
(1)
n
+
2
λ
(2)
n
+
· · ·
(6.5.30)
X
n
= X
(0)
n
+ X
(1)
n
+
2
X
(2)
n
+
· · ·
(6.5.31)
Substituting these expansions in (6.5.28) and comparing terms with like powers of . For
the zeroth power we have
X
(0)
n
+ λ
(0)
n
X
(0)
n
= 0
X
(0)
n
(0) = X
(0)
n
(1) = 0,
which has the unperturbed solution
λ
(0)
n
= (nπ)
2
,
(6.5.32)
X
(0)
n
= sin nπx.
(6.5.33)
For the linear power of , we have
X
(1)
n
+ (nπ)
2
X
(1)
n
=
−
Λ(x) + λ
(1)
n
sin nπx,
(6.5.34)
X
(1)
n
(0) = X
(1)
n
(1) = 0.
(6.5.35)
The inhomogeneous ODE (6.5.34) can be solved by the method of eigenfunction expansion
(see Chapter 8). Let
X
(1)
n
=
∞
m=1
α
m
sin mπx,
(6.5.36)
and
−
Λ(x) + λ
(1)
n
sin nπx =
∞
m=1
Λ
m
sin mπx,
(6.5.37)
where
Λ
m
=
−2
1
0
Λ(x) sin nπx sin mπxdx
− λ
(1)
n
δ
nm
.
(6.5.38)
Substituting (6.5.36) into (6.5.34) we get
∞
m=1
)
−(mπ)
2
+ (nπ)
2
*
α
m
sin mπx =
∞
m=1
Λ
m
sin mπx.
(6.5.39)
Thus
α
m
=
Λ
m
−(mπ)
2
+ (nπ)
2
,
m
= n.
(6.5.40)
143
To find λ
(1)
n
, we multiply (6.5.39) by sin nπx and integrate on the interval (0,1). Thus the
linear order approximation to λ
n
is given by
λ
(1)
n
=
−2
1
0
Λ(x) sin
2
nπxdx.
(6.5.41)
The linear order approximation to X
n
is given by (6.5.36) with the coefficients α
m
given by
(6.5.40). What happens for n = m is left for the reader.
144
Problems
1. The flow down a slightly corrugated channel is given by u(x, y, ) which satisfies
∇
2
u =
−1
in
|y| ≤ h(x, ) = 1 + cos kx
subject to
u = 0
on y =
±h(x, )
and periodic boundary conditions in x.
Obtain the first two terms for u.
2. The functions φ(x, y, ) and λ() satisfy the eigenvalue problem
φ
xx
+ φ
yy
+ λφ = 0
in
0
≤ x ≤ π, 0 + x(π − x) ≤ y ≤ π
subject to
φ = 0
on the boundary.
Find the first order correction to the eigenpair
φ
(0)
1
= sin x sin y
λ
(0)
1
= 2
145
SUMMARY
Theorem For a regular Sturm-Liouville problem
d
dx
p(x)
dX(x)
dx
+ q(x)X(x) + λσ(x)X(x) = 0,
a < x < b.
β
1
X(a) + β
2
X
(a) = 0,
β
3
X(b) + β
4
X
(b) = 0,
the following is true
i. All the eigenvalues λ are real
ii. There exist an infinite number of eigenvalues
λ
1
< λ
2
< . . . < λ
n
< . . .
a. there is a smallest eigenvalue denoted by λ
1
b. λ
n
→ ∞
as
n
→ ∞
iii. Corresponding to each λ
n
there is an eigenfunction X
n
(unique up to an arbitrary
multiplicative constant). X
n
has exactly n
− 1 zeros in the open interval (a, b).
iv. The eigenfunctions form a complete set, i.e. any smooth function f (x) can be represented
as
f (x)
∼
∞
n=1
a
n
X
n
(x).
a
n
=
(
b
a
f (x)X
n
(x)σ(x)dx
(
b
a
X
2
n
(x)σ(x)dx
This infinite series, called generalized Fourier series, converges to
f (x
+
) + f (x
−
)
2
if a
n
are properly chosen.
v. Eigenfunctions belonging to different eigenvalues are orthogonal relative to the weight σ,
i.e.
b
a
σ(x)X
n
(x)X
m
(x)dx = 0,
if
λ
n
= λ
m
.
vi. Any eigenvalue can be related to its eigenfunction by the Rayleigh quotient
λ =
−p(x)X(x)X
(x)
|
b
a
+
(
b
a
{p(x)[X
(x)]
2
− q(x)X
2
(x)
} dx
(
b
a
σ(x)X
2
(x)dx
.
146
7
PDEs in Higher Dimensions
7.1
Introduction
In the previous chapters we discussed homogeneous time dependent one dimensional PDEs
with homogeneous boundary conditions. Also Laplace’s equation in two variables was solved
in cartesian and polar coordinate systems. The eigenpairs of the Laplacian will be used here
to solve time dependent PDEs with two or three spatial variables. We will also discuss the
solution of Laplace’s equation in cylindrical and spherical coordinate systems, thus allowing
us to solve the heat and wave equations in those coordinate systems.
In the top part of the following table we list the various equations solved to this point.
In the bottom part we list the equations to be solved in this chapter.
Equation
Type
Comments
u
t
= ku
xx
heat
1D constant coefficients
c(x)ρ(x)u
t
= (K(x)u
x
)
x
heat
1D
u
tt
− c
2
u
xx
= 0
wave
1D constant coefficients
ρ(x)u
tt
− T
0
(x)u
xx
= 0
wave
1D
u
xx
+ u
yy
= 0
Laplace
2D constant coefficients
u
t
= k(u
xx
+ u
yy
)
heat
2D constant coefficients
u
t
= k(u
xx
+ u
yy
+ u
zz
)
heat
3D constant coefficients
u
tt
− c
2
(u
xx
+ u
yy
) = 0
wave
2D constant coefficients
u
tt
− c
2
(u
xx
+ u
yy
+ u
zz
) = 0
wave
3D constant coefficients
u
xx
+ u
yy
+ u
zz
= 0
Laplace
3D Cartesian
1
r
(ru
r
)
r
+
1
r
2
u
θθ
+ u
zz
= 0
Laplace
3D Cylindrical
u
rr
+
2
r
u
r
+
1
r
2
u
θθ
+
cot θ
r
2
u
θ
+
1
r
2
sin
2
θ
u
φφ
= 0
Laplace
3D Spherical
147
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