Chapter 27
Lecture 31 Sturm-Liouville
Theory
27.1
Boundary value problems and Sturm-Liouville
theory:
• Up till now we have been able to solve partial differential equations
by separating variables which typically reduces the PDE to solving an
eigenvalue problem and some initial value problem (for the Heat and
Wave Equations) or some inhomogeneous boundary value problem (in
the case of Laplace’s Equation).
• The eigenvalue problems thus far have been simple
1.
x + λ2 X = 0
X(0) = 0 = X(L)
2.
X + λ2 X = 0
X (0) = 0 = X (L)
3.
X + λ2 X = 0
→ λn =
nπ
L
→ λn =
2π
X(x + 2π) = X(x) periodic BC
n = 1, 2, . . . Xn = sin
nπ
L
nπx L
n = 0, 1, . . . Xn = cos
→ λn =
X0
nπx nπ
, n = 0, 1, . . .
L
Xn =
= 1.
.
sin nπx
L cos nπx
L
157
L
.
Lecture 31 Sturm-Liouville Theory
• In this section we abstract these problems to a general class of boundary value problems which share a common set of properties. The socalled Sturm-Liouville Problems define a class of eigenvalue problems which include many of the above problems as special cases. The
S − L Problem helps to identify those assumptions that are needed to
define an(delete the word an?) eigenvalue problems with the properties
that we require.
27.2
The regular Sturm-Liouville problem:
p(x)y − q(x)y + λr(x)y = 0 0 < x < α1 y(0) + α2 y (0) = 0 β1 y() + β2 y () = 0
(27.1)
where p, p , q and r are continuous on 0 ≤ x ≤ and p(x) ≥ 0 and r(x) > 0
on 0 ≤ x ≤ .
We define the Sturm-Liouville eigenvalue problem as:
⎫
Ly = λry Ly = −(py ) + qy
⎬
SL
(27.2)
α1 y(0) + α2 y (0) = 0 β1 y() + β2 y () = 0
⎭
p(x) > 0 and r(x) > 0.
Remark 27.1 Note:
1. If p = 1 q = 0 r = 1 α1 = 1 α2 = 0 β1 = 1 β2 = 0 we obtain
Problem (1) above whereas if p = 1 q = 0 r = 1 α1 = 0 α2 = 1
β1 = 0 β2 = 1 we obtain Problem (2) above. Notice that the boundary
conditions for these two problems are specified at separate points and
are called separated BC. The periodic BC X(0) = X(2π) are not
separated so that Problem (3) is not a SL Problem.
2. If p > 0 and r > 0 and < ∞ then the SL Problem is said to be
regular. If p(x) or r(x) is zero for some x or the domain is [0, ∞) then
the problem is singular.
3. There is no loss of generality in the form of Ly = −(py ) + qy since it
is possible to convert a general 2nd order eigenvalue problem
−P (x)y − Q(x)y + R(x)y = λy
(27.3)
to this form by multiplying by an integrating factor μ(x)
−μ(x)P (x)y − μQ(x)y + μ(x)R(x)y = λμ(x)y
158
(27.4)
27.2. THE REGULAR STURM-LIOUVILLE PROBLEM:
but
Ly = −py − p y + qy = λry.
(27.5)
So we can identify that p = μP and p = μQ ⇒ p = μ P + μP = μQ
P Q
− dx
which is a linear 1st order ODE for μ with integrating factor ρ
P
P
μ +
P Q
−
P
P
−
μ = 0 ⇒ Pρ
Q
P
dx
μ =0
ρ
μ=
Q
P
dx
P
.
(27.6)
Example 27.2 Reducing a boundary value problem to SL form:
φ + xφ + λφ = 0
(27.7)
φ(0) = 0 = φ
(27.8)
We bring (27.8) into SL form by multiplying by the integrating factor
1 Q
2
μ = ρ P dx = ρ x dx = ρx /2 P (x) = 1 Q(x) = x R(x) = 1.
p
2
2
2
ρx /2 φ + ρx /2 xφ + λρx /2 φ = 0
(27.9)
2
x /2 x2 /2
φ = λρ
φ
− ρ
p(x) = ρx
2 /2
2 /2
r(x) = ρx
Example 27.3 Convert the equation −y + x4 y 1 = λy to SL form
P = 1 Q = −x4
−x5 /5 Therefore − ρ
μ = ρ−
x4 dx
−x5 /5 4 5 /5
= ρ−x
−x5 /5
y +ρ
x y = λρ
−x5 /5 5
− ρ
y
= λρ−x /5 y.
(27.10)
(27.11)
(27.12)
159
Lecture 31 Sturm-Liouville Theory
27.3
Properties of SL Problems
1. Eigenvalues:
(a) The eigenvalues λ are all real.
(b) There are an ∞ # of eigenvalues λj with λj < λ2 < . . . < λj → ∞
as j → ∞.
α1
β1
(c) λj > 0 provided
< 0,
> 0 q(x) > 0.
α2
β2
2. Eigenfunctions: For each λj there is an eigenfunction φj (x) that is
unique up to a multiplicative const. and which satisfy:
(a) φj (x) are real and can be normalized so that
r(x)φ2j (x) dx = 1.
0
(b) The eigenfunctions corresponding to different eigenvalues are orthogonal with respect to the weight function r(x):
r(x)φj (x)φK (x) dx = 0 j = k.
(27.13)
0
(c) φj (x) has exactly j − 1 zeros on (0, ).
3. Expansion Property: {φj (x)} are complete if f (x) is piecewise
smooth then
f (x) =
where
cn =
∞
cn φn (x)
n=1
r(x)f (x)φn (x) dx
0
0
(27.14)
r(x)φ2n (x) dx
Example 27.4 Robin Boundary Conditions:
α = μ2
X + λX = 0
X (0) = h1 X(0) X () = −h2 X()
(27.15)
X(x) = A cos μx + B sin μx
(27.16)
X (x) = −Aμ sin μx + Bμ cos μx
160
(27.17)
27.3. PROPERTIES OF SL PROBLEMS
BC 1: X (0) = Bμ = h1 X(0) = A A = Bμ/h1 .
BC 2:
X () = −Aμ sin(μ) + Bμ cos(μ) = −h2 X() = −h2 [A cos μ + B sin μ]
(27.18)
2
μ
μ
cos μ + sin μ
(27.19)
⇒ B − sin(μ) + μ cos(μ) = −Bh2
h1
h1
2
h2
μ
(27.20)
B
− + h2 sin μ + μ + μ cos μ = 0.
h1
h1
Therefore
tan(μ) =
μ(h1 + h2 )
.
μ2 − h1 h2
(27.21)
h1 and h2 = 0: Xn = μn cos μn x + sin μn x
h2 = 0
h1 = 0:
Xn =
=
μn
cos μn x + sin μn x
h1
cos μn ( − x)
sin μn (27.22)
(27.23)
h1 → ∞ h2 = 0:
Xn = sin(μn x)
2n − 1 π
μn ∼
2
(27.24)
n = 1, 2, . . .
(27.25)
161