c Anthony Peirce.
Introductory lecture notes on Partial Differential Equations - °
Not to be copied, used, or revised without explicit written permission from the copyright owner.
Lecture 28: Sturm-Liouville Boundary Value Problems
(Compiled 3 March 2014)
In this lecture we abstract the eigenvalue problems that we have found so useful thus far for solving the PDEs to a
general class of boundary value problems that share a common set of properties. The so-called Sturm-Liouville Problems
define a class of eigenvalue problems, which include many of the previous problems as special cases. The S − L Problem
helps to identify those assumptions that are needed to define an eigenvalue problems with the properties that we require.
Key Concepts: Eigenvalue Problems, Sturm-Liouville Boundary Value Problems; Robin Boundary conditions.
Reference Section: Boyce and Di Prima Section 11.1 and 11.2
30 Boundary value problems and Sturm-Liouville theory:
30.1 Eigenvalue problem summary
• We have seen how useful eigenfunctions are in the solution of various PDEs.
• The eigenvalue problems we have encountered thus far have been relatively simple
I: The Dirichlet Problem:
¾
½
X 00 + λ2 X = 0
λn = nπ
L , n = ¡1, 2, .¢. .
=⇒
X(0) = 0 = X(L)
Xn (x) = sin nπx
L
II: The Neumann Problem:
¾
½
X 00 + λ2 X = 0
λn = nπ
L , n = 0,
¡ 1, 2,¢. . .
=⇒
0
0
X (0) = 0 = X (L)
Xn (x) = cos nπx
L
III: The Periodic Boundary Value Problem:

½
X 00 + λ2 X = 0

λn =© nπ
L , n¡ = 0,¢1, 2, .¡. .
¢ª
=⇒
X(−L) = 0 = X(L)

X
(x)
∈
1,
cos nπx
, sin nπx
n
0
0
L
L
X (−L) = 0 = X (L)
IV: Mixed Boundary Value Problem A:
(
¾
λk =
(2k+1)π
,
2L
λk =
(2k+1)π
,
2L
k = 0, 1, 2,´. . .
³
x
Xn (x) = sin (2k+1)π
2L
V: Mixed Boundary Value Problem B:
X 00 + λ2 X = 0
X(0) = 0 = X 0 (L)
X 00 + λ2 X = 0
X 0 (0) = 0 = X(L)
=⇒
(
¾
=⇒
k³ = 0, 1, 2,´. . .
Xn (x) = cos (2k+1)π
x
2L
1
2
30.2 The regular Sturm-Liouville problem:
Consider the the following two-point boundary value problem
¡
¢0
p(x)y 0 − q(x)y + λr(x)y = 0 0 < x < `
α1 y(0) + α2 y 0 (0) = 0 β1 y(`) + β2 y 0 (`) = 0
(30.1)
where p, p0 , 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:

0
Ly = λry where Ly = −(py 0 ) + qy

α1 y(0) + α2 y 0 (0) = 0 and β1 y(`) + β2 y 0 (`) = 0

p(x) > 0 and r(x) > 0.
SL
(30.2)
Remark 1 Note:
(1) If p = 1, q = 0, r = 1, α1 = 1, α2 = 0, β1 = 1, β2 = 0 we obtain Problem (I) above whereas if p = 1,
q = 0, r = 1, α1 = 0, α2 = 1, β1 = 0, β2 = 1, we obtain Problem (II) 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 (III) is not technically 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 0 ) + qy since it is possible to convert a general 2nd order
eigenvalue problem
−P (x)y 00 − Q(x)y 0 + R(x)y = λy
(30.3)
to this form by multiplying by an integrating factor µ(x)
−µ(x)P (x)y 00 − µQ(x)y 0 + µ(x)R(x)y = λµ(x)y
(30.4)
but expanding the differential operator we obtain
Ly = −py 00 − p0 y 0 + qy = λry.
(30.5)
Thus comparing (30.5) and (30.4) we can make the following identifications: p = ZµP and p0 = µQ ⇒ p0 =
Q
P0
− dx)
µ0 P + µP 0 = µQ which is a linear 1st order ODE for µ with integrating factor exp(
P
P
µ
0
µ +
Q
P0
−
P
P
¶
h
µ = 0 ⇒ Pe
−
R
Q
P
dx
i0
µ =0
µ=
e
R
Q
P
P
dx
.
(30.6)
Example 30.1 Reducing a boundary value problem to SL form:
φ00 + xφ0 + λφ = 0
(30.7)
φ(0) = 0 = φ
(30.8)
Sturm-Liouville two-point boundary value problems
3
We bring (30.7) into SL form by multiplying by the integrating factor
R
2
1 R Q dx
e P
= e x dx = ex /2 , P (x) = 1,
P
2
2
2
ex /2 φ00 + ex /2 xφ0 + λex /2 φ = 0
³ 2
´0
2
− ex /2 φ0 = λex /2 φ
µ=
p(x) = ex
2
/2
r(x) = ex
2
Q(x) = x,
R(x) = 1.
(30.9)
/2
Example 30.2 Convert the equation −y 00 + x4 y 1 = λy to SL form
P = 1,
Q = −x4 ,
Therefore − e−x
5
µ = e−
R
x4 dx
= e−x
5
5
/5 00
/5
5
y + e−x /5 x4 y 0 = λe−x /5
¢0
¡
5
5
− e−x /5 y 0 = λe−x /5 y.
(30.10)
(30.11)
(30.12)
30.3 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:
Z`
r(x)φ2j (x) dx = 1.
(a) φj (x) are real and can be normalized so that
0
(b) The eigenfunctions corresponding to different eigenvalues are orthogonal with respect to the weight function r(x):
Z`
r(x)φj (x)φK (x) dx = 0
j 6= k.
(30.13)
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
where
∞
P
f (x)
=
cn φn (x)
n=1
R`
r(x)f (x)φn (x) dx
cn
=
0
R`
0
r(x)φ2n (x) dx
(30.14)
4
Example 30.3 Robin Boundary Conditions:
X 00 + λX = 0,
λ = µ2
0
X (0) = h1 X(0), X 0 (`) = −h2 X(`)
(30.15)
where h1 ≥ 0 and h2 ≥ 0.
X(x) = A cos µx + B sin µx
(30.16)
X 0 (x) = −Aµ sin µx + Bµ cos µx
(30.17)
BC 1: X 0 (0) = Bµ = h1 X(0) = A A = Bµ/h1 .
BC 2: X 0 (`) = −Aµ sin(µ`) + Bµ cos(µ`) = −h2 X(`) = −h2 [A cos µ` + B sin µ`]
·
¸
·
¸
µ2
µ
⇒ B − sin(µ`) + µ cos(µ`) = −Bh2
cos µ` + sin µ`
h1
h1
B
Therefore
(30.18)
½µ
¶
µ
¶
¾
µ2
h2
−
+ h2 sin µ` + µ + µ cos µ` = 0.
h1
h1
(30.19)
·
¸
µ(h1 + h2 )
tan(µ`) =
.
µ2 − h1 h2
(30.20)
2
tan(µ l) & (µ(h1+h2))/(µ −h1h2)
Case I: h1 and h2 6= 0
Xn =
µn
h1
cos µn x + sin µn x, and µn ∼ nπ/` as n → ∞
Case I: h1 and h2 nonzero
5
0
−5
0
2
tan(µ l) & (µ(h1+h2))/(µ −h1h2)
Case II: h1 6= 0 and h2 = 0
µn
Xn =
cos µn x + sin µn x
h1
cos µn (` − x)
=
sin µn `
(30.21)
(30.22)
5
µ
10
Case II: h1 nonzero and and h2=0
4
2
0
−2
−4
0
5
µ
10
Sturm-Liouville two-point boundary value problems
h2 6= 0
Case III: h1−>∞ and h2 nonzero
5
2
tan(µ l) & (µ(h1+h2))/(µ −h1h2)
Case III: h1 → ∞
5
Xn = sin(µn x)
·µ
¶ ¸
2n + 1 π
µn ∼
2
`
(30.23)
n = 0, 1, 2, . . . as n → ∞ (30.24)
0
−5
0
5
µ
10
30.4 Appendix: Some proofs for Sturm-Liouville Theory
30.4.1 Lagrange’s Identity:
Z`
`
`
(vLu − uLv) dx = −p(x)u0 v|0 + p(x)uv 0 |0 .
0
Proof: Let u and v be any sufficiently differentiable functions, then
Z`
vLu dx =
0
Z` n
o
0
v −(pu0 ) + qu dx
(30.25)
0
Z`
Z`
=
`
−vpu0 |0
0
0
Z`
=
+
(30.26)
0
0
`
−vpu0 |0
uqv dx
u pv dx +
+
`
upv 0 |0
+
n
o
0
u −(pv 0 ) + qv dx
(30.27)
uLv dx.
(30.28)
0
Z`
Z`
`
`
vLu dx = −pvu0 |0 + puv 0 |0 +
Therefore
0
¤
0
Now suppose that u and v both satisfy the SL boundary conditions. I.E.
α1 u(0) + α2 u0 (0) = 0
α1 v(0) + α2 v 0 (0) = 0
β1 u(`) + β2 u0 (`) = 0
β1 v(`) + β2 v 0 (`) = 0
(30.29)
then
Z`
Z`
uLv dx = −p(`)u0 (`)v(`) + p(`)u(`)v 0 (`)
vLu dx −
0
0
+p(0)u0 (0)v(0) − p(0)u(0)v 0 (0)
µ
¶¾
½
β1
β1
= p(`) + u(`)v(`) + u(`) − v(`)
β2
β2
µ
¶¾
½
α1
α1
+p(0) − u(0)v(0) − u(0) − v(0)
α2
α2
= 0.
Z`
Thus
Z`
vLu dx =
0
uLv dx whenever u and v satisfy the SL boundary condition.
0
(30.30)
(30.31)
(30.32)
(30.33)
(30.34)
6
Observations:
Z`
Z`
• If L and BC are such that
vLu dx =
0
uLv dx then L is said to be self-adjoint.
0
Z`
• Notation: if we define (f, g) =
f (x)g(x) dx then we may write (v, Lv) = (u, Lv).
0
30.4.2 Proofs using Lagrange’s Identity:
(1a) The λj are real: Let Ly = λry (1) α1 y(0) + α2 y 0 (0) = 0 β1 y(`) + β2 y 0 (`) = 0. Take the conjugate of (1)
Lȳ = λ̄rȳ. By Lagrange’s Identity:
0 = (ȳ, Ly) − (y, Lȳ)
(30.35)
= (ȳ, rλy) − (y, rλ̄ȳ)
Z`
Z`
= ȳ(x)rλy(x) dx − y(x)r(x)λ̄ȳ(x) dx
0
(30.36)
(30.37)
0
Z`
= (λ − λ̄)
¯
¯2
r(x)¯y(x)¯ dx
(30.38)
0
2
Since r(x)|y(x)| ≥ 0 it follows that λ = λ̄ ⇒ λ is real.
0
(1c) λj > 0 provided α1 /α2 < 0 β1 /β2 > 0 and q(x) > 0. Consider Ly = −(py 0 ) + qy = λry (SL) and multiply
(SL) by y and integrate from 0 to `:
Z`
Z`
0 0
(y, Ly) =
2
−(py ) y + qy dx = λ
0
R`
Therefore λ =
£
¤2
r(x) y(x) dx
(30.39)
0
0
−(py 0 ) y + qy 2 dx
0
R`
this is known as Rayleigh’s Quotient.
ry 2
dx
0
[−py 0 y]`0 +
R`
0
R`
=
p(y 0 )2 + qy 2 dx
ry 2
(30.40)
dx
0
£
¤2
£
¤2 R`
2
α1
+p(`) ββ12 y(`) − p(0) α
y(0) + p(y 0 ) + qy 2 dx
2
=
0
R`
.
(30.41)
ry 2 dx
0
Therefore λ > 0 since the RHS is all positive.
Note: If q(x) ≡ 0 and α1 = 0 = β1 then with y 0 (0) = 0 = y 0 (`) we have nontrivial eigenfunction y(x) = 1 and
eigenvalue λ = 0.
Sturm-Liouville two-point boundary value problems
7
(2b) Eigenfunctions corresponding to different eigenvalues are orthogonal. Consider two distinct eigenvalues λj 6= λk λj : Lφj = rλj φj and λk : Lφk = rλk φk . Then
0 = (φk , Lφj ) − (φj , Lφk ) by Lagrange’s Identity
(30.42)
= (φk , rλj φj ) − (φj , rλk φk )
Z`
= (λj − λk ) r(x)φk (x)φj (x) dx
(30.43)
(30.44)
0
now λj 6= λk implies that
Z`
r(x)φk (x)φj (x) dx = 0.
(30.45)
0
(3) The eigenfunctions form a complete set: It is difficult to prove the convergence of the eigenfunction series
expansion for f (x) that is piecewise smooth. However, if we assume the expansion converges then it is a simple
∞
X
matter to use orthogonality to determine the coefficients in the expansion: Let f (x) =
cn φn (x).
n=1
Z`
f (x)φm (x)r(x) dx =
∞
X
n=1
0
Z`
cn
r(x)φm (x)φn (x) dx
(30.46)
0
orthogonality implies
R`
cm =
r(x)f (x)φm (x) dx
0
R`
0
£
¤2
r(x) φm (x) dx
.
(30.47)