Chapter 30
Lecture 34 — Sturm
Liouville Theory
30.1
Properties of SL Problems:
1. Eigenvalue Properties:
• (a) the eigenvalues λ are all real.
• (b) there are an infinite number of eigenvalues λj with λ1 <
λ2 < . . . λj → ∞ as j → ∞.
• (c) λj > 0 provided
α1
α2
< 0,
β1
β2
> 0 and q(x) > 0.
2. Eigenfunction Properties: Corresponding to each eigenvalue λj there
is an eigenfunction φj (x) that is unique up to multiplication by a
constant, and which satisfy:
• (a) φj (x) are real and can be normalized
0
r(x)φ2j (x) dx = 1.
• (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
if j = k.
(30.1)
0
• (c) φj (x) has exactly j − 1 zeros on 0 < x < .
171
Lecture 34 — Sturm Liouville Theory
3. Expansion Property: The eigenfunctions φj (x) form a complete set
so that any piecewise smooth function f (x) can be expanded as a
generalized Fourier Series:
f (x) =
∞
cn φn (x)
(30.2)
n=1
by orthogonality:
cn =
0
r(x)f (x)φn (x) dx
0
30.2
.
(30.3)
r(x)φ2n (x) dx
Lagrange’s Identity:
Lagrange’s Identity
(vLu − uLv) dx = −p(x)u v 0 + p(x)uv 0
0
.
is fundamental to the development of S-L Theory.
Proof: Let u and v be any sufficiently differentiable functions; then
vLu dx =
v −(pu ) + qu dx
0
(30.4)
0
=
−vpu 0
+
u pv dx +
0
=
uqv dx
(30.5)
0
−vpu 0 + upv 0 +
u −(pv ) + qv dx(30.6)
0
vLu dx =
Therefore
−pvu 0
0
+
puv 0
+
uLv dx.
(30.7)
0
Now suppose that u and v both satisfy the SL boundary conditions. I.E.
α1 u(0) + α2 u (0) = 0
α1 v(0) + α2 v (0) = 0
172
β1 u() + β2 u () = 0
β1 v() + β2 v () = 0
(30.8)
30.3. PROOFS TO SELECTED PROPERTIES:
then
vLu dx −
0
uLv dx = −p()u ()v() + p()u()v ()
0
(30.10)
+p(0)u (0)v(0) − p(0)u(0)v (0)
β1
β1
= p() + u()v() + u() − v()
(30.11)
β2
β2
α1
α1
(30.12)
+p(0) − u(0)v(0) − u(0) − v(0)
α2
α2
= 0.
(30.13)
Thus
vLu dx =
0
condition.
Note:
uL?v dx whenever u and v satisfy the SL boundary
0
• If L and BC are such that
uLv dx then L is said to be
vLu dx =
0
0
self-adjoint. Notation if we define (f, g) =
f (x)g(x) dx then we
0
may write (v, Lv) = (u, Lv).
30.3
(30.9)
Proofs to selected properties:
(1a) λ is real: Let Ly = λry (1) α1 y(0) + α2 y (0) = 0 β1 y() + β2 y () = 0.
Take the conjugate of (1) Lȳ = λ̄rȳ. By Lagrange’s Identity:
0 = (ȳ, Ly) − (y, Lȳ)
(30.14)
= (ȳ, rλy) − (y, rλ̄ȳ)
=
ȳ(x)rλy(x) dx − y(x)r(x)λ̄ȳ(x) dx
0
(30.15)
(30.16)
0
= (λ − λ̄)
2
r(x)y(x) dx
(30.17)
0
173
Lecture 34 — Sturm Liouville Theory
Since r(x)|y(x)|2 ≥ 0 it follows that λ = λ̄ ⇒ λ is real.
(1c) λj > 0 provided α1 /α2 < 0 β1 /β2 > 0 and q(x) > 0. Consider
Ly = −(py ) + qy = λry (SL) and multiply (SL) by y and integrate from 0
to :
(y, Ly) =
2
−(py ) y + qy dx = λ
0
Therefore λ =
0
2
r(x) y(x) dx
(30.18)
0
−(py ) y + qy 2 dx
this is known as Rayleigh’s Quotient.
ry 2 dx
0
[−py y]0 +
=
p(y )2 + qy 2 dx
0
(30.19)
ry 2 dx
0
β1 2
2 +p() β2 y() − p(0) αα12 y(0) + p(y )2 + qy 2 dx
=
0
. (30.20)
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 = y () we have
nontrivial eigenfunction y(x) = 1 and eigenvalue λ = 0.
(2b) Eigenfunctions corresponding to different eigenvalues are orthogonal. Consider two distinct eigenvalues λj = λk λj : Lφj = rλj φj and
λk : Lφk = rλk φk . Then
0 = (φK , Lφj ) − (φj , LφK )
by Lagrange’s Identity
= (φK , rλj φj ) − (φj , rλK φK )
= (λj − λK ) r(x)φK (x)φj (x) dx
(30.21)
(30.22)
(30.23)
0
now λj = λk?K implies that
r(x)φk (x)φj (x) dx = 0.
0
174
(30.24)
30.3. PROOFS TO SELECTED PROPERTIES:
(3) Expansion Property 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 matter to use
orthogonality to determine the coefficients in the expansion: Let f (x) =
∞
cn φn (x).
n=1
f (x)φm (x)r(x) dx =
∞
n=1
0
cn
r(x)φm (x)φn (x) dx
(30.25)
0
orthogonality implies
cm =
0
r(x)f (x)φm (x) dx
0
2
r(x) φm (x) dx
.
(30.26)
175