LECTURE 22: THE STURM-LIOUVILLE EIGENVALUE PROBLEMS II
MINGFENG ZHAO
August 12, 2015
The Sturm-Liouville eigenvalue problem
Consider the following regular Sturm-Liouville eigenvalue problem:


 Ly := −[p(x)y]0 + q(x)y = λr(x)y,








 α1 y(a) + α2 y 0 (a) = 0,
(1)



β1 y(b) + β2 y 0 (b) = 0,







 p(x) > 0, r(x) > 0, a ≤ x ≤ b.
a < x < b,
Then
• Eigenvalues:
– All eigenvalues of (1) are real numbers.
– There are infinity many of eigenvalues {λn }∞
n=1 with λ1 < λ2 < λ3 < · · · and lim λn = ∞.
n→∞
α1
β1
– λn > 0 for all n ≥ 1 if
< 0,
> 0 and q(x) > 0 for a < x < b.
α2
β2
• Eigenfunctions:
– For each eigenvalue λn , there is an eigenfunction φn (x) that is unique up to a multiplicative constant.
Z b
– φn (x)’s are real functions and can be normalized so that
r(x)|φn (x)|2 dx = 1.
a
– The eigenfunctions corresponding to difference eigenvalues are orthogonal with respect to the weight function r(x), that is,
Z
b
r(x)φn (x)φm (x) dx = 0,
for all n 6= m.
a
– For each n ≥ 1, φn (x) has exactly n − 1 zeros on (a, b).
• Expansion Property: For any “nice” function f (x) on [a, b], then f (x) is an “infinite” linear combination of
{φn (x)}∞
n=1 , that is,
f (x) =
∞
X
n=1
1
cn φn (x).
2
MINGFENG ZHAO
Moreover, we have
Rb
f (x)r(x)φn (x) dx
cn = Ra b
,
r(x)|φn (x)|2 dx
a
for all n ≥ 1.
Example 1. Consider the following variable coefficient hear conduction problem:







(2)
ut = (x2 ux )x − u,
1 < x < 2, t > 0,
BC : u(1, t) = u(2, t) = 0,






IC :
u(x, 0) = f (x),
t > 0,
1 ≤ x ≤ 2.
Let u(x, t) = X(x)T (t) be a non-zero separated solution to the problem:



(3)
ut = (x2 ux )x ,
1 < x < 2, t > 0,

 BC : u(1, t) = u(2, t) = 0,
t > 0.
Then we get XT 0 = (x2 X 0 )0 T − XT , that is,
−
T0 + T
(x2 X 0 )0
=−
= λ.
X
T
It’s easy to see that λx = λt = 0, then λ is a constant. Since u(1, t) = u(2, t) = 0 for t > 0, then X(0) = X(2) = 0,
that is, we should consider the following eigenvalue problem:

 −[x2 X 0 ]0 = λX,
(4)
 X(1) = X(2) = 0.
Since −[x2 X 0 ]0 = λX, then x2 X 00 + 2xX 0 + λX = 0. Let X(x) = xr be a solution to x2 X 00 + 2xX 0 + λX = 0, then
r2 + r + λ = 0, that is,
r1,2 =
−1 ±
√
2
1 − 4λ
.
1
1
• If 1 − 4λ = 0, then r1 = r2 = − 12 , that is, the general solution to −[x2 X 0 ]0 = λX is X(x) = C1 x− 2 + C2 x− 2 ln x.
1
1
Since X(1) = 0, then C1 = 0 and X(x) = C2 x− 2 ln x. Since X(2) = 0, then C2 2− 2 ln 2 = 0, that is, C2 = 0. So
X(x) = 0.
1
± iµ, that is, the general solution to −[x2 X 0 ]0 = λX is
2
1
sin(µ ln x). Since X(1) = 0, then C1 = 0 and X(x) = C2 x− 2 sin(µ ln x).
• If 1 − 4λ = −4µ2 for some µ > 0, then r1,2 = −
1
1
X(x) = C1 x− 2 cos(µ ln x) + C2 x− 2
1
Since X(2) = 0, then C2 2− 2 sin(µ ln 2) = 0. In order to find a non-zero solution, we need sin(µ ln 2) = 0, that
nπ
1
nπ
1 nπ 2
is, µ ln 2 = nπ for some n ∈ N. So µ =
,λ= +
and X(x) = Cx− 2 sin
ln x .
ln 2
4
ln 2
ln 2
LECTURE 22: THE STURM-LIOUVILLE EIGENVALUE PROBLEMS II
3
1
• If 1 − 4λ = 4µ2 for some µ > 0, then r1,2 = − ± µ, that is, the general solution to −[x2 X 0 ]0 = λX is X(x) =
2
i
h 1
1
− 12 +µ
− 12 −µ
C1 x
+ C2 x
. Since X(1) = 0, then C1 + C2 = 0, that is, C2 = −C1 and X(x) = C1 x− 2 +µ − x− 2 −µ .
h 1
i
1
Since X(2) = 0, then C1 2− 2 +µ − 2− 2 −µ = 0. Since µ > 0, then C1 = 0. So X(x) = 0.
nπ 2
1
and corresponding eigenvalues are Xn (x) =
In summary, the eigenvalues of (4) are λn =
+
4
ln 2
1
nπ
Cx− 2 sin
ln x with n ≥ 1.
ln 2
nπ 2
1 nπ 2
− 54 +( ln
t
0
2)
Now for any λn = +
, since T + (λn + 1)T = 0, then T (t) = Ce
. So for any n ≥ 1, we find a
4
ln 2
non-zero solution to (3):
nπ
2
1
− 5 + nπ
t
un (x, t) = e 4 ( ln 2 ) x− 2 sin
ln x .
ln 2
It’s easy to see that (4) is a Sturm-Liouville eigenvalue problem with the weight function 1, by the properties of
Sturm-Liouville problem, then
f (x) =
∞
X
1
fn x− 2 sin
n=1
nπ
ln 2
ln x ,
where
R2
fn =
Let u(x, t) =
∞
X
1
Z 2
nπ
nπ
f (x)x− 2 sin ln
1
2
2 ln x dx
=
ln x dx,
f (x)x− 2 sin
R2
2
nπ
ln 2 1
ln 2
x−1 sin ln
2 ln x dx
1
1
fn e
−
( lnnπ2 )
5
4+
2
t −1
2
x
sin
nπ
n=1
ln 2
∞
X
fn un (x, t), then u(x, t) is the solution to (2).
ln x =
n=1
Example 2. Let 0 < α < 2π be a constant, consider the following Laplace equation:

1
1


urr + ur + 2 uθθ = 0, 1 < r < 2, 0 < θ < α


r
r



(5)
BC : u(1, θ) = u(2, θ) = 0, 0 ≤ θ ≤ α,







u(r, 0) = 0, u(r, α) = f (r), 1 ≤ r ≤ 2.
Let u(r, θ) = R(r)Θ(θ) be a non-zero solution to

1
1


urr + ur + 2 uθθ = 0, 1 < r < 2, 0 < θ < π,


r
r



(6)
BC : u(1, θ) = u(2, θ) = 0, 0 ≤ θ ≤ α,







u(r, 0) = 0, 1 ≤ r ≤ 2.
1
1
Then R00 Θ + R0 Θ + 2 RΘ00 = 0, that is,
r
r
−
for all n ≥ 1.
R00 + 1r R0
Θ00
=
= λ.
1
Θ
r2 R
4
MINGFENG ZHAO
It’s easy to see that λr = λθ = 0, that is, λ is a constant. Since u(1, θ) = u(2, θ) = 0 for 0 < θ ≤ π, then
R(1) = R(2) = 0. That is, R(r) satisfies:


 R00 + 1 R0 + λ R = 0,
r
r2

 R(1) = R(2) = 0.
(7)
1
λ
1
Since R00 + R0 + 2 R = 0, then −[rR0 ]0 = −rR00 − R0 = λ · R, which is a form of the Sturm-Liouville eigenvalue
r
r
r
1
1
λ
µ
problem with the weight function . Let R(r) = r be a solution to R00 + R0 + 2 R = 0, then
r
r
r
0
= µ(µ − 1)rµ−2 + µrµ−2 + λrµ−2
= rµ−2 [µ(µ − 1) + µ + λ]
= rµ−2 [µ2 + λ].
So we get µ2 + λ = 0, that is,
√
µ1,2 = ± −λ.
λ
1 0
R + 2 R = 0 is R(r) = C1 + C2 ln r. Since
r
r
R(1) = 0, then C1 = 0 and R(r) = C2 ln r. Since R(2) = 0, then C2 ln 2 = 0, that is, C2 = 0. So we get
• If λ = 0, then µ1 = µ2 = 0. So the general solution to R00 +
R(r) = 0.
1 0
λ
R + 2 R = 0 is R(r) =
r
r
C1 cos(β ln r) + C2 sin(β ln r). Since R(1) = 0, then C1 = 0 and R(r) = C2 sin(β ln r). Since R(2) = 0, then
• If λ = β 2 for some β > 0, then µ1,2 = ±βi. So the general solution to R00 +
C2 sin(β ln 2) = 0. In order to find a non-zero solution R(r), then sin(β ln 2) = 0, which implies that β ln 2 = nπ
nπ 2
nπ
nπ
for some n ∈ N, that is, β =
and λ =
. Moreover, we have R(r) = C sin
ln r .
ln 2
ln 2
ln 2
1
λ
• If λ = −β 2 for some β > 0, then µ1,2 = ±β. So the general solution to R00 + R0 + 2 R = 0 is R(r) =
r
r
C1 rβ +C2 r−β . Since R(1) = 0, then C1 +C2 = 0 and R(r) = C1 [rβ −r−β ]. Since R(2) = 0, then C1 [2β −2−β ] = 0.
Since β > 0, then C1 = 0. So R(r) = 0.
In summary, all eigenvalues for (7) are λn =
nπ 2
ln 2
and corresponding eigenfunctions are Rn (r) = C sin
nπ
ln 2
ln r
with n ∈ N.
nπ 2
nπ
nπ
, since Θ00 − λΘ = 0, then Θ(θ) = C1 e ln 2 θ + C2 e− ln 2 θ . Since u(r, 0) = 0 for 1 ≤ r ≤ 2, then
ln 2
nπ
nπ Θ(0) = 0, which implies that C1 + C2 = 0 and Θ(θ) = C1 e ln 2 θ − e− ln 2 θ .
For each λ =
In summary, for any n ∈ N, we find a non-zero solution to (6):
nπ
nπ
nπ un (r, θ) = e ln 2 θ − e− ln 2 θ sin
ln r .
ln 2
LECTURE 22: THE STURM-LIOUVILLE EIGENVALUE PROBLEMS II
5
By the properties of Sturm-Liouville eigenvalue problem, then
f (r) =
∞
X
fn sin
nπ
ln 2
n=1
where
R2
fn =
1
nπ
1
ln 2 ln r · r dr
nπ
1
ln 2 ln r · r dr
f (r) sin
R2
1
sin2
=
2
ln 2
ln r ,
2
Z
f (r) sin
1
1
ln r · dr,
ln 2
r
nπ
for all n ≥ 1.
∞
X
∞
nπ
X
nπ θ
nπ − ln
θ
ln
2
2
Let u(r, θ) =
an e
−e
sin
ln r =
an un (r, θ) be the solution to (5), then
ln 2
n=1
n=1
f (r) = u(r, α) =
∞
X
nπ
nπ
nπ
ln r .
an e ln 2 α − e− ln 2 α sin
ln 2
n=1
So we get
nπ
nπ
fn = an e ln 2 α − e− ln 2 α .
That is,
an =
e
nπ
ln 2 α
fn
2
nπ α
=
nπ
nπ
−
α
− e ln 2
ln 2 e ln 2 − e− ln 2 α
Z
2
f (r) sin
1
1
ln r · dr,
ln 2
r
nπ
for all n ≥ 1.
Remark 1. Notice that
Z
1
2
sin2
1
ln r · dr
ln 2
r
nπ
Z
ln 2
=
0
=
sin2
nπ x dx
ln 2
Let x = ln r
2
.
ln 2
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca