LECTURE 21: THE FINITE DIFFERENCE METHODS III AND THE STURM-LIOUVILLE
EIGENVALUE PROBLEMS
MINGFENG ZHAO
August 11, 2015
Finite difference approximation for the wave equation
Consider the wave equation with initial-boundary value conditions:


utt = 4uxx + x2 , 0 ≤ x ≤ 1, t ≥ 0,





BC : u(0, t) = 2t, ux (1, t) = 3t, t > 0,






IC : u(x, 0) = x2 + 1, ut (x, 0) = x3 , 0 ≤ x ≤ 1.
Use the central difference in t and the central difference in x, then
u(x, t + ∆t) + u(x, t − ∆t) − 2u(x, t)
(∆t)2
= utt (x, t) + O((∆t)2 )
u(x + ∆x, t) + u(x − ∆x, t) − 2u(x, t)
(∆x)2
= uxx (x, t) + ((∆x)2 ).
Then we get
u(x, t + ∆t) + u(x, t + ∆t) − 2u(x, t)
u(x + ∆x, t) + u(x − ∆x, t) − 2u(x, t)
=4·
+ x2 .
(∆t)2
(∆x)2
That is,
u(x, t + ∆t) = 2u(x, t) − u(x, t − ∆t) +
4(∆t)2
· [u(x + ∆x, t) + u(x − ∆x, t) − 2u(x, t)] + x2 (∆t)2 + O((∆t)2 , (∆x)2 ).
(∆x)2
For any T > 0 and integers M, N ≥ 1, let ∆t =
(1)
uk+1
= 2ukn − uk−1
+
n
n
T
L
, ∆x = , tk = k∆t, xn = n∆x and ukn = u(xn , tk ), then
M
N
4(∆t)2 k
[u
− 2ukn + ukn−1 ] + x2n (∆t)2 .
(∆x)2 n+1
For the initial value condition, we have
(2)
u0n = x2n + 1.
1
2
MINGFENG ZHAO
Use the central finite approximation for ut (x, 0), we have
u(x, ∆t) − u(x, −∆t)
= ut (x, 0) + O((∆t)2 ).
2∆t
That is, we have
u(x, ∆t) − u(x, −∆t)
= x3 + O((∆t)2 ).
2∆t
That is,
u(x, −∆t) = u(x, ∆t) − 2x3 ∆t + O((∆t)2 ).
So we have
1
3
u−1
n = un − 2xn ∆t.
(3)
For the boundary value condition, we have
uk0 = 2tk .
(4)
Use the central finite approximation for ux (1, t), we have
u(1 + ∆x, t) − u(1 − ∆x, t)
= ux (1, t) + O((∆x)2 ).
2∆x
That is,
u(1 + ∆x, t) − u(1 − ∆x, t)
= 3t + O((∆x)2 ).
2∆x
Then we get
u(1 + ∆x, t) = u(1 − ∆x, t) + 6t∆x + O((∆x)2 ).
So we have
ukN +1 = ukN −1 + 6tk ∆x.
(5)
So the finite difference approximation for the problem is the following:
• By (2), we get u0n for 0 ≤ n ≤ N . By (5), we get u0N +1 . So we have u0n for 0 ≤ n ≤ N + 1.
• By (1) and (3), we get
u1n = 2u0n − (u1n − 2x3n ∆t) +
4(∆t)2 0
[u
− 2u0n + u0n−1 ] + x2n (∆t)2 .
(∆x)2 n+1
So
u1n = u0n + x3n ∆t +
2(∆t)2 0
[u
− 2u0n + u0n−1 ] + x2n (∆t)2 .
(∆x)2 n+1
Hence we can get u1n for 0 ≤ n ≤ N . By (5), we get u1N +1 . So we have u1n for 0 ≤ n ≤ N + 1.
LECTURE 21: THE FINITE DIFFERENCE METHODS III AND THE STURM-LIOUVILLE EIGENVALUE PROBLEMS
3
• By (1), we get u2n for 0 ≤ n ≤ N . By (5), we get u2N +1 . So we have u2n for 0 ≤ n ≤ N + 1.
• Keep going.
The Sturm-Liouville eigenvalue problem
A Sturm-Liouville eigenvalue problem has the form:
(6)


Ly := −[p(x)y 0 ]0 + q(x)y = λr(x)y,





α1 y(a) + α2 y 0 (a) = 0,






β1 y(b) + β2 y 0 (b) = 0.
a < x < b,
We say that the Sturm-Liouville eigenvalue problem (6) is regular if p(x) > 0 and r(x) > 0 for a ≤ x ≤ b; we say it’s
singular if otherwise.
Theorem 1. Any differential equation −P (x)y 00 − Q(x)y 0 + R(x)y = λy is equivalent to a differential equation of the
form −[p(x)y 0 ]0 + q(x)y = 0.
Proof. Use the method of the integrating factor: Multiply r(x) on the both sides of −P (x)y 00 − Q(x)y 0 + R(x)y = ry,
then
−r(x)P (x)y 00 − r(x)Q(x)y 0 + r(x)R(x)y = λr(x)y.
The above equation is of the form −[p(x)y 0 ]0 + q(x)y = λr(x)y if and only if
d
[r(x)P (x)] = r(x)Q(x).
dx
So we get r0 (x)P (x) + r(x)P 0 (x) = r(x)Q(x), that is, r0 (x) = r(x) ·
r(x) = e
R
Q(x)−P 0 (x)
P (x)
dx
=
Q(x) − P 0 (x)
, which implies that
P (x)
1 R
e
P (x)
Q(x)
P (x)
dx
.
Example 1. Reduce the following boundary value problem to a Sturm-Liouville form:
φ00 + xφ0 + λφ = 0,
φ(0) = φ(1) = 0.
4
MINGFENG ZHAO
Since φ00 + xφ0 + λφ = 0, then −φ00 − xφ0 = λφ. Multiply r(x) on the both sides, then
−r(x)φ00 − xr(x)φ0 = λr(x)φ.
The above equation is a Sturm-Liouville form, then r0 (x) = xr(x), which implies that r(x) = e
x2
2
. So we get
h x2 i0
x2
− e 2 φ0 = λe 2 φ,
which is a regular Sturm-Liouville eigenvalue problem with p(x) = e
x2
2
, q(x) = 0 and r(x) = e
x2
2
.
Example 2. Reduce the following boundary value problem to a Sturm-Liouville form:
−y 00 + x4 y 0 = λy,
y(0) = y(1) = 0.
Multiply r(x) on the both sides of −y 00 + x4 y 0 = λy, then
−r(x)y 00 + x4 r(x)y 0 = λr(x)y.
The above equation is a Sturm-Liouville form, then −r0 (x) = x4 r(x), which implies that r(x) = e−
x5
5
. So we get
h x5 i0
x5
− e− 5 y 0 = λe− 5 y,
which is a regular Sturm-Liouville eigenvalue problem with p(x) = e−
x5
5
, q(x) = 0 and r(x) = e−
x5
5
.
Properties of the regular Sturm-Liouville eigenvalue problem
Consider the following regular Sturm-Liouville eigenvalue problem:



Ly := −[p(x)y 0 ]0 + q(x)y = λr(x)y,








 α1 y(a) + α2 y 0 (a) = 0,
(7)



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 (7) are real numbers.
– There are infinity many of eigenvalues {λn }∞
n=1 with λ1 < λ2 < λ3 < · · · and lim λn = ∞.
n→∞
LECTURE 21: THE FINITE DIFFERENCE METHODS III AND THE STURM-LIOUVILLE EIGENVALUE PROBLEMS
– λn > 0 for all n ≥ 1 if
5
α1
β1
< 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
r(x)|φn (x)|2 dx = 1.
– φn (x)’s are real functions and can be normalized so that
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
cn φn (x).
n=1
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 3. Consider the following initial-boundary value problem for the heat equation:


ut = c2 uxx , 0 < x < L, t > 0,





(8)
BC : u(0, t) = 0, ux (L, t) + u(L, t), t > 0,






IC : u(x, 0) = f (x), 0 ≤ x ≤ L.
Let u(x, t) = X(x)T (t) be a non-zero separated solution to the problem:



ut = c2 uxx , 0 < x < L, t > 0,
(9)

 BC : u(0, t) = 0, u (L, t) + u(L, t) = 0, t > 0.
x
Then we get XT 0 = c2 X 00 T , that is,
−
X 00
T0
= − 2 = λ.
X
c T
It’s easy to see that λx = λt = 0, then λ is a constant. Since u(0, t) = 0 and ux (L, t) + u(L, t) = 0 for t > 0, then
X(0) = 0 and X 0 (L) + X(L) = 0, that is, we should consider the following eigenvalue problem:

 −X 00 = λX,
(10)
 X(0) = 0, X 0 (L) + X(L) = 0.
Let X(x) = erx be a solution to −X 00 = λX, then r2 + λ = 0.
6
MINGFENG ZHAO
• If λ = 0, then r1 = r2 = 0, that is, the general solution to −X 00 = λX = 0 is X(x) = C1 + C2 x. Since X(0) = 0,
then C1 = 0 and X(x) = C2 x. Since X 0 (x) = C2 and X 0 (L) + X(L) = 0, then C2 + C2 L = 0, which implies
that C2 = 0. So X(x) = 0.
• If λ = µ2 for some µ > 0, then r1,2 = ±iµ, that is, the general solution to −X 00 = λX is X(x) = C1 cos(µx) +
C2 sin(µx). Since X(0) = 0, then C1 = 0 and X(x) = C2 sin(µx). Since X 0 (x) = µC2 cos(µx) and X 0 (L) +
X(L) = 0, then µC2 cos(µL)+C2 sin(µL) = 0. In order to find a non-zero solution, we need µ cos(µL)+sin(µL) =
0. Since µ > 0, then we must have cos(µL) 6= 0, which implies that µ + tan(µL) = 0. So there is a sequence
{µ}∞
n=1 such that 0 < µ1 < µ2 < µ3 < · · · such that µn + tan(µn L) = 0 for all n ≥ 1 and lim µn = ∞. For
n→∞
each µn , we have Xn (x) = C sin(µn x).
• If λ = −µ2 for some µ > 0, then r1,2 = ±µ, that is, the general solution to −X 00 = λX is X(x) = C1 eµx +C2 e−µx .
Since X(0) = 0, then C1 + C2 = 0 and X(x) = C1 [eµx − e−µx ]. Since X 0 (x) = C1 [µeµx + µe−µx ] and
X 0 (L) + X(L) = 0, then
C1 [µeµL + µe−µL ] + C1 [eµL − e−µL ] = 0.
So we get
eµL − e−µL
= 0.
C1 µ + µL
e + e−µL
Since µ > 0 and L > 0, then we must have C1 = 0. So X(x) = 0.
In summary, the eigenvalues of (10) are λn = µ2n and corresponding eigenvalues are Xn (x) = C sin(µn x) with n ≥ 1.
Now for any λn , since T 0 + c2 λn T = 0, then T (t) = Ce−c
2
2
un (x, t) = e−c
µ2n t
µ2n t
. So for any n ≥ 1, we find a non-zero solution to (9):
sin(µn x).
It’s easy to see that (10) is a Sturm-Liouville eigenvalue problem, by the properties of Sturm-Liouville problem, then
f (x) =
∞
X
fn sin(µn x),
n=1
where
RL
fn =
Let u(x, t) =
∞
X
2
fn e−c
µ2n t
sin(µn x) =
n=1
∞
X
f (x) sin(µn x) dx
,
RL 2
sin (µn x) dx
0
0
for all n ≥ 1.
fn un (x, t), then u(x, t) is the solution to (8).
n=1
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