Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 7 docx

for later use.

2π
0
xy dt =

2π
0

a
0
2
+
∞

n=1

a
n
cos(nt) + b
n
sin(nt)


c
0
2
+
∞


n=1

c
n
cos(nt) + d
n
sin(nt)


dt
=

2π
0

a
0
c
0
4
+
∞

n=1

a
n
c
n
cos

2
(nt) + b
n
d
n
sin
2
(nt)


dt
= π

1
2
a
0
c
0
+
∞

n=1
(a
n
c
n
+ b
n
d

n
)

In the arclength parametrization we have

dx
ds

2
+

dy
ds

2
= 1.
In terms of t = 2πs/L this is

dx
dt

2
+

dy
dt

2
=


L
2π

2
.
We integrate this identity on [0, 2π].
L
2
2π
=

2π
0


dx
dt

2
+

dy
dt

2

dt
= π

∞


n=1

(nb
n
)
2
+ (−na
n
)
2

+
∞

n=1

(nd
n
)
2
+ (−nc
n
)
2


= π
∞


n=1
n
2
(a
2
n
+ b
2
n
+ c
2
n
+ d
2
n
)
L
2
= 2π
2
∞

n=1
n
2
(a
2
n
+ b
2

n
+ c
2
n
+ d
2
n
)
1414
We assume that the curve is parametrized so that the area is positive. (Reversing the orientation changes the sign
of the area as defined above.) The area is
A =

2π
0
x
dy
dt
dt
=

2π
0

a
0
2
+
∞


n=1

a
n
cos(nt) + b
n
sin(nt)


∞

n=1

nd
n
cos(nt) − nc
n
sin(nt)


dt
= π
∞

n=1
n(a
n
d
n
− b

n
c
n
)
Now we find an upper bound on the area. We will use the inequality |ab| ≤
1
2
|a
2
+ b
2
|, which follows from expanding
(a − b)
2
≥ 0.
A ≤
π
2
∞

n=1
n

a
2
n
+ b
2
n
+ c

2
n
+ d
2
n

≤
π
2
∞

n=1
n
2

a
2
n
+ b
2
n
+ c
2
n
+ d
2
n

We can express this in terms of the perimeter.
=

L
2
4π
L
2
≥ 4πA
Now we determine the curves for which L
2
= 4πA. To do this we find conditions for which A is equal to the upper
bound we obtained for it above. First note that
∞

n=1
n

a
2
n
+ b
2
n
+ c
2
n
+ d
2
n

=
∞


n=1
n
2

a
2
n
+ b
2
n
+ c
2
n
+ d
2
n

1415
implies that all the coefficients except a
0
, c
0
, a
1
, b
1
, c
1
and d

1
are zero. The constraint,
π
∞

n=1
n(a
n
d
n
− b
n
c
n
) =
π
2
∞

n=1
n

a
2
n
+ b
2
n
+ c
2

n
+ d
2
n

then becomes
a
1
d
1
− b
1
c
1
= a
2
1
+ b
2
1
+ c
2
1
+ d
2
1
.
This implies that d
1
= a

1
and c
1
= −b
1
. a
0
and c
0
are arbitrary. Thus curves for which L
2
= 4πA have the
parametrization
x(t) =
a
0
2
+ a
1
cos t + b
1
sin t, y(t) =
c
0
2
− b
1
cos t + a
1
sin t.

Note that

x(t) −
a
0
2

2
+

y(t) −
c
0
2

2
= a
2
1
+ b
2
1
.
The curve is a circle of radius

a
2
1
+ b
2

1
and center (a
0
/2, c
0
/2).
Solution 28.20
1. The Fourier sine series has the form
x(1 − x) =
∞

n=1
a
n
sin(nπx).
The norm of the eigenfunctions is

1
0
sin
2
(nπx) dx =
1
2
.
The coefficients in the expansion are
a
n
= 2


1
0
x(1 − x) sin(nπx) dx
=
2
π
3
n
3
(2 − 2 cos(nπ) − nπ sin(nπ))
=
4
π
3
n
3
(1 − (−1)
n
).
1416
Thus the Fourier sine series is
x(1 − x) =
8
π
3
∞

n=1
odd n
sin(nπx)

n
3
=
8
π
3
∞

n=1
sin((2n − 1)πx)
(2n − 1)
3
.
The Fourier cosine series has the form
x(1 − x) =
∞

n=0
a
n
cos(nπx).
The norm of the eigenfunctions is

1
0
1
2
dx = 1,

1

0
cos
2
(nπx) dx =
1
2
.
The coefficients in the expansion are
a
0
=

1
0
x(1 − x) dx =
1
6
,
a
n
= 2

1
0
x(1 − x) cos(nπx) dx
= −
2
π
2
n

2
+
4 sin(nπ) − nπ cos(nπ)
π
3
n
3
= −
2
π
2
n
2
(1 + (−1)
n
)
Thus the Fourier cosine series is
x(1 − x) =
1
6
−
4
π
2
∞

n=1
even n
cos(nπx)
n

2
=
1
6
−
1
π
2
∞

n=1
cos(2nπx)
n
2
.
1417
-1
-0.5 0.5
1
-0.2
-0.1
0.1
0.2
-1
-0.5 0.5
1
-0.2
-0.1
0.1
0.2

Figure 28.13: The odd and even periodic extension of x(1 − x), 0 ≤ x ≤ 1.
The Fourier sine series converges to the odd periodic extension of the function. Since this function is C
1
,
continuously differentiable, we know that the Fourier coefficients must decay as 1/n
3
. The Fourier cosine series
converges to the even periodic extension of the function. Since this function is only C
0
, continuous, the Fourier
coefficients must decay as 1/n
2
. The odd and even periodic extensions are shown in Figure 28.13. The sine series
is better because of the faster convergence of the series.
2. (a) We substitute x = 0 into the cosine series.
0 =
1
6
−
1
π
2
∞

n=1
1
n
2
∞


n=1
1
n
2
=
π
2
6
1418
(b) We substitute x = 1/2 into the cosine series.
1
4
=
1
6
−
1
π
2
∞

n=1
cos(nπ)
n
2
∞

n=1
(−1)
n

n
2
= −
π
2
12
(c) We substitute x = 1/2 into the sine series.
1
4
=
8
π
3
∞

n=1
sin((2n − 1)π/2)
(2n − 1)
3
∞

n=1
(−1)
n
(2n − 1)
3
= −
π
3
32

1419
Chapter 29
Regular Sturm-Liouville Problems
I learned there are troubles
Of more than one kind.
Some come from ahead
And some come from behind.
But I’ve bought a big bat.
I’m all ready, you see.
Now my troubles are going
To have troubles with me!
-I Had Trouble in Getting to Solla Sollew
-Theodor S. Geisel, (Dr. Suess)
29.1 Derivation of the Sturm-Liouville Form
Consider the eigenvalue problem on the finite interval [a . . . b],
p
2
(x)y

+ p
1
(x)y

+ p
0
(x)y = µy,
1420
subject to the homogeneous unmixed boundary conditions
α
1

y(a) + α
2
y

(a) = 0, β
1
y(b) + β
2
y

(b) = 0.
Here the coefficient functions p
j
are real and continuous and p
2
> 0 on the interval [a . . . b]. (Note that if p
2
were
negative we could multiply the equation by (−1) and replace µ by −µ.) The parameters α
j
and β
j
are real.
We would like to write this problem in a form that can be used to obtain qualitative information about the problem.
First we will write the operator in self-adjoint form. We divide by p
2
since it is non-vanishing.
y

+

p
1
p
2
y

+
p
0
p
2
y =
µ
p
2
y.
We multiply by an integrating factor.
I = exp


p
1
p
2
dx

≡
e
P (x)
e

P (x)

y

+
p
1
p
2
y

+
p
0
p
2
y

=
e
P (x)
µ
p
2
y

e
P (x)
y




+
e
P (x)
p
0
p
2
y =
e
P (x)
µ
p
2
y
For notational convenience, we define new coefficient functions and parameters.
p =
e
P (x)
, q =
e
P (x)
p
0
p
2
, σ =
e
P (x)

1
p
2
, λ = −µ.
Since the p
j
are continuous and p
2
is positive, p, q, and σ are continuous. p and σ are positive functions. The problem
now has the form,
(py

)

+ qy + λσy = 0,
subject to the same boundary conditions,
α
1
y(a) + α
2
y

(a) = 0, β
1
y(b) + β
2
y

(b) = 0.
This is known as a Regular Sturm-Liouville problem. We will devote much of this chapter to studying the properties of

this problem. We will encounter many results that are analogous to the properties of self-adjoint eigenvalue problems.
1421
Example 29.1.1
d
dx

ln x
dy
dx

+ λxy = 0, y(1) = y(2) = 0
is not a regular Sturm-Liouville problem since ln x vanishes at x = 1.
Result 29.1.1 Any eigenvalue problem of the form
p
2
y

+ p
1
y

+ p
0
y = µy, for a ≤ x ≤ b,
α
1
y(a) + α
2
y


(a) = 0, β
1
y(b) + β
2
y

(b) = 0,
where the p
j
are real and continuous and p
2
> 0 on [a, b], and the α
j
and β
j
are real can be
written in the form of a regular Sturm-Liouville problem,
(py

)

+ qy + λσy = 0, on a ≤ x ≤ b,
α
1
y(a) + α
2
y

(a) = 0, β
1

y(b) + β
2
y

(b) = 0.
29.2 Properties of Regular Sturm-Liouville Problems
Self-Adjoint. Consider the Regular Sturm-Liouville equation.
L[y] ≡ (py

)

+ qy = −λσy.
1422
We see that the operator is formally self-adjoint. Now we determine if the problem is self-adjoint.
v|L[u] − L[v]|u = v|(pu

)

+ qu − (pv

)

+ qv|u
= [vpu

]
b
a
− v


|pu

 + v|qu − [pv

u]
b
a
+ pv

|u

 − qv|u
= [vpu

]
b
a
− [pv

u]
b
a
= p(b)

v(b)u

(b) − v

(b)u(b)


+ p(a)

v(a)u

(a) − v

(a)u(a)

= p(b)

v(b)

−
β
1
β
2

u(b) −

−
β
1
β
2

v(b)u(b)

+ p(a)


v(a)

−
α
1
α
2

u(a) −

−
α
1
α
2

v(a)u(a)

= 0
Above we used the fact that the α
i
and β
i
are real.

α
1
α
2


=

α
1
α
2

,

β
1
β
2

=

β
1
β
2

Thus L[y] subject to the boundary conditions is self-adjoint.
Real Eigenvalues. Let λ be an eigenvalue with the eigenfunction φ. We start with Green’s formula.
φ|L[φ] − L[φ]|φ = 0
φ| − λσφ − −λσφ|φ = 0
−λφ|σ|φ + λφ|σ|φ = 0
(λ − λ)φ|σ|φ = 0
Since φ|σ|φ > 0, λ − λ = 0. Thus the eigenvalues are real.
1423
Infinite Number of Eigenvalues. There are an infinite of ei genvalues which have no finite cluster point. This

result is analogous to the result that we derived for self-adjoint eigenvalue problems. When we cover the Rayleigh
quotient, we will find that there is a least eigenvalue. Since the eigenvalues are distinct and have no finite clus ter point,
λ
n
→ ∞ as n → ∞. Thus the eigenvalues form an ordered sequence,
λ
1
< λ
2
< λ
3
< ··· .
Orthogonal Eigenfunctions. Let λ and µ be two distinct eigenvalues with the eigenfunctions φ and ψ. Green’s
formula states
ψ|L[φ] − L[ψ]|φ = 0.
ψ| − λσφ −−µσψ|φ = 0
−λψ|σ|φ + µψ|σ|φ = 0
(µ − λ)ψ|σ|φ = 0
Since the eigenvalues are distinct, ψ|σ|φ = 0. Thus eigenfunctions corresponding to distinct eigenvalues are orthogonal
with respect to σ.
Unique Eigenfunctions. Let λ be an eigenvalue. Suppose φ and ψ are two independent eigenfunctions corre-
sponding to λ.
L[φ] + λσφ = 0, L[ψ] + λσψ = 0
We take the difference of ψ times the first equation and φ times the second equation.
ψL[φ] − φL[ψ] = 0
ψ(pφ

)

− φ(pψ


)

= 0
(p(ψφ

− ψ

φ))

= 0
p(ψφ

− ψ

φ) = const
In order to satisfy the boundary conditions, the constant must be zero.
p(ψφ

− ψ

φ) = 0
1424
Since p > 0 the second factor vanishes.
ψφ

− ψ

φ = 0
φ


ψ
−
ψ

φ
ψ
2
= 0
d
dx

φ
ψ

= 0
φ
ψ
= const
φ and ψ are not i nde pendent. Thus each eigenvalue has a unique, (to within a multiplicative constant), eigenfunction.
Real Eigenfunctions. If λ is an eigenvalue with eigenfunction φ, then
(pφ

)

+ qφ + λσφ = 0.
We take the complex conjugate of this equation.

pφ




+ qφ + λσφ = 0.
Thus φ is also an eigenfunction corresponding to λ. Are φ and φ independent functions, or do they just differ by a
multiplicative constant? (For example,
e
ıx
and
e
−ıx
are independent functions, but ıx and −ıx are dependent.) From
our argument on unique eigenfunctions, we see that
φ = (const)φ.
Since φ and φ only differ by a multipl icative constant, the eigenfunctions can be chosen so that they are real-valued
functions.
1425
Rayleigh’s Quotient. Let λ be an eigenvalue with the eigenfunction φ.
φ|L[φ] = φ| − λσφ
φ|(pφ

)

+ qφ = −λφ|σ|φ

φpφ


b
a
− φ


|p|φ

 + φ|q|φ = −λφ|σ|φ
λ =
−

pφφ


b
a
+ φ

|p|φ

 − φ|q|φ
φ|σ|φ
This is known as Rayleigh’s quotient. It is useful for obtaining qualitative information about the eigenvalues.
Minimum Property of Rayleigh’s Quotient. Note that since p, q, σ and φ are bounded functions, the Rayleigh
quotient is bounded below. Thus there is a least eigenvalue. If we restrict u to be a real continuous function that
satisfies the boundary conditions, then
λ
1
= min
u
−[puu

]
b

a
+ u

|p|u

 − u|q|u
u|σ|u
,
where λ
1
is the least eigenvalue. This form allows us to get upper and lower bounds on λ
1
.
To derive this formula, we first write it in terms of the operator L.
λ
1
= min
u
−u|L[u]
u|σ|u
Since u is continuous and satisfies the boundary conditions, we can expand u in a series of the eigenfunctions.
−
u|L[u]
u|σ|u
= −


∞
n=1
c

n
φ
n


L [

∞
m=1
c
m
φ
m
]



∞
n=1
c
n
φ
n


σ



∞

m=1
c
m
φ
m

= −


∞
n=1
c
n
φ
n


−

∞
m=1
c
m
λ
m
σφ
m




∞
n=1
c
n
φ
n


σ



∞
m=1
c
m
φ
m

1426
We assume that we can interchange summation and integration.
=

∞
n=1

∞
m=1
c
n

c
m
λ
n
φ
m
|σ|φ
n


∞
n=1

∞
m=1
c
n
c
m
φ
m
|σ|φ
n

=

∞
n=1
|c
n

|
2
λ
n
φ
n
|σ|φ
n


∞
n=1
|c
n
|
2
φ
n
|σ|φ
n

≤ λ
1

∞
n=1
|c
n
|
2

φ
n
|σ|φ
n


∞
n=1
|c
n
|
2
φ
n
|σ|φ
n

= λ
1
We see that the minimum value of R ayleigh’s q uotient is λ
1
. The minimum is attained when c
n
= 0 for all n ≥ 2, that
is, when u = c
1
φ
1
.
Completeness. The set of the eigenfunctions of a regular Sturm-Liouville problem is complete. That is, any piecewise

continuous function defined on [a, b] can be expanded in a series of the eigenfunctions,
f(x) ∼
∞

n=1
c
n
φ
n
(x),
where the c
n
are the generalized Fourier coefficients,
c
n
=
φ
n
|σ|f
φ
n
|σ|φ
n

.
Here the sum is convergent in the mean. For any fixed x, the sum converges to
1
2
(f(x
−

)+f(x
+
)). If f (x) is continuous
and satisfies the boundary conditions, then the convergence is uniform.
1427
Result 29.2.1 Properties of regular Sturm-Liouville proble ms.
• The eigenvalues λ are real.
• There are an infinite number of eigenvalues
λ
1
< λ
2
< λ
3
< ··· .
There is a least eigenvalue λ
1
but there is no greatest eigenvalue, (λ
n
→ ∞ as n → ∞).
• For each eigenvalue, there is one unique, (to within a multiplicative constant), eigenfunc-
tion φ
n
. The eigenfunctions can be chosen to be real-valued. (Assume the φ
n
following
are real-valued.) The eigenfunction φ
n
has exactly n − 1 zeros in the open interval
a < x < b.

• The eigenfunctions are orthogonal with respect to the weighting function σ(x).

b
a
φ
n
(x)φ
m
(x)σ(x) dx = 0 if n = m.
• The eigenfunctions are complete. Any piecewise continuous function f(x) defined on
a ≤ x ≤ b can be expanded in a series of eigenfunctions
f(x) ∼
∞

n=1
c
n
φ
n
(x),
where
c
n
=

b
a
f(x)φ
n
(x)σ(x) dx


b
a
φ
2
n
(x)σ(x) dx
.
The sum converges to
1
2
(f(x
−
) + f(x
+
)).
• The eigenvalues can be related to the eigenfunctions with a formula known as the
Rayleigh quotient.
λ
n
=
−pφ
n
dφ
n
dx



b

a
+

b
a

p

dφ
n
dx

2
− qφ
2
n

dx

b
a
φ
2
n
σ dx
1428
Example 29.2.1 A simple example of a Sturm-Liouville problem is
d
dx


dy
dx

+ λy = 0, y(0) = y(π) = 0.
Bounding The Least Eigenvalue. The Rayleigh quotient for the first eigenvalue is
λ
1
=

π
0
(φ

1
)
2
dx

π
0
φ
2
1
dx
.
Immediately we see that the eigenvalues are non-negative. If

π
0
(φ


1
)
2
dx = 0 then φ = (const). The only constant that
satisfies the boundary conditions is φ = 0. Since the trivial solution is not an eigenfunction, λ = 0 is not an eigenvalue.
Thus all the eigenvalues are positive.
Now we get an upper bound for the first eigenvalue.
λ
1
= min
u

π
0
(u

)
2
dx

π
0
u
2
dx
where u is continuous and satisfies the boundary conditions. We choose u = x(x − π) as a trial function.
λ
1
≤


π
0
(u

)
2
dx

π
0
u
2
dx
=

π
0
(2x − π)
2
dx

π
0
(x
2
− πx)
2
dx
=

π
3
/3
π
5
/30
=
10
π
2
≈ 1.013
1429
Finding the Eigenvalues and Eigenfunctions. We consider the cases of negative, zero, and positive eigenvalues
to check our results above.
λ < 0. The general solution is
y = c
e
√
−λx
+d
e
−
√
−λx
.
The only solution that satisfies the boundary conditions is the trivial solution, y = 0. Thus there are no negative
eigenvalues.
λ = 0. The general solution is
y = c + dx.
Again only the trivial solution satisfies the boundary conditions, so λ = 0 is not an eigenvalue.

λ > 0. The general solution is
y = c cos(
√
λx) + d sin(
√
λx).
We apply the boundary conditions.
y(0) = 0 → c = 0
y(π) = 0 → d sin(
√
λπ) = 0
The nontrivial solutions are
√
λ = n = 1, 2, 3, . . . y = d sin(nπ).
Thus the eigenvalues and eigenfunctions are
λ
n
= n
2
, φ
n
= sin(nx), for n = 1, 2, 3, . . .
We can verify that this example satisfies all the prop erties listed in Result 29.2.1. Note that there are an infinite number
of eigenvalues. There is a least eigenvalue λ
1
= 1 but there is no greatest eigenvalu e. For each eigenvalue, there is one
eigenfunction. The n
th
eigenfunction sin(nx) has n −1 zeroes in the interval 0 < x < π.
1430

Since a series of the eigenfunctions is the familiar Fourier sine series, we know that the eigenfunctions are orthogonal
and complete. We check Rayleigh’s quotient.
λ
n
=
−pφ
n
dφ
n
dx



π
0
+

π
0

p

dφ
n
dx

2
− qφ
2
n


dx

π
0
φ
2
n
σ dx
=
−sin(nx)
d(sin(nx))
dx



π
0
+

π
0


d(sin(nx))
dx

2

dx


π
0
sin
2
(nx)dx
=

π
0
n
2
cos
2
(nx) dx
π/2
= n
2
Example 29.2.2 Consider the eigenvalue problem
x
2
y

+ xy

+ y = µy, y(1) = y(2) = 0.
Since x
2
> 0 on [1 . . . 2], we can write this problem in terms of a regular Sturm-Liouville eigenvalue problem. We divide
by x

2
.
y

+
1
x
y

+
1
x
2
(1 − µ)y = 0
We multiply by the integrating factor exp(

1
x
dx) = exp(ln x) = x and make the substitution, λ = 1 − µ to obtain
the Sturm-Liouville form.
xy

+ y

+ λ
1
x
y = 0
(xy


)

+ λ
1
x
y = 0
We see that the eigenfunctions will be orthogonal with respect to the weighting function σ = 1/x.
1431
The Rayleigh quotient is
λ =
−

pφφ


b
a
+ φ

|x|φ


φ|
1
x
|φ
=
φ

|x|φ



φ|
1
x
|φ
.
If φ

= 0, then only the trivial solution, φ = 0, satisfies the boundary conditions. Thus the eigenvalues λ are positive.
Returning to the original problem, we see that the eigenvalues, µ, satisfy µ < 1. Since this is an Euler equation, we
can find solutions with the substitution y = x
α
.
α(α − 1) + α + 1 − µ = 0
α
2
+ 1 − µ = 0
Note that µ < 1.
α = ±ı

1 − µ
The general solution is
y = c
1
x
ı
√
1−µ
+ c

2
x
−ı
√
1−µ
.
We know that the eigenfunctions can be written as real functions. We rewrite the solution.
y = c
1
e
ı
√
1−µ ln x
+c
2
e
−ı
√
1−µ ln x
An equivalent form is
y = c
1
cos(

1 − µ ln x) + c
2
sin(

1 − µ ln x).
We apply the boundary conditions.

y(1) = 0 → c
1
= 0
y(2) = 0 → sin(

1 − µ ln 2) = 0
→

1 − µ ln 2 = nπ, for n = 1, 2, . . .
1432
Thus the eigenvalues and eigenfunctions are
µ
n
= 1 −

nπ
ln 2

2
, φ
n
= sin

nπ
ln x
ln 2

for n = 1, 2, . . .
29.3 Solving Differential Equations With Eigenfunction Expansions
Linear Algebra. Consider the eigenvalue problem,

Ax = λx.
If the matrix A has a complete, orthonormal set of eigenvectors {xi
k
} with eigenvalues {λ
k
} then we can represent
any vector as a linear combination of the eigenvectors.
y =
n

k=1
a
k
xi
k
, a
k
= xi
k
· y
y =
n

k=1
(xi
k
· y) xi
k
This property allows us to solve the inhomogeneous equation
Ax − µx = b. (29.1)

Before we try to solve this equation, we should consider the existence/uniqueness of the solution. If µ is not an
eigenvalue, then the range of L ≡ A − µ is R
n
. The problem has a unique solution. If µ is an eigenvalue, then the
null space of L is the span of the eigenvectors of µ. That is, if µ = λ
i
, then nullspace(L) = span(xi
i
1
, xi
i
2
, . . . , xi
i
m
).
({xi
i
1
, xi
i
2
, . . . , xi
i
m
} are the eigenvalues of λ
i
.) If b is orthogonal to nullspace(L) then Equation
29.1 has a solution,
but it is not unique. If y is a solution then we can add any linear combination of {xi

i
j
} to obtain another solution.
Thus the solutions have the form
x = y +
m

j=1
c
j
xi
i
j
.
1433
If b is not orthogonal to nullspace(L) then Equation 29.1 has no solution.
Now we solve Equation 29.1. We assume that µ is not an eigenvalue. We expand the solution x and the inhomo-
geneity in the orthonormal eigenvectors.
x =
n

k=1
a
k
xi
k
, b =
n

k=1

b
k
xi
k
We substitute the expansions into Equation 29.1.
A
n

k=1
a
k
xi
k
− µ
n

k=1
a
k
xi
k
=
n

k=1
b
k
xi
k
n


k=1
a
k
λ
k
xi
k
− µ
n

k=1
a
k
xi
k
=
n

k=1
b
k
xi
k
a
k
=
b
k
λ

k
− µ
The solution is
x =
n

k=1
b
k
λ
k
− µ
xi
k
.
Inhomogeneous Boundary Value Problems. Consider the self-adjoint eigenvalue problem,
Ly = λy, a < x < b,
B
1
[y] = B
2
[y] = 0.
If the problem has a complete, orthonormal set of eigenfunctions {φ
k
} with eigenvalues {λ
k
} then we can represent
any square-integrable function as a linear combination of the eigenfunctions.
f =


k
f
k
φ
k
, f
k
= φ
k
|f =

b
a
φ
k
(x)f(x) dx
f =

k
φ
k
|fφ
k
1434
This property allows us to solve the inhomogeneous differential equation
Ly −µy = f, a < x < b, (29.2)
B
1
[y] = B
2

[y] = 0.
Before we try to solve this equation, we should consider the existence/uniqueness of the solution. If µ is not an
eigenvalue, then the range of L −µ is the space of square-integrable functions. The problem has a unique solution. If µ
is an eigenvalue, then the null space of L is the span of the eigenfunctions of µ. That is, if µ = λ
i
, then nullspace(L) =
span(φ
i
1
, φ
i
2
, . . . , φ
i
m
). ({φ
i
1
, φ
i
2
, . . . , φ
i
m
} are the eigenvalues of λ
i
.) If f is orthogonal to nullspace(L − µ) then
Equation 29.2 has a solution, but it i s not unique. If u is a solution then we can add any linear combination of {φ
i
j

}
to obtain another solution. Thus the solutions have the form
y = u +
m

j=1
c
j
φ
i
j
.
If f is not orthogonal to nullspace(L −µ) then Equation 29.2 has no solution.
Now we solve Equation 29.2. We assume that µ is not an eigenvalue. We expand the solution y and the inhomo-
geneity in the orthonormal eigenfunctions.
y =

k
y
k
φ
k
, f =

k
f
k
φ
k
It would be handy if we could substitute the expansions into Equation 29.2. However, the expansion of a function is

not necessarily differentiable. Thus we demonstrate that since y is C
2
(a . . . b) and satisfies the boundary conditions
B
1
[y] = B
2
[y] = 0, we are justified in substituting it into the differential equation. In particular, we will show that
L[y] = L


k
y
k
φ
k

=

k
y
k
L [φ
k
] =

k
y
k
λ

k
φ
k
.
To do this we will use Green’s identity. If u and v are C
2
(a . . . b) and satisfy the boundary conditions B
1
[y] = B
2
[y] = 0
then
u|L[v] = L[u]|v.
1435
First we assume that we can differentiate y term-by-term.
L[y] =

k
y
k
λ
k
φ
k
Now we directly expand L[y] and show that we get the same result.
L[y] =

k
c
k

φ
k
c
k
= φ
k
|L[y]
= L[φ
k
]|y
= λ
k
φ
k
|y
= λ
k
φ
k
|y
= λ
k
y
k
L[y] =

k
y
k
λφ

k
The series representation of y may not be differentiable, but we are justified in applying L term-by-term.
Now we substitute the expansions into Equation 29.2.
L


k
y
k
φ
k

− µ

k
y
k
φ
k
=

k
f
k
φ
k

k
λ
k

y
k
φ
k
− µ

k
y
k
φ
k
=

k
f
k
φ
k
y
k
=
f
k
λ
k
− µ
The solution is
y =

k

f
k
λ
k
− µ
φ
k
1436
Consider a second order, inhomogeneous problem.
L[y] = f(x), B
1
[y] = b
1
, B
2
[y] = b
2
We will expand the solution in an orthogonal basis.
y =

n
a
n
φ
n
We would like to substitute the series into the differential equation, but in general we are not allowed to di fferentiate
such series. To get around this, we use integration by parts to move derivatives from the solution y, to the φ
n
.
Example 29.3.1 Consider the problem,

y

+ αy = f(x), y(0) = a, y(π) = b,
where α = n
2
, n ∈ Z
+
. We expand the solution in a cosine series.
y(x) =
y
0
√
π
+
∞

n=1
y
n

2
π
cos(nx)
We also expand the inhomogeneous term.
f(x) =
f
0
√
π
+

∞

n=1
f
n

2
π
cos(nx)
We multiply the differential equation by the orthonormal functions and integrate over the interval. We neglect the
1437