640:527: Homework 16: Solutions
17.7, Problem 1 (c).
The Sturm Liouville problem is
y
00
+
λy
= 0,
y
0
(0) = 0,
y
0
(
L
) = 0. In
the standard notation,
p
= 1,
q
= 0 and
w
= 1. For any function
φ
satisfying the boundary
conditions
φ
0
(0) =
φ
0
(
L
) = 0,
p
(1)
φ
(1)
φ
0
(1)

p
(0)
φ
(0)
φ
0
(0) = 0, and, since
q
= 0, Theorem 17.7.2
implies that there are no negative eigenvalues. It is easy to check that the constant function
˜
φ
= 1
satisﬁes the equation and its boundary condition for
λ
= 0, and hence
λ
= 0 is an eigenvalues. The
positive eigenvalues are easily found to be
λ
n
= (
nπ/L
)
2
for integers
n
≥
1, with corresponding,
unnormalized eigenfunctions
˜
φ
n
(
x
) = cos((
nπ/L
)
x
). With normalization using the inner product
h
h,g
i
=
R
L
0
h
(
x
)
g
(
x
)
dx
, the eigenfunctions are
φ
0
(
x
) =
1
√
L
,
φ
n
(
x
) =
2
√
L
cos((
nπ/L
)
x
)
,
n
≥
1
.
Any function
f
with

f

<
∞
has the expansion
h
f,φ
0
i
φ
0
(
x
) +
∞
X
1
h
f,φ
n
i
φ
n
(
x
)
=
a
0
+
∞
X
1
a
n
cos((
nπ/L
)
x
)
.
where
a
0
=
1
L
Z
L
0
f
(
x
)
dx,
a
n
=
2
L
Z
L
0
f
(
x
)cos((
nπ/L
)
x
)
dx.
We obtain therefore the half range Fourier cosine series in this problem. For the function
f
(
x
),
where
f
(
x
) = 1, for 0
≤
x
≤
L/
2 and
f
(
x
) = 0, for
L/
2
< x
≤
L
, calculation of the coe±cients leads
to the eigenfunction expansion
1
2
+
2
π
∞
X
1
sin(
nπ/
2)
n
cos((
nπ/L
)
x
) =
1
2
+
2
π
∞
X
k
=0
(

1)
k
2
k
+ 1
cos(((2
k
+ 1)
π/L
)
x
)
.
Problem 1 (d).
The problem is
y
00
+
λy
= 0,
y
0
(0) = 0,
y
(
L
) +
y
0
(
L
) = 0. This is a SturmLiouville
problem with
p
= 1,
q
= 0 and
w
= 1. For any function
φ
satisfying the boundary conditions
φ
0
(0) = 0 and
φ
(
L
) +
φ
0
(
L
) = 0,
p
(
L
)
φ
(
L
)
φ
0
(
L
)

p
(0)
φ
(0)
φ
0
(0) =

φ
(
L
)
2
≤
0, Since
q
= 0 also,
Theorem 17.7.2 implies that there are no negative eigenvalues. It is easy to check that
λ
= 0 is not
an eigenvalue.
The positive eigenvalues are found by substituting the general solution
A
cos(
x
√
λ
)+
B
sin(
x
√
λ
)
into the boundary conditions and looking for a condition which ensures there is a nontrivial solution
for
A
and
B
. In this case, substitution in the boundary conditions gives
√
λB
= 0
(
A
+
√
λB
)cos(
L
√
λ
) + (
B

√
λA
)sin(
L
√
λ
) = 0
.