d
dx
(
(1 +
x
2
)
y
′
)
′
+
λ
(1 +
x
2
)
y
= 0
,
y
′
(0) =
y
′
(1) = 0
.
Thus:
p
(
x
) = 1 +
x
2
,
q
(
x
) = 0,
σ
(
x
) = 1 +
x
2
.
(b)
(5 points)
Explain why there canNOT be any negative eigenvalues.
Solution.
The Rayleigh quotient works out to
λ
=
∫
1
0
(1 +
x
2
)
y
′
(
x
)
2
dx
∫
1
0
(1 +
x
2
)
y
(
x
)
2
dx
,
which can clearly not be negative.
(c)
(5 points)
Can
λ
= 0 be an eigenvalue?
JUSTIFY!
Solution.
The answer is YES.
y
= 1 is a non-zero solution of the problem for
λ
= 0.
3. Solve by separation of variables the following problems. You do not have to go through a detailed analysis
of any Sturm-Liouville problem that you already solved several times before, assuming you remember what
you did. Or have it on your cheat sheet. But explain from where it comes! (For example, Write, “As seen in
class, the solution of this problem is given by. . . ”)
(a)
(20 points)
u
t
=
u
xx
,
0
< x < π,
t >
0
,
u
(0
, t
)
=
0
,
u
(
π, t
) = 0
,
t >
0
u
(
x,
0)
=
3 sin 5
x
−
4 sin 7
x,
0
< x < π.
Solution.
The solution of the PDE solving the boundary conditions obtained by separation of variables is
u
(
x, t
) =
∞
∑
n
=1
c
n
e
−
n
2
t
sin
nx.
The
c
n
’s must satisfy
3 sin 5
x
−
4 sin 7
x
=
∞
∑
n
=1
c
n
sin
nx.
Thus
c
5
= 3
, c
7
=
−
4,
c
n
= 0 if
n
̸
= 5
,
7. The solution is
u
(
x, t
) = 3
e
−
25
t
sin 5
x
−
4
e
−
49
t
sin 7
x.
(b)
(20 points)
u
t
=
4
u
xx
,
0
< x <
1
,
t >
0
,
u
x
(0
, t
)
=
0
,
u
(1
, t
) = 0
,
t >
0
u
(
x,
0)
=
5 cos
9
πx
2
,
0
< x <
1
.
Solution.
The solution of the PDE solving the boundary conditions obtained by separation of vari-
ables is
u
(
x, t
) =
∞
∑
n
=1
c
n
e
−
(2
n
−
1)
2
π
2
t/
4
cos
(2
n