Centre for Mathematical Sciences
Mathematics, Faculty of Science
Solutions - Set 7
1. a) Eigenvalues λn =
(2n+1)2 π 2
.
4
Normalized eigenfunctions un (x) =
√
2 cos
(2n+1)π
x
2
.
b)
∞
X
(−1)n · 4
(2n + 1)π
1=
cos
x .
(2n + 1)π
2
n=0
c)
2
x −1=
∞
X
(−1)n+1 · 32
n=0
(2n + 1)3 π 3
cos
(2n + 1)π
x .
2
d) The convergence is uniform for g but not for f . g is in the domain of the operator (g ∈ C 2 ,
g 0 (0) = 0, g(1) = 0), but not f (since f (1) 6= 0). Note that the series for f doesn’t even
converge pointwise to f at x = 1 (since the partial sums sN satisfy sN (1) = 0).
2. We have
Z
1
|g(x)|2 dx =
0
∞
X
|cn |2 ,
n=0
where
√
16 2
.
(2n + 1)3 π 3
n+1
cn = (−1)
Hence,
∞
X
n=0
1
π6
=
(2n + 1)6
512
Z
1
(x2 − 1)2 dx =
0
π6
.
960
3. We have Lfn = n2 fn and thus
kLfn k = n2 kfn k.
It follows that
kL(kfn k−1 fn )k =
kLfn k
= n2 → ∞
kfn k
as n → ∞. But this means that
kLk = sup kLf k = ∞,
kf k=1
so that L is unbounded.
Please, turn over!
4. If λ 6= 0 is an eigenvalue and u a corresponding eigenvector, then A(λ−1 u) = u, so u ∈
Ran(A). Hence,
∞
X
u=
huj , uiuj .
j=0
If λ 6= αj for all j, then huj , ui = 0 for all j (eigenvectors corresponding to different eigenvalues
are orthogonal). Hence we obtain the contradiction u = 0. Thus λ = αn for some n. Since
limj→∞ αj = 0, there are only finitely many j for which αj = αn . Thus
X
u=
huj , uiuj ,
j : αj =αn
where the sum is finite.
If Ran(A), the eigenfunction expansion holds for any u ∈ H. If Au = 0, we have that
huj , ui = 0 for all j since αj 6= 0. But then u = 0, so 0 is not an eigenvalue.
Alternatively, we can use the following argument which doesn’t rely on the eigenfunction expansion. Let {vn } be a sequence in H with Avn → u as n → ∞. Then
kuk2 = hu, ui = lim hu, Avn i = lim hAu, vn i = 0.
n→∞
n→∞
Hence, u = 0 and 0 is not an eigenvalue.
5. Suppose that u0 is a solution of −u00 + qu = 0 with two zeros c < d. Set q0 (x) = q(x) and
q1 (x) ≡ 0, so that q1 (x) < q0 (x) for all x. By Sturm’s comparison theorem, any solution of
the equation u00 = 0 then has a zero of between c and d. But this gives a contradiction since
u1 (x) ≡ 1 is a solution with no zeros.
6. We use Sturm’s comparison theorem. Write the problem as −y 00 + q1 (x)y = 0, with q1 (x) =
−ex . Then q1 (x) ≤ −e0 = −1 on [0, π]. Hence we can take −y 00 − y = 0 as our comparison
equation (q0 (x) = −1). The solution y0 (x) = sin x vanishes at 0 and π, implying that any
solution y1 of y 00 + ex y = 0 has a zero in (0, π) since q1 (x) 6≡ q0 (x).
7. Let λ ≥ 4 and set
q1 (x) = −
5
+4 .
1 + x2
Then q1 (x) < q0 (x) on [π, 2π], where q0 (x) ≡ −4. The function u1 (x) = sin(2x) solves the
equation −u00 − 4u = 0. It vanishes at π, 3π/2 and 2π. By Sturm’s comparison theorem, any
solution of −u00 + q1 (x)u = 0 must therefore have a zero in (π, 3π/2) and a zero in (3π/2, 2π).
Thus, any eigenfunction corresponding to an eigenvalue λ ≥ 4 has at least two zeros in the
interval (π, 2π). If we list the eigenvalues in increasing order, λ0 < λ1 < λ2 < · · · , then Thm
5.17 implies that un (corresponding eigenfunction) has precisely n zeros in (π, 2π). But this
means that λ0 and λ1 are both < 4, since otherwise, the corresponding eigenfunctions would
have at least two zeros.