M2AA1 Diffferential Equations: Problem Sheet 5
1. Solve the following boundary value problem: y 00 + y = x2 , y(0) = 0, y(π/2) = 1.
Answer: The general solution of the differential equation is −2 + x2 + c1 cos(x) +
c2 sin(x). The boundary conditions give −2 + c1 = 0 and −2 + (π/2)2 + c2 = 1. This
has a unique solution.
2. Show that the following boundary value problem has no solution: y 00 + y = x, y(0) +
y 0 (0) = 0, y(π/2) − y 0 (π/2) = π/2.
Answer: The general solution of the differential equation is x + c1 cos(x) + c2 sin(x).
The boundary conditions give c1 + 1 + c2 = 0 and π/2 + c2 − (1 − c1 ) = π/2. This
pair of equations don’t have a solution.
3. Find the eigenvalues λn and eigenfunctions yn for the equation y 00 + λy = 0 with
boundary values y 0 (0) = 0 and y 0 (1) = 0.
Answer: Write λ = µ2 . Then y(x) = c1 cos(µx) + c2 sin(µx) is a solution of the
differential equation. y 0 (0) = 0 implies c2 = 0. y 0 (1) = 0 implies that µ = nπ for
some integer n. So the eigenvalues are λn = (nπ)2 and the eigenfunctions yn (x) =
cos((nπ)x).
4. Verify that for the Sturm-Liouville problem (xy 0 )0 + (λ/x)y = 0 with y 0 (1) = 0 and
x
y 0 (e2π ) = 0 the eigenvalues are λn = n2 /4 and the eigenfunctions cos( n log
2 ).
Answer: Apply the substitution t = log x so x(t) = et and v(t) = y(et ). Then v 0 =
et y 0 (et ) = xy 0 (et ) and v 00 = x0 y 0 (et ) + xet y 00 (et ) = xy 0 + x2 y 00 . So (xy 0 )0 + (λ/x)y = 0
corresponds to v 00 +λv = 0 and v 0 (0) = 0 and v 0 (2π) = 0. This has solutions λ = n2 /4
and yn (t) = cos((n/2)t). This gives the required eigenfunctions.
5. Take y 00 + qy = 0 with q(x) < 0. Show that, unless y ≡ 0, one has that y has at most
one zero.
(Hint: show that if y has zeros at a, b then there is a point x ∈ (a, b) so that
y 00 (x) · y(x) > 0.)
Answer: Assume that y(a) = y(b) = 0 and y(x) > 0 for x ∈ (a, b). Since y 00 + qy = 0
and q < 0, we get y 00 (x) > 0 for all x ∈ (a, b). Since y(a) = y(b) = 0, y takes a
maximum for some x ∈ (a, b) and so there exists a point x ∈ (a, b) so that y 0 (x) = 0
and y 00 (x) ≤ 0. This contradicts y 00 (x) > 0 for all x ∈ (a, b).
If y(x) < 0 for x ∈ (a, b) then the same proof applies, replacing > by < and maximum
by minimum.
6. Let < v, w > be an inner product on a vector space. Then a matrix
P A is called
self-adjoint (or symmetric) if < Av, w >=< v, Aw >. Let < v, w >= v · w where ·
represents the usual inner product.
1
(a) Assume that A real, then show that A is self-adjoint if and only if Atr = A
where Atr is the transpose of A.
tr
(b) Show that - in general - A is self-adjoint if and only if A = A (i.e. A
obtained from A by taking the complex conjugate and the transpose of A.
tr
is
(c) Bonus question: If < w, w > is another inner product, then A self-adjoint might
not coincide with Atr = A.
P P
P P
Answer (a,b): < Av, w >= Av·w̄ = i ( j aij vj )w̄i = j ( i aij w̄i )vj = v·Ātr w̄ and
< v, Aw >= v · Aw = v · Aw̄. A is self-adjoint if and only if if < Av, w >=< v, Aw >
for all v, w so if and only if v · Ātr w̄ = v · Aw̄ for all v, w. Since this holds for all v, w
it holds in particular for v = ei and w = ej for each i, j. So (Ātr )ij = ei · Ātr w̄j =
ei · Aēj = Aij for each i, j. In other words, this is equivalent to Ātr = A. When A is
real this is equivalent to Atr = A.
2