MSO-203 M ASSIGNMENT 2

IIT, KANPUR

4th October , 2023

′′
Problem 1: Prove that any solution of Airy’s equation y + xy = 0 has an
infinite number of zeros on (0, ∞), and at most one zero on (−∞, 0).
Solution First notice that the function f (x) = x ≥ 1, ∀x ≥ 1. Hence applying
Sturm comparison theorem with the equation
′′
u + u = 0 on [1, ∞).

We know sin(x) solves the second equations. Since sin(x) has infinitely many
zeros in [1, ∞), this implies that Airy’s equation has also infinitely many zeros
in [1, ∞).

For the second part, if possible, let a, b with a < b < 0 are two distinct zeros
of Airy’s equation on (−∞, 0). Consider the problem:
′′
u (x) + (b + δ)u(x) = 0, on x ∈ (a − 1, b + δ), (1)
√
where we can always choose δ such that b + δ < 0. Notice u(x) = e −(b+δ)x
solves the above solution and which does not have any zero on the entire real
line. Since,

q2 (x) = b + δ ≥ q1 (x) = x, on b + δ ≥ x ≥ a − 1,
we can apply Strum√ comparison principal, which then implies that there exist
a zero of u(x) = e −(b+δ)x in (a, b). This is a contradiction.

Note: It was necessary that we have to choose δ > 0 such that b + δ < 0,
because if it is strictly positive, then sin and cos function will appear as solution
of (1) that may have a zero in (a, b), that will not lead to any contradiction.

Problem 2 : Consider the following problem

2 2
z ′′ (t) + (et + 1)z(t) = t(et + 1) in R.

Let Z1 denotes a solution (not identically zero function) of the above problem.
Then show that the set
{t | Z1 (t) = t}
has infinitely many element.
Solution : Consider the new function

y(t) = Z1 (t) − t.

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Notice that it is sufficient to prove that there are infinitely many zero’s of the
function y. For this purpose, let us find out the equation that the function y
satisfies. Using ′′ ′′
Z1 (t) = y (t)
it is easy to see that
2
y ′′ (t) + (et + 1)y(t) = 0 in R.
2
Let p(t) = et + 1 and q(t) = 1 in the following
′′ 2
y (t) + (et + 1)y(t) = 0 in R

also consider ′′
y (t) + y(t) = 0 in R.
2
Notice et + 1 = p(t) > 1 = q(t). Now we want to apply Strum Comparison
theorem. Note g(t) = sin(t) solves the last equation which vanishes (zeros) at
the point t = nπ, n ∈ Z. Between two consecutive zeros of g, lies a zero of y1
and hence are infinitely in number.

Problem 3 : Let r be a continuous function on real line. Prove that any

nontrivial solution of ′′
y + r(x)y = 0
on a finite interval has at most a finite number of zeros.
Solution: Let us consider an arbitrary finite interval (a, b). Assume, if possible,
there exists infinitely many zeros of y in (a, b). Since r is a continuous on [a, b]
and [a, b] is closed and bounded interval, this implies r is a bounded function of
[a, b]. Therefore there exist M > 0 such that

r(x) ≤ M, ∀x ∈ [a, b].

Now consider the problem

′′
u + M u = 0, on (a, b).
√
We know u(x) = sin( M X) is a solution of the above equation. Applying
Sturm comparison
√ theorem, it implies, between any two zeros of y lies a zero of
u(x) = sin( M X). Since√we have assumed y has infinitely many zeros in (a, b)
this implies u(x) = sin( M X) has also infinitely many zeros in (a, b). This
clearly contradiction.

Problem 4 : Show that λ = 0, can never be an eigenvalue of the following

Eigenvalue problem:
 3 ′
− et u′ (t) = λu, on (0, 1),

2
 u′ (0) = 0 = u(1).
Solution : Assume (0, u), if possible, be an eigenpair. Multiply the equation
by u, and then integrate by parts to obtain
Z 1
t3 ′ 1 3
e u (t)u(t) 0
− et |u′ (t)|2 dt = 0.
0
This implies Z 1
3
et |u′ (t)|2 dt = 0.
0
3
Since et ̸= 0, ∀t, this implies |u′ (t)|2 = 0, ∀t ∈ (0, 1). Therefore u has to be
a constant function. But since u(1) = 0, the value of the constant has to be
zero. This contradicts the definition of eigenfunction (that it has to a non zero
function).

Problem 5: Find the solution of the following eigenvalue problem:

′′
a) y (t) + λy(t) = 0 on (0, L), y(0) = y(L) = 0, for fixed L > 0.
′′
b) y (t) + λy(t) = 0 on (0, 1), y(0) = y(1), y ′ (0) = y ′ (1).
′
y ′ (t)

c) t + (λ + 1) y(t) π
t3 = 0 on (1, e ), y(1) = 0, y(eπ ) = 0.
′′
d) y (t) + 8y ′ (t) + (λ + 16)y(t) = 0 on (0, π), y(0) = 0 = y(π).
′′
e) y (t) + λy(t) = 0 on (0, 1), y(0) = y ′ (1) = 0.

Solution(a). Given eigenvalue problem is

′′
y (t) + λy(t) = 0 on (0, L), (2)
y(0) = y(L) = 0.

The characteristic equation is given by

m2 + λ = 0. (3)

(Case 1.) Let us first consider the case when λ = 0. Then

m = 0 =⇒ y(t) = A + Bt

for some constant A, B ∈ R. Applying the boundary condition we get

y(0) = 0 =⇒ A = 0,
y(L) = 0 =⇒ BL = 0 =⇒ B = 0 (since L ̸= 0),
=⇒ y(t) ≡ 0.

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 This contradicts the definition of eigenfunction. Recall that eigenfunction by
definition has to be a non-zero function. Therefore λ = 0 can never be an
eigenvalue for the above problem.
(Case 2.) Let λ < 0. Assume that λ = −k 2 for some k ̸= 0. Putting this in
(3) we get

m = ±k
=⇒ y(t) = Aekt + Be−kt

is the expression for any general solution of (2) in this case. Incorporating the
Boundary conditions again, we get

y(0) = 0 =⇒ A + B = 0,
y(L) = 0 =⇒ AekL + Be−kL = 0
=⇒ A(ekL − e−kL ) = 0.

If A ̸= 0 then

e2kL = 1 = e0 =⇒ k = 0 (since L ̸= 0). (4)

This is a contradiction to the assumption that k ̸= 0. Hence A = 0 and

B = −A = 0. Like previous case y(t) ≡ 0 becomes the only available choice of
eigenfunction, which cannot be the case.
(case 3) Let λ > 0. Assume that λ = k 2 , k ̸= 0. In this case the general
solution of (2) is given by

y(t) = A cos(kt) + B sin(kt). (5)

Boundary condition gives

y(0) = 0 =⇒ A + 0 = 0 =⇒ A = 0,
=⇒ y(t) = B sin(kt),
y(L) = 0 =⇒ B sin(kL) = 0.

Since we want to avoid B = 0, this implies

sin(kL) = 0
mπ
=⇒ k = (m ∈ Z \ {0}),
L
where Z is the set of integers. There are infinitely many choices for such k.
The set of eigenvalues are

m2 π 2
λm = , m ∈ N \ {0}.
L2
[Note that in the last line we have changed m ∈ N from m ∈ Z because same
eigenvalue will be counted (eg. λ1 , λ−1 are same).]

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 The eigenfunctions “ym (t)” corresponding to the eigenvalue “λm ” is given
by  mπ 
ym (t) = B sin t , m ∈ {1, 2, 3, . . .}.
L
Solution (b). Given eigenvalue problem is
′′
y (t) + λy(t) = 0 on (0, 1), (6)
′ ′
y(0) = y(1), y (0) = y (1).
Like the previous problem the characteristic equation is given by
m2 + λ = 0. (7)
(Case 1.) Let us first consider the case when λ = 0. Then
y(t) = A + Bt
for some constant A, B ∈ R. Applying the boundary conditions we get
y(0) = y(1) =⇒ B = A + B =⇒ A = 0,
=⇒ y(t) ≡ B (constant function),
satisfies the second boundary condition. Hence λ = 0 is an eigenvalue with
non-zero constant function as the eigenfunction.
(Case 2.) Let λ < 0. Then λ = −k 2 for some k ̸= 0. The general solution is
y(t) = Aekt + Be−kt .
The Boundary conditions gives
y(0) = y(1) =⇒ A + B = Aek + Be−k ,
=⇒ A(1 − ek ) + B(1 − e−k ) = 0. (8)
Second boundary condition gives
y ′ (0) = y ′ (1)
=⇒ k(A − B) = k(Aek − Be−k )
=⇒ A − B = Aek − Be−k (since k ̸= 0)
=⇒ A(1 − ek ) + B(e−k − 1) = 0 (9)
Equation (8) and (9) can be written as
1 − ek 1 − e−k
    
A 0
= . (10)
1 − ek e−k − 1 B 0
Let us calculate,
1 − ek 1 − e−k
   
1 1
det = (1 − ek )(1 − e−k ) det
1 − ek e−k − 1 1 −1
= −2(1 − ek )(1 − e−k ) ̸= 0 (as k ̸= 0).

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Therefore from (10) we have

A = B = 0.

Therefore λ < 0 cannot be an eigenvalue.

(case 3) Let λ > 0. Assume that λ = k 2 , k ̸= 0. In this case the general
solution is given by
y(t) = A sin(kt) + B cos(kt). (11)
Boundary condition gives

y(0) = y(1) =⇒ B = A sin k + B cos k, (12)

and,

y ′ (0) = y ′ (1)
=⇒ A cos(k0) − B sin(k0) = A cos k − B sin k
=⇒ A = A cos k − B sin k. (13)

(12) and (13) together in form of system of equation can be written as

    
sin k cos k − 1 A 0
= . (14)
cos k − 1 − sin k B 0

 
sin k cos k − 1
det = − sin2 k − (cos k − 1)2 = −2 + 2 cos k. (15)
cos k − 1 − sin k

We will get non-trivial solutions of A, B only if (15) becomes 0, that is,

cos k = 1
=⇒ k = ±2nπ, n ∈ N \ {0}
=⇒ λn = 4n2 π 2 .

The eigenfunction corresponding to the eigenvalue λn is given by

yn (t) = A cos(2nπt)
ψn (t) = B sin(2nπt), n ∈ N.

This means that there are two linearly independent eigenfunctions “yn ” and
“ψn ” corresponding to the eigenvalue λn .
Solution (c). For this problem we want to change the independent variable
first to reduce the problem into a constant coefficient problem (linear) for which
we know how to find the general solution.

Put log t = z, and

define ψ(z) = y(t),

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where y(t) satisfies 3c (the given problem).
dy dψ dz 1 dψ
= =
dt dz dt t dz
d2 y 1 dψ 1 d2 ψ
2
=− 2 + 2 2.
dt t dz t dz
Using this we get
 y ′ (t) ′ λ + 1
+ y(t) = 0
t t3
′′
y ′ (t) y (t) y(t)
=⇒ − 2 + + (λ + 1) 3 = 0
t t t
dψ 1 1 d2 ψ ψ(z)
=⇒ − 2 3
+ 3 2 + (λ + 1) 3 = 0
dz t t dz t
d2 ψ dψ
=⇒ −2 + (λ + 1)ψ(z) = 0 (since t > 0), (16)
dz 2 dz
and the initial condition changes as

at t = 1, z=0
π
t=e , z=π
=⇒ ψ(0) = ψ(π) = 0.

Look that we know the general solution of (16) . Characteristic equation is

given by
√
(m − 1)2 + λ = 0 ⇐⇒ m = 1 ± i λ
√ √
=⇒ “ez cos λz” and “ez sin λz”

are two linearly independent solution of (16). Thus the general solution is
√ √
ψ(z) = Aez cos( λz) + Bez sin( λz).

The initial conditions are

ψ(0) = ψ(π) = 0.

Now

ψ(0) = 0 =⇒ A = 0
√
ψ(π) = 0 =⇒ Beπ sin( λπ) = 0.

Since eπ ̸= 0 and we want B ̸= 0,

√
=⇒ sin( λπ) = 0 = sin(nπ), n ∈ Z
=⇒ λ = n2 , n ∈ N \ {0}.

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 Since for distinct values of natural numbers one obtains different eigenvalue, so
it is more appropriate to index the set of eigenfunctions as ψn for each n ∈ N,

ψn (z) = Bez sin(nz) for some constant B ∈ R \ {0}.

Now going back to our original coordinate system “t” by using the relation
log t = z, we get

yn (t) = ψn (z) = Bez sin(nz) = Bt sin(n log t)

=⇒ yn (t) = Bt sin(n log t),

are the countable sequence of eigenfunctions for the above problem.

SOLUTION OF PART 3(d) and 3(e) are kept as exercise for you.

Problem 6: Show that the following family of functions

n o
sin(nπ log(t)} ,
n∈N

are orthogonal. The domain of definition for each function in the above family
is assumed to be (1, e). Are they orthonormal ? If not, can it be turned in to a
orthonormal family ?
Hint: Try finding eigenvalues and eigenfunctions for the following problem:
′′
t2 y (t) + ty ′ (t) + λy = 0, on (1, e), y(1) = y(e) = 0.
Further Hint: Change the independent variable t = log(x) to reduce the above
problem to a more known problem.