Math 334

Assignment 7 — Solutions

1. Chebyshev’s differential equation is

(1 − x2)y′′ − xy′ + α2y = 0,

where α is a constant.

(a) Find two linearly independent power series solutions valid for |x| < 1.

(b) Show that if α = n is a non–negative integer, then there is a polynomial solution of degree n.These polynomials, when properly normalised, are called Chebyshev polynomials.

(c) Find a polynomial solution for each of the cases α = n = 0, 1, 2, 3.

Solution

(a) The point x = 0 is an ordinary point so we look for a solution of the form y(x) =∑

∞

n=0 anxn.Plug the series into the equation to get

∞∑

n=2

n(n − 1)anxn−2 −∞∑

n=2

n(n − 1)anxn −∞∑

n=1

nanxn +∞∑

n=0

α2anxn = 0.

Upon re-arrangement we get

2a2 + α2a0 + (6a3 − a1 + α2a0) x

+∞∑

n=2

{(n + 2)(n + 1)an+2 − [n(n − 1) + n − α2]an}xn = 0.

Equating the powers of x yields

2a2 + α2a0 = 0,

6a3 − a1 + α2a0 = 0,

(n + 2)(n + 1)an+2 − (n2 − α2)an = 0, n = 2, 3, . . . .

Solving for the an’s we get

a2 = −α2

2a0, a3 =

1 − α2

3!a1, an+2 =

n2 − α2

(n + 2)(n + 1)an, n = 2, 3, . . . .

Writing out the first few explicitly yields:

a4 =(22 − α2)(−α2)

4!a0, a5 =

(32 − α2)(1− α2)

5!a1, a6 =

(42 − α2)(22 − α2)(−α2)

6!a0.

A pattern clearly emerges:

a2n =[(2n − 2)2 − α2][(2n − 4)2 − α2] · · · (22 − α2)(−α2)

(2n)!a0, n = 1, 2, . . . ,

a2n+1 =[(2n − 1)2 − α2][(2n − 3)2 − α2] · · · (32 − α2)(1− α2)

(2n + 1)!a1, n = 1, 2, . . . .

Hence, two nontrivial solutions are given by:

ϕ1(x) = 1 +∞∑

n=1

[(2n − 2)2 − α2][(2n − 4)2 − α2] · · · (22 − α2)(−α2)

(2n)!x2n,

ϕ2(x) = x +∞∑

n=1

[(2n − 1)2 − α2][(2n − 3)2 − α2] · · · (32 − α2)(1− α2)

(2n + 1)!x2n+1.

1

Math 334 Assignment 7 — Solutions 2

(b) From the recurrence relation we get

an+2 =n2 − α2

(n + 2)(n + 1)an =⇒

a2n+2 =(2n)2 − α2

(2n + 2)(2n + 1)a2n,

a2n+3 =(2n + 1)2 − α2

(2n + 3)(2n + 2)a2n+1.

Now suppose that α is either an even or an odd integer. Thus,

α = 2N =⇒ a2N+2 = 0 =⇒ a2n+2 = 0 for n > N

which implies that ϕ1 is a polynomial of degree 2N , whereas

α = 2N + 1 =⇒ a2N+3 = 0 =⇒ a2n+3 = 0 for n > N

which implies that ϕ2 is a polynomial of degree 2N + 1.

(c) For the cases α = 0, 1, 2, 3 we have

α = 0 =⇒ a2n = 0 for n > 1 =⇒ ϕ1(x) = 1,

α = 1 =⇒ a2n+1 = 0 for n > 1 =⇒ ϕ2(x) = x,

α = 2 =⇒ a2n = 0 for n > 2 =⇒ ϕ1(x) = 1 − 2x2,

α = 3 =⇒ a2n+1 = 0 for n > 2 =⇒ ϕ2(x) = x − 4

5x3.

2. For each of the following differential equations, locate and classify its singular points.

(a) x3(x− 1)y′′ − 2(x− 1)y′ + 3xy = 0.

(b) x2(x2 − 1)2y′′ − x(x− 1)y′ + 2y = 0.

Solution

(a) Write the differential equation in standard form: y′′ − 2

x3y′ +

3

x2(x − 1)y = 0.

We have P (x) = − 2

x3and Q(x) =

3

x2(x− 1), so the singular points occur at x = 0, 1.

i. At x = 0 we have limx→0

xP (x) = limx→0

−2

x2which does not exist. Therefore x = 0 is an irregular

singular point of the differential equation.

ii. At x = 1 we have limx→1

(x− 1)P (x) = limx→1

−2(x − 1)

x3= 0

and limx→1

(x − 1)2Q(x) = limx→1

3(x− 1)

x2= 0. Therefore x = 1 is a regular singular point of the

differential equation.

(b) Write the differential equation in standard form: y′′ − 1

x(x + 1)(x2 − 1)y′ +

2

x2(x2 − 1)2y = 0.

We have

P (x) =−1

x(x + 1)(x2 − 1)

Q(x) =2

x2(x2 − 1)2

=⇒ x = 0,±1 are singular points

Math 334 Assignment 7 — Solutions 3

At x = 0 we have

limx→0

xP (x) = limx→0

−1

(x + 1)(x2 − 1)= 1

limx→0

x2Q(x) = limx→0

2

(x2 − 1)2= 2

=⇒ x = 0 is a regular singular point

At x = +1 we have

limx→1

(x − 1)P (x) = limx→1

−1

x(x + 1)2= −1

4

limx→1

(x− 1)2Q(x) = limx→1

2

x2(x + 1)2=

1

2

=⇒ x = 1 is a regular singular point

At x = −1 we have

limx→1

(x+1)P (x) = limx→1

−1

x(x2 − 1)which does not exist ∴ x = −1 is an irregular singular point

3. Find the indicial equation and its roots for each of the following differential equations:

(a) x3y′′ + (cos 2x− 1)y′ + 2xy = 0;

(b) 4x2y′′ + (2x4 − 5x)y′ + (3x2 + 2)y = 0.

Solution

(a) A0 = limx→0

A(x) = limx→0

xP (x) = limx→0

cos 2x − 1

x2= lim

x→0

−2 sin 2x

2x= −2.

B0 = limx→0

B(x) = limx→0

x2Q(x) = limx→0

2 = 2.

The indicial equation is r2 + (A0 − 1)r + B0 = 0 i.e. r2 − 3r + 2 = 0. The roots are r = 1, 2.

(b) A0 = limx→0

A(x) = limx→0

xP (x) = limx→0

2x3 − 5

4= −5

4.

B0 = limx→0

B(x) = limx→0

x2Q(x) = limx→0

3x2 + 2

4= 2.

The indicial equation is r2 − 9

4r +

1

2= 0. The roots are r = 2,

1

4.

4. For each of the following equations, verify that the origin is a regular singular point and calculate twoindependent Frobenius series solutions:

(a) 4xy′′ + 2y′ + y = 0;

(b) 2xy′′ + (3 − x)y′ − y = 0.

Solution

(a) Write the differential equation in standard form: y′′ − 1

2xy′ +

1

4xy = 0.

We have P (x) = − 1

2xand Q(x) =

1

4x, so x = 0 is a singular point. Taking limits we get

limx→0

xP (x) = −1/2 and limx→0

x2Q(x) = 0. Therefore x = 0 is a regular singular point. We look for

a solution of the form y(x) =∑

∞

n=0 anxn+r. Plug the series into the equation to get

∞∑

n=0

4(n + r)(n + r − 1)anxn+r−1 +∞∑

n=0

2(n + r)anxn+r−1 +∞∑

n=0

anxn+r = 0.

Math 334 Assignment 7 — Solutions 4

Upon re-arrangement we get

(4r(r − 1) + 2r)a0 xr−1 +∞∑

n=0

{[4(n + r + 1)(n + r) + 2(n + r + 1)]an+1 + an}xn+r = 0.

Equating the powers of x yields

(4r2 − 2r)a0 = 0, an+1 =−an

2(n + r + 1)[2(n + r) + 1], n = 0, 1, . . . .

But a0 6= 0 implies that r = 0, 1/2. Solving for the an’s we get

for r = 0 for r =1

2

an+1 =−an

2(n + 1)(2n + 1)an+1 =

−an

(2n + 3)(2n + 2)

a1 =−a0

2a1 =

−a0

3!

a2 =−a1

4 · 3 =a0

4!a2 =

−a1

5 · 4 =a0

5!

a3 =−a2

6 · 5 =−a0

6!a3 =

−a2

7 · 6 =−a0

7!...

...

an =(−1)n

(2n)!a0 an =

(−1)n

(2n + 1)!a0

Two linearly independent Frobenius series (with a0 = 1) are given by

y1(x) =

∞∑

n=0

anxn =

∞∑

n=0

(−1)n

(2n)!xn, y2(x) =

∞∑

n=0

anxn+ 1

2 =√

x

∞∑

n=0

(−1)n

(2n + 1)!xn.

(b) Write the differential equation in standard form: y′′ +3 − x

2xy′ − 1

2xy = 0.

We have P (x) =3 − x

2xand Q(x) = − 1

2x, so x = 0 is a singular point. Taking limits we get

limx→0

xP (x) = 3/2 and limx→0

x2Q(x) = 0. Therefore x = 0 is a regular singular point. We look for a

solution of the form y(x) =∑

∞

n=0 anxn+r. Plug the series into the equation to get

∞∑

n=0

2(n + r)(n + r − 1)anxn+r−1 + 3

∞∑

n=0

(n + r)anxn+r−1 −∞∑

n=0

(n + r)anxn+r −∞∑

n=0

anxn+r = 0.

Upon re-arrangement we get

(2r(r − 1) + 3r)a0 xr−1 +

∞∑

n=0

{[(n + r + 1)(2n + 2r + 3)]an+1 − (n + r − 1)an}xn+r = 0.

Equating the powers of x yields

(4r2 + r)a0 = 0, an+1 =an

2n + 2r + 3, n = 0, 1, . . . .

Math 334 Assignment 7 — Solutions 5

But a0 6= 0 implies that r = 0,−1/2. Solving for the an’s we get

for r = 0 for r = −1

2

an+1 =an

2n + 3an+1 =

an

2(n + 1)

a1 =a0

3a1 =

a0

2

a2 =a1

5=

a0

5 · 3 =a0 · 4 · 2

5 · 4 · 3 · 2 =22 2!

5!a0 a2 =

a1

4=

a0

4 · 2 =a0

22 2!

a3 =a2

7=

a0

7 · 5 · 3 =a0 · 6 · 4 · 2

7 · 6 · 5 · 4 · 3 · 2 =23 3!

7!a0 a3 =

a2

6=

a0

6 · 22 2!=

a0

23 3!...

...

an =2n n!

(2n + 3)!a0 an =

a0

2n n!

Two linearly independent Frobenius series (with a0 = 1) are given by

y1(x) =∞∑

n=0

anxn =∞∑

n=0

2n n!

(2n + 3)!xn, y2(x) =

∞∑

n=0

anxn− 1

2 =1√x

∞∑

n=0

xn

2n n!.

5. Consider the differential equationx3y′′ + xy′ − y = 0.

(a) Show that x = 0 is an irregular singular point.

(b) Use the fact that y1(x) = x is a solution to find a second independent solution y2(x).

(c) Show that the solution y2(x) found in part (b) cannot be expressed as a Frobenius series.

Solution

(a) We have P (x) = 1/x2 so limx→0

xP (x) = limx→0

(1/x) does not exist. Therefore x = 0 is an irregular

singular point.

(b) We know that y1(x) = x is a solution. A second linearly independent solution is given by

y2(x) = y1(x)

∫

e−R

P (x)dx

y21(x)

dx = x

∫

e−R

dx

x2

x2dx =

∫

e−1/x

x2dx = −xe−1/x.

(c) A Frobenius series is of the form∑

∞

n=0 anxn+r for some r ∈ R. The minimum exponent of x is r

which occurs for n = 0. Using the Taylor series expansion for the exponential function et =∞∑

n=0

tn

n!we see that

y2(x) = −xe−1/x = −x∞∑

n=0

x−n

n!=

∞∑

n=0

x−n+1

n!.

Since there is no minimum exponent of x in this series, y2(x) is not expressible as a Frobeniusseries.