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Math 334
Assignment 7 — Solutions
- Chebyshev’s differential equation is
(1 − x^2 )y′′^ − xy′^ + α^2 y = 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=
n(n − 1)anxn−^2 −
∑^ ∞
n=
n(n − 1)anxn^ −
∑^ ∞
n=
nanxn^ +
∑^ ∞
n=
α^2 anxn^ = 0.
Upon re-arrangement we get 2 a 2 + α^2 a 0 + (6a 3 − a 1 + α^2 a 0 ) x
∑^ ∞
n=
{(n + 2)(n + 1)an+2 − [n(n − 1) + n − α^2 ]an}xn^ = 0.
Equating the powers of x yields 2 a 2 + α^2 a 0 = 0, 6 a 3 − a 1 + α^2 a 0 = 0, (n + 2)(n + 1)an+2 − (n^2 − α^2 )an = 0, n = 2, 3 ,.... Solving for the an’s we get a 2 = − α
2 2 a^0 ,^ a^3 =
1 − α^2 3! a^1 ,^ an+2^ =^
n^2 − α^2 (n + 2)(n + 1) an,^ n^ = 2,^3 ,.... Writing out the first few explicitly yields: a 4 = (
(^2) − α (^2) )(−α (^2) ) 4! a^0 ,^ a^5 =
(3^2 − α^2 )(1 − α^2 ) 5! a^1 ,^ a^6 =
(4^2 − α^2 )(2^2 − α^2 )(−α^2 ) 6! a^0. A pattern clearly emerges: a 2 n = [(2n^ −^ 2)
(^2) − α (^2) ][(2n − 4) (^2) − α (^2) ] · · · (2 (^2) − α (^2) )(−α (^2) ) (2n)! a^0 ,^ n^ = 1,^2 ,... , a 2 n+1 = [(2n^ −^ 1)
(^2) − α (^2) ][(2n − 3) (^2) − α (^2) ] · · · (3 (^2) − α (^2) )(1 − α (^2) ) (2n + 1)! a^1 ,^ n^ = 1,^2 ,.... Hence, two nontrivial solutions are given by:
ϕ 1 (x) = 1 +
∑^ ∞
n=
[(2n − 2)^2 − α^2 ][(2n − 4)^2 − α^2 ] · · · (2^2 − α^2 )(−α^2 ) (2n)! x
2 n,
ϕ 2 (x) = x +
∑^ ∞
n=
[(2n − 1)^2 − α^2 ][(2n − 3)^2 − α^2 ] · · · (3^2 − α^2 )(1 − α^2 ) (2n + 1)! x
2 n+1.
(b) From the recurrence relation we get
an+2 = n
(^2) − α 2 (n + 2)(n + 1) an^ =⇒
a 2 n+2 = (2n)
(^2) − α 2 (2n + 2)(2n + 1) a^2 n, a 2 n+3 = (2n^ + 1)
(^2) − α 2 (2n + 3)(2n + 2) a^2 n+1. Now suppose that α is either an even or an odd integer. Thus, α = 2N =⇒ a 2 N+2 = 0 =⇒ a 2 n+2 = 0 for n > N which implies that ϕ 1 is a polynomial of degree 2N, whereas α = 2N + 1 =⇒ a 2 N+3 = 0 =⇒ a 2 n+3 = 0 for n > N which implies that ϕ 2 is a polynomial of degree 2N + 1. (c) For the cases α = 0, 1 , 2 , 3 we have α = 0 =⇒ a 2 n = 0 for n > 1 =⇒ ϕ 1 (x) = 1, α = 1 =⇒ a 2 n+1 = 0 for n > 1 =⇒ ϕ 2 (x) = x, α = 2 =⇒ a 2 n = 0 for n > 2 =⇒ ϕ 1 (x) = 1 − 2 x^2 , α = 3 =⇒ a 2 n+1 = 0 for n > 2 =⇒ ϕ 2 (x) = x − 45 x^3.
- For each of the following differential equations, locate and classify its singular points.
(a) x^3 (x − 1)y′′^ − 2(x − 1)y′^ + 3xy = 0. (b) x^2 (x^2 − 1)^2 y′′^ − x(x − 1)y′^ + 2y = 0.
Solution (a) Write the differential equation in standard form: y′′^ − (^) x^23 y′^ + (^) x (^2) (x^3 − 1) y = 0. We have P (x) = − (^) x^23 and Q(x) = (^) x (^2) (x^3 − 1) , so the singular points occur at x = 0, 1.
i. At x = 0 we have lim x→ 0 xP (x) = lim x→ 0 − x^22 which does not exist. Therefore x = 0 is an irregular singular point of the differential equation. ii. At x = 1 we have lim x→ 1 (x − 1)P (x) = lim x→ 1 −2(x x^3 − 1)= 0 and lim x→ 1 (x − 1)^2 Q(x) = lim x→ 1 3(x x^ − 2 1)= 0. Therefore x = 1 is a regular singular point of the differential equation. (b) Write the differential equation in standard form: y′′^ − (^) x(x + 1)(^1 x (^2) − 1) y′^ + (^) x (^2) (x (^22) − 1) 2 y = 0. We have P (x) = (^) x(x + 1)(−^1 x (^2) − 1) Q(x) = (^) x (^2) (x (^22) − 1) 2
=⇒^ x^ = 0,^ ±1 are singular points
Upon re-arrangement we get
(4r(r − 1) + 2r)a 0 xr−^1 +
∑^ ∞
n=
{[4(n + r + 1)(n + r) + 2(n + r + 1)]an+1 + an}xn+r^ = 0.
Equating the powers of x yields
(4r^2 − 2 r)a 0 = 0, an+1 = (^) 2(n + r + 1)[2(−ann + r) + 1] , n = 0, 1 ,....
But a 0 6 = 0 implies that r = 0, 1 /2. Solving for the an’s we get
for r = 0 for r =^12 an+1 = (^) 2(n + 1)(2−ann + 1) an+1 = (^) (2n + 3)(2−ann + 2)
a 1 = − 2 a 0 a 1 = − 3!a^0 a 2 = − 4 ·a 13 = a 4!^0 a 2 = − 5 a· 4 1 = a 5!^0 a 3 = − 6 ·a 25 = − 6!a^0 a 3 = − 7 a· 6 2 = − 7!a^0 .. .
an = (−1)
n (2n)! a^0 an^ =^
(−1)n (2n + 1)!a^0 Two linearly independent Frobenius series (with a 0 = 1) are given by
y 1 (x) =
∑^ ∞
n=
anxn^ =
∑^ ∞
n=
(−1)n (2n)! x
n, y 2 (x) =
∑^ ∞
n=
anxn+^12 =
x
∑^ ∞
n=
(−1)n (2n + 1)!x
n.
(b) Write the differential equation in standard form: y′′^ +^3 2 −x^ x y′^ − (^21) x y = 0.
We have P (x) =^3 2 −x^ x and Q(x) = − (^21) x, so x = 0 is a singular point. Taking limits we get xlim→ 0 xP^ (x) = 3/2 and lim x→ 0 x^2 Q(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=
2(n + r)(n + r − 1)anxn+r−^1 + 3
∑^ ∞
n=
(n + r)anxn+r−^1 −
∑^ ∞
n=
(n + r)anxn+r^ −
∑^ ∞
n=
anxn+r^ = 0.
Upon re-arrangement we get
(2r(r − 1) + 3r)a 0 xr−^1 +
∑^ ∞
n=
{[(n + r + 1)(2n + 2r + 3)]an+1 − (n + r − 1)an}xn+r^ = 0.
Equating the powers of x yields (4r^2 + r)a 0 = 0, an+1 = (^2) n + 2anr + 3 , n = 0, 1 ,....
But a 0 6 = 0 implies that r = 0, − 1 /2. Solving for the an’s we get
for r = 0 for r = − (^12) an+1 = (^2) na n+ 3 an+1 = (^) 2(na + 1)n a 1 = a 30 a 1 = a 20
a 2 = a 51 = 5 a ·^0 3 = 5 a ·^0 4 · ·^4 3 · ·^2 2 =^2
5! a^0 a^2 =^
a 1 4 =^
a 0 4 · 2 =^
a 0 22 2! a 3 = a 72 = (^7) ·a 50 · 3 = (^7) ·a 60 ·· 56 ·^ · 4 4 ·· 32 · 2 =^2
7! a^0 a^3 =^
a 2 6 =^
a 0 6 · 22 2! =^
a 0 23 3! .. .
an = 2
n (^) n! (2n + 3)!a^0 an^ =^
a 0 2 n^ n! Two linearly independent Frobenius series (with a 0 = 1) are given by
y 1 (x) =
∑^ ∞
n=
anxn^ =
∑^ ∞
n=
2 n^ n! (2n + 3)!x
n, y 2 (x) =
∑^ ∞
n=
anxn−^12 = √^1 x
∑^ ∞
n=
xn 2 n^ n!.
- Consider the differential equation x^3 y′′^ + xy′^ − y = 0. (a) Show that x = 0 is an irregular singular point. (b) Use the fact that y 1 (x) = x is a solution to find a second independent solution y 2 (x). (c) Show that the solution y 2 (x) found in part (b) cannot be expressed as a Frobenius series. Solution (a) We have P (x) = 1/x^2 so lim x→ 0 xP (x) = lim x→ 0 (1/x) does not exist. Therefore x = 0 is an irregular singular point. (b) We know that y 1 (x) = x is a solution. A second linearly independent solution is given by
y 2 (x) = y 1 (x)
∫ (^) e− R^ P (x)dx y^21 (x) dx^ =^ x
∫ (^) e− R^ xdx 2 x^2 dx^ =
∫ (^) e− 1 /x x^2 dx^ =^ −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=
tn n! we see that y 2 (x) = −xe−^1 /x^ = −x
∑^ ∞
n=
x−n n! =
∑^ ∞
n=
x−n+ n!. Since there is no minimum exponent of x in this series, y 2 (x) is not expressible as a Frobenius series.
