Series Solutions Near a Regular Singular Point

Series Solutions Near a Regular Singular Point MATH 365 Ordinary Differential Equations J. Robert Buchanan Department of Mathematics Fall 2018

Background We will find a power series solution to the equation: P(t)y + Q(t)y + R(t)y = 0. We will assume that t 0 is a regular singular point. This implies: 1. P(t 0 ) = 0, (t t 0 )Q(t) 2. lim exists, t t0 P(t) (t t 0 ) 2 R(t) 3. lim exists. t t0 P(t)

Simplification If t 0 0 then we can make the change of variable x = t t 0 and the ODE: P(x + t 0 )y + Q(x + t 0 )y + R(x + t 0 )y = 0. has a regular singular point at x = 0. From now on we will work with the ODE P(x)y + Q(x)y + R(x)y = 0 having a regular singular point at x = 0.

Assumptions (1 of 2) Since the ODE has a regular singular point at x = 0 we can define x Q(x) P(x) which are analytic at x = 0 and = x p(x) and x 2 R(x) P(x) = x 2 q(x) xq(x) lim x 0 P(x) x 2 R(x) lim x 0 P(x) = lim x 0 (x p(x)) = p 0 = lim x 0 (x 2 q(x)) = q 0.

Assumptions (2 of 2) Furthermore since x p(x) and x 2 q(x) are analytic, x p(x) = x 2 q(x) = p n x n q n x n for all ρ < x < ρ with ρ > 0.

Re-writing the ODE The second order linear homogeneous ODE can be written as 0 = P(x)y + Q(x)y + R(x)y = y + Q(x) P(x) y + R(x) P(x) y = x 2 y + x 2 Q(x) P(x) y + x 2 R(x) P(x) y = x 2 y + x[x p(x)]y + [x 2 q(x)]y = x 2 y + x [p 0 + p 1 x + + p n x n + ] y + [q 0 + q 1 x + + q n x n + ] y.

Special Case: Euler s Equation If p n = 0 and q n = 0 for n 1 then 0 = x 2 y + x [p 0 + p 1 x + + p n x n + ] y + [q 0 + q 1 x + + q n x n + ] y = x 2 y + p 0 xy + q 0 y which is Euler s equation.

General Case When p n 0 and/or q n 0 for some n > 0 then we will assume the solution to x 2 y + x[x p(x)]y + [x 2 q(x)]y = 0 has the form y(x) = x r a n x n = a n x r+n, an Euler solution multiplied by a power series.

Solution Procedure Assuming y(x) = x r 1. the values of r, a n x n we must determine: 2. a recurrence relation for a n, 3. the radius of convergence of a n x n.

Example (1 of 8) Consider the following ODE for which x = 0 is a regular singular point. 4x y + 2y + y = 0 Assuming y(x) = a n x r+n is a solution, determine the values of r and a n for n 0.

Example (1 of 8) Consider the following ODE for which x = 0 is a regular singular point. 4x y + 2y + y = 0 Assuming y(x) = a n x r+n is a solution, determine the values of r and a n for n 0. y (x) = y (x) = (r + n)a n x r+n 1 (r + n)(r + n 1)a n x r+n 2

Example (2 of 8) 0 = 4x y + 2y + y = 4x (r + n)(r + n 1)a n x r+n 2 + 2 (r + n)a n x r+n 1 = = + a n x r+n 4(r + n)(r + n 1)a n x r+n 1 + 2(r + n)a n x r+n 1 + a n x r+n [4(r + n)(r + n 1) + 2(r + n)] a n x r+n 1 + a n x r+n

Example (3 of 8) 0 = = = [4(r + n)(r + n 1) + 2(r + n)] a n x r+n 1 + a n x r+n 2a n (r + n)(2r + 2n 1)x r+n 1 + a n x r+n n=1 2a n (r + n)(2r + 2n 1)x r+n 1 + a n 1 x r+n 1

Example (4 of 8) 0 = 2a n (r + n)(2r + 2n 1)x r+n 1 + a n 1 x r+n 1 = 2a 0 r(2r 1)x r 1 + + a n 1 x r+n 1 n=1 = 2a 0 r(2r 1)x r 1 + n=1 2a n (r + n)(2r + 2n 1)x r+n 1 n=1 [2a n (r + n)(2r + 2n 1) + a n 1 ] x r+n 1 n=1

Example (5 of 8) 0 = 2a 0 r(2r 1)x r 1 + [2a n (r + n)(2r + 2n 1) + a n 1 ] x r+n 1 n=1 This implies 0 = r(2r 1) (the indicial equation) and 0 = 2a n (r + n)(2r + 2n 1) + a n 1 Thus we see that r = 0 or r = 1 2 and the recurrence relation is a n = a n 1, for n 1. (2r + 2n)(2r + 2n 1)

Example, Case r = 0 (6 of 8) When r = 0 the recurrence relation becomes: a n = a n = The series coefficients are thus a n 1 (2r + 2n)(2r + 2n 1) a n 1 2n(2n 1). a 1 = a 0 (2)(1) = a 0 2! a 2 = a 1 (4)(3) = a 0 4! a 3 = a 2 (6)(5) = a 0 6!. a n = ( 1)n a 0. (2n)!

Example, Case r = 0 (6 of 8) When r = 0 the recurrence relation becomes: a n = a n = The series coefficients are thus Thus y 1 (x) = a n 1 (2r + 2n)(2r + 2n 1) a n 1 2n(2n 1). a 1 = a 0 (2)(1) = a 0 2! a 2 = a 1 (4)(3) = a 0 4! a 3 = a 2 (6)(5) = a 0 6!. a n = ( 1)n a 0. (2n)! ( 1) n a 0 x n+0 = (2n)! ( 1) n a 0 ( x) 2n = a 0 cos x. (2n)!

Example, Case r = 1/2 (7 of 8) When r = 1/2 the recurrence relation becomes: a n = a n = Thus the series coefficients are a n 1 (2r + 2n)(2r + 2n 1) a n 1 (2n + 1)2n. a 1 = a 0 (3)(2) = a 0 3! a 2 = a 1 (5)(4) = a 0 5! a 3 = a 2 (7)(6) = a 0 7!. a n = ( 1)n a 0 (2n + 1)!.

Example, Case r = 1/2 (7 of 8) When r = 1/2 the recurrence relation becomes: a n = a n = Thus the series coefficients are Thus y 2 (x) = ( 1) n a 0 (2n + 1)! x n+1/2 = a n 1 (2r + 2n)(2r + 2n 1) a n 1 (2n + 1)2n. a 1 = a 0 (3)(2) = a 0 3! a 2 = a 1 (5)(4) = a 0 5! a 3 = a 2 (7)(6) = a 0 7!. a n = ( 1)n a 0 (2n + 1)!. ( 1) n a 0 (2n + 1)! ( x) 2n+1 = a 0 sin x.

Example (8 of 8) We should verify that the general solution to 4x y + 2y + y = 0 is y(x) = c 1 cos x + c 2 sin x.

Remarks This technique just outlined will succeed provided r 1 r 2 and r 1 r 2 n Z. If r 1 = r 2 or r 1 r 2 = n Z then we can always find the solution corresponding to the larger of the two roots r 1 or r 2. The second (linearly independent) solution will have a more complicated form involving ln x.

Homework Read Section 5.6 Exercises: 3, 6, 11 13

General Case: Method of Frobenius [ ] Given x 2 y + x [x p(x)] y + x 2 q(x) y = 0 where x = 0 is a regular singular point and x p(x) = p n x n and x 2 q(x) = q n x n are analytic at x = 0, we will seek a solution to the ODE of the form y(x) = a n x r+n where a 0 0 (for simplicity s sake often we choose a 0 = 1).

Substitute into the ODE 0 = x 2 (r + n)(r + n 1)a n x r+n 2 = [ ] [ ] + x p n x n (r + n)a n x r+n 1 + q n x n a n x r+n (r + n)(r + n 1)a n x r+n [ ] [ ] + p n x n (r + n)a n x r+n + q n x n a n x r+n

Collect Like Powers of x 0 = a 0 r(r 1)x r + a 1 (r + 1)rx r+1 + + (p 0 + p 1 x + )(a 0 rx r + a 1 (r + 1)x r+1 + ) + (q 0 + q 1 x + )(a 0 x r + a 1 x r+1 + ) = a 0 [r(r 1) + p 0 r + q 0 ] x r + [a 1 (r + 1)r + p 0 a 1 (r + 1) + p 1 a 0 r + q 0 a 1 + q 1 a 0 ] x r+1 + = a 0 [r(r 1) + p 0 r + q 0 ] x r + [a 1 ((r + 1)r + p 0 (r + 1) + q 0 ) + a 0 (p 1 r + q 1 )] x r+1 +

Indicial Equation If we define F(r) = r(r 1) + p 0 r + q 0 then the ODE can be written as 0 = a 0 F(r)x r + [a 1 F(r + 1) + a 0 (p 1 r + q 1 )] x r+1 + [a 2 F(r + 2) + a 0 (p 2 r + q 2 ) + a 1 (p 1 (r + 1) + q 1 )] x r+2 + The equation 0 = F(r) = r(r 1) + p 0 r + q 0 is called the indicial equation. The solutions are called the exponents of singularity.

Recurrence Relation The coefficients of x r+n for n 1 determine the recurrence relation: n 1 0 = a n F(r + n) + a k (p n k (r + k) + q n k ) k=0 n 1 k=0 a n = a k (p n k (r + k) + q n k ) F(r + n) provided F(r + n) 0.

Exponents of Singularity By convention we will let the roots of the indicial equation F(r) = 0 be r 1 and r 2.

Exponents of Singularity By convention we will let the roots of the indicial equation F(r) = 0 be r 1 and r 2. When r 1 and r 2 R we will assign subscripts so that r 1 r 2.

Exponents of Singularity By convention we will let the roots of the indicial equation F(r) = 0 be r 1 and r 2. When r 1 and r 2 R we will assign subscripts so that r 1 r 2. Consequently the recurrence relation where r = r 1, n 1 k=0 a n (r 1 ) = a k (p n k (r 1 + k) + q n k ) F(r 1 + n) is defined for all n 1.

Case: r 1 r 2 / N If r 1 r 2 n for any n N then r 1 r 2 + n for any n N and consequently F(r 2 + n) 0 for any n N.

Case: r 1 r 2 / N If r 1 r 2 n for any n N then r 1 r 2 + n for any n N and consequently F(r 2 + n) 0 for any n N. Consequently the recurrence relation where r = r 2, n 1 k=0 a n (r 2 ) = a k (p n k (r 2 + k) + q n k ) F(r 2 + n) is defined for all n 1.

Example Find the indicial equation, exponents of singularity, and discuss the nature of solutions to the ODE x 2 y x(2 + x)y + (2 + x 2 )y = 0 near the regular singular point x = 0.

Solution x(2 + x) p 0 = lim x x 0 x 2 = lim (2 + x) = 2 x 0 q 0 = lim x 0 x 2 2 + x 2 x 2 = lim x 0 (2 + x 2 ) = 2

Solution x(2 + x) p 0 = lim x x 0 x 2 = lim (2 + x) = 2 x 0 q 0 = lim x 0 x 2 2 + x 2 x 2 The indicial equation is then = lim x 0 (2 + x 2 ) = 2 r(r 1) + ( 2)r + 2 = 0 r 2 3r + 2 = 0 (r 2)(r 1) = 0.

Solution x(2 + x) p 0 = lim x x 0 x 2 = lim (2 + x) = 2 x 0 q 0 = lim x 0 x 2 2 + x 2 x 2 The indicial equation is then = lim x 0 (2 + x 2 ) = 2 r(r 1) + ( 2)r + 2 = 0 r 2 3r + 2 = 0 (r 2)(r 1) = 0. The exponents of singularity are r 1 = 2 and r 2 = 1. Consequently we have one solution of the form ) y 1 (x) = x (1 2 + a n x n. n=1

Recurrence Relation y(x) = y (x) = y (x) = a n x n+2 (assume a 0 = 1) (n + 2)a n x n+1 (n + 2)(n + 1)a n x n Substitute into the ordinary differential equation and re-index. 0 = x 2 y x(2 + x)y + (2 + x 2 )y = 2(a 1 a 0 )x 3 + [(n + 1)(n a n a n 1 ) + a n 2 ] x n+2 a 1 = 1 n=2 a n = a n 1 n a n 2 for n 2. n(n + 1)

First Several Terms in Solution a n = a n 1 n a n 2 n(n + 1) a 0 = 1 a 1 = 1 a 2 = 1 3 a 3 = 1 36 a 4 = 7 720 a 5 = 31 10800 y 1 (x) = x 2 + x 3 + x 4 3 + x 5 36 7x 6 720 31x 7 10800 +

Case: r 1 = r 2 Equal Exponents of Singularity (1 of 4) When the exponents of singularity are equal then F(r) = (r r 1 ) 2.

Case: r 1 = r 2 Equal Exponents of Singularity (1 of 4) When the exponents of singularity are equal then F(r) = (r r 1 ) 2. We have a solution to the ODE of the form ) y 1 (x) = x (1 r + a n (r)x n. n=1 Differentiating this solution and substituting into the ODE yields 0 = F(r)x r [ ] n 1 + a n F(r + n) + a k (p n k (r + k) + q n k ) n=1 = (r r 1 ) 2 x r. k=0 when a n solves the recurrence relation. x r+n

Case: r 1 = r 2 Equal Exponents of Singularity (2 of 4) Recall: for the ODE x 2 y + x[x p(x)]y + [x 2 q(x)]y = 0 we can define the linear operator L[y] = x 2 y + x[x p(x)]y + [x 2 q(x)]y so that the ODE can be written compactly as L[y] = 0. Consider the infinite series solution to the ODE, φ(r, x) = x [1 r + a n (r)x ]. n n=1 Note: since the coefficients of the series depend on r we denote the solution as φ(r, x).

Case: r 1 = r 2 Equal Exponents of Singularity (3 of 4) We have seen that φ(r 1, x) solve the ODE and now we claim that φ r (r 1, x) solves it as well. L [ φ r ] (r 1, x) = r [L [φ] (r, x)] r=r 1 = r = = 0 [( (r r 1 ) 2 x r )] r=r 1 [ 2(r r 1 )x r + (r r 1 ) 2 (ln x)x r ] r=r 1

Case: r 1 = r 2 Equal Exponents of Singularity (4 of 4) φ(r, x) y 2 (x) = r r=r1 ( )] = [x r 1 + a n (r)x n r n=1 r=r ) 1 = (ln x)x (1 r + a n (r)x n + x r n=1 = (ln x)y 1 (x) + x r 1 a n(r 1 )x n n=1 n=1 a n(r)x n r=r 1

Example (1 of 9) Find the general solution to the ODE: x y + y + x y = 0 near the regular singular point x = 0.

Example (1 of 9) Find the general solution to the ODE: x y + y + x y = 0 near the regular singular point x = 0. lim x x 0 ( ) 1 x ) ( lim x 2 x x 0 x = 1 = p 0 = 0 = q 0 Thus the indicial equation is F(r) = r(r 1) + r = r 2 = 0 and the exponents of singularity are r 1 = r 2 = 0.

Example (2 of 9) Assume y(x) = a n x r+n, differentiate, and substitute into the given ODE.

Example (2 of 9) Assume y(x) = a n x r+n, differentiate, and substitute into the given ODE. 0 = x y + y + x y 0 = x (r + n)(r + n 1)a n x r+n 2 + (r + n)a n x r+n 1 = = + x a n x r+n (r + n)(r + n 1)a n x r+n 1 + (r + n)a n x r+n 1 + a n x r+n+1 (r + n) 2 a n x r+n 1 + a n x r+n+1

Example (3 of 9) 0 = = (r + n) 2 a n x r+n 1 + a n x r+n+1 (r + n) 2 a n x r+n 1 + a n 2 x r+n 1 = r 2 x r 1 + a 1 (r + 1) 2 x r + + a n 2 x r+n 1 n=2 = r 2 x r 1 + a 1 (r + 1) 2 x r + n=2 (r + n) 2 a n x r+n 1 n=2 n=2 [(r + n) 2 a n + a n 2 ] x r+n 1

Example (4 of 9) 0 = r 2 x r 1 + a 1 (r + 1) 2 x r + n=2 [(r + n) 2 a n + a n 2 ] x r+n 1 The exponents of singularity are r 1 = r 2 = 0. The recurrence relation is a n (r) = a n 2(r) (r + n) 2. a 1 = 0 which implies a 2n+1 = 0 for all n N.

Example (5 of 9) When r = 0, and a 0 = 1, a 2 = a 0 2 2 = 1 2 2 (1!) 2 a 4 = a 2 4 2 = 1 2 4 (2!) 2 a 6 = a 4 6 2 = 1 2 6 (3!) 2. a 2n = ( 1)n 2 2n (n!) 2 thus y 1 (x) = 1 + n=1 ( 1) n x 2n 2 2n (n!) 2 = 1 + n=1 ( 1) n ( x ) 2n. (n!) 2 2

Example (6 of 9) Now find the second solution. a n (r) = a n 2(r) (r + n) 2 a n(r) = a n 2 (r)(r + n)2 a n 2 (r)2(r + n) (r + n) 4 = a n 2 (r)(r + n) 2a n 2(r) (r + n) 3 a n(0) = 2a n 2(0) n a n 2 (0) n 3

Example (7 of 9) Since a 2n+1 (r) = 0 for all n N then a 2n+1 (r) = 0 for all n N. Since a 0 = 1 then a 0 = 0.

Example (8 of 9) Recall the recurrence relation for n 2: If n = 2 then If n = 4 then a n(0) = 2a n 2(0) n a n 2 (0) n 3 a 2 (0) = 2a 0 2a 0 2 3 = 2 2 3 = 1 2 2 = (1) 1 2 2 (1!) 2. a 4 (0) = 2a 2 4a 2 4 3 = a 2 2a 2 2 4 (2!) ( ( )) 1 1 = 2 4 (2!) 2 2 2 1 (1) 2 2 (1!) ( 2 = 1 + 1 ) 1 2 2 4 (2!) 2

Example (9 of 9) Thus a 6 (0) = 2a 4 6a 4 6 3 = a 4 3a 4 6 2 (3) ( = 1 + 1 2 + 1 ) 1 3 2 6 (3!) 2. a 2n (0) = ( 1)n+1 n k=1 1 k 2 2n (n!) 2 y 2 (x) = (ln x)y 1 (x) + n=1 ( ( 1) n+1 n k=1 1 k (n!) 2 ) (x ) 2n. 2

Homework Read Section 5.6 Exercises: 13

The Story So Far (1 of 3) Considering the second-order linear, homogeneous ODE: P(x)y + Q(x)y + R(x)y = 0 where x 0 = 0 is a regular singular point. This implies P(x 0 ) = 0 and lim (x x 0 ) Q(x) x x 0 P(x) lim (x x 0 ) 2 R(x) x x 0 P(x) = lim x 0 x p(x) = p 0 = lim x 0 x 2 q(x) = q 0.

The Story So Far (2 of 3) Define the polynomial F(r) = r(r 1) + p 0 r + q 0, then r(r 1) + p 0 r + q 0 = 0 is called the indicial equation and the roots r 1 r 2 are called the exponents of singularity. If r 1 r 2 N then we have a fundamental set of solutions of the form [ ] y 1 (x) = x r 1 1 + a n (r 1 )x n y 2 (x) = x r 2 [ 1 + n=1 ] a n (r 2 )x n. n=1

The Story So Far (3 of 3) If r 1 = r 2 then we have a fundamental set of solutions of the form [ ] y 1 (x) = x r 1 1 + a n (r 1 )x n n=1 y 2 (x) = y 1 (x) ln x + x r 1 a n(r 1 )x n. n=1

The Story So Far (3 of 3) If r 1 = r 2 then we have a fundamental set of solutions of the form [ ] y 1 (x) = x r 1 1 + a n (r 1 )x n n=1 y 2 (x) = y 1 (x) ln x + x r 1 a n(r 1 )x n. n=1 Now we may take up the final case when r 1 r 2 N.

Case: r 1 r 2 = N N The second solution has the form [ y 2 (x) = a y 1 (x) ln x + x r 2 a = lim r r2 (r r 2 )a N (r) 1 + and ] c n (r 2 )x n n=1 where c n (r 2 ) = d dr [(r r 2)a n (r)] r=r2. We can assume a 0 = 1 for simplicity.

Example (1 of 8) Find the general solution to the ODE x y y = 0 with regular singular point at x = 0.

Example (1 of 8) Find the general solution to the ODE x y y = 0 with regular singular point at x = 0. ( ) 0 lim x x 0 x ( ) 1 lim x 2 x 0 x = 0 = p 0 = 0 = q 0 Thus the indicial equation is F(r) = r(r 1) and the exponents of singularity are r 1 = 1 and r 2 = 0.

Example (2 of 8) Assume a solution of the form y(x) = a n x n+r with a 0 = 1. 0 = x (r + n)(r + n 1)a n x r+n 2 a n x r+n = = (r + n)(r + n 1)a n x r+n 1 a n x r+n (r + n)(r + n 1)a n x r+n 1 a n 1 x r+n 1 = r(r 1)x r 1 + n=1 [(r + n)(r + n 1)a n a n 1 ] x r+n 1 n=1

Example (3 of 8) Recurrence relation for n 1: a n 1 (r) a n (r) = (r + n)(r + n 1) a n (1) = a n 1(1) n(n + 1) If a 0 = 1 then and y 1 (x) = x a n (1) = [ 1 + 1 n!(n + 1)! n=1 ] x n. n!(n + 1)!

Example (4 of 8) According to the formula of Frobenius [ ] y 2 (x) = a y 1 (x) ln x + x r 2 1 + c n (r 2 )x n. n=1 a = lim r r2 (r r 2 )a N (r) = lim r 0 r a 1 (r) = lim r 0 r 1 r(r + 1) = lim r + 1 = 1 r 0 1

Example (5 of 8) c 1 (r 2 ) = d dr [(r r 2)a 1 (r)] r=r2 [ ] r a0 c 1 (0) = d dr r(r + 1) ] [ a0 r=0 = d dr r + 1 = d [ ] 1 dr r + 1 = 1 r=0 r=0

Example (6 of 8) c 2 (r 2 ) = d dr [(r r 2)a 2 (r)] r=r2 c 2 (0) = d dr [r a 2(r)] r=0 = d [ ] r a 1 (r) dr (r + 1)(r + 2) = d [ ] r a 0 dr r(r + 1) 2 (r + 2) = d [ ] 1 dr (r + 1) 2 (r + 2) = 5 4 r=0 r=0 r=0

Example (7 of 8) In general c 3 (r 2 ) = d dr [(r r 2)a 3 (r)] r=r2 c 3 (0) = d [ ] 1 dr (r + 1) 2 (r + 2) 2 (r + 3) = 5 18 r=0 c n (r 2 ) = d dr [(r r 2)a n (r)] r=r2 c n (0) = d [ ] 1 dr (r + 1) 2 (r + 2) 2 (r + n 1) 2 (r + n) ( 1 + 2n ) n 1 k=1 1 k = (n!) 2 r=0

Example (8 of 8) So the second solution has the form ( 1 + 2n ) n 1 k=1 1 k y 2 (x) = y 1 (x) ln x + 1 (n!) 2 n=1 x n.

Homework Read Section 5.7 Exercise: 1, 5, 9, 15, 19