Series Solutions Near an Ordinary Point

Series Solutions Near an Ordinary Point MATH 365 Ordinary Differential Equations J. Robert Buchanan Department of Mathematics Spring 2018

Ordinary Points (1 of 2) Consider the second order linear homogeneous ODE: P(t)y + Q(t)y + R(t)y = 0 where P, Q, and R are polynomials. Definition A point t 0 such that P(t 0 ) 0 is called an ordinary point. If P(t 0 ) = 0 then t 0 is called a singular point.

Ordinary Points (2 of 2) P(t)y + Q(t)y + R(t)y = 0 If t 0 is an ordinary point, then by continuity there exists an interval (a, b) containing t 0 on which P(t) 0 for all t (a, b). Thus the functions p(t) = Q(t) R(t) and q(t) = are defined P(t) P(t) and continuous on (a, b) and the ODE can be written as y + p(t)y + q(t)y = 0. If the initial conditions are y(t 0 ) = y 0 and y (t 0 ) = y 0 then there exists a unique solution to the ODE satisfying the initial conditions.

Power Series Solutions (1 of 5) Consider the ODE y 4y = 0 and find a power series solution with positive radius of convergence centered at an ordinary point.

Power Series Solutions (1 of 5) Consider the ODE y 4y = 0 and find a power series solution with positive radius of convergence centered at an ordinary point. Let t 0 = 0 be the ordinary point for simplicity and y(t) = a n t n

Power Series Solutions (2 of 5) 0 = y 4y

Power Series Solutions (2 of 5) 0 = y 4y = n(n 1)a n t n 2 4 a n t n n=2

Power Series Solutions (2 of 5) 0 = y 4y = n(n 1)a n t n 2 4 a n t n = = n=2 (n + 2)(n + 1)a n+2 t n 4 a n t n [a n+2 (n + 2)(n + 1) 4a n ] t n

Power Series Solutions (2 of 5) 0 = y 4y = n(n 1)a n t n 2 4 a n t n = = n=2 (n + 2)(n + 1)a n+2 t n 4 a n t n [a n+2 (n + 2)(n + 1) 4a n ] t n 0 = a n+2 (n + 2)(n + 1) 4a n (for n = 0, 1, 2,...) a n+2 = 2 2 a n (n + 2)(n + 1) This last equation is called a recurrence relation.

Power Series Solutions (3 of 5) Let a 0 be arbitrary, then a 2 = a 4 = a 6 = 2 2 (2)(1) a 0 = 22 2! a 0 2 2 (4)(3) a 2 = 24 4! a 0 2 2 (6)(5) a 4 = 26 6! a 0. a 2n = 22n (2n)! a 0.

Power Series Solutions (4 of 5) Let a 1 be arbitrary, then a 3 = a 5 = a 7 =. a 2n+1 = 2 2 (3)(2) a 1 = 22 3! a 1 2 2 (5)(4) a 3 = 24 5! a 1 2 2 (7)(6) a 5 = 26 7! a 1 2 2n (2n + 1)! a 1.

Power Series Solutions (5 of 5) Thus the general solution to y 4y = 0 can be written as y(t) = a 2n t 2n + a 2n+1 t 2n+1 = a 0 = a 0 2 2n (2n)! t2n + a 1 (2t) 2n (2n)! + a 1 2 = a 0 cosh(2t) + a 1 2 sinh(2t) 2 2n (2n + 1)! t2n+1 (2t) 2n+1 (2n + 1)! We can confirm this series converges for all t R.

Airy s Equation (1 of 6) Find a power series solution about the ordinary point t 0 = 0 to Airy s equation: y t y = 0.

Airy s Equation (1 of 6) Find a power series solution about the ordinary point t 0 = 0 to Airy s equation: y t y = 0. y(t) = y (t) = y (t) = a n t n na n t n 1 n=1 n(n 1)a n t n 2. n=2 Substitute into the ODE.

Airy s Equation (2 of 6) 0 = y ty = n(n 1)a n t n 2 t a n t n = n=2 (n + 2)(n + 1)a n+2 t n a n t n+1 = 2a 2 + (n + 2)(n + 1)a n+2 t n a n 1 t n n=1 n=1 = 2a 2 + [(n + 2)(n + 1)a n+2 a n 1 ] t n n=1

Airy s Equation (3 of 6) Thus 0 = 2a 2 0 = (n + 2)(n + 1)a n+2 a n 1 (for n = 1, 2,....) From the last equation we obtain the recurrence relation: a n+2 = a n 1 (n + 2)(n + 1).

Airy s Equation (3 of 6) Thus 0 = 2a 2 0 = (n + 2)(n + 1)a n+2 a n 1 (for n = 1, 2,....) From the last equation we obtain the recurrence relation: a n+2 = a n 1 (n + 2)(n + 1). Since a 2 = 0 then a 5 = a 8 = a 11 = = 0.

Airy s Equation (4 of 6) Let a 0 be arbitrary then a 3 = a 0 2 3 a 6 = a 3 5 6 = a 0 6 5 3 2 a 9 = a 6 8 9 = a 0 9 8 6 5 3 2. a 3n = a 0 (2)(3)(5)(6) (3n 4)(3n 3)(3n 1)(3n).

Airy s Equation (5 of 6) Let a 1 be arbitrary then a 4 = a 1 3 4 a 7 = a 4 6 7 = a 1 7 6 4 3 a 7 a 10 = 9 10 = a 1 10 9 7 6 4 3. a 3n+1 = a 1 (3)(4)(6)(7) (3n 3)(3n 2)(3n)(3n + 1).

Airy s Equation (6 of 6) Thus the solution to Airy s equation is y(t) = a 0 [1 + n=1 + a 1 [t + y ty = 0 ] t 3n (2)(3)(5)(6) (3n 4)(3n 3)(3n 1)(3n) n=1 = a 0 y 1 (t) + a 1 y 2 (t). t 3n+1 (3)(4)(6)(7) (3n 3)(3n 2)(3n)(3n + 1) ]

Illustration y 10 5-10 -5 5 10 x -5 y 1 (t) y 2 (t) -10

Homework Read Section 5.2 Exercises: 1, 2, 3, 5, 6, 21

Review We have determined a method for solving an ODE of the form P(t)y + Q(t)y + R(t)y = 0 where P, Q, and R are polynomials. The solution is a power series of the form y(t) = a n (t t 0 ) n where t 0 is an ordinary point.

Review We have determined a method for solving an ODE of the form P(t)y + Q(t)y + R(t)y = 0 where P, Q, and R are polynomials. The solution is a power series of the form y(t) = where t 0 is an ordinary point. a n (t t 0 ) n Today we extend this work to a broader range of functions than polynomial P, Q, and R.

Differentiation of Power Series Suppose φ(t) = a n (t t 0 ) n has a positive radius of convergence, then we may differentiate the series term-by-term. n φ (n) (t) φ (n) (t 0 ) 0 a n (t t 0 ) n a 0 1 2. m na n (t t 0 ) n 1 a 1 n=1 n(n 1)a n (t t 0 ) n 2 2a 2 n=2. n(n 1) (n m + 1)a n (t t 0 ) n m n=m. m!a m

Solution to an ODE Suppose φ(t) solves the ODE y + p(t)y + q(t)y = 0 then 0 = φ (t) + p(t)φ (t) + q(t)φ(t) φ (t) = p(t)φ (t) q(t)φ(t)

Solution to an ODE Suppose φ(t) solves the ODE y + p(t)y + q(t)y = 0 then 0 = φ (t) + p(t)φ (t) + q(t)φ(t) φ (t) = p(t)φ (t) q(t)φ(t) φ (t 0 ) = p(t 0 )φ (t 0 ) q(t 0 )φ(t 0 )

Solution to an ODE Suppose φ(t) solves the ODE y + p(t)y + q(t)y = 0 then 0 = φ (t) + p(t)φ (t) + q(t)φ(t) φ (t) = p(t)φ (t) q(t)φ(t) φ (t 0 ) = p(t 0 )φ (t 0 ) q(t 0 )φ(t 0 ) 2!a 2 = p(t 0 )a 1 q(t 0 )a 0.

Solution to an ODE Suppose φ(t) solves the ODE y + p(t)y + q(t)y = 0 then 0 = φ (t) + p(t)φ (t) + q(t)φ(t) φ (t) = p(t)φ (t) q(t)φ(t) φ (t 0 ) = p(t 0 )φ (t 0 ) q(t 0 )φ(t 0 ) 2!a 2 = p(t 0 )a 1 q(t 0 )a 0. Thus we may find a 2 in terms of a 0 and a 1.

Continued Differentiation (1 of 2) With patience we can find higher order terms in the series solution through continued differentiation. φ (t) = p(t)φ (t) q(t)φ(t) φ (t) = p (t)φ (t) p(t)φ (t) q (t)φ(t) q(t)φ (t) φ (t 0 ) = p (t 0 )φ (t 0 ) p(t 0 )φ (t 0 ) q (t 0 )φ(t 0 ) q(t 0 )φ (t 0 ) 3!a 3 = p (t 0 )a 1 2p(t 0 )a 2 q (t 0 )a 0 q(t 0 )a 1

Continued Differentiation (2 of 2) We can proceed by repeated differentiation to find a 4, a 5,... provided:

Continued Differentiation (2 of 2) We can proceed by repeated differentiation to find a 4, a 5,... provided: p(t) and q(t) have derivatives of all orders, and

Continued Differentiation (2 of 2) We can proceed by repeated differentiation to find a 4, a 5,... provided: p(t) and q(t) have derivatives of all orders, and we can show the resulting power series converges.

Example Assuming that y = φ(t) is a solution to the IVP: y + t 2 y + (sin t)y = 0 y(0) = 1 y (0) = 1 find the first four nonzero terms in the power series representation of φ(t).

Example Assuming that y = φ(t) is a solution to the IVP: y + t 2 y + (sin t)y = 0 y(0) = 1 y (0) = 1 find the first four nonzero terms in the power series representation of φ(t). a 0 = 1 a 1 = 1 a 2 = 0 a 3 = 1 3! a 4 = 1 3!

Analytic Functions If p(t) and q(t) are analytic functions at t 0, in other words have Taylor series expansions about t 0 which converge to p(t) and q(t) respectively then p and q will have derivatives of all orders at t 0. p(t) = q(t) = p n (t t 0 ) n q n (t t 0 ) n

Ordinary and Singular Points Revisited Suppose P(t)y + Q(t)y + R(t)y = 0 y + Q(t) P(t) y + R(t) P(t) y = 0 y + p(t)y + q(t)y = 0 where p(t) = Q(t) P(t) and q(t) = R(t) P(t)

Ordinary and Singular Points Revisited Suppose where P(t)y + Q(t)y + R(t)y = 0 y + Q(t) P(t) y + R(t) P(t) y = 0 y + p(t)y + q(t)y = 0 p(t) = Q(t) P(t) and q(t) = R(t) P(t) If p(t) and q(t) are analytic at t 0, then we say that t 0 is an ordinary point of the ODE. Otherwise t 0 is a singular point.

Main Result Theorem If t 0 is an ordinary point of the ODE P(t)y + Q(t)y + R(t)y = 0 then the general solution of the ODE is y(t) = a n (t t 0 ) n = a 0 y 1 (t) + a 1 y 2 (t) where a 0 and a 1 are arbitrary and y 1 and y 2 are linearly independent series solutions that are analytic at t 0. Further the radius of convergence for each of y 1 and y 2 is at least as large as the minimum of the radii of convergence for the series p(t) = Q(t)/P(t) and q(t) = R(t)/P(t).

Radius of Convergence If p(t) = Q(t)/P(t) and q(t) = R(t)/P(t) and p(t) and q(t) are analytic at t 0 then from the theory of complex variables we have the result that the radius of convergence of p(t) (and similarly q(t)) is at least as large as the minimum distance from t 0 to any root of P(t) in the complex plane.

Example The value t 0 = 1 is an ordinary point of the ODE t 2 y + (1 + t)y + 3(ln t)y = 0. Find the radius of convergence of p(t) = 1 + t t 2 q(t) = 3 ln t t 2. and

Example The value t 0 = 0 is an ordinary point of the ODE (1 + t 4 )y + 4ty + y = 0. Find the radius of convergence of p(t) = q(t) = 1 1 + t 4. 4t 1 + t 4 and

Example The value t 0 = 0 is an ordinary point of the ODE (1 + t 4 )y + 4ty + y = 0. Find the radius of convergence of p(t) = q(t) = 1 1 + t 4. Using Euler s Identity: t 4 + 1 = 0 t 4 = 1 = e i(2n+1)π t = e i(2n+1)π/4 2 2 t = ± 2 ± i 2. 4t 1 + t 4 and

Illustration 1.0 i y 0.5-1.0-0.5 0.5 1.0 x -0.5

Example The value t 0 = 1 is an ordinary point of the ODE (1 + t 4 )y + 4ty + y = 0. Find the radius of convergence of p(t) = q(t) = 1 1 + t 4. 4t 1 + t 4 and

Illustration 1.0 i y 0.5-1.0-0.5 0.5 1.0 x -0.5

Homework Read Section 5.3 Exercises: 1 7 odd, 10, 11, 22 29