Chapter 3Power series solutions of second order ODE 3.1 Introduction In the methods we have discussed until now, there no general method we have seen to solve the linear differential equations with variable coefficients. Therefore in this chapter we introduce at a method called Power series series method and also a more general method called Frobenius method to solve linear differential equations with variable coefficients. Consider the following second order homogeneous linear differential equations with variable coefficients d2 y dy a2 (x) 2 + a1 (x) + a0 (x)y = 0 (3.1) dx dx 3.1.1 Ordinary & singular points : The x0 in an interval I is said to be an ordinary point of the differential equation (3.1) if a2 (x0 ) 6= 0, else it is called a singular point. 3.1.2 Regular singular point : At a singular point of the interval, if we rewrite the equation (3.1) as d2 y a1 (x) dy a0 (x) + + y = 0 dx2 a2 (x) dx a2 (x) Or more precisely, if x = x0 is a singular point of the differential equation and if we rewrite it as d2 y p1 (x) dy p0 (x) 2 + + y = 0 (3.2) dx (x − x0 ) dx (x − x0 )2 such that p1 and p0 are analytic in the neighborhood of x = x0 , then the point x0 is called a regular singular point. 31 32 Sanyasiraju V S S Yedida
[email protected] Example : Find the singular points and then classify their regularity for the equation (1 − x2 )y ′′ − 2xy ′ + n(1 + n)y = 0, n is a constant Since 1 − x2 = 0 for x = ±1, therefore x = 1 and x = −1 are the two singular points of the given equation. To show that x = 1 is a regular singular point, rewrite the given equation as 1 −2x ′ 1 (1 − x)n(1 + n) y ′′ + y + 2 y=0 1−x1+x (1 − x) 1+x Therefore, we have −1 −2x 2(x − 1 + 1) (x − 1) p1 (x) = = − = −(1 + (x − 1)) 1 + 1+x x−1+2 2 2 (x − 1) (x − 1) = −(1 + (x − 1)) 1 − + − ... 2 4 (x − 1) (x − 1)2 = − 1+ − + ... 2 4 and −1 (1 − x)n(1 + n) 1−x (x − 1) p0 (x) = = n(n + 1) 1+ 1+x 2 2 2 (x − 1) (x − 1) (x − 1) = −n(n + 1) 1− + − ... 2 2 4 (x − 1) (x − 1)2 = n(n + 1) − + −... 2 4 Therefore, x = 1 is a regular singular point, similarly, to show that x = −1 is also a regular singular point we rewrite the given equation as 1 −2x ′ 1 (1 + x)n(1 + n) y ′′ + y + 2 y=0 1+x1−x (1 + x) 1−x Therefore, we have −1 −2x 2(x + 1 − 1) 1 p1 (x) = = = (1 − (x + 1)) 1 − 1−x x+1−2 1+x 2 (x + 1) (x + 1) = (1 − (x + 1)) 1 + + + ... 2 4 (x + 1) (x + 1)2 = 1− − − ... 2 4 Lecture Notes MA2020 Differential Equations 33 and −1 (1 − x)n(1 + n) (x + 1) p0 (x) = = n(n + 1)((x + 1) − 2) 1 − 1+x 2 2 (x + 1) (x + 1) = −n(n + 1)(x + 1) 1 + + + ... 2 4 (x + 1)2 (x + 1)3 = −n(n + 1) (x + 1) + + + ... 2 4 Hence x = −1 is also a regular singular point. 3.1.3 Problems : Find the singular points for the following equations and then classify them 1. x2 y ′′ + axy ′ + by = 0, a, b are constant 2. x2 y ′′ + (x + x2 )y ′ − y = 0 3. x3 (x − 2)y ′′ + x3 y ′ + 6y = 0 4. x2 y ′′ + sin xy ′ + cos xy = 0 For linear equations, power series solutions can be obtained at ordinary and regular singular points. The power series solution is an infinite series appears in the form ∞ X am (x − x0 )m = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + . . . + (3.3) m=0 where a0 , a1 . . . are constants and x0 is also a constant called the center of the series. Some simple power series which we know are ∞ 1 X = xm = 1 + x + x2 + . . . ( |x| < 1, it is the geometric series) 1−x m=0 ∞ x X xm x2 e = = 1+x+ + ... (3.4) m=0 m! 2! ∞ X (−1)m x2m x2 x4 cos x = = 1− + −... m=0 (2m)! 2! 4! ∞ X (−1)m x2m+1 x3 x5 sin x = = x− + − ... m=0 (2m + 1)! 3! 5! 34 Sanyasiraju V S S Yedida
[email protected] 3.1.4 Some Simple Examples : We demonstrate the power series method through some simple example problems for which we already know the answer. 1. Consider y ′ − y = 0 for which y = cex , where c is a constant, is the general solution. To obtain the same through power series method, consider ∞ X ∞ X m y = am x ⇒ y = ′ am mxm−1 m=0 m=1 X∞ ∞ X ⇒ y′ − y = am mxm−1 − am xm = 0 m=1 m=0 X∞ ⇒ am+1 (m + 1)xm − am xm = 0 m=0 X∞ ⇒ (am+1 (m + 1) − am ) xm = 0 m=0 that is, a1 − a0 = 0 ⇒ a1 = a0 a1 a0 2a2 − a1 = 0 ⇒ a2 = = 2 2 a2 a0 3a3 − a2 = 0 ⇒ a3 = = 3 3! ... x2 x3 ⇒ y = a0 1+x+ + + . . . = a0 ex 2 3! 2 2. y ′ − 2xy = 0 for which y = cex is the solution. ∞ X ∞ X m y = am x ⇒ y = ′ am mxm−1 m=0 m=1 X∞ ∞ X ⇒ y ′ − 2xy = am mxm−1 − 2 am xm+1 = 0 m=1 m=0 X∞ ∞ X m ⇒ am+1 (m + 1)x − 2 am−1 xm = 0 m=0 m=1 ∞ X ⇒ a1 + (am+1 (m + 1) − 2am−1 ) xm = 0 m=1 Lecture Notes MA2020 Differential Equations 35 that is, ⇒ a1 = 0 2a2 − 2a0 = 0 ⇒ a2 = a0 3a3 − 2a1 = 0 ⇒ a3 = 0 a2 a0 4a4 − 2a2 = 0 ⇒ a4 = = 2 2 ... x4 x6 2 2 ⇒ y = a0 1 + x + + + . . . = a0 ex 2! 3! 3. y ′′ + y = 0 for which y = c1 cos x + c2 sin x is the solution. ∞ X ∞ X ∞ X y = am xm ⇒ y ′ = am mxm−1 , y ′′ = am m(m − 1)xm−2 m=0 m=1 m=1 X∞ ∞ X ⇒ y ′′ + y = am m(m − 1)xm−2 + am xm = 0 m=2 m=0 X∞ ∞ X m ⇒ am+2 (m + 2)(m + 1)x + am xm = 0 m=0 m=1 that is, ⇒ a1 = 0 a0 2a2 + a0 = 0 ⇒ a2 = − 2 a1 3 2a3 + a1 = 0 ⇒ a3 = − 3! a0 4 3a4 + a2 = 0 ⇒ a4 = 4! ... a0 a1 a0 ⇒ y = a0 + a1 x − x2 − x3 + x4 + . . . 2 3! 4! 3 2 4 x5 x x x = a0 1 − + + . . . + a1 x − + + ... 2 4! 3! 5! = a0 cos x + a1 sin x The important aspect of the power series solutions is the convergence of the series solution obtained in this procedure. Fortunately, whenever p1 and p0 are analytic in some interval then the power series is convergent with a radius of convergence R. Further it also allows us to do point wise addition, point wise multiplication, point wise differentiation and point wise integration. 36 Sanyasiraju V S S Yedida
[email protected] 3.1.5 Example : Find the power series solution for (4 + x2)y ′′ − 6xy ′ + 8y = 0 with origin as the center ∞ X y= am xm m=0 X∞ ∞ X ⇒ y′ = am mxm−1 = am+1 (m + 1)xm m=1 m=0 X∞ ∞ X m−2 ⇒ y = ′′ am m(m − 1)x = am+2 (m + 1)(m + 2)xm m=2 m=0 2 ′′ ′ (4 + x )y − 6xy + 8y = 0 ∞ X ∞ X ∞ X ⇒ (4 + x2 ) am+2 (m + 1)(m + 2)xm − 6x am+1 (m + 1)xm − 8 am xm = 0 m=0 m=0 m=0 ∞ X ∞ X ⇒ (4(m + 1)(m + 2)am+2 + 8am ) xm + (m + 1)(m + 2)am+2 xm+2 m=0 m=0 ∞ X −6 am+1 (m + 1)xm+1 = 0 m=0 ∞ X ∞ X ∞ X ⇒ (4(m + 1)(m + 2)am+2 + 8am ) xm + (m − 1)(m)am xm − 6 am (m)xm = 0 m=0 m=2 m=1 ∞ X ⇒ (8a2 + 8a0 ) + (24a3 + 8a1 )x + (4(m + 1)(m + 2)am+2 + 8am + (m − 1)(m)am ) xm m=2 ∞ X −6a1 x − 6 am (m)xm = 0 m=2 ∞ X 4(m + 1)(m + 2)am+2 + (m2 − 7m + 8)am xm = 0 ⇒ (8a2 + 8a0 ) + (24a3 + 2a1 )x + m=2 Comparing the coefficients of xm on both sides gives 8a2 + 8a0 = 0 ⇒ a2 = −a0 a1 24a3 + 2a2 = 0 ⇒ a3 = − 12 2a2 a0 4.3.4a4 − 2a2 = 0 ⇒ a4 = =− 48 24 4a3 a1 4.4.5a5 − 2a5 = 0 ⇒ a5 = =− 80 240 Lecture Notes MA2020 Differential Equations 37 Finally, substituting a0 , a1 , a2 , . . . in the definition of y gives ∞ X a1 3 a0 4 a1 5 y = am xm = a0 + a1 x − a0 x2 − x − x − x + ... m=0 12 24 240 x4 x6 x3 x5 x7 2 = a0 1 − x − − − . . . + a1 x − − − − ... 24 720 12 240 20160 = a0 y1 + a1 y2 4 x6 3 x5 x7 where y1 = 1 − x2 − x24 − 720 − . . . and y2 = x − x12 − 240 − 20160 − . . . are 1 0 linearly independent because the Wronskian w(y1, y2 )at x=0 = = 1. There- 0 1 fore, y1 and y2 are linearly independent, hence the obtained solution is the general solution of the given problem. 3.1.6 Example : Find the power series solution for y ′′ + xy ′ + y = 0 with x0 = 2 as the center ∞ X y= am (x − 2)m m=0 X∞ ∞ X m−1 ⇒ y = ′ am m(x − 2) = am+1 (m + 1)(x − 2)m m=1 m=0 X∞ ∞ X ⇒ y ′′ = am m(m − 1)(x − 2)m−2 = am+2 (m + 1)(m + 2)(x − 2)m m=2 m=0 y ′′ + (x − 2 + 2)y ′ + y = 0 ∞ ∞ ∞ ! X X X ⇒ am+2 (m + 1)(m + 2) + (x − 2 + 2) am+1 (m + 1) + am (x − 2)m = 0 m=0 m=0 m=0 ∞ X X∞ ⇒ ((m + 1)(m + 2)am+2 + 2(m + 1)am+1 + am ) (x − 2)m + (m + 1)am+1 (x − 2)m+1 = 0 m=0 m=0 X∞ X∞ ⇒ ((m + 1)(m + 2)am+2 + 2(m + 1)am+1 + am ) (x − 2)m + (m)am (x − 2)m = 0 m=0 m=1 ∞ X ⇒ 2a2 + 2a1 + a0 + ((m + 1)(m + 2)am+2 + 2(m + 1)am+1 + (m + 1)am ) (x − 2)m = 0 m=1 Comparing the coefficients of xm on both sides gives 1 2a2 + 2a1 + a0 = 0 ⇒ a2 = −a1 − a0 2 1 1 1 a0 1 1 1 6a3 + 4a2 + 2a1 = 0 ⇒ a3 = − a2 − a1 = − (−a1 − ) − a1 = a1 + a0 3 3 3 2 3 3 3 1 1 1 1 1 1 a0 1 7 12a4 + 6a3 + 3a2 = 0 ⇒ a4 = − a3 − a2 = − ( a1 + a0 ) − (−a1 − ) = a1 − a0 2 4 2 3 3 4 2 12 24 38 Sanyasiraju V S S Yedida
[email protected] Finally, substituting a0 , a1 , a2 , . . . in the definition of y gives ∞ X y = am (x − 2)m m=0 1 2 1 1 3 1 7 = a0 + a1 (x − 2) − a1 + a0 (x − 2) + a1 + a0 (x − 2) + a1 − a0 (x − 2)4 + . . . 2 3 3 12 24 3 4 1 (x − 2) 7(x − 2) = a0 1 − (x − 2)2 + − − ... 2 3 24 (x − 2)3 (x − 2)4 2 +a1 (x − 2) − (x − 2) + − + ... 3 12 = a0 y1 + a1 y2 where (x − 2)3 7(x − 2)4 1 2 y1 = 1 − (x − 2) + − −... 2 3 24 and (x − 2)3 (x − 2)4 2 y2 = (x − 2) − (x − 2) + − + ... 3 12 1 0 are linearly independent because the Wronskian w(y1, y2 )at x=2 = = 1. 0 1 Therefore, y1 and y2 are linearly independent, hence the obtained solution is the general solution of the given problem. 3.2 Legendre Equation The linear second order equation given by (1 − x2 )y ′′ − 2xy ′ + n(n + 1)y = 0 (3.5) is called the Legendre equation for a constant n. x = ±1 are regular singular points for this equation, and hence there exists a power series solution for |x| < 1. To find Lecture Notes MA2020 Differential Equations 39 the power series solution, assume ∞ X y= am xm m=0 X∞ ∞ X m−1 ⇒ y = ′ am mx = am+1 (m + 1)xm m=1 m=0 X∞ ∞ X ⇒ y ′′ = am m(m − 1)xm−2 = am+2 (m + 1)(m + 2)xm m=2 m=0 (1 − x2 )y ′′ − 2xy ′ + n(n + 1)y = 0 ∞ ∞ ∞ ! X X X ⇒ (1 − x2 ) am+2 (m + 1)(m + 2) − 2x am+1 (m + 1) + n(n + 1) am xm = 0 m=0 m=0 m=0 ∞ ! X ⇒ am+2 (m + 1)(m + 2) + n(n + 1)am xm m=0 ∞ X ∞ X − am+2 (m + 1)(m + 2)xm+2 − 2 am+1 (m + 1)xm+1 = 0 m=0 m=0 ∞ ! X ⇒ am+2 (m + 1)(m + 2) + n(n + 1)am xm m=0 ∞ X ∞ X − am (m − 1)(m)xm − 2 am (m)xm = 0 m=2 m=1 Since taking m = 0 in the last term contributes only 0 and similarly m = 0 and m = 1 does not contribute to the earlier term, therefore ∞ ! X ⇒ am+2 (m + 1)(m + 2) + n(n + 1)am xm m=0 ∞ X ∞ X m − am (m − 1)(m)x − 2 am (m)xm = 0 m=0 m=0 ∞ ! X ⇒ am+2 (m + 1)(m + 2) + (n + m + 1)(n − m)am xm = 0 m=0 Comparing the coefficients on both sides of the equation, gives (n + m + 1)(n − m) am+2 = − am , m = 0, 1, 2, . . . , (m + 1)(m + 2) 40 Sanyasiraju V S S Yedida
[email protected] Therefore, we have (n + 1)n a2 = − a0 1.2 (n + 2)(n − 1) a3 = − a1 2.3 (n + 3)(n − 2) (n + 3)(n + 1)n(n − 2) a4 = − a2 = a0 3.4 4! (n + 4)(n − 3) (n + 4)(n + 2)(n − 1)(n − 3) a5 = − a3 = a1 4.5 5! with general terms (−1)m (n + 2m − 1)(n + 2m − 3) . . . (n + 1)n(n − 2) . . . (n − 2m + 2) a2m = a0 (2m)! (−1)m (n + 2m)(n + 2m − 2) . . . (n + 2)(n − 1)(n − 3) . . . (n − 2m + 1) a2m+1 = a1 (2m + 1)! Finally the solution y is given (n + 1)n 2 (n + 3)(n + 1)n(n − 2) 4 y(x) = a0 1 − x + x −... 2! 4! (n + 2)(n − 1) 3 (n + 4)(n + 2)(n − 1)(n − 3) 5 +a1 x − x + x − ... 3! 5! = a0 y1 + a1 y2 where (n + 1)n 2 (n + 3)(n + 1)n(n − 2) 4 y1 = 1 − x + x −... 2! 4! ∞ X (−1)m (n + 2m − 1)(n + 2m − 3) . . . (n + 1)n(n − 2) . . . (n − 2m + 2) 2m = 1+ x m=1 (2m)! (n + 2)(n − 1) 3 (n + 4)(n + 2)(n − 1)(n − 3) 5 y2 = x − x + x − ... 3! 5! ∞ X (−1)m (n + 2m)(n + 2m − 2) . . . (n + 2)(n − 1)(n − 3) . . . (n − 2m + 1) 2m+1 = x− x m=1 (2m + 1)! For y1 , Lt Tm+1 Lt (n + 2m + 1)(n − 2m) 2 = x m → ∞ Tm m → ∞ (2m + 1)(2m + 2) n Lt n − 2 2 = x2 < 1 m = 1+ 2 x m→∞ 2m + 1 2+ m Lecture Notes MA2020 Differential Equations 41 therefore y1 is convergent. Similarly for y2 , Lt Tm+1 Lt (n + 2m + 2)(n − 2m − 1) 2 = x m → ∞ Tm m→∞ (2m + 2)(2m + 3) n−1 Lt n m −2 2 2 = 1+ x =x <1 m→∞ 2m + 2 2 + m3 therefore y2 is also convergent. Further y1 and y2 are linearly independent, hence y = a0 y1 + a1 y2 is the general solution of the Legendre equation for |x| < 1. 3.2.1 Properties of the solutions of the Legendre Equation • Legendre Polynomials : For integer values of n, one of the y1 and y2 is a polynomial and the other is an infinite series. If n is positive and even or negative and odd then y1 is a polynomial solution and similarly, if n is positive and odd or negative and even then y2 is a polynomial solution. Therefore, for integer values of n there exists a polynomial solution, denoted by Pn (x), to the Legendre equation (3.5). The polynomial solutions which satisfy Pn (1) = 1 are called Legendre polynomials. The other infinite series is called Legendre function and is denoted by Qn (x). • n > −1 : If n ≤ −1 then it can be written as n = −(1 + µ) for some µ > 0. We have n(n + 1) = −(1 + µ)(−1 − µ + 1) = µ(µ + 1). • All roots of Pn (x) = 0 lie between -1 and 1. • Some Legendre Polynomials are P0 (x) = 1, P1 (x) = x, P2 (x) = 12 (3x2 − 1), P3 (x) = 12 (5x3 − 3x), P4 (x) = 18 (35x4 − 30x2 + 3) and P5 (x) = 81 (65x5 − 70x3 + 15x). 1 0.5 n=0 n=1 P (x) n=2 0 n=3 n n=4 n=5 −0.5 −1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 x Figure 3.1: Roots of the Legendre Polynomial 42 Sanyasiraju V S S Yedida
[email protected] • Express f (x) = x4 + 3x3 − x2 + 5x − 2 in terms of Legendre polynomials. 1 35x4 − 30x2 + 3 P4 (x) = 8 8 6 3 f (x) = P4 (x) + x2 − + 3x3 − x2 + 5x − 2 35 7 35 8 1 = P4 (x) + 3x3 − x2 + 5x − 2 35 7 1 5x3 − 3x P3 (x) = 2 9 1 73 f (x) = + x − x2 + 5x − 5 7 35 8 6 1 34 73 = P4 (x) + P3 (x) − x2 + x − 35 5 7 5 35 1 2x2 − 1 P2 (x) = 2 8 6 2 1 34 73 f (x) = P4 (x) + P3 (x) − P2 (x) + + x− 35 5 21 21 5 35 P1 (x) = x, P0 (x) = 1 8 6 2 34 224 f (x) = P4 (x) + P3 (x) − P2 (x) + P1 (x) − P0 (x) 35 5 21 5 105 3.2.2 Rodrigue’s formula 1 dn The Legendre polynomials can be obtained from Pn (x) = n!2n dxn (x2 − 1)n dV V = (x2 − 1)n ⇒ V1 = = n(x2 − 1)n−1 2x ⇒ V1 = 2nx(x2 − 1)n−1 dx (1 − x2 )V1 + 2nxV = 0 differentiate w.r.t x (1 − x2 )V2 − 2x(1 − n)V1 + 2nV = 0 differentiate w.r.t x (1 − x2 )V3 − 2x(2 − n)V2 + 2(2n − 1)V1 = 0 differentiate w.r.t x (1 − x2 )V4 − 2x(3 − n)V3 + 2(3n − 3)V2 = 0 differentiate w.r.t x (1 − x2 )V5 − 2x(4 − n)V4 + 2(4n − 6)V3 = 0 differentiate w.r.t x ... 2 ¯ (1 − x )Vn+2 − 2x(n + 1 − n)Vn+1 + 2 (n(n + 1) − 1 − 2 − 3 . . . − n) Vn = 0 (1 − x2 )y ′′ − 2xy ′ + n(n + 1)y = 0 Lecture Notes MA2020 Differential Equations 43 d n 2 n Therefore, Vn+2 = C dx n (x − 1) satisfies the Legendre equation. But, we have Pn (1) = 1, therefore n n d 2 n d n n 1 = C (x − 1) = C (x − 1) (x + 1) dxn x=1 dxn x=1 = C (n!(x + 1)n + terms of n − 1 ) 1 = C n! 2n ⇒ C = n 2 n! 3.2.3 Rodrigue’s formula as a summation PN (−1)r (2n−2r)! Pn (x) = r=0 2n r! (n−r)! (n−2r)! x n−2N where N is n2 or n−1 2 depending on n is even or odd. From Binomial theorem we have n n X X n! (x2 − 1)n = nCr (x2 )n−r (−1)r = (−1)r (x)2n−2r r=0 r=0 r! (n − r)! n ! n n 1 d 1 d X n! Pn (x) = n (x2 − 1) = n (−1)r (x)2n−2r 2 n! dxn 2 n! dxn r=0 r! (n − r)! n ! n X 1 d = (−1)r n n (x)2n−2r r=0 2 r! (n − r)! dx n X 1 (2n − 2r)! n−2r = (−1)r (x) r=0 2n r! (n − r)! (n − 2r)! N X (−1)r (2n − 2r)! n−2r = n (x) r=0 2 r! (n − r)! (n − 2r)! In the last term n − 2r must be either 1 or 0 due to the derivative with respect to x, all other terms are zero, therefore r = N. 3.2.4 Generating function for Pn (x) 1 1 Prove that (1 − 2xt + t2 )− 2 = (1 − t(2x − t))− 2 = tn Pn (x) P∞ n=0 We know from Binomial expansion that m(m − 1) 2 m(m − 1)(m − 2) 3 (1 + x)m = 1 + mx + x + x + ... 1.2 1.2.3 1 3 1 3 5 1 1 . 2 2 2 . . (1 − z)− 2 = 1+ z+ z + 2 2 2 z3 + . . . 2 2! 3! 2! 4! 6! = 1+ 2 2 z+ 2 4 z2 + 2 6 z3 + . . . (1!) 2 (2!) 2 (3!) 2 1 2! 4! 2 (1 − t(2x − t))− 2 = 1+ t(2x − t) + t (2x − t)2 + . . . (1!)2 22 (2!)2 24 (2n − 2r)! (2n)! n 2 (2n−2r) t(n−r) (2x − t)(n−r) + . . . + t (2x − t)n + . . . ((n − r)!) 2 (n!)2 22n 44 Sanyasiraju V S S Yedida
[email protected] Collecting the tn terms from tn−r (2x − t)n−r gives (2n − 2r)! = 2 (2n−2r) t(n−r) (n − r)Cr (−t)r (2x)n−2r ((n − r)!) 2 (2n − 2r)! (n − r)! = (−1)r tn (2x)n−2r ((n − r)!)2 2(2n−2r) r! (n − 2r)! (2n − 2r)! (n − r)! = 2 (2n−2r) (−1)r tn (2x)n−2r ((n − r)!) 2 r! (n − 2r)! (−1)r (2n − 2r)! = n xn−2r tn 2 r! (n − r)! (n − 2r)! Now collecting all such terms gives N X (−1)r (2n − 2r)! n−2r (x) = tn Pn (x) r=0 2n r! (n − r)! (n − 2r)! n n−1 where, N = 2 or 2 depending on n is even or odd. 3.2.5 Numerical Example Using the generating function of the Legendre polynomial, compute the values of Pn (1), Pn (−1) and Pn (0). Fixing the value of x as 1, -1 and 0 gives the required values. 3.2.6 Recurrence formulae for Pn (x) 1. (n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x) We have ∞ 1 X (1 − 2xt + t2 )− 2 = tn Pn (x) n=0 Differentiating with respect t gives ∞ 1 3 X − (1 − 2xt + t2 )− 2 (−2x + 2t) = ntn−1 Pn (x) 2 n=0 ∞ 2 − 21 X 2 (x − t)(1 − 2xt + t ) = (1 − 2xt + t ) ntn−1 Pn (x) n=0 ∞ X ∞ X (x − t) tn Pn (x) = (1 − 2xt + t2 ) ntn−1 Pn (x) n=0 n=0 n Equating the coefficients of t on both sides gives the required relation 2. nPn (x) = xPn′ (x) − Pn−1 ′ (x) Differentiating the generating function w.r.to x gives ∞ 2 − 23 X t(1 − 2xt + t ) = tn Pn′ (x) n=0 Lecture Notes MA2020 Differential Equations 45 Differentiating the generating function w.r.to t gives ∞ 2 − 32 X (x − t)(1 − 2xt + t ) = ntn−1 Pn (x) n=0 Dividing one with the other gives ∞ X ∞ X n (x − t) t Pn′ (x) =t ntn−1 Pn (x) n=0 n=0 Comparing the coefficients of tn gives the required relation 3. (2n + 1)Pn (x) = Pn+1′ (x) − Pn−1 ′(x) Differentiating the recurrence relation (1) with respect to x and then using xPn′ (x) from (2) gives the required relation 4. Pn′ (x) = xPn−1 ′ (x) + nPn−1 (x) Differentiating w.r.to x and rewriting (1) gives (n + 1)Pn′ (x) = (2n + 1)Pn (x) + (n + 1)xPn′ (x) + n xPn′ (x) − Pn−1 ′ (x) Using the relation (2) and replacing n with n − 1 gives the required relation 5. (1 − x2 )Pn′ (x) = n (Pn−1 (x) − xPn (x)) Using recurrence relations (2) and (4) gives the required relation 3.2.7 Orthogonality of the Legendre Polynomials The Legendre polynomials Pn (x) satisfy the following orthogonal property Z 1 0, m 6= n Pm (x) Pn (x) dx = 2 (3.6) −1 2n+1 , m=n Proof : Assume that Pm (x) is the solution of (1 − x2 )u′′ − 2xu′ + m(m + 1)u = 0 and Pn (x) is the solution of (1 − x2 )v ′′ − 2xv ′ + n(n + 1)v = 0. Multiplying the first one with v and second one with u and then subtracting gives (for m 6= n) d (1 − x2 )(u′v − uv ′) + (m − n)(m + n + 1)uv = 0 dx Integrating both sides with respect to x in the limits -1 to 1 gives Z 1 1 Z 1 uv dx = (1 − x2 )(u′v − uv ′) −1 = 0 ⇒ (m−n)(m+n+1) Pm (x) Pn (x) dx = 0, (m 6= n) −1 −1 46 Sanyasiraju V S S Yedida
[email protected] For m = n, from Rodrigue’s formula we can write Z 1 Z 1 n 2 2 (n!2 ) Pn (x) dx = D n (x2 − 1)n D n (x2 − 1)n dx −1 −1 Z 1 D (x − 1)n D n−1 (x2 − 1)n−1 1 − n 2 D n+1 (x2 − 1)n D n−1 (x2 − 1)n dx = −1 −1 Z 1 = − D n+1 (x2 − 1)n D n−1 (x2 − 1)n dx −1 Integrating n − 1 times gives Z 1 n = (−1) D 2n (x2 − 1)n (x2 − 1)n dx −1 Z 1 Z 1 n 2 n = (−1) (2n)! (x − 1) dx = 2(2n!) (1 − x2 )n dx −1 0 Z π 2 = 2(2n)! cos2n+1 θdθ dx 0 2n (2n − 2) . . . 4 2 = 2(2n)! (2n + 1) (2n − 1) . . . 2 1 (2n (2n − 2) . . . 4 2)2 2 = 2(2n)! = (n!2n )2 (2n + 1)! 2n + 1 1 2 Z Pn2 (x) dx = −1 2n + 1 3.2.8 Fourier - Legendre Expansion If f (x) is a function defined from x = −1 to 1 then wee can write f (x) = ∞ P n=0 cn Pn (x). To find the coefficients cn multiply Pn (x) on both sides and integrate with respect to x in the limits -1 to 1 and finally using the orthogonality condition of Legendre polynomials gives Z 1 Z 1 cn f (x) Pn (x) dx = cn Pn2 (x) = −1 −1 2n + 1 2n+1 R1 Therefore, cn = n −1 f (x) Pn (x) dx. Chapter 4 Frobeneous Method Power series method to solve homogeneous linear second order equations with variable coefficients at regular singular points generally results in to trivial solution, therefore alternatively, the Frobeneous method is used to solve such equations at regular singu- lar points. To describe the method, let us consider the general second order equation (x − x0 )2 y ′′ + (x − x0 )p1 (x)y ′ + p0 (x)y = 0 where x = x0 is a regular singular point and p1 (x) and p0 (x) are some analytic functions that is, there exists some power series expansions for them, about x0 , given by p0 (x) = p00 + p10 (x − x0 ) + p20 (x − x0 )2 + . . . p1 (x) = p01 + p11 (x − x0 ) + p21 (x − x0 )2 + . . . Finally, assume (a0 6= 0) ∞ X y = am xm+k m=0 X∞ y′ = am (m + k)xm+k−1 m=0 X∞ y ′′ = am (m + k)(m + k − 1)xm+k−2 m=0 Substituting them in the given equation gives ∞ ∞ ! ∞ X X X am (m + k)(m + k − 1)(x − x0 )m+k + pl1 (x − x0 )l am (m + k)(x − x0 )m+k m=0 l=0 m=0 ∞ ! ∞ X X + pl0 (x − x0 )l am (x − x0 )m+k = 0 l=0 m=0 47 48 Sanyasiraju V S S Yedida
[email protected] Collecting the coefficients of the lowest powers of (x − x0 ), which is a quadratic polynomial in k, called as indicial polynomial, and equating it to zero gives k(k − 1) + p01 k + p00 = 0 (4.1) The indicial equation (4.1) has two roots and depending on the nature of these two roots the general solution of the given equation is written 1. Case 1 : The roots are real, distinct and not differ by an integer In this case, the two linearly independent solutions can be obtained as in the power series method, and the general solution will looks like y = C1 y1 + C2 y2 . Numerical Example 1: Find the general solution of 2x2 y ′′ + xy ′ − (x2 + 1)y = 0 Let the solution be written as ∞ X y = am xm+k c0 6= 0 m=0 X∞ y′ = am (m + k)xm+k−1 m=0 X∞ y ′′ = am (m + k)(m + k − 1)xm+k−2 m=0 ⇒ 2x y + xy ′ − (x2 + 1)y 2 ′′ ∞ X X∞ 2 am (m + k)(m + k − 1)xm+k + am (m + k)xm+k m=0 m=0 ∞ X − (x2 + 1) am xm+k = 0 m=0 The indicial equation is given by (m = 0), 2k 2 − k − 1 = 0, Therefore, the two roots are k = 1, − 21 Similarly, equating the coefficients of xk+1 to zero gives a1 (2k 2 + 3k) = 0, since (2k 2 + 3k) 6= 0 for k = 1, − 21 , therefore a1 must be zero. Now the summation can be written as ∞ X ∞ X m+k (2(m + k)(m + k − 1) + (m + k) − 1) am x − am xm+k+2 = 0 m=2 m=0 X∞ X∞ (2(m + k)(m + k − 1) + (m + k) − 1) am xm+k − am−2 xm+k = 0 m=2 m=2 ∞ X ((2(m + k)(m + k − 1) + (m + k) − 1) am − am−2 ) xm+k = 0 m=2 ∞ X (((m + k)(2m + 2k − 1) − 1) am − am−2 ) xm+k = 0 m=2 Lecture Notes MA2020 Differential Equations 49 Equating the coefficients of xm+k gives ((m + k)(2m + 2k − 1) − 1) am − am−2 = 0 or am−2 am = , m≥2 (m + k)(2m + 2k − 1) − 1 Further, since a1 = 0, therefore a3 = a5 = . . . = 0 and for k = 1, we have am−2 a0 a2 a0 am = , ⇒ a2 = , a4 = = ,... (m + 1)(2m + 1) − 1 14 44 616 Similarly, for k = − 21 , we have am−2 a0 a2 a0 am = , ⇒ a2 = , a4 = = ,... (m + 1/2)(2m − 2) − 1 2 20 40 Therefore, the two linearly independent solutions are x2 x4 y1 (x) = a0 x 1 + + + ... 14 616 x2 x4 − 12 y2 (x) = a0 x 1+ + + ... 2 40 Numerical Example 2: Find the general solution of 9x(1 − x)y ′′ − 12y ′ + 4y = 0 Let the solution be written as ∞ X y = am xm+k c0 6= 0 m=0 X∞ y′ = am (m + k)xm+k−1 m=0 X∞ y ′′ = am (m + k)(m + k − 1)xm+k−2 m=0 9x(1 − x)y ′′ − 12y ′ + 4y = 0 X∞ ∞ X m+k−1 ⇒ (9(m + k)(m + k − 1) − 12(m + k))am x + (−9(m + k) + 4)am xm+k = 0 m=0 m=0 X∞ ∞ X ⇒ 3(m + k)(3m + 3k − 7)am xm+k−1 + (−9(m + k) + 4)am xm+k = 0 m=0 m=0 50 Sanyasiraju V S S Yedida
[email protected] The indicial equation is given by (coefficient of xk−1 ), 9k 2 − 21k = 0, Therefore, the two roots are k = 0, 37 Now the summation can be written as ∞ X ∞ X m+k−1 3(m + k)(3m + 3k − 7)am x + (−9(m + k) + 4)am xm+k = 0 m=1 m=0 ∞ X X∞ 3(m + k + 1)(3m + 3k − 4)am+1 xm+k + (−9(m + k) + 4)am xm+k = 0 m=0 m=0 ∞ X (3(m + k + 1)(3m + 3k − 4)am+1 + (−9(m + k) + 4)am ) xm+k = 0 m=0 Equating the coefficients of xm+k gives 3(m + k + 1)(3m + 3k − 4)am+1 + (−9(m + k) + 4)am = 0 or −9(m + k) + 4 am+1 = am , m≥0 3(m + k + 1)(3m + 3k − 4) Finally, the coefficients can be written as 3k + 1 a1 = a0 3(k + 1) 3k + 4 (3k + 4)(3k + 1) a2 = a1 = 2 a0 3(k + 2) 3 (k + 2)(k + 1) 3k + 7 (3k + 7)(3k + 4)(3k + 1) a3 = a2 = 3 a0 3(k + 3) 3 (k + 3)(k + 2)(k + 1) ... For, k = 0, we have 1 14 147 a1 = a0 , a2 = a0 , a3 = a0 , . . . 3 36 369 Therefore, 1 14 2 147 3 y1 (x) = a0 1+ x+ x + x + ... 3 36 369 Similarly, for k = 37 , we have 7 8 8 11 2 8 11 14 3 y2 (x) = a0 x 1 + x + 3 x + x + ... 10 10 13 10 13 16 Numerical Example 3: Find the general solution of 2x2 y ′′ + xy ′ − (x2 + 1)y = 0 Lecture Notes MA2020 Differential Equations 51 The roots of the indicial equation are 1, − 12 . The two linearly independent solutions are x2 x4 y1 (x) = a0 x 1 + + + ... 14 616 x2 x4 − 12 y2 (x) = a0 x 1+ + + ... 2 40 2. Case II : The roots of the indicial equation are equal In this case the second linearly independent solution is obtained, after obtaining y1 (x), taking partial derivative of y1 with respect to k and substituting k = k1 . Numerical Example : Find the general solution of xy ′′ + y ′ − xy = 0 Let the solution be written as ∞ X y = am xm+k c0 6= 0 m=0 X∞ y′ = am (m + k)xm+k−1 m=0 X∞ y ′′ = am (m + k)(m + k − 1)xm+k−2 m=0 ⇒ xy ′′ + y ′ − xy ∞ X ∞ X m+k−1 m+k−1 am (m + k)(m + k − 1)x + am (m + k)) x − am xm+k+1 = 0 m=0 m=0 ∞ X ∞ X (m + k)2 am xm+k−1 − am xm+k+1 = 0 m=0 m=0 The indicial equation is given by (m = 0), k 2 = 0, Therefore, the two roots are k = 0, 0 Similarly, equating the coefficients of xk is equal to zero gives a1 (k 2 + 1) = 0, since (k 2 + 1) 6= 0 for k = 0, therefore a1 must be zero. Now the summation can be written as ∞ X ∞ X 2 m+k−1 (m + k) am x − am xm+k+1 = 0 m=2 m=0 ∞ X X∞ (m + k + 2)2 am+2 xm+k+1 − am xm+k+1 = 0 m=0 m=0 ∞ X (m + k + 2)2 am+2 − am xm+k+1 = 0 m=0 52 Sanyasiraju V S S Yedida
[email protected] Equating the coefficients of xm+k gives (m + k + 2)2 am+2 − am = 0 or am am+2 = m≥0 (m + k + 2)2 Further, since a1 = 0, therefore a3 = a5 = . . . = 0. For m = 2, 4, . . . a0 a2 a0 a2 = 2 , ⇒ a4 = 2 = (k + 2) (k + 4) (k + 2) (k + 4)2 2 a4 a0 a6 = 2 = ,... (k + 6) (k + 2) (k + 4)2 (k + 6)2 2 Substituting them in the series gives x2 x4 x6 k y1 = a0 x 1 + + + + ... (k + 2)2 (k + 2)2 (k + 4)2 (k + 2)2 (k + 4)2 (k + 6)2 (4.2) Finally, fixing k = 0 gives x2 x4 x6 y1 = a0 1 + 2 + 2 2 + 2 2 2 + . . . 2 2 4 2 4 6 To compute the second linearly independent solution, substitute (4.2) in the given equation. (This results into a remainder of the indicial polynomial as the coefficients in the other terms are computed by substituting them in the equation), that is, we get xy ′′ + y ′ − xy = a0 k 2 xk−1 Differentiating with respect to k gives 2 d d dy = a0 2kxk−1 + k 2 xk−1 ln(x) x 2+ −x dx dx dk dy Since, the right hand side of the above equation is zero at k = 0, therefore dk is also a solution of the given equation. Therefore, differentiating (4.2) with respect to k and then using k = 0 in the resultant function gives the other linearly independent solution. x2 x4 x6 dy1 k = a0 x ln(x) 1 + + + + ... dk (k + 2)2 (k + 2)2 (k + 4)2 (k + 2)2 (k + 4)2 (k + 6)2 2x2 k 4 2 2 + a0 x − −x + + ... (k + 2)3 (k + 2)3 (k + 4)2 (k + 2)3 (k + 2)2 2x2 k 4 1 1 = ln(x)y1 (x) + a0 x − − 2x + + ... (k + 2)3 (k + 2)3 (k + 4)2 (k + 2)3 (k + 2)2 Now substituting k = 0 gives 2 3x4 dy1 x y2 (x) = = ln(x)y1 (x) − a0 + + ... dk k=0 4 128 Lecture Notes MA2020 Differential Equations 53 3. Case III : The roots of the indicial equation differ by an integer. There can be three possibilities in this case, they are • The two roots of the indicial equation may two linearly independent solu- tions as in the Case I • One root of the indicial equation itself may give the complete solution with two parameters and the other root of the indicial equation may produce only a dependent solution. • One root of the indicial equation may produce indeterminant terms how- ever by modifying the coefficient, it may be manageable to eliminate the indeterminacy. In this case the other root is obtained by the method fol- lowed in the Case II Now let us look at some examples of Case III. Numerical Example 1: Find the general solution of x(x + 1)y ′′ + 3xy ′ + y = 0 Let the solution be written as ∞ X y = am xm+k c0 6= 0 m=0 X∞ y′ = am (m + k)xm+k−1 m=0 X∞ y ′′ = am (m + k)(m + k − 1)xm+k−2 m=0 ⇒ x(x + 1)y ′′ + 3xy ′ + y ∞ X X∞ am (m + k)(m + k − 1)xm+k−1 + ((m + k)(m + k − 1) + 3(m + k) + 1) am xm+k = 0 m=0 m=0 The indicial equation is given by (m = 0), k(k − 1) = 0, Therefore, the two roots are k = 0, 1, which are differ by an integer Now the summation can be written as ∞ X ∞ X m+k−1 am (m + k)(m + k − 1)x + ((m + k)(m + k − 1) + 3(m + k) + 1) am xm+k = 0 m=1 m=0 ∞ X ((m + k + 1)(m + k)am+1 + ((m + k)(m + k + 2) + 1)am ) xm+k = 0 m=0 Equating the coefficients of xm+k to zero gives (for m ≥ 0) (m + k)(m + k + 2) + 1 (m + k + 1)2 (m + k + 1) am+1 = − am , = − am = − am (m + k + 1)(m + k) (m + k + 1)(m + k) (m + k) 54 Sanyasiraju V S S Yedida
[email protected] k+1 a1 = − a0 k k+2 k+2 a2 = − a1 = a0 k+1 k k+3 k+3 a3 = − a2 = a0 k+2 k ... Therefore, the solution can be written as k k+1 k+2 2 k+3 3 y1 (x) = a0 x 1 − x+ x − x + ... k k k In the above solution all the terms after the first one are equal to ∞ as k = 0, However if we modify a0 = kc0 , in general replace a0 with c0 (x − x0 ), then the above solution can be written as y1 (x) = c0 xk k − (k + 1)x + (k + 2)x2 − (k + 3)x3 + . . . For k = 0 we have y1 (x) = c0 −x + 2x2 − 3x3 + . . . = −c0 x(1 + x)−2 Similarly, for k = 1 we have (k = 1 is the root of the modified indicial equation) y2 (x) = c0 x 1 − 2x + 3x2 − 4x3 + . . . = c0 x(1 + x)−2 Therefore, y1 and y2 are not linearly independent, to obtained the other solution take the partial derivative of y1 with respect to k, to get y2 (x) = c0 y1 ln(x) + c0 xk 1 − x + x2 − x3 + . . . = c0 y1 ln(x) + c0 xk (1 + x)−1 Substituting k = 0 gives y2 (x) = c0 y1 ln(x) + c0 (1 + x)−1 Numerical Example 2: Find the general solution of x2 y ′′ + 6xy ′ + (6 + x2 )y = 0 Let the solution be written as ∞ X y = am xm+k c0 6= 0 m=0 X∞ y′ = am (m + k)xm+k−1 m=0 X∞ y ′′ = am (m + k)(m + k − 1)xm+k−2 m=0 Lecture Notes MA2020 Differential Equations 55 x2 y ′′ + 6xy ′ + (6 + x2 )y = 0 X∞ ∞ X ⇒ ((m + k)(m + k + 5) + 6)am xm+k + am xm+k+2 = 0 m=0 m=0 The indicial equation is given by (k + 2)(k + 3) = 0, Therefore, the two roots are k = −2, −3, and k = −3 gives two linearly independent solutions. Now the summation can be written as ∞ X ((m + k + 2)(m + k + 7) + 6)am+2 + am )xm+k+2 = 0 m=0 Equating the coefficients of xm+k to zero gives (for m ≥ 0) am am+2 = − (m + k + 2)(m + k + 7) + 6 a0 a2 = − (k + 2)(k + 7) + 6 a0 a3 = − (k + 3)(k + 8) + 6 a2 a0 a4 = − = (k + 4)(k + 9) + 6 ((k + 4)(k + 9) + 6)((k + 2)(k + 7) + 6) a3 a1 a5 = − = (k + 5)(k + 10) + 6 ((k + 5)(k + 10) + 6)((k + 3)(k + 8) + 6) ... For k = −3. we have a0 a1 a0 a1 a2 = − , a3 = − , a4 = , a5 = ,... 2 6 24 120 The general solution is x2 x4 x3 x5 −3 y(x) = x a0 1 − + + . . . + a1 x − + + ... 2! 4! 3! 5! cos x sin x = a0 3 + a1 3 x x 56 Sanyasiraju V S S Yedida
[email protected] 4.1 Bessel’s Function 4.1.1 Gamma Function Gamma function is defined as Z∞ Γ(ν) = e−x xν−1 dx ν > 0 (4.3) 0 Z∞ ∞ Γ(1) = e−x x0 dx = −e−x 0 = 1 (4.4) 0 Z∞ Z∞ ∞ Γ(ν + 1) = e−x xν dx = −xν e−x 0 + e−x xν−1 dx = νΓ(ν) (4.5) 0 0 Or (4.6) Γ(ν + 1) Γ(ν) = (4.7) ν That is, for integer values of ν = n it works as a factorial function, therefore, the Gamma can be taken as the generalization of the Factorial function (to the non-zero, non-negative real numbers). Further, if Γ(ν) is known in an interval, say, 1 < ν ≤ 2 then it can be computed in the interval 2 < ν ≤ 3, 3 < ν ≤ 4 so on and similarly, 0 < ν ≤ 1, that is for all ν except for ν = 0 and negative integers. Therefore, in general, the values of the Gamma function can be obtained using Γ(r + 1) Γ(r + 2) Γ(r + k + 1) Γ(r) = = = ··· = r r(r + 1) r(r + 1)(r + 2) · · · (r + k) Γ(r + k + 1) Γ(r) = , (r 6= 0, −1, −2, · · · ) r(r + 1)(r + 2) · · · (r + k) k may be chosen as a smallest integer such that r + k + 1 > 0. A typical graph of the Gamma function is shown in the Figure 4.1.1. √ One special value of the Gamma function is Γ(1/2) = π. Proof : Consider Z∞ 1 1 Γ = e−x x( 2 −1) dx 2 0 2 Take the transformation x = u Z∞ 2 = 2 e−u du 0 Lecture Notes MA2020 Differential Equations 57 Gamma Function 25 20 15 Γ(x) 10 5 0 −5 −10 −15 −4 −3 −2 −1 0 1 2 3 4 5 x Figure 4.1: Gamma Function squaring the terms will give 2 Z∞ Z∞ 1 2 2 Γ = 4 e−u e−v du dv 2 0 0 changing in to polar coordinates will give π Z2 Z∞ 2 = 4 e−r r dr dθ 0 0 π Z 2 = 2 1 dθ = π 0 √ That is, Γ(1/2) = π. 4.1.2 Bessel’s equation of order ν d2 y dy x2 2 + x + (x2 − ν 2 )y = 0 (4.8) dx dx Frobeneous method can be used to solve the Bessel’s equation near x = 0. Consider ∞ X y = am xm+k c0 6= 0 m=0 X∞ y′ = am (m + k)xm+k−1 m=0 X∞ y ′′ = am (m + k)(m + k − 1)xm+k−2 m=0 58 Sanyasiraju V S S Yedida
[email protected] d2 y dy ⇒ x2 2 + x + (x2 − ν 2 )y = 0 dx dx ∞ X ∞ 2 m+k X ⇒ am (m + k)(m + k − 1) + (m + k) − ν x + am xm+k+2 = 0 m=0 m=0 The indicial equation is given by (m = 0), k(k − 1) + k − ν 2 = 0, Therefore, the two roots are k = ±ν Similarly, equating the coefficients of xk+1 to zero gives a1 ((k + 1)2 − ν 2 ) = 0, since (k + 1)2 − ν 2 6= 0 for k = ±ν, therefore a1 must be zero. Now the summation can be written as ∞ X ∞ X 2 m+k am xm+k+2 (m + k)(m + k − 1) + (m + k) − ν am x + = 0 m=2 m=0 ∞ X X∞ (m + k)2 − ν 2 am xm+k + am−2 xm+k = 0 m=2 m=2 ∞ X (m + k)2 − n2 am + am−2 xm+k = 0 m=2 Equating the coefficients of xm+k gives (m + k)2 − ν 2 am + am−2 = 0 or am−2 am = , m≥2 (m + k)2 − ν 2 Further, since a1 = 0, therefore a3 = a5 = . . . = 0 and for k = 1, we have −a0 a2 a0 a2 = 2 2 , a4 = − 2 2 = ,... (k + 2) − ν (k + 4) − ν ((k + 2) − ν )((k + 4)2 − ν 2 ) 2 2 Therefore, the solution can be written as x2 x4 k y(x) = a0 x 1 − + + ... (k + 2)2 − ν 2 ((k + 2)2 − ν 2 )((k + 4)2 − ν 2 ) For k = ±n we get x2 x4 n y1 (x) = a0 x 1 − + −... (4.9) 4(ν + 1) 42 2!(ν + 1)(ν + 2) x2 x4 n y2 (x) = a0 x 1 − + − ... (4.10) 4(−ν + 1) 42 2!(−ν + 1)(−ν + 2) 1. Case(i) : n is not zero or an integer, then (4.9) and (4.10) gives two linearly independent solutions and the general solution to the Bessel’s equation y is 1 y = c1 y1 + c2 y2 . Further if a0 = 2ν Γ(ν+1) then (4.9) is called Bessel function of Lecture Notes MA2020 Differential Equations 59 first kind of order ν and is denoted by Jν (x). It’s expression is given by xν 1 1 x 2 1 x 4 Jν (x) = − + − · ·(4.11) · 2 Γ(ν + 1) 1! Γ(ν + 2) 2 2! Γ(ν + 3) 2 ∞ x ν+2r X 1 = (−1)r (4.12) r=0 2 r! Γ(ν + r + 1) Similarly ∞ x −ν+2r X 1 J−ν (x) = (−1)r (4.13) r=0 2 r! Γ(−ν + r + 1) 2. Case(ii) : ν is zero, In this case since y2 = y1 , therefore, y2 is obtained using y2 = ∂y 1 ∂k k=0 . 3. Case(iii) : ν = n is an integer, then y2 fails to give a solution for positive n and similarly y1 fails to give for negative n. In such cases, the second solution is obtained by assuming y2 (x) = u(x) y1 (x). Substituting y2 in Bessel’s equation gives x2 u′′ Jn + 2x2 u′Jn′ + xu′ Jn = 0 dividing with x2 u′Jn gives d xu′ Jn2 = 0 ⇒ xu′ Jn2 = A dx A dx Z ′ ⇒ u =; 2 ⇒ u=A +B xJn xJn2 Therefore, the second solution is y2 (x) = Jn (x) x(J dx(x))2 which is called Bessel R n function of second kind order n (or Neumann function) and is denoted by Yn . Observation J−n (x) = (−1)n Jn (x), (n is an integer) We have ∞ x −n+2r X 1 J−n (x) = (−1)r r=0 2 r! Γ(−n + r + 1) Observe that, Γ(−n + r + 1) = (r − n)! and it will tend to ∞ for all negative r − n + 1 (Γ function is not defined for 0 and negative integers) , therefore, the first n − 1 terms in the above series are zero and it can be re-written as ∞ x −n+2r X 1 J−n (x) = (−1)r r=n 2 r! Γ(−n + r + 1) 60 Sanyasiraju V S S Yedida
[email protected] Replacing r with r + n gives ∞ x n+2r X 1 J−n (x) = (−1)r+n r=0 2 (r + n)! Γ(−n + r + n + 1) ∞ x n+2r X 1 = (−1)r+n r=0 2 (r + n)! r! ∞ x n+2r X r+n 1 = (−1) r=0 2 Γ(n + r + 1) r! n = (−1) Jn (x) 4.1.3 Recurrence Relations d 1. (xν Jν (x)) = xν Jν−1 (x) dx We have ∞ x ν+2r X 1 xν Jν (x) = xν (−1)r r=0 2 r! Γ(ν + r + 1) ∞ X 1 1 = (−1)r x2ν+2r ν+2r r=0 r! Γ(ν + r + 1) 2 ∞ d ν X 2(ν + r) 1 (x Jν (x)) = (−1)r x2ν+2r−1 ν+2r dx r=0 r! Γ(ν + r + 1) 2 ∞ x ν+2r−1 X (ν + r) = xν (−1)r r=0 2 r! Γ(ν + r + 1) ∞ x (ν−1)+2r X 1 = xν (−1)r r=0 2 r! Γ((ν − 1) + r + 1) ν = x Jν−1 (x) d 2. dx (x−ν Jν (x)) = −x−ν Jν+1 (x) Lecture Notes MA2020 Differential Equations 61 We have ∞ x ν+2r X 1 −ν x Jν (x) = x −ν (−1)r r=0 2 r! Γ(ν + r + 1) ∞ X 1 1 = (−1)r x2r ν+2r r=0 r! Γ(ν + r + 1) 2 ∞ d X 2r 1 (−1)r x2r−1 x−ν Jν (x) = n+2r dx r=0 r! Γ(ν + r + 1) 2 ∞ x ν+2r−1 X r = x −ν (−1)r r=0 2 r! Γ(ν + r + 1) ∞ x ν+2r−1 X r = x−ν (−1)r (∵ r = 0 contributes only zero r=1 2 r! Γ(ν + r + 1) ∞ x n+2(r−1)+1 −ν X r−1 1 = −x (−1) r=1 2 (r − 1)! Γ(n + (r − 1) + 2) ∞ x ν+2r+1 −ν X r 1 = −x (−1) (replace r − 1 with r r=0 2 r! Γ(ν + r + 2) ∞ x ν+2r+1 X 1 = −x−ν (−1)r r=0 2 r! Γ(ν + 1 + r + 1) = −x−ν Jν+1 (x) x 3. Jν (x) = 2ν (Jν−1 (x) + Jν+1 (x)) From first and second recurrence relations, we have d ν (x Jν (x)) = xν Jν−1 (x) dx d x−ν Jν (x) = −x−ν Jν+1 (x) dx or x Jν (x) + νxν−1 Jν (x) = xν Jν−1 (x) ν ′ x−ν Jν′ (x) − νx−ν−1 Jν (x) = −x−ν Jν+1 (x) or ′ ν Jν (x) + Jν (x) = Jν−1 (x) x ν −Jν′ (x) + Jν (x) = Jν+1 (x) x x ⇒ Jν (x) = (Jν−1 (x) + Jν+1 (x)) 2ν 1 ⇒ Jν′ (x) = (Jν−1 (x) − Jν+1 (x)) 2 62 Sanyasiraju V S S Yedida
[email protected] ν 4. Jν′ (x) = J (x) x ν − Jν+1 (x) Comes directly from recurrence relation (2) 2ν 5. Jν+1 (x) = x ν J (x) − Jν−1 (x) Comes directly from recurrence relation (3) 4.1.4 Observation J0 (x) = 0 has no complex roots but an infinite number of real positive roots. Its first four roots are x = 2.4, 5.52, 8.65 and 11.79 4.1.5 Compute J0 (x), J1 (x) and J1/2(x) ∞ x ν+2r X 1 Jν (x) = (−1)r r=0 2 r! Γ(ν + r + 1) For ν = 0 ∞ x 2r X r 1 J0 (x) = (−1) r=0 2 r! Γ(r + 1) 1 x 2 1 x 4 1 x 6 = 1− + − +··· 1! 2 (2!)2 2 (3!)2 2 For ν = 1 ∞ x 2r+1 X 1 J1 (x) = (−1)r r=0 2 r! Γ(r + 2) x 1 x 2 1 x 4 1 x 6 = 1− + − +··· 2 1! 2! 2 (2! 3!) 2 (3! 4!) 2 1 For ν = 2 ∞ x 2r+1/2 X 1 J1/2 (x) = (−1)r r=0 2 r! Γ(r + 1 + 1/2) x 1/2 1 1 x 2 1 x 4 1 x 6 = − + − +··· 2 Γ(3/2) 1!Γ(5/2) 2 2!Γ(7/2) 2 3!Γ(9/2) 2 x 1/2 1 1 x 2 1 x 4 = − + − +··· 2 1/2 Γ(1/2) 1! 1/2 3/2 Γ(1/2) 2 2 5/2 3/2 1/2 Γ(1/2) 2 x 1/2 1 2 2x2 2x4 = − + −+··· 2 Γ(1/2) 1! 3! 5! √ ! x3 x5 √ 2 x = √ − + − +··· ∵ Γ(1/2) = π xπ 1! 3! 5! r 2 = sin x Similarly xπ r 2 J−1/2 (x) = cos x xπ Lecture Notes MA2020 Differential Equations 63 4.1.6 Express J5 (x) in terms of J1(x) and J0(x) We have x 2ν Jν (x) = (Jν−1 (x) + Jν+1 (x)) ⇒ Jν+1 (x) = Jν (x) − Jν−1 (x) 2ν x 2 J2 (x) = J1 (x) − J0 (x) x 4 8 4 J3 (x) = J2 (x) − J1 (x) = 2 − 1 J1 (x) − J0 (x) x x x 6 48 8 24 J4 (x) = J3 (x) − J2 (x) = − J1 (x) + 1 − 2 J0 (x) x x3 x x 8 384 72 12 192 J5 (x) = J4 (x) − J3 (x) = − 2 − 1 J1 (x) + − 3 J0 (x) x x4 x x x 4.1.7 Prove the following q 2 3−x2 1. J 5 (x) = xπ x2 sin x − x3 cos x 2 We have 2ν Jν+1 = Jν (x) − Jν−1 (x) x 2 . 3/2 J 5 (x) = J 3 +1 (x) = J3/2 (x) − J(3/2−1) (x) 2 2 x 2 . 3/2 2 . 1/2 = J1/2 (x) − J(1/2−1) (x) − J1/2 (x) x x 3 1 = J1/2 (x) − J(−1/2) (x) − J1/2 (x) x x 3 − x2 3 = 2 J1/2 (x) − J(−1/2) (x) rx x 2 2 3−x 3 = sin x − cos x xπ x2 x 2. Jν′′ = 41 (Jν−2 − 2Jν + Jν+2 ) We have 1 Jν′ = (Jν−1 − Jν+1 ) 2 1 1 1 Jν′′ = (Jν−2 − Jν ) − (Jν − Jν+2 ) 2 2 2 1 = (Jν−2 − 2Jν + Jν+2 ) 4 64 Sanyasiraju V S S Yedida
[email protected] d (x Jν Jν+1 ) = x Jν2 − Jν+1 2 3. dx d d (x−ν Jν ) (xν+1 Jν+1 ) (x Jν Jν+1 ) = dx dx d d x−ν Jν (xν+1 Jν+1 ) + (x−ν Jν ) xν+1 Jν+1 = dx dx = −x−ν Jν+1 (xν+1 Jν+1 ) + (x−ν Jν ) xν+1 Jν = x Jν2 − Jν+1 2 2 R 4. J3 dx = c − J2 (x) − x J1 (x) Z Z J3 dx = x2 (x−2 J3 ) dx Z 2 −2 d = −x (x J2 ) + 2x x−2 J2 dx (∵ dx (x−ν Jν (x)) = −x−ν Jν+1 (x) Z = −J2 + 2 x−1 J2 dx = −J2 + 2(−x−1 J1 ) 2 = c − J2 (x) − J1 (x) x x2 x J02 (x) dx = (J02 (x) + J12 (x)) R 5. 2 x2 x2 x2 2 Z Z Z x J02 (x) dx = J02 (x) − 2J0 J0′ dx = J (x) − x2 J0 J0′ dx 2 2 2 0 x2 Z = J02 (x) + (x J0 ) (x J1 ) dx (∵ d dx (x−0 J0 (x)) = −x−0 J0+1 (x) 2 x2 d Z = J02 (x) + (x J1 ) (x J1 ) dx (∵ d dx (x J1 ) = x J0 2 dx x2 x2 2 = J02 (x) + J (x) 2 2 1 6. Using Rolle’s theorem, prove that between two consecutive Zeros of J0 (x) there is precisely one zero of J1 (x). Rolls theorem : For a continuous and closed function f in an interval I, if f ′ exists at every point of I and f (a) = f (b) = 0 then there exists a point c ∈ I such that f ′ (c) = 0. If a and b are any two consecutive roots of J0 (x), then we have J0 (a) = J0 (b) = 0. Since J0 satisfies the conditions required for the Rolle’s theorem, there must ex- ists a c, (a < c < b) such that J0′ (c) = 0. Lecture Notes MA2020 Differential Equations 65 d d But we have dx (x−ν Jν ) = −x−ν Jν+1 ⇒ dx (x−0 J0 ) = −x−0 J0+1 ⇒ d dx (J0 ) = −J1 . therefore J1 (c) = 0. d d Similarly using the rule dx (xν Jν ) = xν Jν−1 ⇒ dx (x1 J1 ) = x1 J0 ⇒ d dx (x J1 ) = xJ0 . That is two consecutive zeros of J1 contains a zero of J0 (the result: Zeros of J1 = 0 and also the zeros of xJ1 = 0 are used). 4.1.8 Equations Reducible to Bessel’s Equation d y 2 dy 1. x2 dx 2 2 2 2 + x dx + (k x − ν )y = 0. 2 Take the transformation t = kX gives the equation t2 ddt2y + t dy dt + (t2 −ν 2 )y = 0, whose solution is y(t) = c1 Jν (t) + c2 J−ν (t), if ν is not an integer and y(t) = c1 Jn (t) + c2 Yn (t), if ν = n is an integer. Therefore, y(x) = c1 Jν (kx) + c2 J−ν (kx)(t) or y(x) = c1 Jn (kx) + c2 Yn (kx)(t) are the solutions of the given equation. 2 d y dy 2 2. x dx 2 + a dx + k xy = 0. Taking y = xν z and dividing with xν−1 and using 2ν + a = 1 gives d2 z dz x2 2 + x + (k 2 x2 − ν 2 )z = 0 dx dx 2 d y dy 2 r 3. x dx 2 + c dx + k x y = 0. Taking x = tm gives d2 y dy t 2 + (1 − m + cm) + (km)2 tmr+m−1 y = 0 dt dt 2 r+2c−1 Fixing mr + m − 1 = 1 that is m = r+1 and a = 1 − m + cm = r+1 gives dy dy t 2 + a + (km)2 ty = 0 dt dt 4.1.9 Orthogonality of the Bessel Functions Z 1 0, α 6= β x Jn (αx) Jn (βx) dx = 1 2 (4.14) 0 J (α), 2 n+1 α=β where α and β are the roots of Jn (x) = 0. Consider d2 u du x2 2 + x + (α2 x2 − n2 )u = 0 dx dx 2 d v dv x2 2 + x + (β 2 x2 − n2 )v = 0 dx dx 66 Sanyasiraju V S S Yedida
[email protected] Multiply the first with xv and second with ux and then subtracting one from the other gives 2 d2 v du du dv x 2 v−u 2 + v −u + (α2 − β 2 )xuv = 0 dx dx dx dx d du dv x v −u = (β 2 − α2 )xuv dx dx dx Integrating both sides from 0 to 1 Z1 1 2 2 du dv du dv (β − α ) x uv dx = x v −xu = v −u dx dx 0 dx dx x=1 0 Using u = Jn (αx), v = Jn (βx), u′ = αJn′ (αx) and v ′ = βJn′ (βx) (for α 6= β ) gives Z1 2 2 dJn (αx) dJn (βx) (β − α ) x Jn (αx)Jn (βx) dx = Jn (βx) − Jn (αx) dx dx x=1 0 Z1 (α Jn (βx) Jn′ (αx) − β Jn (αx) Jn′ (βx))x=1 x Jn (αx)Jn (βx) dx = (β 2 − α2 ) 0 Z1 α Jn (β) Jn′ (α) − β Jn (α) Jn′ (β) x Jn (αx)Jn (βx) dx = β 2 − α2 0 = 0 ( since α and β are the roots of Jn (x) = 0 For the case with α = β, let us assume α is a root of Jn (x) = 0 and β is a variable approaching α, that is Z1 lt lt α Jn (β) Jn′ (α) x Jn (αx)Jn (βx) dx = β→α β→α β 2 − α2 0 lt α Jn′ (β) Jn′ (α) = by L’Hospital’s rule β→α 2β 1 ′ 2 = (Jn (α)) 2 1 ′ 2 = Jn+1 (α) ∵ Jn′ = nx Jn − Jn+1 & Jn (α) = 0 2 4.1.10 Observation Za a2 ′ 2 x Jn2 (αx) dx = Jn+1 (α a) 2 0 Lecture Notes MA2020 Differential Equations 67 4.1.11 Fourier-Bessel Expansion If f (x) is a continuous function having finite number of oscillations in an interval (0, a) then we can write ∞ X f (x) = ck Jn (αk x) (4.15) k=1 where α1 , α2 , · · · are the roots of Jn (x) = 0. Proof : Multiplying both sides with xJn (αr x) and integrating in the limits 0 to a gives Za Za x f (x) Jn (αr x) dx = cr x Jn2 (αr x) dx (due to orthogonality) 0 0 2 a 2 = cr J (aαr ) 2 n+1 Za 2 cr = 2 2 x f (x) Jn (αr x) dx, for r = 1, 2, 3, · · · a Jn+1 (αr x) 0 Note: 1. When f (x) and f ′ (x) have only finite number of jump discontinuities, the above series can be shown to converge however such an exercise is beyond the scope of this notes and hence is not included here. 2. At x = 1, the above series converges to 0 irrespective of the function f . Simi- larly, at x = 0, it converges to zero again for n > 0 and converges to f (0+) if n = 0. 4.1.12 Problem Expand f (x) = x2 in the interval 0 < x < 2 in terms of Jn (αn x) where αn are the roots of J2 (2αn ) = 0. Answer : ∞ X x2 = ck J2 (αk x) k=1 where Z2 2 x3 J3 (αn x) 2 2 2 4 cn = 2 x x J2 (αn x) dx = 2 = 2 J3 (2 αn ) 2 J3 (2 αn ) αn 0 αn J3 (2 αn ) 0 ∞ 2 X J2 (αk x) x = 4 αn J3 (2 αn ) k=1 68 Sanyasiraju V S S Yedida
[email protected] 4.1.13 Problem Expand 1 0 ≤ x < 21 1 f (x) = x = 12 2 1 0 2 <x≤1 in terms of J0 (αn x) where αn are the positive roots of J0 (x). Answer : ∞ X f (x) = ck J0 (αk x) k=1 then any coefficient cn , n = 1, 2, · · · can be computed using αn 3 2 Z2 2 x J3 (αn x) 2 t dt cn = = J0 (t) 22 J3 (2 αn ) αn 0 (J1 (αn ))2 αn αn 0 αn 2 2 1 d Z = 2 (t J1 (t)) dt (J1 (αn )) αn2 dt 0 2 1 αn 2 1 αn α n = 2 2 (t J1 (t))0 = 2 2 2 J1 (J1 (αn )) αn (J1 (αn )) αn 2 2 αn J1 2 = αn (J1 (αn ))2 J1 α2n X∞ f (x) = 2 J0 (αn x) k=1 α n (J 1 (α n )) Chapter 5 Sturm - Liouvelle Problems 5.1 Introduction A differential equation of the form d dy r(x) + (q(x) + λp(x))y = 0 (5.1) dx dx with continuous functions r(x), r ′(x), q(x), p(x)(> 0), ( p(x) is either positive or neg- ative but don’t change its sign in the interval I ) defined in an interval I = [a, b], is called a Sturm-Liouvelle problem. In addition if we have conditions dy K1 y(a) + K2 = 0 dx x=a dy L1 y(b) + L2 = 0 (5.2) dx x=b not both K1 & K2 are zero and similarly, not both L1 & L2 are zero, then the differential equation (5.1) along with the conditions (5.2) is called Sturm-Liouvelle boundary value problem (SLBVP). If r(a) = 0 or r(b) = 0 then such a problem is called a singular SLBVP else it is called a regular SLBVP. 5.2 Some Examples of the Sturm-Liouvelle Equa- tions d2 y 1. dx2 + λy = 0 2. Legendre equation d2 y dy (1 − x2 ) 2 − 2x + n(n + 1)y = 0 dx dx which can also be written as d 2 dy (1 − x ) + n(n + 1)y = 0 dx dx 69 70 Sanyasiraju V S S Yedida
[email protected] The eigenvalues in this case are λ = n(n + 1) and the eigenfunctions are the corresponding Legendre Polynomials. 3. Bessel’s equation d2 y dy t2 2 + t + (t2 − ν 2 )y = 0 dt dt It is equivalent to 2 d dy ν x + − + λx y = 0 (5.3) dx dx x after taking the transformation t = kx. The trivial solution y ≡ 0 is always a solution of any Sturm - Liouvelle problem (also SLBVP) however, the interest is to find some non-trivial (non-zero) solutions. Such non-trivial solutions exists for SLBVP, under some simple conditions, for some real values of λ, called eigenvalues of the problem, and the corresponding solutions are called eigenfucntions of the problem. 5.3 Orthogonality of the Eigenfunctions Consider the SL problem (5.1) in an interval x ∈ [a, b] with ym and yn are two eigenfucntions corresponding to the eigenvalues λm and λn (λm 6= λn ), respectively. Then ym and yn are orthogonal in the interval (a, b) with respect to the weight function p(x). Further for singular SLBVP, one of the boundary conditions of (5.2) ( condition at x = a if r(a) = 0 and similarly at x = b if r(b) = 0 ) is not required. Finally for r(a) = r(b), the boundary conditions can be made periodic, that is, y(a) = y(b) and y ′(a) = y ′(b). Proof : Let ym and yn be eigenfunctions of the equations d dym r(x) + (q(x) + λp(x))ym = 0 dx dx d dyn r(x) + (q(x) + λp(x))yn = 0 dx dx respectively. These equations can be rearranged to ′ (λm − λn )pym yn = ym (ryn′ )′ − yn (rym ′ ′ ) = (ym (ryn′ ) − yn (rym ′ )) Integrating on both sides with respect x in the interval a to b gives Zb b (λm − λn ) p ym yn dx = (r(ym yn′ − yn ym ′ ))a (5.4) a = r(b) (ym (b)yn′ (b) − yn (b)ym ′ (b)) − r(a) (ym (a)yn′ (a) − yn (a)ym ′ (a)) Lecture Notes MA2020 Differential Equations 71 1. Case (i) : r(a) = r(b) = 0. Then the right hand side of (5.4) is zero giving the required orthogonality condition for the eigenfunctions. 2. Case (ii) : r(a) = 0, r(b) 6= 0. (such a problem is called Singular Sturm- Liouvelle Porblem ) From the boundary conditions (5.2) at x = b one can write ′ L1 ym (b) + L2 ym (b) = 0 ′ L1 yn (b) + L2 yn (b) = 0 if the coefficients L1 and L2 are not zero then one can come up with either L1 (ym (b)yn′ (b) − yn (b)ym ′ (b)) = 0, or L2 (ym (b)yn′ (b) − yn (b)ym ′ (b)) = 0 since the coefficients are not zero gives (ym (b)yn (b) − yn (b)ym (b)) 0. Using this in (5.4) ′ ′ gives the required orthogonality condition for the eigenfunctions. (Proof is similar even if one of L1 and L2 is zero, anyway both can’t be zero from the statement of theorem) 3. Case (iii) : r(a) 6= 0, r(b) = 0. Proof is similar to case (ii). 4. Case (iv) : r(a) 6= 0, r(b) 6= 0. Use the proofs of both case (ii) and case (iii). 5. Case (v) : r(a) = r(b). (such a problem is called Periodic Sturm-Liouvelle Porblem ) Clearly, the right hand side of (5.4) is zero if we replace the condition (5.4) with periodic conditions as given (before that, the procedure adopted above can be used once again to prove the orthogonality). 5.3.1 Orthogonality of the Bessel Functions Since equation (5.3) is a (singular) Sturm-Liouvelle equation (it is singular because r(x) = x = 0 at the initial point x = 0), therefore the orthogonality condition is given over the solutions of the equation (5.3), that is over Jn (kx) for any integer n. The Bessel orthogonality condition is given by Za x Jn (ki x) Jn (kj x) dx = 0 0 αin αjn where ki = a and kj = a , αin , αjn are any distinct roots of Jn (kx) = 0. 5.3.2 Example Problems 1. Find a SL problem for which eigenfunctions are 1, cos x, cos 2x, · · · 2. Find a SL problem for which eigenfunctions are 1, cos mπx L , sin mπx L , m = 1, 2, · · · . ( y + λy = 0, y(−π) = y(π), y (−π) = y (π).) ′′ ′ ′ 3. Find the eigenvalues and eigenfucntions of the following SLBVP 72 Sanyasiraju V S S Yedida
[email protected] (a) y ′′ + λy = 0, y(0) = 0, y ′(l) = 0 (b) (xy ′ )′ + λx y = 0, y(1) = 0, y(s) = 0 4. Expand f (x) = x, 0 < x < 1, in terms of the Bessel functions J1 (ki1 x), i = 1, 2, 3, · · · where ki1 are roots of J1 (x) = 0, then prove that ∞ X J1 (ki1 x) x = , 0<x<1 i=1 k J (k i1 2 i1 ) 2 5.4 Eigenvalues of a Sturm-Liuovelle Problem are Real If p, q, r and r ′ are real valued continuous functions in the interval I = [a, b], and fur- ther, if p(x) doesn’t change its sign in I, then the eigenvalues of the Sturm-Liuovelle problem λ are real. Proof : Let us assume the eigenvalues are complex taking the form λ = a + i b and the corresponding eigenfunctions are y(x) = u(x) + i v(x). Substituting the y(x) in the SLP (5.1) and equating the real imaginary part to zero gives (r u′ )′ + (q + ap)u − bpv = 0 (r v ′ )′ + (q + ap)v + bpu = 0 Multiplying the first with v and the second with −u and adding gives −b(u2 + v 2 )p = u(r v ′ )′ − v(r u′ )′ = (u(r v ′ ) − v(r u′ )) ′ Integrating in the limits a ≤ x ≤ b gives the right hand side of the above equation is zero (proved in the orthogonality) but left side integral is not equal to zero as (u2 +v 2 ) is positive and p(x) doesn’t change its sign in a ≤ x ≤ b, therefore b = 0 proving the λ as a real quantity.