ODE Homework 6 5.2. Series Solutions Near an Ordinary Point, Part I 1. Seek power series solution of the equation y + k 2 x 2 y = 0, k a constant about the the point x 0 = 0. Find the recurrence relation; also find the first four terms in each of two solutions y 1 and y 2 (unless the series terminates sooner). By evaluating the Wronskian W (y 1, y 2 )(x 0 ), show that y 1 and y 2 form a fundamental set of solutions. If possible, find the general terms in each solutions. [ 5.2 #4] Sol. Let y = a n x n. Then y = n(n 1)a n x n 2 = (n + 2)(n + 1)a n+2 x n Substituting into the equation we have (n + 2)(n + 1)a n+2 x n + k 2 x 2 a n x n = 0 Before proceeding, write It follows that k 2 x 2 2a 2 + 6a x + a n x n = k 2 a n 2 x n [(n + 2)(n + 1)a n+2 + k 2 a n 2 ]x n = 0 We obtain a 2 = a = 0 and the recurrence relation k 2 a n+2 = (n + 2)(n + 1) a n 2, n = 2,, According to the recurrence relation we have that a 4m+2 = a 4m+ = 0, m = 0, 1, 2,, and a 4 = k2 4 a 0, a 8 = k2 7 8 a 4 = a 5 = k2 4 5 a 1, a 9 = k2 8 9 a 5 = k 4 4 7 8 a 0, k 4 4 5 8 9 a 1,
By setting a 0 = 1, a 1 = 0 and a 0 = 0, a 1 = 1 respectively, we get y 1 (x) = 1 k2 12 x4 + k4 672 x8 k6 88704 x12 + y 2 (x) = x k2 20 x5 + k4 1440 x9 k6 224640 x1 + It is easy to see that W (y 1, y 2 )(0) = 1 0 0 1 = 1, which shows that y 1, y 2 form a fundamental set of solutions. Moreover, so k 2 a 4m = 4m(4m 1) a 4m 4 = = k 2 a 4m+1 = 4m(4m + 1) a 4m = = y 1 (x) = 1 + y 2 (x) = x + m=1 m=1 In fact, y 1 and y 2 can be written as ( 1) m k 2m 4 7 8 (4m 1) 4m a 0 ( 1) m k 2m 4 7 8 (4m 1) 4m x4m ( 1) m k 2m 4 5 8 9 (4m )(4m + 1) a 1 ( 1) m k 2m 4 5 8 9 (4m )(4m + 1) x4m+1 ( 1) m k 2m Γ ( ) 4 y 1 (x) = 1+ m=1 2 4m m!γ ( )x 4m ( 1) m k 2m Γ ( ) 5 4, y 2 (x) = x+ 4m+ 4 m=1 2 4m m!γ ( 4m+5 4 2. Seek power series solution of the equation (1 + x 2 )y 4xy + 6y = 0 about the the point x 0 = 0. Find the recurrence relation; also find the first four terms in each of two solutions y 1 and y 2 (unless the series terminates sooner). By evaluating the Wronskian W (y 1, y 2 )(x 0 ), show that y 1 and y 2 form a fundamental set of solutions. If possible, find the general terms in each solutions. [ 5.2 #9] )x 4m+1
Sol. Let y = a n x n. Then y = y = na n x n 1 = (n + 1)a n+1 x n n(n 1)a n x n 2 = (n + 2)(n + 1)a n+2 x n Substituting into the equation we have (1+x 2 ) (n+2)(n+1)a n+2 x n 4x (n+1)a n+1 x n +6 a n x n = 0 Before proceeding, write x 2 x (n + 2)(n + 1)a n+2 x n = (n + 1)a n+1 x n = n(n 1)a n x n na n x n It follows that 2(a 0 +a 2 )+2(a 1 +a )x+ [(n+2)(n+1)a n+2 +n(n 1)a n 4na n +6a n ]x n = 0 We obtain a 2 = a 0, a = 1 a 1 and the recurrence relation a n+2 = n(n + 1) 4n + 6 (n + 2)(n + 1) (n 2)(n ) a n = (n + 1)(n + 2) a n, n = 2,, Observe that for n = 2, we get a 4 = a 5 = 0, and thus a n = 0, n 4. Hence the general solution for the equation is of the form y(x) = a 0 + a 1 x a 0 x 2 1 a 1x Letting a 0 = 1, a 1 = 0, and a 0 = 0, a 1 = 1 we get respectively y 1 (x) = 1 x 2, y 2 (x) = x 1 x The Wronskian W (y 1, y 2 )(x) = 1 x 2 x x 6x 1 x 2 = (x2 + 1) 2
Hence W (y 1, y 2 )(0) = 1 which shows that y 1, y 2 form a fundamental set of solutions.. Seek power series solution of the equation 2y + (x + 1)y + y = 0 about the the point x 0 = 2. Find the recurrence relation; also find the first four terms in each of two solutions y 1 and y 2 (unless the series terminates sooner). By evaluating the Wronskian W (y 1, y 2 )(x 0 ), show that y 1 and y 2 form a fundamental set of solutions. If possible, find the general terms in each solutions. [ 5.2 #14] Sol. Let y = a n (x 2) n. Then y = y = na n (x 2) n 1 = (n + 1)a n+1 (x 2) n n(n 1)a n (x 2) n 2 = (n + 2)(n + 1)a n+2 (x 2) n Substituting into the equation we have 2 (n+2)(n+1)a n+2 (x 2) n +(x+1) (n+1)a n+1 (x 2) n + a n (x 2) n = 0 Before proceeding, write (x + 1) (n + 1)a n+1 (x 2) n = (x 2 + ) = = (n + 1)a n+1 (x 2) n (n + 1)a n+1 (x 2) n+1 + (n + 1)a n+1 (x 2) n na n (x 2) n + (n + 1)a n+1 (x 2) n It follows that 4a 2 +a 1 +a 0 + [2(n+2)(n+1)a n+2 +na n +(n+1)a n+1 +a n ](x 2) n = 0
We obtain the recurrence relation a n+2 = 2(n + 2) a n + n+1 2(n + 2)(n + 1) a n, n = 0, 1, 2 According to the recurrence relation, we have a 2 = 4 a 1 4 a 0 a = 1 2 a 2 1 a 1 = 1 24 a 1 + 8 a 0 a 4 = 8 a 5 24 a 2 = 9 64 a 1 + 1 64 a 0. By setting a 0 = 1, a 1 = 0 and a 0 = 0, a 1 = 1 respectively, we get y 1 (x) = 1 4 (x 2)2 + 8 (x 2) + 1 64 (x 2)4 + y 2 (x) = (x 2) 4 (x 2)2 + 1 24 (x 2) + 9 64 (x 2)4 + It is easy to see that W (y 1, y 2 )(2) = 1 0 0 1 = 1, which shows that y 1, y 2 form a fundamental set of solutions. 4. (a) By making the change of variable x 1 = t and assuming that y has a Taylor series in powers of t, find two series solutions of y + (x 1) 2 y + (x 2 1)y = 0 in powers of x 1. (b) Show that you obtain the same result by assuming that y has a Taylor series in powers of x 1 and also expressing the coefficient x 2 1 in powers of x 1. [ 5.2 #19] Sol. (a) Let x 1 = t, then x = t + 1. Let z(t) = y(t + 1), then the equation can be transformed to z + t 2 z + (t 2 + 2t)z = 0
Let z = a n t n. Then z = z = na n t n 1 = (n + 1)a n+1 t n n(n 1)a n t n 2 = (n + 2)(n + 1)a n+2 t n Substituting into the equation we have (n+2)(n+1)a n+2 t n +t 2 Before proceeding, write t 2 (t 2 + 2t) (n + 1)a n+1 t n = a n t n = (n+1)a n+1 t n +(t 2 +2t) (n 1)a n 1 t n a n 2 t n + 2a n 1 t n a n t n = 0 It follows that 2a 2 +2(a 0 +a )t+ [(n+2)(n+1)a n+2 +(n 1)a n 1 +a n 2 +2a n 1 ]t n = 0 We obtain a 2 = 0, a = 1 a 0 and the recurrence relation a n+2 = 1 n + 2 a 1 n 1 (n + 2)(n + 1) a n 2, n = 2, According to the recurrence relation, we have a 4 = 1 4 a 1 1 12 a 0 a 5 = 1 5 a 2 1 20 a 1 = 1 20 a 1 a 6 = 1 6 a 1 0 a 2 = 1 18 a 0 a 7 = 1 7 a 4 1 42 a = 1 28 a 1 + 5 252 a 0.
By setting a 0 = 1, a 1 = 0 and a 0 = 0, a 1 = 1 respectively, we get z 1 (t) = 1 1 t 1 12 t4 + 1 18 t6 + 5 252 t7 + z 2 (t) = t 1 4 t4 1 20 t5 + 1 28 t7 + or y 1 (x) = 1 1 (x 1) 1 12 (x 1)4 + 1 18 (x 1)6 + 5 252 (x 1)7 + y 2 (x) = (x 1) 1 4 (x 1)4 1 20 (x 1)5 + 1 28 (x 1)7 + (b) Since the Taylor series for x 2 1 about x = 1 is x 2 1 = 2(x 1) + (x 1) 2 Then the equation becomes y + (x 1) 2 y + [2(x 1) + (x 1) 2 ]y = 0 which is identical to the transformed equation with t = x 1. 5. The Hermite Equation. The equation y 2xy + λy = 0, < x <, where λ is a constant, is known as the Hermite equation. It is an important equation in mathematical physics. (a) Find the first four terms in each of two solutions about x = 0 and show that they form a fundamental set of solutions. (b) Observe that if λ is a nonnegative even integer, then one or the other of the series solutions terminates and becomes a polynomial. Find the polynomial solutions for λ = 0, 2, 4, 6, 8, and 10. Note that each polynomial is determined only up to a multiplicative constant. (c) The Hermite polynomial H n (x) is defined as the polynomial solution of the Hermite equation with λ = 2n for which the coefficient of x n is 2 n. Find H 0 (x),, H 5 (x). [ 5.2 #21] Sol.
(a) Let y = a n x n. Then y = y = na n x n 1 = (n + 1)a n+1 x n n(n 1)a n x n 2 = (n + 2)(n + 1)a n+2 x n Substituting into the equation we have (n+2)(n+1)a n+2 x n 2x (n+1)a n+1 x n +λ a n x n = 0 Before proceeding, write It follows that 2a 2 + λa 0 + 2x (n + 1)a n+1 x n = 2na n x n [(n + 2)(n + 1)a n+2 2na n + λa n ]x n = 0 We obtain the recurrence relation Thus a n+2 = a = 2 1 λ 2 2n λ (n + 2)(n + 1) a n, n = 0, 1, 2, a 2 = λ 2 a 0, a 4 = 2 2 λ a 2 = 4 a 1, a 5 = 2 λ a = 4 5 More precisely, we have λ(λ 2 2) a 0, 2 4 (λ 2 1)(λ 2 ) a 1, 2 4 5 2(2m 2) λ a 2m = 2m(2m 1) a m λ(λ 4) (λ 4m + 4) 2m 2 = = ( 1) a 0 (2m)! 2(2m 1) λ a 2m+1 = 2m(2m + 1) a m (λ 2)(λ 6) (λ 4m + 2) 2m 1 = = ( 1) (2m + 1)! a 1
By setting a 0 = 1, a 1 = 0 and a 0 = 0, a 1 = 1 respectively, we get y 1 (x) = 1 λ 2! x2 + = 1 + λ(λ 4) x 4 4! λ(λ 4)(λ 8) x 6 + 6! m λ(λ 4) (λ 4m + 4) ( 1) (2m)! m=1 y 2 (x) = x λ 2 x (λ 2)(λ 6) + x 5 (λ 2)(λ 6)(λ 10) x 7 +! 5! 7! m (λ 2)(λ 6) (λ 4m + 2) = x + ( 1) x 2m+1 (2m + 1)! m=1 It is easy to see that W (y 1, y 2 )(0) = 1 0 0 1 = 1, which shows that y 1, y 2 form a fundamental set of solutions. (b) If λ = 2k with k = 0, 1, 2,,, then the recurrence relation becomes 2n 2k a n+2 = (n + 2)(n + 1) a n, n = 0, 1, 2, x 2m which implies that a k+2 = 0 and hence a k+2l = 0, l = 0, 1, 2,. More precisely, if k = 2j, then y 1 (x) becomes a polynomial with degree at most k and if k = 2j + 1, then y 2 (x) becomes a polynomial with degree at most k. In particular, we obtain the polynomial solutions corresponding to λ = 0, 2, 4, 6, 8, 10 are λ = 0 y 1 (x) = 1, λ = 2 y 2 (x) = x λ = 4 y 1 (x) = 1 2x 2, λ = 6 y 2 (x) = x 2 x λ = 8 y 1 (x) = 1 4x 2 + 4 x4 λ = 10 y 2 (x) = x 4 x + 4 15 x5 (c) If λ = 2n. By letting a 0 = a 1 = 1, we have that m 2n(2n 4) (2n 4m + 4) a 2m = = ( 1) (2m)! m (2n 2)(2n 6) (2n 4m + 2) a 2m+1 = = ( 1) (2m + 1)! for m = 1, 2,, n. It follows that the coefficient of 2 xn, in y 1 and y 2, is { ( 1) l 4 l l! for n = 2l (2l)! a n = ( 1) l 4l l! for n = 2l + 1 (2l+1)! a 0 a 1
Hence, we have that { 2 n ( 1) k l (2l)! 4 H n (x) = y l l! 1(x) = ( 1) k l (2l)! y l! 1 (x) for n = 2l 2 n ( 1) l (2l+1)! y 4 l l! 2 (x) = ( 1) k l 2 (2l+1)! y l! 2 (x) for n = 2l + 1 Therefore, the Hermite polynomials H 0 (x),, H 5 (x) are H 0 (x) = 1, H 2 (x) = 4x 2 2, H 4 (x) = 16x 4 48x 2 + 12, H 1 (x) = 2x H (x) = 8x 12x H 5 (x) = 2x 5 160x + 120x 5.. Series Solutions Near an Ordinary Point, Part II 6. The Chebyshev Equation. The Chebyshev differential equation is (1 x 2 )y xy + α 2 y = 0, where α is a constant. (a) Determine two solutions in powers of x for x < 1 and show that they form a fundamental set of solutions. (b) Show that if α is a nonnegative integer n, then there is a polynomial solution of degree n. These polynomials, when properly normalized, are called the Chebyshev polynomials. They are very useful in problems that require a polynomial approximation to a function defined on 1 x 1. (c) Find a polynomial solution for each of the cases α = n = 0, 1, 2,. [ 5. #10] Sol. (a) Let y = a n x n. Then y = y = na n x n 1 = (n + 1)a n+1 x n n(n 1)a n x n 2 = (n + 2)(n + 1)a n+2 x n Substituting into the equation we have (1 x 2 ) (n+2)(n+1)a n+2 x n x (n+1)a n+1 x n +α 2 a n x n = 0
It follows that (2a 2 +α 2 a 0 )+(6a +(α 2 1)a 1 )x+ [(n+2)(n+1)a n+2 +(α 2 n 2 )a n ]x n = 0 We obtain the recurrence relation n 2 α 2 a n+2 = (n + 2)(n + 1) a n, n = 0, 1, 2, According to the recurrence relation we have that a 2 = α2 2! a 0, a 4 = 22 α 2 4 a 2 = α2 (2 2 α 2 ) a 0, 4! a = 12 α 2 a 1, a 5 = 2 α 2! 5 4 a 5 = (2 α 2 )(1 2 α 2 ) a 1, 5! That is, a 2m = α2 (2 2 α 2 ) ((2m 2) 2 α 2 ) a 0, m = 1, 2, (2m)! a 2m+1 = (12 α 2 )( 2 α 2 ) ((2m 1) 2 α 2 ) a 1, m = 1, 2, (2m + 1)! Therefore we get two linearly independent solutions by setting (a 0, a 1 ) = (1, 0) and (a 0, a 1 ) = (0, 1) respectively α 2 (2 2 α 2 ) ((2m 2) 2 α 2 ) y 1 (x) = 1 x 2m (2m)! m=1 (1 2 α 2 )( 2 α 2 ) ((2m 1) 2 α 2 ) y 2 (x) = x + x 2m+1 (2m + 1)! m=1 (b) If α = n a nonnegative integer. For n = 2k, then k α 2 (2 2 α 2 ) ((2m 2) 2 α 2 ) y 1 (x) = P n (x) = 1 x 2m (2m)! m=1 is a polynomial with degree 2k = n. If n = 2k + 1, then k (1 2 α 2 )( 2 α 2 ) ((2m 1) 2 α 2 ) y 2 (x) = x+ x 2m+1 (2m + 1)! m=1 is a polynomial with degree 2k + 1 = n. (c) The polynomial solutions corresponding to α = 0, 1, 2, are α = 0 P 0 (x) = 1, α = 1 P 1 (x) = x α = 2 P 2 (x) = 1 2x 2, α = P (x) = x 4 x
7. The Legendre Equation. The Legendre equation is (1 x 2 )y 2xy + α(α + 1)y = 0. The point x = 0 is an ordinary point of the equation, and the distance from the origin to the nearest zero of P (x) = 1 x 2 is 1. Hence the radius of convergence of series solutions about x = 0 is at leat 1. Also notice that we need to consider only α > 1 because if α 1, then the substituting α = (1 + γ), where γ 0, leads to the Legendre equation (1 x 2 )y 2xy + γ(γ + 1)y = 0. Show that two solutions of the Legendre equation for x < 1 are α(α + 1) y 1 (x) = 1 x 2 α(α 2)(α + 1)(α + ) + x 4 2! 4! m α (α 2m + 2)(α + 1) (α + 2m 1) + ( 1) (2m)! m= (α 1)(α + 2) y 2 (x) = x x (α 1)(α )(α + 2)(α + 4) + x 5! 5! m (α 1) (α 2m + 1)(α + 2) (α + 2m) + ( 1) (2m + 1)! [ 5. #22] m= Proof. Let y = a n x n. Then y = y = na n x n 1 = (n + 1)a n+1 x n n(n 1)a n x n 2 = (n + 2)(n + 1)a n+2 x n x 2m x 2m+1 Substituting into the equation we have (1 x 2 ) (n+2)(n+1)a n+2 x n 2x (n+1)a n+1 x n +α(α+1) a n x n = 0 It follows that (2a 2 + α(α + 1)a 0 ) + (6a + (α 1)(α + 2)a 1 )x + [(n + 2)(n + 1)a n+2 + (α n)(α + n + 1)a n ]x n = 0
We obtain the recurrence relation a n+2 = (α n)(α + n + 1) a n, n = 0, 1, 2, (n + 2)(n + 1) According to the recurrence relation we have that α(α + 1) α(α 2)(α + 1)(α + ) a 2 = a 0, a 4 = a 0, 2! 4! (α 1)(α + 2) (α 1)(α )(α + 2)(α + 4) a = a 1, a 5 = a 1,! 5! That is, m α (α 2m + 2)(a + 1) (α + 2m 1) a 2m = ( 1) a 0, m = 1, 2, (2m)! m (α 1) (α 2m + 1)(a + 2) (α + 2m) a 2m+1 = ( 1) a 1, m = 1, 2, (2m + 1)! Therefore we get two linearly independent solutions by setting (a 0, a 1 ) = (1, 0) and (a 0, a 1 ) = (0, 1) respectively α(α + 1) y 1 (x) = 1 x 2 α(α 2)(α + 1)(α + ) + x 4 2! 4! m α (α 2m + 2)(α + 1) (α + 2m 1) + ( 1) (2m)! m= (α 1)(α + 2) y 2 (x) = x x (α 1)(α )(α + 2)(α + 4) + x 5! 5! m (α 1) (α 2m + 1)(α + 2) (α + 2m) + ( 1) (2m + 1)! m= x 2m x 2m+1 5.4. Euler Equations; Regular Singular Points 8. Find all singular points of the equation x 2 (1 x 2 )y + 2 x y + 4y = 0 and determine whether each one is regular or irregular. [ 5.4 #19] Sol. Rewrite the equation as the form y + 2 4 x (1 x 2 ) y + x 2 (1 x 2 ) y = 0
2 and let p(x) =, q(x) = 4. The singular points x (1 x 2 ) x 2 (1 x 2 ) are x = 0, ±1. At x = 0 lim xp(x) = 2 x 0 x 2 (1 x 2 ) which does not exist, so x = 0 is an irregular singular point. At x = 1, 2 lim (x + 1)p(x) = lim x 1 x 1 x (x 1) = 1 lim (x + x 1 1)2 q(x) = lim x 1 4(x + 1) x 2 (x 1) = 0 so x = 1 is an regular singular point. At x = 1, 2 lim(x 1)p(x) = lim x 1 x 1 x (x + 1) = 1 lim (x 4(x 1) x 1 1)2 q(x) = lim x 1 x 2 (x + 1) = 0 so x = 1 is an regular singular point. 9. Find all singular points of the equation x 2 y + xy + (x 2 ν 2 )y = 0 Bessel equation and determine whether each one is regular or irregular. [ 5.4 #22] Sol. Rewrite the equation as the form y + 1 x y + x2 ν 2 x 2 y = 0 and let p(x) = 1 x, q(x) = x2 ν 2 x 2. The only singular point is x = 0. At x = 0, lim xp(x) = lim 1 = 1 x 0 x 0 lim x 0 x2 q(x) = lim(x 2 ν 2 ) = ν 2 x 0 so x = 0 is an regular singular point. 10. Find all singular points of the equation y + ln x y + xy = 0 and determine whether each one is regular or irregular. [ 5.4 #29] Sol. Since ln x is singular at x = 0, so the only singular point is x = 0. Furthermore, since ln x is not analytic at x = 0, so
does the function x ln x. Hence x = 0 is an irregular singular point. 11. Find all singular points of the equation x 2 y + 2(e x 1)y + (e x cos x)y = 0 and determine whether each one is regular or irregular. [ 5.4 #0] Sol. Rewrite the equation as the form y + 2(ex 1) y + e x cos x y = 0 x 2 x 2 and let p(x) = 2(ex 1), q(x) = e x cos x. The only singular point x 2 x 2 is x = 0. At x = 0, 2(e x 1) lim xp(x) = lim x 0 x 0 x lim x 0 x2 q(x) = lim e x cos x = 1 x 0 so x = 0 is an regular singular point. 12. Find all singular points of the equation 2e x = lim x 0 1 = 2 x 2 y ( sin x)y + (1 + x 2 )y = 0 and determine whether each one is regular or irregular. [ 5.4 #2] Sol. Rewrite the equation as the form y sin x y + 1 + x2 y = 0 x 2 x 2 and let p(x) = sin x, q(x) = 1+x2. The only singular point is x 2 x 2 x = 0. At x = 0, sin x cos x lim xp(x) = lim = lim = x 0 x 0 x x 0 1 lim x 0 x2 q(x) = lim(1 + x 2 ) = 1 x 0 so x = 0 is an regular singular point. 1. Consider the Euler equation x 2 y + αxy + βy = 0. Find conditions on α and β so that: (a) All solutions approach zero as x 0. (b) All solutions are bounded as x 0. (c) All solutions approach zero as x. (d) All solutions are bounded as x. (e) All solutions are bounded both as x 0 and as x.
[ 5.4 #9] Sol. Note that the general solution of the equation in any interval containing the origin is determined by the roots r 1 and r 2 of the equation F (r) = r(ρ 1) + αr + β = r 2 + (α 1)r + β that is, r 1 + r 2 = 1 α, r 1 r 2 = β, more precisely And r 1, r 2 = (α 1) ± (α 1) 2 4β 2 c 1 x r 1 + c 2 x r 2, if r 1, r 2 R, r 1 r 2 y(x) = c 1 x r 1 + c 2 x r 1 ln x, if r 1, r 2 R, r 1 = r 2 c 1 x λ cos ( µ ln x ) + c 2 x λ sin ( µ ln x ), if r 1, r 2 = λ ± iµ where λ = α 1 2, µ = 4β (α 1)2. (a) For solutions to approach zero as x 0, we need that r 1, r 2 > 0 α < 1, β > 0, if r 1, r 2 R, r 1 r 2 r 1 = r 2 > 0 α < 1, β > 0, if r 1, r 2 R, r 1 = r 2 λ > 0 α < 1, if r 1, r 2 = λ ± iµ Thus, if α < 1 and β > 0, the all solutions approach zero as x 0. (b) For solutions are bounded as x 0, we need that r 1, r 2 0 α < 1, β 0, if r 1, r 2 R, r 1 r 2 r 1 = r 2 > 0 α < 1, β > 0, if r 1, r 2 R, r 1 = r 2 λ > 0 or λ = 0, µ 0 α < 1 or α = 1, β > 0, if r 1, r 2 = λ ± iµ Thus, if α < 1 and β 0 or α = 1 and β > 0, the all solutions are bounded as x 0. (c) For solutions to approach zero as x, we need that r 1, r 2 < 0 α > 1, β > 0, if r 1, r 2 R, r 1 r 2 r 1 = r 2 < 0 α > 1, β > 0, if r 1, r 2 R, r 1 = r 2 λ < 0 α > 1, if r 1, r 2 = λ ± iµ Thus, if α > 1 and β > 0, the all solutions approach zero as x. (d) For solutions are bounded as x, we need that r 1, r 2 0 α > 1, β 0, if r 1, r 2 R, r 1 r 2 r 1 = r 2 < 0 α > 1, β > 0, if r 1, r 2 R, r 1 = r 2 λ < 0 or λ = 0, µ 0 α > 1 or α = 1, β > 0, if r 1, r 2 = λ ± iµ Thus, if α > 1 and β 0 or α = 1 and β > 0, the all solutions are bounded x.
(e) For solutions are bounded both as x 0 and as x, we need that impossible, if r 1, r 2 R, r 1 r 2 impossible, if r 1, r 2 R, r 1 = r 2 λ = 0, µ 0 α = 1, β > 0, if r 1, r 2 = λ ± iµ Thus, if α = 1 and β > 0 and, the all solutions are bounded both as x 0 and as x. 14. Using the method of reduction of order, show that if r 1 is a repeated root of r(r 1) + αr + β = 0, then x r 1 and x r 1 ln x are solutions of x 2 y + αxy + βy = 0 for x > 0. [ 5.4 #40] Proof. Assume that y = v(x)x r 1, then y = x r 1 v +r 1 x r 1 1 v, y = x r 1 v +2r 1 x r 1 1 v +r 1 (r 1 1)x r 1 2 v Substituting into the equation, we get x r 1+2 v + [2r 1 + α]x r 1+1 v + [r 1 (r 1 1) + αr 1 + β]x r 1 v = 0 If r 1 is a repeated root of r(r 1) + αr + β = 0, then r 1 (r 1 1) + αr 1 + β = 0, 2r 1 + α 1 = 0 Hence v satisfies the equation x r 1+2 v + x r 1+1 v = x r 1+1 [xv + v ] = 0 and this immediately implies that v(x) = ln x, since x > 0. Therefore the second solution for the Euler equation is x r 1 ln x. 15. Show that the point x = 0 is a regular singular point of the equation 2x 2 y + xy (1 + x)y = 0 Try to find solutions of the form a n x n. Show that (except for constant multiplies) there are no nonzero solutions of this form in our problem. Thus, in this case, the general solution can not be found in this manner. This is typical of equations with singular points. [ 5.4 #42]
Proof. Rewrite the equation as the form y + 2x y 1 + x 2x y = 0 2 and let p(x) = 1+x, q(x) =. The only singular point is 2x 2x 2 x = 0. At x = 0, lim xp(x) = lim x 0 x 0 2 = 2 1 + x lim x 0 x2 q(x) = lim = 1 x 0 2 2 so x = 0 is a regular singular point. Let y = a n x n. Substituting into the equation we have 2x 2 (n+2)(n+1)a n+2 x n +x (n+1)a n+1 x n (1+x) a n x n = 0 It follows that a 0 + (2a 1 a 0 )x + [2n(n 1)a n + na n a n a n 1 ]x n = 0 We obtain a 0 = 0, a 1 = 1 2 a 0 = 0 and the recurrence relation a n = 1 (2n 1)(n + 1) a n 1, n = 2,, Thus we can conclude that a n = 0, n = 0, 1, 2,. Hence y(x) = 0 is the only solution that can be obtained. 5.5. Series Solutions Near a Regular Singular Point, Part I 16. (a) Show that the equation ( x 2 y + xy + x 2 1 ) y = 0 9 has a regular singular point at x = 0. (b) Determine the indicial equation, the recurrence relation, and the roots of the indicial equation. (c) Find the series solution (x > 0) corresponding to the largest root. (d) If the roots are unequal and do not differ by an integer, find the series solution corresponding to the smaller root also. [ 5.5 #2] Sol.
(a) Rewrite the equation as the form y + 1 x y + 9x2 1 9x 2 y = 0 and let p(x) = 1, q(x) = 9x2 1. The only singular point is x 9x 2 x = 0. At x = 0, lim xp(x) = lim 1 = 1 x 0 x 0 9x 2 1 lim x 0 x2 q(x) = lim x 0 9 so x = 0 is a regular singular point. (b) Let y = a n x n+r. Then y = (n+r)a n x n+r 1, y = = 1 9 (n+r)(n+r 1)a n x n+r 2 Substituting into the equation we have ( (n+r)(n+r 1)a n x n+r + (n+r)a n x n+r + x 2 1 ) a n x n+r = 0 9 It follows that ( r(r 1) + r 1 ) ( a 0 x r + (r + 1)r + (r + 1) 1 ) a 1 x r+1 9 9 [( + (n + r)(n + r 1) + (n + r) 1 ] )a n + a n 2 x n+r = 0 9 We obtain the indicial equation r 2 1 9 = 0 with roots r = ± 1. For either value of r it is necessary to take a 1 = 0 in order that the coefficient of x r+1 be zero. Hence we get the recurrence relation 1 a n+2 = (n + r) 2 1 a n 2, n = 2,, 9 (c) For r = 1, we have 1 a n+2 = ( ) n + 1 2 a n 2 = 1 9 a n 2 n ( n + 2 ), n = 2,,
According to the recurrence relation and a 1 = 0, we have that a 2m+1 = 0, m = 0, 1, 2,. For n = 2m, we have that a 2m 2 a 2m 4 a 2m = 2 2 m ( ) = m + 1 2 4 m(m 1) ( ) m + )( 1 m 1 + 2 ( 1) m a 0 = = 2 2m m! ( ( ) m + ) 1 1 + 1 By setting a 0 = 1, we get ( y 1 (x) = x 1 ( 1) m ) 1 + 2 2m m! ( ( )x m + ) 2m 1 1 + 1 = x 1 ( 1 + m=1 ( 1) m Γ ( ) 4 ) 2 2m m!γ ( )x 2m m+4 m=1 (d) Since r 2 = 1 r 1 = 1 and r 1 r 2 = 2 is not an integer, we can calculate a second series solution corresponding to r = 1. For r = 1, we have 1 a n+2 = ( ) n 1 2 a n 2 = a n 2 1 n ( ), n = 2,, n 2 9 According to the recurrence relation and a 1 = 0, we have that a 2m+1 = 0, m = 0, 1, 2,. For n = 2m, we have that a 2m 2 a 2m 4 a 2m = 2 2 m ( ) = m 1 2 4 m(m 1) ( ) m )( 1 m 1 2 ( 1) m a 0 = = 2 2m m! ( ( ) m ) 1 1 1 By setting a 0 = 1, we get ( y 1 (x) = x 1 ( 1) m ) 1 + 2 2m m! ( ( )x m ) 2m 1 1 1 = x 1 ( 1 + m=1 m=1 17. The Legendre equation of order α is ( 1) m Γ ( ) 2 ) 2 2m m!γ ( )x 2m m+2 (1 x 2 )y 2xy + α(α + 1)y = 0. It was shown that x = ±1 are regular singular points. (a) Determine the indicial equation and its roots for the point x = 1.
(b) Find a series solution in powers of x 1 for x 1 > 0. Hint: Write 1 + x = 2 + (x 1) and x = 1 + (x 1). Alternatively, make the change of variable x 1 = t and determine a series solutions in powers of t. [ 5.5 #11] Sol. (a) By letting x 1 = t and let u(t) = y(t + 1), the Legendre equation can be transformed to (t 2 + 2t)u (t) + 2(t + 1)u (t) α(α + 1)u(t) = 0 Note that 2(t + 1) lim t t 0 t(t + 2) = lim 2(t + 1) t 0 t + 2 lim t 0 t 2 α(α + 1) t(t + 2) = lim t 0 = 1 α(α + 1)t t + 2 = 0 so t = 0 is a regular singular point for above equation. Let u = a n t n+r. Then u = (n+r)a n t n+r 1, u = (n+r)(n+r 1)a n t n+r 2 Substituting into the equation we have (n + r)(n + r 1)a n t n+r + 2(n + r)(n + r 1)a n t n+r 1 + 2(n + r)a n t n+r + It follows that 2(n + r)a n t n+r 1 α(α + 1)a n x n+r = 0 [2r(r 1) + 2r]a 0 t r 1 ( ) + 2(n + r + 1) 2 a n+1 + [(n + r)(n + r + 1) α(α + 1)]a n t n+r = 0 We obtain the indicial equation 2r 2 = 0 with repeated roots r = 0. (b) Since r = 0, the recurrence relation becomes a n+1 = α(α + 1) n(n + 1) 2(n + 1) 2 a n, n = 0, 1, 2,
According to the recurrence relation, we have that α(α + 1) (n 1)n a n = a 2n 2 n 1 [α(α + 1) (n 1)n][α(α + 1) (n 2)(n 1)] = 2 2( n(n 1) ) 2 a n 2 = = [α(α + 1) n(n 1)] [α(α + 1) 2 1] α(α + 1) 2 n( n! ) 2 a 0 By setting a 0 = 0 and replacing t = x 1, we get the series solution for the Legendre equation is y 1 (x) = [α(α + 1) n(n 1)] [α(α + 1) 2 1] α(α + 1) 2 n( n! ) 2 (x 1) n Note that since r = 0 is repeated root for the indicial equation, there will be only one series solution of the form y = a n (x 1) n+r. 18. The Chebyshev equation of order α is (1 x 2 )y xy + α 2 y = 0, where α is a constant. (a) Show that x = 1 and x = 1 are regular singular points and find the exponents at each of these singularities. (b) Find two solutions about x = 1. [ 5.5 #12] Sol. (a) Rewrite the equation of the form y x 1 x 2 y + α2 1 x 2 y = 0 and let p(x) = x, q(x) = α2. There are two singular 1 x 2 1 x 2 points which are x = ±1. At x = 1, p 0 = lim x 1 (x 1)p(x) = lim x 1 x x + 1 = 1 2 q 0 = lim x 1 (x 1) 2 q(x) = lim x 1 α(x 1) (x + 1) = 0
At x = 1, p 0 = lim (x + 1)p(x) = lim x 1 x 1 q 0 = lim x 1 (x + 1)2 q(x) = lim x 1 x x 1 = 1 2 α(x + 1) (x 1) = 0 Hence x = ±1 are regular singular points. Note that the indicial equation is given by ( r(r 1) + p 0 r + q 0 = r r 1 ) = 0 2 with distinct real roots r = 0, r = 1 2. (b) Set t = x 1 and u(t) = y(t + 1), the equation can be transformed as (t 2 + 2t)u + (t + 1)u α 2 u = 0 By a. we know that t = 0 is a regular singular point for above equation. Let u = a n t n+r. Substituting into the equation we have (n + r)(n + r 1)a n t n+r + + (n + r)a n t n+r + 2(n + r)(n + r 1)a n t n+r 1 (n + r)a n t n+r 1 That is, ( (n+r) 2 a n t n+r + 2(n+r) n+r 1 ) a n t n+r 1 2 It follows that [ ( 2r r 1 2 + )] a 0 t r 1 α 2 a n x n+r = 0 α 2 a n t n+r = 0 [ ( 2(n + r + 1) n + r + 1 ) ] a n+1 + [(n + r) 2 α 2 ]a n t n+r = 0 2 we obtain r(2r 1)a 0 = 0 and the recurrence relation a n+1 = α 2 (n + r) 2 2(n + r + 1) ( n + r + 1 2 )a n, n = 0, 1, 2,
Assuming that a 0 0, then for r 1 = 1, according to the 2 recurrence relation, a n = 4α2 (2n 1) 2 4n(2n + 1) a n 1 = [4α2 (2n 1) 2 ][4α 2 (2n ) 2 ] a 4 2 n 2 n(n 1)(2n + 1)(2n 1) = = [4α2 (2n 1) 2 ][4α 2 (2n ) 2 ] [4α 2 1 2 ] a 4 n 0 n!(2n + 1)(2n 1) 5 = [4α2 (2n 1) 2 ][4α 2 (2n ) 2 ] [4α 2 1 2 ] 4 n n! (2n+1)! 2 n n! = [4α2 (2n 1) 2 ][4α 2 (2n ) 2 ] [4α 2 1 2 ], n = 1, 2, 2 n (2n + 1)! On the other hand, for r = 0, according the recurrence relation, a n = α2 (n 1) 2 n(2n 1) a n 1 = [α2 (n 1) 2 ][α 2 (n 2) 2 ] a n 2 n(n 1)(2n 1)(2n ) = = [α2 (n 1) 2 ][α 2 (n 2) 2 ] [α 2 1 2 ] α 2 a 0 n!(2n 1)(2n ) 1 = [α2 (n 1) 2 ][α 2 (n 2) 2 ] [α 2 1 2 ] α 2 (2n 1)! n! 2 n 1 (n 1)! = 2n 1 [α 2 (n 1) 2 ][α 2 (n 2) 2 ] [α 2 1 2 ] α 2, n = 1, 2 n(2n 1)! By letting a 0 = 1 and replacing t = x 1, we get two linearly independent solutions of the equation y 1 (x) = ( x 1 1 + y 2 (x) = 1 + [4α 2 (2n 1) 2 ][4α 2 (2n ) 2 ] [4α 2 1 2 ] ) (x 1) n 2 n (2n + 1)! 2 n 1 [α 2 (n 1) 2 ][α 2 (n 2) 2 ] [α 2 1 2 ] α 2 (x 1) n n(2n 1)! 19. The Bessel equation of order zero is x 2 y + xy + x 2 y = 0. (a) Show that x = 0 is a regular singular point. (b) Show that the roots of the indicial equation are r 1 = r 2 = 0.
(c) Show that one solution for x > 0 is ( 1) n x 2n J 0 (x) = 1 + 2 2n (n!). 2 (d) Show that the series for J 0 (x) converges for all x. The function J 0 is known as the Bessel function of the first kind of order zero. [ 5.5 #14] Proof. (a) Rewrite the equation of the form y + 1 x y + y = 0 and let p(x) = 1, q(x) = 1. The only singular points is x x = 0. At x = 0, p 0 = lim xp(x) = lim 1 = 1 x 0 x 0 q 0 = lim x 2 q(x) = lim x 2 = 0 x 0 x 1 Hence x = 0 is a regular singular point. (b) Let y = a n x n+r. Then y = (n+r)a n x n+r 1, y = (n+r)(n+r 1)a n x n+r 2 Substituting into the equation we have (n+r)(n+r 1)a n x n+r + (n+r)a n x n+r + a n x n+r+2 = 0 It follows that [r(r 1)+r]a 0 x r +[r(r+1)+(r+1)]a 1 x r+1 + [(n+r) 2 a n +a n 2 ]x n+r = 0 The indicial equation is r 2 = 0 with repeated real roots r 1 = r 2 = 0. (c) For r = 0, it is necessary to take a 1 = 0 in order that the coefficient of x r+1 be zero. Note that the recurrence relation is of the form a n = a n 2, n = 2,, n2
The fact that a 1 = 0 implies that a 2m+1 = 0, m = 0, 1, 2,. And a 2m = a 2m 2 2 2 m 2 = a 2m 4 2 4 m 2 (m 1) 2 = = ( 1)m a 0 2 2m (m!) 2 By setting a 0 = 1, we obtain one solution for x > 0 of the equation ( 1) n x 2n J 0 (x) = 1 + 2 2n (n!) 2 (d) Let a n = ( 1)n, n = 1, 2,. Then 2 2n (n!) 2 lim a ( 1) n+1 n+1 = lim 2 2n+2 ((n+1)!) 2 n a n n = lim ( 1) n 2 2n (n!) 2 n 1 4(n + 1) 2 = 0 which shows that the radius of convergence of J 0 (x) is infinite, that is, J 0 (x) converges for all x. 20. the Bessel equation of order one is x 2 y + xy + (x 2 1)y = 0. (a) Show that x = 0 is a regular singular point. (b) Show that the roots of the indicial equation are r 1 = 1 and r 2 = 1. (c) Show that one solution for x > 0 is J 1 (x) = x 2 ( 1) n x 2n (n + 1)!n!2 2n. (d) Show that the series for J 1 (x) converges for all x. The function J 1 is known as the Bessel function of the first kind of order one. (e) Show that it is impossible to determine a second solution of the form [ 5.5 #16] Proof. x 1 b n x n, x > 0.
(a) Rewrite the equation of the form y + 1 x y + x2 1 x 2 y = 0 and let p(x) = 1, q(x) = x2 1. The only singular points is x x 2 x = 0. At x = 0, p 0 = lim xp(x) = lim 1 = 1 x 0 x 0 q 0 = lim x 2 q(x) = lim(x 2 1) = 1 x 0 x 1 Hence x = 0 is a regular singular point. (b) Let y = a n x n+r. Then y = (n+r)a n x n+r 1, y = (n+r)(n+r 1)a n x n+r 2 Substituting into the equation we have (n+r)(n+r 1)a n x n+r + (n+r)a n x n+r + a n x n+r+2 a n x n+r = 0 [(n + r) 2 1]a n x n+r + a n 2 x n+r = 0 It follows that ( ) (r 2 1)a 0 x r +[(r+1) 2 1]a 1 x r+1 + [(n+r) 2 1]a n +a n 2 x n+r = 0 The indicial equation is r 2 1 = 0 with distinct real roots r 1 = 1, r 2 = 1. (c) For either r = ±1, it is necessary to take a 1 = 0 in order that the coefficient of x r+1 be zero. Note that the recurrence relation is of the form a n 2 a n =, n = 2,, (n + r) 2 1 The fact that a 1 = 0 implies that a 2m+1 = 0, m = 0, 1, 2,. If r = 1, then a 2m 2 a 2m = (2m + 1) 2 1 = a 2m 2 4m(m + 1) = a 2m 4 4 2 m 2 (m + 1)(m 1) a 2m 6 = 4 m(m + 1) (m 1)m (m 2)(m 1) = = ( 1)m a 0 4 m m!(m + 1)!
By setting a 0 = 1, we obtain one solution for x > 0 of the 2 equation J 1 (x) = x 2 (d) Let a n = lim a n+1 = lim n a n n ( 1) n x 2n (n + 1)!n!2 2n. ( 1)n., n = 0, 1,. Then (n+1)!n!2 2n ( 1) n+1 (n+2)!(n+1)!2 2n+2 ( 1) n (n+1)!n!2 2n = lim n 1 4(n + 2)(n + 1) = 0 which shows that the radius of convergence of J 0 (x) is infinite, that is, J 1 (x) converges for all x. (e) For r = 1 the recurrence relation becomes a n 2 a n =, n = 2,, (n 1) 2 1 If n = 2 the coefficient of a 2 zero and we cannot calculate a 2. Consequently it is not possible to find a series solution of the form x 1 b n x n.