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ODE Homework 6 5.2. Series Solutions Near an Ordinary Point, Part I 1. Seek power series solution of the equation y 00 + k 2 x2 y = 0, k a constant about the the point x0 = 0. Find the recurrence relation; also find the first four terms in each of two solutions y1 and y2 (unless the series terminates sooner). By evaluating the Wronskian W (y1 , y2 )(x0 ), show that y1 and y2 form a fundamental set of solutions. If possible, find the general terms in each solutions. [§5.2 #4] ∞ P Sol. Let y = an xn . Then n=0 y 00 = ∞ X n(n − 1)an xn−2 = n=2 ∞ X (n + 2)(n + 1)an+2 xn n=0 Substituting into the equation we have ∞ ∞ X X n 2 2 (n + 2)(n + 1)an+2 x + k x an x n = 0 n=0 n=0 Before proceeding, write 2 2 k x ∞ X n an x = n=0 ∞ X k 2 an−2 xn n=2 It follows that ∞ X 2a2 + 6a3 x + [(n + 2)(n + 1)an+2 + k 2 an−2 ]xn = 0 n=2 We obtain a2 = a3 = 0 and the recurrence relation an+2 = − k2 an−2 , n = 2, 3, · · · (n + 2)(n + 1) According to the recurrence relation we have that a4m+2 = a4m+3 = 0, m = 0, 1, 2, · · · , and k2 k2 k4 a0 , a8 = − a4 = a0 , · · · 3·4 7·8 3·4·7·8 k2 k2 k4 a5 = − a1 , a9 = − a5 = a1 , · · · 4·5 8·9 4·5·8·9 a4 = − By setting a0 = 1, a1 = 0 and a0 = 0, a1 = 1 respectively, we get k2 4 k4 8 k 6 12 x + x − x + ··· 12 672 88704 k2 5 k4 9 k6 y2 (x) = x − x + x − x13 + · · · 20 1440 224640 1 0 = 1, which shows It is easy to see that W (y1 , y2 )(0) = 0 1 that y1 , y2 form a fundamental set of solutions. Moreover, y1 (x) = 1 − k2 (−1)m k 2m a4m−4 = · · · = a0 4m(4m − 1) 3 · 4 · 7 · 8 · · · (4m − 1) · 4m k2 (−1)m k 2m =− a4m−3 = · · · = a1 4m(4m + 1) 4 · 5 · 8 · 9 · · · (4m − 3)(4m + 1) a4m = − a4m+1 so ∞ X (−1)m k 2m y1 (x) = 1 + x4m 3 · 4 · 7 · 8 · · · (4m − 1) · 4m m=1 y2 (x) = x + ∞ X (−1)m k 2m x4m+1 4 · 5 · 8 · 9 · · · (4m − 3)(4m + 1) m=1 In fact, y1 and y2 can be written as ∞ ∞ X X (−1)m k 2m Γ 54 4m+1 (−1)m k 2m Γ 34 4m x , y2 (x) = x+ x y1 (x) = 1+ 4m m!Γ 4m+3 4m m!Γ 4m+5 2 m=1 2 m=1 4 4 2. Seek power series solution of the equation (1 + x2 )y 00 − 4xy 0 + 6y = 0 about the the point x0 = 0. Find the recurrence relation; also find the first four terms in each of two solutions y1 and y2 (unless the series terminates sooner). By evaluating the Wronskian W (y1 , y2 )(x0 ), show that y1 and y2 form a fundamental set of solutions. If possible, find the general terms in each solutions. [§5.2 #9] Sol. Let y = ∞ P an xn . Then n=0 y0 = 00 y = ∞ X nan xn−1 = n=1 ∞ X ∞ X (n + 1)an+1 xn n=0 n(n − 1)an x n−2 n=2 = ∞ X (n + 2)(n + 1)an+2 xn n=0 Substituting into the equation we have 2 (1+x ) ∞ X n (n+2)(n+1)an+2 x −4x n=0 ∞ X n (n+1)an+1 x +6 n=0 ∞ X an x n = 0 n=0 Before proceeding, write x 2 ∞ X n (n + 2)(n + 1)an+2 x = n=0 ∞ X n(n − 1)an xn n=2 ∞ ∞ X X nan xn (n + 1)an+1 xn = x n=1 n=0 It follows that ∞ X 2(3a0 +a2 )+2(a1 +3a3 )x+ [(n+2)(n+1)an+2 +n(n−1)an −4nan +6an ]xn = 0 n=2 We obtain a2 = −3a0 , a3 = − 13 a1 and the recurrence relation an+2 = − n(n + 1) − 4n + 6 (n − 2)(n − 3) an = − an , n = 2, 3, · · · (n + 2)(n + 1) (n + 1)(n + 2) Observe that for n = 2, 3 we get a4 = a5 = 0, and thus an = 0, ∀ n ≥ 4. Hence the general solution for the equation is of the form 1 y(x) = a0 + a1 x − 3a0 x2 − a1 x3 3 Letting a0 = 1, a1 = 0, and a0 = 0, a1 = 1 we get respectively y1 (x) = 1 − 3x2 , 1 y2 (x) = x − x3 3 The Wronskian 3 1 − 3x2 x − x3 W (y1 , y2 )(x) = −6x 1 − x2 = (x2 + 1)2 Hence W (y1 , y2 )(0) = 1 which shows that y1 , y2 form a fundamental set of solutions. 3. Seek power series solution of the equation 2y 00 + (x + 1)y 0 + 3y = 0 about the the point x0 = 2. Find the recurrence relation; also find the first four terms in each of two solutions y1 and y2 (unless the series terminates sooner). By evaluating the Wronskian W (y1 , y2 )(x0 ), show that y1 and y2 form a fundamental set of solutions. If possible, find the general terms in each solutions. [§5.2 #14] ∞ P Sol. Let y = an (x − 2)n . Then n=0 0 y = y 00 = ∞ X n−1 nan (x − 2) n=1 ∞ X ∞ X = (n + 1)an+1 (x − 2)n n=0 n(n − 1)an (x − 2)n−2 = ∞ X (n + 2)(n + 1)an+2 (x − 2)n n=0 n=2 Substituting into the equation we have ∞ ∞ ∞ X X X an (x−2)n = 0 (n+1)an+1 (x−2)n +3 (n+2)(n+1)an+2 (x−2)n +(x+1) 2 n=0 n=0 n=0 Before proceeding, write ∞ X (x + 1) (n + 1)an+1 (x − 2)n n=0 ∞ X = (x − 2 + 3) (n + 1)an+1 (x − 2)n n=0 ∞ ∞ X X n+1 = (n + 1)an+1 (x − 2) +3 (n + 1)an+1 (x − 2)n = n=0 ∞ X n=0 nan (x − 2)n + 3 n=1 ∞ X (n + 1)an+1 (x − 2)n n=0 It follows that ∞ X 4a2 +3a1 +3a0 + [2(n+2)(n+1)an+2 +nan +3(n+1)an+1 +3an ](x−2)n = 0 n=1 We obtain the recurrence relation an+2 = − 3 n+3 an+1 − an , n = 0, 1, 2 · · · 2(n + 2) 2(n + 2)(n + 1) According to the recurrence relation, we have 3 a2 = − a1 − 4 1 a3 = − a2 − 2 3 a4 = − a3 − 8 .. . 3 a0 4 1 1 3 a1 = a1 + a0 3 24 8 5 9 1 a2 = a1 + a0 24 64 64 By setting a0 = 1, a1 = 0 and a0 = 0, a1 = 1 respectively, we get 3 1 3 y1 (x) = 1 − (x − 2)2 + (x − 2)3 + (x − 2)4 + · · · 4 8 64 3 1 9 2 y2 (x) = (x − 2) − (x − 2) + (x − 2)3 + (x − 2)4 + · · · 4 24 64 1 0 = 1, which shows It is easy to see that W (y1 , y2 )(2) = 0 1 that y1 , y2 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 00 + (x − 1)2 y 0 + (x2 − 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 x2 − 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 00 + t2 z 0 + (t2 + 2t)z = 0 Let z = ∞ P an tn . Then n=0 0 ∞ X 00 n=1 ∞ X z = z = n−1 nan t ∞ X = (n + 1)an+1 tn n=0 n−2 n(n − 1)an t = n=2 ∞ X (n + 2)(n + 1)an+2 tn n=0 Substituting into the equation we have ∞ ∞ ∞ X X X (n+1)an+1 tn +(t2 +2t) an tn = 0 (n+2)(n+1)an+2 tn +t2 n=0 n=0 n=0 Before proceeding, write t 2 ∞ X n (n + 1)an+1 t = (t + 2t) (n − 1)an−1 tn n=2 n=0 2 ∞ X ∞ X n an t = n=0 ∞ X n=2 n an−2 t + ∞ X 2an−1 tn n=1 It follows that ∞ X 2a2 +2(a0 +3a3 )t+ [(n+2)(n+1)an+2 +(n−1)an−1 +an−2 +2an−1 ]tn = 0 n=2 We obtain a2 = 0, a3 = − 13 a0 and the recurrence relation an+2 = − 1 1 an−1 − an−2 , n = 2, 3 · · · n+2 (n + 2)(n + 1) According to the recurrence relation, we have 1 a4 = − a1 − 4 1 a5 = − a2 − 5 1 a6 = − a3 − 6 1 a7 = − a4 − 7 .. . 1 a0 12 1 1 a1 = − a1 20 20 1 1 a2 = a0 30 18 1 1 5 a3 = a1 + a0 42 28 252 By setting a0 = 1, a1 = 0 and a0 = 0, a1 = 1 respectively, we get 1 1 1 5 7 z1 (t) = 1 − t3 − t4 + t6 + t + ··· 3 12 18 252 1 1 1 z2 (t) = t − t4 − t5 + t7 + · · · 4 20 28 or 1 1 1 5 y1 (x) = 1 − (x − 1)3 − (x − 1)4 + (x − 1)6 + (x − 1)7 + · · · 3 12 18 252 1 1 1 y2 (x) = (x − 1) − (x − 1)4 − (x − 1)5 + (x − 1)7 + · · · 4 20 28 (b) Since the Taylor series for x2 − 1 about x = 1 is x2 − 1 = 2(x − 1) + (x − 1)2 Then the equation becomes y 00 + (x − 1)2 y 0 + [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 00 − 2xy 0 + 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 Hn (x) is defined as the polynomial solution of the Hermite equation with λ = 2n for which the coefficient of xn is 2n . Find H0 (x), · · · , H5 (x). [§5.2 #21] Sol. (a) Let y = ∞ P an xn . Then n=0 0 y = y 00 = ∞ X nan x n=1 ∞ X n−1 = ∞ X (n + 1)an+1 xn n=0 n(n − 1)an xn−2 = n=2 ∞ X (n + 2)(n + 1)an+2 xn n=0 Substituting into the equation we have ∞ ∞ ∞ X X X n n (n+2)(n+1)an+2 x −2x (n+1)an+1 x +λ an x n = 0 n=0 n=0 n=0 Before proceeding, write 2x ∞ X n (n + 1)an+1 x = n=0 ∞ X 2nan xn n=1 It follows that ∞ X [(n + 2)(n + 1)an+2 − 2nan + λan ]xn = 0 2a2 + λa0 + n=1 We obtain the recurrence relation an+2 = 2n − λ an , n = 0, 1, 2, · · · (n + 2)(n + 1) Thus λ 2·2−λ λ(λ − 2 · 2) a2 = − a0 , a 4 = a2 = a0 , · · · 2 3·4 2·3·4 2·1−λ 2·3−λ (λ − 2 · 1)(λ − 2 · 3) a3 = a1 , a 5 = a3 = a1 , · · · 2·3 4·5 2·3·4·5 More precisely, we have a2m = a2m+1 2(2m − 2) − λ λ(λ − 4) · · · (λ − 4m + 4) a2m−2 = · · · = (−1)m a0 2m(2m − 1) (2m)! 2(2m − 1) − λ (λ − 2)(λ − 6) · · · (λ − 4m + 2) = a2m−1 = · · · = (−1)m a1 2m(2m + 1) (2m + 1)! By setting a0 = 1, a1 = 0 and a0 = 0, a1 = 1 respectively, we get λ 2 λ(λ − 4) 4 λ(λ − 4)(λ − 8) 6 x + x − x + ··· 2! 4! 6! ∞ X λ(λ − 4) · · · (λ − 4m + 4) 2m =1+ (−1)m x (2m)! m=1 y1 (x) = 1 − λ − 2 3 (λ − 2)(λ − 6) 5 (λ − 2)(λ − 6)(λ − 10) 7 x + x − x + ··· 3! 5! 7! ∞ X (λ − 2)(λ − 6) · · · (λ − 4m + 2) 2m+1 x =x+ (−1)m (2m + 1)! m=1 1 0 = 1, which It is easy to see that W (y1 , y2 )(0) = 0 1 shows that y1 , y2 form a fundamental set of solutions. (b) If λ = 2k with k = 0, 1, 2, · · · ,, then the recurrence relation becomes 2n − 2k an+2 = an , n = 0, 1, 2, · · · (n + 2)(n + 1) y2 (x) = x − which implies that ak+2 = 0 and hence ak+2` = 0, ` = 0, 1, 2, · · · . More precisely, if k = 2j, then y1 (x) becomes a polynomial with degree at most k and if k = 2j + 1, then y2 (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 ⇒ y1 (x) = 1, λ = 2 ⇒ y2 (x) = x λ = 4 ⇒ y1 (x) = 1 − 2x2 , λ = 6 ⇒ y2 (x) = x − 32 x3 4 4 2 λ = 8 ⇒ y1 (x) = 1 − 4x + 3 x λ = 10 ⇒ y2 (x) = x − 43 x3 + (c) If λ = 2n. By letting a0 = a1 = 1, we have that 2n(2n − 4) · · · (2n − 4m + 4) a0 (2m)! (2n − 2)(2n − 6) · · · (2n − 4m + 2) = · · · = (−1)m a1 (2m + 1)! a2m = · · · = (−1)m a2m+1 for m = 1, 2, · · · , b n2 c. It follows that the coefficient of xn , in y1 and y2 , is ( 4` `! (−1)` (2`)! for n = 2` an = 4` `! (−1)` (2`+1)! for n = 2` + 1 4 5 x 15 Hence, we have that ( 2n · (−1)k ` (2`)! y1 (x) for n = 2` y (x) = (−1)k ` (2`)! `! 4` `! 1 Hn (x) = y2 (x) for n = 2` + 1 2n · (−1)` (2`+1)! y2 (x) = (−1)k ` 2·(2`+1)! `! 4` `! Therefore, the Hermite polynomials H0 (x), · · · , H5 (x) are H0 (x) = 1, H2 (x) = 4x2 − 2, H4 (x) = 16x4 − 48x2 + 12, H1 (x) = 2x H3 (x) = 8x3 − 12x H5 (x) = 32x5 − 160x3 + 120x 5.3. Series Solutions Near an Ordinary Point, Part II 6. The Chebyshev Equation. The Chebyshev differential equation is (1 − x2 )y 00 − xy 0 + α2 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, 3. [§5.3 #10] Sol. ∞ P (a) Let y = an xn . Then n=0 0 ∞ X 00 n=1 ∞ X y = y = nan x n−1 ∞ X = (n + 1)an+1 xn n=0 n(n − 1)an x n=2 n−2 = ∞ X (n + 2)(n + 1)an+2 xn n=0 Substituting into the equation we have 2 (1−x ) ∞ X n=0 n (n+2)(n+1)an+2 x −x ∞ X n=0 n (n+1)an+1 x +α 2 ∞ X n=0 an x n = 0 It follows that ∞ X (2a2 +α a0 )+(6a3 +(α −1)a1 )x+ [(n+2)(n+1)an+2 +(α2 −n2 )an ]xn = 0 2 2 n=2 We obtain the recurrence relation n2 − α2 an+2 = an , n = 0, 1, 2, · · · (n + 2)(n + 1) According to the recurrence relation we have that α2 22 − α2 α2 (22 − α2 ) a0 , a 4 = a2 = − a0 , · · · 2! 4·3 4! 32 − α2 (32 − α2 )(12 − α2 ) 12 − α2 a1 , a 5 = a5 = a1 , · · · a3 = 3! 5·4 5! That is, a2 = − α2 (22 − α2 ) · · · ((2m − 2)2 − α2 ) a2m = − a0 , m = 1, 2, · · · (2m)! (12 − α2 )(32 − α2 ) · · · ((2m − 1)2 − α2 ) a2m+1 = a1 , m = 1, 2, · · · (2m + 1)! Therefore we get two linearly independent solutions by setting (a0 , a1 ) = (1, 0) and (a0 , a1 ) = (0, 1) respectively y1 (x) = 1 − ∞ X α2 (22 − α2 ) · · · ((2m − 2)2 − α2 ) 2m x (2m)! m=1 ∞ X (12 − α2 )(32 − α2 ) · · · ((2m − 1)2 − α2 ) 2m+1 y2 (x) = x + x (2m + 1)! m=1 (b) If α = n a nonnegative integer. For n = 2k, then k X α2 (22 − α2 ) · · · ((2m − 2)2 − α2 ) 2m y1 (x) = Pn (x) = 1− x (2m)! m=1 is a polynomial with degree 2k = n. If n = 2k + 1, then k X (12 − α2 )(32 − α2 ) · · · ((2m − 1)2 − α2 ) 2m+1 x y2 (x) = x+ (2m + 1)! m=1 is a polynomial with degree 2k + 1 = n. (c) The polynomial solutions corresponding to α = 0, 1, 2, 3 are α = 0 ⇒ P0 (x) = 1, α = 1 ⇒ P1 (x) = x 2 α = 2 ⇒ P2 (x) = 1 − 2x , α = 3 ⇒ P3 (x) = x − 43 x3 7. The Legendre Equation. The Legendre equation is (1 − x2 )y 00 − 2xy 0 + 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 − x2 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 − x2 )y 00 − 2xy 0 + γ(γ + 1)y = 0. Show that two solutions of the Legendre equation for |x| < 1 are α(α + 1) 2 α(α − 2)(α + 1)(α + 3) 4 y1 (x) = 1 − x + x 2! 4! ∞ X α · · · (α − 2m + 2)(α + 1) · · · (α + 2m − 1) 2m (−1)m + x (2m)! m=3 (α − 1)(α + 2) 3 (α − 1)(α − 3)(α + 2)(α + 4) 5 x + x 3! 5! ∞ X (α − 1) · · · (α − 2m + 1)(α + 2) · · · (α + 2m) 2m+1 + (−1)m x (2m + 1)! m=3 y2 (x) = x − [§5.3 #22] ∞ P Proof. Let y = an xn . Then n=0 y0 = 00 y = ∞ X nan xn−1 = n=1 ∞ X ∞ X (n + 1)an+1 xn n=0 n(n − 1)an x n−2 n=2 = ∞ X (n + 2)(n + 1)an+2 xn n=0 Substituting into the equation we have ∞ ∞ ∞ X X X (1−x2 ) (n+2)(n+1)an+2 xn −2x (n+1)an+1 xn +α(α+1) an x n = 0 n=0 n=0 n=0 It follows that (2a2 + α(α + 1)a0 ) + (6a3 + (α − 1)(α + 2)a1 )x ∞ X + [(n + 2)(n + 1)an+2 + (α − n)(α + n + 1)an ]xn = 0 n=2 We obtain the recurrence relation (α − n)(α + n + 1) an+2 = − an , n = 0, 1, 2, · · · (n + 2)(n + 1) According to the recurrence relation we have that α(α − 2)(α + 1)(α + 3) α(α + 1) a0 , a 4 = − a0 , · · · 2! 4! (α − 1)(α + 2) (α − 1)(α − 3)(α + 2)(α + 4) a3 = − a1 , a 5 = a1 , · · · 3! 5! That is, a2 = − α · · · (α − 2m + 2)(a + 1) · · · (α + 2m − 1) a0 , m = 1, 2, · · · (2m)! (α − 1) · · · (α − 2m + 1)(a + 2) · · · (α + 2m) a1 , m = 1, 2, · · · = (−1)m (2m + 1)! a2m = (−1)m a2m+1 Therefore we get two linearly independent solutions by setting (a0 , a1 ) = (1, 0) and (a0 , a1 ) = (0, 1) respectively α(α + 1) 2 α(α − 2)(α + 1)(α + 3) 4 x + x 2! 4! ∞ X α · · · (α − 2m + 2)(α + 1) · · · (α + 2m − 1) 2m + (−1)m x (2m)! m=3 y1 (x) = 1 − (α − 1)(α + 2) 3 (α − 1)(α − 3)(α + 2)(α + 4) 5 x + x 3! 5! ∞ X (α − 1) · · · (α − 2m + 1)(α + 2) · · · (α + 2m) 2m+1 (−1)m x + (2m + 1)! m=3 y2 (x) = x − 5.4. Euler Equations; Regular Singular Points 8. Find all singular points of the equation 2 x2 (1 − x2 )y 00 + y 0 + 4y = 0 x and determine whether each one is regular or irregular. [§5.4 #19] Sol. Rewrite the equation as the form y 00 + x3 (1 2 4 y0 + 2 y=0 2 −x ) x (1 − x2 ) 2 and let p(x) = x3 (1−x 2 ) , q(x) = are x = 0, ±1. At x = 0 4 . x2 (1−x2 ) The singular points 2 x→0 − x2 ) which does not exist, so x = 0 is an irregular singular point. At x = −1, −2 = −1 lim (x + 1)p(x) = lim 3 x→−1 x→−1 x (x − 1) −4(x + 1) lim (x + 1)2 q(x) = lim 2 =0 x→−1 x→−1 x (x − 1) so x = −1 is an regular singular point. At x = 1, −2 = −1 lim (x − 1)p(x) = lim 3 x→1 x→−1 x (x + 1) 4(x − 1) lim (x − 1)2 q(x) = lim 2 =0 x→1 x→−1 x (x + 1) so x = 1 is an regular singular point. lim xp(x) = x2 (1 9. Find all singular points of the equation x2 y 00 + xy 0 + (x2 − 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 1 x2 − ν 2 y=0 y 00 + y 0 + x x2 and let p(x) = x1 , q(x) = x = 0. At x = 0, x2 −ν 2 . x2 The only singular point is lim xp(x) = lim 1 = 1 x→0 x→0 2 lim x q(x) = lim (x2 − ν 2 ) = −ν 2 x→0 x→0 so x = 0 is an regular singular point. 10. Find all singular points of the equation y 00 + ln |x|y 0 + 3xy = 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 x2 y 00 + 2(ex − 1)y 0 + (e−x cos x)y = 0 and determine whether each one is regular or irregular. [§5.4 #30] Sol. Rewrite the equation as the form 2(ex − 1) 0 e−x cos x y 00 + y + y=0 x2 x2 x −x x . The only singular point and let p(x) = 2(ex2−1) , q(x) = e xcos 2 is x = 0. At x = 0, 2(ex − 1) 2ex lim xp(x) = lim = lim =2 x→0 x→0 x→0 1 x lim x2 q(x) = lim e−x cos x = 1 x→0 x→0 so x = 0 is an regular singular point. 12. Find all singular points of the equation x2 y 00 − (3 sin x)y 0 + (1 + x2 )y = 0 and determine whether each one is regular or irregular. [§5.4 #32] Sol. Rewrite the equation as the form 3 sin x 0 1 + x2 y + y=0 x2 x2 2 x , q(x) = 1+x . The only singular point is and let p(x) = − 3 sin x2 x2 x = 0. At x = 0, 3 sin x 3 cos x lim xp(x) = − lim = − lim = −3 x→0 x→0 x→0 x 1 lim x2 q(x) = lim (1 + x2 ) = 1 y 00 − x→0 x→0 so x = 0 is an regular singular point. 13. Consider the Euler equation x2 y 00 + αxy 0 + 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 #39] Sol. Note that the general solution of the equation in any interval containing the origin is determined by the roots r1 and r2 of the equation F (r) = r(ρ − 1) + αr + β = r2 + (α − 1)r + β that is, r1 + r2 = 1 − α, r1 r2 = β, more precisely p −(α − 1) ± (α − 1)2 − 4β r1 , r2 = 2 And if r1 , r2 ∈ R, r1 6= r2 c1 |x|r1 + c2 |x|r2 , r1 r1 c |x| + c |x| ln |x|, y(x) = 1 2 if r1 , r2 ∈ R, r1 = r2 λ λ c1 |x| cos µ ln |x| + c2 |x| sin µ ln |x| , if r1 , r2 = λ ± iµ , µ = 4β − (α − 1)2 . where λ = − α−1 2 (a) For solutions to approach zero as x → 0, we need that r1 , r2 > 0 ⇒ α < 1, β > 0, if r1 , r2 ∈ R, r1 6= r2 r1 = r2 > 0 ⇒ α < 1, β > 0, if r1 , r2 ∈ R, r1 = r2 λ > 0 ⇒ α < 1, if r1 , r2 = 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 r1 , r2 ≥ 0 ⇒ α < 1, β ≥ 0, if r1 , r2 ∈ R, r1 6= r2 r1 = r2 > 0 ⇒ α < 1, β > 0, if r1 , r2 ∈ R, r1 = r2 λ > 0 or λ = 0, µ 6= 0 ⇒ α < 1 or α = 1, β > 0, if r1 , r2 = 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 r1 , r2 < 0 ⇒ α > 1, β > 0, if r1 , r2 ∈ R, r1 6= r2 r1 = r2 < 0 ⇒ α > 1, β > 0, if r1 , r2 ∈ R, r1 = r2 λ < 0 ⇒ α > 1, if r1 , r2 = 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 r1 , r2 ≤ 0 ⇒ α > 1, β ≥ 0, if r1 , r2 ∈ R, r1 6= r2 r1 = r2 < 0 ⇒ α > 1, β > 0, if r1 , r2 ∈ R, r1 = r2 λ < 0 or λ = 0, µ 6= 0 ⇒ α > 1 or α = 1, β > 0, if r1 , r2 = 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 r1 , r2 ∈ R, r1 6= r2 impossible, if r1 , r2 ∈ R, r1 = r2 λ = 0, µ 6= 0 ⇒ α = 1, β > 0, if r1 , r2 = 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 r1 is a repeated root of r(r − 1) + αr + β = 0, then xr1 and xr1 ln x are solutions of x2 y 00 + αxy 0 + βy = 0 for x > 0. [§5.4 #40] Proof. Assume that y = v(x)xr1 , then y 0 = xr1 v 0 +r1 xr1 −1 v, y 00 = xr1 v 00 +2r1 xr1 −1 v 0 +r1 (r1 −1)xr1 −2 v Substituting into the equation, we get xr1 +2 v 00 + [2r1 + α]xr1 +1 v 0 + [r1 (r1 − 1) + αr1 + β]xr1 v = 0 If r1 is a repeated root of r(r − 1) + αr + β = 0, then r1 (r1 − 1) + αr1 + β = 0, 2r1 + α − 1 = 0 Hence v satisfies the equation xr1 +2 v 00 + xr1 +1 v 0 = xr1 +1 [xv 00 + v 0 ] = 0 and this immediately implies that v(x) = ln x, since x > 0. Therefore the second solution for the Euler equation is xr1 ln x. 15. Show that the point x = 0 is a regular singular point of the equation 2x2 y 00 + 3xy 0 − (1 + x)y = 0 ∞ P Try to find solutions of the form an xn . Show that (except n=0 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 1+x 3 y 00 + y 0 − y=0 2x 2x2 3 , q(x) = − 1+x . The only singular point is and let p(x) = 2x 2x2 x = 0. At x = 0, 3 3 lim xp(x) = lim = x→0 x→0 2 2 1 1 +x =− lim x2 q(x) = − lim x→0 x→0 2 2 ∞ P so x = 0 is a regular singular point. Let y = an xn . Substin=0 tuting into the equation we have ∞ ∞ ∞ X X X n 2 n an x n = 0 (n+1)an+1 x −(1+x) (n+2)(n+1)an+2 x +3x 2x n=0 n=0 n=0 It follows that −a0 + (2a1 − a0 )x + ∞ X [2n(n − 1)an + 3nan − an − an−1 ]xn = 0 n=2 We obtain a0 = 0, a1 = 12 a0 = 0 and the recurrence relation 1 an−1 , n = 2, 3, · · · (2n − 1)(n + 1) Thus we can conclude that an = 0, ∀ n = 0, 1, 2, · · · . Hence y(x) = 0 is the only solution that can be obtained. an = 5.5. Series Solutions Near a Regular Singular Point, Part I 16. (a) Show that the equation 1 x2 y 00 + xy 0 + x2 − 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 1 9x2 − 1 y 00 + y 0 + y=0 x 9x2 9x2 −1 . 9x2 and let p(x) = x1 , q(x) = x = 0. At x = 0, The only singular point is lim xp(x) = lim 1 = 1 x→0 x→0 9x2 − 1 1 =− x→0 x→0 9 9 so x = 0 is a regular singular point. ∞ P (b) Let y = an xn+r . Then lim x2 q(x) = lim n=0 0 y = ∞ X (n+r)an x n+r−1 ∞ X (n+r)(n+r−1)an xn+r−2 ,y = 00 n=2 n=0 Substituting into the equation we have ∞ ∞ ∞ X X X n+r n+r 2 1 (n+r)(n+r−1)an x + (n+r)an x + x − an xn+r = 0 9 n=0 n=0 n=0 It follows that 1 r+1 1 r a0 x + (r + 1)r + (r + 1) − a1 x r(r − 1) + r − 9 9 ∞ h i X 1 + (n + r)(n + r − 1) + (n + r) − an + an−2 xn+r = 0 9 n=2 We obtain the indicial equation r2 − 1 =0 9 with roots r = ± 13 . For either value of r it is necessary to take a1 = 0 in order that the coefficient of xr+1 be zero. Hence we get the recurrence relation an+2 = − 1 an−2 , n = 2, 3, · · · (n + r)2 − 19 (c) For r = 13 , we have an+2 = − n+ 1 1 2 3 − 1 9 an−2 = −an−2 , n = 2, 3, · · · n n + 23 According to the recurrence relation and a1 = 0, we have that a2m+1 = 0, m = 0, 1, 2, · · · . For n = 2m, we have that −a2m−2 a2m−4 = a2m = 2 1 4 2 m m+ 3 2 m(m − 1) m + 13 m − 1 + 23 (−1)m a0 = · · · = 2m 2 m! m + 13 · · · 1 + 31 By setting a0 = 1, we get ∞ X 1 y1 (x) = x 3 1 + (−1)m 2m x 22m m! m + 13 · · · 1 + 13 m=1 ∞ X (−1)m Γ 34 1 2m = x3 1 + x 22m m!Γ 3m+4 3 m=1 (d) Since r2 = − 13 6= r1 = 13 and r1 − r2 = 32 is not an integer, we can calculate a second series solution corresponding to r = − 31 . For r = − 31 , we have an+2 = − 1 n− 1 2 3 − 1 9 an−2 = −an−2 , n = 2, 3, · · · n n − 23 According to the recurrence relation and a1 = 0, we have that a2m+1 = 0, m = 0, 1, 2, · · · . For n = 2m, we have that −a2m−2 a2m−4 = a2m = 2 1 4 2 m m− 3 2 m(m − 1) m − 13 m − 1 − 23 (−1)m a0 = · · · = 2m 2 m! m − 13 · · · 1 − 13 By setting a0 = 1, we get ∞ X 1 y1 (x) = x 3 1 + (−1)m 2m x 22m m! m − 13 · · · 1 − 13 m=1 ∞ X (−1)m Γ 23 − 13 2m =x 1+ x 22m m!Γ 3m+2 3 m=1 17. The Legendre equation of order α is (1 − x2 )y 00 − 2xy 0 + 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 (t2 + 2t)u00 (t) + 2(t + 1)u0 (t) − α(α + 1)u(t) = 0 Note that 2(t + 1) 2(t + 1) = lim =1 t→0 t→0 t + 2 t(t + 2) α(α + 1)t α(α + 1) = − lim =0 − lim t2 · t→0 t→0 t(t + 2) t+2 lim t · so t = 0 is a regular singular point for above equation. Let ∞ P u= an tn+r . Then n=0 u0 = ∞ ∞ X X (n+r)(n+r−1)an tn+r−2 (n+r)an tn+r−1 , u00 = n=2 n=0 Substituting into the equation we have ∞ ∞ X X (n + r)(n + r − 1)an tn+r + 2(n + r)(n + r − 1)an tn+r−1 n=0 n=0 + ∞ X n+r 2(n + r)an t n=0 + ∞ X n+r−1 2(n + r)an t − n=0 ∞ X α(α + 1)an xn+r = 0 n=0 It follows that [2r(r − 1) + 2r]a0 tr−1 ∞ X + 2(n + r + 1)2 an+1 + [(n + r)(n + r + 1) − α(α + 1)]an tn+r = 0 n=0 We obtain the indicial equation 2r2 = 0 with repeated roots r = 0. (b) Since r = 0, the recurrence relation becomes an+1 = α(α + 1) − n(n + 1) an , n = 0, 1, 2, · · · 2(n + 1)2 According to the recurrence relation, we have that α(α + 1) − (n − 1)n an−1 2n2 [α(α + 1) − (n − 1)n][α(α + 1) − (n − 2)(n − 1)] an−2 = 2 22 n(n − 1) [α(α + 1) − n(n − 1)] · · · [α(α + 1) − 2 · 1] · α(α + 1) = ··· = a0 2 2n n! an = By setting a0 = 0 and replacing t = x − 1, we get the series solution for the Legendre equation is ∞ X [α(α + 1) − n(n − 1)] · · · [α(α + 1) − 2 · 1] · α(α + 1) y1 (x) = (x−1)n 2 n 2 n! n=0 Note that since r = 0 is repeated root for the indicial equation, there will be only one series solution of the form ∞ P y= an (x − 1)n+r . n=0 18. The Chebyshev equation of order α is (1 − x2 )y 00 − xy 0 + α2 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 00 − α2 x 0 y + y=0 1 − x2 1 − x2 2 x α and let p(x) = − 1−x 2 , q(x) = 1−x2 . There are two singular points which are x = ±1. At x = 1, x 1 = x→1 x→1 x + 1 2 α(x − 1) q0 = lim (x − 1)2 q(x) = lim =0 x→1 x→1 (x + 1) p0 = lim (x − 1)p(x) = lim At x = −1, 1 x = x→−1 x→−1 x − 1 2 α(x + 1) q0 = lim (x + 1)2 q(x) = lim =0 x→−1 x→−1 (x − 1) p0 = lim (x + 1)p(x) = lim Hence x = ±1 are regular singular points. Note that the indicial equation is given by 1 =0 r(r − 1) + p0 r + q0 = r r − 2 with distinct real roots r = 0, r = 12 . (b) Set t = x − 1 and u(t) = y(t + 1), the equation can be transformed as (t2 + 2t)u00 + (t + 1)u0 − α2 u = 0 By a. we know that t = 0 is a regular singular point for ∞ P above equation. Let u = an tn+r . Substituting into the n=0 equation we have ∞ X n+r (n + r)(n + r − 1)an t + ∞ X 2(n + r)(n + r − 1)an tn+r−1 n=0 n=0 + ∞ X n=0 n+r (n + r)an t + ∞ X n=0 n+r−1 (n + r)an t − ∞ X α2 an xn+r = 0 n=0 That is, ∞ ∞ ∞ X X 1 n+r−1 X 2 n+r 2 n+r (n+r) an t + 2(n+r) n+r− an t − α an t =0 2 n=0 n=0 n=0 It follows that h 1 i r−1 a0 t 2r r − 2 ∞ X 1 + 2(n + r + 1) n + r + an+1 + [(n + r)2 − α2 ]an tn+r = 0 2 n=0 we obtain r(2r − 1)a0 = 0 and the recurrence relation an+1 = α2 − (n + r)2 an , n = 0, 1, 2, · · · 2(n + r + 1) n + r + 21 Assuming that a0 6= 0, then for r1 = 12 , according to the recurrence relation, 4α2 − (2n − 1)2 [4α2 − (2n − 1)2 ][4α2 − (2n − 3)2 ] an−1 = an−2 4n(2n + 1) 42 n(n − 1)(2n + 1)(2n − 1) [4α2 − (2n − 1)2 ][4α2 − (2n − 3)2 ] · · · [4α2 − 12 ] a0 = ··· = 4n n!(2n + 1)(2n − 1) · · · 5 · 3 [4α2 − (2n − 1)2 ][4α2 − (2n − 3)2 ] · · · [4α2 − 12 ] = 4n n! (2n+1)! 2n n! an = = [4α2 − (2n − 1)2 ][4α2 − (2n − 3)2 ] · · · [4α2 − 12 ] , n = 1, 2, · · · 2n (2n + 1)! On the other hand, for r = 0, according the recurrence relation, α2 − (n − 1)2 [α2 − (n − 1)2 ][α2 − (n − 2)2 ] an−1 = an−2 n(2n − 1) n(n − 1)(2n − 1)(2n − 3) [α2 − (n − 1)2 ][α2 − (n − 2)2 ] · · · [α2 − 12 ] · α2 = ··· = a0 n!(2n − 1)(2n − 3) · · · 3 · 1 [α2 − (n − 1)2 ][α2 − (n − 2)2 ] · · · [α2 − 12 ] · α2 = (2n−1)! n! 2n−1 (n−1)! an = = 2n−1 [α2 − (n − 1)2 ][α2 − (n − 2)2 ] · · · [α2 − 12 ] · α2 , n = 1, 2 · · · n(2n − 1)! By letting a0 = 1 and replacing t = x − 1, we get two linearly independent solutions of the equation y1 (x) = √ ∞ X [4α2 − (2n − 1)2 ][4α2 − (2n − 3)2 ] · · · [4α2 − 12 ] n x−1 1+ (x − 1) 2n (2n + 1)! n=1 y2 (x) = 1 + ∞ X 2n−1 [α2 − (n − 1)2 ][α2 − (n − 2)2 ] · · · [α2 − 12 ] · α2 n=1 n(2n − 1)! 19. The Bessel equation of order zero is x2 y 00 + xy 0 + x2 y = 0. (a) Show that x = 0 is a regular singular point. (b) Show that the roots of the indicial equation are r1 = r2 = 0. (x − 1)n (c) Show that one solution for x > 0 is J0 (x) = 1 + ∞ X (−1)n x2n n=1 22n (n!)2 . (d) Show that the series for J0 (x) converges for all x. The function J0 is known as the Bessel function of the first kind of order zero. [§5.5 #14] Proof. (a) Rewrite the equation of the form 1 y 00 + y 0 + y = 0 x and let p(x) = x1 , q(x) = 1. The only singular points is x = 0. At x = 0, p0 = lim xp(x) = lim 1 = 1 x→0 x→0 2 q0 = lim x q(x) = lim x2 = 0 x→0 x→1 Hence x = 0 is a regular singular point. ∞ P (b) Let y = an xn+r . Then n=0 0 y = ∞ X (n+r)an x n=0 n+r−1 ∞ X (n+r)(n+r−1)an xn+r−2 ,y = 00 n=2 Substituting into the equation we have ∞ ∞ ∞ X X X n+r n+r (n+r)(n+r−1)an x + (n+r)an x + an xn+r+2 = 0 n=0 n=0 n=0 It follows that ∞ X [r(r−1)+r]a0 xr +[r(r+1)+(r+1)]a1 xr+1 + [(n+r)2 an +an−2 ]xn+r = 0 n=2 The indicial equation is r2 = 0 with repeated real roots r1 = r2 = 0. (c) For r = 0, it is necessary to take a1 = 0 in order that the coefficient of xr+1 be zero. Note that the recurrence relation is of the form an−2 an = − 2 , n = 2, 3, · · · n The fact that a1 = 0 implies that a2m+1 = 0, ∀ m = 0, 1, 2, · · · . And a2m = − a2m−2 a2m−4 (−1)m a0 = = · · · = 22 m2 24 m2 (m − 1)2 22m (m!)2 By setting a0 = 1, we obtain one solution for x > 0 of the equation J0 (x) = 1 + ∞ X (−1)n x2n n=1 22n (n!)2 (−1)n , 22n (n!)2 n = 1, 2, · · · . Then (−1)n+1 a 22n+2 ((n+1)!)2 1 n+1 = lim lim =0 = lim n (−1) 2 n→∞ n→∞ an 22n (n!)2 n→∞ 4(n + 1) (d) Let an = which shows that the radius of convergence of J0 (x) is infinite, that is, J0 (x) converges for all x. 20. the Bessel equation of order one is x2 y 00 + xy 0 + (x2 − 1)y = 0. (a) Show that x = 0 is a regular singular point. (b) Show that the roots of the indicial equation are r1 = 1 and r2 = −1. (c) Show that one solution for x > 0 is ∞ x X (−1)n x2n J1 (x) = . 2 n=0 (n + 1)!n!22n (d) Show that the series for J1 (x) converges for all x. The function J1 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 ∞ X −1 x bn x n , x > 0. n=0 [§5.5 #16] Proof. (a) Rewrite the equation of the form 1 x2 − 1 y 00 + y 0 + y=0 x x2 2 and let p(x) = x1 , q(x) = x x−1 2 . The only singular points is x = 0. At x = 0, p0 = lim xp(x) = lim 1 = 1 x→0 x→0 2 q0 = lim x q(x) = lim (x2 − 1) = −1 x→0 x→1 Hence x = 0 is a regular singular point. ∞ P (b) Let y = an xn+r . Then 0 y = n=0 ∞ X (n+r)an x n+r−1 ∞ X ,y = (n+r)(n+r−1)an xn+r−2 00 n=0 n=2 Substituting into the equation we have ∞ ∞ ∞ ∞ X X X X n+r n+r n+r+2 (n+r)(n+r−1)an x + (n+r)an x + an x − an xn+r = 0 n=0 n=0 ∞ X n=0 [(n + r)2 − 1]an xn+r + ∞ X n=0 an−2 xn+r = 0 n=2 n=0 It follows that 2 r 2 (r −1)a0 x +[(r+1) −1]a1 x r+1 ∞ X [(n+r)2 −1]an +an−2 xn+r = 0 + n=2 2 The indicial equation is r − 1 = 0 with distinct real roots r1 = 1, r2 = −1. (c) For either r = ±1, it is necessary to take a1 = 0 in order that the coefficient of xr+1 be zero. Note that the recurrence relation is of the form an−2 an = − , n = 2, 3, · · · (n + r)2 − 1 The fact that a1 = 0 implies that a2m+1 = 0, ∀ m = 0, 1, 2, · · · . If r = 1, then a2m−2 a2m−2 a2m−4 a2m = − =− = 2 2 2 (2m + 1) − 1 4m(m + 1) 4 m (m + 1)(m − 1) a2m−6 =− 3 4 m(m + 1) · (m − 1)m · (m − 2)(m − 1) (−1)m a0 = ··· = m 4 m!(m + 1)! By setting a0 = 12 , we obtain one solution for x > 0 of the equation ∞ x X (−1)n x2n J1 (x) = . 2 n=0 (n + 1)!n!22n (−1)n ., (n+1)!n!22n n = 0, 1, · · · . Then (−1)n+1 a 2n+2 1 (n+2)!(n+1)!2 n+1 = lim lim =0 = lim n (−1) n→∞ n→∞ n→∞ an 4(n + 2)(n + 1) (n+1)!n!22n (d) Let an = which shows that the radius of convergence of J0 (x) is infinite, that is, J1 (x) converges for all x. (e) For r = −1 the recurrence relation becomes an−2 an = − , n = 2, 3, · · · (n − 1)2 − 1 If n = 2 the coefficient of a2 zero and we cannot calculate a2 . Consequently it is not possible to find a series solution ∞ P of the form x−1 bn x n . n=0

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