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Methods MAE294A/SIO203A: in Applied Mechanics Fall Quarter 2000 Solutions IV. Question 3.24 made the homework too long. I have nevertheless included it in my marking, so as not to penalize people who spent a lot of time working on it. I will take the length of this homework into account when working out nal grades. 3.3 The ode w + f (x)w + g(z)w = 0 may be rewritten as wtt + 2 1 1 2 f (t 1 ) wt + 4 g(t 1 )w = 0 t t t using the transformation z = t 1 . The classi cation of can be deduced from this equation, knowing the behavior of f and g for small t (i.e. large z). The point z = is an ordinary point if the coef cients of wt and w are analytic at t = 0, i.e. if 2z z 2 f (z) and z 4 g(z) have expansions of the form n for large z. n=0 an z The point z = is a regular singular point if f and g take the following form f (z) = z 1 n=0 fn z n and g(z) = z 2 n=0 gn z n . Otherwise the point z = is an irregular singular point. (a) (b) (c) x= x=0 x= x=0 x=1 x= x=0 x= x=0 x= x= x= x = 1 x= ISP RSP ISP RSP RSP RSP RSP ISP RSP ISP ISP ISP RSP ISP (d) (e) (f) (g) (h) if (a = 0 or b = 0) & c = 0: OP if (a = 0 & c = b + 1) or (b = 0 & c = a + 1): OP if (a = 0 & b = 1) or (a = 1 & b = 0): OP if (a = 0 & b = 0): OP 1 if ( = 0 & 2 = 4 ): OP if (h = 0 & = 0): RSP if = 0: RSP; if ( = 0 & = 0): OP 1 3.4 We need the results d3 d3 = t6 3 + ; 3 dx dt (a) (b) (c) (d) (e) (f) x=0 ISP RSP OP ISP RSP ISP x= RSP RSP ISP RSP ISP ISP d4 d4 = t8 4 + . 4 dx dt 3.6 (a) The origin is an ordinary point. Substituting in the Taylor series y = n n=0 an x leads to the recurrence relation an+2 = 2n 8 an . (n + 2)(n + 1) The odd series (which is the relevant one for these boundary conditions) terminates, and the resulting polynomial is y = 4(1 4x2 + 4 x4 ). 3 (b) The even series does not terminate. However, the recurrence relation can be solved to give a2n+1 = = = (4n 10)(4n 12) . . . 2( 2)( 6) a1 (2n + 1)(2n) . . . 2 2n (2n 6)(2n 8) . . . 2 (2n 5)(2n 7) . . . 1( 1)( 3) a1 (2n + 1)! (2n 6)(2n 8) . . . 2 3 2n (2n 5)! a1 , 2n 3 (2n + 1)!(n 3)! where the last equality is only valid for n 3. Hence the series may be written as 25 (2n 5)! 3 y = 4x 4x + x + 96 x2n+1 . 5 (2n + 1)!(n 3)! n=3 n=0 (c) The origin is an ordinary point. Substituting in the Taylor series y = an xn leads to the recurrence relation an+2 = (n + 4)(n 3) an . (n + 2)(n + 1) The odd series, which is the one we need, terminates. The resulting solution is y = 3x 5x3 . 2 n=0 (d) The origin is an ordinary point. Substituting in the Taylor series y = an xn leads to the recurrence relation an+3 = an an+1 (n + 3)(n + 2) for n 0 and 2a2 = a0 . This is a three-term recurrence relation. We are given a1 = 0, so the resulting series takes the form y = 1 1 x2 + 1 x3 + . 2 6 4 Note that u = w, v = u and w = v. Then (u3 + v 3 + w3 3uvw) = 3u2 u + 3v 2 v + 3w2 w 3u vw 3uv w 3uvw = 3u2 w + 3v 2 u + 3w2 v 3vw2 3wu2 3uv 2 which vanishes. Hence the quantity in brackets is constant. At x = 0, u = 1 and v = w = 0. This gives u3 + v 3 + w3 3uvw = 1. 5 (a) The origin is a regular singular point. Substituting in the series y = n+s with a0 = 0 gives the recurrence relation n=0 an x an+1 = (n + s + 2)(n + s 1) an . 2(n + s + 1)2 The recurrence relation can be solved to give an = ( 1)n (n + s + 1)(n + s) . . . (s + 2)(n + s 1)(n + s) . . . (s 1) a0 . 2(n + s)2 (n + s 1)2 . . . (s + 1)2 The indicial equation is s2 = 0 with double root 0. Hence the second solution will not be of Frobenius form. For the rst solution with s = 0, the recurrence relation is an+1 = (n + 2)(n 1) an . 2(n + 1)2 This series terminates and the solution is the polynomial y1 = 1 + x. The second solution with s = 0 is given by y2 = = y(s; x) = lim an (s)xn+s s 0 s s 0 s n=0 lim (1 + x) log x + n=0 s 0 lim an (s) s xn . 3 The second solution is proportional to 1 2 (1 + x) log x+2 1. x n=0 (b) The origin is a regular singular point. Substituting in the series y = an xn+s with a0 = 0 gives the recurrence relation an+1 = 1 an . (n + s + 1)(n + s) The recurrence relation can be solved to give an = ( 1)n a0 . (n + s)(n + s 1) . . . (s + 1)(n + s)(n + s 1) . . . s The indicial equation is s(s 1) = 0 with roots 0 and 1. These roots differ by an integer and substituting s = 0 into the recurrence relation with n = 0 shows that the second solution will not be of Frobenius form. For the rst solution with s = 1, the recurrence relation is an = ( 1)n an . (n + 1)!n! The rst solution hence takes the form y1 = x 1 ( x)n = x1/2 J1 (2x1/2 ). (n + 1)!n! n=0 The second solution with s = 0 is is given by y2 = lim s y(s; x) = lim san (s)xn+s s 0 s s n=0 1/2 s 0 = (1 + x)x J1 (2x 1/2 )+ n=0 s 0 lim (san (s)) s xn . The second solution is proportional to x1/2 Y1 (2x1/2 ) + x1/2 J1 (2x1/2 ). (c) The origin is an ordinary point. Substituting in the series y = gives the equation n=0 an xn n(n 1)an xn 2 + n=2 xm m! m=1 an = xn 0, n=0 using the series expansion for ex . Now the double sum may be rewritten as xm m! m=1 p 1 an xn = n=0 p=1 k=0 ak (p k)! xp . 4 The ode then takes the form n 1 (n + 2)(n + 1)an+2 + n=0 k=0 ak xn = 0, (n k)! which leads to the recurrence relation n 1 an+2 = k=0 ak (n k)! with a2 = 0. The rst two coef cients are arbitrary, so the solution is 1 y = a0 (1 6 x3 14 24 x + ) + a1 (x 14 12 x + ). n=0 (d) The origin is a regular singular point. Substituting in the series y = an xn+s with a0 = 0 gives the recurrence relation an+1 = n+s+1 an n+s or an = n+s a0 . s The indicial equation is s(s 1) = 0 with roots 0 and 1. These roots differ by an integer, and substituting s = 0 into the recurrence relation with n = 0 shows that the second solution will not be of Frobenius form. The rst solution with s = 1 is y1 = x n=0 (n + 1)xn = x . (1 x)2 The second solution with s = 0 is given by y2 = sy(s; x) = lim (n + s)xn+s s 0 s s 0 s n=0 lim = log x n=0 nxn + n=0 xn = log x x 1 + . 2 (1 x) 1 x n=0 (e) The origin is a regular singular point. Substituting in the series y = an xn+s with a0 = 0 gives the recurrence relation an+3 = 1 an (n + s + 3)(2n + 2s + 3) and a1 = a2 = 0. The indicial equation is s(2s 3) = 0 with roots 0 and 3 . These roots do not 2 differ by an integer, so there will be two solutions of Frobenius form. For the rst solution with s = 0, a3n = = 1 ( 1)n a3n 3 = a0 3n(6n 3) 3n(3n 3) . . . 3(6n 3)(6n 9) . . . 3 ( 1)n 1 2n(2n 2) . . . 2 ( 1)n 2n a = 2n a0 . 2n n! (2n 1)(2n 3) . . . 1 2n(2n 2) . . . 2 0 3 3 (2n)! 5 The rst solution hence takes the form y1 = n=0 ( 1)n 2n 32n (2n)! x3n = cos ( 2x3/2 /3). For the second solution with s = 3 , 2 a3n = = 1 ( 1)n a3n 3 = a0 3n(6n + 3) 3n(3n 3) . . . 3(6n + 3)(6n 3) . . . 3 ( 1)n 1 2n(2n 2) . . . 2 ( 1)n 2n a0 = 2n a0 . 32n n! (2n + 1)(2n 1) . . . 1 2n(2n 2) . . . 2 3 (2n + 1)! The second solution hence takes the form y2 = x3/2 n=0 ( 1)n 2n x3n = sin ( 2x3/2 /3). 2n (2n + 1)! 3 (f) The origin is a regular singular point. Substituting in the series y = an xn+s with a0 = 0 and expanding sin x and cos x gives a dif cult recurrence relation, while the indicial equation has roots 0 and 3, which differ by an integer. However, y = cos x is a solution, so we may reduce the order of the equation by one by writing y = u cos x. The resulting equation for u is n=0 u 2 u = 0, sin x cos x which has the solution u = tan2 x. Integrating once again gives u = tan x x. Hence the two solutions are y1 = cos x = 1 1 x2 + 2 which corresponds to s = 0 and y2 = sin x x cos x = 1 x3 3 15 30 x 14 24 x + + n=0 which corresponds to s = 3. Both solutions are Frobenius series. (g) The origin is an ordinary point. Substituting in the series y = gives the recurrence relation an+4 = 1 an (n + 4)(n + 3) and a2 = a3 = 0. an xn For the even series, the recurrence relation may be written a4n = = 1 1 a4n 4 = a0 (4n + 4)(4n 1) 4n(4n 4) . . . 4(4n 1)(4n 5) . . . 3 ( 3 ) 4 42n n! (n + 3 ) 4 6 The even series hence takes the form y1 = ( 3 ) 4 n! (n + 3 ) 4 n=0 x 2 4n . For the odd series, the recurrence relation may be written a4n+1 = = 1 1 a4n 3 = a1 (4n + 1)4n (4n + 1)(4n 3) . . . 5(4n)(4n 4) . . . 4 ( 5 ) 4 5 42n n! (n + 4 ) The odd series hence takes the form y2 = x ( 5 ) 4 n! (n + 5 ) 4 n=0 x 2 4n . These two solutions are proportional to x1/2 I 1/4 ( 1 x2 ) and x1/2 I 1/4 ( 1 x2 ) respec2 2 tively, where I is the modi ed Bessel function of order . (h) The origin is a regular singular point. Substituting in the series y = n+s with a0 = 0 gives the recurrence relation n=0 an x an+1 = (n + s + 2)(n + s 2) an . (2n + 2s + 1)(n + s + 1) 1 The indicial equation is s(2s 1) = 0 with roots 0 and 2 . These roots do not differ by an integer, so there will be two solutions of Frobenius form. For the rst solution with s = 1 , 2 an+1 = (n + 5 )(n 3 ) 2 2 an . (2n + 3)(n + 1) This is hard to simplify. For the second solution with s = 0, an+1 = (n + 2)(n 2) an . (2n + 1)(n + 1) This series terminates and the solution is the polynomial y2 = a0 (1 + 4x + 2x2 ). The rst solution is proportional to x1/2 (1 + x) 2 + x. (i) The origin is a regular singular point. Substituting in the series y = n+s with a0 = 0 gives the recurrence relation n=0 an x an+1 = 1 an . n+s+ 1 2 7 The indicial equation is s(s 1 ) = 0 with roots 0 and 1 . These roots do not 2 2 differ by an integer, so there will be two solutions of Frobenius form. For the rst solution with s = 1 , 2 an+1 = 1 an n+1 or an = 1 a0 . n! The rst solution hence takes the form y1 = x1/2 For the second solution with s = 0, an+1 = 1 an n+s+ 1 2 or an = 1 a0 , ( 1 )n 2 xn = x1/2 ex . n! n=0 where (a)n = a(a+1)(a+2) . . . (a+n 1) and (a)0 = 1 (this is the Pochhammer symbol). The second solution hence takes the form y2 = xn . (1) n=0 2 n These solutions may be written y = M (1, 1 , x) and y = U (1, 1 , x), and are (Kummer) 2 2 con uent hypergeometric functions. 8

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