Lista 6 - Boyce - Cálculo 4 2022 2

Lista de exercicios 6 — Equacdo de Bessel (do livro de Boyce e Di Prima) Problems In each of Problems | through 3, show that the given differential 9. In this section we showed that one solution of Bessel’s equation equation has a regular singular point at x = 0, and determine two of order zero solutions for x > 0. Lily] =x2y" +xy’ +2x2y =0 Ll xy" + 2xy' + xy =0 is Jo, where Jo(x) is given by equation (7) with aj = 1. According to 2. x?y"+3xy'+(1+x)y =0 Theorem 5.6.1, a second solution has the form (x > 0) 3. x?y"+xy' + 2xy =0 os , 4. Find two solutions (not multiples of each other) of the Bessel ya(x) = Jo(x) Inx + »— bnx". 3 n=1 equation of order — 2 a. Show that Co Co 9 = _ n n ey" bxy! 4 G _ i) =0, x>0. Lyx) = Sona = box" + So nbax 4 n=2 n=1 CO 5. Show that the Bessel equation of order one-half + S- b,x"t? 42x Ji(x). (34) 1 n=1 xy" +xy+(x7-—])y=0, x>0 4 b. Substituting the series representation for Jo(x) in equation (34), show that can be reduced to the equation 00 Way <0 bix + 2box? + S- (n> by + bya)" n=3 by the change of dependent variable y = x~!/2v(x). From this, _ =>} (=1)"2nx7” (35) conclude that y,(x) =x7!/cosx and y.(x) =x7!/*sinx are ~ 2nr(n!j2 ~ solutions of the Bessel equation of order one-half. nl 6. Show directly that the series for Jo(x), equation (7), converges c. Note that only even powers of x appear on the right-hand absolutely for all x. side of equation (35). Show that b} = b3 = bs = --- = 0, 7. Show directly that the series for J; (x), equation (27), converges by = — and that absolutely for all x and that Jj(x) = —J;(x). 21!) 8. Consider the Bessel equation of order v (2n)2boy + by_2 = Ge n=2.3.4.... n! xy" + xy +(x2-v7)y=0, x>0, Deduce that where v is real and positive. a. Show that x = 0 is a regular singular point and that the roots by = to (: +4 5) and be = to (: 4 I 4 :) of the indicial equation are v and —v. 2? 4° 2 2? 4? 6? 2 3 b. Corresponding to the larger root v, show that one solution is The general solution of the recurrence relation is —| n+l H, 1 x\? 1 x\4 by = on Substituting for b, in the expression for ma =x" [1-5 (5) +55 5 2 h(n!) W(il+v)\2 21 +v)(2+v) \2 yo(x), we obtain the solution given in equation (10). 10. Find a second solution of Bessel’s equation of order one by oo 1m 2m computing the c,(72) and a of equation (24) of Section 5.6 according + S> ED (5) . to the formulas (19) and (20) of that section. Some guidelines along 3 the way of this calculation are the following. First, use equation “mil tv). (mtv) \2 he way of this calculati he following. Fi quation (24) of this section to show that a,;(—1) and a\( —1) are 0. Then show that c. If 2v is not an integer, show that a second solution is c;(—1) = 0 and, from the recurrence relation, that c,(—1) = 0 for 5 , 2= 3,5, ... . Finally, use equation (25) to show that (x) _y 1 1 x 7 1 x ao xX) =x — —— | = ————_——~ [ = =_— y ld—v)\2 21 —v)(2—-v) \2 a0) =~ Cry)’ ao ar) = ————_—————., oo 2m (r+ Dr +3)(7+3)(7 +5) +> nk m\(1—v)---(m—v)y \2 : and that a (=1)"ao . A_(1) = hm & 3. Note that y;(x) — 0 as x — 0, and that y.(x) is unbounded as (r+1)---(r+2m—1)(r +3) ---(r +2m+4+1) x > 0. d. Verify by direct methods that the power series in the Then show that expressions for y;(x) and y2(x) converge absolutely for all x. (-1)""! (Hn + Hn-1) Also verify that y is a solution, provided only that v is not an C2m(—1) = ~~ 2mm\in —b! —p! » mot. integer.