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Unformatted text preview: Unit B: 8.2 Series Solutions Neor Ordinary Points 631 With an appropriate choice (made separately for each m of the arbitrary
constants ('0 (it even) or c. (it odd). the nthdegree polynomial solution of Leg
cndre‘s equation of order n. > (1 — .\')3y" — 2):)" + nu: + 13y = 0. (25.)
is denoted by Putx) and is called the Legendre polynomial of degree n. It is cas tomary (for a reason indicated in Problem 32) to choose the arbitrnr} constant so
that the coeiiicient of x" in P,,(x) is (2n)!/ [2”(11!)3]. It then turns out that N.
 (—1)“(2n—2l'<)l JR ) P ‘1 = —__‘ _ I uh) k:t)ZJ'kHH—kiiUl—anr [ l where N = {In/2]]. the integral part of n/2. The ﬁrst six Legendre polynomials are FUDGE i. P1(x)=.\'. I
P_;(.\‘) = 3(5x3 — 3x}. 1 . 1 t
Pm) = §(35x4 ~ 303’ + 3). t°5(.\') = §(63.\'5 — 70):" + 15.1 t. and their graphs are shown in Fig. 8.2.2. FIGURE 8.2.2. Graphs y = P,,(.\') of the Legendre poiyiomials
for n = l. 2. 3. 4. and 5. The graphs are distinguished h) the fact
that all] n zeros of Pn(.\‘t lie in the interval — < .1' < I. Problems m! solutions in powers of .\' of the differential equa 7. (x: + My" — 71')" + My = 0
Problems I through 15. Smte the ret'urn’nc'c’ relation 8, (3 __\3)_1" — _\y’ 4. my = (t . . ._rl .,
guaranteed rndms of murertzwu‘e in each (am. A: 9. (X _ 1”." + alJ.’ + '3}. = n __ my" + 4x}; + 2}, : 0 '10. 3):" + I)" ~ 4y 2 U
+ 3U." + hw + 3.1. = 0 I]. 5y” — 3X)" + 10y = 0
30.! + y = 0 [2. y" — xi)" — by = (l
+ I )y" + my + 43' = 0 13. .r" + 3'2)" + lo = 0
"  3)}"' + 21’)" = 0 [1. y" + .\'_t' = 0 [an Air)‘ aquarier .. In?" — 61')" + : 15. '1'" + .\':_" = O 632 Use power series to solve the initial value problems in
Problems 16 and 17. 16. (1+ x2)y” + 2xy’ — 2y = 0; y(0) = 0, y’(0) =1 [17. y” +xy’ —2y 2 0; y(0) = 1. y’(0) = 0 Solve the initial value problems in Problems 18 through 22.
First make a substitution of the form t = x — a, then ﬁnd a
solution cht" of the transformed diﬁerential equation. State
the interval of values ofx for which Theorem I of this section
guarantees convergence. 18.
19.
20. 21.
22. y” + (x 1)y’+ y = 0; y(1)= 2. y’(1)= 0 (ZxXZW” —6(X —1)y'4y = 0; y(1)= 0. y’(1) =1 (x2 — 6x + 10)y”  4(X — 3)y’ + 6y = 0; y(3) = 2,
VB) = 0 (4x2 +16x +17)y” = 8y; y(—2) = 1, y'(—2) = 0 (x2 + 6x)y” + (3x + 9)y’ — 3y = 0; y(—3) = 0, y’(—3) = 2 In Problems 23 through 26, ﬁnd a three—term recurrence rela
tion for solutions of the form y = chx". Then ﬁnd the ﬁrst
three nonzero terms in each of two linearly independent solu
tions. 23.
24.
25.
26.
27. y"+(1+x)y=0 (x2 — l)y” + 2xy’ + 2xy = 0 yu + x2yl +x2y = 0 (1+ x3)y" + xfy = 0 Solve the initial value problem y” +xy’ + (2x2 + l)y = 0; y(0)=1,y’(0)= —1. Determine sufﬁciently many terms to compute y(1/2)
accurate to four decimal places. In Problems 28 through 30, ﬁnd the ﬁrst three nonzero terms
in each of two linearly independent solutions of the form y = chx”. Substitute known Taylor series for the analytic
functions and retain enough terms to compute the necessary
coejﬁcients. 28. y” + e‘xy = O 29. (cos x)y” + y = O 30. xy" + (sin x)y’ + xy = 0 31. Derive the recurrence relation in (21) for the Legendre 32. equation.
Follow the steps outlined in this problem to establish
Rodrigues’s formula dn
__ (x2 _ 1)" PM) : n!2” dx" for the nth—degree Legendre polynomial. (a) Show that
v = (x2 — 1)” satisﬁes the differential equation (1 — x2)v’ + 2nxv = 0.
Differentiate each side of this equation to obtain (1 — x2)v” + 2(n — l)xv’ + 2nv = 0. Unit B: Chapter 8 Power Series Methods 33. 34. (b) Differentiate each side of the last equation n times in
succession to obtain (1 — x2)v(”+2) — 2xv("+1) + n(n + 1)v(") = 0. Thus it = v‘") = D"(x2 — 1)” satisﬁes Legendre’s equa
tion of order n. (c) Show that the coefﬁcient of x”
in u is (2n)!/n!; then state why this proves Rodrigues
formula. (Note that the coefﬁcient of x" in P,,(x) is
(2n)!/ [2"(nl)2] .) The Hermite equation of order a is
y” — 2xy' + 2ay = 0.
(a) Derive the two power series solutions °° 2’"a(a—2)(a—2m+2) .
= 1 __1 III .17!
y] + Z( ) (2m)! x m=l and °° 2mm — l)(a — 3)    (a — 2m +1) M]
+21 1) (2m+1)! x '
Show that yl is a polynomial if a is an even inte
ger, whereas y; is a polynomial if a is an odd integer.
(b) The Hermite polynomial of degree n is denoted by
H,,(x). It is the nthdegree polynomial solution of Her
mite‘s equation, multiplied by a suitable constant so that
the coefﬁcient of x” is 2". Show that the ﬁrst six Hermite
polynomials are mm 2 1, H2(x) = 4x2 — 2, Hm) = 16x4 — 48x2 + 12,
H5(x) = 32x5 — 160x3 + 120x. H(x) = 2x,
H3(x) = 8x3 —12x, A general formula for the Hermite polynomials is 2 d" .2
Hn(x) = (—1)"e" (w dx"
Verify that this formula does in fact give an nthdegree
polynomial. It is interesting to use a computer algebra
system to investigate the conjecture that (for each n) the
zeros of the Hermite polynomials H,,(x) and H,,+1 are
“interlaced”—that is, the n zeros of H" line in the n
bounded open intervals whose endpoints are successive
pairs of zeros of HM].
The discussion following Example 4 in Section 8.1 sug—
gests that the differential equation y” + y = 0 could be
used to introduce and deﬁne the familiar sine and cosine
functions. In a similar fashion. the Airy equation H y =xy Unit B: 8.3 Regular Singular Points 647 with analytic coefﬁcient functions, in order to investigate the possible existence of
series solutions we ﬁrst write the equation in the standard form > y" + P(x)y’ + Q(x)y = 0. If P(x) and Q(x) are both analytic at x = 0, then x = 0 is an ordinary point‘ and
the equation has two linearly independent power series solutions. Otherwise, x = 0 is a singular point, and we next write the differential equation
in the form ‘ X X“ If p(x) and q(x) are both analytic at x = 0, then x = 0 is a regular singular point.
In this case we ﬁnd the two exponents r, and r2 (assumed real, and with r] 2 r;)
by solving the indicial equation > r(r—l)+por+q0=0, where p0 : p(O) and an = (1(0). There always exists a Frobenius series solution
y = x’ﬁZanx" associated with the larger exponent n, and if r] * r: is not an
integer, the existence of a second Frobenius series solution y; = x’1 anx" is also
guaranteed. 8.3 Pr0b19m§ ._ In Problems I through 8, determine whether x = O is an _o_r£li— l6. x3(l — x)y” + (3x + 2)y’ + xy = O
nary point, a t‘egulat"sitzgztlar point, or an irregular singular point. If it is a regular singular point, ﬁnd the exponents of Find ""0 linear .’ (fOr the differential equation at x = 0. X > 0) of each of the diﬁerential equations in Problems 17 through 26.
1. xy" + (x — x3)y’ + (sin x)y = 0 2. xy”+x2y’+(e"— l)y:0 lrl7. 4xy”+2y'+y=0
3. xzy"+(cosx)y’+xy=0 18. nyn+3;y —y=0
C4. 3x3y”+2x3y’+(l —x3)y=0 ‘19 ny ‘y ‘y:0 20. 3xy" + 2y’ + 2y 2 0
i721. 2x3y" + xy' — (l + 2x3)y = 0
22. 2x331” + xy’ — (3 — 2x3)y = O
23. 6x3y" + 7xy’ — (x2 + 2)y =
24. 3x3y” + 2xy' + Xzy = 0 25. 2xy”+ (l +x)y’+y = O 26. 2xy” + (l — 2x3)y’ — 4xy = 0 5. x(l +x)y”+2y’+3xy=0 6. x2(l —xz)y”+2xy’—2y=0 7. xly”+(6sinx)y’+6y=0 8. (6x+2x3)y”+21xy’+9(x3—l)y=0 If x = a 95 0 is a singular point of a secondorder linear dif—
ferential equation, then the substitution t = x—a transforms it
into a differential equation having I = 0 as a singular point.
We then attribute to the original equation at x = a the behav—
ior of the new equation at t = 0. Classifv (as regular or
irregular) the singular points of the differential equations in
Problems 9 through [6. Use the method of Example 6 to ﬁnd two linearly indepen
dent Frobenius series solutions of the diﬁrerential equations in
Problems 27 through 3], Then construct a graph showing their
graphs for x > 0. 27. xy" + Zy’ + 9xy = O C 9. (l — x)y” + xy’ + xzy = 0 (28. xy" + zyr _ 4x}, = 0
10 (1— X)?" + (2X * 2))" + y = 0 “29. 4xy” + 8y’ + xy = 0
ll. (l—x2)y”—2xy’+12y=0 30. xyu_y:+4x3y=0
12 (X  2W” + 30‘  2):)" + W = 0 31. 4x1y" — 4xy’ + (3 — 4x3)y = 0
13. (x2 —4)y” + (x — 2)y’ + (X + 21v 2 0
14. (x2 — 9)3y” + (x2 + 9) y’ + (X1 + 4) y = 0 In Problems 32 through 34, ﬁnd the ﬁrst three nonzero terms ‘ 15. (x — 2)3y” — (xZ — 4)y’ + (x + 2)y = 0 of each of two linearly independent Frobenius series solutions. Unit B: 8.4 Method of Frobenius: The Exceptional Cases 663 Note that the technique of reduction of order readily yields the ﬁrst several terms of
the series, but does not provide a recurrence relation that can be used to determine
the general term of the series. With a computation similar to that shown in Example 3 (but more complicated
—see Problem 21). the method of substitution can be used to derive the solution (—1)"(Hn + I‘In—lezn—l
23"nl(n—1).' ‘ l
_ 1 _ _ E j]
y3(x) y.(x) nx x + (3 ) 11:!
where H,, is deﬁned in (44) for n g 1; H0 : 0. The reader can verify that the terms
shown in Eq. (50) agree with 3
y:(x) = ZJMX) +y3(x). (52) The most commonly used linearly independent [of J.] solution of Bessel’s equation
of order 1 is the combination 2 2
Y: (x) = —(r  In 2)yt(X) + —y3(.\')
72' 72' _ 2 x 1 °° (—1)"<H,,+H,,_t)x3"—'
_; (/+ln§>J1(X)—;+Z 22,1"!("_1)! . (53)
l ":1 Examples 3 and 4 illustrate two methods of ﬁnding the solution in the logarith mic cases—direct substitution and reduction of order. A third alternative is outlined
in Problem 19. 8.4 Problems In Problems 1 through 8, either apply the method of Example 12. X3)” + xzy' — 2y = 0 l to ﬁnd two linearly independent Frobenius series solutions, 13, x3 y” + (2 x3 _ 3x)y’ + 3 y = 0
or ﬁnd one such solution and show (as in Example 2) that a 14. x1 y" + x“ + My _ 4y : 0
second such solution does not exist. 15_ Begin with l. xy”+(3—x)y’—y=0 2 4 .6
2. Xy”+(5—x)y’—y=0 J”(x)=]_£+x__ " +...r ‘
3. xy"+(5+3x)y’+3y=0 4 64 2304 t
[4. 5Xy" + (30 + 3x)y’ + 3y = 0 Using the method of reduction of order. derive the second
5 X)?" — (4 + le/ + 3y = 0 linearly independent solution '
6. 2xy” — (6 + 2x)y’ + y = 0
7. xzy" + (2x + 3x3)y’ — 2y = 0 x2 3x4 1 1x6
[8. x(] — X)y” — 3y’ +2y = 0 W“) = “mm” I — E +13284 _ ' "
I‘n Problems 9 through 14. ﬁrst ﬁnd the ﬁrst four nonzero 0f Bessel5 equation 0f order [em
terms in a Frobenius series solution of the given (liﬂerential 16 Find 1W0 “nearly independent FrObenlUS series SOlUtiOUS l
equation. Then use the reduction of order technique (as in Of 36556” equation 0f order Example 4) to ﬁnd the logarithmic term and the ﬁrst three
nonzero terms in a second linearly independent solution. Xzyl/ + "y, + (X: _ y = 0‘
9 X}‘"+,V' —Xy = 0 [\17. (3) Verify that y[(x) = xe" is one solution of
10. x3y”—xy’+(x2+l)y=0 l/ 11. xzy” + (x2 — 3x)y’ +4y = 0 xly” — x(l +.\')y’ + y = 0. 664 18. £19. (b) Note that r[ = r3 = 1. Substitute x
y; = y] lnx + Z b,,x"+l 11:] in the diﬂferential equation to deduce that b. = —l and that 1
"bit — bn—l = __’
n. (c) Substitute b,, = c,,/n! in this recurrence relation and
conclude from the result that C" = —H,,. Thus the second
solution is for n g 2. x yg(x) = xe" lnx — 2 n=1 H XIIH
H n! Consider the equation xy" — y = 0. which has exponents
r. = l and r3 = 0 at x = 0. (a) Derive the Frobenius
series solution x x”
“WOO = Z n.'(n— l)! 11:! (b) Substitute y; = Cy. In x + Z bnx” n:() in the equation xy” — y = 0 to derive the recurrence
relation
2n + l n(n+l)bn+1 ‘ bu = ‘W ' Conclude from this result that a second solution is ac y:(—\‘) = y[(x)lnx+ 1 — 2 ":1 II HI] + Hll—l
n! (n — 1)! Suppose that the differential equation
L[y] = xzy" + xp(x)y’ + q(x)y = 0 (54) has equal exponents r. = r: at the regular singular point
x = 0. so that its indicial equation is 42m = (r—r.)2 =0. Let C“ = l and deﬁne c,,(r) for n g l by using Eq. (9):
that is. Lll(r:c0‘ci‘ “wot—l) n = _ c (r) ¢(r + n) ( )
Then define the function y(x,r) of x and r to be
y(x. r) = Z 0.,(r)x"”. (56)
":0 (a) Deduce from the discussion preceding Eq. (9) that L[y(x.r)] = xr(r—r.)2t (57) Unit 8: Chapter 8 Power Series Methods 20. 21. Hence deduce that DC y] = )‘(xrl) = Z 6,.(r1)x"+" "=0 (58) is one solution of Eq. (54). (b) Differentiate Eq. (57)
with respect to r to show that a '1
L[yr(x~rl J] = a [x'(r — r1)‘] : 0. r=rl Deduce that y: = y,(x.r]) is a second solution of
Eq. (54). (c) Differentiate Eq. (58) with respect to r to
show that so y2 = y1 In x + x" Z c:,(r, )X". 11:1 (59) Use the method of Problem 19 to derive both the solutions
in (38) and (45) of Bessel’s equation of order zero. The
following steps outline this computation. (a) Take co = 1;
show that Eq. (55) reduces in this case to (r+1)2c.(r)= 0 and cit—3(r) 1 (n+r) cll(r) : _ for n g (b) Next show that c1(0) = c’](0) = 0. and then deduce
from (60) that c,,(0) = 4(0) : O for n odd. Hence
you need to compute c,,(0) and c,’,(0) only for 11 even.
(c) Deduce from (60) that (—1)"
(r + 2)3(r + 4)2    (r + 2n)? CIHU‘) : With r = r1 = 0 in (58), this gives J(,(x). (d) Diﬁ‘erentiate
(61) to show that (—1>"+'H,. 6211(0) = 111(n)2 Substitution of this result in (59) gives the second solu
tion in (45). Derive the logarithmic solution in (51) of Bessel‘s equa
tion of order 1 by the method of substitution. The fol
lowing steps outline this computation. (3) Substitute y: = CJ[(X) ll’lX‘i'X—I + anxn> n=l in Bessel‘s equation to obtain —b1 + x + 2th — 1)b,.+i + b.. Jx" ":2 3° _ n ZnH
+C [HEM] =0. 23"(n+1)!n! (62) n=l 674 FIGURE 8.5.6. The positive zeros y“, yng. y,,3. Unit B: Chapter 8 Power Series Methods of the Bessel function J,,(x). 8.5 Problems 1. Differentiate termwise the series for J0(x) to show directly
that .1600 = —J1(x) (another analogy with the cosine and
sine functions). . (a) Deduce from Eqs. (10) and (12) that l35(2n—l) 2n F (n + =
(b) Use the result of part (a) to verify the formulas in
Eq. (19) for J1/3(x) and J_./2(x), and construct a ﬁgure
showing the graphs of these functions. . (3) Suppose that m is a positive integer. Show that 25—8(3m—l)r 3.. (> (b) Conclude from part (a) and Eq. (13) that F(m+%)= J—1/3(X) = (x/2)—/3 x (_l)m3mx2m
[~(g) 1+222mmg.2.5...(3m_1) ' 3 m=l . Apply Eqs. (19), (26), and (27) to show that 2 .
— (smx — xcos x) J w =
3/(X) ﬁx} 2 .
—1(cosx + xsmx). J '7 = —
3/(x) ﬁx Construct a ﬁgure showing the graphs of these two func
tions. . Express J4(x) in terms of J0(x) and J1(x).
. Derive the recursion formula in Eq. (2) for Bessel’s equation. . Verify the identity in (23) by termwise differentiation.
. Deduce the identities in Eqs. (24) and (25) from those in Eqs. (22) and (23). . Verify that the substitution 1 = ax transforms the para metric Bessel equation in (28) into the equation in (29). 10. Show that 4J‘;'(x) = Jp_3(x) — 2Jp(x) + Jp+2(x). 11. Use the relation F(x +1) 2 x1"(x) to deduce from Eqs. (13) and (14) that if p is not a negative integer,
then Jp(x) : (x/2)P 1+2 (_l)m(X/2)2m F(p+l) "1:1m!(p+1)(p+2)~(p+m) ‘
This form is more convenient for the computation of Jp(x)
because only the single value F(p+1) of the gamma func
tion is required. . Use the series of Problem 11 to ﬁnd y(0) = lingy(x) if
x_. J 7 J_ '7
y(x) = x2 [ 5/'(x)+ Suﬁ] . Jl/2(x)+J—l/2(X) Use a computer algebra system to graph y(x) for x near
0. Does the graph corroborate your value of y(0)‘? Any integral of the form Ix”’.l,,(x)dx can be evaluated in
terms of Bessel functions and the indeﬁnite integral JJn(x)dx. The latter integral cannot be simpliﬁed further, but the func tion J(,(t)dt is tabulated in Table 11.] of Abramowit: and
Stegun. Use the identities in Eqs. (22) and (23) to evaluate
the integrals in Problems 13 through 2]. 13. Jleo(x)dx 14. Jx3J(,(x)dx
x4J0(x) dx l6. xJ.(x)dx
18. 20 Jx3Jl(x)dx J3(x)dx ...
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