Series Solutions of ODEs. Special Functions

Transcription

1 CHAPTER 5 Series Solutions of ODEs. Specil Functions In the previous chpters, we hve seen tht liner ODEs with constnt coefficients cn be solved by lgebric methods, nd tht their solutions re elementry functions known from clculus. For ODEs with vrible coefficients the sitution is more complicted, nd their solutions my be nonelementry functions. Legendre s, Bessel s, nd the hypergeometric equtions re importnt ODEs of this kind. Since these ODEs nd their solutions, the Legendre polynomils, Bessel functions, nd hypergeometric functions, ply n importnt role in engineering modeling, we shll consider the two stndrd methods for solving such ODEs. The first method is clled the power series method becuse it gives solutions in the form of power series 0 x 2 x 2 3 x 3 Á. The second method is clled the Frobenius method nd generlizes the first; it gives solutions in power series, multiplied by logrithmic term ln x or frctionl power x r, in cses such s Bessel s eqution, in which the first method is not generl enough. All those more dvnced solutions nd vrious other functions not ppering in clculus re known s higher functions or specil functions, which hs become technicl term. Ech of these functions is importnt enough to give it nme nd investigte its properties nd reltions to other functions in gret detil (tke look into Refs. [GenRef], [GenRef0], or [All] in App. ). Your CAS knows prcticlly ll functions you will ever need in industry or reserch lbs, but it is up to you to find your wy through this vst terrin of formuls. The present chpter my give you some help in this tsk. COMMENT. You cn study this chpter directly fter Chp. 2 becuse it needs no mteril from Chps. 3 or 4. Prerequisite: Chp. 2. Section tht my be omitted in shorter course: 5.5. References nd Answers to Problems: App. Prt A, nd App Power Series Method The power series method is the stndrd method for solving liner ODEs with vrible coefficients. It gives solutions in the form of power series. These series cn be used for computing vlues, grphing curves, proving formuls, nd exploring properties of solutions, s we shll see. In this section we begin by explining the ide of the power series method. 67

2 68 CHAP. 5 Series Solutions of ODEs. Specil Functions From clculus we remember tht power series (in powers of x x 0 ) is n infinite series of the form () m0 m (x x 0 ) m 0 (x x 0 ) 2 (x x 0 ) 2 Á. Here, x is vrible. 0,, 2, Á re constnts, clled the coefficients of the series. x 0 is constnt, clled the center of the series. In prticulr, if x 0 0, we obtin power series in powers of x (2) m0 m x m 0 x 2 x 2 3 x 3 Á. We shll ssume tht ll vribles nd constnts re rel. We note tht the term power series usully refers to series of the form () [or (2)] but does not include series of negtive or frctionl powers of x. We use m s the summtion letter, reserving n s stndrd nottion in the Legendre nd Bessel equtions for integer vlues of the prmeter. EXAMPLE Fmilir Power Series re the Mclurin series x x m x x 2 Á ( ƒ x ƒ, geometric series) m0 e x m0 cos x m0 x m 2 x x m! 2! x 3 3! Á () m x 2m x 2 (2m)! 2! x 4 Á 4! sin x m0 () m x 2m x x 3 (2m )! 3! x 5 5! Á. Ide nd Technique of the Power Series Method The ide of the power series method for solving liner ODEs seems nturl, once we know tht the most importnt ODEs in pplied mthemtics hve solutions of this form. We explin the ide by n ODE tht cn redily be solved otherwise. EXAMPLE 2 Power Series Solution. Solve yr y 0. Solution. In the first step we insert (2) y 0 x 2 x 2 3 x 3 Á m0 m x m

3 SEC. 5. Power Series Method 69 nd the series obtined by termwise differentition (3) yr 2 2 x 3 3 x 2 Á m m m x m into the ODE: ( 2 2 x 3 3 x 2 Á ) ( 0 x 2 x 2 Á ) 0. Then we collect like powers of x, finding ( 0 ) (2 2 )x (3 3 2 )x 2 Á 0. Equting the coefficient of ech power of x to zero, we hve 0 0, 2 2 0, , Á. Solving these equtions, we my express, 2, Á in terms of, which remins rbitrry: 0 0, !, !, Á. With these vlues of the coefficients, the series solution becomes the fmilir generl solution y 0 0 x 0 2! x 2 0 3! x 3 Á 0 x x 2 2! x 3 3! b 0e x. Test your comprehension by solving y 0 cos x sin x. ys y 0 by power series. You should get the result We now describe the method in generl nd justify it fter the next exmple. For given ODE (4) ys p(x)yr q(x)y 0 we first represent p(x) nd q(x) by power series in powers of x (or of x x 0 if solutions in powers of x x 0 re wnted). Often p(x) nd q(x) re polynomils, nd then nothing needs to be done in this first step. Next we ssume solution in the form of power series (2) with unknown coefficients nd insert it s well s (3) nd (5) ys # 2 3 x 4 # 3 4 x 2 Á m2 m(m ) m x m2 into the ODE. Then we collect like powers of x nd equte the sum of the coefficients of ech occurring power of x to zero, strting with the constnt terms, then tking the terms contining x, then the terms in x 2, nd so on. This gives equtions from which we cn determine the unknown coefficients of (3) successively. EXAMPLE 3 A Specil Legendre Eqution. The ODE ( x 2 )ys 2xyr 2y 0 occurs in models exhibiting sphericl symmetry. Solve it.

4 70 CHAP. 5 Series Solutions of ODEs. Specil Functions Solution. Substitute (2), (3), nd (5) into the ODE. ( x 2 )ys gives two series, one for ys nd one for x 2 ys. In the term 2xyr use (3) nd in 2y use (2). Write like powers of x verticlly ligned. This gives ys x 2 4 x x x 4 Á x 2 ys 2 2 x x x 4 Á 2xyr 2 x 4 2 x x x 4 Á 2y x 2 2 x x x 4 Á. Add terms of like powers of x. For ech power x 0, x, x 2, Á by [0] (constnt terms), [] (first power of x), nd so on: equte the sum obtined to zero. Denote these sums Sum Power Equtions [0] [x 0 ] 2 0 [] [x] 3 0 [2] [x 2 ] , [3] [x 3 ] 5 0 since 3 0 [4] [x 4 ] , ( 3) This gives the solution y x 0 ( x 2 3 x 4 5 x 6 Á ). 0 nd remin rbitrry. Hence, this is generl solution tht consists of two solutions: x nd x 2 3 x 4 5 x 6 Á. These two solutions re members of fmilies of functions clled Legendre polynomils P nd Legendre functions ; here we hve nd x 2 3 x 4 5 x 6 Á n (x) Q n (x) x P (x) Q (x). The minus is by convention. The index is clled the order of these two functions nd here the order is. More on Legendre polynomils in the next section. Theory of the Power Series Method The nth prtil sum of () is (6) s n (x) 0 (x x 0 ) 2 (x x 0 ) 2 Á n (x x 0 ) n where n 0,, Á. If we omit the terms of from (), the remining expression is s n (7) R n (x) n (x x 0 ) n n2 (x x 0 ) n2 Á. This expression is clled the reminder of () fter the term n (x x 0 ) n. For exmple, in the cse of the geometric series x x 2 Á x n Á we hve s 0, R 0 x x 2 x 3 Á, s x, R x 2 x 3 x 4 Á, s 2 x x 2, R 2 x 3 x 4 x 5 Á, etc.

5 SEC. 5. Power Series Method 7 In this wy we hve now ssocited with () the sequence of the prtil sums s 0 (x), s (x), s 2 (x), Á. If for some x x this sequence converges, sy, then the series () is clled convergent t x x, the number s(x ) is clled the vlue or sum of () t x, nd we write Then we hve for every n, lim s n(x ) s(x ), n: s(x ) m0 m (x x 0 ) m. (8) s(x ) s n (x ) R n (x ). If tht sequence diverges t x x, the series () is clled divergent t x x. In the cse of convergence, for ny positive P there is n N (depending on P) such tht, by (8) (9) ƒ R n (x ) ƒ ƒ s(x ) s n (x ) ƒ P for ll n N. Geometriclly, this mens tht ll s n (x ) with n N lie between s(x ) P nd s(x ) P (Fig. 04). Prcticlly, this mens tht in the cse of convergence we cn pproximte the sum s(x ) of () t by s n (x ) s ccurtely s we plese, by tking n lrge enough. x s(x ) ε s(x ) Fig. 04. Inequlity (9) s(x ) + ε Where does power series converge? Now if we choose x x 0 in (), the series reduces to the single term 0 becuse the other terms re zero. Hence the series converges t x 0. In some cses this my be the only vlue of x for which () converges. If there re other vlues of x for which the series converges, these vlues form n intervl, the convergence intervl. This intervl my be finite, s in Fig. 05, with midpoint x 0. Then the series () converges for ll x in the interior of the intervl, tht is, for ll x for which (0) ƒ x x 0 ƒ R nd diverges for ƒ x x 0 ƒ R. The intervl my lso be infinite, tht is, the series my converge for ll x. Divergence Convergence Divergence R R x 0 R x 0 x 0 + R Fig. 05. Convergence intervl (0) of power series with center x 0

6 72 CHAP. 5 Series Solutions of ODEs. Specil Functions The quntity R in Fig. 05 is clled the rdius of convergence (becuse for complex power series it is the rdius of disk of convergence). If the series converges for ll x, we set R (nd >R 0). The rdius of convergence cn be determined from the coefficients of the series by mens of ech of the formuls () () R ^ lim ^ lim m: 2m ƒ m ƒ (b) R m: ` m m ` provided these limits exist nd re not zero. [If these limits re infinite, then () converges only t the center.] EXAMPLE 4 Convergence Rdius R,, 0 For ll three series let m : m0 e x m0 x m0 x 0 x m 2 x x m! 2! Á m >(m )!, ` ` : 0, R m >m! m x m x x 2 Á m, ` `, R m m!x m x 2x 2 Á m (m )!, ` ` m :, R 0. m m! Convergence for ll x (R ) is the best possible cse, convergence in some finite intervl the usul, nd convergence only t the center (R 0) is useless. When do power series solutions exist? Answer: if p, q, r in the ODEs (2) ys p(x)yr q(x)y r(x) hve power series representtions (Tylor series). More precisely, function f (x) is clled nlytic t point x x 0 if it cn be represented by power series in powers of x x 0 with positive rdius of convergence. Using this concept, we cn stte the following bsic theorem, in which the ODE (2) is in stndrd form, tht is, it begins with the ys. If your ODE begins with, sy, h(x)ys, divide it first by h(x) nd then pply the theorem to the resulting new ODE. THEOREM Existence of Power Series Solutions If p, q, nd r in (2) re nlytic t x x 0, then every solution of (2) is nlytic t x x 0 nd cn thus be represented by power series in powers of x x 0 with rdius of convergence R 0. The proof of this theorem requires dvnced complex nlysis nd cn be found in Ref. [A] listed in App.. We mention tht the rdius of convergence R in Theorem is t lest equl to the distnce from the point x x 0 to the point (or points) closest to x 0 t which one of the functions p, q, r, s functions of complex vrible, is not nlytic. (Note tht tht point my not lie on the x-xis but somewhere in the complex plne.)

7 SEC. 5. Power Series Method 73 Further Theory: Opertions on Power Series In the power series method we differentite, dd, nd multiply power series, nd we obtin coefficient recursions (s, for instnce, in Exmple 3) by equting the sum of the coefficients of ech occurring power of x to zero. These four opertions re permissible in the sense explined in wht follows. Proofs cn be found in Sec Termwise Differentition. A power series my be differentited term by term. More precisely: if y(x) m (x x 0 ) m m0 converges for ƒ x x 0 ƒ R, where R 0, then the series obtined by differentiting term by term lso converges for those x nd represents the derivtive yr of y for those x: yr(x) m m m (x x 0 ) m ( ƒ x x 0 ƒ R). Similrly for the second nd further derivtives. 2. Termwise Addition. Two power series my be dded term by term. More precisely: if the series (3) m (x x 0 ) m nd m0 m0 b m (x x 0 ) m hve positive rdii of convergence nd their sums re f (x) nd g(x), then the series m0 ( m b m )(x x 0 ) m converges nd represents f (x) g(x) for ech x tht lies in the interior of the convergence intervl common to ech of the two given series. 3. Termwise Multipliction. Two power series my be multiplied term by term. More precisely: Suppose tht the series (3) hve positive rdii of convergence nd let f (x) nd g(x) be their sums. Then the series obtined by multiplying ech term of the first series by ech term of the second series nd collecting like powers of x x 0, tht is, 0 b 0 ( 0 b b 0 )(x x 0 ) ( 0 b 2 b 2 b 0 )(x x 0 ) 2 Á ( 0 b m b m Á m b 0 )(x x 0 ) m m0 converges nd represents f (x)g(x) for ech x in the interior of the convergence intervl of ech of the two given series.

8 74 CHAP. 5 Series Solutions of ODEs. Specil Functions 4. Vnishing of All Coefficients ( Identity Theorem for Power Series. ) If power series hs positive rdius of convergent convergence nd sum tht is identiclly zero throughout its intervl of convergence, then ech coefficient of the series must be zero. PROBLEM SET 5.. WRITING AND LITERATURE PROJECT. Power Series in Clculus. () Write review (2 3 pges) on power series in clculus. Use your own formultions nd exmples do not just copy from textbooks. No proofs. (b) Collect nd rrnge Mclurin series in systemtic list tht you cn use for your work. 2 5 REVIEW: RADIUS OF CONVERGENCE Determine the rdius of convergence. Show the detils of your work m0 m0 m0 m0 (m )mx m () m k x 2m m x 2m (2m )! 2 3 b m x 2m 6 9 SERIES SOLUTIONS BY HAND Apply the power series method. Do this by hnd, not by CAS, to get feel for the method, e.g., why series my terminte, or hs even powers only, etc. Show the detils. 6. ( x)yr y 7. yr 2xy 8. xyr 3y k ( const) 9. ys y SERIES SOLUTIONS Find power series solution in powers of x. Show the detils. 0. ys yr xy 0. ys yr x 2 y 0 2. ( x 2 )ys 2xyr 2y 0 3. ys ( x 2 )y 0 4. ys 4xyr (4x 2 2)y 0 5. Shifting summtion indices is often convenient or necessry in the power series method. Shift the index so tht the power under the summtion sign is x m. Check by writing the first few terms explicity. s(s ) s 2 x s, s2 ( p )! x p4 6 9 CAS PROBLEMS. IVPs Solve the initil vlue problem by power series. Grph the prtil sums of the powers up to nd including x 5. Find the vlue of the sum s (5 digits) t x. 6. yr 4y, y(0).25, x ys 3xyr 2y 0, y(0), yr(0), x ( x 2 )ys 2xyr 30y 0, y(0) 0, yr(0).875, x (x 2)yr xy, y(0) 4, x CAS Experiment. Informtion from Grphs of Prtil Sums. In numerics we use prtil sums of power series. To get feel for the ccurcy for vrious x, experiment with sin x. Grph prtil sums of the Mclurin series of n incresing number of terms, describing qulittively the brekwy points of these grphs from the grph of sin x. Consider other Mclurin series of your choice p p Fig. 06. CAS Experiment 20. sin x nd prtil sums s 3, s 5, s 7 x

9 SEC. 5.2 Legendre s Eqution. Legendre Polynomils P n (x) Legendre s Eqution. Legendre Polynomils P n (x) Legendre s differentil eqution () ( x 2 )ys 2xyr n(n )y 0 (n constnt) is one of the most importnt ODEs in physics. It rises in numerous problems, prticulrly in boundry vlue problems for spheres (tke quick look t Exmple in Sec. 2.0). The eqution involves prmeter n, whose vlue depends on the physicl or engineering problem. So () is ctully whole fmily of ODEs. For n we solved it in Exmple 3 of Sec. 5. (look bck t it). Any solution of () is clled Legendre function. The study of these nd other higher functions not occurring in clculus is clled the theory of specil functions. Further specil functions will occur in the next sections. Dividing () by x 2, we obtin the stndrd form needed in Theorem of Sec. 5. nd we see tht the coefficients 2x>( x 2 ) nd n(n )>( x 2 ) of the new eqution re nlytic t x 0, so tht we my pply the power series method. Substituting (2) y m0 m x m nd its derivtives into (), nd denoting the constnt n(n ) simply by k, we obtin ( x 2 ) m(m ) m x m2 2x m m x m k m x m 0. By writing the first expression s two seprte series we hve the eqution m2 m m0 m(m ) m x m2 m(m ) m x m 2m m x m k m x m 0. m2 m2 m m0 It my help you to write out the first few terms of ech series explicitly, s in Exmple 3 of Sec. 5.; or you my continue s follows. To obtin the sme generl power x s in ll four series, set m 2 s (thus m s 2) in the first series nd simply write s insted of m in the other three series. This gives (s 2)(s ) s2 x s s(s ) s x s 2s s x s k s x s 0. s0 s2 s s0 ADRIEN-MARIE LEGENDRE ( ), French mthemticin, who becme professor in Pris in 775 nd mde importnt contributions to specil functions, elliptic integrls, number theory, nd the clculus of vritions. His book Éléments de géométrie (794) becme very fmous nd hd 2 editions in less thn 30 yers. Formuls on Legendre functions my be found in Refs. [GenRef] nd [GenRef0].

10 76 CHAP. 5 Series Solutions of ODEs. Specil Functions (Note tht in the first series the summtion begins with s 0.) Since this eqution with the right side 0 must be n identity in x if (2) is to be solution of (), the sum of the coefficients of ech power of x on the left must be zero. Now x 0 occurs in the first nd fourth series only, nd gives [remember tht k n(n ) ] (3) 2 # 2 n(n ) 0 0. x occurs in the first, third, nd fourth series nd gives (3b) 3 # 2 3 [2 n(n )] 0. The higher powers x 2, x 3, Á occur in ll four series nd give (3c) (s 2)(s ) s2 [s(s ) 2s n(n )] s 0. The expression in the brckets [ Á ] cn be written (n s)(n s ), s you my redily verify. Solving (3) for 2 nd (3b) for 3 s well s (3c) for s2, we obtin the generl formul (4) (n s)(n s ) s2 s (s 2)(s ) (s 0,, Á ). This is clled recurrence reltion or recursion formul. (Its derivtion you my verify with your CAS.) It gives ech coefficient in terms of the second one preceding it, except for nd, which re left s rbitrry constnts. We find successively 0 n(n ) 2 0 2! (n 2)(n 3) 4 4 # 2 3 (n 2)n(n )(n 3) 4! 0 (n )(n 2) 3 3! (n 3)(n 4) 5 5 # 3 4 (n 3)(n )(n 2)(n 4) 5! nd so on. By inserting these expressions for the coefficients into (2) we obtin (5) y(x) 0 y (x) y 2 (x) where (6) n(n ) (n 2)n(n )(n 3) y (x) x 2 x 4 2! 4! Á (n )(n 2) (n 3)(n )(n 2)(n 4) (7) y 2 (x) x x 3 x 5 3! 5! Á.

11 SEC. 5.2 Legendre s Eqution. Legendre Polynomils P n (x) 77 These series converge for ƒ x ƒ (see Prob. 4; or they my terminte, see below). Since (6) contins even powers of x only, while (7) contins odd powers of x only, the rtio y >y 2 is not constnt, so tht y nd y 2 re not proportionl nd re thus linerly independent solutions. Hence (5) is generl solution of () on the intervl x. Note tht x re the points t which x 2 0, so tht the coefficients of the stndrdized ODE re no longer nlytic. So it should not surprise you tht we do not get longer convergence intervl of (6) nd (7), unless these series terminte fter finitely mny powers. In tht cse, the series become polynomils. Polynomil Solutions. Legendre Polynomils The reduction of power series to polynomils is gret dvntge becuse then we hve solutions for ll x, without convergence restrictions. For specil functions rising s solutions of ODEs this hppens quite frequently, leding to vrious importnt fmilies of polynomils; see Refs. [GenRef], [GenRef0] in App.. For Legendre s eqution this hppens when the prmeter n is nonnegtive integer becuse then the right side of (4) is zero for s n, so tht n2 0, n4 0, n6 0, Á. Hence if n is even, y (x) reduces to polynomil of degree n. If n is odd, the sme is true for y 2 (x). These polynomils, multiplied by some constnts, re clled Legendre polynomils nd re denoted by P n (x). The stndrd choice of such constnts is done s follows. We choose the coefficient of the highest power x n s (8) n (2n)! (n positive integer) 2 n (n!) 2 # 3 # 5 Á (2n ) n! (nd n if n 0). Then we clculte the other coefficients from (4), solved for s in terms of, tht is, s2 n P n (x) (9) (s 2)(s ) s (n s)(n s ) s2 (s n 2). The choice (8) mkes p n () for every n (see Fig. 07); this motivtes (8). From (9) with s n 2 nd (8) we obtin n(n ) n2 2(2n ) n(n ) n # (2n)! 2(2n ) 2 n (n!) 2 Using (2n)! 2n(2n )(2n 2)! in the numertor nd n! n(n )! nd n! n(n )(n 2)! in the denomintor, we obtin n(n )2n(2n )(2n 2)! n2 2(2n )2 n n(n )! n(n )(n 2)!. n(n )2n(2n ) cncels, so tht we get (2n 2)! n2 2 n (n )! (n 2)!.

12 78 CHAP. 5 Series Solutions of ODEs. Specil Functions Similrly, (n 2)(n 3) n4 n2 4(2n 3) (2n 4)! 2 n 2! (n 2)! (n 4)! nd so on, nd in generl, when n 2m 0, (0) n2m () m (2n 2m)! 2 n m! (n m)! (n 2m)!. The resulting solution of Legendre s differentil eqution () is clled the Legendre polynomil of degree n nd is denoted by P n (x). From (0) we obtin () P n (x) M m0 () m (2n 2m)! 2 n m! (n m)! (n 2m)! x n2m (2n)! 2 n (n!) x (2n 2)! n 2 2 n! (n )! (n 2)! x n2 Á where M n>2 or (n )>2, whichever is n integer. The first few of these functions re (Fig. 07) P 0 (x), P (x) x () P 2 (x) 2 (3x 2 ), P 3 (x) 2 (5x 3 3x) P 4 (x) 8 (35x 4 30x 2 3), P 5 (x) 8 (63x 5 70x 3 5x) nd so on. You my now progrm () on your CAS nd clculte P n (x) s needed. P n (x) P 0 P P 4 x P 3 P 2 Fig. 07. Legendre polynomils

13 SEC. 5.2 Legendre s Eqution. Legendre Polynomils P n (x) 79 The Legendre polynomils P n (x) re orthogonl on the intervl x, bsic property to be defined nd used in mking up Fourier Legendre series in the chpter on Fourier series (see Secs..5.6). PROBLEM SET LEGENDRE POLYNOMIALS AND FUNCTIONS. Legendre functions for n 0. Show tht (6) with n 0 gives P 0 (x) nd (7) gives (use x 2 x 2 3 x 3 Á ln ( x) ) y 2 (x) x 3 x 3 5 x 5 Á 2 Verify this by solving () with n 0, setting z yr nd seprting vribles. 2. Legendre functions for n. Show tht (7) with n gives y 2 (x) P (x) x nd (6) gives y x 2 3 x 4 5 x 6 Á 3. Specil n. Derive (r) from (). 4. Legendre s ODE. Verify tht the polynomils in (r) stisfy (). 5. Obtin nd. P 6 x x ln 2 x. P CAS PROBLEMS 6. Grph P 2 (x), Á, P 0 (x) on common xes. For wht x (pproximtely) nd n 2, Á, 0 is ƒ P n (x) ƒ 2? 7. From wht n on will your CAS no longer produce fithful grphs of P n (x)? Why? 8. Grph Q 0 (x), Q (x), nd some further Legendre functions. 9. Substitute s x s s x s s2 x s2 into Legendre s eqution nd obtin the coefficient recursion (4). 0. TEAM PROJECT. Generting Functions. Generting functions ply significnt role in modern pplied mthemtics (see [GenRef5]). The ide is simple. If we wnt to study certin sequence ( f n (x)) nd cn find function G(u, x) n0 f n (x)u n, ln x x. we my obtin properties of ( f n (x)) from those of G, which genertes this sequence nd is clled generting function of the sequence. () Legendre polynomils. Show tht (2) G(u, x) 2 2xu u 2 P n (x)u n is generting function of the Legendre polynomils. Hint: Strt from the binomil expnsion of > v, then set v 2xu u 2, multiply the powers of 2xu u 2 out, collect ll the terms involving u n, nd verify tht the sum of these terms is P n (x)u n. (b) Potentil theory. Let A nd A 2 be two points in spce (Fig. 08, r 2 0). Using (2), show tht r 2r 2 r 2 2 2r r 2 cos u r 2 P m (cos u) r m r b. 2 m0 This formul hs pplictions in potentil theory. ( Q>r is the electrosttic potentil t A 2 due to chrge Q locted t A. And the series expresses >r in terms of the distnces of A nd A 2 from ny origin O nd the ngle u between the segments OA nd OA 2.) 0 θ r 2 r A Fig. 08. Tem Project 0 r n0 (c) Further pplictions of (2). Show tht P n (), P n () () n, nd P 2n (0) () n # # 3 Á, P 2n (0) 0 (2n )>[2 # 4 Á (2n)]. 5 FURTHER FORMULAS. ODE. Find solution of ( 2 x 2 )ys 2xyr n(n )y 0, 0, by reduction to the Legendre eqution. 2. Rodrigues s formul (3) 2 Applying the binomil theorem to (x 2 ) n, differentiting it n times term by term, nd compring the result with (), show tht (3) P n (x) 2 n n! A 2 d n dx n [(x 2 ) n ]. 2 OLINDE RODRIGUES (794 85), French mthemticin nd economist.

14 80 CHAP. 5 Series Solutions of ODEs. Specil Functions 3. Rodrigues s formul. Obtin (r) from (3). 4. Bonnet s recursion. 3 Differentiting (3) with respect to u, using (3) in the resulting formul, nd compring coefficients of u n, obtin the Bonnet recursion. (4) (n )P n (x) (2n )xp n (x) np n (x), where n, 2, Á. This formul is useful for computtions, the loss of significnt digits being smll (except ner zeros). Try (4) out for few computtions of your own choice. 5. Associted Legendre functions P k n (x) re needed, e.g., in quntum physics. They re defined by (5) P k n (x) ( x 2 ) k>2 dk p n (x) dx k nd re solutions of the ODE (6) ( x 2 )ys 2xyr q(x)y 0 where q(x) n(n ) k 2 >( x 2 ). Find P (x), P P 2, nd P 2 2 (x), 2 (x) 4 (x) nd verify tht they stisfy (6). 5.3 Extended Power Series Method: Frobenius Method Severl second-order ODEs of considerble prcticl importnce the fmous Bessel eqution mong them hve coefficients tht re not nlytic (definition in Sec. 5.), but re not too bd, so tht these ODEs cn still be solved by series (power series times logrithm or times frctionl power of x, etc.). Indeed, the following theorem permits n extension of the power series method. The new method is clled the Frobenius method. 4 Both methods, tht is, the power series method nd the Frobenius method, hve gined in significnce due to the use of softwre in ctul clcultions. THEOREM Frobenius Method Let b(x) nd c(x) be ny functions tht re nlytic t x 0. Then the ODE () ys b(x) x c(x) yr x 2 y 0 hs t lest one solution tht cn be represented in the form (2) y(x) x r m x m x r ( 0 x 2 x 2 Á ) m0 ( 0 0) where the exponent r my be ny (rel or complex) number (nd r is chosen so tht 0 0). The ODE () lso hs second solution (such tht these two solutions re linerly independent) tht my be similr to (2) (with different r nd different coefficients) or my contin logrithmic term. (Detils in Theorem 2 below.) 3 OSSIAN BONNET (89 892), French mthemticin, whose min work ws in differentil geometry. 4 GEORG FROBENIUS (849 97), Germn mthemticin, professor t ETH Zurich nd University of Berlin, student of Krl Weierstrss (see footnote, Sect. 5.5). He is lso known for his work on mtrices nd in group theory. In this theorem we my replce x by x x 0 with ny number x 0. The condition 0 0 is no restriction; it simply mens tht we fctor out the highest possible power of x. The singulr point of () t x 0 is often clled regulr singulr point, term confusing to the student, which we shll not use.

15 SEC. 5.3 Extended Power Series Method: Frobenius Method 8 For exmple, Bessel s eqution (to be discussed in the next section) ys x yr x2 v 2 x 2 b y 0 (v prmeter) is of the form () with b(x) nd c(x) x 2 v 2 nlytic t x 0, so tht the theorem pplies. This ODE could not be hndled in full generlity by the power series method. Similrly, the so-clled hypergeometric differentil eqution (see Problem Set 5.3) lso requires the Frobenius method. The point is tht in (2) we hve power series times single power of x whose exponent r is not restricted to be nonnegtive integer. (The ltter restriction would mke the whole expression power series, by definition; see Sec. 5..) The proof of the theorem requires dvnced methods of complex nlysis nd cn be found in Ref. [A] listed in App.. Regulr nd Singulr Points. A regulr point of the ODE The following terms re prcticl nd commonly used. is point the ODE x 0 ys p(x)yr q(x)y 0 t which the coefficients p nd q re nlytic. Similrly, regulr point of ~ h (x)ys p ~ (x)yr(x) q ~ (x)y 0 ~ ~ is n x t which h re nlytic nd (so wht we cn divide by h ~, p ~, q ~ 0 h (x 0 ) 0 nd get the previous stndrd form). Then the power series method cn be pplied. If x 0 is not regulr point, it is clled singulr point. Indicil Eqution, Indicting the Form of Solutions We shll now explin the Frobenius method for solving (). Multipliction of () by gives the more convenient form x 2 (r) x 2 ys xb(x)yr c(x)y 0. We first expnd b(x) nd c(x) in power series, b(x) b 0 b x b 2 x 2 Á, c(x) c 0 c x c 2 x 2 Á or we do nothing if b(x) nd c(x) re polynomils. Then we differentite (2) term by term, finding yr(x) (m r) m x mr x r 3r 0 (r ) x Á 4 m0 (2*) ys(x) (m r)(m r ) m x mr2 m0 x r2 3r(r ) 0 (r )r x Á 4.

16 82 CHAP. 5 Series Solutions of ODEs. Specil Functions By inserting ll these series into (r) we obtin (3) x r [r(r ) 0 Á ] (b 0 b x Á ) x r (r 0 Á ) (c 0 c x Á ) x r ( 0 x Á ) 0. We now equte the sum of the coefficients of ech power x r, x r, x r2, Á to zero. This yields system of equtions involving the unknown coefficients m. The smllest power is x r nd the corresponding eqution is [r(r ) b 0 r c 0 ] 0 0. Since by ssumption, the expression in the brckets [ Á 0 0 ] must be zero. This gives (4) r (r ) b 0 r c 0 0. This importnt qudrtic eqution is clled the indicil eqution of the ODE (). Its role is s follows. The Frobenius method yields bsis of solutions. One of the two solutions will lwys be of the form (2), where r is root of (4). The other solution will be of form indicted by the indicil eqution. There re three cses: Cse. Distinct roots not differing by n integer, 2, 3, Á. Cse 2. A double root. Cse 3. Roots differing by n integer, 2, 3, Á. Cses nd 2 re not unexpected becuse of the Euler Cuchy eqution (Sec. 2.5), the simplest ODE of the form (). Cse includes complex conjugte roots r nd r 2 r becuse r r 2 r r 2i Im r is imginry, so it cnnot be rel integer. The form of bsis will be given in Theorem 2 (which is proved in App. 4), without generl theory of convergence, but convergence of the occurring series cn be tested in ech individul cse s usul. Note tht in Cse 2 we must hve logrithm, wheres in Cse 3 we my or my not. THEOREM 2 Frobenius Method. Bsis of Solutions. Three Cses Suppose tht the ODE () stisfies the ssumptions in Theorem. Let r nd r 2 be the roots of the indicil eqution (4). Then we hve the following three cses. Cse. Distinct Roots Not Differing by n Integer. A bsis is (5) y (x) x r ( 0 x 2 x 2 Á ) nd (6) y 2 (x) x r 2 (A 0 A x A 2 x 2 Á ) with coefficients obtined successively from (3) with r r nd r r 2, respectively.

17 SEC. 5.3 Extended Power Series Method: Frobenius Method 83 Cse 2. Double Root r r 2 r. A bsis is (7) y (x) x r ( 0 x 2 x 2 Á ) [r 2 ( b 0)] (of the sme generl form s before) nd (8) y 2 (x) y (x) ln x x r (A x A 2 x 2 Á ) (x 0). Cse 3. Roots Differing by n Integer. A bsis is (9) y (x) x r ( 0 x 2 x 2 Á ) (of the sme generl form s before) nd (0) y 2 (x) ky (x) ln x x r 2 (A 0 A x A 2 x 2 Á ), where the roots re so denoted tht r r 2 0 nd k my turn out to be zero. Typicl Applictions Techniclly, the Frobenius method is similr to the power series method, once the roots of the indicil eqution hve been determined. However, (5) (0) merely indicte the generl form of bsis, nd second solution cn often be obtined more rpidly by reduction of order (Sec. 2.). EXAMPLE Euler Cuchy Eqution, Illustrting Cses nd 2 nd Cse 3 without Logrithm For the Euler Cuchy eqution (Sec. 2.5) x 2 ys b 0 xyr c 0 y 0 ( b 0, c 0 constnt) substitution of y x r gives the uxiliry eqution r(r ) b 0 r c 0 0, which is the indicil eqution [nd y x r is very specil form of (2)!]. For different roots r, r 2 we get bsis y, nd for double root r we get bsis x r, x r x r, y 2 x r 2 ln x. Accordingly, for this simple ODE, Cse 3 plys no extr role. EXAMPLE 2 Illustrtion of Cse 2 (Double Root) Solve the ODE () x(x )ys (3x )yr y 0. (This is specil hypergeometric eqution, s we shll see in the problem set.) Solution. Writing () in the stndrd form (), we see tht it stisfies the ssumptions in Theorem. [Wht re b(x) nd c(x) in ()?] By inserting (2) nd its derivtives (2*) into () we obtin (2) (m r)(m r ) m x mr (m r)(m r ) m x mr m0 m0 3 (m r) m x mr (m r) m x mr m x mr 0. m0 m0 m0

18 84 CHAP. 5 Series Solutions of ODEs. Specil Functions The smllest power is to zero we hve x r, occurring in the second nd the fourth series; by equting the sum of its coefficients [r(r ) r] 0 0, thus r 2 0. Hence this indicil eqution hs the double root r 0. First Solution. We insert this vlue r 0 into (2) nd equte the sum of the coefficients of the power x s to zero, obtining s(s ) s (s )s s 3s s (s ) s s 0 thus. Hence 0 2 Á s s, nd by choosing 0 we obtin the solution y (x) x m x m0 ( ƒ x ƒ ). Second Solution. We get second independent solution y 2 by the method of reduction of order (Sec. 2.), substituting y 2 uy nd its derivtives into the eqution. This leds to (9), Sec. 2., which we shll use in this exmple, insted of strting reduction of order from scrtch (s we shll do in the next exmple). In (9) of Sec. 2. we hve p (3x )>(x 2 x), the coefficient of yr in () in stndrd form. By prtil frctions, Hence (9), Sec. 2., becomes 3x p dx x(x ) dx 2 x b dx 2 ln (x ) ln x. x ur U y 2 e p dx (x )2 (x ) 2 x x, u ln x, y 2 uy ln x x. y nd y 2 re shown in Fig. 09. These functions re linerly independent nd thus form bsis on the intervl 0 x (s well s on x ). y y y x Fig. 09. Solutions in Exmple 2 EXAMPLE 3 Cse 3, Second Solution with Logrithmic Term Solve the ODE (3) (x 2 x)ys xyr y 0. Solution. Substituting (2) nd (2*) into (3), we hve (x 2 x) (m r)(m r ) m x mr2 x (m r) m x mr m x mr 0. m0 m0 m0

19 SEC. 5.3 Extended Power Series Method: Frobenius Method 85 We now tke, x, nd x inside the summtions nd collect ll terms with power nd simplify lgebriclly, x 2 x mr (m r ) 2 m x mr (m r)(m r ) m x mr 0. m0 m0 In the first series we set m s nd in the second m s, thus s m. Then (4) (s r ) 2 s x sr (s r )(s r) s x sr 0. s0 s The lowest power is (tke s in the second series) nd gives the indicil eqution x r r(r ) 0. The roots re r nd r 2 0. They differ by n integer. This is Cse 3. First Solution. From (4) with r r we hve This gives the recurrence reltion 3s 2 s (s 2)(s ) s 4x s 0. s0 s s 2 (s 2)(s ) s (s 0,, Á ). Hence successively. Tking, we get s first solution y x r 0, 2 0, Á 0 0 x. Second Solution. Applying reduction of order (Sec. 2.), we substitute y 2 y u xu, yr 2 xur u nd ys 2 xus 2ur into the ODE, obtining (x 2 x)(xus 2ur) x(xur u) xu 0. xu drops out. Division by x nd simplifiction give (x 2 x)us (x 2)ur 0. From this, using prtil frctions nd integrting (tking the integrtion constnt zero), we get us ur x 2 x 2 x 2 x x, x ln ur ln 2 2 x 2. Tking exponents nd integrting (gin tking the integrtion constnt zero), we obtin ur x x 2 x x 2, u ln x x, y 2 xu x ln x. y y 2 y 2 nd re linerly independent, nd hs logrithmic term. Hence nd constitute bsis of solutions for positive x. y y 2 The Frobenius method solves the hypergeometric eqution, whose solutions include mny known functions s specil cses (see the problem set). In the next section we use the method for solving Bessel s eqution.

20 86 CHAP. 5 Series Solutions of ODEs. Specil Functions PROBLEM SET 5.3. WRITING PROJECT. Power Series Method nd Frobenius Method. Write report of 2 3 pges explining the difference between the two methods. No proofs. Give simple exmples of your own. 2 3 FROBENIUS METHOD Find bsis of solutions by the Frobenius method. Try to identify the series s expnsions of known functions. Show the detils of your work (x 2) 2 ys (x 2)yr y 0 xys 2yr xy 0 xys y 0 xys (2x )yr (x )y 0 xys 2x 3 yr (x 2 2)y 0 ys (x )y 0 xys yr xy 0 2x(x )ys (x )yr y 0 xys 2yr 4xy 0 xys (2 2x)yr (x 2)y 0 x 2 ys 6xyr (4x 2 6)y 0 xys ( 2x)yr (x )y 0 4. TEAM PROJECT. Hypergeometric Eqution, Series, nd Function. Guss s hypergeometric ODE 5 is (5) Here,, b, c re constnts. This ODE is of the form p 2 ys p yr p 0 y 0, where p 2, p, p 0 re polynomils of degree 2,, 0, respectively. These polynomils re written so tht the series solution tkes most prcticl form, nmely, (6) x( x)ys [c ( b )x]yr by 0. y (x) b ( )b(b ) x x 2! c 2! c(c ) ( )( 2)b(b )(b 2) x 3. 3! c(c )(c 2) Á This series is clled the hypergeometric series. Its sum y (x) is clled the hypergeometric function nd is denoted by F(, b, c; x). Here, c 0,, 2, Á. By choosing specific vlues of, b, c we cn obtin n incredibly lrge number of specil functions s solutions of (5) [see the smll smple of elementry functions in prt (c)]. This ccounts for the importnce of (5). () Hypergeometric series nd function. Show tht the indicil eqution of (5) hs the roots r 0 nd r 2 c. Show tht for r 0 the Frobenius method gives (6). Motivte the nme for (6) by showing tht F (,, ; x) F(, b, b; x) F (,, ; x) x. (b) Convergence. For wht or b will (6) reduce to polynomil? Show tht for ny other, b, c ( c 0,, 2, Á ) the series (6) converges when ƒ x ƒ. (c) Specil cses. Show tht Find more such reltions from the literture on specil functions, for instnce, from [GenRef] in App.. (d) Second solution. Show tht for r 2 c the Frobenius method yields the following solution (where c 2, 3, 4, Á ): ( c )(b c ) y 2 (x) x c x (7)! (c 2) ( c )( c 2)(b c )(b c 2) x 2 2! (c 2)(c 3) Show tht ( x) n F (n, b, b; x), ( x) n nxf ( n,, 2; x), rctn x xf( 2,, 3 2 ; x 2 ) rcsin x xf( 2, 2, 3 2 ; x 2 ), ln ( x) xf(,, 2; x), x ln x 2xF( 2,, 3 2 ; x 2 ). y 2 (x) x c F( c, b c, 2 c; x). (e) On the generlity of the hypergeometric eqution. Show tht (8) (t 2 At B) ## y (Ct D)y # Ky 0 Á b. 5 CARL FRIEDRICH GAUSS ( ), gret Germn mthemticin. He lredy mde the first of his gret discoveries s student t Helmstedt nd Göttingen. In 807 he becme professor nd director of the Observtory t Göttingen. His work ws of bsic importnce in lgebr, number theory, differentil equtions, differentil geometry, non-eucliden geometry, complex nlysis, numeric nlysis, stronomy, geodesy, electromgnetism, nd theoreticl mechnics. He lso pved the wy for generl nd systemtic use of complex numbers.

21 SEC. 5.4 Bessel s Eqution. Bessel Functions J (x) 87 # with y dy>dt, etc., constnt A, B, C, D, K, nd t HYPERGEOMETRIC ODE At B (t t )(t t 2 ), t t 2, cn be reduced to the hypergeometric eqution with independent vrible Find generl solution in terms of hypergeometric functions. x t t 5. 2x( x)ys ( 6x)yr 2y 0 t 2 t 6. x( x)ys ( 2 2x)yr 2y 0 nd prmeters relted by Ct D c(t 2 t ), 7. 4x( x)ys C b, K b. From this you see tht (5) 8. 4(t 2 3t 2)y ## yr 2y # 8y 0 y 0 is normlized form of the more generl (8) nd tht vrious cses of (8) cn thus be solved in terms 9. 2(t 2 5t 6)y ## (2t 3)y # 8y 0 of hypergeometric functions t( t)y ## ty # y Bessel s Eqution. Bessel Functions J (x) One of the most importnt ODEs in pplied mthemtics in Bessel s eqution, 6 () x 2 ys xyr (x 2 2 )y 0 where the prmeter (nu) is given rel number which is positive or zero. Bessel s eqution often ppers if problem shows cylindricl symmetry, for exmple, s the membrnes in Sec.2.9. The eqution stisfies the ssumptions of Theorem. To see this, divide () by x 2 to get the stndrd form ys yr>x ( 2 >x 2 )y 0. Hence, ccording to the Frobenius theory, it hs solution of the form (2) y(x) m0 m x mr ( 0 0). Substituting (2) nd its first nd second derivtives into Bessel s eqution, we obtin m0 (m r)(m r ) m x mr m0 (m r) m x mr We equte the sum of the coefficients of to zero. Note tht this power corresponds to m s in the first, second, nd fourth series, nd to m s 2 in the third series. Hence for s 0 nd s, the third series does not contribute since m 0. m0 x sr m x mr2 2 m x mr 0. m0 x sr 6 FRIEDRICH WILHELM BESSEL ( ), Germn stronomer nd mthemticin, studied stronomy on his own in his spre time s n pprentice of trde compny nd finlly becme director of the new Königsberg Observtory. Formuls on Bessel functions re contined in Ref. [GenRef0] nd the stndrd tretise [A3].

22 88 CHAP. 5 Series Solutions of ODEs. Specil Functions For s 2, 3, Á ll four series contribute, so tht we get generl formul for ll these s. We find () r(r ) 0 r (s 0) (3) (b) (c) (r )r (r ) 2 0 (s r)(s r ) s (s r) s s2 2 s 0 (s ) (s 2, 3, Á ). From (3) we obtin the indicil eqution by dropping, (4) (r )(r ) 0. The roots re r ( 0) nd r 2. Coefficient Recursion for r r v. For r, Eq. (3b) reduces to (2 ) 0. Hence 0 since 0. Substituting r in (3c) nd combining the three terms contining gives simply s 0 (5) (s 2)s s s2 0. Since nd, it follows from (5) tht 3 0, 5 0, Á 0 0. Hence we hve to del only with even-numbered coefficients with s 2m. For s 2m, Eq. (5) becomes Solving for 2m (2m 2)2m 2m 2m2 0. gives the recursion formul s (6), m, 2, Á 2m. 2 2 m( m) 2m2 From (6) we cn now determine 2, 4, Á successively. This gives nd so on, nd in generl () (v 2) ! ( )( 2) (7) 2m () m 0 2 2m m! ( )( 2) Á (m), m, 2, Á. Bessel Functions J n (x) for Integer n Integer vlues of v re denoted by n. This is stndrd. For nthe reltion (7) becomes () m 0 (8) m, 2, Á 2m. 2 2m m! (n )(n 2) Á (n m),

23 SEC. 5.4 Bessel s Eqution. Bessel Functions J (x) 89 0 is still rbitrry, so tht the series (2) with these coefficients would contin this rbitrry fctor 0. This would be highly imprcticl sitution for developing formuls or computing vlues of this new function. Accordingly, we hve to mke choice. The choice 0 would be possible. A simpler series (2) could be obtined if we could bsorb the growing product (n )(n 2) Á (n m) into fctoril function (n m)! Wht should be our choice? Our choice should be (9) 0 2 n n! becuse then n! (n ) Á (n m) (n m)! in (8), so tht (8) simply becomes () m (0) m, 2, Á 2m. 2 2mn m! (n m)!, By inserting these coefficients into (2) nd remembering tht c 0, c 3 0, Á prticulr solution of Bessel s eqution tht is denoted by J n (x): we obtin () J n (x) x n m0 () m x 2m 2 2mn m! (n m)! (n 0). J n (x) is clled the Bessel function of the first kind of order n. The series () converges for ll x, s the rtio test shows. Hence J n (x) is defined for ll x. The series converges very rpidly becuse of the fctorils in the denomintor. EXAMPLE Bessel Functions J 0 (x) nd J (x) For n 0 we obtin from () the Bessel function of order 0 (2) () m x 2m 2 J 0 (x) 2 2m (m!) x (!) x (2!) x (3!) Á 2 m0 which looks similr to cosine (Fig. 0). For n we obtin the Bessel function of order (3) J (x) m0 () m x 2m, 2 2m m! (m )! x 2 x 3 2 3! 2! x ! 3! x ! 4! Á which looks similr to sine (Fig. 0). But the zeros of these functions re not completely regulrly spced (see lso Tble A in App. 5) nd the height of the wves decreses with incresing x. Heuristiclly, n 2 >x 2 in () in stndrd form [() divided by x 2 ] is zero (if n 0) or smll in bsolute vlue for lrge x, nd so is yr>x, so tht then Bessel s eqution comes close to ys y 0, the eqution of cos x nd sin x; lso yr>x cts s dmping term, in prt responsible for the decrese in height. One cn show tht for lrge x, (4) J n (x) B 2 px cos x np 2 p 4 b where is red symptoticlly equl nd mens tht for fixed n the quotient of the two sides pproches s x :.

24 90 CHAP. 5 Series Solutions of ODEs. Specil Functions J J x Fig. 0. Bessel functions of the first kind J 0 nd J Formul (4) is surprisingly ccurte even for smller x (0). For instnce, it will give you good strting vlues in computer progrm for the bsic tsk of computing zeros. For exmple, for the first three zeros of J 0 you obtin the vlues (2.405 exct to 3 decimls, error 0.049), (5.520, error 0.022), (8.654, error 0.05), etc. Bessel Functions J (x) for ny 0. Gmm Function We now proceed from integer n to ny 0. We hd 0 >(2 n n!) in (9). So we hve to extend the fctoril function n! to ny 0. For this we choose (5) 0 2 ( ) with the gmm function ( ) defined by (6) ( ) e t t dt 0 ( ). (CAUTION! Note the convention on the left but in the integrl.) Integrtion by prts gives ( ) e t t `. 0 e t t dt 0 () This is the bsic functionl reltion of the gmm function (7) ( ) (). Now from (6) with 0 nd then by (7) we obtin () e t dt e t ` 0 () 0 nd then (2) # ()!, (3) 2() 2! nd in generl 0 0 (8) (n ) n! (n 0,, Á ).

25 SEC. 5.4 Bessel s Eqution. Bessel Functions J (x) 9 Hence the gmm function generlizes the fctoril function to rbitrry positive. Thus (5) with n grees with (9). Furthermore, from (7) with given by (5) we first hve () m 2m. 2 2m m! ( )( 2) Á (m)2 ( ) Now (7) gives ( )( ) ( 2), ( 2)( 2) ( 3) nd so on, so tht ( )( 2) Á (m)( ) ( m ). Hence becuse of our (stndrd!) choice (5) of (9) 2m () m. 2 2m m! ( m ) the coefficients (7) re simply With these coefficients nd r r we get from (2) prticulr solution of (), denoted by J (x) nd given by (20) J (x) x () m x 2m. m0 2 2m m! ( m ) J (x) is clled the Bessel function of the first kind of order. The series (20) converges for ll x, s one cn verify by the rtio test. Discovery of Properties from Series 0 Bessel functions re model cse for showing how to discover properties nd reltions of functions from series by which they re defined. Bessel functions stisfy n incredibly lrge number of reltionships look t Ref. [A3] in App. ; lso, find out wht your CAS knows. In Theorem 3 we shll discuss four formuls tht re bckbones in pplictions nd theory. 0 THEOREM Derivtives, Recursions The derivtive of J (x) with respect to x cn be expressed by J (x) or J (x) by the formuls (2) () [x J (x)]r x J (x) (b) [x J (x)]r x J (x). Furthermore, J (x) nd its derivtive stisfy the recurrence reltions (2) (c) (d) J (x) J (x) 2 x J (x) J (x) J (x) 2Jr (x).

26 92 CHAP. 5 Series Solutions of ODEs. Specil Functions PROOF () We multiply (20) by nd tke under the summtion sign. Then we hve We now differentite this, cncel fctor 2, pull out, nd use the functionl reltionship ( m) ( m)( m) [see (7)]. Then (20) with insted of shows tht we obtin the right side of (2). Indeed, (x J )r m0 (b) Similrly, we multiply (20) by, so tht in (20) cncels. Then we differentite, cncel 2m, nd use m! m(m )!. This gives, with m s, (x J )r m x x J (x) m0 x 2 () m 2(m )x 2m2 2 2m m! ( m ) x () m x 2m () m x 2m2 2 2m m! ( m ). Eqution (20) with insted of nd s insted of m shows tht the expression on the right is x J (x). This proves (2b). (c), (d) We perform the differentition in (2). Then we do the sme in (2b) nd multiply the result on both sides by x 2. This gives (*) x J x Jr x J (b*) x J x Jr x J. Substrcting (b*) from (*) nd dividing the result by (b*) nd dividing the result by x gives (2d). x x 2 x x m0 2 2m (m )! ( m ) s0 x () m x 2m 2 2m m! ( m). () s x 2s 2 2s s! ( s 2). gives (2c). Adding (*) nd EXAMPLE 2 Appliction of Theorem in Evlution nd Integrtion Formul (2c) cn be used recursively in the form for clculting Bessel functions of higher order from those of lower order. For instnce, J 2 (x) 2J (x)>x J 0 (x), so tht J 2 cn be obtined from tbles of J 0 nd J (in App. 5 or, more ccurtely, in Ref. [GenRef] in App. ). To illustrte how Theorem helps in integrtion, we use (2b) with 3 integrted on both sides. This evlutes, for instnce, the integrl A tble of J 3 I 2 J (x) 2 x J (x) J (x) 2 x 3 J 4 (x) dx x 3 J 3 (x) 2 (on p. 398 of Ref. [GenRef]) or your CAS will give you. 8 J 3(2) J 3 () # Your CAS (or humn computer in precomputer times) obtins J 3 from (2), first using (2c) with 2, tht is, J then (2c) with, tht is, J 2 2x 3 4x J 2 J, J J 0. Together,

27 SEC. 5.4 Bessel s Eqution. Bessel Functions J (x) 93 This is wht you get, for instnce, with Mple if you type int ( Á ). And if you type evlf(int ( Á )), you obtin , in greement with the result ner the beginning of the exmple. Bessel Functions with Hlf-Integer Are Elementry We discover this remrkble fct s nother property obtined from the series (20) nd confirm it in the problem set by using Bessel s ODE. EXAMPLE 3 Elementry Bessel Functions with. The Vlue ( 2, 3 2, 5 2, Á 2 ) We first prove (Fig. ) 2 I x 3 (4x (2x J J 0 ) J ) J (2) 2J 0 (2) J (2)4 38J () 4J 0 () J ()4 8 J (2) 4 J 0 (2) 7J () 4J 0 (). J J (22) 2 () J >2 (x) B px sin x, (b) J 2 >2(x) cos x. B px The series (20) with 2 is J >2 (x) x The denomintor cn be written s product AB, where (use (6) in B) here we used (proof below) m0 () m x 2m 2 2m>2 m! (m 3 2) 2 B x m0 () m x 2m 2 2m m! (m 3 2). A 2 m m! 2m(2m 2)(2m 4) Á 4 # 2, B 2 m (m 3 2) 2 m (m 2)(m 2) Á 3 2 # 2( 2) (2m )(2m ) Á 3 # # p ; (23) ( 2) p. The product of the right sides of A nd B cn be written Hence AB (2m )2m (2m ) Á 3 # 2 # p (2m )!p. 2 J >2 (x) B px m0 () m x 2m (2m )! 2 sin x. B px 0 2π 4π 6π x Fig.. Bessel functions J >2 nd J >2

28 94 CHAP. 5 Series Solutions of ODEs. Specil Functions This proves (22). Differentition nd the use of (2) with now gives This proves (22b). From (22) follow further formuls successively by (2c), used s in Exmple 2. We finlly prove ( 2) p by stndrd trick worth remembering. In (5) we set t u 2. Then dt 2u du nd 2 b e t t >2 dt 2 e u2 du. 0 2 [x J >2 (x)]r B 2 p cos x x >2 J >2 (x). We squre on both sides, write v insted of u in the second integrl, nd then write the product of the integrls s double integrl: 2 2 b 4 e u2 du 0 e v2 dv 4 0 e (u2 v 2) du dv. 0 0 We now use polr coordintes r, u by setting u r cos u, v r sin u. Then the element of re is du dv r dr du nd we hve to integrte over r from 0 to nd over u from 0 to p>2 (tht is, over the first qudrnt of the uv-plne): 2 p>2 2 b 4 e r 2 r dr du 4 # p 2 e r 2 r dr 2 2 b 2 er ` p By tking the squre root on both sides we obtin (23). Generl Solution. Liner Dependence For generl solution of Bessel s eqution () in ddition to J we need second linerly independent solution. For not n integer this is esy. Replcing by in (20), we hve (24) J (x) x () m x 2m. m0 2 2m m! (m ) Since Bessel s eqution involves, the functions J nd J re solutions of the eqution for the sme. If is not n integer, they re linerly independent, becuse the first terms in (20) nd in (24) re finite nonzero multiples of x nd x. Thus, if is not n integer, generl solution of Bessel s eqution for ll x 0 is 2 y(x) c J (x) c 2 J (x) This cnnot be the generl solution for n integer n becuse, in tht cse, we hve liner dependence. It cn be seen tht the first terms in (20) nd (24) re finite nonzero multiples of x nd x, respectively. This mens tht, for ny integer n, we hve liner dependence becuse (25) J n (x) () n J n (x) (n, 2, Á ).

29 SEC. 5.4 Bessel s Eqution. Bessel Functions J (x) 95 PROOF To prove (25), we use (24) nd let pproch positive integer n. Then the gmm function in the coefficients of the first n terms becomes infinite (see Fig. 553 in App. A3.), the coefficients become zero, nd the summtion strts with m n. Since in this cse (m n ) (m n)! by (8), we obtin (26) () m x 2m n J n (x) mn 2 2mn m! (m n)! s0 () ns x 2sn 2 2sn (n s)! s! (m n s). The lst series represents () n J n (x), s you cn see from () with m replced by s. This completes the proof. The difficulty cused by (25) will be overcome in the next section by introducing further Bessel functions, clled of the second kind nd denoted by. Y PROBLEM SET 5.4. Convergence. Show tht the series () converges for ll x. Why is the convergence very rpid? 2 0 ODES REDUCIBLE TO BESSEL S ODE This is just smple of such ODEs; some more follow in the next problem set. Find generl solution in terms of J nd J or indicte when this is not possible. Use the indicted substitutions. Show the detils of your work. 2. x 2 ys xyr (x 2 49)y xys yr 4 y 0 (x z) 4. ys (e 2x 9)y 0 (e x z) 5. Two-prmeter ODE x 2 ys xyr (l 2 x 2 2 )y 0 (lx z) x 2 ys 4 (x 3 4) y 0 (y ux, x z) x 2 ys xyr 4 (x 2 )y 0 (x 2z) (2x ) 2 ys 2(2x )yr 6x(x )y 0 (2x z) xys (2 )yr xy 0 (y x u) x 2 ys ( 2)xyr 2 (x 2 2 )y 0 (y x u, x z). CAS EXPERIMENT. Chnge of Coefficient. Find nd grph (on common xes) the solutions of ys kx yr y 0, y(0), yr(0) 0, for k 0,, 2, Á, 0 (or s fr s you get useful grphs). For wht k do you get elementry functions? Why? Try for noninteger k, prticulrly between 0 nd 2, to see the continuous chnge of the curve. Describe the chnge of the loction of the zeros nd of the extrem s k increses from 0. Cn you interpret the ODE s model in mechnics, thereby explining your observtions? 2. CAS EXPERIMENT. Bessel Functions for Lrge x. () Grph J for n 0, Á n (x), 5 on common xes. (b) Experiment with (4) for integer n. Using grphs, find out from which x x n on the curves of () nd (4) prcticlly coincide. How does x n chnge with n? (c) Wht hppens in (b) if n 2? (Our usul nottion in this cse would be.) (d) How does the error of (4) behve s function of x for fixed n? [Error exct vlue minus pproximtion (4).] (e) Show from the grphs tht J 0 (x) hs extrem where J (x) 0. Which formul proves this? Find further reltions between zeros nd extrem. 3 5 ZEROS of Bessel functions ply key role in modeling (e.g. of vibrtions; see Sec. 2.9). 3. Interlcing of zeros. Using (2) nd Rolle s theorem, show tht between ny two consecutive positive zeros of J n (x) there is precisely one zero of J n (x). 4. Zeros. Compute the first four positive zeros of J 0 (x) nd J (x) from (4). Determine the error nd comment. 5. Interlcing of zeros. Using (2) nd Rolle s theorem, show tht between ny two consecutive zeros of J 0 (x) there is precisely one zero of J (x). 6 8 HALF-INTEGER PARAMETER: APPROACH BY THE ODE 6. Elimintion of first derivtive. Show tht y uv with v(x) exp ( 2 p(x) dx) gives from the ODE ys p(x)yr q(x)y 0 the ODE us 3q(x) 4 p(x) 2 2 pr(x)4 u 0, not contining the first derivtive of u.