THE METHOD OF FROBENIUS WITHOUT THE MESS AND THE MYSTERY. William F. Trench Trinity University San Antonio, Texas 1

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1 THE METHO OF FROBENIUS WITHOUT THE MESS AN THE MYSTERY William F. Trench Trinity University San Antonio, Texas Introduction In this note we try to simplify a messy topic usually covered in an elementary differential equations course. The method of Frobenius is a way to lve equations of the form x 2. 0 C x/y 00 C x.ˇ0 C ˇx/y 0 C. 0 C x/y 0; () where i; ˇi, and i are constants and 0 0. Important examples are x 2 y 00 C xy 0 C.x 2 2 /y 0 and xy 00 C. x/y 0 C y 0; where and are constants, and multiplying the second equation through by x puts it in the form (). The first is Bessel s equation, named after the German mathematician and astronomer Frederick Wilhelm Bessel ( ), and the second is Laguerre s equation, named after the French mathematician Edward Nicholas Laguerre ( ), who contributed extensively to the theory of infinite series. The method is named after its inventor, Ferdinand George Frobenius (849 97), a profesr at the University of Berlin. Frobenius developed it in his doctoral thesis, written under the direction of Karl Weierstrass (85-897), one of the founders of the theory of functions of a complex variable. Frobenius studied more general problems (for example, see [] [4]), but most equations treated by his method in elementary courses can be written as in (). If ˇ 0 then () reduces to 0x 2 y 00 C ˇ0xy C 0 y 0; which is called Euler s equation in honor of Leonhard Euler ( ), a Swiss mathematician who made important contributions to many areas of mathematics and is considered to be one of the greatest mathematicians of all time. We don t need the method of Frobenius to lve Euler s equation. In any elementary differential equations textbook you ll see that its general lution is determined by the zeros of p 0.r/ 0r.r / C ˇ0r C 0 0.r r /.r r 2 /; which is called the indicial polynomial of (). Specifically, the general lution of Euler s equation is ( c x r C c 2 x r 2 if r r 2 ; y x r.c C c 2 ln x/ if r r 2 : wtrench@trinity.edu; mailing address: 659 Hopkinton Road, Hopkinton, NH 03229

2 Therefore we assume that at least one of, ˇ, and is nonzero, and we define p.r/ r.r / C ˇr C : In most textbooks it is assumed that r, r 2, and the constants in () are real numbers, but this doesn t simplify matters, we ll allow them to be complex. If r r 2 is not an integer then the method of Frobenius yields lutions of the form (a) X X y a n x ncr and (b) y 2 b n x ncr 2 ; where the series converge in me interval containing the origin. If r r 2 then (a) and (b) are the same, and finding a second lution that isn t just a multiple of the first is mewhat complicated and treated incompletely in most textbooks. Worse, if r r 2 C k where k is a nonnegative integer then () has a lution of the form (a), but not (b), and methods usually presented for finding a second lution are a mystery to most students. We will try to clarify all three cases. For brevity, when we say that a y 2 is a second lution of (), we mean that y 2 is not a constant multiple of y. Then every lution of () can be written as y c y Cc 2 y 2 where c and c 2 are constants. 2 Preliminaries For convenience, let Ly x 2. 0 C x/y 00 C x.ˇ0 C ˇx/y 0 C. 0 C x/y; we want find y such that Ly 0. A series y.x; r/ X a n.r/x ncr where r and a 0.r/, a.r/,..., are independent of x can be differentiated term by term with respect to x, X X y 0.x; r/.ncr/a n.r/x ncr and y 00.x; r/.ncr/.ncr /a n.r/x ncr 2 : Therefore ix 2 y 00.x; r/ C ˇixy 0.x; r/ C i y.x; r/ X p i.n C r/a n.r/x ncr ; i ; 2; 2

3 Ly.x; r/ X X p 0.n C r/a n.r/x ncr C p.n C r/a n x ncrc X X p 0.n C r/a n.r/x ncr C p.n C r p 0.r/a 0.r/x r C n /a n.r/x ncr X Œp 0.n C r/a n C p.n C r /a n x ncr :(2) n We want to find r and a 0.r/, a.r/,..., that Ly.x; r/ 0 for all x. For a big step in the right direction, let r be a complex number such that p 0.n C r/ is nonzero for all positive integers n, and define a 0.r/ and a n.r/ p.n C r / a n.r/; n : (3) p 0.n C r/ Then (2) and (3) imply that Applying the second equality in (3) repeatedly yields Ly.x; r/ p 0.r/x r : (4) a.r/ p.r/ p 0.r C / ; a 2.r/ p.r C /p.r/ p 0.r C 2/p 0.r C / ; and, for n 2, a n.r/. / n p.n C r / p.r C /p.r/ p 0.n C r/ p 0.r C 2/p 0.r C / ; which we write more compactly as Q n a n.r/. / n j p.j C r / Q n j p ; n 0; (5) 0.j C r/ where 0Y p.j C r j We ll see that it is useful to define / 0Y p 0.j C r/ : r/ y.x; r/ ln x C X an 0.r/xnCr : n (Since a 0.r/ for all r, a0 0.r/ 0.) Then (4) implies that L.x; @r Ly.x; r/ p0 0.r/xr C p 0.r/x r ln x: (6) 3

4 3 Case : r r 2 an integer If r r 2 is not an integer then p 0.nCr / and p 0.nCr 2 / are both nonzero for all n. Therefore (4) implies that Ly.x; r / Ly.x; r 2 / 0 (since p.r / p.r 2 / 0), and y y.x; r / y 2 y.x; r 2 / are linearly independent lutions of (). Example For the equation r, and r 2 =3. From (3), X a n.r /x ncr (7) X a n.r 2 /x ncr 2 (8) 3x 2 y 00 C x. C x/y 0. C 3x/y 0; p 0.r/.r /.3r C /; p.r/ r 3; a n.r/ n C r 4.n C r /.3n C 3r C / a n.r/: Hence a n./ n 3 n.3n C 4/ a n./ and a n. =3/ y x C 2 7 x2 C 70 x3 and y 2 X. / n 3 n nš 3n 3 3n.3n 4/ a n. =3/; n ; j 3j 3 A x n =3 : 3j 4 4 Case 2: r r 2 If r r 2 then (7) and (8) define the same lution, which we will call y. However, in this case p.r/ 0.r r / 2 and p 0.r/ 2 0.r r /; p.r / p 0.r / 0: Therefore (6) implies that is a second lution if (). y 2.x; r / y ln x C X an 0.r /x ncr n 4

5 a 0 0 To use this result we must compute an 0.r / for all n 0. Since a 0.r/ for all r,.r/ 0. If p 0.j C r/ 0 and p.r C j / 0; j m; (the first inequality for all j with r ), then (5) implies that ln ja n.r/j ln p.j C r / ˇ p 0.j C r/ ˇ j ifferentiating this yields where.ln jp.j Cr /j ln jp 0.j Cr/j/; n m: j an 0.r/ a n.r/ b j.r/ (9) j b j.r/ p 0.j C r/p 0.j C r / p0 0.j C r/p.j C r / ; j m: (0) p.j C r /p 0.j C r/ Hence, a 0 n.r/ a n.r/ b j.r/; j m: () j Therefore, if p 0.r/ 0.r r 0 / 2 and p.j C r / 0 for all j, we can let m be arbitrary in (0), 0 X y 2 a n.r b j.r / A ncr is a second lution of (). Example 2 For the equation and r r 2. From (3) and (5), Therefore j x 2. C 2x/y 00 C x.3 C 5x/y 0 C. 2x/y 0 p 0.r/.r C / 2 ; p.r/.r C 2/.2r /; a n.r/ a n.r/. / n 2n C 2r 3 n C r C a n.r/; n 0; n Y j a n. /. /n nš 2j C 2r 3 j C r C ; n 0:.2j 5/; n 0; j 5

6 and If a n.r/ 0 then a 0 n.r/ a n.r/ Hence ln ja n.r/j 0 X. / n nš 5/ A x n : j.ln j2j C 2r 3j ln jj C r C j/ ; j 2 2j C 2r 3 j a 0 n. / a n. / n j C r C j j 5 j.2j 5/ ; 5.j C r C /.2j C 2r 3/ Q 0 X n y 2 y ln x C 5. / n j.2j A x n : nš j.2j 5/ The method we used to obtain () is called logarithmic differentiation. It doesn t work if p contains a factor of the form r r m where m is a nonnegative integer, since this puts a zero in the denominator of (0) with r r and j m C. If m is the least nonnegative integer for which p has this property, then either or j p.r/.r r m/ where 0 (2) p.r/.r r m/.r s/ where 0 and s r C `; 0 ` m : (3) In either case (5) implies that 8 Q n ˆ<. / n j p.j C r / Q a n.r / n j ˆ: p if 0 j m; 0.j C r / 0 if n > m; y mx a n.r /x ncr : To find y 2 we must compute an 0.r / for all n. We can use logarithmic differentiation as in (9) and (0) for n m, since p.j C r / 0 for j m. If n > m then (5) and (2) or (3) imply that a n.r/.r r /b n.r/ if n > m; 6

7 with b n.r/. / n bqn j p.j C r / Q n j p ; 0.j C r/ where the b over the product symbol in the numerator indicates that the factor.r r / in p.m C r/ if n > m is omitted. Therefore an 0.r/.b n.r/ C.r r /bn 0.r//; n > m; a 0 n.r / b n.r /; n > m: If (3) holds with s r C q where q is an integer m, then b n.r/ has the factor r r if n > q, a 0 n.r / 0 if n > q. Example 3 For the equation and r r 2. From (3) and (5), x 2. x/y 00 C x.3 2x/y 0 C. C 2x/y 0; p 0.r/.r C / 2 ; p.r/.r /.r C 2/; a n.r/ n C r 2 n C r C a n.r/; a n.r/ j j C r 2 j C r C ; n 0: In particular, a. / a.r/ r r C 2 and a 2.r/ r.r /.r C 2/.r C 3/ ; 2, a 2. /, and a n. / 0 if n 3. Therefore y. x/ 2 =x: Al, a 0. / a0 2. / Therefore a 0.r/ 3.r C 2/ 2 and a 0 2.r/ 6.r 2 C 2r /.r C 2/ 2.r C 3/ 2 ; 3. If n 3, then a n.r/.r C /b n.r/ where b n.r/ a 0 n. / b n. / y 2 y ln x C 3 3x C 2 r.r /.n C r /.n C r/.n C r C / ; 2.n 2/.n /n ; n 3: X n3.n 2/.n /n xn : 7

8 5 Case 3: r r 2 a positive integer If r r 2 k is a positive integer then p 0.n C r / 0 for n, y in (7) is a lution of (). However, y 2 in (8) is undefined, since p 0.r 2 C k/ p 0.r / 0. In this case we construct a second lution of () as a linear combination of X w.x; r / y.x/ ln x C an 0.r /x ncr and Xk w 2.x/ c n x ncr 2 ; n ; (4) for suitably chosen c 0,..., c k. From (6), Lw p0 0.r / 0.0/.r r 2 / k 0; (5) As in the proof of (2), Lw 2.x/ Xk kx p 0.n C r 2 /c n x ncr 2 C p.n C r 2 /c n x ncr 2C Xk p 0.n C r 2 /c n x ncr 2 C Therefore, if we define kx p.n C r 2 /c n x ncr 2 n Xk p 0.r 2 /c 0 x r 2 C Œp 0.n C r 2 /c n C p.n C r 2 /c n x ncr 2 Cp.k C r 2 /c k x kcr 2 : c 0 and c n p.n C r 2 / c n ; n k p 0.n C r 2 / (we can t let n k because k C r 2 r and p 0.r / 0), then Q n c n. / n j p.j C r 2 / Q n j p ; 0 n k ; (6) 0.j C r 2 / and, since p 0.r 2 / 0, Lw 2 p.k C r 2 /c k x kcr 2 p.r /c k x r : This and (5) imply that if C is any constant, then Therefore, if L.w 2 C Cw / Œp.r /c k C Ck 0 x r : C p.r /c k k 0 then y 2 w 2 C Cw is a lution of (). (7) 8

9 Example 4 For the equation x 2 y 00 C x. 2x/y 0.4 C x/y 0; p 0.r/.r 2/.r C 2/; p.r/ 2r ; r 2, r 2 2, and k r r 2 4. From (3) and (5), and if r a n.r/ 2. Therefore a n.r/ 2j C 2r.j C r 2/.j C r C 2/ a n.r/; n ; (8) j 2j C 2r ; n 0; (9).j C r 2/.j C r C 2/ From (6), a n.2/ nš y X c n nš j nš j 2j C 3 j C 4 ; n 0; j 2j C 3 A x nc2 : j C 4 2j 5 j 4 ; 0 j 3; c 0 ; c ; c 2 4 ; and c 3 2 : Therefore, from (4), w 2 x 2 C x C 4 x2 2 x3 : From (7) and (8) with r 2 and r 2 2, C p./c 3 4 From (9) and logarithmic differentiation, =6: a 0 n.r/ 2a n.r/ a 0 n.2/ j j 2 C j.2r / C r 2 r C 4.j C r 2/.j C r C 2/.2j C 2r / ; n ; 2a n.2/ j j 2 C 3j C 6 j.j C 4/.2j C 3/ ; n : 9

10 Therefore y 2 x 2 C x C 4 x2 References 6 y ln x C 8 X n nš 2 x3 0 2j C 3 j 2 C 3j C 6 A x nc2 : j C 4 j.j C 4/.2j C 3/ j [] A. R. Forsyth, Theory of ifferential Equations, Part III, Ordinary Linear Equations, Cambridge University Press, [2] E. L. Ince, Ordinary ifferential Equations, Courier over Publications, 956. [3] H. Jeffreys and B. S. Jeffreys, Methods of Mathematical Physics, Third Edition, Cambridge University Press, 956. [4] E. T. Whittaker and G. N. Watn, A Course of Modern Analysis, Fourth Edition, Cambridge University Press, 952. j 0