The KMO Method for Solving Non-homogenous, m th Order Differential Equations

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1 The KMO Method for Solvng Non-homogenous, m th Order Dfferental Equatons Davd Krohn Danel Marño-Johnson John Paul Ouyang March 14, 2013 Abstract Ths paper shows a smple tabular procedure for fndng the partcular soluton to dfferental equatons of the form: a j d j y j = P (x)eαx (1) Ths procedure reduces the dervatves of the product of a general polynomal and an exponental to rows of constants representng the coeffcents of the terms. The rows are each multpled by a j and summed to produce a m th order dfferental equaton of the general polynomal. The underlyng mechancs of ths converson and ts formulac mplcaton are then dscussed. Usng the formula derved, the partcular soluton for equaton 1 wll be found. Ths procedure s dfferent than the method of undetermned coeffcents because whle the method of undetermned coeffcent requres substtuton of a product of a polynomal and an exponental nto the dfferental equaton mmedately, ths procedure s derved from the examnaton of the substtuton of the product of any fucton and an exponental. Ths allows for a rcher understandng of the relatonshp between the dfferental equaton n y and the dfferental equaton n Q. Ultmately ths method s better than the method of undetermned coeffcents because t s more straghtforward. 1 Introducton Dfferental equatons are a very old feld of research. In modern tmes much attenton s beng spent on nonlnear equatons due to the ncrease n computatonal abltes and geometrc technques. Lnear dfferental equatons are well understood especally those wth constant coeffcents. Of these are the ones lke equaton 1. The establshed method for solvng these equatons s the method of undetermned coeffcents. Ths procedure nvolves the substtuton of partcular soluton y p = e αx x s n u=1 c ux u or y p = e αx Q(x) where s s the number of tmes α s a root of the characterstc polynomal. Ths leads to a to a system of equatons whch yeld c u. We nstead consder the correspondng dfferental equaton n Q frst by means of a tabular procedure, then by means of

2 a formula derved. Couplng ths formula wth trangular matrx nverson and a more general procedure we show the partcular soluton and n nduced map κ. 2 Example Applcaton of the Tabular Procedure We consder the followng dfferental equaton: 2 d2 y dy + y = (5x + 3)e 2x (2) The general soluton to ths equaton s found by addng the partcular soluton to the soluton of the correspondng homogenous equaton. The homogenous soluton s of the form: (wth c 1 and c 2 constants) y h = c 1 e λ1x + c 2 e λ2x Where c 1 and c 2 are constants. By the quadratc equaton, the solutons of the characterstc equaton may be found: λ 1,2 = 3 ± 3 2 4(2)(1) 2(2) The KMO method s concerned exclusvely wth fndng the partcular soluton. Thus henceforth let y mean y p. Usually y s found wth the method of undetermned coeffcents as detaled n [1]. The method of undetermned coeffcents s somewhat tme consumng as t often nvolves arduous algebra. The tabular procedure begns wth a soluton of the form y = Q(x)e αx where Q s a polynomal. In the case of equaton 2 ths s y = Q(x)e 2x. Usng a general polynomal elmnates the tral and error that sometmes occurs n the method of undetermned coeffcents. We take dervatves of y: y = Q(x)e 2x dy = dq e 2x 2Q(x)e 2x d 2 y 2 = d2 Q 2 e 2x 4 dq e 2x + 4Q(x)e 2x Now we place the coeffcents nto a table: Q(x) dq d 2 Q 2 y dy d 2 y

3 The tabular format s useful for examnng the coeffcents because when the dervatves of y are summed along wth ther coeffcents, the coeffcents of terms wth lke order dervatves of Q are summed together. For example, when the y term s added to the frst dervatve terms, the coeffcent 1 of Q(x)e 2x n y s added to the coeffcent 2 of Q(x)e 2x n dy. However, before the summaton the coeffcents of the dfferental equaton n y must be accounted for. Now we multply each of the rows by the correspondng coeffcents of the dervatves of y: Q(x) dq d 2 Q 2 1 y dy d2 y The sum of the components n each column produces the coeffcents for the dervatves of Q n the correspondng dfferental equaton n Q. Snce all terms contan the same exponental term, whch s never zero, t may be dvded out yeldng: 2 d2 Q 2 5dQ + 3Q(x) = 5x + 3 (3) In equaton 3, the thrd term s the polynomal of hghest degree and by equvalence must be of frst degree. Ths means wth Q(x) = ax + b, dq = a, and d2 Q = 0. Substtuton nto equaton 3 2 yelds 5a + 3ax + 3b = 5x + 3. By algebra a = and b = 9. Thus y = ( 5 3 x )e 2x. 3 Removal of the Exponental The coeffcents of the dfferental equaton n Q are: b = ( ) j a j α j Proof. The dfferental equaton may be compacted by vewng the left sde of the equaton as the nner product of a and y D where a = [a 0, a 1,..., a m ] and y D = [ d0 y, d1 y 0... dm y 1 ]. Thus, equaton m 1 reduces to the followng equaton: < a, y D >= P (x)e αx (4) The goal, usng ths new notaton, s to fnd the value of b n the dfferental equaton < b, Q D >= P (x). Snce takng successve dervatves of a product of functons derves ths equaton, a rule for the n th dervatves of ths product would be useful. The Lebnz rule s just such a rule [2]. The Lebnz rule states that dn n f(x)g(x) = n =0 ( n 3 ) d f d n g where ( n ) s the bnomal coeffcent. n

4 If f s let to be Q and g s let to be e αx, then by the Lebnz rule, Substtuton nto equaton 4 yelds the followng equaton: d j y j = j ( j ) d Q =0 α j e αx. [ j ( ) j d Q < a, αj e αx] >= P (x)e αx 1n+1 =0 As demonstrated n the prevous specfc example, the next step to obtan a dfferental equaton of the form < b, Q D >= P (x) s to dvde both sdes by a factor of e αx. Explctly wrtng out the sum mpled by the nner product allows ths dvson. The combnaton of the sum defnng elements of y D and the sum that s the nner product yelds the followng double summaton over ndces and j: j a j =0 ( ) j d Q αj e αx = P (x)e αx The e αx may be now be dvded out because t s not summed over. j a j =0 ( ) j d Q αj = P (x) The a j can be moved nsde the nner summaton snce ether way all terms wth j are summed by the frst summaton. The nner summaton may be summed to m nstead snce terms ndexed by > j vansh due to the bnomal coeffcent. Ths s useful snce as a result the orders of summaton may now be swtched due to the commutatve property of summaton. A smple example of ths 2 =1 a j = (a 11 + a 21 ) + (a 12 + a 22 ) and 2 =1 2 j=1 a j = property s the equavalence of 2 j=1 (a 11 + a 12 ) + (a 21 + a 22 ) due to regroupng. Wth ths new orderng, d Q the nner summaton to yeld the followng equaton: =0 d Q ( ) j a j α j = P (x) may be moved outsde of Ths left sde of ths equaton may be recognzed as the nner product of Q D and b where b s the nner summaton. 3.1 Applcaton to Equaton 2 The vector a may be quckly seen to be [1, 3, 2], α = 2, and m = 2. Thus the expresson for b s as follows: 2 ( ) j a j ( 2) j The calculatons of the coeffcent are as follows: b 0 = (1 1 1) + (1 3 ( 2)) + (1 2 4) = 3 4

5 b 1 = 0 + (1 3 1) + (2 2 ( 2)) = 5 Thus the result s the same as equaton 3. b 2 = (1 2 1) = 2 4 Partcular Soluton of Equaton 4 The partcular soluton s Q(x)e αx. The order of Q s n + p where p s the ndex of the frst nonvanshng b j and n s the order of P. Thus Q(x) = n+p u=0 c ux u. From u equals 0 to p 1, c u = 0. Let T = { Z, n m + p < n} and R = { Z + {0}, 0 n m + p}. Also for future convenence let a j = (+j)!! b j. The leadng coeffcent of Q s the leadng coeffcent of P dvded by a p n; the remanng coeffcents are determned recursvely [3] the followng way. For t such that n t T, c p+ = 1 (k a p t =1 ap+ c p++ ) and for t such that n t R, c p+ = 1 (k a p n T =1 ap+ c p++ ). The ks are coeffcents of P and n T s the order of the dfferental equaton mnus p. Proof. Let P n (x) specfy the order of P as n. The converted dfferental equaton of m th order may now be explctly wrtten as the followng: b j d j Q j = P n(x) (5) The order of Q deps on the value of j such that b j s the frst non-vanshng coeffcent. Ths may be attrbuted to the fact that f j = 3 then the lowest order dervatve s d3 Q meanng 3 that f n for P n (x) s 4, d3 Q must be of order 4 snce there are no hgher order polynomals beng 3 summed. For maxmum generalty, the partcular soluton wll assume that j = p. The maxmum order of Q s n + p snce the p th dervatve of Q reduces to a polynomal of degree n, whch s the same order as P. Ths also mples c u for u < p are zero snce they do not appear n the equaton and therefore are not determned by t. Consequently, Q(x) = n+p =0 c x where the constant coeffcents, c, are to be found. Snce dj Q = n+p j (+j)! j =0! c +j x, substtuton nto equaton 5 yelds m j=p b j n+p j =0 (+j)!! c +j x or more smply: j=p n+p j =0 a j c +jx Ths s due to the fact that as prevously noted, vectors ndexed by the dummy varable of the outer sum may be moved nto the nner sum. The product of (+j)!! and b j produces a j as prevously defned. Ths double summaton may be magned as summaton over the parallelogram of ntegers pctured (see fgure 1). The th row represents the th order term of the resultng polynomal on the left hand sde of the equaton. The unknown varable n queston s c. As Q s dfferentated, the c u coeffcents are pushed back to terms of lower degrees, ths means that f a c u n column h s on row t, n column 5

6 Fgure 1: Regon over whch the summaton takes place h + 1, that c u s on row t 1. Thus the components of c are on the dagonals pctured (see fgure 2). Regardless of m, p, or n, a p nc n+p s not summed wth any other coeffcent and thus solvng for c n+p s smple. It s the soluton to a p nc n+p = k n whch s c n+p = kn a. There are three p n possbltes for the rest of the summaton. These possbltes dep on shape of the regon over whch the summatons take place. If p = 0 and m n, the regon s trangular. If p = m, the regon s the rectangular strp of a m s wth s = 0... n. If otherwse, the regon s trapezodal. If p = m t s easy to see that the c u are solved n the same way as c n+p. For trangles and trapezods, even though there s a summaton takng place, the soluton for the thrd from last coeffcent deps on the last and second from last coeffcents. Ths s due to the fact that the dagonals dep on the same coeffcent. The trangular regon T may be seen to occur from = n + p m + 1 to n snce for less than that all terms of order p or more are beng summed. The rectangular regon R s thus n + p m f t exsts. The sum of a row n T that s t unts from the top s the sum of the p th term to the t th term from t and the sum of these coeffcents s the t th term from the leadng coeffcent of P. Mathematcally, ths equaton s a p c n+p t + t =1 ap+ c n+p t+ = k. As prevously dscussed, the c u n the summaton are already known; therefore wth algebra we may solve for c n+p t. That s, subtracton by the summaton and dvson by a p yelds c p+ = 1 (k a p t =1 ap+ c p++ ). The expresson for c u whos ndces are n R are smlar except that the upper ndex of the sum has no depence on t rather s fxed by the number of js from p such that p plus ths number s m. Ths s n T. Thus the c u may be determned recursvely by c p+ = 1 (k a p t =1 ap+ c p++ ) and c p+ = 1 (k a p n T =1 ap+ c p++ ). 6

7 Fgure 2: One-dmensonal subregons over whch u s constant 4.1 Applcaton to Equaton 2 A quck examnaton of order and vanshng coeffcents yelds that m = 2, n = 1, and p = 0. As a result, the leadng coeffcent of Q s c 1 and the regon over whch the summaton takes place s the trangular one pctured (see fgure 3). Snce c p+n = kn a p n and the leadng coeffcent of P s 5, c 1 = 5 a 0 1 b 0 = 3, a 0 1 = 1(3) = 3. By calculaton c 1 = 5 3. Snce c p+ = 1 a p c 0 = 1 (3 a 1 a 0( ). a0 0 = 3 and a 1 0 = 5 so therefore c 0 = 4 3 Q(x) = 5 3 x The soluton s the same as prevously shown. 5 Concluson. Snce a j = (+j)!! b j and (k t =1 ap+ c p++ ),. Combnng ths nformaton, The KMO method provdes a quck alternatve to solvng dfferental equatons by usng the formulas derved. Substtuton of parameters a, k and α nto the formulae quckly yeld the partcular soluton to equaton 4 n the form of a vector c. In ths sense KMO s a map κ : R m R n R = R m+n+1 R n+j. Furthermore, these formulae may be programed nto a computer wth fractonal arthmetc to obtan exact solutons. Included s the MATLAB code for κ wth trangular summaton regons. 6 Aknowledgements We would lke to thank Dr. Bell, our teacher, for hs support and encouragement. At the same tme, we would lke to thank our parents and the Heghts School for ther patence and generosty. 7

8 Fgure 3: Regon over whch the summaton takes place for equaton 2 References [1] W.E. Boyce, R.C. DPrma, Elementary Dfferental Equatons and Boundary Value Problems. John Wley and Sons, New York, 6th Edton, [2] A.B. Israel, R. Glbert, Computer Supported Calculus, Sprnger Wen, New York, [3] R.L. Devaney, A Frst Course n Chaotc Dynamcal Systems, Perseus Books, Boston, A MATLAB code for κ functon [ c ] = kappa(a, k, alpha) m = numel(a) - 1; n = numel(k) - 1; % calculate b b = zeros(1, m + 1); for =0:m for j=:m b(+1) = b(+1) + nchoosek(j,)*a(j+1)*(alphaˆ(j-)); % get j*, nt, and the order of Q jstar = fnd(b, 1, frst )-1; nt = m - jstar; c = zeros(1,n + jstar + 1); % calculate c 8

9 c(jstar + n + 1) = (1/(b(jstar + 1)*factoral(jstar + n))/factoral(n))*(k(n + 1)); for t=1:nt - 1 sum = 0; for =1:t; sum = sum + (c(jstar + n - t + + 1)*(b(jstar )* factoral(jstar + n - t + ))/factoral(n - t)); c(jstar + n - t + 1) = (1/(b(jstar + 1)*factoral(jstar + n -t))/factoral(n - t))* (k(n - t + 1)-sum); 9