CHAPTER XXX. ADDITIONS TO CHAPTER XVII. [THE present Chapter consists of additions to Chapter XVII. Art. 1 was intended to follow Chap. XVII. Art. 1.] 1. The theory of the solution of linear differential equations in a series flows very beautifully from their symbolical expression. It is usual in treating this subject to assume the form of the series, and deduce from the differential equation the law of its coefficients; but the symbolical form of the differential equation determines in reality the form of the solution as well as the law of derivation of its successive terms. Let us begin with the binomial equation f(D) u-f(D) ero u = 0. Operating on both sides with {f(D)}~1, we have u-$(D) €"°u={ƒ.(D)}~10, Now {f(D)0 will be determined by the solution of a linear differential equation with constant coefficients, and will be necessarily of the form AP+BQ+ CR + ..., in which A, B, C, ... are arbitrary constants, and P, Q, R,..... are functions of the independent variable. We have then {1 − (D) €o} u = AP+BQ + CR+ ..., therefore u={1-$ (D) € ̃o¿ ̄1 (AP+BQ + CR+ ...). Now let us represent (D) e" by p; then u = (1 − p)1 (AP+BQ + CR + ...) = = (1 + p + p2 + p3 + ...) (AP+BQ + CR+ ...) · A (1 + p + p2 + p3 + ...) P +B(1+p+ p2 + p3 + ...) Q + C(1+p+ p2 + p3 + ...) R +... Represent the first line of the above expression by u,, then since = we have mr0 † €TMre (D+mr) $ (D+ mr − r) ...... $(D+r), u1 = A {P+e”o† (D+ r) P+ e2ro † (D+2r) $ (D + r) P +€3ro $ (D+ 3r) $ (D + 2r) $ (D + r) P+ ...}, in which it only remains to perform the operations indicated by (D+r), by $ (D + 2r) $ (D+r), ... on the function P. Let us in the first place suppose the symbolic function f(D) to be of the form (D − a) (D −b).....; then = Here Peo. Hence substituting in the above expression for u, and observing that ƒ (D) en2 = ƒ (n) e1o, we find u1 = Acao {1+$(a+r) e1o + $ (a + 2r) $ (a + r) €21o + .....}, or, since e=x, u1 = Axa {1 + $(a + r) x* + $(a + 2r) † (a +r) x2r + ... }; and u = Axa {1 + $ (a + r) x2 + $(a + 2r) † (a + r) x2r + .....} + Bx3 {1 + $ (b + r) x2 + $ (b + 2r) $(b + r) x2 + ...} + ..., the solution sought. Consider now the general equation fo(D) u+f, (D) e°u + ... + ƒn (D) e1o u =0. Here we have, representing fm (D) by pm (D), fo (D) {1 + $1 (D) €o + ... + $n (D) €1o } u = {ƒ.(D)}~10 ; therefore u = {1 + $1 (D) e° + Þn (D) €1o}−1 { ƒ。(D)}-10. ...... Here we have first to determine {f(D)}-10, then to determine the effect of the operation represented by {1 + $1 (D) €o + ...... Þr (D) eno}-1 upon this. Now {f(D)0 is given by the solution of a linear differential equation with constant coefficients, and will therefore be of the form AP+BQ+CR+ ..., A, B, C,... being arbitrary constants, and P, Q, R, . tions of 0. it ... func may be shewn by a process of actual symbolical division, B. D.E. II. 12 attending to the laws of combination of symbols, that the expression may be expanded in the form F. (D) +F, (D) e° + F2(D) €2o + .... To determine the functions F(D), F,(D),.............. we may proceed as follows. From the equation 1={1+1(D)e°+...+$„(D) e1o} {F ̧ (D)+F, (D) e°+F2(D) e2o+...} = F(D)+{F, (D) + 6, (D) F, (D − 1)}+ Hence therefore and so on. F(D) = 1, F1(D) + 4, (D) F (D − 1) = 0 ; F(D)=-6, (D) F (D-1), (1). Hence F(D), F(D), ...... are determined in succession. The general law is as follows: the coefficient of me in the second member of (1), when m is greater than 1, is Fm (D) + Þ1 (D) Fm-1 (D − 1) + 1⁄2 (D) Fm-2 (D − 2) +..... (2), whence F(D) = −6, (D) Fm (D-1) — 6, (D) Fm-2 (D − 2) —.... m-1 − By this formula the successive values of F (D) can be deduced from those of Fm-1 (D), Fm-2(D),.... Combining the above results we obtain thus for u the expression u = {1+ F, (D) e° + F2 (D) €2o + ....} {AP+BQ + ...} = 2 = A{P+ F1(D) c° P+F2 (D) e2o P+ ....} 20 + B{ Q+F, (D) e° Q + F2 (D) €2o Q + ....} Let us in applying this expression first suppose that the factors of f(D) are real and unequal, so that f(D) is of the form (D-a) (D-b) (D-c).... Further, let us suppose that no two of the quantities a, b, c, .... differ by an integer. or, since A{ea® + F1(D) e(a+1)0 + F2(D) €(a+2)0 + ....} +B{e1o + F ̧ (D) €(3+1)0 + F ̧ (D) €(0+2)0 + ....} + 2 u = A{ea® + F', (a + 1) e(a+1)0 + F2 (a + 2) €(a+2)0 + ....} Hence, replacing e by x, a+1 u = A{xa + F1(a + 1) x2+1 + F2 (a + 2) xa+2 +. + B{x3 + F1 (b + 1) 1⁄23+1 1 2 •} + F2 (b + 2) 20+2 + ....} 2 In (2) replace in like manner D by a+i and we have, putting i for m, Ua+i+1 (a + i) Ua+i-1 + $2 (a + i) Ua+i-2 + ...... = 0. 1 Put m for a + i, thus 2 Um + Þ1 (m) Um-1 + $2 (M) Um-2 + ...... = = 0. This agrees with the law established in [there is no reference in the manuscript, but the law intended appears to be that given in Chap. XVII. Art. 9.] |