On a Linear Differential Equation of the Second Order

On a Linear Differential Equation of the Second OrderAuthor(s): Thomas CraigSource: American Journal of Mathematics, Vol. 8, No. 2 (Feb., 1886), pp. 180-192Published by: The Johns Hopkins University PressStable URL: http://www.jstor.org/stable/2369299 .

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On a Linear Differentiatl Equation of the Second Order.

BY THOMAS CRAIG.

I.

The differential equation to be studied is d2y Aj1X2 + A12X + Als dy + A21x4 + A22X3 + A23X2 + A24X + A25 ,

dx2 XS (1-,_X2) dx x2 (1 - x2)2 y

or for brevity d$2 P + P2.

The coefficients P1 and P2 have for points of (polar) discontinluity X0, W. x 1, x= -1.

The domain of the point x = 0 is a circle of radius unity having the origin as center, that of the point x 1 is a circle of radius unity having the point x = 1 as center, and similarly for the point x - 1. The domain of the point x = co is the entire infinite plane lying outside the circle having the origin as center and radiuis unity. These several domains will be denoted by Co, C1 C1,-1,

C., and the portion of the plane common to two domains Ci and C will be denoted by Cj. The fundamental integrals of the equation in the domains C, and Cj must coincide in the common domain C6Y.

In the domain of the point x = 0 write the differential equation in the form dy_ Q dy+ Q2

dx2 x dx- The fundamnental determinant equation for the point x = 0 is

r (r -1) - rQj (0) -Q2 (?) = ? or, on substituting for Q, (0) and Q2 (0) their values

r (r - 1) + rA13 + A25 the roots of which are

; - (1-A13) + V(1 -A13)2- 4A25 r2 = 1 { (1 - A13) - (1 - A13)2 - 4A2 .



CRAIG: On a Linear Differential Equation of the Second Order. 181

Assume as a special case A13 = 1, A25 = 0, then the roots are each equal to zero, and consequently the fundamental integrals of the equation in the domain of x = O are Yi = ii (x),

Y2; -P21 (x) + P22 (x) log X. Where (1p, p21, p22 are, in the domain C0, uniform and continuous functions of x and do not vanish for x = 0; and where cp. differs only by a constant factor from q 11.

Take the first integral and write for brevity pl = u. Being a uniform and continuous function of x in the region of the point x = 0 and not vanishing for x = 0, we have for u the form ao

U6 _ EGX2ff a=O

a convergent series in which C0 is not equal to zero. Substitute u in the differential equation and develop (1 - x)-' and (1 - x2)-2 in series going

du according to ascending powers of x. The product of d and the development of (1 - x2)-' is 00

E A X XT = o

where A k =,+=l+ e(n+ k+?1)ii7r

A + e,

~~kOCk, k=l

and multiplying this by the factor - (Ajjx2 + A12x + 1),

we have for the first term in the right-hand side of the differential equation d u _ (Alux2 + A12x + 1) du (A21X3 + A22X92 + A23 + A24) dX2 1-X2 dx (1X-2)2 U

the value = 0 - I~2 [A1An -2 + A12An-1 + An] x;.

n=O

Multiplying u by the development of (1 - x2)-2, we have

~~ ~1+e,(n+k)iff(n-k+ ) n=0 k==0

or, writing for brevity

k-0



182 CRAIG: On a Linear Diferential Equation of the Second Order.

n=oo

simply > n

n=O

On multiplying by the remaining factor - (A21x3 + A22x2 + A23X + A24),

the second term on the right-hand side of the differential equation is n = oo

- > [A21Bn_3 + A22B-2 + A23B?,-l + A24B.] Xn.

n=O

The whole equation is now w = 00 n- =oo

n (n + 1)O +1 = - 0 , [A11An-2 + A12An-1 + An] e n=0 n=O

It = 00

- [A2Bn-3 + A22B -2 + A23Bn-1 + A24Bj] Xn n=O

Equating the coefficients of xn in this and we obtain a series of equations for the determination of C0, Cl, C2 . . . . It is clear that ultimately every C will be merely Co multiplied by a determinate constant. We have for CO +1 after transposing one term to the left-hand side of the equation

k=n-3

(n + +) C,X1 - A21 +e(nj?lcif (1+k + ? C k=O n l-2

1+22 2?k)+7T (- 2+k +1)k

0 n-1

- A23 E 1 + e(n - (n_- +-i + i) C,k

0

0

n

A24 1 + e(n +

kn+ k ) Ck

-1 A11~ ~ ~ e(1 - I 2 -) fi

7

n

A12 1+ e(n+ k)i7T

7t

~ 1+e(00+k +1) 7kC



CRAIG: On a Linecar Dfferential Equation of the Second Order. 183

In particular, C1 - - A24 C0

C2 -= [-A23 + A24 + A12A24] C0

C3 - [2A12A23 - 3A12A24 2A42A24 + 5A23A24 - A 4A4 A24 4A24 4A22],

etc. If we assume

A24o , A23 - 1, A22 =0, A21 =1, A13 1, A12-0, A11=-3, the differential equation becomes

d2y 1- 3x2 dy 1 _ _ c +

dx (1 - X) (IX 1-x which has for onle integral the elliptic integral, K, of the first kind with modulus x. The values of the constants (comnputed fromn the genera] formula) are readily found to be

c0, c1=o0, c2= c01, c3 = O, C4=C0 , Go2-2 42 s=o2 , (-/6 CO 2 2 etc.

Assuming then Co-- we have for Yi the value

12 1. 32 12.32.52 6 yl - - [1+ 22 . X2 + 2242 + -2-42 62 +*** Y 2L 22 ~22.42X 2~2 2.6

which is the kiown development of K in terms of the modulus. I have similarly verified the general formula in the case of the differential

equation, givilig the elliptic integral of the second kind. Another relation between five consecutive coefficients is at once obtained if

we multiply the given equation through by x (1 - x2)2 and then equate coefficients of Xn: this is (n + 1)2Cl?+ 1 + (nA12+ A24) C11+ [-2 (n - 2)(n - 1) + (n - 1)(All - 1) + A23] Cv, -1

+ [A22- (n- 2)A12] C_2+ [(n- 3)(n- 4)- (n- 3)A,, + A21] C-3 = 0, or say 1D (n, C) - 0. Considered as a function of n we may write

dl(n,C) _2 (it + 1)Cn+ + A12CQ, + [1 + A11-4

-A12 C71-2 + [2 }3) - ( +All)] C.,t For the second integral of the equation we have

Y2 +(X) + u log x, where (p (x) is a unliform continuous function of x in the domain C0 aild does not vanish for x = 0, we have then

) (x) = ? C x'

l =O 0



184 CRAIG: On a Linear Differential Equation of the Second Order.

Substituting in the differential equation and equating coefficients of xe, we find as before the relation 2(n+ l)Cz +Al2Cn,,+ [1 +A1 -4(n - 1)] C-l A]2-2

+ [2 (n - 3)- (1 + All)] Cn3 + (n + 1)2c,+1 + (nAj2 + A24) c, + [- 2 (n - 2)(n - 1) + (n - )(All -1) + A23]C-l1 + [A22 - (n - 2) Aj2]cn-2 + [(n - 3)(n - 4) - (n - 3) All + A2l]c,3 = 0,

or, as this may obviously be written

'1(n, c) + d(P (n, C) (D n, ) + dn 0

A simple application of the formula3 is to the differential equation d2-y+ 1-(a+,G+1)x dy a, Y 2 dx' + w(1 - ) dxs . (1 - )Y

of which a solution is the hypergeometric series u = F(oa, 1 x). Here we have All=-(a+13+ 1), A12=--(a+),

A21 = acl, A22 =aj, A23- ag3, A24 - 03. Applying the formula for C1, C2, C3... and writing C0 = 1, we find readily

G, a. ji a2 a(a +1)R(l@+ )

1.1 1' 1.2.1.2

C a (a + 1)(a + 2)Q9( + 1)(P+ 2), etc.

Also for the coefficients cl, c2 *

Cil= a3 [co + 1j + 2 1

c1- d[Co++-+ -2(1+) 1]tc c2 1

a 2 12 [cO +

a + + 1 + 1

1-2 1 + etc. 1.21. a a +1 P P +1 2J We have now found a system of fundamental integrals of the equation

+2y Aj1x2 + A12X + A13 dy (A21x4 + A22X8 + A23x2 + A24X + A25) 0 dx2 x (1 X2) dx x2 (1 _X2)2

in the domain of the point x = 0, and for this case the roots of the fundamental determinant equation were taken to be equal and each equal to zero, t. e. we made A13 = 1 and A25 = 0. Removing this restriction now, we have for the roots of the fundamental determinant equation

r= ~{(1 --A13) + V/(1-A13)2 -4A25} =- w (1 -A13) - V(1 -A13)2 4A25}

or say r1 ? +%/3 r=a - s/V .




We will assume first that i3, . e. (1 - A13)2 - 4A25 is not equal to zero; now two cases arise, either r, - r2 is zero or an integer or the same difference is neither zero nor an integer. By hypothesis the difference cannot be zero, so we have only to consider the cases when V/(1 A13)2 4A25 is and is not an integer.

Assume first that V/(1- A13)2- 4A 5 is not an integer, then the fundamental integrals will be of the form

Yi Xr

p1k (X),

112 Xr2 P2 (X)

Where 41 (x) and 192 (x) are uniform and continuous functionls of x in the domain C. and are not equal to zero for x = 0. We have therefore

n_= oo

(p1 (X), = U, = A z,x, AO not = zero.

n =0

The differential equationi for u is

du Q1 du Q2 dxi x d

writing Q Q'(x)_ ( - Q x

and substituting for Q1 and Q2 their values we have for the differential equation in ut

d2+ du = 2_ rx(Auj+ A13- 2r) + A12X + 2i] du dx + (A13 + 2r) dx L 1-2 d

-[ X3 [A21-rAnj + r (r-1)] + E2[A22-rAj2] 1

L + x [A23-rAi3 + rA1- 2r (r- 1)] + [A24 + rAI2]] (1_ SX)2

Substituting now for u its value, and equating the coefficients of xn, we hlave, after some easy reductions,

(n + 1)(n + A13) An+1 [- A21 + rA11-r-(r- 1)] Bn3 + [EA12 -- A22] Bn-2

+ [rA13- rA1- A23 + 2r(r- 1)] Bn-l [A2 + rA12] B. + [- All - A13 + 2r] An-2 Al2Al -I 2rAn,

where

k=0

k = 0 kOn

' o e(n +k) i 0+1)A

A -21 2 '+J. lAk 1

VOL. VIII




(It will be noticed that the coefficient A25 does not appear explicitly in the above formula.)

In particular make r = 0, A13 1; then for n- 0 we have

Al - A24AO,

for n= 1 A2 ( A23 + A24 + A12A24) A0 etc., results agreeing with those obtained above.

If we compute the constants A fromn the formula, they will all be found to be of the form 4, = r.Ot

then giving to r the values r, and r2 we find the two functions above desigulated as qp (x) and cp2 (x): the fundamental integrals of the differential equation in the domain (C0 are therefore Yi = - (Pi (x),

!12 X Xr2 (2 (x),

in the case where r1 and r2 are different and their difference is not an integer. A particular case of the above formula is when

All= -(oc+g+ 1), A12=y a -3 1, A,3-y, A21= ag, A22=a,3, A23 -ag, A24 -ag3, A25 0 .

The roots of the fundamental determinant equation are now r= 0,

= 1-,Y and the differential equation is

d + [ al+ -dx+ x(- Y = ?;

which, when y is not a negative integer, has for the integral corresponding to

r, = 0 the hypergeometric series F (a, @, y, x) and, corresponding to r2= 1 -, when 2 - is not a negative integer, the

hypergeometric series xl--F (a + 1 - 3+ 1 y, 2 -y), x).

Substituting the above values of the A's in the formula we have first for r=0 A O= aS

A2 = a

(a +2 1z) ; (-++ )-etc.

For the case of r2 = 1 -y7 the formula is similarly verified. The relation between five consecutive coefficients is found by clearing the

equation of fractions; doing this and substituting for n its value, we find



CRAIG: On a Linear Differential Equttion of the Second Order. 187

{n (t + 1) + (n + 1)(A13 + 2r) + [A25 + -A13 + r (r-1)] }AA 1

+ {nA12 + A24 + rA12} An

+ 1- 2(n- 1)(n- 2) + (n- 1)(All -A13- 4rA13) + [A23 + r (All - A13)- 2r (r - 1)] An + { - (n - 2) A12 + A22 - rA,2, ~An-2

+--{(- 3)(n- 4) + [-All+ 2r](n -3)+ [A21 -rAll+ 3(- 1)]}A 3= 0.

Take flow the case where r1 - r2 is ani integer, i. e. V/(1 - A13)2- 4A25 =integer,

then of course 4A25 must be equal to the sum of two squares, one of whiclh is an integer.

The two integrals now are of the form

Y1= -X' >Pii

112 = Z2 [(P21 + (P22 log XI j or say Yi = X;u

1/2 = x2 [r + ' log x], since ),22 olnly differs from-i qp,j by a constant factor, and since (Cy2 is an integral as well as 112

The first of these integrals is of course the same as the one obtained above. To finid the second substitute in the differential equation: the coefficient of log x will be zero, and we have left

z dv A + [

X2 + Aj2x + Als + 2r dv

+ [A2X4 A22X3 -- A23X + A24X + A25 + r [A 'X2 + Aj2x + A1s3] + (r -1)]}

x2 (-X2T x 2(I1-x2) x2

- {2du + [2y-1 + Allx2 + A12X + A13] }

or d2v (A,, - 2r) x3 + A12X2 + (A13 + 23 ) x dv d.X2 1 .S XI dx

+ { x4 [A21 - -Al + r (r - 1)] + x3 [A22 - rA,2] + x2 [A23 + rAll - rA13- 2r(r - 1)]

+ x [A24+ ?rA12] } (1 _ 2)2

du (All - 2r+1) X2+Aj2x +Als+ 2r - 1

Multiplying out by (1 - x2)2: writing n?== oo 71~fl co

'a -c = =i oo t6 E lS9x V - Ax1t




and equating coefficients of X + 1 we have for the determination of the coefficients in v, the relation

in (n + 1) + (n + 1)(A13 + 2r) + [A25 + rA13 + r (r -1)] } &+

+ JnAj2+ A24+rAl2} I k

+ {-2 (n- i)(t- 2) + (n- 1)(All-A13- 4rA13) + A23+r(1-1) 2 r 1)}A-

+ -A12 (n- 2) + A22-r A12} An-2

+ t(n- 3)(n- 4) + [- All + 2r](n- 3) + A21- rA1j + r (r-1)} k-3

- [- 2(n + 1) -(A13+ 2r- 1)] An+1 _A12An + [4 (n-1)-All + A13 + 4r - 2] An-i + A12An 2- 2 (n - 3) Aa-3.

Here we must of course replace r by r2, in order to find the values of A.

The relation satisfied by the five consecutive coefficients An + 1 An7 X n-1,

A4,-2 An-3 is what the above equation becomes when on the left-hand side A is replaced by A, and the right-hand side is made equal to zero; using the same notation as that employed when r1 = r2 = 0, this may be written

(D(n, 2;,A) O ,

and consequently the relation giving the 8's may be written

cP(n, r2, S) + dn J-(n, r, A) 0.

(It will be observed that the coefficients of A. + 1 and .+1 are integers, since 2r = 1 - A13 -integer.)

Consider now the domain C1 of the point x = 1. The fundamental determinant equation for this point is

r (r-1) + r All+A+A3l+A2l?A22+ A2+ A24+ A25 = 0;

or writing A1l + A12 + A13 __

2

A21+A22+A23+A24+A25 j

42 (1- J - - J2 .2 -rq(1-JD+J2= O,

the roots of which are } 2 W { 1-J + V/(1-J1) -4J2J 7-2 J {l J-/( J24X2.

Suppose, first, J, = 0 and J2= 0, i. e., rl1, r2 0.

The integrals of the differential equation are nowv of the form Y'= (x- 1)u Y2 = v + 't log (x - I)




where i and v are in the domain C1 uniform and continuous functions of x, which for x - 1 do not vanish. We have then

mC., (x -

00=

V C- e (x- 1)'.

n=0

The differential equition becomes naow

11y Allx + All + A12 d1/ A21x3 + (A21 + A0,) xz + (A21 + A22 + A2 dx2 x (1+ x) dx L3

+ (A21 + A22 + A23 + A2D41 ? + Q2 (X_ 1)(1+)2 J X

or d2y _ Allx- A13 dy + A23_ +(A21+ A2) X2 +(A21+ A22 + A23) X -A25 dX2 -x (x +1) dx X2 (X -1)(X +1)2

Change the variable by the transformation x = x' -+ .

Change also, for brevity, the notation for the coefficienits, by writing A1l - a1, A1-A3 = A13

A21 = a2, 4A21 + A22= /2, 6A21 + 3A22 + A23 - 72' 3A21 + 2A22 + A23 - A25 =2

and the equation becomes d2y - aX I + 81 d+ a2x'3 + 192X'2 + r2x + 82

dx'2 (x' + 1)(x' + 2) dx' XI (x' + 1)2(.I + 2)2 Y) also Yi = x u

Y2 = v +u log ',

, = X

0

V = Cnx v 1 =

For convenience, again write the equation in the form

d 2y PI dy P2 dx2 dxI+ x 2Y

where I (ax' + ?1 (X + 1)(X +2)

a+ i2x + i2' ? 202)

(I + 1)2(.+)



190 CRAIG: On a Linear Dift*rential Eqatcttiot of the Secondc Order.

It is at onice seen that the roots of the fundamiental determinant equation for the above differential equation are, in the domuaini of x'- 0, r = 1, r2 =-, as they should be. Write nlow q = x'u; substituting in

d2y _ alx' + dS dy + a2x" + A2X'2 + Y2a' + d2

dx2 (v'+ 1)(x'+ 2) dx x (J+1)2x'+2)2 T 2

we have for i the equation

br u F(a, 2) xl + (8,- 6) x - 4 du dxl L x. (x + 1)(x + 2) - dx

+ [(i+a2X4+(Pi+3ai+oI)X8+(;+2- + )2+( N12 r(al + a2) 4 : a 1 21 02 Xa + 1 u 2} 2

- S~~~~~~2

(X + 1)2(X + 2)2 ...w

I write, for convenience (as nlo confusion is likely to arise), x instead of x'.

Clearing this of fractions, substituting for u its value, and equating the coefficients of x7 ?1, we find, without much difficulty, the relation

4 (n + 1)(n + 2) Q+l + [1+l2n (g 1) - z (2S1 - 24) -(281 + 82)1 Cn

+ [13 (n - 1)(n - 2)- (n - 1)(2ai + 31 - 26)- (72 + 2a, + 3z1)] CU-1

+ [6 (n - 2)(n- 3) - (z - 2)Q(1 + 3ca - 12)- (P2 + 3ax + S1)1 C.2

+ E(n -3)(n -4) -(a, 2)(n -3) - (a, + a,)] Q,-3 0 . As an example of this, take the case of

a,=- 1, ~ 8=2,

a2= 1) g3=4, y2--4; 82

The differenitial equation in y now becomes +12 y I dy 1

d,X2 x+lx 'dx (X +1)2 X

tlle inategrals of which are y1 = Sill log (x + 1),

= cos log (x + 1).

The first of these is the onie to be considered. Expanding siln log (x + 1) in a series going accordinig to ascending powers of x, we have

sin log (x + 1) =u0 + u1x4 +4 'tx2 + * * . where i (n + 1) u,+ + n (2n - 1) utn + (n2 - 2n + 2) un, = 0; furtlher, since for x = 0 we have sin log (x + 1) = 0, it follows that v, = 0.

The next few coefficients are u, (which is unity-but may obviously be

replaced by an arbitrary constant in the expressioni for thle integral), 'u - i,, t3- Ul u4 0, u5 =--u1, u6 -8 1, etc.

We ought now to have

CO - t, C- =-?tU2, C2 U13, C3- 0, C, = u15, etc.1



CRAIG: On a Linear DifeZirential Equction of the Secondcl Ordcer. 191

and by substitution of the above values of ac,, ...2 in the formula, it is easily seen that these conditions are all satisfied. Another relation conrnecting

the coefficients C in this particular case is found by expanding + a (nd

in the differential equation d2( xu) 1 d (xu) x _

d2 + 1? d ++

and equating to zero the coefficient of Xn: thliS iS

k=n-1

(+ 1)(n+ 2)) +& ()n +7(-1 [5 (7 + 1) Ck+1+ 2(n-k)C7j] h^=O k-O

+ (_)n + h,(gt _70+ 1) Ck 0

or, replacing n by n - 1 and C by it,, + r we have for the relation coninecting the coefficients ul , Zt2 * *

t(n+ 1)u?1+l + > (_)(n+ -2) [5(c + 1)uk+2 + 2((u-k+ 1) Uk+1]

k =-0 -=nw-l

+ ( ) n+k-1 (9n, 7e) 0t . +I = 0

k =O

Having found the values of the coefficients C, and conisequently the value of the funletion u, it is only necessary to replace x by x - 1 in order to get the first integral y = (XG- 1)u

of the original differential equation. The second integral is in general of the form y2 v + ?u6 log (x -1), or, chalnginig the variable again, simply

Y2 = v + ? ,log x,

where it and v go according to ascenlding powers of x. In tlhe simple case just noted, viz. d2y 1 dy 1

dx2 +(x +1) dx (x +1)2YO

there is evidently nio logarithmn-the two integrals beinig in fact, as already mentioned, y = sin log (x + i), (= xii),

112- cos log (x+1) thie coefficienits in the developrmenits being in each case conniiected by the relation

n? (n + 1) Unl + n (2n -1) un + (t2 - 2n + 2) utt,,1 0,

with, in the case of Y1, i = 0.



192 CRAIG: On a Linear Differenti(al EquaGCtion of the Seconnd Order.

To determin-e whether or not logarithms exist in the case at present treated, i. e. when the roots of the fundamenital determinant equation are r1 - 1, r2 = 0, and where consequently r1 -)r2 = 1, wve form the differential equation satisfied by the second derivatives of the integrals yi and y2; according as there exist or do not exist negative roots of the fundanmental determinant equation belonging to this new differential equationi, there exist or do not, exist logarithms in the original equation in y.

Resuming now the differential equation d2y aix + __ dy + a9x3 + j2 + r 2X + ?'2

dx2 (x + 1)(x + 2) dx x (x + 1)'(x + 2)2 Y'

write =v + logUx (here, as before, x is written for x'). After some easy reductions, we find for v the equation

d2V dv [x (x + l )(x + 2)]' d XI (X + 1 )(X + 2) (a,x + S)d-x (ca2x3 + 2'x' + y,x + 82) V

2x (x + 1)2(x + 2)2 dm + [(x + 1)2(x + 2)2 + x (x + 1)(x + 2)(a,x + S1)] it. dx

Substituting for ut and v their values, viz.,

6 = E =S cn V 0

anld equatin-g coefficients of Xn +'1 we have for the relation connecting the coefficients C and e

4n (n + 1)c,?+I+ [12n (n- 1)- 2j1n -2]c + [ 1 3 (n- 1)(t- 2) -(in- 1)(2x1 + 3j1)-y2]c c

+ [6 (n-2)(n -3)- (t-2)(331 + ^1)- 32] el -2

+ [(n - 3)(n-4)- a, (n- 3)-a2] cn 3

=4(2nz+1)Gn+j+[ 24n+12+2S1] Q1+ [-26(n-1) + 13+2uj+31] Cl,_ + [- 12(n- 2) + 6 + 3cx + 8i] Ca 2 + [- 2(n 3) + (1 + a,)] Cn-3

In the particular case already referred to, viz. 2 + 1X + 1 0

dx.' (x +1) dlx (+)

(i.e a1 -1, 1 -2, a2- -1 4 2---, 72= 4, ( 0), the initegral

which we are in search of is siiymply the function v, consequently the values of the constants c are found by writing the left-lhand memiiber of the above equation

0; it is easy to see then that the coefficients c of the functionl v are the same as those in the development of cos log (x + 1).



On a Linear Differential Equation of the Second Order

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