CIRCULAR CYLINDERS UNDER CERTAIN PRACTICAL SYSTEMS OF LOAD. 153 
Eliminate either clujclz or chvldr between ( 6 ) and (7) : it is found that either of 
these quantities satisfies the partial difierential equation 
(S3+D^)2y = 0.(8). 
The whole problem of the determination of the elastical equililnuum of the circidar 
cylinder under any symmetrical system of stress depends therefore on the solution of 
this difierential equation. 
§ 4. Solution of the Differential Equation. 
The difierential equation 
(^2 q_ p) 2 ^ y ^ 0 
is really identical with Laplace’s equation in cylindrical co-ordinates, namely, 
1 cl rd.Y . 1 cPY , cPY 
-- L - 4. — — 0 
r cir dr ^ S d(f>^ ^ dz- 
Suppose V independent of cf) and difierentiate with regard to r, we have 
(.52 D3) clY/clr = 0 . 
If therefore V be such a solution of Laplace’s equation, y = clY/dr will be a solution 
of the given difierential equation. For our jjurpose, however, it will be simpler to 
proceed from the equation itself 
Assume a typical solution 
l/i — 1^] • ■Zi, 
where is a function of r only, a function of 2 only. 
AVe find easily 
+ + = 0 . 
cPZi 
dz^ 
-h = 0 
(9), 
(10). 
The solutions of (9) are of the form 
Ij {kr) and Kj (^r), 
,.n+2i 
T — V__ 
’ 7 (s) n (.s -I- n) ’ 
where 
K,fx) = (— 1 )' 
1.3. {2n — 1) d'“cos {x siiili (/>) 
J g cosh®" (f) 
d(f) . 
(See Gray and Mathews, ‘ Bessel’s Functions,’ pp. 66-7.) 
VOL. CXCVIII.—a. 
X 
