188 MR JOHN DOUGALL ON AN ANALYTICAL THEORY OF 
For a distribution of force of finite intensity per unit area or per unit volume, the ) 
potential energy of that part of the solid within the cylinder is clearly finite. The 
energy of the remaining part of the solid can be determined from the forms of § 29. 
Now, the energy between the cylinders p=a, p =p is the integral of 
(Upp +2 pw + W pz) 
taken over the belt of the cylinder p=p cut off by the plate, diminished by the 
corresponding integral for p=a. Hence the condition of finiteness of the whole 
potential energy is simply that the value of the integral for the surface p=p tends to 
zero as p tends to infinity. This condition is obviously satisfied by all the partial 
solutions of § 29, except those which have been already singled out as conveying a finite 
resultant stress. It is also satisfied by one of the latter class, namely, that which 
conveys a couple in the plane of the plate. 
Hence, when force is applied to a circumscribed portion of the solid, a solution giving 
finite potential energy will exist provided the force either constitutes an equilibrating 
system, or reduces to a couple in the plane of the plate. It does not follow, however, 
even for an equilibrating application of force, that a solution will exist giving vanishing 
displacements at infinity. We need only point to the solutions of § 29, II. (vi) and 
(iv), (v) with m=2. This being so, it may be of interest to write down a few more 
details of those solutions which rank in importance next to the solutions of finite — 
resultant stress. 
I. (ii) with m= 2. 
Be IS SU) Tp eos 10 | Pb { (a — 7)p~* + (a — 3)32’p i cos 2w 
v=] ~ 24114 (3-a)zp-? | sin 2u 
po _ OF ED) 2 —4 5 
P —= p?+(a— 3)3z*p \ sin 20 
w= (3—a)zp~ cos 2m { » ( ) 
This solution occurs with coefficient (X,x7,— Yyy,)/327uh. 
I. (iv) with m=2 is obtained by writing sin 2, —cos 2 for cos 2, sin 2 in the 
preceding, and the coefficient is (Xyy,+ Y,2,)/327uh. | 
I. (v) with m=0. 
— | a 
= 0 —2 
V0 - = A Coefficient = {22a +¥i) + (a3) Z; | [32m 
w=0 | 
I]. (iv) with m=2. F=+4 cos 2. 
pe (—s ay 2a \p-9 cos 2w 19? Bele cos 2m +a term in p™* 
6 2p 2 
go sin dort € aa i) p-Ssin Qo} 2@— 2+ sin Qu 
4 6 2 4 
w=” " i cos 2w + ee 2B + i’) p-” cos 2w = 2(2 —h’)p3 cos 2w. 
