OF WHICH IS MOVING ROTATIONALLY AND PART IRROTATIONALLY. 387 
Now 
dA _ ca(y— Y) 
dy a+ b ’ 
dA — cbx 
dx a + b 
inside the cylinder. 
At the surface of the cylinder, — and — have the same values whether calculated 
dx dy 
from the value of A inside it or outside it. 
In order to have the rotational motion continuous with the irrotational motion, it is 
necessary that all over the surface of the cylinder 
f %-Y) ca(y—Y) 
b 2 a + b 
cbx 
a + b’ 
But these equations cannot be satisfied unless a=b, Y=0. 
The solution may, however, be completed for a finite portion of the plane of x, y 
outside the cylinder by means of Example III. which follows 
In Example III., put cy = 0, this will make 
\=A of Example 111.==/.—^- log (\/a 2 +e+ v / & + e)+^^ 2 (^ 3 ” (y— Y) 2 ) 
_ f(a°~ 
ab(a i 
where 
s 8 ■ (y- Y) 2 
a? + e J 2 + e 
+ & 2 ) 
- b 2 ) 
s/ (a 3 +e)(6 3 4-e) 
ar 
(y- Y) ! 
a 2 + e 
V 2, + e 
Therefore the current function A of this example is 
x-A=/AV 5 log (v/oa+€+ 
2/ 
cib & 7 a? — V- 
ab(a 2 — b 2 ) v A 1 7 \<P + e 6 2 + e 
— £cY 
This is equivalent to the form in the abstract printed in the Proceedings. 
For the motion to be possible, it must be supposed to be confined to a cylinder of 
finite section, appropriate surface conditions being supplied at the surface of the 
bounding cylinder. 
11. Example II. In the preceding example it was shown that none of the methods 
would apply for the whole of space surrounding the rotationally moving liquid. 
Knowing Kirchhoff’s investigation of the rotating elliptic vortex cylinder, let the 
form of the equation given in Art. 6 be considered, and the following simple case of 
it taken. 
(d * l_d 
\d/r r dr 
1 d 3 
r 2 dd 2 
Trr X — =C 
3 D 2 
