of Edinburgh, Session 1869 - 70 . 
89 
A similar investigation applies to equipotential lines. The image 
of a point in a rectilineal equipotential line is the same in position 
as the image in a stream line, but of opposite sign. No source 
can lie on an equipotential line. Hence, to show that for right 
equipotential line the points must image two by two, we have only 
to remove all sources on one side of the line, placing equal sources 
of the same sign at their images. The line is still equipotential, 
therefore we may suppose it charged to constant potential, and all 
sources removed. Hence all stream lines become rectilineal, which 
is absurd. Similarly if a circle is equipotential, the sources must 
balance about it tw r o by two, i.e., must be in a straight line with 
its centre, at distances to which the radius is mean proportional— 
otherwise we can find a system reducible to a single point at the 
centre of the circle, and in which all stream lines are rectilineal. 
Hence, no incomplete system can have a rectilineal or circular 
potential line. 
Points of Inflexion occur at all points on the locus- 
d 2 u 
d 2 u 0 d 2 u dii , 
- + 2- — + 
dx 2 dxdy dx 
_ t dy 
dy 2 dx 
= 0 
(17) 
Remembering that 
d 2 u 
dx 2 
d 2 u 
df 
sin 2 0 
*,2 
we can readily bring (17) into the form— 
v /sin 20\ ^ /cos (0 + 0')\ _ v /cos 2 O' 
'sin (0 + 0') 
(o+jn\ = 
rr' J 
= 0, 
or, 
2 sin (2 0 - O' - 0 ") _ q 
9 f II 
PT X 
■ ( 18 ). 
In this last expression O' and 0" may assume the value 0. 
The radius of curvature may be similarly expressed, but such 
expressions can hardly have a practical application. 
The cases of practical interest are mainly those where the number 
of sources is small. We have already examined the cases of two 
