MR. J. H. MICHELL OX THE TJlfc^ORY OF FREE STREAM LLNES. 
399 
must have clYjdq = 0 along all these portions. Along a free stream line the pressure 
is constant, since it must be equal to the pressure on the liquid which is at rest. 
Now, in steady motion, we have Bernoulli’s pressure equation 
constant. 
Therefoi'e, along a free stream line 
is constant, and, therefoi’e, V is 
constant. 
All the portions of ^ = 0 which do not correspond to plane boundaries correspond 
to free stream lines, for which V is constant. 
Lastly 
'^Y+ 
dxj ^ 
dj)_ 
dyj 
is zero at reentrant angles of the boundary, and also at points 
where stream lines branch. 
For these points V is infinite and positive. 
In all the cases we shall consider, these infinite points will be along = 0, so that 
a stream line only branches at the boundary. 
We have now reduced the problem to finding a function V which satisfies Laplace’s 
equation ; is finite and continuous in that half of the plane u for which q is 
positive ; is constant along parts of ^ = 0 ; along the other parts satisfies clYjdq — 0 ; 
and at points along it is + • 
This problem has an obvious solution. 
It is plain that V is merely the potential due to conductors coinciding with those 
parts of ^ = 0 for which V is constant, and having that constant as potential, together 
with masses at the points for which V is infinite. 
The general solution of this has been given in Problem 11 , 
Let U be the conjugate of V, so that U + iV — Iq). Then, translating 
Problem 11 . into the present notation we have 
d(U -h W) _ {u - UrY’-' 
ddp + iq) {u — q),y 
and for a jooint mass two factors of the denominator coincide. 
Write this 
d(U -h iV) 
dll 
= ./'(«)» 
SO that 
U -f ^ V = \f{u)dij. 
Now 
V - log 
dz 
dw 
dv'' ’ 
U = ^'log 
therefore 
