622 Prof. T. H. Havelock on an Integral Equation 
plane of yz be a thin rigid barrier which is made to move 
parallel to Oy with a velocity V(t). 
Since the equation of fluid motion is 
Ov 070 
ot = Coe 3 ay Reactant tess el (1) 
with the boundary conditions v=0 for c=h and v= V(¢) for 
«=0, we may write down the solution as in a similar problem 
in the conduction of heat ; we have, for 2>0 and ¢>0, 
2 2nimv 
se 2 h? 
The frictional force, per unit area, on the plane of yz is 
the value of 2u(0v/d#) for 2=0, counting both sides of the 
p'ane ; if we suppose the plane to start from rest, so that 
V (0) =0, this gives, after integrating by parts, 
t ~o 
uit) Vi(r) {1+ 230 Vdr, . (8) 
0 1 
PPA os : n2n2y7/h2 9 
é sin V(a)e dr, . (2) 
0 
In the class of problems we are considering, V(t) is the 
function to be determined and it is the forces on the plane 
which are given. Asa first example, consider motion under 
gravity. Suppose that the plane of yz is vertical and that it 
hag amass o per unit area; we require the motion of the 
plane as it falls under gravity, starting from rest and having 
fixed parallel walls ata distance A on either side. Using (3), 
the equation of motion of the plane can be put at once into 
the form 
t (a) 
Vit (not) V(r) {14235 > rut-nil?} dry. (4) 
0 1 
This is an integral equation of Poisson’s type, which can be 
solved for V'(¢) in the following manner. 
3. In the paper already quoted, Whittaker considers an 
equation 
Het (Hoeee—sls=/), - . . ©) 
in which the nucleus 5 the sum of exponentials, or 
x(x) =Pert?+Qet+....+Ver . . . (6) 
The solution is obtained as 
pa) =f) "F)K(e— ds, @ 
where the solving function is also a sum of w exponentials, or 
I(@) SNe BES Sona SEIN, 656 0 (©) 
178 
