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 Y(t). 

 Since the equation of fluid motion is 



§* =J, ^> (1) 



with the boundary conditions v = for x = h and r = V(7) for 

 .r = 0, we may write down the solution as in a similar problem 

 in the conduction of heat ; we have, for x>0 and t>0, 



n=i n" n Jo 



The frictional force, per unit area, on the plane of yz is 

 the value of 2/jL(duj da?) for a?=0, counting both sides of the 

 plane ; if we suppose the plane to start from rest, so that 

 V(0)— 0, this gives, after integrating by parts, 



(2 M //*)fV(T{ 



Jo 



l + 2Xe- n ^-^-^ hn -}dr. . . (3) 



In the class of problems we are considering, Y(t) is the 

 function to be determined and it is the forces on the plane 

 which are given. As a first example, consider motion under 

 gravity. Suppose that the plane of yz is vertical and that it 

 has a mass a per unit area ; we require the motion of the 

 plane as it falls under gravity, starting from rest and having 

 fixed parallel walls at a distance h on either side. Using (3), 

 the equation of motion of the plane can be put at once into 

 the form 



YXt) + (2fil*h)VY'(T){l + 2Xe- n ^ t -^ k2 }dT = fj. (4) 

 Jo i 



This is an integral equation of Poisson's type, which can be 

 solved for Y'{t) in the following manner. 



3. In the paper already quoted, Whittaker considers an 

 equation 



*(«)+! <f>{s)>c(x-s)ds = f v v), ... (5) 

 Jo 



in which the nucleus is the sum of /* exponentials, or 



K{z)=Yer x + Qei x + + Ye vx . ... (6) 



The solution is obtained as 



<l>(z)=f(.v)-\ X f(s)K(x-s)ds, ... (7) 



Jo 



where the solving function is also a sum of jx exponentials, or 

 K(ci')=Ae*'-f Be^+ + NV* . . . (8) 



