Wave Motion due to a Submerged Obstacle. 522 
Here ¢ is the velocity potential after (2) has been added to (1); thus the 
equations for a and 8 are 
ca®Ke—"S—Ka = ia B (6) 
tacwe—"T + iga—gB+ ycare-"F+ wa = 0 
From these we obtain the expressions for X; and 7, namely 
X,= eaten ( git oe ite e*F-) sin nx dx, (7) 
0 Jk—o°+tyo 
-/@ 2 } : 
n = caret { ei iesiaione e-*F sin «x de. (8) 
0 Jk—o"+ tuo 
The expression for X, can be divided into two parts 
: [2 3 , Co —K(F-”) gj ] 
es —eate=t | e-"(F-9) gin rode — Dear | STAAL LL (9) 
0 0 G—two—gJK 
If we regard X, as the image of the oscillating cylinder in the free 
surface, we see from the form of the first integral in (9) that part of the 
image is a negative doublet at the image point (0, f). We obtain next the 
velocity potential of the motion produced by a sudden small displacement of 
the cylinder, and we take this to be equivalent to a momentary doublet of 
constant strength. Suppose then that at a time + a doublet is suddenly 
created, maintained constant for a time 67, and then annihilated. The 
velocity potential at any subsequent time ¢ is given by a Fourier synthesis of 
the preceding results for an oscillating cylinder, and we have 
= = i ea) [g]do, (10) 
where [] is the sum of (1) and (9), omitting the factor et. 
Carrying out this integration for the value of ¢ in (1) and for the first part 
of (9) gives simply the momentary doublet at the centre of the cylinder and 
the negative doublet at the image point. These doublets last for a short 
time 67; the subsequent fluid motion is contributed by the second part of 
(9). For this we have to evaluate the real part of 
io) eta (t—7) 
de ; t—7T>0. (11) 
We obtain the value by contour integration; further we simplify the 
result by neglecting y?. We shall make w zero ultimately, but we must 
retain it sufficiently to keep the integrals convergent ; however, at one or 
two stages, superfluous terms may be omitted when it is clear that the final 
limiting values will not be affected. We find for (11) the value 
mm eH) sin fxV(t—7)} KV 
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