Ogilvie 



some good justification (or rationalization) for proceeding this way, 

 but it is really an open question. 



With such restrictions and reservations expressed, we can 

 write down a "solution" of the above problem. Define: 



^ = Re (yz)}; $(x,y) = Re{F(z)}; 



Note that: 



fo(^) = '^q{^.^)-> k(x) = g[ ^o,(x,0)]'^ 



Then the solution is given by: 



F'(z)=-i5-j dsp'(s)j ^Jl^exp[-i^ duk(u)J. 



The t, integral is a contour integral starting at x = - oo , located 

 entirely in the lower half- space. It should pass above the location 

 of the singularity in k(z). This solution represents no disturbance 

 at the upstream infinity, as one would expect. 



Far downstream, this solution can be approximated: 



F'(z) ^ 2ie''" \ ds p"(s) exp fi/cs - i \ du[k(u) - k\\ , 



where K = g/U . Then, from (5-22), we obtain the wave shape far 

 aft of the body: 



H(x) «>^ - — \ ds p"(s) sin[ /C(x - s) + K(s)] , 



g J.QD 



where 



K(s) = \ du [ k(u) - k] . 



Calculation of the wave resistance is then very simple in principle. 

 (In practice, it is a very tedious calculation.) Note that the expres- 

 sion for the wave shape downstream does not require knowledge of 



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