1. First-Order Solution of the Boundary Value Problem. 



This solution results from the superposition of the velocity poten- 

 tials for individual waves: 



(-1-1 g^l ]r V 



(j)*- -"(x.y.t) = — - e V cos (k^x - co^t + 6^) 



^^^2 k V 

 + -^ e 2^ cos (k^x - w^t + &2^ • (E-15) 



Check the solution: 

 ,V'^ - 0. 



-~ — ->- because of exponential function. 



y ->- _ 00 9y J^ 



8^(j)'^^^ 3(j)*^^^ , k,y ,, . X -, 

 — ^^—2- + g 3 = - gf^^A^e 1' cos (k^x - w^t + 6^) 



o t 



k V 



- ga)„A-e 2^ cos (k^x - w t + (5 ) 



k t 

 + g {cj,A e 1 cos(k,x - 0) t + 6 ) 



k V 

 + co^A^e 2-'^ cos (k^x - co^t + ^o-' -^ ~ ^' 



Therefore, this is a solution. 

 Surface elevation then becomes: 



- k^ sin(k X - w t + 6 ). (E-16) 



To prepare for the second-order solution, construct the right-hand 

 side of the free-surface boundary condition (E-14) : 



•■2 at 3y ^ g^2 ■*■ 2 8y 3x ax3t 



- 2 ^ ~378t-1^^0 = i ^g^ ^""^^^ - "1^ " ^P 



+ gA^ sinCk-X - cj t + 6 )}{0} - 2{- w.A. sin(k x - (jj,t + 6,) (E-17) 



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