2 

 - 0) A2 sinCk-x - w t + 62)^^^"! ^1 ^°s ^^-[^ - ^-[^ "^ '^i) 



2 

 + cj A cosCk^x - u^t + 62) } - 2{a),A^ cos (k x - a),t + 6,) 



2 

 + cj-A^ cosCk-X - (ii^^ ■•■ <S2)}><{w, A, sinfk x - w t + 6^) 



2 

 + u) A2 sinCk^x = w t + iS ) } = 0. 



Since this condition is homogeneous, the first-order potential is 

 the solution to the second-order problem. 



Second-Order Results. 



The free-surface elevation will be modified when terms of second 

 order are included: 



r-y^ 1 sJl) ;^2 fl) (1) 2 (1) 2 



n LX.tj 2 ^ 3t 3y9t ^ 2g *■ 9x -* *■ 9y ^ ' ' q 



= + ^ (gA sin(k X - (jj,t + 6 ) + gA2 sin(k x - m^^ + 152) } x 



g 



2 2 



{A, oj sin(k x - w t + 6 ) + A2 0^2 sin(k2X - cjj2t + ^2) } 



2 

 - J- i[- w A, sin(k X - w t + <5^) - ^^^^^2 sin(k2X - W2t + 62)] 



2 

 + [(JJ-iA, cos (k X - cj^t + "5,) + ^^2^2 '^°^(^^2^ ~ '^2^ "^ '^2-^-' ^ 



(2) 2 2 2 



gn ■'(x,t) = 0)^ A^ sin (k^x - u^^t + 6^) 



2 

 + CO A,A„ sin(k„x - oj^t + 62) sin(k^x - w t + 6^^) 



2 

 + 0) A,A„ sin(k.x - u) t + 6.) sin(k2X - 0^2^ * ^2^ 



2 2 2 

 + 0)2 ^2 sin (k2X - W2t + &2^ 



2 {o). A, sin (k^x - co^^t + 6^) 



+ 2a),u)2A,A sin(k x - w,t + 6j^)sin(k2X - ^2^ + ^2^ 



152 



