Long Waves in Canals and Standing Waves in Closed Basins 207 



the middle of the canal. For the equations (VI. 100) one can select a solution 

 in the form of the equation (VI. 15). Then (VI. 18) is replaced by 



aU-fV = KgZ , 



dZ 



and 



oV+fU = -g dr 



d LdZ\ , /<r 2 -/ 2 2 . , fx dh\ 7 



(VI. 104) 



whereas for the boundary conditions at the walls of the canal 



*(f +?*)-• 



replaces (VI. 19) for y — ±a, if the canal has the width of —a to -\-a. 

 With the boundary condition for y = —a, we can derive from these 



equations 



+y 



hV = * £-</*/*» J (^ -gh\Ze-v*l°»dy (VI. 105) 



—a 



and the boundary condition for y — -\-a takes the form 



+a 



f H ~ gh ) Ze ~ {fKh)yd y = ° • ( YL 106 > 



As Z is always positive, this equation is equivalent to the condition 



t 



x~2 



a = glh, (VI. 107) 



in which h x is a kind of an average value between the largest and the smallest 

 value of h. 



When the profile of the canal slopes up on both sides, there must at least 

 be two values of y, for which h becomes = h x ; let these be y x and y 2 . Then 

 we see that hVe~ {fHla)y increases from zero at —a to a maximum at y = y l9 

 then decreases to a minimum at y — y 2 , and finally becomes again zero on 

 the other bank for y = -\-a. Therefore, V must change its sign between 

 y x and y 2 , so that Kis positive for — a < y < y x and negative for y x < y < +a, 

 i.e. on both sides of the canal the transverse motions are in an opposite 

 direction. 



Although with the above-mentioned equations the problem is solved, we 

 cannot gain an insight into the entire oscillatory process until we calculate 

 special cases. If a = +/, which would be applicable to a semi-diurnal wave 

 at the Pole, or to a diurnal wave at 30° latitude, one obtains from (VI. 104) 



g dy 



