156 Dr. E/. A. Houstoun on tlie Damping of 



As a solution we assume 



u = e tat Z sin mx sin ny, 

 where Z is a function of z alone. Substitution in (6) gives 



(i<rvn 2 -v 2 )Z+griA h Z dz - ^^| =0. . . (7) 



Z must be zero when ^ = and 'dZ/'dz must be zero when 



z — h. The solution satisfying these conditions is 



Z = coshA;/i — cosh k(h — z) (8) 



On substituting (8) in (7) we find that 



a 2 -vian 2 + viak 2 = 0. . . (9) 

 and 2 



(* 2 -ghm 2 )-vian 2 + ^ta,nhkh=0. . . . (10) 



In order to satisfy the boundary conditions we must put 

 m = qirla and n = r7r/5, where q and r are positive integers. 

 From (9) and (10) it is obvious that both h and a are 

 complex. Put G=p + itc ; k is then the reciprocal of the 

 modulus of decay 



The complete value of n is given by 



.w=real part SB^'^sin^ sin ^— sin —~ [cosh kh — cosh k{li—z)~\. (11) 



a 



We shall now investigate the rate of decay of the different 

 terms of the above series. Reserving the justification until 

 later, we shall assume that the term vicn 2 may be neglected 

 in equations (9) and (10) and that tanh hk may be put equal 

 to unity. 



In place of (9) and (10) we have then 



* 2 =V' (12) 



and (a 2 -g7im 2 ) + ^ = (13) 



If p + ik be substituted for a in the above equations, and k 2 

 be neglected in comparison with p 2 , we find that tc is given 

 by the equation 



It is easy to show that this equation has always one, and 

 never more than one, real positive root. This positive root 

 gives the rate of damping, since that rate must be positive ; 

 its reciprocal is the modulus of decay. 



