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Dynamics  of  Inviscid  Fluids                                                              67


              4.1        Helmholtz  Instability

                 Now  we  will  study  the  stability  of  an  inviscid,  incompressible,  parallel
             fluid  flow,  containing  a  velocity  discontinuity,  following  [22].  Precisely,

              we  will  suppose  that,  above  the  Oz  axis,  the  fluid  moves  with  a  uniform
              velocity  U  in  the  positive  sense  and,  below,  it  moves  with  a  uniform
             velocity  of  equal  magnitude  but  in  the  opposite  sense.  In  this  case,  the
              Oz  axis  represents  a  discontinuity  surface  for  the  velocity  and  it  is  the
             site  of  a  vortex  sheet  of  uniform  circulation  2U  per  unit  of  width.  We
             remember  that  the  circulation  is



                                                       =  [veds


             where V  is  the  magnitude  of  the  velocity  of  the  fluid  and  ds  is  the  arc
             element  along  a  closed  curve  encircling  the  vortex.
                 Such  a  vortex  sheet  is  unstable  1e.,  if  a  displacement  happens  the
             sheet  will  go  away  and  will  not  return  to  its  initial  position.  This  could  be
             shown  by  analytical  studies,  considering  small  sinusoidal  perturbations.

             Here  we  will  numerically  analyze  the  time  evolution  of  such  perturba-
             tions.
                 We  divide  the  vortex  sheet  into  segments  of  equal  length  A  on  Ox  and
             each  segment  will  be  divided  into  m  equispaced  discrete  vortices.  As
             the  total  circulation  per  unit  length  is  2U,  each  discrete  vortex  has  the

             circulation  2UA/m.  We  will  suppose  that  at  the  initial  moment  these
             vortices  are  displaced  from  their  initial  positions  y,  =  0  to  the  positions

                                                2
                                ne  =  asin (|  nee        k=---—2,-1,0,1,::-                         (2.2)



                 Let  us  consider  the  row  of  vortices  containing  the  vortices  k,  k  +m,
              k  +  2m,...  The  complex  potential  generated  by  this  row  is
                                                                                              _
                               oo      )
                                                                                                 =e)
                  wr(z)  =  »  a  mn  log(z  —  z,  —  nA)  =  i—-  log  sin  m  (2  ;
                                                                                             =  24)

                                                                                              DN
              Thus  the  complex  potential  generated  by  all  the  m  rows  which  compose
              the  sheet  is


                            w(2)  =  Dol )=  Di  tg  sin EE)




                 Replacing  this  potential  in  the  relation

                                                     dw            ;
                                                    —  =u-—iv,
                                                     dz
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