anchor.diff:          x(i) = (i-2)*dx
       end do
   
  -    x(1)=x(pasx)
  -    x(2)=x(pasx+1)
  -
  -    do j=1,pasy      
  -        y(j)=-Ly+(j-1)*dy
  +    x(1) = x(pasx)
  +    x(2) = x(pasx+1)
  +
  +    do j = 1, pasy      
  +        y(j) = -Ly+(j-1)*dy
       end do
   
       temp1 = g*h0/f/L/U
  -    do i=1,pasx+1
  +    do i = 1, pasx+1
           temp2 = 0.01*sin(pi*x(i)/Lx)
  -        do j=1,pasy
  +        do j = 1,pasy
               if(abs(y(j)).LT.1) then
  -                eta(i,j)=temp1*temp2-0.5*y(j)**2*(1+temp2)
  +                eta(i,j) = temp1*temp2 - 0.5*y(j)**2*(1+temp2)
               else
  -                eta(i,j)=temp1*temp2+(0.5-abs(y(j)))*(1+temp2)
  +                eta(i,j) = temp1*temp2 + (0.5-abs(y(j)))*(1+temp2)
               endif
           end do
       end do
  
positive.code:     endif
  
      write(*,*)
      write(*,*)'Un pas de temps =',T*dt,'sec.'
  
      ! ------------------------------------------------------------------      
      !                      Calcul du profil initial
  
negative.code:     kmoy  = 0.
      save  = 1
  
      write(*,*) 'Ly/L =', Ly, '(', (3./4. + g*h0/(2.*f*U*L)), ')'
      write(*,*) 'Bu   =', Bu,'Ro   =', Ro
      write(*,*) 'dx   =', dx, ' ; dy   =', dy 
      write(*,*) 'pasx =', pasx,' ; pasy =', pasy
      write(*,*) 'T    =', T, 'dH   =', f*L*U/g
      write(*,*) 'dt (max) =', dt, 'sec.' 
      write(*,*) 'Le canal fait ', 2*Lx*L, 'mtres de long'
      write(*,*) '           et ', 2*Ly*L, 'mtres de large'
      write(*,*) 'tol (1D-6)?'
      read(*,*) tol
      write(*,*) 'w (1.825)?'
  