double precision function f3(problem,x,y,n,c0)

      implicit none

      double precision x,y,c0
      integer problem,n
      double precision pi
      pi = 3.141592653589793

c n=1: y=0 -diff(u,[y])
c n=2: x=1  diff(u,[x])
c n=3: y=1  diff(u,[y])
c n=4: x=0 -diff(u,[x])

      if(problem .eq. 1) then
         if(n .eq. 1) then
            f3 = -x*exp(x*y)                     
         else if(n .eq. 2) then
            f3 =  y*exp(x*y)         
         else if(n .eq. 3) then
            f3 =  x*exp(x*y)         
         else
            f3 = -y*exp(x*y)         
         endif         
      else if (problem .eq. 2) then
         if(n .eq. 1) then
            f3 = -2*sin(c0*pi*x)**2*sin(c0*pi*y)*cos(c0*pi*y)*c0*pi
         else if(n .eq. 2) then
            f3 =  2*sin(c0*pi*x)*sin(c0*pi*y)**2*cos(c0*pi*x)*c0*pi
         else if(n .eq. 3) then
            f3 =  2*sin(c0*pi*x)**2*sin(c0*pi*y)*cos(c0*pi*y)*c0*pi
         else
            f3 = -2*sin(c0*pi*x)*sin(c0*pi*y)**2*cos(c0*pi*x)*c0*pi
         endif         
      else if (problem .eq. 3) then
         f3 = 0
      else
         write(*,*)'Problem for f3() not implemented.'
         f3 = 0
      endif
      
      return
      end