I am studying about the identification of immersed obstacles in fluid flows by using a traction vector (stress vector) on the boundary as an inverse problem procedure. The domain is rectangular and the right edge is the outflow boundary. At this outflow boundary sigma(u,p).n=0

I just want to ask you about the traction vector(sigma). I want to correct that on the outflow-boundary this traction vector is exactly zero. All traction vectors are zero on the whole boundary in my freefem++ if I write the boundaries as follows.

trac1xneg=int1d(Th,1)(-Mu*(dx(uy)+dy(ux)));( First component of traction vector on bottom)

trac2xneg=int1d(Th,1)(p-2*Mu*(dy(uy))); (Second component of traction vector on bottom)

trac1ypoz=int1d(Th,3)(-p+2*Mu*(dx(ux))); (fist component of the traction vector on outflow)

trac2ypoz=int1d(Th,3)(Mu*(dx(uy)+dy(ux))); (second component of the traction vector on outflow)

trac1xpoz=int1d(Th,1)(Mu*(dx(uy)+dy(ux)));( First component of traction vector on top)

trac2xpoz=int1d(Th,1)(-p+2*Mu*(dy(uy))); (Second component of traction vector on top)

trac1yneg=int1d(Th,2)(p-2*Mu*(dx(ux))); (First component of traction vector on inlet)

trac2yneg=int1d(Th,2)(-Mu*(dx(uy)+dy(ux)));(Second component of traction vector on inlet)

If I define it as a function, the traction vector on outflow and also other boundaries is nonzero.

func trac1xneg=-Mu*(dx(uy)+dy(ux));

func trac2xneg=p-2*Mu*(dy(uy));

func trac1ypoz=-p+2*Mu*(dx(ux));

func trac2ypoz=Mu*(dx(uy)+dy(ux));

func trac1xpoz=Mu*(dx(uy)+dy(ux));

func trac2xpoz=-p+2*Mu*(dy(uy));

func trac1yneg=p-2*Mu*(dx(ux));

func trac2yneg=-Mu*(dx(uy)+dy(ux));

Could you give me a suggestion to correct that traction is exactly zero on the just outflow boundary? How can I show it?

Thank you