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-2Mu(dy(uy))); (Second component of traction vector on bottom)
trac1ypoz=int1d(Th,3)(-p+2Mu(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+2Mu(dy(uy))); (Second component of traction vector on top)
trac1yneg=int1d(Th,2)(p-2Mu(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-2Mu(dy(uy));
func trac1ypoz=-p+2Mu(dx(ux));
func trac2ypoz=Mu*(dx(uy)+dy(ux));
func trac1xpoz=Mu*(dx(uy)+dy(ux));
func trac2xpoz=-p+2Mu(dy(uy));
func trac1yneg=p-2Mu(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