- #1
bugatti79
- 794
- 1
Homework Statement
Folks, determine the weak form given Navier Stokes eqns for 2d flow of viscous incompressible fluids
##\displaystyle uu_x+vu_y=-\frac{1}{\rho} P_x+\nu(u_{xx}+u_{yy})## (1)
##\displaystyle uv_x+vv_y=-\frac{1}{\rho} P_y+\nu(v_{xx}+v_{yy})##
##\displaystyle u_x+v_y=0##
all defining the domain ##\Omega##
with boundary conditions
##u=u_0##, ##v=v_0## on ##\Gamma_1##
##\displaystyle \nu(u_xn_x+u_y*n_y)-\frac{1}{\rho} Pn_x= \hat t_x##
##\displaystyle \nu(v_xn_x+v_y*n_y)-\frac{1}{\rho} Pn_y= \hat t_y## both on ##\Gamma_2## where
##n_x## and ##n_y## are the direction cosines.
Homework Equations
The Attempt at a Solution
Just focusing on (1). If we set (1)=0, multiply by weight function ##w_1## and set it up as an integral over the domain we have something
##\displaystyle 0=\int_\Omega w_1[uu_x+vu_y+\frac{1}{\rho} P_x-\nu(u_{xx}+u_{yy}]dxdy##
Using the definition of gradient theorems ##\int_\Omega w G_x dxdy=-\int_\Omega w_x G dxdy+ \int_\Gamma n_x wGds## for the third last term ie
##-\nu\int_\Omega u_{xx} dxdy## and letting ##G=u_x## we get
## \displaystyle \int_\Omega w_{1x} u_x dxdy - \int_{\Gamma_2} w_1 u_x n_x ds##
and similarly
## \displaystyle \int_\Omega w_{1y} u_y dxdy - \int_{\Gamma_2} w_1 u_y n_y ds##
but the boundary term answer is given as
##\displaystyle -\int_{\Gamma_2} w_1 \hat t_x ds##
The ##t_x## is similar to what I calculate but contains the additional term ##-\frac{1}{\rho} Pn_x##. Where does that come out of?