# Weak form of Navier Stokes Equation

1. Aug 21, 2012

### bugatti79

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
3. 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?