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Weak form of Navier Stokes Equation

  1. Aug 21, 2012 #1
    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?
     
  2. jcsd
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