Weak form of Navier Stokes Equation

[bugatti79]

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?

 

[mark726]

Thank you for your question. The weak form of the Navier-Stokes equations for 2D flow of viscous incompressible fluids can be determined by following a similar approach to what you have already done. However, there are a few important points to keep in mind.

Firstly, it is important to note that the Navier-Stokes equations are a system of partial differential equations, meaning that they have both a spatial and temporal dependence. Therefore, when determining the weak form, we must take into account both the spatial and temporal derivatives.

Secondly, the weak form is typically derived by multiplying the equations by a test function, rather than a weight function as you have done. The test function should satisfy certain regularity conditions and should also be zero on the boundary where Dirichlet boundary conditions are applied.

With these points in mind, the weak form for the Navier-Stokes equations can be determined as follows:

1. Multiply the first equation by a test function $w$ and integrate over the domain:

$\displaystyle \int_\Omega w \left(u_t + uu_x + vu_y + \frac{1}{\rho}P_x - \nu(u_{xx} + u_{yy})\right) dxdy = 0$

2. Use the product rule for differentiation to expand the convective term:

$\displaystyle \int_\Omega w \left(u_t + u_xu + v_yu + \frac{1}{\rho}P_x - \nu(u_{xx} + u_{yy})\right) dxdy = 0$

3. Use the definition of the divergence to expand the pressure gradient term:

$\displaystyle \int_\Omega w \left(u_t + u_xu + v_yu - \frac{1}{\rho}\nabla \cdot P - \nu(u_{xx} + u_{yy})\right) dxdy = 0$

4. Integrate by parts to move the spatial derivatives onto the test function:

$\displaystyle \int_\Omega w_tu + wu_xu + wv_yu - w\nu(u_{xx} + u_{yy}) dxdy + \int_{\Gamma_2} w \frac{1}{\rho} Pn_x ds = 0$

5. Use the boundary conditions to eliminate the first three terms on the left-hand side