Boundary conditions treatment

In this section, the treatment of several types of boundary conditions is explained in details. First, the boundary $\partial\Omega$ of the computational domain $\Omega$ is decomposed into several disjoint parts. On each part a different boundary condition is prescribed. Following the notation from FEMFLUID implementation, the following boundary parts are considered:

• $\Gamma_{Z}$ - fixed wall,
  
  
  

  $$\mathbf{v}=0$$

• $\Gamma_{D}$ - inlet/Dirichlet part of boundary,
  
  
  

  $$\mathbf{v}=\mathbf{v}_D$$

• $\Gamma_{W_t}$ - moving part of boundary with velocity $\mathbf{w}_{D}$,$\\ [3mm]$
  

  $$\mathbf{v}=\mathbf{w}_D$$

  In this section a simplified analysis is carried only for Navier-Stokes system of equations. This boundary condition is employed only on moving meshes which follows the motion of the boundary $\Gamma_{Wt}$. The numerical simulation on moving meshes can be treated with the aid of ALE formulation.
• $\Gamma_O$ - outlet,$\\ [3mm]$
  

  do-nothing boundary condition (see, e.g., [4]),

• $\Gamma_{SYM}$ - symmetry condition,$\\ [3mm]$
  
  
  

  $$\mathbf{v}\cdot \mathbf{n}=0$$

  $$\frac{\partial (\mathbf{v}\cdot \mathbf{t})}{\partial \mathbf{n}} =0$$

  
  $\Gamma_{SYM_y}$ - symmetry condition for boundary with the fixed $y$-coordinate $y=const$, i.e. the normal vector is $\mathbf{n}=(0, \pm 1)$,
  $\Gamma_{SYM_x}$ - symmetry condition for boundary with the fixed $x-$coordinate $x=const$, i.e. the normal vector is $\mathbf{n}=( \pm 1, 0)$.

In order to describe the weak formulation we start from incompressible system of Navier-Stokes equations. Here, the system of Navier-Stokes equations in the form

$$\frac{\partial \mathbf{v}}{\partial t} + (\mathbf{v}\cdot \nabla) \mathbf{v}- \nu \triangle \mathbf{v} + \nabla p = 0,$$ (1)

$$\nabla \cdot \mathbf{v} = 0,$$

is considered. In agreement the following boundary conditions are prescribed on mutually disjoint parts of $\partial\Omega=\Gamma_D \cup \Gamma_Z \cup \Gamma_{Wt} \cup \Gamma_O \cup \Gamma_{S} \cup \Gamma_{Sx} \cup \Gamma_{Sy}$. Here, we list the prescribed boundary conditions

$$\mathbf{v}(\mathbf{x}) = \mathbf{v}_D(\mathbf{x}), \qquad \mathbf{x}\in \Gamma_{D},$$

$$\mathbf{v}(\mathbf{x}) = 0, \qquad\qquad \mathbf{x}\in \Gamma_{0},$$

$$\mathbf{v}(\mathbf{x}) = \mathbf{w}_{D}(\mathbf{x}), \qquad \mathbf{x}\in \Gamma_{Wt},$$

$$-\nu \frac{\partial \mathbf{v}}{\partial \mathbf{n}} + (p-p_0) \mathbf{n}+ \frac12 (\mathbf{v}\cdot \mathbf{n})^{-} \, \mathbf{v} = 0, \qquad \qquad \ \mathbf{x}\in \Gamma_{O},$$ (2)

$$\mathbf{v}_1(\mathbf{x}) = 0, \quad \frac{\partial \mathbf{v}_2}{\partial \mathbf{n}} (\mathbf{x}) = 0, \qquad \qquad \mathbf{x}\in \Gamma_{SYM_x},$$

$$\frac{\partial \mathbf{v}_1}{\partial \mathbf{n}} (\mathbf{x}) = 0, \quad \mathbf{v}_2(\mathbf{x}) = 0, \qquad \qquad \mathbf{x}\in \Gamma_{SYM_y},$$

$$\mathbf{v}\cdot \mathbf{n}= 0 , \quad \frac{\partial (\mathbf{v}\cdot \mathbf{t})}{\partial \mathbf{n}} = 0, \qquad \qquad \mathbf{x}\in \Gamma_{SYM},$$

where $\mathbf{n}$ is the unit outward normal to the boundary of $\partial\Omega$ and $\tt$ is the unit tangent vector on the boundary of $\partial\Omega$ (oriented in such a way, that the domain $\Omega$ is on the left-hand side).

Condition (2) can be replaced by the well known do-nothing condition

$$-\nu \frac{\partial \mathbf{v}}{\partial \mathbf{n}} + (p-p_0) \mathbf{n}= 0, \qquad \qquad \ \mathbf{x}\in \Gamma_{O}.$$ (3)

If the boundary $\Gamma_{O}$ is really the outlet part of the boundary (t.j. $\mathbf{v}\cdot \mathbf{n}> 0$), both conditions are equal. If this is not the case, the boundary condition (2) rejects the kinetic energy coming from the outlet part of boundary.

Further, the space of test functions needs to be chosen in agreement with the prescribed boundary conditions. The following conditions are considered :

$$\mathbf{N}(\mathbf{x})=0 \mathbf{x}\in \Gamma_D \cup \Gamma_0 \cup \Gamma_{Wt},$$ a) (4)

$$N_1(\mathbf{x})=0 \mathbf{x}\in \Gamma_{SYM_x},$$ b)

$$N_2(\mathbf{x})=0 \mathbf{x}\in \Gamma_{SYM_y},$$ c)

$$(\mathbf{N}\cdot \mathbf{n}) (\mathbf{x})=0 \mathbf{x}\in \Gamma_{SYM}.$$ d)

The space $V$ is defined by

$$V=\left\{ \mathbf{N}\in H^1(\Omega)^2,\qquad \mathbf{N}\mbox{ satisfies (\ref{eq:TFNbc}a-d)} \right\}.$$

Now, take a test function $\mathbf{N}\in V$, multiply system (1) and integrate over $\Omega$. For simplicity we denote

$$\Gamma_4 = \Gamma_{O} \cup \Gamma_{SYM} \cup \Gamma_{SYM_x} \cup \Gamma_{SYM_y}.$$

With the fact that $\mathbf{N}\equiv 0$ on $\partial\Omega\setminus \Gamma_4$ and by applying Green's theorem we get step by step

$$\int_\Omega -\nu \mathbf{v}\cdot \mathbf{N}dx = \int_{\Gamma_4 } ... ... \mathbf{N}d S + {\nu \int_\Omega \nabla \mathbf{v}\cdot \nabla \mathbf{N}dx },$$ (5)

and similary

$$\int_\Omega \nabla p \cdot \mathbf{N}dx = \int_{\Gamma_{4}} p \mathbf{n}\cdot \mathbf{N}d S - {\int_\Omega p \nabla \cdot \mathbf{N}dx}.$$ (6)

Further, for the convective term ( $\nabla \cdot \mathbf{v}= 0$ from cotinuity equation)

$$\int_\Omega (\mathbf{v}\cdot \nabla) \mathbf{v}\cdot \mathbf{N}dx = c (\mathbf{v}; \mathbf{v},\mathbf{N}) - \int_{\Omega} \frac12 (... ...ma_{4}} \frac12 (\mathbf{v}\cdot \mathbf{n}) \, \mathbf{v}\cdot \mathbf{N}d S =$$

$$= c (\mathbf{v}; \mathbf{v},\mathbf{N}) + \int_{\Gamma_{4}} \frac12 (\mathbf{v}\cdot \mathbf{n}) \, \mathbf{v}\cdot \mathbf{N}d S$$ (7)

where the tri-linear skew-symmetric form $c$ reads

$${ c (\mathbf{u}; \mathbf{v},\mathbf{w})}= {\int_\Omega \left[ \fr... ...f{w}- \frac12 (\mathbf{u}\cdot \nabla) \mathbf{w}\cdot \mathbf{v}\right] dx }.$$

Finally, with the use of Eqs. 5, 6, 7 we get the weak formulation of Navier-Stokes equations

$$\Bigl(\partial_t \mathbf{v}, \mathbf{N}\Bigr)_\Omega + \nu \Bigl(... ...ot \mathbf{v}, q\Bigr)_\Omega + {\cal B} ( \mathbf{v}, p ; \mathbf{N},q ) = 0$$

where on the left hand side there is - except the standard terms - also the boundary integral

$${\cal B} ( \mathbf{v}, p ; \mathbf{N},q ) = \int_{\Gamma_4} \lef... ...n}+ \frac12 (\mathbf{v}\cdot \mathbf{n}) \mathbf{v}\right) \cdot \mathbf{N}dS.$$

This form needs to rewritten with respect to the boundary conditions. Consider now the different parts of the boundary $\Gamma_4$: first on $\Gamma_{O}$

$$\int_{\Gamma_{O}} \left(-\nu \frac{\partial \mathbf{v}}{\partial ... ...ac12 (\mathbf{v}\cdot \mathbf{n})^{+} \, \mathbf{v}+ p_0 \right) \mathbf{N}dS,$$

then on $\Gamma_{SYM_x}$

$$\int_{\Gamma_{SYM_x}} \left(-\nu \frac{\partial \mathbf{v}}{\part... ...rac12 (\mathbf{v}\cdot \mathbf{n}) \mathbf{v}\right) \mathbf{N}dS \qquad \qquad =$$

$$= \int_{\Gamma_{SYM_x}} \left(-\nu {\frac{\partial \mathbf{v}_2 }... ...\mathbf{n}_2} + \frac12 \mathbf{v}_1 \cdot \mathbf{v}_2 \right) \mathbf{N}_2 dS = 0,$$

and similary on $\Gamma_{SYM_y}$

$$\int_{\Gamma_{SYM_x}} \left(-\nu \frac{\partial \mathbf{v}}{\part... ...f{n}+ \frac12 (\mathbf{v}\cdot \mathbf{n}) \mathbf{v}\right) \mathbf{N}dS = 0.$$

Finally, on $\Gamma_{SYM}$ we write

$$\mathbf{N}= \underbrace{(\mathbf{N}\cdot \mathbf{n})}_{=0} \mathb... ...athbf{N}\cdot \mathbf{t}) \mathbf{t}= (\mathbf{N}\cdot \mathbf{t}) \mathbf{t},$$

so that

$$\int_{\Gamma_{SYM}} \left(-\nu \frac{\partial \mathbf{v}}{\partia... ...rac12 (\mathbf{v}\cdot \mathbf{n}) \mathbf{v}\right) \mathbf{N}dS \qquad \qquad =$$

$$\int_{\Gamma_{SYM}} \left(-\nu \frac{\partial (\mathbf{v}\cdot \m... ...bf{n}} + p (\mathbf{n}\cdot \mathbf{t}) \right) (\mathbf{N}\cdot \mathbf{t}) dS = 0.$$

The following terms are included in term $\cal B$

$${\cal B} ( \mathbf{v}, p ; \mathbf{N},q ) = \int_{\Gamma_{O}} \l... ...ac12 (\mathbf{v}\cdot \mathbf{n})^{+} \, \mathbf{v}+ p_0 \right) \mathbf{N}dS,$$

which differs from the case of do-nothing boundary condition, where

$${\cal B} ( \mathbf{v}, p ; \mathbf{N},q ) = \int_{\Gamma_{O}} \left(\frac12 (\mathbf{v}\cdot \mathbf{n}) \, \mathbf{v}+ p_0 \right) \mathbf{N}dS.$$