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Navier-Stokes equample continued
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Aug 09, 2019
by
Praetorius, Simon
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Navier-Stokes equation
======================
We consider the incompressible Navier-Stokes equation in a rectangular domain $
`\Omega=(0,Lx)\times(0,Ly)`
$,
We consider the incompressible Navier-Stokes equation in a rectangular domain $
`\Omega=(0,L
_
x)\times(0,L
_
y)`
$,
```
math
\partial_t\mathbf{u}
- \nabla\cdot\big(\nu\mathbf{D}(\mathbf{u})\big) + (\mathbf{u}\cdot\nabla)
\mathbf{u} + \nabla p = \mathbf{f},\qquad\text{ in }\Omega \\
\varrho(
\partial_t\mathbf{u}
+ (\mathbf{u}\cdot\nabla)\mathbf{u}) - \nabla\cdot\big(2\nu\mathbf{D}(
\mathbf{u}
)\big)
+ \nabla p = \mathbf{f},\qquad\text{ in }\Omega \\
\nabla\cdot\mathbf{u} = 0
```
with velocity $
`\mathbf{u}`
$ and pressure $
`p`
$, w.r.t. to boundary conditions $
`\mathbf{u}=\mathbf{g}`
$ on $
`\partial\Omega`
$.
with velocity $
`\mathbf{u}`
$, pressure $
`p`
$, dynamic viscosity $
`\nu`
$,
rate-of-strain tensor $
`\mathbf{D}(\mathbf{u})=\frac{1}{2}(\nabla\mathbf{u}+\nabla\mathbf{u}^\top)`
$,
and external volume force $
`\mathbf{f}`
$, w.r.t. to boundary conditions $
`\mathbf{u}=\mathbf{g}`
$ on $
`\partial\Omega`
$.
For a coarse overview, see also
[
Wikipedia
](
https://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations
)
.
In a weak variational formulation, we try to find $
`\mathbf{u}(t,\cdot)\in V:=H^1(\Omega)^d`
$ and $
`p(t,\cdot)\in Q:=L^2(\Omega)`
$, such that
```
math
\int_\Omega \varrho(\partial_t\mathbf{u} + (\mathbf{u}\cdot\nabla)\mathbf{u})\cdot\mathbf{v} + \nu\nabla\mathbf{u}:\nabla\mathbf{v} - p\nabla\cdot\mathbf{v}\,\text{d}\mathbf{x} = \int_\Omega \mathbf{f}\cdot\mathbf{v}\,\text{d}\mathbf{x}\qquad\forall \mathbf{v}\in V, \\
\int_\Omega q\nabla\mathbf{u}\,\text{d}\mathbf{x} = 0\qquad \forall q\in Q.
```