Stokes equations and change of variable
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Let $Omega$ be domain of $mathbbR^n$ and $Phi : Omega to Phi(Omega)$ a deformation. Consider the Stokes equations written in the deformed configuration
beginalign
- 2mu operatornamediv(D(u)) + nabla p &= f, quad textin Phi(Omega) \
operatornamediv(u) &= 0, quad textin Phi(Omega) \
endalign
where $u$ is the velocity of the fluid, $p$ the pressure, $mu>0$ is the constant viscosity, $f$ is an external force and $D$ is the operator defined by
$$
D(u) = frac12 (nabla u + nabla u^T).
$$
How can the Stokes equations be written in the domain $Omega$ using a change of variable ?
pde divergence change-of-variable linear-pde
asked Aug 16 at 21:58
PeteAgor
164
add a comment |Â
up vote
3
down vote
favorite
Let $Omega$ be domain of $mathbbR^n$ and $Phi : Omega to Phi(Omega)$ a deformation. Consider the Stokes equations written in the deformed configuration
beginalign
- 2mu operatornamediv(D(u)) + nabla p &= f, quad textin Phi(Omega) \
operatornamediv(u) &= 0, quad textin Phi(Omega) \
endalign
where $u$ is the velocity of the fluid, $p$ the pressure, $mu>0$ is the constant viscosity, $f$ is an external force and $D$ is the operator defined by
$$
D(u) = frac12 (nabla u + nabla u^T).
$$
How can the Stokes equations be written in the domain $Omega$ using a change of variable ?
pde divergence change-of-variable linear-pde
asked Aug 16 at 21:58
PeteAgor
164
You can get proper formatting for operators like $operatornamediv$ using e.g.operatornamediv.
– joriki
Aug 16 at 22:58
I'm confused, if $operatornamediv = 0$ in the deformed configuration, then doesn't the first equation reduce to $nabla p = f$ as it is also in the same deformed configuration?
– Mattos
Aug 17 at 7:47
I forgot the symetric gradient operator in the divergence operator. I edited the question.
– PeteAgor
Aug 17 at 8:09
Now asked on MathOverflow: Change of variable for the Stokes equations.
– Martin Sleziak
Aug 17 at 16:15
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $Omega$ be domain of $mathbbR^n$ and $Phi : Omega to Phi(Omega)$ a deformation. Consider the Stokes equations written in the deformed configuration
beginalign
- 2mu operatornamediv(D(u)) + nabla p &= f, quad textin Phi(Omega) \
operatornamediv(u) &= 0, quad textin Phi(Omega) \
endalign
where $u$ is the velocity of the fluid, $p$ the pressure, $mu>0$ is the constant viscosity, $f$ is an external force and $D$ is the operator defined by
$$
D(u) = frac12 (nabla u + nabla u^T).
$$
How can the Stokes equations be written in the domain $Omega$ using a change of variable ?
pde divergence change-of-variable linear-pde
asked Aug 16 at 21:58
PeteAgor
164
Let $Omega$ be domain of $mathbbR^n$ and $Phi : Omega to Phi(Omega)$ a deformation. Consider the Stokes equations written in the deformed configuration
beginalign
- 2mu operatornamediv(D(u)) + nabla p &= f, quad textin Phi(Omega) \
operatornamediv(u) &= 0, quad textin Phi(Omega) \
endalign
where $u$ is the velocity of the fluid, $p$ the pressure, $mu>0$ is the constant viscosity, $f$ is an external force and $D$ is the operator defined by
$$
D(u) = frac12 (nabla u + nabla u^T).
$$
How can the Stokes equations be written in the domain $Omega$ using a change of variable ?
pde divergence change-of-variable linear-pde
asked Aug 16 at 21:58
PeteAgor
164
edited Aug 17 at 8:07
asked Aug 16 at 21:58
PeteAgor
164
asked Aug 16 at 21:58
PeteAgor
164
asked Aug 16 at 21:58
PeteAgor
164
164
You can get proper formatting for operators like $operatornamediv$ using e.g.operatornamediv.
– joriki
Aug 16 at 22:58
I'm confused, if $operatornamediv = 0$ in the deformed configuration, then doesn't the first equation reduce to $nabla p = f$ as it is also in the same deformed configuration?
– Mattos
Aug 17 at 7:47
I forgot the symetric gradient operator in the divergence operator. I edited the question.
– PeteAgor
Aug 17 at 8:09
Now asked on MathOverflow: Change of variable for the Stokes equations.
– Martin Sleziak
Aug 17 at 16:15
add a comment |Â
You can get proper formatting for operators like $operatornamediv$ using e.g.operatornamediv.
– joriki
Aug 16 at 22:58
I'm confused, if $operatornamediv = 0$ in the deformed configuration, then doesn't the first equation reduce to $nabla p = f$ as it is also in the same deformed configuration?
– Mattos
Aug 17 at 7:47
I forgot the symetric gradient operator in the divergence operator. I edited the question.
– PeteAgor
Aug 17 at 8:09
Now asked on MathOverflow: Change of variable for the Stokes equations.
– Martin Sleziak
Aug 17 at 16:15
You can get proper formatting for operators like $operatornamediv$ using e.g.
operatornamediv.– joriki
Aug 16 at 22:58
You can get proper formatting for operators like $operatornamediv$ using e.g.
operatornamediv.– joriki
Aug 16 at 22:58
I'm confused, if $operatornamediv = 0$ in the deformed configuration, then doesn't the first equation reduce to $nabla p = f$ as it is also in the same deformed configuration?
– Mattos
Aug 17 at 7:47
I'm confused, if $operatornamediv = 0$ in the deformed configuration, then doesn't the first equation reduce to $nabla p = f$ as it is also in the same deformed configuration?
– Mattos
Aug 17 at 7:47
I forgot the symetric gradient operator in the divergence operator. I edited the question.
– PeteAgor
Aug 17 at 8:09
I forgot the symetric gradient operator in the divergence operator. I edited the question.
– PeteAgor
Aug 17 at 8:09
Now asked on MathOverflow: Change of variable for the Stokes equations.
– Martin Sleziak
Aug 17 at 16:15
Now asked on MathOverflow: Change of variable for the Stokes equations.
– Martin Sleziak
Aug 17 at 16:15
add a comment |Â
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You can get proper formatting for operators like $operatornamediv$ using e.g.
operatornamediv.– joriki
Aug 16 at 22:58
I'm confused, if $operatornamediv = 0$ in the deformed configuration, then doesn't the first equation reduce to $nabla p = f$ as it is also in the same deformed configuration?
– Mattos
Aug 17 at 7:47
I forgot the symetric gradient operator in the divergence operator. I edited the question.
– PeteAgor
Aug 17 at 8:09
Now asked on MathOverflow: Change of variable for the Stokes equations.
– Martin Sleziak
Aug 17 at 16:15