Obstruction to Navier-Stokes blowup with cylindrical symmetry
Is there a known obstruction to cylindrically symmetric solutions (with swirl) of 3D Navier-Stokes blowing up in finite time ?
ap.analysis-of-pdes navier-stokes
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Is there a known obstruction to cylindrically symmetric solutions (with swirl) of 3D Navier-Stokes blowing up in finite time ?
ap.analysis-of-pdes navier-stokes
add a comment |
Is there a known obstruction to cylindrically symmetric solutions (with swirl) of 3D Navier-Stokes blowing up in finite time ?
ap.analysis-of-pdes navier-stokes
Is there a known obstruction to cylindrically symmetric solutions (with swirl) of 3D Navier-Stokes blowing up in finite time ?
ap.analysis-of-pdes navier-stokes
ap.analysis-of-pdes navier-stokes
asked 11 hours ago
Jean Duchon
2,640415
2,640415
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You certainly know this one, but some readers could ignore it. The fact that NS iw globally well-posed in 2D is due to the so-called Ladyzhenskaia inequality
$$|f|_4^2le c|f|_2|nabla f|_2.$$
The fact that this does not hold in 3D is the reason why there is a 1M-dollars problem...
But if the flow is axisymmetric, even with swirl, and if the domain is a container between two cylinders ($0<r_0<sqrt{x^2+y^2}<r_1<infty$), then LI is still valid, and the solution exists, is unique and smooth whenever the initial energy is finite.
To summarize, the only major difficulty is when the domain reaches the symmetry axis.
By the way, I have proved (see my notes at the Compte Rendus in 1991 and 1999) that in absence of viscosity, that is for the Euler equation, the vorticity of such a fluid (incompressible, axisymmetric, with swirl), generically increases linearly in time. If the flow exists globally, of course.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You certainly know this one, but some readers could ignore it. The fact that NS iw globally well-posed in 2D is due to the so-called Ladyzhenskaia inequality
$$|f|_4^2le c|f|_2|nabla f|_2.$$
The fact that this does not hold in 3D is the reason why there is a 1M-dollars problem...
But if the flow is axisymmetric, even with swirl, and if the domain is a container between two cylinders ($0<r_0<sqrt{x^2+y^2}<r_1<infty$), then LI is still valid, and the solution exists, is unique and smooth whenever the initial energy is finite.
To summarize, the only major difficulty is when the domain reaches the symmetry axis.
By the way, I have proved (see my notes at the Compte Rendus in 1991 and 1999) that in absence of viscosity, that is for the Euler equation, the vorticity of such a fluid (incompressible, axisymmetric, with swirl), generically increases linearly in time. If the flow exists globally, of course.
add a comment |
You certainly know this one, but some readers could ignore it. The fact that NS iw globally well-posed in 2D is due to the so-called Ladyzhenskaia inequality
$$|f|_4^2le c|f|_2|nabla f|_2.$$
The fact that this does not hold in 3D is the reason why there is a 1M-dollars problem...
But if the flow is axisymmetric, even with swirl, and if the domain is a container between two cylinders ($0<r_0<sqrt{x^2+y^2}<r_1<infty$), then LI is still valid, and the solution exists, is unique and smooth whenever the initial energy is finite.
To summarize, the only major difficulty is when the domain reaches the symmetry axis.
By the way, I have proved (see my notes at the Compte Rendus in 1991 and 1999) that in absence of viscosity, that is for the Euler equation, the vorticity of such a fluid (incompressible, axisymmetric, with swirl), generically increases linearly in time. If the flow exists globally, of course.
add a comment |
You certainly know this one, but some readers could ignore it. The fact that NS iw globally well-posed in 2D is due to the so-called Ladyzhenskaia inequality
$$|f|_4^2le c|f|_2|nabla f|_2.$$
The fact that this does not hold in 3D is the reason why there is a 1M-dollars problem...
But if the flow is axisymmetric, even with swirl, and if the domain is a container between two cylinders ($0<r_0<sqrt{x^2+y^2}<r_1<infty$), then LI is still valid, and the solution exists, is unique and smooth whenever the initial energy is finite.
To summarize, the only major difficulty is when the domain reaches the symmetry axis.
By the way, I have proved (see my notes at the Compte Rendus in 1991 and 1999) that in absence of viscosity, that is for the Euler equation, the vorticity of such a fluid (incompressible, axisymmetric, with swirl), generically increases linearly in time. If the flow exists globally, of course.
You certainly know this one, but some readers could ignore it. The fact that NS iw globally well-posed in 2D is due to the so-called Ladyzhenskaia inequality
$$|f|_4^2le c|f|_2|nabla f|_2.$$
The fact that this does not hold in 3D is the reason why there is a 1M-dollars problem...
But if the flow is axisymmetric, even with swirl, and if the domain is a container between two cylinders ($0<r_0<sqrt{x^2+y^2}<r_1<infty$), then LI is still valid, and the solution exists, is unique and smooth whenever the initial energy is finite.
To summarize, the only major difficulty is when the domain reaches the symmetry axis.
By the way, I have proved (see my notes at the Compte Rendus in 1991 and 1999) that in absence of viscosity, that is for the Euler equation, the vorticity of such a fluid (incompressible, axisymmetric, with swirl), generically increases linearly in time. If the flow exists globally, of course.
answered 10 hours ago
Denis Serre
29.1k791195
29.1k791195
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