Obstruction to Navier-Stokes blowup with cylindrical symmetry
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Is there a known obstruction to cylindrically symmetric solutions (with swirl) of incompressible 3D Navier-Stokes blowing up in finite time ?
EDIT: in the whole space $mathbb R^3$, I forgot to say.
Same question for Euler (ideal fluid) flow, with everything (velocity, vorticity, pressure) vanishing at $infty$.
ap.analysis-of-pdes navier-stokes
asked Dec 31 '18 at 14:37
Jean DuchonJean Duchon
2,655415
add a comment |
Is there a known obstruction to cylindrically symmetric solutions (with swirl) of incompressible 3D Navier-Stokes blowing up in finite time ?
EDIT: in the whole space $mathbb R^3$, I forgot to say.
Same question for Euler (ideal fluid) flow, with everything (velocity, vorticity, pressure) vanishing at $infty$.
ap.analysis-of-pdes navier-stokes
asked Dec 31 '18 at 14:37
Jean DuchonJean Duchon
2,655415
You probably know about this preprint, but some readers may not [ arxiv.org/abs/1802.09936 ]
– Alex Gavrilov
Jan 1 at 11:44
add a comment |
Is there a known obstruction to cylindrically symmetric solutions (with swirl) of incompressible 3D Navier-Stokes blowing up in finite time ?
EDIT: in the whole space $mathbb R^3$, I forgot to say.
Same question for Euler (ideal fluid) flow, with everything (velocity, vorticity, pressure) vanishing at $infty$.
ap.analysis-of-pdes navier-stokes
asked Dec 31 '18 at 14:37
Jean DuchonJean Duchon
2,655415
Is there a known obstruction to cylindrically symmetric solutions (with swirl) of incompressible 3D Navier-Stokes blowing up in finite time ?
EDIT: in the whole space $mathbb R^3$, I forgot to say.
Same question for Euler (ideal fluid) flow, with everything (velocity, vorticity, pressure) vanishing at $infty$.
ap.analysis-of-pdes navier-stokes
ap.analysis-of-pdes navier-stokes
asked Dec 31 '18 at 14:37
Jean DuchonJean Duchon
2,655415
asked Dec 31 '18 at 14:37
Jean DuchonJean Duchon
2,655415
edited Jan 1 at 10:06
Jean Duchon
asked Dec 31 '18 at 14:37
Jean DuchonJean Duchon
2,655415
asked Dec 31 '18 at 14:37
Jean DuchonJean Duchon
2,655415
asked Dec 31 '18 at 14:37
Jean DuchonJean Duchon
2,655415
2,655415
You probably know about this preprint, but some readers may not [ arxiv.org/abs/1802.09936 ]
– Alex Gavrilov
Jan 1 at 11:44
add a comment |
You probably know about this preprint, but some readers may not [ arxiv.org/abs/1802.09936 ]
– Alex Gavrilov
Jan 1 at 11:44
You probably know about this preprint, but some readers may not [ arxiv.org/abs/1802.09936 ]
– Alex Gavrilov
Jan 1 at 11:44
You probably know about this preprint, but some readers may not [ arxiv.org/abs/1802.09936 ]
– Alex Gavrilov
Jan 1 at 11:44
add a comment |
1 Answer
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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 ($0<r_0<sqrtx^2+y^2<r_1<infty$), 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 Dec 31 '18 at 15:31
Denis SerreDenis Serre
29.1k791196
I couldn't acess your note of 1991, but in that of 1999, it is restricted to a domain between two cylinders, right? If so, my question extended to swirling cylindrically symmetric Euler flows remains. I personally believe blowup may occur in that setting, and a key step is the (unproven yet) existence of a stationary self-similar weak Euler solution with dissipation $cdelta$.
– Jean Duchon
Jan 1 at 10:21
@JeanDuchon. This sounds reasonnable. Yes, my result holds only when the domain is between two cylinder. It ignores the singularity at the axis.
– Denis Serre
Jan 1 at 20:05
add a comment |
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1 Answer
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1 Answer
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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 ($0<r_0<sqrtx^2+y^2<r_1<infty$), 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 Dec 31 '18 at 15:31
Denis SerreDenis Serre
29.1k791196
I couldn't acess your note of 1991, but in that of 1999, it is restricted to a domain between two cylinders, right? If so, my question extended to swirling cylindrically symmetric Euler flows remains. I personally believe blowup may occur in that setting, and a key step is the (unproven yet) existence of a stationary self-similar weak Euler solution with dissipation $cdelta$.
– Jean Duchon
Jan 1 at 10:21
@JeanDuchon. This sounds reasonnable. Yes, my result holds only when the domain is between two cylinder. It ignores the singularity at the axis.
– Denis Serre
Jan 1 at 20:05
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 ($0<r_0<sqrtx^2+y^2<r_1<infty$), 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 Dec 31 '18 at 15:31
Denis SerreDenis Serre
29.1k791196
I couldn't acess your note of 1991, but in that of 1999, it is restricted to a domain between two cylinders, right? If so, my question extended to swirling cylindrically symmetric Euler flows remains. I personally believe blowup may occur in that setting, and a key step is the (unproven yet) existence of a stationary self-similar weak Euler solution with dissipation $cdelta$.
– Jean Duchon
Jan 1 at 10:21
@JeanDuchon. This sounds reasonnable. Yes, my result holds only when the domain is between two cylinder. It ignores the singularity at the axis.
– Denis Serre
Jan 1 at 20:05
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 ($0<r_0<sqrtx^2+y^2<r_1<infty$), 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 Dec 31 '18 at 15:31
Denis SerreDenis Serre
29.1k791196
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 ($0<r_0<sqrtx^2+y^2<r_1<infty$), 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 Dec 31 '18 at 15:31
Denis SerreDenis Serre
29.1k791196
answered Dec 31 '18 at 15:31
Denis SerreDenis Serre
29.1k791196
answered Dec 31 '18 at 15:31
Denis SerreDenis Serre
29.1k791196
answered Dec 31 '18 at 15:31
Denis SerreDenis Serre
29.1k791196
29.1k791196
I couldn't acess your note of 1991, but in that of 1999, it is restricted to a domain between two cylinders, right? If so, my question extended to swirling cylindrically symmetric Euler flows remains. I personally believe blowup may occur in that setting, and a key step is the (unproven yet) existence of a stationary self-similar weak Euler solution with dissipation $cdelta$.
– Jean Duchon
Jan 1 at 10:21
@JeanDuchon. This sounds reasonnable. Yes, my result holds only when the domain is between two cylinder. It ignores the singularity at the axis.
– Denis Serre
Jan 1 at 20:05
add a comment |
I couldn't acess your note of 1991, but in that of 1999, it is restricted to a domain between two cylinders, right? If so, my question extended to swirling cylindrically symmetric Euler flows remains. I personally believe blowup may occur in that setting, and a key step is the (unproven yet) existence of a stationary self-similar weak Euler solution with dissipation $cdelta$.
– Jean Duchon
Jan 1 at 10:21
@JeanDuchon. This sounds reasonnable. Yes, my result holds only when the domain is between two cylinder. It ignores the singularity at the axis.
– Denis Serre
Jan 1 at 20:05
I couldn't acess your note of 1991, but in that of 1999, it is restricted to a domain between two cylinders, right? If so, my question extended to swirling cylindrically symmetric Euler flows remains. I personally believe blowup may occur in that setting, and a key step is the (unproven yet) existence of a stationary self-similar weak Euler solution with dissipation $cdelta$.
– Jean Duchon
Jan 1 at 10:21
I couldn't acess your note of 1991, but in that of 1999, it is restricted to a domain between two cylinders, right? If so, my question extended to swirling cylindrically symmetric Euler flows remains. I personally believe blowup may occur in that setting, and a key step is the (unproven yet) existence of a stationary self-similar weak Euler solution with dissipation $cdelta$.
– Jean Duchon
Jan 1 at 10:21
@JeanDuchon. This sounds reasonnable. Yes, my result holds only when the domain is between two cylinder. It ignores the singularity at the axis.
– Denis Serre
Jan 1 at 20:05
@JeanDuchon. This sounds reasonnable. Yes, my result holds only when the domain is between two cylinder. It ignores the singularity at the axis.
– Denis Serre
Jan 1 at 20:05
add a comment |
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You probably know about this preprint, but some readers may not [ arxiv.org/abs/1802.09936 ]
– Alex Gavrilov
Jan 1 at 11:44