Obstruction to Navier-Stokes blowup with cylindrical symmetry












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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







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    4














    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







    share|cite|improve this question

























      4












      4








      4







      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







      share|cite|improve this question













      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






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      asked 11 hours ago









      Jean Duchon

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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
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            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.






            share|cite|improve this answer


























              4














              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.






              share|cite|improve this answer
























                4












                4








                4






                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.






                share|cite|improve this answer












                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 10 hours ago









                Denis Serre

                29.1k791195




                29.1k791195






























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