When is a vector field the curl of another vector field?












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Under what conditions does a given vector field $bf X$ on some open subset $U subseteq mathbb{R}^3$ satisfy ${bf X} = text{curl } {bf Y}$ for some vector field $bf Y$ on $U$?






multivariable-calculus vector-analysis







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  • $begingroup$
    Do you mean, Under what conditions does a given vector field $bf X$ on some open subset $U subseteq mathbb{R}^3$ satisfy ${bf X} = text{curl } {bf Y}$ for some vector field $bf Y$ on $U$?
    $endgroup$
    – Travis
    Mar 3 '15 at 9:10










  • $begingroup$
    Yes. It is question based on Mathematical Physics.
    $endgroup$
    – user3575652
    Mar 3 '15 at 9:12
















0












$begingroup$


Under what conditions does a given vector field $bf X$ on some open subset $U subseteq mathbb{R}^3$ satisfy ${bf X} = text{curl } {bf Y}$ for some vector field $bf Y$ on $U$?






multivariable-calculus vector-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean, Under what conditions does a given vector field $bf X$ on some open subset $U subseteq mathbb{R}^3$ satisfy ${bf X} = text{curl } {bf Y}$ for some vector field $bf Y$ on $U$?
    $endgroup$
    – Travis
    Mar 3 '15 at 9:10










  • $begingroup$
    Yes. It is question based on Mathematical Physics.
    $endgroup$
    – user3575652
    Mar 3 '15 at 9:12














0












0








0


1



$begingroup$


Under what conditions does a given vector field $bf X$ on some open subset $U subseteq mathbb{R}^3$ satisfy ${bf X} = text{curl } {bf Y}$ for some vector field $bf Y$ on $U$?






multivariable-calculus vector-analysis







share|cite|improve this question











$endgroup$




Under what conditions does a given vector field $bf X$ on some open subset $U subseteq mathbb{R}^3$ satisfy ${bf X} = text{curl } {bf Y}$ for some vector field $bf Y$ on $U$?






multivariable-calculus vector-analysis




multivariable-calculus vector-analysis






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edited Dec 24 '18 at 1:46









Travis

60.6k767147




60.6k767147










asked Mar 3 '15 at 9:07









user3575652user3575652

87112




87112












  • $begingroup$
    Do you mean, Under what conditions does a given vector field $bf X$ on some open subset $U subseteq mathbb{R}^3$ satisfy ${bf X} = text{curl } {bf Y}$ for some vector field $bf Y$ on $U$?
    $endgroup$
    – Travis
    Mar 3 '15 at 9:10










  • $begingroup$
    Yes. It is question based on Mathematical Physics.
    $endgroup$
    – user3575652
    Mar 3 '15 at 9:12


















  • $begingroup$
    Do you mean, Under what conditions does a given vector field $bf X$ on some open subset $U subseteq mathbb{R}^3$ satisfy ${bf X} = text{curl } {bf Y}$ for some vector field $bf Y$ on $U$?
    $endgroup$
    – Travis
    Mar 3 '15 at 9:10










  • $begingroup$
    Yes. It is question based on Mathematical Physics.
    $endgroup$
    – user3575652
    Mar 3 '15 at 9:12
















$begingroup$
Do you mean, Under what conditions does a given vector field $bf X$ on some open subset $U subseteq mathbb{R}^3$ satisfy ${bf X} = text{curl } {bf Y}$ for some vector field $bf Y$ on $U$?
$endgroup$
– Travis
Mar 3 '15 at 9:10




$begingroup$
Do you mean, Under what conditions does a given vector field $bf X$ on some open subset $U subseteq mathbb{R}^3$ satisfy ${bf X} = text{curl } {bf Y}$ for some vector field $bf Y$ on $U$?
$endgroup$
– Travis
Mar 3 '15 at 9:10












$begingroup$
Yes. It is question based on Mathematical Physics.
$endgroup$
– user3575652
Mar 3 '15 at 9:12




$begingroup$
Yes. It is question based on Mathematical Physics.
$endgroup$
– user3575652
Mar 3 '15 at 9:12










3 Answers
3






active

oldest

votes


















2












$begingroup$

If the second de Rham cohomology $H^2(U)$ of the underlying set $U$ is trivial (which is true, e.g., for all contractible sets $U$), then a necessary and sufficient condition is that $$text{div } {bf X} = 0,$$ which follows more or less from the definition of cohomology and the relationship between the $text{curl}$ and $text{div}$ operators and the exterior derivative operator.



If $H^2(U)$ is not trivial, then $text{div } {bf X} = 0$ is a necessary but not a sufficient condition.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Given $mathbf{X}=(X_1,X_2,X_3)$, a vector field $mathbf{Y}=(Y_1,Y_2,Y_3)$ such that $mathbf{X}=text{curl }mathbf{Y}$ will satisfy:



    $$begin{cases}
    X_1&=dfrac{partial Y_3}{partial y}-dfrac{partial Y_2}{partial z}\
    X_2&=dfrac{partial Y_1}{partial z}-dfrac{partial Y_3}{partial x}\
    X_3&=dfrac{partial Y_2}{partial x}-dfrac{partial Y_1}{partial y}
    end {cases}$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      The condition $text{div }{bf X} = 0$ is obviously necessary. If the domain is star-shaped, is also sufficient and there is an explicit formula. Supposing wlog that that the domain is star-shaped respect to the origin:
      $${bf Y}({bf r}) = int_0^1{bf X}(t{bf r})times t{bf r},dt.$$
      See http://www.maths.tcd.ie/~houghton/231/Notes/LN/231.I.6.pdf or http://www.maths.tcd.ie/~houghton/231/Notes/ChrisFord/vp.ps






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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        If the second de Rham cohomology $H^2(U)$ of the underlying set $U$ is trivial (which is true, e.g., for all contractible sets $U$), then a necessary and sufficient condition is that $$text{div } {bf X} = 0,$$ which follows more or less from the definition of cohomology and the relationship between the $text{curl}$ and $text{div}$ operators and the exterior derivative operator.



        If $H^2(U)$ is not trivial, then $text{div } {bf X} = 0$ is a necessary but not a sufficient condition.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          If the second de Rham cohomology $H^2(U)$ of the underlying set $U$ is trivial (which is true, e.g., for all contractible sets $U$), then a necessary and sufficient condition is that $$text{div } {bf X} = 0,$$ which follows more or less from the definition of cohomology and the relationship between the $text{curl}$ and $text{div}$ operators and the exterior derivative operator.



          If $H^2(U)$ is not trivial, then $text{div } {bf X} = 0$ is a necessary but not a sufficient condition.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            If the second de Rham cohomology $H^2(U)$ of the underlying set $U$ is trivial (which is true, e.g., for all contractible sets $U$), then a necessary and sufficient condition is that $$text{div } {bf X} = 0,$$ which follows more or less from the definition of cohomology and the relationship between the $text{curl}$ and $text{div}$ operators and the exterior derivative operator.



            If $H^2(U)$ is not trivial, then $text{div } {bf X} = 0$ is a necessary but not a sufficient condition.






            share|cite|improve this answer









            $endgroup$



            If the second de Rham cohomology $H^2(U)$ of the underlying set $U$ is trivial (which is true, e.g., for all contractible sets $U$), then a necessary and sufficient condition is that $$text{div } {bf X} = 0,$$ which follows more or less from the definition of cohomology and the relationship between the $text{curl}$ and $text{div}$ operators and the exterior derivative operator.



            If $H^2(U)$ is not trivial, then $text{div } {bf X} = 0$ is a necessary but not a sufficient condition.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 3 '15 at 9:21









            TravisTravis

            60.6k767147




            60.6k767147























                1












                $begingroup$

                Given $mathbf{X}=(X_1,X_2,X_3)$, a vector field $mathbf{Y}=(Y_1,Y_2,Y_3)$ such that $mathbf{X}=text{curl }mathbf{Y}$ will satisfy:



                $$begin{cases}
                X_1&=dfrac{partial Y_3}{partial y}-dfrac{partial Y_2}{partial z}\
                X_2&=dfrac{partial Y_1}{partial z}-dfrac{partial Y_3}{partial x}\
                X_3&=dfrac{partial Y_2}{partial x}-dfrac{partial Y_1}{partial y}
                end {cases}$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Given $mathbf{X}=(X_1,X_2,X_3)$, a vector field $mathbf{Y}=(Y_1,Y_2,Y_3)$ such that $mathbf{X}=text{curl }mathbf{Y}$ will satisfy:



                  $$begin{cases}
                  X_1&=dfrac{partial Y_3}{partial y}-dfrac{partial Y_2}{partial z}\
                  X_2&=dfrac{partial Y_1}{partial z}-dfrac{partial Y_3}{partial x}\
                  X_3&=dfrac{partial Y_2}{partial x}-dfrac{partial Y_1}{partial y}
                  end {cases}$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Given $mathbf{X}=(X_1,X_2,X_3)$, a vector field $mathbf{Y}=(Y_1,Y_2,Y_3)$ such that $mathbf{X}=text{curl }mathbf{Y}$ will satisfy:



                    $$begin{cases}
                    X_1&=dfrac{partial Y_3}{partial y}-dfrac{partial Y_2}{partial z}\
                    X_2&=dfrac{partial Y_1}{partial z}-dfrac{partial Y_3}{partial x}\
                    X_3&=dfrac{partial Y_2}{partial x}-dfrac{partial Y_1}{partial y}
                    end {cases}$$






                    share|cite|improve this answer









                    $endgroup$



                    Given $mathbf{X}=(X_1,X_2,X_3)$, a vector field $mathbf{Y}=(Y_1,Y_2,Y_3)$ such that $mathbf{X}=text{curl }mathbf{Y}$ will satisfy:



                    $$begin{cases}
                    X_1&=dfrac{partial Y_3}{partial y}-dfrac{partial Y_2}{partial z}\
                    X_2&=dfrac{partial Y_1}{partial z}-dfrac{partial Y_3}{partial x}\
                    X_3&=dfrac{partial Y_2}{partial x}-dfrac{partial Y_1}{partial y}
                    end {cases}$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 3 '15 at 9:26









                    DemostheneDemosthene

                    4,21311033




                    4,21311033























                        0












                        $begingroup$

                        The condition $text{div }{bf X} = 0$ is obviously necessary. If the domain is star-shaped, is also sufficient and there is an explicit formula. Supposing wlog that that the domain is star-shaped respect to the origin:
                        $${bf Y}({bf r}) = int_0^1{bf X}(t{bf r})times t{bf r},dt.$$
                        See http://www.maths.tcd.ie/~houghton/231/Notes/LN/231.I.6.pdf or http://www.maths.tcd.ie/~houghton/231/Notes/ChrisFord/vp.ps






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          The condition $text{div }{bf X} = 0$ is obviously necessary. If the domain is star-shaped, is also sufficient and there is an explicit formula. Supposing wlog that that the domain is star-shaped respect to the origin:
                          $${bf Y}({bf r}) = int_0^1{bf X}(t{bf r})times t{bf r},dt.$$
                          See http://www.maths.tcd.ie/~houghton/231/Notes/LN/231.I.6.pdf or http://www.maths.tcd.ie/~houghton/231/Notes/ChrisFord/vp.ps






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            The condition $text{div }{bf X} = 0$ is obviously necessary. If the domain is star-shaped, is also sufficient and there is an explicit formula. Supposing wlog that that the domain is star-shaped respect to the origin:
                            $${bf Y}({bf r}) = int_0^1{bf X}(t{bf r})times t{bf r},dt.$$
                            See http://www.maths.tcd.ie/~houghton/231/Notes/LN/231.I.6.pdf or http://www.maths.tcd.ie/~houghton/231/Notes/ChrisFord/vp.ps






                            share|cite|improve this answer









                            $endgroup$



                            The condition $text{div }{bf X} = 0$ is obviously necessary. If the domain is star-shaped, is also sufficient and there is an explicit formula. Supposing wlog that that the domain is star-shaped respect to the origin:
                            $${bf Y}({bf r}) = int_0^1{bf X}(t{bf r})times t{bf r},dt.$$
                            See http://www.maths.tcd.ie/~houghton/231/Notes/LN/231.I.6.pdf or http://www.maths.tcd.ie/~houghton/231/Notes/ChrisFord/vp.ps







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 3 '15 at 9:34









                            Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

                            34.4k42871




                            34.4k42871






























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