$x_i frac{partial f(x)}{partial x_i} = f(x)g_i(x)$ for all $i$












0












$begingroup$


What can we say about the function $f:mathbb{R}^k rightarrow mathbb{R}$ if for all $i=1,...,k$ and all $x in mathbb{R}^k$ we have



$x_ifrac{partial f(x)}{partial x_i} = f(x)g_i(x)$



Is there actually any restriction we can make other than that $f$ is differentiable?






ordinary-differential-equations functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    $f(x)=0$ implies $x_ifrac {partial f (x)} {partial x_i}=0$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 5 '18 at 7:31


















0












$begingroup$


What can we say about the function $f:mathbb{R}^k rightarrow mathbb{R}$ if for all $i=1,...,k$ and all $x in mathbb{R}^k$ we have



$x_ifrac{partial f(x)}{partial x_i} = f(x)g_i(x)$



Is there actually any restriction we can make other than that $f$ is differentiable?






ordinary-differential-equations functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    $f(x)=0$ implies $x_ifrac {partial f (x)} {partial x_i}=0$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 5 '18 at 7:31
















0












0








0





$begingroup$


What can we say about the function $f:mathbb{R}^k rightarrow mathbb{R}$ if for all $i=1,...,k$ and all $x in mathbb{R}^k$ we have



$x_ifrac{partial f(x)}{partial x_i} = f(x)g_i(x)$



Is there actually any restriction we can make other than that $f$ is differentiable?






ordinary-differential-equations functional-equations







share|cite|improve this question











$endgroup$




What can we say about the function $f:mathbb{R}^k rightarrow mathbb{R}$ if for all $i=1,...,k$ and all $x in mathbb{R}^k$ we have



$x_ifrac{partial f(x)}{partial x_i} = f(x)g_i(x)$



Is there actually any restriction we can make other than that $f$ is differentiable?






ordinary-differential-equations functional-equations




ordinary-differential-equations functional-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 7:36









Martin Rosenau

1,156139




1,156139










asked Dec 5 '18 at 7:21









HRSEHRSE

233110




233110












  • $begingroup$
    $f(x)=0$ implies $x_ifrac {partial f (x)} {partial x_i}=0$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 5 '18 at 7:31




















  • $begingroup$
    $f(x)=0$ implies $x_ifrac {partial f (x)} {partial x_i}=0$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 5 '18 at 7:31


















$begingroup$
$f(x)=0$ implies $x_ifrac {partial f (x)} {partial x_i}=0$.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 7:31






$begingroup$
$f(x)=0$ implies $x_ifrac {partial f (x)} {partial x_i}=0$.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 7:31












3 Answers
3






active

oldest

votes


















1












$begingroup$

Consider $phi=ln|f|$, then $$frac{∂phi(x)}{∂x_i}=frac{g_i(x)}{x_i}.$$ By the Schwarz lemma, the mixed second derivatives do not depend on the order of derivation, thus
$$
frac1{x_i}frac{∂g_i(x)}{∂x_j}=frac1{x_j}frac{∂g_j(x)}{∂x_i}
$$
is a necessary condition for $f$ to exist.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If $g_i(x)=-1$ then you have that



    $frac{partial}{partial x_i}(x_i f(x))=0$



    So a solution is



    $f(x)=frac{1}{prod_{k=1}^n x_k}$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      We can say nearly nothing. Let $f(x)$ never equals to zero. Then we can take $g_i(x)=frac{x_ifrac{partial f(x)}{partial x_i}}{f(x)}$






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        Consider $phi=ln|f|$, then $$frac{∂phi(x)}{∂x_i}=frac{g_i(x)}{x_i}.$$ By the Schwarz lemma, the mixed second derivatives do not depend on the order of derivation, thus
        $$
        frac1{x_i}frac{∂g_i(x)}{∂x_j}=frac1{x_j}frac{∂g_j(x)}{∂x_i}
        $$
        is a necessary condition for $f$ to exist.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Consider $phi=ln|f|$, then $$frac{∂phi(x)}{∂x_i}=frac{g_i(x)}{x_i}.$$ By the Schwarz lemma, the mixed second derivatives do not depend on the order of derivation, thus
          $$
          frac1{x_i}frac{∂g_i(x)}{∂x_j}=frac1{x_j}frac{∂g_j(x)}{∂x_i}
          $$
          is a necessary condition for $f$ to exist.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Consider $phi=ln|f|$, then $$frac{∂phi(x)}{∂x_i}=frac{g_i(x)}{x_i}.$$ By the Schwarz lemma, the mixed second derivatives do not depend on the order of derivation, thus
            $$
            frac1{x_i}frac{∂g_i(x)}{∂x_j}=frac1{x_j}frac{∂g_j(x)}{∂x_i}
            $$
            is a necessary condition for $f$ to exist.






            share|cite|improve this answer









            $endgroup$



            Consider $phi=ln|f|$, then $$frac{∂phi(x)}{∂x_i}=frac{g_i(x)}{x_i}.$$ By the Schwarz lemma, the mixed second derivatives do not depend on the order of derivation, thus
            $$
            frac1{x_i}frac{∂g_i(x)}{∂x_j}=frac1{x_j}frac{∂g_j(x)}{∂x_i}
            $$
            is a necessary condition for $f$ to exist.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 5 '18 at 10:20









            LutzLLutzL

            56.9k42055




            56.9k42055























                0












                $begingroup$

                If $g_i(x)=-1$ then you have that



                $frac{partial}{partial x_i}(x_i f(x))=0$



                So a solution is



                $f(x)=frac{1}{prod_{k=1}^n x_k}$






                share|cite|improve this answer











                $endgroup$


















                  0












                  $begingroup$

                  If $g_i(x)=-1$ then you have that



                  $frac{partial}{partial x_i}(x_i f(x))=0$



                  So a solution is



                  $f(x)=frac{1}{prod_{k=1}^n x_k}$






                  share|cite|improve this answer











                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    If $g_i(x)=-1$ then you have that



                    $frac{partial}{partial x_i}(x_i f(x))=0$



                    So a solution is



                    $f(x)=frac{1}{prod_{k=1}^n x_k}$






                    share|cite|improve this answer











                    $endgroup$



                    If $g_i(x)=-1$ then you have that



                    $frac{partial}{partial x_i}(x_i f(x))=0$



                    So a solution is



                    $f(x)=frac{1}{prod_{k=1}^n x_k}$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 5 '18 at 7:59

























                    answered Dec 5 '18 at 7:52









                    Federico FalluccaFederico Fallucca

                    1,85018




                    1,85018























                        0












                        $begingroup$

                        We can say nearly nothing. Let $f(x)$ never equals to zero. Then we can take $g_i(x)=frac{x_ifrac{partial f(x)}{partial x_i}}{f(x)}$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          We can say nearly nothing. Let $f(x)$ never equals to zero. Then we can take $g_i(x)=frac{x_ifrac{partial f(x)}{partial x_i}}{f(x)}$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            We can say nearly nothing. Let $f(x)$ never equals to zero. Then we can take $g_i(x)=frac{x_ifrac{partial f(x)}{partial x_i}}{f(x)}$






                            share|cite|improve this answer









                            $endgroup$



                            We can say nearly nothing. Let $f(x)$ never equals to zero. Then we can take $g_i(x)=frac{x_ifrac{partial f(x)}{partial x_i}}{f(x)}$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 5 '18 at 9:10









                            MinzMinz

                            951127




                            951127






























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