smooth map derivative along Euler vector field












0












$begingroup$


let $p$ be a non zero integer. Let $U = mathbb{R}^3 setminus{0}$ and $f$ be a smooth map $f: Urightarrow mathbb{R}: forall (x, y, z) ∈ U , forall t>0:$



$f (tx, ty, tz) = t^p f (x, y, z)$



Let $E$ be Euler's field on $U: E=xpartial_x+ypartial_y+zpartial_z$



Find $E(f)$.



I don't know where to begin. Thank you






differential-geometry







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    let $p$ be a non zero integer. Let $U = mathbb{R}^3 setminus{0}$ and $f$ be a smooth map $f: Urightarrow mathbb{R}: forall (x, y, z) ∈ U , forall t>0:$



    $f (tx, ty, tz) = t^p f (x, y, z)$



    Let $E$ be Euler's field on $U: E=xpartial_x+ypartial_y+zpartial_z$



    Find $E(f)$.



    I don't know where to begin. Thank you






    differential-geometry







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      let $p$ be a non zero integer. Let $U = mathbb{R}^3 setminus{0}$ and $f$ be a smooth map $f: Urightarrow mathbb{R}: forall (x, y, z) ∈ U , forall t>0:$



      $f (tx, ty, tz) = t^p f (x, y, z)$



      Let $E$ be Euler's field on $U: E=xpartial_x+ypartial_y+zpartial_z$



      Find $E(f)$.



      I don't know where to begin. Thank you






      differential-geometry







      share|cite|improve this question









      $endgroup$




      let $p$ be a non zero integer. Let $U = mathbb{R}^3 setminus{0}$ and $f$ be a smooth map $f: Urightarrow mathbb{R}: forall (x, y, z) ∈ U , forall t>0:$



      $f (tx, ty, tz) = t^p f (x, y, z)$



      Let $E$ be Euler's field on $U: E=xpartial_x+ypartial_y+zpartial_z$



      Find $E(f)$.



      I don't know where to begin. Thank you






      differential-geometry




      differential-geometry






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 21 '18 at 16:08









      PerelManPerelMan

      661313




      661313






















          2 Answers
          2






          active

          oldest

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          1












          $begingroup$

          Hint: notice that $E(f)$ is the same as the directional derivative of $f$ along the vector $(x,y,z)$. More precisely,
          $$ E(f)(x,y,z) = lim_{tto 0} frac{f(x+tx,y+ty,z+tz)-f(x,y,z)}{t}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! that would give $E(f)(x,y,z)= p.f(x,y,z)$
            $endgroup$
            – PerelMan
            Dec 21 '18 at 16:36



















          2












          $begingroup$

          The flow of $E$ is $phi_t(x,y,z)=(e^tx,e^ty,e^tz)$, $f(phi_t(x,y,z))=f(e^tx,e^ty,e^tz)=e^{pt}f(x,y,z)$ implies that=t $E(f)={dover{dt}}_{t=0}f(phi_t(x,y,z))={dover{dt}}_{t=0}e^{pt}f(x,y,z)=pf(x,y,z).$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Hint: notice that $E(f)$ is the same as the directional derivative of $f$ along the vector $(x,y,z)$. More precisely,
            $$ E(f)(x,y,z) = lim_{tto 0} frac{f(x+tx,y+ty,z+tz)-f(x,y,z)}{t}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you! that would give $E(f)(x,y,z)= p.f(x,y,z)$
              $endgroup$
              – PerelMan
              Dec 21 '18 at 16:36
















            1












            $begingroup$

            Hint: notice that $E(f)$ is the same as the directional derivative of $f$ along the vector $(x,y,z)$. More precisely,
            $$ E(f)(x,y,z) = lim_{tto 0} frac{f(x+tx,y+ty,z+tz)-f(x,y,z)}{t}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you! that would give $E(f)(x,y,z)= p.f(x,y,z)$
              $endgroup$
              – PerelMan
              Dec 21 '18 at 16:36














            1












            1








            1





            $begingroup$

            Hint: notice that $E(f)$ is the same as the directional derivative of $f$ along the vector $(x,y,z)$. More precisely,
            $$ E(f)(x,y,z) = lim_{tto 0} frac{f(x+tx,y+ty,z+tz)-f(x,y,z)}{t}.$$






            share|cite|improve this answer









            $endgroup$



            Hint: notice that $E(f)$ is the same as the directional derivative of $f$ along the vector $(x,y,z)$. More precisely,
            $$ E(f)(x,y,z) = lim_{tto 0} frac{f(x+tx,y+ty,z+tz)-f(x,y,z)}{t}.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 21 '18 at 16:11









            SvanNSvanN

            2,0661422




            2,0661422












            • $begingroup$
              Thank you! that would give $E(f)(x,y,z)= p.f(x,y,z)$
              $endgroup$
              – PerelMan
              Dec 21 '18 at 16:36


















            • $begingroup$
              Thank you! that would give $E(f)(x,y,z)= p.f(x,y,z)$
              $endgroup$
              – PerelMan
              Dec 21 '18 at 16:36
















            $begingroup$
            Thank you! that would give $E(f)(x,y,z)= p.f(x,y,z)$
            $endgroup$
            – PerelMan
            Dec 21 '18 at 16:36




            $begingroup$
            Thank you! that would give $E(f)(x,y,z)= p.f(x,y,z)$
            $endgroup$
            – PerelMan
            Dec 21 '18 at 16:36











            2












            $begingroup$

            The flow of $E$ is $phi_t(x,y,z)=(e^tx,e^ty,e^tz)$, $f(phi_t(x,y,z))=f(e^tx,e^ty,e^tz)=e^{pt}f(x,y,z)$ implies that=t $E(f)={dover{dt}}_{t=0}f(phi_t(x,y,z))={dover{dt}}_{t=0}e^{pt}f(x,y,z)=pf(x,y,z).$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              The flow of $E$ is $phi_t(x,y,z)=(e^tx,e^ty,e^tz)$, $f(phi_t(x,y,z))=f(e^tx,e^ty,e^tz)=e^{pt}f(x,y,z)$ implies that=t $E(f)={dover{dt}}_{t=0}f(phi_t(x,y,z))={dover{dt}}_{t=0}e^{pt}f(x,y,z)=pf(x,y,z).$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                The flow of $E$ is $phi_t(x,y,z)=(e^tx,e^ty,e^tz)$, $f(phi_t(x,y,z))=f(e^tx,e^ty,e^tz)=e^{pt}f(x,y,z)$ implies that=t $E(f)={dover{dt}}_{t=0}f(phi_t(x,y,z))={dover{dt}}_{t=0}e^{pt}f(x,y,z)=pf(x,y,z).$






                share|cite|improve this answer









                $endgroup$



                The flow of $E$ is $phi_t(x,y,z)=(e^tx,e^ty,e^tz)$, $f(phi_t(x,y,z))=f(e^tx,e^ty,e^tz)=e^{pt}f(x,y,z)$ implies that=t $E(f)={dover{dt}}_{t=0}f(phi_t(x,y,z))={dover{dt}}_{t=0}e^{pt}f(x,y,z)=pf(x,y,z).$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 21 '18 at 16:17









                Tsemo AristideTsemo Aristide

                59k11445




                59k11445






























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