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Are the components of a Cauchy-Riemann mapping harmonic?


Cauchy-Riemann implies analyticityConsequence of Riemann mapping theoremCharacterization of Cauchy-Riemann operatorCauchy-Riemann equation and differentiability in $mathbb R^2$Are the Cauchy Riemann conditions sufficient for analyticityCauchy Riemann equation and Harmonic ConditionGiven u and v are harmonic in some region R prove the followingGeneralization of the Cauchy-Riemann equationsConverse of the Cauchy-Riemann conditionProving that the composition of a harmonic function and a Cauchy-Riemann mapping is harmonic













0












$begingroup$



Let $mathcalO$ be an open subset of the plane $mathbbR^2$ and
let the mapping $F : mathcalO rightarrow mathbbR^2$ be
represented by $F(x, y) = (u(x, y), v(x, y))$ for $(x, y)$ in
$mathcalO$. Then, we say the mapping $F : mathcalO rightarrow
mathbbR^2$
is called a Cauchy-Riemann mapping provided that
each of the functions $u : mathcalO rightarrow mathbbR$ and $v
: mathcalO rightarrow mathbbR$
has continuous second-order
partial derivatives and $$fracpartial upartial x(x, y) =
fracpartial vpartial y(x, y) hspace1em text and
hspace1em fracpartial upartial y = -fracpartial
vpartial x(x, y)$$



for all $(x, y)$ in $mathcalO$.




Is it necessarily true that $u(x, y)$ and $v(x, y)$ are harmonic?










share|cite|improve this question









$endgroup$
















    0












    $begingroup$



    Let $mathcalO$ be an open subset of the plane $mathbbR^2$ and
    let the mapping $F : mathcalO rightarrow mathbbR^2$ be
    represented by $F(x, y) = (u(x, y), v(x, y))$ for $(x, y)$ in
    $mathcalO$. Then, we say the mapping $F : mathcalO rightarrow
    mathbbR^2$
    is called a Cauchy-Riemann mapping provided that
    each of the functions $u : mathcalO rightarrow mathbbR$ and $v
    : mathcalO rightarrow mathbbR$
    has continuous second-order
    partial derivatives and $$fracpartial upartial x(x, y) =
    fracpartial vpartial y(x, y) hspace1em text and
    hspace1em fracpartial upartial y = -fracpartial
    vpartial x(x, y)$$



    for all $(x, y)$ in $mathcalO$.




    Is it necessarily true that $u(x, y)$ and $v(x, y)$ are harmonic?










    share|cite|improve this question









    $endgroup$














      0












      0








      0





      $begingroup$



      Let $mathcalO$ be an open subset of the plane $mathbbR^2$ and
      let the mapping $F : mathcalO rightarrow mathbbR^2$ be
      represented by $F(x, y) = (u(x, y), v(x, y))$ for $(x, y)$ in
      $mathcalO$. Then, we say the mapping $F : mathcalO rightarrow
      mathbbR^2$
      is called a Cauchy-Riemann mapping provided that
      each of the functions $u : mathcalO rightarrow mathbbR$ and $v
      : mathcalO rightarrow mathbbR$
      has continuous second-order
      partial derivatives and $$fracpartial upartial x(x, y) =
      fracpartial vpartial y(x, y) hspace1em text and
      hspace1em fracpartial upartial y = -fracpartial
      vpartial x(x, y)$$



      for all $(x, y)$ in $mathcalO$.




      Is it necessarily true that $u(x, y)$ and $v(x, y)$ are harmonic?










      share|cite|improve this question









      $endgroup$





      Let $mathcalO$ be an open subset of the plane $mathbbR^2$ and
      let the mapping $F : mathcalO rightarrow mathbbR^2$ be
      represented by $F(x, y) = (u(x, y), v(x, y))$ for $(x, y)$ in
      $mathcalO$. Then, we say the mapping $F : mathcalO rightarrow
      mathbbR^2$
      is called a Cauchy-Riemann mapping provided that
      each of the functions $u : mathcalO rightarrow mathbbR$ and $v
      : mathcalO rightarrow mathbbR$
      has continuous second-order
      partial derivatives and $$fracpartial upartial x(x, y) =
      fracpartial vpartial y(x, y) hspace1em text and
      hspace1em fracpartial upartial y = -fracpartial
      vpartial x(x, y)$$



      for all $(x, y)$ in $mathcalO$.




      Is it necessarily true that $u(x, y)$ and $v(x, y)$ are harmonic?







      real-analysis complex-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 29 at 15:33







      user658798



























          2 Answers
          2






          active

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          1












          $begingroup$

          Yes. For instance,
          $$
          fracpartial^2 upartial^2 x + fracpartial^2 upartial^2 y =
          fracpartial partial xfracpartial upartial x + fracpartial partial yfracpartial upartial y
          = fracpartial partial xfracpartial vpartial y - fracpartial partial yfracpartial vpartial x = fracpartial^2 vpartial xpartial y - fracpartial^2 vpartial xpartial y = 0.
          $$

          The calculation for $v$ is similar.






          share|cite|improve this answer









          $endgroup$




















            0












            $begingroup$

            The real and imaginary parts of a holomorphic function are harmonic. Source: Cartan's Elementary Theory of Analytic Functions.






            share|cite|improve this answer









            $endgroup$













              Your Answer





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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

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              1












              $begingroup$

              Yes. For instance,
              $$
              fracpartial^2 upartial^2 x + fracpartial^2 upartial^2 y =
              fracpartial partial xfracpartial upartial x + fracpartial partial yfracpartial upartial y
              = fracpartial partial xfracpartial vpartial y - fracpartial partial yfracpartial vpartial x = fracpartial^2 vpartial xpartial y - fracpartial^2 vpartial xpartial y = 0.
              $$

              The calculation for $v$ is similar.






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                Yes. For instance,
                $$
                fracpartial^2 upartial^2 x + fracpartial^2 upartial^2 y =
                fracpartial partial xfracpartial upartial x + fracpartial partial yfracpartial upartial y
                = fracpartial partial xfracpartial vpartial y - fracpartial partial yfracpartial vpartial x = fracpartial^2 vpartial xpartial y - fracpartial^2 vpartial xpartial y = 0.
                $$

                The calculation for $v$ is similar.






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  Yes. For instance,
                  $$
                  fracpartial^2 upartial^2 x + fracpartial^2 upartial^2 y =
                  fracpartial partial xfracpartial upartial x + fracpartial partial yfracpartial upartial y
                  = fracpartial partial xfracpartial vpartial y - fracpartial partial yfracpartial vpartial x = fracpartial^2 vpartial xpartial y - fracpartial^2 vpartial xpartial y = 0.
                  $$

                  The calculation for $v$ is similar.






                  share|cite|improve this answer









                  $endgroup$



                  Yes. For instance,
                  $$
                  fracpartial^2 upartial^2 x + fracpartial^2 upartial^2 y =
                  fracpartial partial xfracpartial upartial x + fracpartial partial yfracpartial upartial y
                  = fracpartial partial xfracpartial vpartial y - fracpartial partial yfracpartial vpartial x = fracpartial^2 vpartial xpartial y - fracpartial^2 vpartial xpartial y = 0.
                  $$

                  The calculation for $v$ is similar.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 29 at 15:38









                  Ernie060Ernie060

                  2,940719




                  2,940719





















                      0












                      $begingroup$

                      The real and imaginary parts of a holomorphic function are harmonic. Source: Cartan's Elementary Theory of Analytic Functions.






                      share|cite|improve this answer









                      $endgroup$

















                        0












                        $begingroup$

                        The real and imaginary parts of a holomorphic function are harmonic. Source: Cartan's Elementary Theory of Analytic Functions.






                        share|cite|improve this answer









                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          The real and imaginary parts of a holomorphic function are harmonic. Source: Cartan's Elementary Theory of Analytic Functions.






                          share|cite|improve this answer









                          $endgroup$



                          The real and imaginary parts of a holomorphic function are harmonic. Source: Cartan's Elementary Theory of Analytic Functions.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Mar 29 at 15:36









                          avsavs

                          3,894515




                          3,894515



























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