Harmonic holomorphic function in Ω












0












$begingroup$


$Ω$ is simply connected in $C$, $u$ is a harmonic function in $Ω$ , $v$ in $Ω$



$$v(x,y) = int_0^1 (yu{Tiny x} (tx,ty)-xu{Tiny y} (tx,ty)) dt$$



Prove that there exists a holomorphic function $u+iv$ in $Ω$



I have the solution - can someone explain me the equalities to where i put the orange question marks? I dont understand why the first part of the term disappears.



Photo/solution






holomorphic-functions







share|cite|improve this question









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    0












    $begingroup$


    $Ω$ is simply connected in $C$, $u$ is a harmonic function in $Ω$ , $v$ in $Ω$



    $$v(x,y) = int_0^1 (yu{Tiny x} (tx,ty)-xu{Tiny y} (tx,ty)) dt$$



    Prove that there exists a holomorphic function $u+iv$ in $Ω$



    I have the solution - can someone explain me the equalities to where i put the orange question marks? I dont understand why the first part of the term disappears.



    Photo/solution






    holomorphic-functions







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      $Ω$ is simply connected in $C$, $u$ is a harmonic function in $Ω$ , $v$ in $Ω$



      $$v(x,y) = int_0^1 (yu{Tiny x} (tx,ty)-xu{Tiny y} (tx,ty)) dt$$



      Prove that there exists a holomorphic function $u+iv$ in $Ω$



      I have the solution - can someone explain me the equalities to where i put the orange question marks? I dont understand why the first part of the term disappears.



      Photo/solution






      holomorphic-functions







      share|cite|improve this question









      $endgroup$




      $Ω$ is simply connected in $C$, $u$ is a harmonic function in $Ω$ , $v$ in $Ω$



      $$v(x,y) = int_0^1 (yu{Tiny x} (tx,ty)-xu{Tiny y} (tx,ty)) dt$$



      Prove that there exists a holomorphic function $u+iv$ in $Ω$



      I have the solution - can someone explain me the equalities to where i put the orange question marks? I dont understand why the first part of the term disappears.



      Photo/solution






      holomorphic-functions




      holomorphic-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 4 '18 at 10:29









      malilinimalilini

      44




      44






















          1 Answer
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          0












          $begingroup$

          Let $g(t):= tu_y(tx,ty)$, then



          $ int_0^1 frac{d}{dt}(tu_y(tx,ty)) dt = int_0^1 g'(t) dt = g(1)-g(0) = u_y(x,y).$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks, this i can understand. but what happened with $$-u{Tiny y} (tx,ty) $$
            $endgroup$
            – malilini
            Dec 4 '18 at 10:53












          • $begingroup$
            I didnt think about tagging you @Fred
            $endgroup$
            – malilini
            Dec 4 '18 at 19:45













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          1 Answer
          1






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          active

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          0












          $begingroup$

          Let $g(t):= tu_y(tx,ty)$, then



          $ int_0^1 frac{d}{dt}(tu_y(tx,ty)) dt = int_0^1 g'(t) dt = g(1)-g(0) = u_y(x,y).$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks, this i can understand. but what happened with $$-u{Tiny y} (tx,ty) $$
            $endgroup$
            – malilini
            Dec 4 '18 at 10:53












          • $begingroup$
            I didnt think about tagging you @Fred
            $endgroup$
            – malilini
            Dec 4 '18 at 19:45


















          0












          $begingroup$

          Let $g(t):= tu_y(tx,ty)$, then



          $ int_0^1 frac{d}{dt}(tu_y(tx,ty)) dt = int_0^1 g'(t) dt = g(1)-g(0) = u_y(x,y).$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks, this i can understand. but what happened with $$-u{Tiny y} (tx,ty) $$
            $endgroup$
            – malilini
            Dec 4 '18 at 10:53












          • $begingroup$
            I didnt think about tagging you @Fred
            $endgroup$
            – malilini
            Dec 4 '18 at 19:45
















          0












          0








          0





          $begingroup$

          Let $g(t):= tu_y(tx,ty)$, then



          $ int_0^1 frac{d}{dt}(tu_y(tx,ty)) dt = int_0^1 g'(t) dt = g(1)-g(0) = u_y(x,y).$






          share|cite|improve this answer









          $endgroup$



          Let $g(t):= tu_y(tx,ty)$, then



          $ int_0^1 frac{d}{dt}(tu_y(tx,ty)) dt = int_0^1 g'(t) dt = g(1)-g(0) = u_y(x,y).$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 10:40









          FredFred

          44.7k1846




          44.7k1846












          • $begingroup$
            thanks, this i can understand. but what happened with $$-u{Tiny y} (tx,ty) $$
            $endgroup$
            – malilini
            Dec 4 '18 at 10:53












          • $begingroup$
            I didnt think about tagging you @Fred
            $endgroup$
            – malilini
            Dec 4 '18 at 19:45




















          • $begingroup$
            thanks, this i can understand. but what happened with $$-u{Tiny y} (tx,ty) $$
            $endgroup$
            – malilini
            Dec 4 '18 at 10:53












          • $begingroup$
            I didnt think about tagging you @Fred
            $endgroup$
            – malilini
            Dec 4 '18 at 19:45


















          $begingroup$
          thanks, this i can understand. but what happened with $$-u{Tiny y} (tx,ty) $$
          $endgroup$
          – malilini
          Dec 4 '18 at 10:53






          $begingroup$
          thanks, this i can understand. but what happened with $$-u{Tiny y} (tx,ty) $$
          $endgroup$
          – malilini
          Dec 4 '18 at 10:53














          $begingroup$
          I didnt think about tagging you @Fred
          $endgroup$
          – malilini
          Dec 4 '18 at 19:45






          $begingroup$
          I didnt think about tagging you @Fred
          $endgroup$
          – malilini
          Dec 4 '18 at 19:45




















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