Finding the harmonic conjugate of $T(x,y)= e^{-y} sin x$?












4












$begingroup$


I try the following two method on finding the harmonic conjugate of $T(x,y)= e^{-y} sin x$ :



Method 1 : by a method of the book Complex Variables and Applications by Brown and Churchill, Chapter 9 Section 104, $$v(x,y) = int_{(0,0)}^{(x,y)} -u_t(s, t) ds + u_s(s, t) dt = int_{(0,0)}^{(x,y)} e^{-y} sin x dx + e^{-y} cos x dy = -e^{-y} cos x - e^{-y} cos x +C = -2e^{-y} cos x +C$$ which $C$ can be chosen to be zero.



Method 1 : Since $T(x,y)= e^{-y} sin x$ and its harmonic conjugate must be the real and imaginary parts of an analytic function, respectively, so it can be $f(z)=-ie^{iz} =e^{-y} sin x - ie^{-y} cos x.$



So why there is an extra $2$ coefficient in Method 1? Where did I do wrong?



Added. Probably the Method 1 is wrong. It comes from if $F_x(x,y) = P(x,y), F_y(x,y) = Q(x,y)$ holds in here. For example the method gives the correct answer for $u=xy$ but again it give a wrong answer for $u=x^3-3xy^2:$ that is $v=6x^2y-y^3$; but the correct one is $v=3x^2y-y^3$. But it looks impossible such a famous book to make a mistake. (???)






complex-analysis proof-verification examples-counterexamples







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    4












    $begingroup$


    I try the following two method on finding the harmonic conjugate of $T(x,y)= e^{-y} sin x$ :



    Method 1 : by a method of the book Complex Variables and Applications by Brown and Churchill, Chapter 9 Section 104, $$v(x,y) = int_{(0,0)}^{(x,y)} -u_t(s, t) ds + u_s(s, t) dt = int_{(0,0)}^{(x,y)} e^{-y} sin x dx + e^{-y} cos x dy = -e^{-y} cos x - e^{-y} cos x +C = -2e^{-y} cos x +C$$ which $C$ can be chosen to be zero.



    Method 1 : Since $T(x,y)= e^{-y} sin x$ and its harmonic conjugate must be the real and imaginary parts of an analytic function, respectively, so it can be $f(z)=-ie^{iz} =e^{-y} sin x - ie^{-y} cos x.$



    So why there is an extra $2$ coefficient in Method 1? Where did I do wrong?



    Added. Probably the Method 1 is wrong. It comes from if $F_x(x,y) = P(x,y), F_y(x,y) = Q(x,y)$ holds in here. For example the method gives the correct answer for $u=xy$ but again it give a wrong answer for $u=x^3-3xy^2:$ that is $v=6x^2y-y^3$; but the correct one is $v=3x^2y-y^3$. But it looks impossible such a famous book to make a mistake. (???)






    complex-analysis proof-verification examples-counterexamples







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      I try the following two method on finding the harmonic conjugate of $T(x,y)= e^{-y} sin x$ :



      Method 1 : by a method of the book Complex Variables and Applications by Brown and Churchill, Chapter 9 Section 104, $$v(x,y) = int_{(0,0)}^{(x,y)} -u_t(s, t) ds + u_s(s, t) dt = int_{(0,0)}^{(x,y)} e^{-y} sin x dx + e^{-y} cos x dy = -e^{-y} cos x - e^{-y} cos x +C = -2e^{-y} cos x +C$$ which $C$ can be chosen to be zero.



      Method 1 : Since $T(x,y)= e^{-y} sin x$ and its harmonic conjugate must be the real and imaginary parts of an analytic function, respectively, so it can be $f(z)=-ie^{iz} =e^{-y} sin x - ie^{-y} cos x.$



      So why there is an extra $2$ coefficient in Method 1? Where did I do wrong?



      Added. Probably the Method 1 is wrong. It comes from if $F_x(x,y) = P(x,y), F_y(x,y) = Q(x,y)$ holds in here. For example the method gives the correct answer for $u=xy$ but again it give a wrong answer for $u=x^3-3xy^2:$ that is $v=6x^2y-y^3$; but the correct one is $v=3x^2y-y^3$. But it looks impossible such a famous book to make a mistake. (???)






      complex-analysis proof-verification examples-counterexamples







      share|cite|improve this question











      $endgroup$




      I try the following two method on finding the harmonic conjugate of $T(x,y)= e^{-y} sin x$ :



      Method 1 : by a method of the book Complex Variables and Applications by Brown and Churchill, Chapter 9 Section 104, $$v(x,y) = int_{(0,0)}^{(x,y)} -u_t(s, t) ds + u_s(s, t) dt = int_{(0,0)}^{(x,y)} e^{-y} sin x dx + e^{-y} cos x dy = -e^{-y} cos x - e^{-y} cos x +C = -2e^{-y} cos x +C$$ which $C$ can be chosen to be zero.



      Method 1 : Since $T(x,y)= e^{-y} sin x$ and its harmonic conjugate must be the real and imaginary parts of an analytic function, respectively, so it can be $f(z)=-ie^{iz} =e^{-y} sin x - ie^{-y} cos x.$



      So why there is an extra $2$ coefficient in Method 1? Where did I do wrong?



      Added. Probably the Method 1 is wrong. It comes from if $F_x(x,y) = P(x,y), F_y(x,y) = Q(x,y)$ holds in here. For example the method gives the correct answer for $u=xy$ but again it give a wrong answer for $u=x^3-3xy^2:$ that is $v=6x^2y-y^3$; but the correct one is $v=3x^2y-y^3$. But it looks impossible such a famous book to make a mistake. (???)






      complex-analysis proof-verification examples-counterexamples




      complex-analysis proof-verification examples-counterexamples






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 18 '18 at 2:21







      72D

















      asked Dec 14 '18 at 3:44









      72D72D

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          $begingroup$

          You did not compute the line integral correctly.



          For $u(x,y)=e^{-y}sin x$, we have $u_{,1}(x,y)=dfrac{partial u(x,y)}{partial x}=e^{-y}cos x$ and $u_{,2}(x,y)=dfrac{partial u(x,y)}{partial y}=-e^{-y}sin x$. Hence taking the straight-line path $gammacolontauin[0,1]mapsto(xtau,ytau)$ joining $(0,0)$ to $(x,y)$, we have



          begin{align*}
          &int_{(0,0)}^{(x,y)} (-u_{,2},u_{,1})(s, t)cdot(mathrm{d}s,mathrm{d}t)\
          &=int_0^1 (-u_{,2},u_{,1})(gamma(tau))cdotdotgamma(tau),mathrm{d}tau\
          &=int_0^1 (-u_{,2},u_{,1})(xtau,ytau)cdotdot(x,y),mathrm{d}tau\
          &=int_0^1 (e^{-ytau}sin (xtau),e^{-ytau}cos xtau)cdotdot(x,y),mathrm{d}tau\
          &=int_0^1 (xe^{-ytau}sin (xtau)+ye^{-ytau}cos xtau),mathrm{d}tau\
          &=Big[-e^{-ytau}cos(xtau)Big]_{tau=0}^{tau=1}=1-e^{-y}cos x.
          end{align*}






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            1 Answer
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            1 Answer
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            active

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            active

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            1





            +50







            $begingroup$

            You did not compute the line integral correctly.



            For $u(x,y)=e^{-y}sin x$, we have $u_{,1}(x,y)=dfrac{partial u(x,y)}{partial x}=e^{-y}cos x$ and $u_{,2}(x,y)=dfrac{partial u(x,y)}{partial y}=-e^{-y}sin x$. Hence taking the straight-line path $gammacolontauin[0,1]mapsto(xtau,ytau)$ joining $(0,0)$ to $(x,y)$, we have



            begin{align*}
            &int_{(0,0)}^{(x,y)} (-u_{,2},u_{,1})(s, t)cdot(mathrm{d}s,mathrm{d}t)\
            &=int_0^1 (-u_{,2},u_{,1})(gamma(tau))cdotdotgamma(tau),mathrm{d}tau\
            &=int_0^1 (-u_{,2},u_{,1})(xtau,ytau)cdotdot(x,y),mathrm{d}tau\
            &=int_0^1 (e^{-ytau}sin (xtau),e^{-ytau}cos xtau)cdotdot(x,y),mathrm{d}tau\
            &=int_0^1 (xe^{-ytau}sin (xtau)+ye^{-ytau}cos xtau),mathrm{d}tau\
            &=Big[-e^{-ytau}cos(xtau)Big]_{tau=0}^{tau=1}=1-e^{-y}cos x.
            end{align*}






            share|cite|improve this answer









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              1





              +50







              $begingroup$

              You did not compute the line integral correctly.



              For $u(x,y)=e^{-y}sin x$, we have $u_{,1}(x,y)=dfrac{partial u(x,y)}{partial x}=e^{-y}cos x$ and $u_{,2}(x,y)=dfrac{partial u(x,y)}{partial y}=-e^{-y}sin x$. Hence taking the straight-line path $gammacolontauin[0,1]mapsto(xtau,ytau)$ joining $(0,0)$ to $(x,y)$, we have



              begin{align*}
              &int_{(0,0)}^{(x,y)} (-u_{,2},u_{,1})(s, t)cdot(mathrm{d}s,mathrm{d}t)\
              &=int_0^1 (-u_{,2},u_{,1})(gamma(tau))cdotdotgamma(tau),mathrm{d}tau\
              &=int_0^1 (-u_{,2},u_{,1})(xtau,ytau)cdotdot(x,y),mathrm{d}tau\
              &=int_0^1 (e^{-ytau}sin (xtau),e^{-ytau}cos xtau)cdotdot(x,y),mathrm{d}tau\
              &=int_0^1 (xe^{-ytau}sin (xtau)+ye^{-ytau}cos xtau),mathrm{d}tau\
              &=Big[-e^{-ytau}cos(xtau)Big]_{tau=0}^{tau=1}=1-e^{-y}cos x.
              end{align*}






              share|cite|improve this answer









              $endgroup$
















                1





                +50







                1





                +50



                1




                +50



                $begingroup$

                You did not compute the line integral correctly.



                For $u(x,y)=e^{-y}sin x$, we have $u_{,1}(x,y)=dfrac{partial u(x,y)}{partial x}=e^{-y}cos x$ and $u_{,2}(x,y)=dfrac{partial u(x,y)}{partial y}=-e^{-y}sin x$. Hence taking the straight-line path $gammacolontauin[0,1]mapsto(xtau,ytau)$ joining $(0,0)$ to $(x,y)$, we have



                begin{align*}
                &int_{(0,0)}^{(x,y)} (-u_{,2},u_{,1})(s, t)cdot(mathrm{d}s,mathrm{d}t)\
                &=int_0^1 (-u_{,2},u_{,1})(gamma(tau))cdotdotgamma(tau),mathrm{d}tau\
                &=int_0^1 (-u_{,2},u_{,1})(xtau,ytau)cdotdot(x,y),mathrm{d}tau\
                &=int_0^1 (e^{-ytau}sin (xtau),e^{-ytau}cos xtau)cdotdot(x,y),mathrm{d}tau\
                &=int_0^1 (xe^{-ytau}sin (xtau)+ye^{-ytau}cos xtau),mathrm{d}tau\
                &=Big[-e^{-ytau}cos(xtau)Big]_{tau=0}^{tau=1}=1-e^{-y}cos x.
                end{align*}






                share|cite|improve this answer









                $endgroup$



                You did not compute the line integral correctly.



                For $u(x,y)=e^{-y}sin x$, we have $u_{,1}(x,y)=dfrac{partial u(x,y)}{partial x}=e^{-y}cos x$ and $u_{,2}(x,y)=dfrac{partial u(x,y)}{partial y}=-e^{-y}sin x$. Hence taking the straight-line path $gammacolontauin[0,1]mapsto(xtau,ytau)$ joining $(0,0)$ to $(x,y)$, we have



                begin{align*}
                &int_{(0,0)}^{(x,y)} (-u_{,2},u_{,1})(s, t)cdot(mathrm{d}s,mathrm{d}t)\
                &=int_0^1 (-u_{,2},u_{,1})(gamma(tau))cdotdotgamma(tau),mathrm{d}tau\
                &=int_0^1 (-u_{,2},u_{,1})(xtau,ytau)cdotdot(x,y),mathrm{d}tau\
                &=int_0^1 (e^{-ytau}sin (xtau),e^{-ytau}cos xtau)cdotdot(x,y),mathrm{d}tau\
                &=int_0^1 (xe^{-ytau}sin (xtau)+ye^{-ytau}cos xtau),mathrm{d}tau\
                &=Big[-e^{-ytau}cos(xtau)Big]_{tau=0}^{tau=1}=1-e^{-y}cos x.
                end{align*}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 18 '18 at 7:41









                user10354138user10354138

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                7,4322925






























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