Harmonic conjugate of $ln(|z|)$












3














Find the harmonic conjugate of $u$. $u = u(z) = ln(|z|)$ so $u(z) = ln(sqrt{x^2 + y^2})$



I am trying to find now its harmonic conjugate I did all the math:



I got two solutions though.
One is $v(z) = arctan(y/x) + C$ if I solve $partial u/partial x = -partial v/partial y$ & other is
$v(z) = - arctan(x/y) + C$ if I solved $partial v/partial y = partial u/partial x$



can I have two solutions, or is only one solution correct?






complex-analysis







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  • Since I just answered another, I thought I would see if I could answer this one for you as well.
    – Stephen Montgomery-Smith
    Feb 23 '15 at 0:30
















3














Find the harmonic conjugate of $u$. $u = u(z) = ln(|z|)$ so $u(z) = ln(sqrt{x^2 + y^2})$



I am trying to find now its harmonic conjugate I did all the math:



I got two solutions though.
One is $v(z) = arctan(y/x) + C$ if I solve $partial u/partial x = -partial v/partial y$ & other is
$v(z) = - arctan(x/y) + C$ if I solved $partial v/partial y = partial u/partial x$



can I have two solutions, or is only one solution correct?






complex-analysis







share|cite|improve this question
























  • Since I just answered another, I thought I would see if I could answer this one for you as well.
    – Stephen Montgomery-Smith
    Feb 23 '15 at 0:30














3












3








3


2





Find the harmonic conjugate of $u$. $u = u(z) = ln(|z|)$ so $u(z) = ln(sqrt{x^2 + y^2})$



I am trying to find now its harmonic conjugate I did all the math:



I got two solutions though.
One is $v(z) = arctan(y/x) + C$ if I solve $partial u/partial x = -partial v/partial y$ & other is
$v(z) = - arctan(x/y) + C$ if I solved $partial v/partial y = partial u/partial x$



can I have two solutions, or is only one solution correct?






complex-analysis







share|cite|improve this question















Find the harmonic conjugate of $u$. $u = u(z) = ln(|z|)$ so $u(z) = ln(sqrt{x^2 + y^2})$



I am trying to find now its harmonic conjugate I did all the math:



I got two solutions though.
One is $v(z) = arctan(y/x) + C$ if I solve $partial u/partial x = -partial v/partial y$ & other is
$v(z) = - arctan(x/y) + C$ if I solved $partial v/partial y = partial u/partial x$



can I have two solutions, or is only one solution correct?






complex-analysis




complex-analysis






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share|cite|improve this question













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edited Feb 23 '15 at 0:31









Stephen Montgomery-Smith

17.7k12247




17.7k12247










asked Oct 9 '13 at 5:22









Illustionist

379211




379211












  • Since I just answered another, I thought I would see if I could answer this one for you as well.
    – Stephen Montgomery-Smith
    Feb 23 '15 at 0:30


















  • Since I just answered another, I thought I would see if I could answer this one for you as well.
    – Stephen Montgomery-Smith
    Feb 23 '15 at 0:30
















Since I just answered another, I thought I would see if I could answer this one for you as well.
– Stephen Montgomery-Smith
Feb 23 '15 at 0:30




Since I just answered another, I thought I would see if I could answer this one for you as well.
– Stephen Montgomery-Smith
Feb 23 '15 at 0:30










2 Answers
2






active

oldest

votes


















2














$arctan(y/x) + arctan(x/y) = pm fracpi2 $. So in essence, your two solutions are the same.



Another way to get your solution: $ln(z)$ is analytic, and if $z = r e^{itheta}$, then $ln(r e^{itheta}) = ln r + itheta$. And $theta$ is the angle $z$ makes to the $x$-axis, that is, $arctan(y/x)$. You can get the other solutions by considering, for example, $ln(iz)$, or more generally $ln(az)$ where $|a| = 1$.






share|cite|improve this answer































    1














    A harmonic function is a twice differentiable function that satisfies Laplace's equation. Recall that for a function to be differentiable, it must first be continuous.



    As you found out by going through the calculations, the harmonic conjugate $v(z)$, if it exists, must be either $v(z) = arctan(y/x) + C_1(x)$ (where $C_1$ is a constant with respect to $y$ but is possibly a non-constant function of $x$) or $v(z) = arctan(x/y) + C_2(y)$.



    However, neither choice is continuous on $mathbb{R}^2$ or $mathbb{C}$, so neither choice is harmonic. Therefore $u(z)$ does not have a harmonic conjugate.



    To see that neither choice is continuous, consider the limit $(x, y) to (0, 0)$ of $arctan(y/x)$ and $arctan(x/y)$ along the lines $y = x$ and $y = -x$. Along the first line, they are both equal to $arctan(1) = sqrt{3}$; along the second line they are both equal to $arctan(-1) = -sqrt{3}$.






    share|cite|improve this answer





















    • Well, $u(z)$ itself is discontinuous at $0$, so it is not surprising that $v$ is discontinuous there. You will get a harmonic conjugate on any simply connected domain where $u$ is defined.
      – GEdgar
      Feb 23 '15 at 0:37













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    $arctan(y/x) + arctan(x/y) = pm fracpi2 $. So in essence, your two solutions are the same.



    Another way to get your solution: $ln(z)$ is analytic, and if $z = r e^{itheta}$, then $ln(r e^{itheta}) = ln r + itheta$. And $theta$ is the angle $z$ makes to the $x$-axis, that is, $arctan(y/x)$. You can get the other solutions by considering, for example, $ln(iz)$, or more generally $ln(az)$ where $|a| = 1$.






    share|cite|improve this answer




























      2














      $arctan(y/x) + arctan(x/y) = pm fracpi2 $. So in essence, your two solutions are the same.



      Another way to get your solution: $ln(z)$ is analytic, and if $z = r e^{itheta}$, then $ln(r e^{itheta}) = ln r + itheta$. And $theta$ is the angle $z$ makes to the $x$-axis, that is, $arctan(y/x)$. You can get the other solutions by considering, for example, $ln(iz)$, or more generally $ln(az)$ where $|a| = 1$.






      share|cite|improve this answer


























        2












        2








        2






        $arctan(y/x) + arctan(x/y) = pm fracpi2 $. So in essence, your two solutions are the same.



        Another way to get your solution: $ln(z)$ is analytic, and if $z = r e^{itheta}$, then $ln(r e^{itheta}) = ln r + itheta$. And $theta$ is the angle $z$ makes to the $x$-axis, that is, $arctan(y/x)$. You can get the other solutions by considering, for example, $ln(iz)$, or more generally $ln(az)$ where $|a| = 1$.






        share|cite|improve this answer














        $arctan(y/x) + arctan(x/y) = pm fracpi2 $. So in essence, your two solutions are the same.



        Another way to get your solution: $ln(z)$ is analytic, and if $z = r e^{itheta}$, then $ln(r e^{itheta}) = ln r + itheta$. And $theta$ is the angle $z$ makes to the $x$-axis, that is, $arctan(y/x)$. You can get the other solutions by considering, for example, $ln(iz)$, or more generally $ln(az)$ where $|a| = 1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 23 '15 at 0:28

























        answered Feb 23 '15 at 0:23









        Stephen Montgomery-Smith

        17.7k12247




        17.7k12247























            1














            A harmonic function is a twice differentiable function that satisfies Laplace's equation. Recall that for a function to be differentiable, it must first be continuous.



            As you found out by going through the calculations, the harmonic conjugate $v(z)$, if it exists, must be either $v(z) = arctan(y/x) + C_1(x)$ (where $C_1$ is a constant with respect to $y$ but is possibly a non-constant function of $x$) or $v(z) = arctan(x/y) + C_2(y)$.



            However, neither choice is continuous on $mathbb{R}^2$ or $mathbb{C}$, so neither choice is harmonic. Therefore $u(z)$ does not have a harmonic conjugate.



            To see that neither choice is continuous, consider the limit $(x, y) to (0, 0)$ of $arctan(y/x)$ and $arctan(x/y)$ along the lines $y = x$ and $y = -x$. Along the first line, they are both equal to $arctan(1) = sqrt{3}$; along the second line they are both equal to $arctan(-1) = -sqrt{3}$.






            share|cite|improve this answer





















            • Well, $u(z)$ itself is discontinuous at $0$, so it is not surprising that $v$ is discontinuous there. You will get a harmonic conjugate on any simply connected domain where $u$ is defined.
              – GEdgar
              Feb 23 '15 at 0:37


















            1














            A harmonic function is a twice differentiable function that satisfies Laplace's equation. Recall that for a function to be differentiable, it must first be continuous.



            As you found out by going through the calculations, the harmonic conjugate $v(z)$, if it exists, must be either $v(z) = arctan(y/x) + C_1(x)$ (where $C_1$ is a constant with respect to $y$ but is possibly a non-constant function of $x$) or $v(z) = arctan(x/y) + C_2(y)$.



            However, neither choice is continuous on $mathbb{R}^2$ or $mathbb{C}$, so neither choice is harmonic. Therefore $u(z)$ does not have a harmonic conjugate.



            To see that neither choice is continuous, consider the limit $(x, y) to (0, 0)$ of $arctan(y/x)$ and $arctan(x/y)$ along the lines $y = x$ and $y = -x$. Along the first line, they are both equal to $arctan(1) = sqrt{3}$; along the second line they are both equal to $arctan(-1) = -sqrt{3}$.






            share|cite|improve this answer





















            • Well, $u(z)$ itself is discontinuous at $0$, so it is not surprising that $v$ is discontinuous there. You will get a harmonic conjugate on any simply connected domain where $u$ is defined.
              – GEdgar
              Feb 23 '15 at 0:37
















            1












            1








            1






            A harmonic function is a twice differentiable function that satisfies Laplace's equation. Recall that for a function to be differentiable, it must first be continuous.



            As you found out by going through the calculations, the harmonic conjugate $v(z)$, if it exists, must be either $v(z) = arctan(y/x) + C_1(x)$ (where $C_1$ is a constant with respect to $y$ but is possibly a non-constant function of $x$) or $v(z) = arctan(x/y) + C_2(y)$.



            However, neither choice is continuous on $mathbb{R}^2$ or $mathbb{C}$, so neither choice is harmonic. Therefore $u(z)$ does not have a harmonic conjugate.



            To see that neither choice is continuous, consider the limit $(x, y) to (0, 0)$ of $arctan(y/x)$ and $arctan(x/y)$ along the lines $y = x$ and $y = -x$. Along the first line, they are both equal to $arctan(1) = sqrt{3}$; along the second line they are both equal to $arctan(-1) = -sqrt{3}$.






            share|cite|improve this answer












            A harmonic function is a twice differentiable function that satisfies Laplace's equation. Recall that for a function to be differentiable, it must first be continuous.



            As you found out by going through the calculations, the harmonic conjugate $v(z)$, if it exists, must be either $v(z) = arctan(y/x) + C_1(x)$ (where $C_1$ is a constant with respect to $y$ but is possibly a non-constant function of $x$) or $v(z) = arctan(x/y) + C_2(y)$.



            However, neither choice is continuous on $mathbb{R}^2$ or $mathbb{C}$, so neither choice is harmonic. Therefore $u(z)$ does not have a harmonic conjugate.



            To see that neither choice is continuous, consider the limit $(x, y) to (0, 0)$ of $arctan(y/x)$ and $arctan(x/y)$ along the lines $y = x$ and $y = -x$. Along the first line, they are both equal to $arctan(1) = sqrt{3}$; along the second line they are both equal to $arctan(-1) = -sqrt{3}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 9 '13 at 7:33









            Kyle

            1,092717




            1,092717












            • Well, $u(z)$ itself is discontinuous at $0$, so it is not surprising that $v$ is discontinuous there. You will get a harmonic conjugate on any simply connected domain where $u$ is defined.
              – GEdgar
              Feb 23 '15 at 0:37




















            • Well, $u(z)$ itself is discontinuous at $0$, so it is not surprising that $v$ is discontinuous there. You will get a harmonic conjugate on any simply connected domain where $u$ is defined.
              – GEdgar
              Feb 23 '15 at 0:37


















            Well, $u(z)$ itself is discontinuous at $0$, so it is not surprising that $v$ is discontinuous there. You will get a harmonic conjugate on any simply connected domain where $u$ is defined.
            – GEdgar
            Feb 23 '15 at 0:37






            Well, $u(z)$ itself is discontinuous at $0$, so it is not surprising that $v$ is discontinuous there. You will get a harmonic conjugate on any simply connected domain where $u$ is defined.
            – GEdgar
            Feb 23 '15 at 0:37




















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