How to find the harmonic conjugate of $v(x,y)=log((x-1)^2 +(y-2)^2)$?












0














So I have found out $$frac{partial v}{partial x} = frac{2x-2}{(x-1)^2 +(y-2)^2}, frac{partial v}{partial y} = frac{2y-4}{(x-1)^2 +(y-2)^2}.$$



Using the Cauchy Riemann equations, I find:
$$frac{partial u}{partial x} = frac{2y-4}{(x-1)^2 +(y-2)^2}, frac{partial u}{partial y} = frac{2-2x}{(x-1)^2 +(y-2)^2}.$$



Then I integrate either one of $partial u/partial x$ or $partial u/partial y$ to then differentiate with respect to the other variable I just integrated with. However, when I integrate $partial u/partial x$ or $partial u/partial y$, I get two different functions: $2arctanfrac{(x-1)}{(y-2)} + g(y)$ and $-2arctanfrac{(y-2)}{(x-1)} + g(x)$, respectively.



What am I doing wrong? Can you have multiple harmonic conjuguates?






complex-analysis harmonic-functions







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    0














    So I have found out $$frac{partial v}{partial x} = frac{2x-2}{(x-1)^2 +(y-2)^2}, frac{partial v}{partial y} = frac{2y-4}{(x-1)^2 +(y-2)^2}.$$



    Using the Cauchy Riemann equations, I find:
    $$frac{partial u}{partial x} = frac{2y-4}{(x-1)^2 +(y-2)^2}, frac{partial u}{partial y} = frac{2-2x}{(x-1)^2 +(y-2)^2}.$$



    Then I integrate either one of $partial u/partial x$ or $partial u/partial y$ to then differentiate with respect to the other variable I just integrated with. However, when I integrate $partial u/partial x$ or $partial u/partial y$, I get two different functions: $2arctanfrac{(x-1)}{(y-2)} + g(y)$ and $-2arctanfrac{(y-2)}{(x-1)} + g(x)$, respectively.



    What am I doing wrong? Can you have multiple harmonic conjuguates?






    complex-analysis harmonic-functions







    share|cite|improve this question



























      0












      0








      0







      So I have found out $$frac{partial v}{partial x} = frac{2x-2}{(x-1)^2 +(y-2)^2}, frac{partial v}{partial y} = frac{2y-4}{(x-1)^2 +(y-2)^2}.$$



      Using the Cauchy Riemann equations, I find:
      $$frac{partial u}{partial x} = frac{2y-4}{(x-1)^2 +(y-2)^2}, frac{partial u}{partial y} = frac{2-2x}{(x-1)^2 +(y-2)^2}.$$



      Then I integrate either one of $partial u/partial x$ or $partial u/partial y$ to then differentiate with respect to the other variable I just integrated with. However, when I integrate $partial u/partial x$ or $partial u/partial y$, I get two different functions: $2arctanfrac{(x-1)}{(y-2)} + g(y)$ and $-2arctanfrac{(y-2)}{(x-1)} + g(x)$, respectively.



      What am I doing wrong? Can you have multiple harmonic conjuguates?






      complex-analysis harmonic-functions







      share|cite|improve this question















      So I have found out $$frac{partial v}{partial x} = frac{2x-2}{(x-1)^2 +(y-2)^2}, frac{partial v}{partial y} = frac{2y-4}{(x-1)^2 +(y-2)^2}.$$



      Using the Cauchy Riemann equations, I find:
      $$frac{partial u}{partial x} = frac{2y-4}{(x-1)^2 +(y-2)^2}, frac{partial u}{partial y} = frac{2-2x}{(x-1)^2 +(y-2)^2}.$$



      Then I integrate either one of $partial u/partial x$ or $partial u/partial y$ to then differentiate with respect to the other variable I just integrated with. However, when I integrate $partial u/partial x$ or $partial u/partial y$, I get two different functions: $2arctanfrac{(x-1)}{(y-2)} + g(y)$ and $-2arctanfrac{(y-2)}{(x-1)} + g(x)$, respectively.



      What am I doing wrong? Can you have multiple harmonic conjuguates?






      complex-analysis harmonic-functions




      complex-analysis harmonic-functions






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      share|cite|improve this question













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      edited Nov 22 '18 at 12:36









      Ennar

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      14.4k32343










      asked Nov 22 '18 at 11:32









      M. Calculator

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          You will not be able to find an harmonic conjugate of $v$ since there is none. That is, there is no function $ucolonmathbb{C}setminus{1+2i}longrightarrowmathbb C$ which is an harmonic conjugate of $v$. You will find a proof here.






          share|cite|improve this answer



















          • 1




            I have deleted my first sentence. On the other hand, I thought that your problem was to find an harmonic conjugate $u$ of $v$. Am I wrong?
            – José Carlos Santos
            Nov 22 '18 at 11:51








          • 1




            No. The fact that $v$ is harmonic only assures that a harmonic conjugate exists localy. But your function is a standard example of function without a global harmonic conjugate.
            – José Carlos Santos
            Nov 22 '18 at 11:56






          • 1




            For each $(x,y)in D_v$ there is an open ball $B$ centered at that point such that $v|_B$ has a harmonic conjugate; that's the meaning of the fact that locally $v$ has a harmonic conjugate. However, there is no harmonic conjugate of $v$ whose domain is $D_v$; that's what it means that there is no harmonic conjugate globally.
            – José Carlos Santos
            Nov 22 '18 at 12:02








          • 1




            Actually, $1+2i$ is a bad choice, since $1+2i$ is the only point outside the domain of $v$. There is no global harmonic conjugate of $v$, but there is one near every point of its domain.
            – José Carlos Santos
            Nov 22 '18 at 12:26






          • 1




            I have no more ways of telling you that $v$ has no harmonic conjugate. I'm off.
            – José Carlos Santos
            Nov 22 '18 at 12:37











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          1 Answer
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          1 Answer
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          active

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          1














          You will not be able to find an harmonic conjugate of $v$ since there is none. That is, there is no function $ucolonmathbb{C}setminus{1+2i}longrightarrowmathbb C$ which is an harmonic conjugate of $v$. You will find a proof here.






          share|cite|improve this answer



















          • 1




            I have deleted my first sentence. On the other hand, I thought that your problem was to find an harmonic conjugate $u$ of $v$. Am I wrong?
            – José Carlos Santos
            Nov 22 '18 at 11:51








          • 1




            No. The fact that $v$ is harmonic only assures that a harmonic conjugate exists localy. But your function is a standard example of function without a global harmonic conjugate.
            – José Carlos Santos
            Nov 22 '18 at 11:56






          • 1




            For each $(x,y)in D_v$ there is an open ball $B$ centered at that point such that $v|_B$ has a harmonic conjugate; that's the meaning of the fact that locally $v$ has a harmonic conjugate. However, there is no harmonic conjugate of $v$ whose domain is $D_v$; that's what it means that there is no harmonic conjugate globally.
            – José Carlos Santos
            Nov 22 '18 at 12:02








          • 1




            Actually, $1+2i$ is a bad choice, since $1+2i$ is the only point outside the domain of $v$. There is no global harmonic conjugate of $v$, but there is one near every point of its domain.
            – José Carlos Santos
            Nov 22 '18 at 12:26






          • 1




            I have no more ways of telling you that $v$ has no harmonic conjugate. I'm off.
            – José Carlos Santos
            Nov 22 '18 at 12:37
















          1














          You will not be able to find an harmonic conjugate of $v$ since there is none. That is, there is no function $ucolonmathbb{C}setminus{1+2i}longrightarrowmathbb C$ which is an harmonic conjugate of $v$. You will find a proof here.






          share|cite|improve this answer



















          • 1




            I have deleted my first sentence. On the other hand, I thought that your problem was to find an harmonic conjugate $u$ of $v$. Am I wrong?
            – José Carlos Santos
            Nov 22 '18 at 11:51








          • 1




            No. The fact that $v$ is harmonic only assures that a harmonic conjugate exists localy. But your function is a standard example of function without a global harmonic conjugate.
            – José Carlos Santos
            Nov 22 '18 at 11:56






          • 1




            For each $(x,y)in D_v$ there is an open ball $B$ centered at that point such that $v|_B$ has a harmonic conjugate; that's the meaning of the fact that locally $v$ has a harmonic conjugate. However, there is no harmonic conjugate of $v$ whose domain is $D_v$; that's what it means that there is no harmonic conjugate globally.
            – José Carlos Santos
            Nov 22 '18 at 12:02








          • 1




            Actually, $1+2i$ is a bad choice, since $1+2i$ is the only point outside the domain of $v$. There is no global harmonic conjugate of $v$, but there is one near every point of its domain.
            – José Carlos Santos
            Nov 22 '18 at 12:26






          • 1




            I have no more ways of telling you that $v$ has no harmonic conjugate. I'm off.
            – José Carlos Santos
            Nov 22 '18 at 12:37














          1












          1








          1






          You will not be able to find an harmonic conjugate of $v$ since there is none. That is, there is no function $ucolonmathbb{C}setminus{1+2i}longrightarrowmathbb C$ which is an harmonic conjugate of $v$. You will find a proof here.






          share|cite|improve this answer














          You will not be able to find an harmonic conjugate of $v$ since there is none. That is, there is no function $ucolonmathbb{C}setminus{1+2i}longrightarrowmathbb C$ which is an harmonic conjugate of $v$. You will find a proof here.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 22 '18 at 11:49

























          answered Nov 22 '18 at 11:41









          José Carlos Santos

          152k22123224




          152k22123224








          • 1




            I have deleted my first sentence. On the other hand, I thought that your problem was to find an harmonic conjugate $u$ of $v$. Am I wrong?
            – José Carlos Santos
            Nov 22 '18 at 11:51








          • 1




            No. The fact that $v$ is harmonic only assures that a harmonic conjugate exists localy. But your function is a standard example of function without a global harmonic conjugate.
            – José Carlos Santos
            Nov 22 '18 at 11:56






          • 1




            For each $(x,y)in D_v$ there is an open ball $B$ centered at that point such that $v|_B$ has a harmonic conjugate; that's the meaning of the fact that locally $v$ has a harmonic conjugate. However, there is no harmonic conjugate of $v$ whose domain is $D_v$; that's what it means that there is no harmonic conjugate globally.
            – José Carlos Santos
            Nov 22 '18 at 12:02








          • 1




            Actually, $1+2i$ is a bad choice, since $1+2i$ is the only point outside the domain of $v$. There is no global harmonic conjugate of $v$, but there is one near every point of its domain.
            – José Carlos Santos
            Nov 22 '18 at 12:26






          • 1




            I have no more ways of telling you that $v$ has no harmonic conjugate. I'm off.
            – José Carlos Santos
            Nov 22 '18 at 12:37














          • 1




            I have deleted my first sentence. On the other hand, I thought that your problem was to find an harmonic conjugate $u$ of $v$. Am I wrong?
            – José Carlos Santos
            Nov 22 '18 at 11:51








          • 1




            No. The fact that $v$ is harmonic only assures that a harmonic conjugate exists localy. But your function is a standard example of function without a global harmonic conjugate.
            – José Carlos Santos
            Nov 22 '18 at 11:56






          • 1




            For each $(x,y)in D_v$ there is an open ball $B$ centered at that point such that $v|_B$ has a harmonic conjugate; that's the meaning of the fact that locally $v$ has a harmonic conjugate. However, there is no harmonic conjugate of $v$ whose domain is $D_v$; that's what it means that there is no harmonic conjugate globally.
            – José Carlos Santos
            Nov 22 '18 at 12:02








          • 1




            Actually, $1+2i$ is a bad choice, since $1+2i$ is the only point outside the domain of $v$. There is no global harmonic conjugate of $v$, but there is one near every point of its domain.
            – José Carlos Santos
            Nov 22 '18 at 12:26






          • 1




            I have no more ways of telling you that $v$ has no harmonic conjugate. I'm off.
            – José Carlos Santos
            Nov 22 '18 at 12:37








          1




          1




          I have deleted my first sentence. On the other hand, I thought that your problem was to find an harmonic conjugate $u$ of $v$. Am I wrong?
          – José Carlos Santos
          Nov 22 '18 at 11:51






          I have deleted my first sentence. On the other hand, I thought that your problem was to find an harmonic conjugate $u$ of $v$. Am I wrong?
          – José Carlos Santos
          Nov 22 '18 at 11:51






          1




          1




          No. The fact that $v$ is harmonic only assures that a harmonic conjugate exists localy. But your function is a standard example of function without a global harmonic conjugate.
          – José Carlos Santos
          Nov 22 '18 at 11:56




          No. The fact that $v$ is harmonic only assures that a harmonic conjugate exists localy. But your function is a standard example of function without a global harmonic conjugate.
          – José Carlos Santos
          Nov 22 '18 at 11:56




          1




          1




          For each $(x,y)in D_v$ there is an open ball $B$ centered at that point such that $v|_B$ has a harmonic conjugate; that's the meaning of the fact that locally $v$ has a harmonic conjugate. However, there is no harmonic conjugate of $v$ whose domain is $D_v$; that's what it means that there is no harmonic conjugate globally.
          – José Carlos Santos
          Nov 22 '18 at 12:02






          For each $(x,y)in D_v$ there is an open ball $B$ centered at that point such that $v|_B$ has a harmonic conjugate; that's the meaning of the fact that locally $v$ has a harmonic conjugate. However, there is no harmonic conjugate of $v$ whose domain is $D_v$; that's what it means that there is no harmonic conjugate globally.
          – José Carlos Santos
          Nov 22 '18 at 12:02






          1




          1




          Actually, $1+2i$ is a bad choice, since $1+2i$ is the only point outside the domain of $v$. There is no global harmonic conjugate of $v$, but there is one near every point of its domain.
          – José Carlos Santos
          Nov 22 '18 at 12:26




          Actually, $1+2i$ is a bad choice, since $1+2i$ is the only point outside the domain of $v$. There is no global harmonic conjugate of $v$, but there is one near every point of its domain.
          – José Carlos Santos
          Nov 22 '18 at 12:26




          1




          1




          I have no more ways of telling you that $v$ has no harmonic conjugate. I'm off.
          – José Carlos Santos
          Nov 22 '18 at 12:37




          I have no more ways of telling you that $v$ has no harmonic conjugate. I'm off.
          – José Carlos Santos
          Nov 22 '18 at 12:37


















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