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I am trying to resolve a question whether a certain function is harmonic or not. If yes, I should find its harmonic conjugate.

The function is $u = \frac{x}{x^2+y^2}$.

I found that it is a harmonic function by using Laplace equation, but I am not sure. How can I find its harmonic conjugate, please help anyone.

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You can do this without calculation. Recall that a holomorphic function is harmonic.

The function $$ f:z\longmapsto \frac{1}{z} $$ is holomorphic, so it is harmonic.

Hence its real part $$ u:z=x+iy\longmapsto \mbox{Re}f(z)=\mbox{Re}\frac{\bar{z}}{|z|^2}=\frac{\mbox{Re} \bar{z}}{|z|^2}=\frac{x}{x^2+y^2} $$ is harmonic.

And the harmonic conjugate of $u$ is $$ v:z=x+iy\longmapsto \mbox{Im} f(z) =\frac{\mbox{Im} \bar{z}}{|z|^2}=-\frac{y}{x^2+y^2}. $$

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  • $\begingroup$ But how do you find $f$ in the first place? In this case it's well-known, but what about the general case where you don't immediately see what $f$ is? $\endgroup$ – Elmar Zander Mar 4 '13 at 13:01
  • $\begingroup$ @ElmarZander In the general case, you do like you said and you solve the differential equation. $\endgroup$ – Julien Mar 4 '13 at 14:46
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To the function to be harmonic its Laplacian should be zero $$ \Delta f = f_{xx}(x,y)+f_{yy}(x,y) = 0 $$ Just check it $$ f_{xx} = \left(\frac {x}{x^2+y^2}\right)_{xx} = \left [\frac {-x^2+y^2}{\left( x^2+y^2\right)^2}\right]_x = \frac {2x(x^2-3y^2)}{\left( x^2+y^2\right)^3} \\ f_{yy} = \left(\frac {x}{x^2+y^2}\right)_{yy} = \left [-\frac {2xy}{\left ( x^2+y^2\right)^3} \right ]_y = -\frac {2x(x^2-3y^2)}{\left( x^2+y^2\right)^3} $$ which means $\Delta f = 0$ hence harmonic.

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If a function $f(z)=u(x,y)+i v(x,y)$, with $z=x+iy$ and $u,v:\mathbb R^2\to\mathbb R$, is holomorphic, it is also harmonic. So, if you can find a $v$, such that the Cauchy-Riemann equations hold, i.e.: \begin{align} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x}, \end{align} you've found your harmonic conjugate.

If you plug in your $u$ into these equation, you get two differential equations that you can solve to get your $v$ up to some (arbitrary) constant of integration.

Edit: The equations you get in this case are: \begin{align} \frac{\partial v}{\partial y} &= \frac{y^2-x^2}{(x^2+y^2)^2} \\ \frac{\partial v}{\partial x} &= \frac{2xy}{(x^2+y^2)^2} \end{align} Each of these you can integrate separately, and you get something like $v(x,y)=v_1(x,y) + f(x)$ from the first, $v(x,y)=v_2(x,y) + g(y)$ from the second equation. The functions $f$ and $g$ are "constants" of integration, which you need to match with corresponding terms in $v_2$ and $v_1$, respectively.

Edit 2: There is also a way to do it without using the CR equations. I knew I read it some time ago, but it took me quite a while to find it again: William T. Shaw "Recovering Holomorphic Functions from Their Real or Imaginary Parts without the Cauchy--Riemann Equations" SIAM Rev., 46(4), 717–728.

You can recover $f$ by $$f(z)=2u\left( \frac{z+\bar a}{2},\frac{z-\bar a}{2i}\right)-\overline{f(a)}$$ where $a$ is an arbitrary point, such that $f$ is holomorphic in a neighborhood of $a$, and you need to extend $u$ here is the extension from the given $u$ to $\mathbb C\times\mathbb C$. Why that works, you would need to read up in the paper I cited. (Or maybe search for Milne-Thomson method, I think it's more or less the same).

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  • $\begingroup$ This is a false statement. If $f$ is holomorphic, then $f$ is harmonic. But not every harmonic function is holomorphic. For instance, antiholomorphic functions are also harmonic. So the Cauchy-Riemann equations need not hold. And what is $v$ anyway? Once you have the harmonic conjugate, you can indeed appeal to Cauchy-Riemann. $\endgroup$ – Julien Mar 4 '13 at 12:40
  • $\begingroup$ @julien Hmm, you are right. It was somehow the wrong way round. Better now? $\endgroup$ – Elmar Zander Mar 4 '13 at 12:59
  • $\begingroup$ Yes, looks good to me. $\endgroup$ – Julien Mar 4 '13 at 14:47
  • $\begingroup$ @ElmarZander will this method help me solve this problem? math.stackexchange.com/questions/398543/… $\endgroup$ – User69127 May 22 '13 at 2:48
  • $\begingroup$ @User69127 You could use it to find that your $u$ is the real part of $f(z)=1+iz+1/z+ci$. Then you could plug in $z^2$ instead of $z$ and take the real part of that function, i.e. take $\Re(f(z^2))$. $\endgroup$ – Elmar Zander May 28 '13 at 8:56

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