Let $D$ be an annulus slit with

$$D= \{a<|z|< b \}$$ excluding $(-b,-a)$.

Show that any harmonic function on $D$ has a harmonic conjugate on $D$.

The hint says to fix $c$ between $a$ and $b$ and then do a line integral along a radius and then along a circular arc. I did this, but I'm not sure why this proves the existence of a harmonic conjugate.

I just integrated the differential $dv$ and used the Cauchy Riemann equations to get a line integral in terms of a harmonic function $u$. But what have I shown?


1 Answer 1


So am I going to try my best to answer this, don't be too harsh on me if it is all wrong!

I think you will want to look at the Cauchy-Riemann equations in polar coordinates, namely:

$$\frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta}$$ and $$\frac{ \partial v}{\partial r} = - \frac{1}{r} \frac{\partial u}{\partial \theta}$$

The reason I say this is because we are going to be integrating using polar coordinates. What we are going to do is given our harmonic function $u(r,\theta)$ we will take partial derivatives with respect to $r$ and $\theta$ giving us $u_r$ and $u_\theta$ respectively. We will then solve for $v$ using the Cauchy Riemann equations and integrating.

To this end, set $$v(r,\theta) = \int_a^r -\frac{1}{t} u_\theta(t,0) dt + \int_0^{\theta} r u_r(r,\psi) d\psi$$

We can then check that $u$ and $v$ satisfy the Cauchy-Riemann equations by taking partial derivatives.

We get: $$\frac{\partial v}{\partial \theta} = r u_r = r \frac{\partial u}{\partial r}$$ by the Fundamental Theorem of calculus and

$$\frac{\partial v}{\partial r} = - \frac{1}{r}u_\theta(r, 0) + \int_0^\theta \frac{\partial}{\partial r} r u_r (r, \psi) d\psi$$ (we can differentiate under the integral sign because $u$ is $C^2$ and the region is compact) $$= \int_0^\theta u_r(r,\psi)d\psi + \int_0^\theta r \frac{\partial^2 u}{\partial r^2} (r, \psi) d\psi$$

Now, $u$ is harmonic, so the Laplace equation in polar coordinates gives us $$\frac{\partial^2 u}{\partial r^2} = -\frac{1}{r}\frac{\partial u}{\partial r} -\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}$$

Hence, the above is $$= \int_0^\theta u_r(r,\psi)d\psi + \int_0^\theta r \left( -\frac{1}{r}u_r(r,\psi) -\frac{1}{r^2}u_{\theta\theta}(r,\psi) \right ) d\psi$$ $$=\int_0^\theta u_r(r,\psi)d\psi - \int_0^\theta u_r(r,\psi)d\psi + \int_0^\theta -\frac{1}{r}u_{\theta\theta}(r,\psi) d\psi= -\frac{1}{r}u_\theta$$ Hence, $$\frac{\partial v}{\partial r} = -\frac{1}{r}\frac{\partial u}{\partial \theta}$$

So $u,v$ satisfy the Cauchy Riemann equations.

  • $\begingroup$ Nice! I did almost the same thing, but how do we know that we've found a function that works for sure? That is to say, if D included $$(-b,-a)$$, what would have went wrong? $\endgroup$
    – user250095
    Apr 23, 2014 at 16:37
  • $\begingroup$ To be honest, I'm not completely sure on the reason why it fails, but I could give an example where it does fail if we don't take out the line segment. The example is $u(z) = \log |z|$ which is harmonic. If we use the above to solve for $v$ we would get $v(z) =\arg(z) = \arctan (y/x)$ which would not be continuous if we don't take out $(-b,-a)$. $\endgroup$ Apr 24, 2014 at 3:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .