Given a harmonic function $u$, it is the real part of some analytic function $f$ whose imaginary part, $v$, is the harmonic conjugate of $u$. Is this relationship symmetric? That is to say is then $u$ the harmonic conjugate of $v$?

Second question, if this harmonic function $u$ is defined on a domain $D$, does that mean its associated analytic function $f$ is also defined on all of $D$?

Final question, if a harmonic function is bounded, does that automatically mean that its harmonic conjugate is also bounded?



i) no, $-u$ is conjugate to $v$. (Easy to see cause $i(u+iv) = -(v-iu)$)

ii) this is true only on simply connected domains. If $u$ is harmonic then the (locally unique) conjugate $v$ such that for one given point $p$, say $v(p) = c$, a prescribed constant, may be obtained locally as $$v(z) -c = \int_0^1 \frac{d}{dt}v(p+t(z-p)) dt $$ since, using the Cauchy Riemann equations ($u_x = v_y$ and $u_y = - v_x$), you can express this integral using the derivatives of $u$. (Instead of a straight line you can use any smooth curve, of course). Once $v$ is known in a neighbourhood it is uniquely determined along any smooth curve on which $u$ is defined (e.g. by an integral as above or by analytic continuation), but the values obtained may depend on the curve chosen. It can be shown that in simply connected domains the path chosen is not relevant, but on, say, an annulus, this is no longer true.

As for iii) I have to admit I don't know, but in general I'd expect the answer to be no. I'd start searching by looking at domains like $$\{(x,y): x>0, f_1(x) < y < f_2(x) \}$$ where $f_1(0) = f_2(0) = f_1'(0) = f_2'(0) = 0$ (so you have a cusp) and then by looking at harmonic $u$ on such domains such that $u_x$ remains bounded as $x\rightarrow 0$ but $u_y\rightarrow \infty$. If you can find such $u$ the above mentioned definition of $v$ will be unbounded near $0$.

  • $\begingroup$ Ok thanks Thomas, I'll take a look at this. $\endgroup$ – Thoth Apr 21 '12 at 17:34

As to the third question I think function $f(z)=\log\frac1{1-z}$ in the unit circle can be an example, since $\Im f(z)$ is bounded there and (using the polar coordinates) $$ \Re f(z)=\sum_{n=1}^\infty r^n\frac{\cos n\varphi}n $$ evidently not. But the sup norm is not well suited for many questions in partial differential equations. If I remember correctly, there is sun an estimate in $L_p$, $1<p<\infty$, due to Hardy and Littlewood.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.