Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was looking at a comprehensive exam, and I found the this question. Can anyone help me out?

If $u$ is a harmonic function, which type of function $f$ is needed so that $f(u)$ is harmonic?

share|cite|improve this question
The standard Laplacian on $\mathbb C$ is a constant multiple of $\partial \bar \partial$, so you can obtain the answer to your question by breaking out the chain rule for Wirtinger derivatives. – Gunnar Þór Magnússon Nov 12 '11 at 17:07
Maybe this might be interesting... – draks ... Mar 30 '12 at 19:35
up vote 3 down vote accepted

Note that $u$ is harmonic if and only if $$\Delta u:=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0.$$ Therefore, if $u$ is harmonic, by chain rule we have $$\Delta f(u)=\frac{df}{du}\Big(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\Big)+\frac{d^2f}{du^2}\Big(\big(\frac{\partial u}{\partial x}\big)^2+\big(\frac{\partial u}{\partial y}\big)^2\Big)=\frac{d^2f}{du^2}\Big(\big(\frac{\partial u}{\partial x}\big)^2+\big(\frac{\partial u}{\partial y}\big)^2\Big).$$ If $u$ is a constant function (which is harmonic), then $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}=0$. It follows from the above formula that $f(u)$ is harmonic for any function $f$. On the other hand, if $u$ is a nonconstant harmonic function, then $\big(\frac{\partial u}{\partial x}\big)^2+\big(\frac{\partial u}{\partial y}\big)^2\neq 0$. Again it follows from the above formula that $f(u)$ is harmonic when $\frac{d^2f}{du^2}=0$, that is, when $f$ is a linear function in $u$: $$f(u)=Au+B,$$ where $A$ and $B$ are constants.

share|cite|improve this answer
Thank you Paul. Should it be $\frac{d^2 f}{du^2}$ instead fo $\frac{d^2 f}{d^2 u}$? – josh Nov 13 '11 at 14:43
Yes, you are right. That was a typo. I corrected it. – Paul Nov 13 '11 at 21:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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