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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?

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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 Magnusson Nov 12 '11 at 17:07
    
Maybe this might be interesting... –  draks ... Mar 30 '12 at 19:35
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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.

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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
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