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My question is related to the Riemann Mapping Theorem. A function $f$ is biharmonic if $\Delta^2f=0$.

Let $D$ be a simply connected domain in $\mathbb{R}^2$ and denote by $B$ the unit ball. Assume that $f:D\to \mathbb{R}$ is a biharmonic function defined on $D$. Can I find a function $\Phi:B\to D$ such that the composition $f\circ \Phi:B\to \mathbb{R}$ is biharmonic?

If so, what can be said about $\Phi$? Can I choose $\Phi$ to be harmonic?

Note that the composition of a biharmonic function with a harmonic function need not be biharmonic in general!

edited May 18 2010 at 15:35

asked May 18 2010 at 15:09

Philipp
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	You probably want to add some more assumptions about the behavior of $\Phi$. For example, do you want $\Phi:B\to D$ to be a bijective diffeomorphism? Without something like that, the answer to your question is clearly 'yes'. Just identify $\mathbb{R}^2$ with $\mathbb{C}$ in the usual way and take $\Phi(z) = az+b$ for constants $a\not=0$ and $b$ in $\mathbb{C}$ suitably chosen so that $\Phi(B)\subset B$. This will certainly work. Also, you may need to add some assumptions about $f$, too, or are you asking for all biharmonic $f$? – Robert Bryant Jan 13 at 15:41
	(continued): oops! I meant to write '$\Phi(B)\subset D$' instead of '$\Phi(B)\subset B$' in the above comment, but I can't change it now. – Robert Bryant Jan 13 at 15:49

woodbass · Answer 1 · 2012-12-30 12:43:49Z UTC

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Why only consider biharmonic instead of $n-$harmonic ($n\geq 2$), say function satisfying $\Delta^n f$=0$?

answered Dec 30 at 12:43

woodbass
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Dear woodbass: since this is not an answer to the question but merely a comment, you should leave it as a comment via the "add comment" option. – Yemon Choi Dec 31 at 1:31

Sorry.Thank you for your notice. – woodbass Dec 31 at 7:45

biharmonic morphisms

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