Let $F$ be the family $f:U\to R^3$ of smooth mappings of the unit disk onto $R^3$ mapping the boundary $T$ onto a smooth Jordan curve $\Gamma\subset R^3$. Form the energy integral $$E[f]=\int_U \|f_x\|^2 +\|f_y\|^2 \, dx \, dy $$ my question is what is the minimum of this functional on $F$?
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$\begingroup$ You want an explicit number ? Do you require $f$ to be equal to a specific parametrization of $\Gamma$ on the boundary of $U$ ? $\endgroup$– Thomas RichardApr 1, 2016 at 7:21
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$\begingroup$ & Thomas Richard. Yes, I need some specific parametrization on the boundary. It seems that the solution is harmonic mapping but I am not sure anymore $\endgroup$– DaglasApr 1, 2016 at 7:34
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1$\begingroup$ If the question you ask is "are minimizers of E harmonic maps ?" then answer is yes. $\endgroup$– Thomas RichardApr 1, 2016 at 7:41
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$\begingroup$ @Thomas the following problem arises, let $\rho(z)=1/(1+|z|^2)$ be the metric which comes from the stereographic projection $f(z)=(2x/(1+|z|^2, ... )$ of $R^2 = C$ onto the Sphere $S^2$. Then $g_{z\bar z} + \partial_w \log(\rho(w))\circ g \cdot g_z g_{\bar z}=0$ is harmonic mapping equation. An example of solution is $g(x+i y)=\tan (x)$. Then $E_\rho[g]= E[f\circ g]$, where $E_\rho =\int_U \rho^2(g(z))(|g_x|^2+|g_y|^2) dx dy$. but $f\circ g$ is not Euclidean harmonic. Probably the problem here is that $g$ is not minimizer of $E_\rho$ ;) $\endgroup$– DaglasApr 1, 2016 at 7:58
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1$\begingroup$ If I am not mistaken when the source space is 2D, the harmonic map equation is conformally invariant, so there must an error in your computation. $\endgroup$– Thomas RichardApr 4, 2016 at 17:53
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1 Answer
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This the classical Plateau problem, the minimum is achieved by a minimal surface which bounds $\Gamma$. It is not trivial, since your functional is invariant by the Mobius group which is not compact, you will find a clear exposition in the Struwe'book: Plateau probelm
