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Assume that $U:\mathbb R^2\to S^2=\{y\in\mathbb R^3:|y|=1\}$ is a smooth solution of the equation $\Delta U+|\nabla U|^2U=0$ in $\mathbb R^2$ with $\int_{\mathbb R^2}|\nabla U|^2\,dx<+\infty$. Prove that $$\frac1{4\pi}\int_{\mathbb R^2}|\nabla U|^2\,dx\in\mathbb N_{\geq0}.\label{1}\tag{$*$}$$

This is a claim taken from Juan Davila, Manuel Del Pino and Juncheng Wei's paper (formula (1.5)). The paper refers to a proof using some ideas from geometry. I'm looking for an analytical proof of \eqref{1}.

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    $\begingroup$ Can you clarify what you mean by an "analytical proof"? The "geometric" proof follows the following steps (A) Lift $U$ conformally to a smooth mapping $\phi: \mathbb{S}^2\to\mathbb{S}^2$. (B) Since $\phi$ is harmonic, it is holomorphic (or antiholomorphic) [due to Wood]. (C) For holomorphic functions the energy is determined by the topological degree. Steps (A) and (C) are basic computations, so it feels to me that what you are asking for is an "analytical proof" of Wood's result. $\endgroup$
    – Willie Wong
    Commented Feb 20 at 6:42
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    $\begingroup$ But there is a problem: the result is necessarily global. Lemaire proved that if the domain is replaced by higher genus surfaces, there exists non-holomorphic harmonic maps into the sphere. So somewhere along the way the topology of $\mathbb{R}^2$ (or $\mathbb{S}^2$) must creep in. So it is unclear to me what you are actually asking. $\endgroup$
    – Willie Wong
    Commented Feb 20 at 6:49
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    $\begingroup$ @ChristianRemling: here $U(x,y) = (U_1(x,y), U_2(x,y), U_3(x,y))$ and $|\nabla U|^2 = \sum_{i} |\partial_x U_i|^2 + |\partial_y U_i|^2$. $\endgroup$
    – Willie Wong
    Commented Feb 20 at 14:55
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    $\begingroup$ By differentiating $U_1^2+U_2^2+U_3^2=1$ twice, we obtain $U\cdot\Delta U+|\nabla U|^2=0$ (for any function $U$), so the equation is only the extra condition that $\Delta U$ always points in the direction of $-U$. $\endgroup$
    – Christian Remling
    Commented Feb 20 at 22:40
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    $\begingroup$ @ChristianRemling Thanks for you comments! The fact that there are no harmonic functions with $\nabla \varphi\in L^2$ follows from: (1) If an $L^2$ function is harmonic in $\mathbb R^d$, then it is identically zero (by the mean value property and Hölder's inequality); (2) If $\varphi$ is harmonic, then $\nabla \varphi$ is also harmonic. $\endgroup$
    – Feng
    Commented Feb 21 at 2:43

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