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Let $\Omega$ be the unit sphere in two Or three dimensions. Why is then $(\nabla u, \nabla w)=(|\nabla u|^2u,w)$ for all testfunctions w in $C_0^\infty(\Omega, R^n)$? How to compute it?

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Welcome to Math.SE. Your question needs some clarification. I see that $w$ is a test function. But what is $u$? What exactly do you want to compute? – user53153 Feb 2 '13 at 17:27
I want to compute the stated identity. – gurfd Feb 2 '13 at 18:23
OK. And $u$ is a weakly harmonic map from $\Omega$ to something? What is the target space of $u$? – user53153 Feb 2 '13 at 21:58
Sorry, i forgot the important information that $u=x/|x|$. – gurfd Feb 3 '13 at 13:00
up vote 1 down vote accepted

I think you mean that $\Omega $ is the unit ball $\{x:|x|<1\}$ rather than the unit sphere $\{x:|x|=1\}$. Indeed, the Laplace equation for $u$ takes the form $\Delta u+|\nabla u|^2 u=0$ (with the Euclidean Laplacian $\Delta$) only when the domain of $u$ is flat and the target is a unit sphere $S^k$ of any dimension.

We can calculate the derivative of $u$ using the product rule and the chain rule: $$ \nabla (|x|^{-1}x) = \nabla (|x|^{-1})\otimes x + |x|^{-1}\nabla x = (-1)|x|^{-3} x\otimes x + |x|^{-1} I $$ where $I$ is the identity matrix. In a simpler form, $$ \nabla u = \frac{|x|^2I-x\otimes x}{|x|^3} \tag{1} $$ With a little algebra we get $$ \begin{align} |\nabla u|^2 &= \frac{1}{|x|^6} \sum_{i,j=1}^n (|x|^2\delta_{ij}-x_ix_j)^2 \\&=\frac{1}{|x|^6} \sum_{i=1}^n \left\{ (|x|^2-x_i^2)^2 + x_i^2 \sum_{j\ne i} x_j^2 \right\} \\&=\frac{1}{|x|^6}\sum_{i=1}^n \left\{ (|x|^2-x_i^2)|x|^2\right\} =\frac{n-1}{|x|^2} \end{align}\tag{2}$$

Note that in two dimensions the expression in (2) is not integrable near the origin. Therefore, in this case $u$ does not belong to $H^1$ and is not weakly harmonic. (Aside: weakly harmonic maps in two dimensions are smooth, by a theorem of Hélein.)

In dimensions $n\ge 3$ we have $u\in W^{2,1}(\Omega)$ because the second-order derivatives of $u$ are homogeneous of degree $-2$. Thus, we can show that the identity holds in the strong sense: $-\Delta u=|\nabla u|^2u$ in $L^1$. Indeed, for any $i=1,\dots,n$ we have $$ \begin{align} \Delta (|x|^{-1}x_i)&=\operatorname{div}\nabla (|x|^{-1}x_i)=\operatorname{div}\left( |x|^{-1}e_i-x_i|x|^{-3}x\right) \\ &= \frac{\partial}{\partial x_i}|x|^{-1} -\sum_{j=1}^n \frac{\partial}{\partial x_j} (x_i|x|^{-3}x_j) \\ &=-\frac{x_i}{|x|^3}-\sum_{j=1}^n x_i\left\{|x|^{-3}-3x_j^2|x|^{-5}\right\}-|x|^{-3}x_i \\&= -(n-1)\frac{x_i}{|x|^3} \end{align} $$ which matches $|\nabla u|^2u$ component by component.

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