Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.