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The question itself is pretty simple. I am running into confusion. Seems like there is a typo in the book. I wanna check myself. Maybe I am doing something wrong.

Suppose we have the function (which I term full hessian): $$ w=\sum_{i,j}f_if_{ij}f_j $$ with $$ i,j=x,y,z $$ where $f_{ij}$ and $f_i$ means partial derivatives with respect to $i$ and $j$, and $i$, respectively.

How do I write $w$ for some function $\rho(r)$ expressed in spherical coordinates? Note that $\rho$ is independent of $\phi$ and $\theta$.

Let's take $\rho(r)=e^{-ar}$ as an example.

Thanks in advance:)

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up vote 2 down vote accepted

You are free to call it full Hessian, but the established name for $w$ is $\infty$-Laplacian. Here is a nice expository article, though being 10 years old, it's a little out of date in the fast-developing subject.

Recall that for any vector $v$ the expression $\sum_{i,j} v_i f_{ij} v_j $ is the directional second derivative of $f$, that is, $\frac{d^2}{dt^2}f(x+tv)$ evaluated at $t=0$. When we plug in $v=\nabla f(x)$, the result is the second derivative along the gradient.

For your function $f(x)=\rho(|x|)$, the gradient is $\rho\,'(|x|)\frac{x}{|x|}$; that is, a radial vector with magnitude $|\rho\,'|$. The second derivative along this vector is $\rho\,''(\rho\,')^2$. This is the $\infty$-Laplacian of a spherically symmetric function.

In particular, we observe that spherically symmetric $\infty$-harmonic functions are precisely the cones $f(x)=a|x|+b$. Comparison with cones is an effective method of gaining control over $\infty$-harmonic functions, as seen in the aforementioned article by Aronsson, Crandall and Juutinen.

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thnx for your nice answer!:) It was really helpful for me. –  molkee Jan 26 '13 at 22:47
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