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Let $f=f_{n,k}$ be a $C^{2}(\mathbb{R}^{n})$ function with compact support such that if $R\leq |x|\leq 2^{k}R$

$$f(x):=\begin{cases} -\log|x| & \text{when } n=2 , \\ |x|^{2-n} & \text{when } n>2. \end{cases}$$ Then, $\Delta f=0$ in $A_{k}=\{x,R<|x|<2^{k}R\}$ with $\lvert\cdot\rvert$ be Euclidean norm $\Delta$ is the Laplace operator as it's a second order differential operator in the $n$-dimensional Euclidean space $$\Delta f = \sum_{i=1}^n \frac {\partial^2 f}{\partial x^2_i}.$$ How can I get $\Delta f =0$? Indeed we know that $g=u\circ v$ then $g'=(u'\circ v ).v'$.
for $f(x)=-\log|x|$ then $\frac{\partial f}{\partial x}=-\left(\frac{1}{|x|}\right)\cdot|x|'=?$

Please respond I'll be grateful for any help offered!

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  1. For the case $n=2$: we have $f(x_1,x_2)=-\frac 12\log(x_1^2+x_2^2)$ on $A_k$, hence $$\partial_{x_1}f(x_1,x_2)=-\frac 12\frac{2x_1}{x_1^2+x_2^2}=-\frac{x_1}{x_1^2+x_2^2}.$$ In a similar way, we can express $\partial_{x_1,x_1}$ and $\partial_{x_2,x_2}$.

  2. For $n\gt 2$, we have $f(x_1,\dots,x_n)=\left(\sum_{i=1}^nx_i^2\right)^{1-n/2}$. The derivative with respect to $x_j$ is $$\partial_jf(x)=\left(1-\frac n2\right)\left(\sum_{i=1}^nx_i^2\right)^{-n/2}(2x_j).$$ Using the formula for the derivative of a product, we can find $\partial_{jj}$.

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  • $\begingroup$ hi Davide i agree with you but i still can't get $\Delta f=0$ $\endgroup$
    – Educ
    Sep 10 '13 at 21:12
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    $\begingroup$ For which case? Can you show your computations? $\endgroup$ Sep 10 '13 at 21:19
  • $\begingroup$ for example case $n=2$: we have $f(x_1,x_2)=-\frac 12\log(x_1^2+x_2^2)$ on $A_k$, hence $$\partial_{x_1}f(x_1,x_2)=-\frac 12\frac{2x_1}{x_1^2+x_2^2}=-\frac{x_1}{x_1^2+x_2^2}.$$ $$\partial_{x_2}f(x_1,x_2)=-\frac 12\frac{2x_2}{x_1^2+x_2^2}=-\frac{x_2}{x_1^2+x_2^2}.$$ $$\partial_{x_1x_2}f(x_1,x_2)=\partial_{x_1}(-\frac{x_2}{(x_1^2+x_2^2)^{2}}).$$ $$\partial_{x_1x_2}f(x_1,x_2)=\frac{2x_1(x_1^2+x_2^2)}{(x_1^2+x_2^2)^{4}} $$ $$\partial_{x_1x_2}f(x_1,x_2)=\frac{2x_1}{(x_1^2+x_2^2)^{3}} $$ $$\partial_{x_2x_1}f(x_1,x_2)=\frac{2x_2}{(x_1^2+x_2^2)^{3}} $$ $\endgroup$
    – Educ
    Sep 10 '13 at 22:11

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