In an exercise I find the request of evaluating the Laplacian of $f(x)=\frac{1}{|x|}$ in $\mathbb{R}^3$. But it exists in a classical sense only if $x\neq 0$ otherwise I have to see it in distributional sense, isn't it?
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Simply put, yes, at $x=0$, Laplacian of $f$ can be defined as a constant times Dirac delta in the sense of distribution. – Shuhao Cao Mar 01 '13 at 21:15
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Yes, but if I want it only in the case $x\neq 0$ I can calculate it in the classical sense, isn't it? – Mario Mar 01 '13 at 21:25
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Yes, component-wise. – Shuhao Cao Mar 01 '13 at 21:32
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What you mean for component-wise? – Mario Mar 01 '13 at 21:36
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I mean $x = (x_1,x_2,x_3)$ having three component in $\mathbb{R}^3$, and you wanna compute each direction's partial derivative. – Shuhao Cao Mar 01 '13 at 21:44
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Yes; I have calculated the partial derivativ in each direction $x_i$ and then I have summed their square! – Mario Mar 01 '13 at 21:49
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I guess you meant the second derivatives. – Maesumi Mar 02 '13 at 03:05
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This is Laplacian of $1/r$. Sum of the second derivatives of $1/r$ wrt the components. The Laplacian of Columb potential, so defined in $3d$, is $0$.
Let $r=\sqrt{x_1^2+x_2^2+x_3^2}$, then ${\partial r \over {\partial x_1}}=x_1/r$. Now ${\partial (1/r) \over \partial x_1 }= -1/r^2 {\partial r \over {\partial x_1}}=-x_1/r^3$. Next ${\partial \over \partial x_1}{\partial (1/r) \over \partial x_1 }={{-r^2+3x_1^2}\over{r^5}} $. The sum of three such terms is ${-3r^2+3(x_1^2+3x_2^2+3x_3^2)} \over {r^5}$ which is $0$.
At the origin it is $-4\pi \delta(r)$, where $\delta $ is the Dirac function.
Maesumi
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