Suppose that $\mathbb{R}^3$, fix $f \in L^1(\mathbb{R}^3;\mathbb{R})$ and let $g \in L^1(\mathbb{R}^3;\mathbb{R}^3)$ satisfies $$ div(g)=\delta_a -f. $$ Then what is $g$? It what is the distributional anti-derivative of this thing?
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You must mean $f \in L^1(\mathbb{R}^3;\mathbb{R})$ instead of $f \in L^1(\mathbb{R}^3;\mathbb{R}^3)$. – md2perpe Oct 04 '19 at 17:29
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Indeed this is correct – AB_IM Oct 06 '19 at 09:37
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In dimension 3, it is classical that $$ \mathrm{div} \left( \frac{\vec x}{\| \vec x \|^3} \right) = 4 \pi \delta_0$$ See for instance $\nabla \cdot \big(\frac{\hat{r}}{r^{2}}\big)$ and Dirac Delta Function
So you can take $$g = \frac{1}{4\pi}\left( \frac{\vec x}{\| \vec x \|^3} - \frac{\vec x-\vec a}{\| \vec x - \vec a\|^3} \right)$$
Héhéhé
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1Let $H(x,y,z) =e^{- (x+y+z)} 1_{x+y+z > 0} \delta(x-y)\delta(z-\frac{x+y+z}{3})$ what is $div(H)$ ? Thus what is $div (H \ast f)$ ? @AIM_BLB – reuns Oct 04 '19 at 15:13
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Let $\vec{G}(\vec{x}) := \frac{1}{4\pi} \frac{\vec x}{\| \vec x \|^3}$. Then $\operatorname{div} \vec{G} = \delta_0$. Now set $\vec{g}_p := \vec{G}*(\delta_{\vec{a}}-f)$ i.e. $$ \vec{g}_p(\vec{x}) = \int \frac{1}{4\pi} \frac{\vec{x}-\vec{x}'}{\| \vec{x}-\vec{x}' \|^3} (\delta_{\vec{a}}(\vec{x}') - f(\vec{x}')) \, d^3x' \\ = \frac{1}{4\pi} \frac{\vec{x}-\vec{a}}{\| \vec{x}-\vec{a} \|^3} - \int \frac{1}{4\pi} \frac{\vec{x}-\vec{x}'}{\| \vec{x}-\vec{x}' \|^3} f(\vec{x}') \, d^3x'. $$ Then $$ \operatorname{div} \vec{g}_p = \operatorname{div}(\vec{G}*(\delta_{\vec{a}}-f)) = (\operatorname{div}\vec{G})*(\delta_{\vec{a}}-f) = \delta_0 * (\delta_{\vec{a}}-f) = \delta_{\vec{a}}-f. $$
But $\vec{g}_p$ is not the only solution. We can to it add any $\vec{g}_h$ such that $\operatorname{div}\vec{h}_h=0.$ All solutions are thus given by $\vec{g} = \vec{g}_p + \vec{g}_h.$
md2perpe
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