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The reason for the lack of proper horizontal spacing in "grad Phi" was that \mathrm rather than \operatorname was used. The latter gives you context-dependent spacing.
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edit approved Jul 11 at 20:38
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edit approved Jul 11 at 20:38
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Solving $X$ for prescribed $\mathrm$\operatorname{div}(X)$ of compact support

Let $f \in C^\infty_0(\mathbb{R}^3)$ be a smooth scalar function of compact support. I would like to find a smooth vector field $X : \mathbb{R}^3 \to \mathbb{R}^3$ satisfying $\mathrm{div}(X) = f$$\operatorname{div}(X) = f$ such that $X$ is also of compact support. Is that possible?

Looking for $X$ of the form $X = \mathrm{grad} \Phi$$X = \operatorname{grad} \Phi$ does not work, since solving the Laplace equation $\Delta \Phi = f$, $\Phi$ will not have compact support.

Solving $X$ for prescribed $\mathrm{div}(X)$ of compact support

Let $f \in C^\infty_0(\mathbb{R}^3)$ be a smooth scalar function of compact support. I would like to find a smooth vector field $X : \mathbb{R}^3 \to \mathbb{R}^3$ satisfying $\mathrm{div}(X) = f$ such that $X$ is also of compact support. Is that possible?

Looking for $X$ of the form $X = \mathrm{grad} \Phi$ does not work, since solving the Laplace equation $\Delta \Phi = f$, $\Phi$ will not have compact support.

Solving $X$ for prescribed $\operatorname{div}(X)$ of compact support

Let $f \in C^\infty_0(\mathbb{R}^3)$ be a smooth scalar function of compact support. I would like to find a smooth vector field $X : \mathbb{R}^3 \to \mathbb{R}^3$ satisfying $\operatorname{div}(X) = f$ such that $X$ is also of compact support. Is that possible?

Looking for $X$ of the form $X = \operatorname{grad} \Phi$ does not work, since solving the Laplace equation $\Delta \Phi = f$, $\Phi$ will not have compact support.

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Solving $X$ for prescribed $\mathrm{div}(X)$ of compact support

Let $f \in C^\infty_0(\mathbb{R}^3)$ be a smooth scalar function of compact support. I would like to find a smooth vector field $X : \mathbb{R}^3 \to \mathbb{R}^3$ satisfying $\mathrm{div}(X) = f$ such that $X$ is also of compact support. Is that possible?

Looking for $X$ of the form $X = \mathrm{grad} \Phi$ does not work, since solving the Laplace equation $\Delta \Phi = f$, $\Phi$ will not have compact support.