Let $A =\mathbb{R}^3 \setminus\{ (0,0,0)\},a,b,c>0$.
Let $F=\frac{(x,y,z}{(a^2x^2+b^2y^2+c^2z^2)^{1.5}}$ be a vector field.
Is there a vector field $G$ such that $curl(G)=F$ ?
I have no idea how to approach this problem , any help is welcome.
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Hint: $\mathrm{div}(\mathrm{curl}(G)) = 0$ for any smooth vector field $G$. – whpowell96 Dec 17 '23 at 17:20
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1In the special case $a=b=c=1$ that vector field describes Coulomb's law and Newton's law of universal gravitation. It is known that there is no vector field $G$ defined on all of $\mathbb R^3\setminus{0}$ whose curl is $F,.$ At least on a ray starting at the origin $G$ must have a singularity. But an expression for $G$ is known. See for example this post. The singularity of $G$ is also known as Dirac string. – Kurt G. Dec 17 '23 at 17:48