Differential equation for a vector potential

From Helmholtz’s theorem, any smooth vector field $$\mathbf{F}$$ that goes to zero at infinite distance can be uniquely decomposed everywhere in the sum of a divergence free component and an irrotational component.

In particular if it's conservative, one can prove that it can be uniquely determined everywhere solving the Poisson equation for its potential: $$$$\Delta\varphi(\mathbf{r})=f(\mathbf{r})$$$$ where $$f$$ is the divergence of $$\mathbf{F}$$. This equation can be found from the definition of scalar potential provided by Helmholtz’s theorem $$$$\varphi(\mathbf{r})=\frac{1}{4\pi}\int_{\mathbb{R}^{n}}\frac{f(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}d\mathbf{r}'$$$$ I was wondering if something similar can be done (and how) for the vector potential, i.e., if it's possible to uniquely determine a solenoidal vector field by solving a differential equation that involves its vector potential and its curl starting from the Helmholtz’s theorem $$$$\mathbf{A}(\mathbf{r}):=\frac{1}{4\pi}\int_{\mathbb{R}^{n}}\frac{\mathbf{C}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}d\mathbf{r}'$$$$ where $$\mathbf{C}$$ is the curl of $$\mathbf{F}$$. So I'm assuming this field goes to zero at infinite distance and it's smooth. In case this is not possible, I would appreciate a proof or a counterexample.

• If you additionally assume that $$F= \nabla\times G , \quad \nabla\cdot G = 0$$ then when you take the curl again and use the double curl identity $\nabla \times (\nabla\times G) = \nabla(\nabla\cdot G)-\Delta G$ you get $$-\Delta G = \nabla\times F$$ which is again a Poisson equation so you can solve for $G$. this is called the Biot-Savart law. Without this assumption you still have an equation but its more complicated Commented Jul 30, 2021 at 10:48
• @CalvinKhor That's the Coulomb gauge. I was hoping for a more general approach.
– Simo
Commented Jul 30, 2021 at 11:26
• well I have no idea what a Coulomb gauge is but it wasn’t in your question despite being a partial answer hence my comment. That’s the end of what I know so gl Commented Jul 30, 2021 at 12:29