Define vector field with known curl and div [duplicate]

So I am trying to write a report on the Helmholtz decomposition theorem on $\mathbb{R}^3$. The theorem states that under certain conditions, every vector field $\textbf{F}:U \subseteq \mathbb{R}^3 \to V \subseteq \mathbb{R}^3$ can be decomposed into a curl-free and a divergence-free component as: \begin{equation} \textbf{F}=-\nabla \phi+\nabla \times \textbf{A} \end{equation} where $\phi$ and $\textbf{A}$ are the correspoding scalar and vector potentials of the field. So I tried to apply the theorem on the following vector fields: \begin{equation} \textbf{F}_1(x,y,z)=(-y,x,0) \quad \text{and} \quad \textbf{F}_2(x,y,z)=(x,-y,0) \end{equation} For the vector field $\textbf{F}_1$ I calculated that: \begin{equation} \nabla \times \textbf{F}_1=(0,0,2) \quad \text{and} \quad \nabla \cdot \textbf{F}_1=0 \end{equation} Therefore it is said that $\textbf{F}_1$ is a divergence-free field. In the same way I calculated that: \begin{equation} \nabla \times \textbf{F}_2=(0,0,0) \quad \text{and} \quad \nabla \cdot \textbf{F}_2=0 \end{equation} Therefore is said to be a curl-free and divergence-free field. Now, Helmholtz decomposition states that the sources of the field which is to be decomposed are defined as: \begin{equation} \phi=\nabla \cdot \textbf{F} \quad \text{και} \quad \textbf{A}=\nabla \times \textbf{F} \end{equation} which I find it to be terribly wrong since if I apply the divergence of a field decomposed by Helmholtz I would have: \begin{equation} \nabla \cdot \textbf{F}=-\nabla^2 \phi \end{equation} which is not $\phi$ but the Laplace equation of $\phi$! Is the notation wrong? I am really confused.

Moreover, if lets say I knew the curl and the divergence of $\textbf{F}_1$ in that particular example would it be possible to construct the whole field $\textbf{F}_1$? Helmholtz says that I can but when I try to do that I get (according to the wrong way): \begin{equation} \nabla \cdot \textbf{F}_1=0 \quad \text{and} \quad \nabla \cdot \textbf{F}_1=\phi \end{equation} Therefore $\phi=0$?? If that is true then $\textbf{F}_1=\nabla \times \textbf{A}=(0,0,0)$ which is obviously not $\textbf{F}_1$.

Finally, the Laplacian term which defines the harmonic field, appears into the second example and somehow the Laplacian is $\phi(x,y)=1/2(x^2-y^2)$. How is this done?

I believe that if I understand the application on the simple examples then I will be able to see the whole picture.

Thank you!

If ${\bf F} = - \nabla \phi + \nabla \times {\bf A}$, we get $\nabla \cdot {\bf F} = - \Delta \phi$. So we want to get $\phi$ by solving a Laplace equation. Then we get $\bf A$ by solving $\nabla \times {\bf A} = {\bf F} + \nabla \phi$.

In your first example, ${\bf F}_1$ is indeed divergence-free, so you can take any $\phi$ whose Laplacian is $0$. You could take $\phi = 0$, but if you prefer you can use any harmonic function. For example, let's take $\phi = x$ which makes $\nabla \phi = [1,0,0]$. So now you want $\nabla \times {\bf A} = [1-y,x,0]$. One possible solution is ${\bf A} = [xz, yz-z,0]$. There are many possible solutions: the decomposition is far from unique.

First of all, the wrong thing that you think is wrong, is totally wrong! :) So, yes, you are right!

If you take the divergence and curl of the following equation

$$ \textbf{F}=-\nabla \phi+\nabla \times \textbf{A} $$

then you will get, respectively

$$\boxed{ \begin{align} \nabla \cdot \textbf{F} &= - \nabla^2 \phi \\ \nabla \times \textbf{F} &= - \nabla^2 \textbf{A} \end{align} }\tag{1}$$

where use has been made of the identitiy

$$\nabla \times \nabla \times \textbf{A} = \nabla (\nabla \cdot \textbf{A})-\nabla^2 \textbf{A}$$

and the assumption

$$\nabla \cdot \textbf{A} =0 $$

which has been made for simplicity.

In conclusion, for a given vector field $\textbf{F}$, you should solve the Poisson equations mentioned in $(1)$ for $\phi$ and $\textbf{A}$.