I encounter a vector differential equation in solving poroelastic flow as follows:
$$\nabla\times\nabla\times\boldsymbol{\Phi}=\mathbf{A}+\nabla(\frac{r^2}{2}\chi)$$ where $\boldsymbol{\Phi}$ is an unknown vector and $\mathbf{r}$ is the position vector. $\mathbf{A}$ is a known vector that satisfies the vector Laplace equation and $\chi$ is a known scalar that satisfies the scalar Laplace equation as follows: $$\nabla^{2}\mathbf{A}=\mathbf{0}, \quad\quad \nabla^{2}\chi=0.$$ My question is: how to find a general solution of $\boldsymbol{\Phi}$ in terms of $\mathbf{A}$, $\chi$, and $\mathbf{r}$. Thank you!
Constructing the general solution to the equation in $\mathbb{R}^3$ is rather simple, however in order to make the solution unique, one would have to specify further constraints on the field $\mathbf{\Phi}$.
Using the Helmholtz decomposition of the field, we write
$$\mathbf{\Phi}=-\nabla E+\nabla\times\mathbf{B}~~\\ ~\\ ~~\nabla\cdot\mathbf{B}=0$$
This allows us to rewrite the equation as
$$\nabla^2(\nabla\times \mathbf{B})=-\left(\mathbf{A}+\nabla\frac{\chi r^2}{2}\right)$$
which shows that the function
$$\mathbf{\Phi}(\mathbf{r})=-\nabla E(\mathbf{r})+\frac{1}{4\pi}\int_V d^3\mathbf{r'}\frac{\mathbf{A}(\mathbf{r'})+\frac{1}{2}\nabla\left(\chi(\mathbf{r')} \mathbf{r'}^2\right)}{|\mathbf{r}-\mathbf{r'}|}$$ is a solution to the problem posed for any scalar function $E(\mathbf{r})$ only if, of course, the source fields obey the additional constraint
$$2\nabla\cdot\mathbf{A}+\nabla^2(\chi r^2)=0$$
This constraint can be subsequently eliminated if one writes
$$\mathbf{A}=-\frac{1}{2}\nabla(\chi \mathbf{r}^2)+\nabla\times\mathbf{\Omega}$$
and then the solution can be written concisely in the form
$$\mathbf{\Phi}(\mathbf{r})=-\nabla E(\mathbf{r})+\frac{1}{4\pi}\nabla \times\int_V d^3\mathbf{r'}\frac{\mathbf{\Omega}(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|}$$
If, additionally, wlog we require that $\nabla\cdot\mathbf{\Omega}=0$, then this field satisfies the Airy equation:
$$\nabla^4\mathbf{\Omega}=0$$
Note that the auxiliary field $\mathbf{\Omega}$ introduces no new constraints on $\chi$ since
$$0=-\nabla^2(\nabla\cdot\mathbf{A})=\nabla^4(\chi r^2/2)=\left(\frac{\mathbf{r}^2\nabla^2}{2}+4\mathbf{r}\cdot\nabla+10\right)\nabla^2\chi$$ which vanishes automatically whenever $\chi$ obeys the Laplace equation.
EDIT: In answer to a question in the comments, with $\boldsymbol{\Omega}$ divergenceless, it is possible to express it in terms of $\mathbf{A}$, since
$$\nabla\times\nabla\times\mathbf{\Omega}=-\nabla^2\Omega=\nabla\times\mathbf{A}$$ and hence $$\mathbf{\Omega}=\frac{1}{4\pi}\int d^3\mathbf{r'}\frac{\nabla\times\mathbf{A}(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|}$$