My question is about non-uniqueness of solutions of an elliptic PDE in $\mathbb R^n$ with source term in a scaling-subcritical space (regular, but with too slow decay at infinity), and with some nice a priori bounds. I have zero background in this kind of problems, so I need help to find the relevant references.

Notation. $H^k(\mathbb R^n)$ denotes the Sobolev space $W^{k,2}(\mathbb R^n)$. The Sobolev space $W^{k,p}(\mathbb R^n)$ is the Banach space of functions in $L^p$ with all derivatives up to order $k$ in $L^p$.

Consider a very simple semilinear elliptic PDE: $$ \left\{\begin{aligned} &-\Delta u=\partial_{x}(u^2+f)& &\text{in } \mathbb R^3,\\ & \,\,\,u(\mathbf x)\to 0\,,& &\mathbf x\to\infty, \end{aligned}\right. $$ where $\mathbf x=(x,y,z)$. The boundary conditions are not rigorous, it is just to describe the kind of solutions we want. The critical Lebesgue spaces with respect to the natural scaling symmetry are: $$ u\in L^{3}(\mathbb R^3), f\in L^{3/2}(\mathbb R^3). $$ I know two ways of finding solutions.

1. Strong solutions. By using the fixed point theorem, it is possible to show that if $f\in L^{3/2}(\mathbb R^3)$ is small enough, then there exists a unique small solution $u\in L^3(\mathbb R^3)$, which moreover satisfies $$ \|\nabla u\|_{L^{3/2}(\mathbb R^3)}\lesssim \|f\|_{L^{3/2}(\mathbb R^3)} $$ (note that $\dot W^{1,3/2}(\mathbb R^3)\hookrightarrow L^3(\mathbb R^3)$).

2. Weak solutions. If I instead consider a scaling-subcritical datum, like $f\in L^2(\mathbb R^3)$, I cannot use the fixed point theorem. However, notice that multiplying the equation by $u$ and integrating by parts, we obtain the a priori bound $$ \|\nabla u\|_{L^2(\mathbb R^3)}\lesssim \|f\|_{L^2(\mathbb R^3)}. $$ Using this a-priori bound it is possible to prove* the existence of weak solutions $u\in \dot H^1(\mathbb R^3)$ for any $f\in L^2(\mathbb R^3)$ (we have $\dot H^1(\mathbb R^3)\hookrightarrow L^6(\mathbb R^3)$). In this case, estimates for the difference of two solutions do not seem to exist, so the uniqueness cannot be proved by standard arguments (this is the only reason why I am calling them “weak solutions”; you can see that they are more regular than strong solutions).

My question

My problem is the following. Is the weak solution unique if $f\in L^2(\mathbb R^3)$? If not, how can one find a counterexample? More generally, in a case where we have a priori bounds like in the PDE I wrote, do we have known non-uniqueness results in case of data decaying too slowly at infinity?

I feel it could be possible that there are data that lead to two different solutions, but I know no technique that would prove that or examples like that in the literature. I would be happy if you could point me to articles about anything that resembles this problem or other similar non-uniqueness problems.

Further remarks.

1. This is not necessarily a regularity prolem: data in $L^2$ are "more regular" than data in $L^{3/2}$. The problem is that data grow more slowly at infinity than data in the critical space. I could ask the same question for data $f\in H^{100}(\mathbb R^3)$, and the answer would still be non trivial. In fact, I would be happy to know what happens with more regularity assumptions.
2. I am fairly sure that a unique continuation property exists for this kind of equations, so this problem is extremely different from those of non-uniqueness of weak solutions of evolution PDE at low regularity (like, e.g., the famous results for Euler equation).

_{*I think there is more than one way of doing that. The way I thought would be to approximate the Cauchy problem with that of the same PDE on a half plane (with Dirichlet boundary conditions), where the "edge" of the plane goes to infinity. On a half plane, you can look at the PDE as an evolution equation via the Poisson kernel, so you can show that a solution exists, and then you can use the uniform bound and weak compactness to show the existence of the solution of the original problem... Probably not the easiest way. I have to say I am not an elliptic person, I am mostly dispersive, sometimes parabolic... In any case, feel free to comment on this point.}