Consider a very simple semilinear elliptic PDE: $$ \left\{\begin{aligned} &-\Delta u=\partial_{x_1}(u^2+f)& &\text{in } \mathbb R^n,\\ & \,\,\,u(x)\to 0\,,& &x\to\infty. \end{aligned}\right. $$$$ \left\{\begin{aligned} &-\Delta u=\partial_{x}(u^2+f)& &\text{in } \mathbb R^3,\\ & \,\,\,u(\mathbf x)\to 0\,,& &\mathbf x\to\infty, \end{aligned}\right. $$ where $x=(x_1,...,x_n)$$\mathbf x=(x,y,z)$. The boundary conditions are not rigorous, it is just to describe the kind of solutions we want. The critical Lebesgue spaces with respect to the natural scaling symmetry are: $$ u\in L^{n}(\mathbb R^n), f\in L^{n/2}(\mathbb R^n). $$$$ u\in L^{3}(\mathbb R^3), f\in L^{3/2}(\mathbb R^3). $$ Assume $n=3$ to fix ideas. II know two ways of finding solutions.
Strong solutions. Assume $n=3$ to fix ideas. ByBy 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 $\dot W^{1,3/2}(\mathbb R^3)\hookrightarrow L^3(\mathbb R^3)$).
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).