Convergence of solutions of a family of elliptic PDEs I'm diving head-first into elliptic PDEs, so please forgive me if my question is one whose answer is known to everyone in the field. (Also, the only similar questions I have found are this one, unfortunately for a different PDE, and that one which, I have to admit, goes a bit over my head.)
Let $B ⊂ ℝ^n$ be a ball and suppose for each $σ > 0$ we're given a 2nd-order elliptic PDE on $ℝ^n ∖ B$ with smooth coefficients $a^σ_{ij}(x)$, $b^σ_i(x)$ and $c^σ(x)$ which also depend smoothly on $σ$. Suppose, also, that for each $σ$ there is a unique solution $u_σ ∈ C^∞$ to said PDE that fulfills $u_σ(x) → 1$ at infinity.
Assume now that for each $k$ the coefficients (and their derivatives) converge in $C^k$ on compact sets to functions $a_ij(x)$, $b_i(x)$ and $c(x)$ as $σ → ∞$ and that there exists a unique solution $u ∈ C^∞$ to the corresponding PDE with coefficients $a_ij(x)$, $b_i(x)$ and $c(x)$, which fulfills $u → 1$.
Question A: (Under what additional conditions) Do we then have $u_σ → u$ on compact sets, too? (Possibly up to taking a subsequence.) If so, how does this follow? (Since I'm a beginner I'd also appreciate references where I could read up on things.)
I suppose the challenge consists in showing uniform convergence of $u_σ$ and its derivatives because then I think one should be able to use a result like this one to commute limits and derivatives and show that the limit of $u_σ$ fulfills the same PDE as $u$(?)
Question B: What is the answer to the above question when instead of an unbounded domain like $ℝ^n ∖ B$ and an asymptotic condition like $u → 1$ we're considering the Dirichlet problem on a bounded domain with sufficiently regular boundary?
 A: I don’t recall the theory well-enough to help you with question A, but I think I can partially answer question B. 
Let $L_{\sigma}$ be the elliptic partial differential operator for $\sigma$ and $L$ be the limit operator. 
We are assuming that all these operators are elliptic with the same positive constant. 
We assume that for some smooth $f$ on the bounded domain with smooth boundary $\Omega$, $L_{\sigma}u_{\sigma}=f$ (the argument is similar if $f$ depends with sufficient regularity on $\sigma$, so that enough $H^k$ norms of $f$ are bounded independently from $\sigma$, but let’s do simple). 
Then $Lu_{\sigma}=f+(L-L_{\sigma})u_{\sigma}=f+R_{\sigma}$. 
Now, when expanding $L-L_{\sigma}$, $\|R_{\sigma}\|_{H^k} \leq o_{\sigma}(1)\|u_{\sigma}\|_{H^{k+2}}$, because $a^{\sigma} \rightarrow a$ and so on. By higher-order elliptic regularity with $L_{\sigma}$, $\|u_{\sigma}\|_{H^{k+2}} \leq C_{\sigma} (\|u_{\sigma}\|_{L^2}+\|f\|_{H^k}) \leq C(\|u_{\sigma}\|_{L^2}+\|f\|_{H^k})$. 
Assume (that’s much milder than anything, but admittedly we’re not starting from scratch) that $u^n=u_{\sigma_n}$ is bounded in $L^2$, then 
$Lu^n=f+o_{H^k}(1)$ for each $k$. 
By (again) higher-order elliptic regularity (but with $L$) it follows $u^n$ bounded in $H^k$ for each $k$, hence by Sobolev injections the sequence is compact in $C^{\infty}$. Any limit point in $C^{\infty}$ of $u^n$ is a classical solution of $Lv=f$. 
If $Lv=0$ has no nontrivial weak solution in $H^1_0$, then this implies that $u^n$ must converge in $C^{\infty}$ (in $H^{\infty}$ actually) to the unique solution of $Lv=f$. 
Note that the assumption of boundedness in $L^2$ is necessary. Indeed, take $L=L_{\sigma}$ with a nonzero weak solution to $Lv=0$. Then take $u_{\sigma}=\sigma^{-1}v$.
I‘m confident the assumption is automatically met as soon as $L_{\sigma} \geq rId$ for some $r > 0$ uniformly in small enough $\sigma$ (because we have a bound $r\|u\|_{L^2} \leq \|L_{\sigma}u\|_{L^2}$). 
I think this condition holds as long as the equation $Lu=0$ has no nontrivial weak solutions, but I am not certain. 
