Poisson modification with regularity Let $u:\mathbb{R}^n\rightarrow \mathbb{R}$ be a $C^\infty$ function. Let $r > 0$ be any radius and let $t < 0$ be some constant. Consider the Poisson modification $\tilde{u}$ of $u$ given by
$$\left\{\begin{array}{ll}
-\Delta \tilde{u} = t & \text{on } B(0,r),\\
\tilde{u} = u & \text{on } \mathbb{R}^n\setminus B(0,r).
\end{array}\right.$$
The function $\tilde{u}$ is not differentiable at $\partial B(0,r)$ in general. Thus $-\Delta \tilde{u}$, viewed as a signed measure, may have some singular part supported on $\partial B(0,r)$.
Now assume that we want to find a function $\tilde{u}$ satisfying
$$\left\{\begin{array}{ll}
-\Delta \tilde{u} \geq t & \text{on }B(0,r),\\
\tilde{u} = u & \text{on }\mathbb{R}^n\setminus B(0,r),\\
-\Delta \tilde{u} \text{ absolutely continuous w.r.t Lebesgue measure}.
\end{array}\right.$$
It is clear that this problem will not always admit a solution. Intuitively, at least some conditions on $\nabla u$ along $\partial B(0,r)$ will be necessary. I am looking for results related to this problem, references or keywords in order to start searching myself.
 A: Clearly $-\Delta u \geq t$ on $\partial B(0,r)$ is necessary (otherwise $\Delta \tilde{u}$ is discontinuous across $\partial B(0,r)$. I claim it is also sufficient. To see this, define
$$f(x) = \max\{-\Delta u(x), t\}\geq t$$
for $x \in B(0,r)$. Then $f\in C^{0,1}(B(0,r))$ and if $-\Delta u(x) \geq t$ for $x \in \partial B(0,r)$, then $f$ attains the boundary values $-\Delta u$ on $\partial B(0,r)$ continuously. 
Now solve the Poisson problem $-\Delta \tilde{u}=f \geq t$ in $B(0,r)$ with boundary condition $u=\tilde{u}$ on $\partial B(0,r)$. By the Schauder theory, $\tilde{u}\in C^{2,1}(B(0,r))$, so $-\Delta \tilde{u}$ is actually Lipschitz continuous across $\partial B(0,r)$. 
It seems this is the best you can do in general. If you had $-\Delta u > t$ on $\partial B(0,r)$, then you might be able to ensure $\tilde{u}$ is smooth, but taking a smooth extension of $-\Delta u$ in place of $f$. 
EDIT: Sorry, just realized I misread your question. I read it as "absolutely continuous" (and missed the wrt Lebesgue mesure). 
