My question is related, but not a duplicate of Asymptotic convergence of the solution to a parabolic pde to the solution of an elliptic pdeSuppose I solve the parabolic PDE:$u_t = \Delta u + f(x,t)$ on $\Omega \times(0,T]$$\nabla u(x,t) \cdot n = 0 $ on $\partial\Omega$$u(x,0) = g(x)$ on $\Omega$and obtain the (unique) steady state solution $u^*(x)$. Assume that the source term in the parabolic PDE satisfies the compatibility condition for a pure neumann problem $\int_{\Omega}f(x,t)= \int_{\Omega}u_t$From the compatibility above, can I claim that it will  also hold as $t \rightarrow \infty$ ? I suppose that the functions $f(x,t)$ and $u_t$ will need to be nice enough for me to take the limit inside the integral ?Now consider the elliptic PDE:$0 = \Delta u + \tilde{f}(x)$ on $\Omega$$\nabla u(x) \cdot n = 0 $ on $\partial\Omega$.This is now a pure Neumann problem.Assume that the steady state value of the source  is used, i.e., $\tilde{f}(x) = \lim_{t \rightarrow \infty} f(x,t)$ is used and that it satisfies the compatibility condition for this pure Neumann problem.Now, I know that if $\hat{u}(x)$ be a solution, then $\hat{u}(x) +c$ for all real numbers $c$, also solves the PDE. I am aware that we fix the constant by demanding  zero mean solution or prescribing the value at a point.So my questions are:Is it possible to prescribe a condition for the elliptic problem  a priori  so that the solution to the elliptic PDE matches the steady state solution $u^*(x)$ of the parabolic PDE ?Can we say anything else about the relation between these two solutions ?
steady state solution from parabolic problem vs solution of elliptic problem
pde;boundary conditions;convergence;parabolic pde;elliptic pde