Mean time until adsorption for a well-mixed bounded random walk that suddenly allows for adsorption I have a random walk on some interval $[0, N]$ with probability $p$ of taking a $+1$ step, probability $(1-p)$ of taking a $-1$ step, and where we have that $p>(1-p)$.  Initially the boundaries are reflecting, i.e. if the walker lands at $0$ or $N$, then it will remain in place until it takes a $+1$ or $-1$ step with probability $p$ and $(1-p)$, respectively.  Let $\pi$ be the stationary distribution of this walk ( Finding the exact stationary distribution for a biased random walk on a bounded interval).
Now, imagine that we allow the walk to proceed for an arbitrary number of time steps until the walker's probability distribution on the interval approaches the stationary distribution.  We then "flip a switch" and specify that the boundary $0$ will be fully adsorbing and that the walker will have a probability of $f(x)$ of being moved to $0$ at any step.  More specifically, we flip a coin to decide $p$ and $q$ as we normally would, then flip another coin to decide whether to adsorb with probability $f(x)$.  Here, $f(x)$ is some function that depends on the position, $x$, of the walker.
What is our mean time until adsorption?  Is there a technical term for what $\pi$ is after we "flip the switch" (some kind of quasi-stationary distribution)?   
For example, is it correct to assume that:
(1) Ignoring $f(x)$, that the probability of adsorbing at $(x=0)$ is $\approx \pi(0)$?
(2) That the probability the walker will be moved to $0$ at any given step can be approximated as:
$P[k \to 0]=\sum_{k=1}^{N} k(f(k)\pi(k))$
And thus that we can approximate the adsorption time as: $\mu(ads)=[(1-P[k \to 0])(1-\pi(0))]^{-1}$
I'm pretty sure, however, that the above approximation is wrong.

Update - I would be happy for a solution that only looks at absorbance at $0$ without consideration of $f(x)$.  Hopefully there can an analytic solution with this simplification?
 A: This is not (yet) an answer: I just try to formalise the question asked by the OP
The stationary distribution before the adsorption process sets in can be obtained from the balance equations $p \pi(k) = (1-p)\pi(k+1)$. It is given by
$$\pi(k) = \alpha^k \pi(0)= \frac{(1-\alpha) \alpha^k}{1-\alpha^{1+N}} ,\qquad \alpha = \frac{p}{1-p}.$$ This distribution serves as the start for the new random walk. The new random walk is given by the rate equations
$$\begin{align}P_{i+1}(0) &= (1-f_1)(1-p) P_i (1) + \sum_{k=1}^N f_k P_i(k)\\
  P_{i+1}(1)&= (1-f_2)(1-p) P_i(2)\\
  P_{i+1} (2\leq j \leq N-1)&= (1-f_{j+1})(1-p) P_i (j+1) + (1-f_{j-1})p P(j-1) \\
  P_{i+1} (N) &= (1-f_{N-1})p P_{i}(N-1) +(1-f_{N})p  P_i(N)
\end{align}$$
where $P_i(k)$ denotes the probability to be at step $i$ at position $k$. The initial condition is $P_0(k)=\pi(k)$.
You are asking for the mean absorbtion time given by
$$\mu=\sum_{i=0}^\infty i P_{i}(0) .$$
We define the transition matrix $M$ via
$\mathbf{P}_{i+1} = M \mathbf{P}_i$ where we collect the probabilities to a vector $\mathbf{P} =(P_0, \ldots, P_N)$. We have also the initial vector of probabilities $\boldsymbol{\pi}$. With this notation, we can write
$$\mu= \sum_{j=0}^\infty e_0 j M^j \boldsymbol{\pi} 
=e_0 \frac{M}{(I-M)^2}  \boldsymbol{\pi} $$
