Imagine I have some Markov process consisting of a biased random walk on the integers, over some interval $[0, L]$, with $+1$ and $-1$ step probabilities of $p$ and $q$, respectively, s.t. $(p + q) = 1$. Here, the walker is reflected at $X = 0$ with probability $p$ (staying in place with probability $(1-p)$), and the walker is reflected at $X=L$ with probability $q$ (staying in place with probability $(1-q)$). We can also imagine the same process where $p=q$ everywhere except at one position where there is some $p \approx \epsilon$ transition probability.

Clarification: Unlike the random walk from (Levin, Peres, and Wilmer), where there is a probability of $\frac{1}{2}$ to fall back to the origin at any point, the random walk I described above can only fall back a step with probability $\frac{p}{q}$. The regime where $\frac{p}{q} = 1 - C$ for small $C$ is of interest to me. I do not understand how $\Theta(1)$ could be asymptotically tight for this?

I strictly initiate the random walk at a fixed position $X = i$ on the interval. How can I analyze the mixing time for the above process. I.e. after how many steps do we have: $||Prob[X=n] - \pi(n)|| \leq C$ for some steady state distribution $\pi$ and some constant $C$? Clearly it can't be $\Theta(N^2)$ like we would expect for an unbiased random walk, since $p \approx \epsilon$ can arbitrarily increase the time to sample states across some "hurdle". I'd expect that the mixing time would have to be an asymptotic function of the lowest probability transition state(s)?