case of \[thm:chan\] separately. First we lay some groundwork that applies to all cases. Recall that we have $$e^{iAt} = e^{ikt}E_k + e^{i{\theta}t} E_{\theta}+ e^{i\tau t} E_\tau,$$ and so we can expand both sides of this equation in terms of the spectral decomposition. Comparing the expressions for $E_k$, $E_{\theta}$ and $E_\tau$ on both sides, we get the following system of equations: $$\begin{aligned}
ce^{ikt} \sqrt{n} & = 1 + xk + y(v-k-1) \label{eqn:flatsrg1} \\
ce^{i{\theta}t} \sqrt{n} & = 1 + x{\theta}+ y(-{\theta}-1) \label{eqn:flatsrg2} \\
ce^{i\tau t} \sqrt{n} & = 1 + x\tau + y(-\tau-1)\label{eqn:flatsrg3}.\end{aligned}$$ These equations are the *characterizing equations*. The graph $X$ admits uniform mixing if and only if a solution to the characterizing equations exists in $t$, $x$, $y$ and $c$ with $|x| = |y| = |c| = 1$.

Chan shows (Lemma 2.2) that if $X$ has at least $5$ vertices, then ${\theta}+ \tau \in \{0,1,2\}$. Moreover, for each value of ${\theta}+\tau$, Chan finds parameter sets $(n,k,{\lambda},\mu)$ in Theorem \[thm:chan\] that can occur and also determines the values of $x$ and $y$ that can occur in the characterizing equations.

In each case, we must also consider if the complement has uniform mixing $ce^{i\bar{A}t} = I + xA + y\bar{A}$. 