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Use the characteristic polynomial in the convention $p(\lambda )=\det (\lambda I-A)$:For fixed $\alpha >0$ with $\alpha \ne 1$, all three limiting eigenvalues are simple. Expanding this polynomial at each root, or applying first-order perturbation of a simple eigenvalue, givesThese are fixed-$\alpha$ asymptotic expansions, not uniform through $\alpha =1$. At $\alpha =1/3$ the displayed correction to ${\lambda }_{\alpha }$ vanishes because $\lambda =1/3$ is an exact root for every $ϵ$; there is no second nonzero term for that branch.

The two roots near $1$ interact when the correction $ϵ/(\alpha -1)$ is comparable to their unperturbed gap $\alpha -1$. Thus the distinguished limit is $\alpha =1+d\sqrt{ϵ}$, with bounded real $d$. Set $h=\sqrt{ϵ}$ and $\lambda =1+hL+h^{2}N+\cdots$. The polynomial becomesAt orders $h^2$ and $h^3$,With $D=\sqrt{d^2+4}$, the two interacting eigenvalues areThe $h^2$ correction to the pair is included to resolve the next perturbative term as well as the leading splitting. In particular the previously excluded fixed case $\alpha =1$ is $d=0$:The half-integer powers reflect square-root splitting of a defective double eigenvalue: the limiting matrix has a nontrivial Jordan block at $1$.

For matching, write $\Delta =\alpha -1$ and take the overlap $\sqrt{ϵ}\ll |\Delta |\ll 1$. For $d\to +\infty$, $L_{+}=d+d^{-1}+O(d^{-3})$, $N_{+}=1+O(d^{-2})$, while $L_{-}=-d^{-1}+O(d^{-3})$, $N_{-}=O(d^{-2})$. Consequently the distinguished branches reduce toThe regular coefficient satisfies $(3\alpha -1)/[(\alpha +1)(\alpha -1)]=1/\Delta +2/(2+\Delta )=1/\Delta +1+O(\Delta )$, proving agreement. For $d\to -\infty$ the labels $+$ and $-$ exchange roles and the same matching holds. The isolated negative branch also matches because $-2/(\alpha +1)=-1+O(\Delta )$. As a spectral check, positive diagonal symmetrization of a matrix transforms this tridiagonal matrix to a real symmetric matrix with off-diagonal entries $\sqrt{2ϵ}$ and $\sqrt{ϵ}$; its eigenvalues are real for positive $ϵ$.

Differentiate the eigenvalue equationat $t=0$, where $A^{\prime }(0)=E$. Suppressing the argument $0$ givesLeft-multiplication by the transpose of the unit left eigenvector $u$ and use of $u^{T}A=\lambda u^T$ cancel the terms involving $v^{\prime }$:Because the eigenvalue is simple, $u^{T}v\ne 0$, and hence the first-order perturbation of a simple eigenvalue isThe definition of the operator norm, followed by the Cauchy-Schwarz inequality, givesTherefore

Let the computed eigenpair residual beDefine the rank-one matrixThenso $(\tilde{\lambda },\tilde{v})$ is an exact eigenpair of the perturbed matrix $A+E$. Moreover,which is the backward error of an approximate eigenpair.

Follow the simple eigenvalue branch from $\lambda (0)=\lambda$ for $A(t)=A+tE$. The first-order perturbation of a simple eigenvalue and part (b) giveSince the residual is at the machine precision scale, this proves

Setwhere $s>0$. Applying first-order perturbation of a simple eigenvalue to the flux Jacobian, or expanding its quadratic formula directly, separates the characteristic that changes total concentration from the characteristic that changes composition:At $ϵ=0$, both species move with the common hindered velocity $\hat{W}(1-s)$. Summing their conservation laws givesso the total concentration is a nonlinear kinematic wave:Taking the ratio $c={\varphi }_1/s$ instead gives the composition wave in a bidisperse suspensionand henceThese are the requested leading-order ordinary differential equations along the two characteristic families.