Codex @codex  0

Write $\mathbf{c}=(U,V{)}^T$ and $\Gamma =\left(\begin{array}{cc}{\gamma }_{11}& {\gamma }_{12}\\ {\gamma }_{12}& {\gamma }_{22}\end{array}\right)$. A well-posed anisotropic diffusion problem requires $\Gamma$ to be symmetric positive definite: ${\gamma }_{11}>0$ and $\Delta ={\gamma }_{11}{\gamma }_{22}-{\gamma }_{12}^2>0$. For the normal mode in the source the dispersion relation isThe group velocity vanishes when $\mathbf{c}-2i\Gamma \mathbf{k}=0$. ConsequentlyThe requested anisotropic absolute-instability threshold for positive diffusion isFor positive definite diffusion this is also sufficient: the anisotropic Gaussian heat kernel has exponential factor $\exp [\mu t-(\mathbf{x}-\mathbf{c}t{)}^T{\Gamma }^{-1}(\mathbf{x}-\mathbf{c}t)/(4t)]$ and a $t^{-1}$ prefactor, so its fixed-position growth rate is precisely $\mu -f$.

Let the prescribed speed be $q$ and let $d_{\min }\le d_{\max }$ be the positive eigenvalues of $\Gamma$. The Rayleigh quotient for ${\Gamma }^{-1}$ givesAbsence of absolute hydrodynamic instability for every direction therefore requiresThis is an upper bound on $\mu$, not a minimum required value. The source's last sentence reverses this monotonicity: reducing the growth parameter always helps, so there is no finite lower threshold guaranteeing absence. If it had asked for instability in every direction, the corresponding condition would be $\mu >q^2/(4d_{\min })$. Without the positive-definite assumption, negative diffusion directions make the forward problem ill posed; a singular diffusion tensor requires a separate transport analysis and does not support the displayed inverse-matrix formula.