 Solution

For $q=\eta =1$, $J_{\ast }$ has eigenvectors $(1,1)$ and $(1,-1)$ with eigenvalues $-1$ and $-1/3$. The Neumann spatial modes are $\cos (j\pi x/L)$, $j=0,1,\dots$. Combining each with those two reaction directions gives temporal ratesEvery linear perturbation therefore decays at least as fast as $e^{-t/3}$. The slowest mode is the spatially uniform antisymmetric change of the two biological populations. ThusSmall nonlinear perturbations have the same bound, not just a formal mode calculation. In one spatial dimension take sufficiently small smooth Neumann data, or small data in a norm such as $H^1$ that controls quadratic products. The linear semigroup decays at rate $1/3$, and the nonlinear remainder is quadratic. Standard small-data variation-of-constants estimates first give decay $e^{-ct}$ for any fixed $c<1/3$. Choose $1/6<c<1/3$; then the remainder is $O(e^{-2ct})$, and its Duhamel convolution with $e^{-(t-s)/3}$ is $O(e^{-t/3})$ because $2c>1/3$. This sharpens the preliminary bound to the displayed rate. Special initial data with no slow-mode component may decay faster, but the general estimate cannot be improved uniformly.