**Proof plan.** Linearise the reaction-diffusion system about the steady state $(u^*, v^*)$. Expanding perturbations in the eigenbasis of $-\Delta$ with Neumann conditions decomposes the linearised system into independent $2 \times 2$ ODE systems, one for each eigenvalue $\mu$ of $-\Delta$. The growth rate $\lambda(\mu)$ is determined by the eigenvalues of the matrix $J - \mu D$. Conditions (i)–(ii) ensure stability at $\mu = 0$; conditions (iii)–(iv) ensure that $\det(J - \mu D) < 0$ for some range of $\mu > 0$, producing a positive growth rate.
**Step 1: Linearisation.** Write $u = u^* + \varepsilon\,U$, $v = v^* + \varepsilon\,V$ with $0 < \varepsilon \ll 1$. Substituting into the system and using $f(u^*, v^*) = 0$, $g(u^*, v^*) = 0$, the $O(\varepsilon)$ terms give
\begin{align*}
\partial_t \begin{pmatrix} U \\ V \end{pmatrix} &= J \begin{pmatrix} U \\ V \end{pmatrix} + D\,\Delta \begin{pmatrix} U \\ V \end{pmatrix}.
\end{align*}
**Step 2: Modal decomposition.** Let $\{\varphi_k\}_{k \ge 0}$ be an $L^2(\Omega)$-orthonormal eigenbasis for $-\Delta$ with homogeneous Neumann conditions, so that $-\Delta\varphi_k = \mu_k\,\varphi_k$ in $\Omega$ and $\partial_\nu\varphi_k = 0$ on $\partial\Omega$, with $0 = \mu_0 \le \mu_1 \le \mu_2 \le \cdots$. Seeking solutions of the form $(U, V)^\top = \zeta\,e^{\lambda t}\,\varphi_k(x)$ with $\zeta \in \mathbb{R}^2$ and substituting:
\begin{align*}
\lambda\,\zeta &= (J - \mu_k\,D)\,\zeta.
\end{align*}
Hence $\lambda$ is an eigenvalue of $M(\mu) := J - \mu\,D$ evaluated at $\mu = \mu_k$.
**Step 3: Dispersion relation.** The characteristic polynomial of $M(\mu)$ is $\lambda^2 - \operatorname{tr}(M(\mu))\,\lambda + \det(M(\mu)) = 0$, where
\begin{align*}
\operatorname{tr}(M(\mu)) &= (J_{11} - \mu\,D_1) + (J_{22} - \mu\,D_2) = \operatorname{tr}(J) - \mu\,(D_1 + D_2), \\
\det(M(\mu)) &= (J_{11} - \mu\,D_1)(J_{22} - \mu\,D_2) - J_{12}\,J_{21}.
\end{align*}
**Step 4: Trace remains negative.**
[claim:Trace Negativity]
$\operatorname{tr}(M(\mu)) < 0$ for all $\mu \ge 0$.
[/claim]
[proof]
$\operatorname{tr}(M(\mu)) = \operatorname{tr}(J) - \mu(D_1 + D_2)$. Since $\operatorname{tr}(J) < 0$ by condition (i) and $D_1 + D_2 > 0$ and $\mu \ge 0$, both terms are non-positive, and the first is strictly negative.
[/proof]
Since $\operatorname{tr}(M(\mu)) < 0$, the quadratic $\lambda^2 - \operatorname{tr}(M)\,\lambda + \det(M) = 0$ has both roots with negative real part if and only if $\det(M(\mu)) > 0$ (by the Routh–Hurwitz criterion for $2 \times 2$ matrices: $\operatorname{tr} < 0$ and $\det > 0$ imply both eigenvalues have negative real part). Hence $\operatorname{Re}(\lambda) > 0$ for some eigenvalue $\lambda$ if and only if $\det(M(\mu)) < 0$.
**Step 5: Determinant as a quadratic in $\mu$.**
Expanding:
\begin{align*}
\det(M(\mu)) &= D_1 D_2\,\mu^2 - (D_2\,J_{11} + D_1\,J_{22})\,\mu + \det(J).
\end{align*}
This is a quadratic in $\mu$ with leading coefficient $D_1 D_2 > 0$ and constant term $\det(J) > 0$ by condition (ii). Hence $\det(M(0)) = \det(J) > 0$ (confirming stability at $\mu = 0$) and $\det(M(\mu)) \to +\infty$ as $\mu \to +\infty$. For $\det(M(\mu)) < 0$ for some $\mu > 0$, the minimum of the quadratic must be negative. The minimum occurs at $\mu^* = \frac{D_2 J_{11} + D_1 J_{22}}{2D_1 D_2}$, which is positive if and only if $D_2 J_{11} + D_1 J_{22} > 0$ (condition (iii)). The minimum value is
\begin{align*}
\det(M(\mu^*)) &= \det(J) - \frac{(D_2 J_{11} + D_1 J_{22})^2}{4D_1 D_2},
\end{align*}
which is negative if and only if $(D_2 J_{11} + D_1 J_{22})^2 > 4D_1 D_2\det(J)$ (condition (iv)).
**Step 6: Range of unstable eigenvalues.** When conditions (iii)–(iv) hold, $\det(M(\mu)) < 0$ for $\mu \in (k_-^2, k_+^2)$, where $k_\pm^2$ are the two roots of $\det(M(\mu)) = 0$:
\begin{align*}
k_\pm^2 &= \frac{(D_2 J_{11} + D_1 J_{22}) \pm \sqrt{(D_2 J_{11} + D_1 J_{22})^2 - 4D_1 D_2\det(J)}}{2D_1 D_2}.
\end{align*}
Both roots are positive (since the quadratic has positive leading coefficient, positive constant term, and negative minimum). For each Neumann eigenvalue $\mu_k \in (k_-^2, k_+^2)$, the matrix $M(\mu_k)$ has $\det(M(\mu_k)) < 0$, so its two real eigenvalues have opposite signs, and the positive one gives an exponentially growing mode.
**Step 7: Consequence for the signs of $J_{11}$ and $J_{22}$.**
[claim:Opposite Signs Of Diagonal Entries]
Conditions (i)–(iii) imply $J_{11}$ and $J_{22}$ have opposite signs, and $D_1 \neq D_2$.
[/claim]
[proof]
Condition (i) gives $J_{11} + J_{22} < 0$. If both $J_{11} \le 0$ and $J_{22} \le 0$, then $D_2 J_{11} + D_1 J_{22} \le 0$, contradicting condition (iii). Hence at least one is positive. Since $J_{11} + J_{22} < 0$, they cannot both be positive. So exactly one is positive and the other is negative. If $D_1 = D_2 =: D$, then condition (iii) gives $D(J_{11} + J_{22}) > 0$, contradicting condition (i). Hence $D_1 \neq D_2$.
[/proof]
