[proofplan] We prove exponential stability of $X^*(\epsilon)$ for small $\epsilon$ by constructing a norm in which the Lyapunov exponent is strictly negative. First, we centre coordinates at the equilibrium and expand to obtain the linearised system plus a higher-order remainder. Then we transform to a special Jordan basis where the off-diagonal coupling is controlled by a free parameter $\nu > 0$. In this basis, we compute the time [derivative](/page/Derivative) of the squared norm $\|Z\|^2$ and show it satisfies $\frac{d}{dt}\|Z\|^2 \le -2\alpha\|Z\|^2$ for $\alpha = \beta - \nu > 0$, where $-\beta$ bounds the real parts of the eigenvalues. Gronwall's inequality then yields the exponential decay estimate. [/proofplan] [step:Centre coordinates at the equilibrium and extract the linearisation] Since the [eigenvalues](/page/Eigenvalue%20and%20Eigenvector) of $A(0)$ have strictly negative real parts and eigenvalues depend continuously on matrix entries, the eigenvalues of $A(\epsilon)$ also have strictly negative real parts for all sufficiently small $|\epsilon|$. Let $-\beta < 0$ be an upper bound on the real parts of the eigenvalues of $A(\epsilon)$: \begin{align*} \max_{j \in \{1, \ldots, n\}} \operatorname{Re}(\lambda_j(\epsilon)) \le -\beta < 0. \end{align*} Define the centred variable: \begin{align*} Y: [0, \infty) &\to \mathbb{R}^n \\ t &\mapsto X(t) - X^*(\epsilon). \end{align*} Differentiating and using $f(X^*(\epsilon), \epsilon) = 0$: \begin{align*} \dot{Y}(t) &= f(Y(t) + X^*(\epsilon), \epsilon) \\ &= f(X^*(\epsilon), \epsilon) + A(\epsilon)\,Y(t) + R(Y(t)) \\ &= A(\epsilon)\,Y(t) + R(Y(t)), \end{align*} where the remainder $R: \mathbb{R}^n \to \mathbb{R}^n$ satisfies $|R(Y)| = o(|Y|)$ as $|Y| \to 0$ (from the Taylor expansion of $f$ about $X^*(\epsilon)$). [guided] The idea is to separate the dynamics into a linear part $A(\epsilon)\,Y$ and a nonlinear remainder $R(Y)$ that is negligible near the equilibrium. The linear part governs the local behaviour: if all eigenvalues of $A(\epsilon)$ have negative real parts, the linear system $\dot{Y} = A(\epsilon)\,Y$ is exponentially stable, and we need to show that the nonlinear remainder does not destroy this stability. Define the centred variable $Y(t) := X(t) - X^*(\epsilon)$. Since $X^*(\epsilon)$ is an equilibrium (constant in time), we have $\dot{Y}(t) = \dot{X}(t) = f(X(t), \epsilon) = f(Y(t) + X^*(\epsilon), \epsilon)$. The equilibrium condition gives $f(X^*(\epsilon), \epsilon) = 0$. Expanding $f$ about $X^*(\epsilon)$ via Taylor's theorem: \begin{align*} f(Y + X^*(\epsilon), \epsilon) = \underbrace{f(X^*(\epsilon), \epsilon)}_{= 0} + \underbrace{\frac{\partial f}{\partial X}\bigg|_{X = X^*(\epsilon)}}_{= A(\epsilon)} Y + R(Y), \end{align*} where the remainder $R: \mathbb{R}^n \to \mathbb{R}^n$ satisfies $|R(Y)| = o(|Y|)$ as $|Y| \to 0$. The remainder captures all quadratic and higher-order terms in the Taylor expansion of $f$ about the equilibrium. Substituting into the ODE yields the centred equation: \begin{align*} \dot{Y}(t) = A(\epsilon)\,Y(t) + R(Y(t)). \end{align*} The eigenvalues of $A(\epsilon)$ depend [continuously](/page/Continuity) on $\epsilon$ because they are roots of the characteristic polynomial $\det(A(\epsilon) - \lambda I) = 0$, whose coefficients are continuous functions of the matrix entries. At $\epsilon = 0$, all eigenvalues have strictly negative real parts by hypothesis. By continuity, there exists $\beta > 0$ such that for all sufficiently small $|\epsilon|$: \begin{align*} \max_{j \in \{1, \ldots, n\}} \operatorname{Re}(\lambda_j(\epsilon)) \le -\beta < 0. \end{align*} This spectral gap $\beta > 0$ is the quantitative input that drives the exponential decay in subsequent steps. [/guided] [/step] [step:Transform to the special Jordan basis with tunable off-diagonal parameter $\nu$] Let $Q$ be an invertible $n \times n$ matrix (possibly complex) and define $Z := Q^{-1}Y$. The equation for $Z$ becomes: \begin{align*} \dot{Z}(t) = Q^{-1}A(\epsilon)\,Q\,Z(t) + Q^{-1}R(QZ(t)) =: \tilde{A}\,Z(t) + \tilde{R}(Z(t)), \end{align*} where $\tilde{A} := Q^{-1}A(\epsilon)\,Q$ and $\tilde{R}(Z) := Q^{-1}R(QZ)$ satisfies $|\tilde{R}(Z)| = o(|Z|)$. We choose $Q$ to place $\tilde{A}$ in **special Jordan form**: the diagonal entries are the eigenvalues $\lambda_1, \ldots, \lambda_n$ of $A(\epsilon)$, and the super-diagonal entries (which are $1$ in the standard Jordan form) are replaced by an arbitrary parameter $\nu > 0$. Concretely, if the standard Jordan basis vectors satisfy $A(\epsilon)\,v_j = \lambda_j\,v_j + \delta_j\,v_{j-1}$ (where $\delta_j \in \{0, 1\}$ indicates Jordan block coupling), define scaled basis vectors $\tilde{v}_j := \nu^{-j}\,v_j$. In this basis: \begin{align*} (\tilde{A})_{jj} = \lambda_j, \quad (\tilde{A})_{j-1,j} = \nu\,\delta_j. \end{align*} The parameter $\nu > 0$ can be chosen arbitrarily small, making the off-diagonal coupling as weak as desired. [guided] Why do we need the special Jordan form rather than the standard one? In the standard Jordan form, the super-diagonal entries are $1$, and the energy estimate would involve a fixed off-diagonal contribution that competes with the eigenvalue decay rate $\beta$. By introducing the parameter $\nu$, we can make this competition arbitrarily one-sided in favour of the diagonal (eigenvalue) terms. Starting from the standard Jordan decomposition $A(\epsilon) = P\,J\,P^{-1}$ where $J$ has blocks of the form $\lambda_j\,I + N$ (with $N$ the nilpotent part having $1$s on the super-diagonal), we rescale: define $D_\nu = \operatorname{diag}(\nu^{-1}, \nu^{-2}, \ldots, \nu^{-n})$ and set $Q = P\,D_\nu^{-1}$. Then: \begin{align*} \tilde{A} = D_\nu\,J\,D_\nu^{-1}. \end{align*} The diagonal entries of $\tilde{A}$ remain $\lambda_j$ (eigenvalues are similarity-invariant). The super-diagonal entries become $\nu \cdot 1 = \nu$ because the rescaling multiplies each super-diagonal entry by $\nu^{j}/\nu^{j-1} = \nu$. So $\tilde{A}$ has the structure: \begin{align*} (\tilde{A})_{jj} = \lambda_j, \quad (\tilde{A})_{j-1,j} = \nu\,\delta_j, \quad \text{all other entries } = 0. \end{align*} The remainder transforms as $\tilde{R}(Z) = Q^{-1}R(QZ)$, and since $R(Y) = o(|Y|)$, we have $|\tilde{R}(Z)| = o(|Z|)$ as well (the linear change of coordinates preserves the asymptotic order). [/guided] [/step] [step:Compute the energy estimate $\frac{d}{dt}\|Z\|^2 \le -2\alpha\|Z\|^2$] Define the squared norm $\|Z\|^2 := \sum_{j=1}^{n} |Z_j|^2$ (using the standard complex modulus if eigenvalues are complex). Define the standard Hermitian inner product $\langle u, v \rangle := \sum_{j=1}^{n} \overline{u}_j \, v_j$ for $u, v \in \mathbb{C}^n$, so that $\|Z\|^2 = \langle Z, Z \rangle$. Differentiating along the trajectory: \begin{align*} \frac{d}{dt}\|Z\|^2 = 2\operatorname{Re}\langle \dot{Z}, Z\rangle = 2\operatorname{Re}\langle \tilde{A}\,Z, Z\rangle + 2\operatorname{Re}\langle \tilde{R}(Z), Z\rangle. \end{align*} Expanding the linear term using the structure of $\tilde{A}$: \begin{align*} 2\operatorname{Re}\langle \tilde{A}\,Z, Z\rangle = 2\sum_{j=1}^{n} \operatorname{Re}(\lambda_j)\,|Z_j|^2 + 2\operatorname{Re}\sum_{j} \nu\,\delta_j\,\overline{Z}_{j-1}\,Z_j. \end{align*} Since $\operatorname{Re}(\lambda_j) \le -\beta$ for all $j$, the diagonal contribution satisfies $2\sum_j \operatorname{Re}(\lambda_j)|Z_j|^2 \le -2\beta\|Z\|^2$. For the off-diagonal terms, the AM--GM inequality $2|a|\,|b| \le |a|^2 + |b|^2$ gives: \begin{align*} \left|2\operatorname{Re}\sum_{j} \nu\,\delta_j\,\overline{Z}_{j-1}\,Z_j\right| \le 2\nu\sum_{j} |Z_{j-1}|\,|Z_j| \le \nu\sum_j (|Z_{j-1}|^2 + |Z_j|^2) \le 2\nu\,\|Z\|^2. \end{align*} Combining, and using $|\tilde{R}(Z)| = o(|Z|)$ so that $|2\operatorname{Re}\langle \tilde{R}(Z), Z\rangle| \le o(1)\|Z\|^2$: \begin{align*} \frac{d}{dt}\|Z\|^2 \le -2\beta\,\|Z\|^2 + 2\nu\,\|Z\|^2 + o(1)\,\|Z\|^2 = -2(\beta - \nu)\,\|Z\|^2 + o(1)\,\|Z\|^2. \end{align*} Choose $\nu < \beta/2$ and set $\alpha := \beta - \nu > 0$. For $\|Z\|$ sufficiently small (i.e., within a neighbourhood where the $o(1)$ term is at most $\alpha$): \begin{align*} \frac{d}{dt}\|Z\|^2 \le -2\alpha\,\|Z\|^2. \end{align*} [guided] The energy method is the heart of the proof. We want to show that $\|Z(t)\|$ decreases exponentially. Define the standard Hermitian inner product $\langle u, v \rangle := \sum_{j=1}^{n} \overline{u}_j \, v_j$ for $u, v \in \mathbb{C}^n$. The squared norm $\|Z\|^2 = \langle Z, Z \rangle = \sum_j |Z_j|^2$ serves as a Lyapunov function. Its time derivative splits into three contributions: 1. **Diagonal (eigenvalue) terms:** $2\sum_j \operatorname{Re}(\lambda_j)|Z_j|^2 \le -2\beta\|Z\|^2$. This is the stabilising force -- it drives the energy down at rate $2\beta$. 2. **Off-diagonal (Jordan) terms:** $2\operatorname{Re}\sum_j \nu\,\delta_j\,\overline{Z}_{j-1}\,Z_j$. These arise from the Jordan block coupling and could potentially destabilise the system. Applying AM--GM: $2\nu|Z_{j-1}|\,|Z_j| \le \nu(|Z_{j-1}|^2 + |Z_j|^2)$, so the total off-diagonal contribution is at most $2\nu\|Z\|^2$. This is why we introduced $\nu$: by choosing $\nu$ small, we ensure the off-diagonal terms are dominated by the diagonal decay. 3. **Nonlinear remainder:** $2\operatorname{Re}\langle \tilde{R}(Z), Z\rangle$. Since $|\tilde{R}(Z)| = o(|Z|)$, by Cauchy--Schwarz $|2\operatorname{Re}\langle \tilde{R}(Z), Z\rangle| \le 2|\tilde{R}(Z)|\,|Z| = o(|Z|^2) = o(1)\|Z\|^2$. Near the origin, this term is negligible. Combining all three: \begin{align*} \frac{d}{dt}\|Z\|^2 \le (-2\beta + 2\nu + o(1))\|Z\|^2. \end{align*} Choose $\nu < \beta/2$. Then for $Z$ in a sufficiently small ball around $0$ (so that the $o(1)$ term is at most $\beta - 2\nu > 0$, leaving the coefficient negative), we get: \begin{align*} \frac{d}{dt}\|Z\|^2 \le -2\alpha\,\|Z\|^2 \end{align*} with $\alpha := \beta - \nu > \beta/2 > 0$. [/guided] [/step] [step:Integrate the differential inequality to obtain exponential decay] If $Z(0) = 0$, then $X_0 = X^*(\epsilon)$ and the estimate $\|X(t) - X^*(\epsilon)\| = 0 \le 0$ is trivially true. For $Z(0) \neq 0$, the energy estimate $\frac{d}{dt}\|Z\|^2 \le -2\alpha\,\|Z\|^2 < 0$ shows $\|Z(t)\|^2$ is strictly decreasing, hence $\|Z(t)\|^2 > 0$ for all $t \ge 0$. Dividing by $\|Z(t)\|^2$ and integrating over $[0, t]$: \begin{align*} \frac{d}{dt}\ln(\|Z(t)\|^2) \le -2\alpha. \end{align*} Integrating: \begin{align*} \ln(\|Z(t)\|^2) - \ln(\|Z(0)\|^2) \le -2\alpha\,t, \end{align*} which gives: \begin{align*} \|Z(t)\| \le \|Z(0)\|\,e^{-\alpha t} \quad \text{for all } t \ge 0. \end{align*} [guided] The differential inequality $\frac{d}{dt}\|Z\|^2 \le -2\alpha\,\|Z\|^2$ is a standard Gronwall-type estimate. If $Z(0) = 0$, the conclusion is trivial ($X_0 = X^*(\epsilon)$ is already at the equilibrium). For $Z(0) \neq 0$, the negative sign of the derivative ensures $\|Z(t)\|^2$ remains strictly positive for all $t \ge 0$, so we may divide by $\|Z(t)\|^2$: \begin{align*} \frac{d}{dt}\ln(\|Z(t)\|^2) = \frac{1}{\|Z(t)\|^2} \frac{d}{dt}\|Z(t)\|^2 \le -2\alpha. \end{align*} Integrating both sides over $[0, t]$: \begin{align*} \ln(\|Z(t)\|^2) - \ln(\|Z(0)\|^2) \le -2\alpha\,t. \end{align*} Exponentiating: $\|Z(t)\|^2 \le \|Z(0)\|^2 \, e^{-2\alpha t}$. Taking square roots: $\|Z(t)\| \le \|Z(0)\| \, e^{-\alpha t}$. This is the exponential decay in the transformed coordinates. [/guided] [/step] [step:Translate back to the original coordinates to conclude] Since $Y = QZ$, define the norm $\|\cdot\|_*$ on $\mathbb{R}^n$ by $\|Y\|_* := \|Q^{-1}Y\| = \|Z\|$. Then: \begin{align*} \|X(t) - X^*(\epsilon)\|_* = \|Y(t)\|_* = \|Z(t)\| \le \|Z(0)\|\,e^{-\alpha t} = \|Y(0)\|_*\,e^{-\alpha t} = \|X_0 - X^*(\epsilon)\|_*\,e^{-\alpha t}. \end{align*} This holds for all $t \ge 0$ provided $X_0$ lies in a sufficiently small neighbourhood of $X^*(\epsilon)$ (the neighbourhood where the $o(1)$ bound on the nonlinear remainder is valid). The estimate holds for all sufficiently small $\epsilon$ (the range where the eigenvalues of $A(\epsilon)$ remain in the left half-plane with real parts below $-\beta$). [guided] The exponential decay $\|Z(t)\| \le \|Z(0)\| e^{-\alpha t}$ was proved in the special Jordan coordinates. We now translate this back to the original coordinates $X(t)$. Recall $Z = Q^{-1}Y$ where $Y(t) = X(t) - X^*(\epsilon)$. The norm $\|Y\|_* := \|Q^{-1}Y\|$ is equivalent to the standard Euclidean norm on $\mathbb{R}^n$ (since $Q$ is invertible, $\|Q^{-1}\|^{-1} |Y| \le \|Y\|_* \le \|Q^{-1}\| |Y|$). In this norm: \begin{align*} \|X(t) - X^*(\epsilon)\|_* = \|Y(t)\|_* = \|Z(t)\| \le \|Z(0)\| e^{-\alpha t} = \|X_0 - X^*(\epsilon)\|_* \, e^{-\alpha t}. \end{align*} The restriction to a small neighbourhood of $X^*(\epsilon)$ is necessary because the $o(1)$ bound on the nonlinear remainder $\tilde{R}$ was used in the energy estimate — specifically, we required $|\tilde{R}(Z)| \le (\alpha - \beta + 2\nu)|Z|$ (or equivalently, the $o(1)$ coefficient at most $\alpha$), which holds only for $\|Z\|$ below some threshold $r > 0$. The solution remains in this ball for all $t \ge 0$ because $\|Z(t)\| \le \|Z(0)\| e^{-\alpha t} \le \|Z(0)\| < r$ provided $\|Z(0)\| < r$. The estimate holds for all sufficiently small $|\epsilon|$ because the spectral gap $\beta > 0$ was obtained from continuity of eigenvalues, which guarantees a uniform $\beta$ on a neighbourhood of $\epsilon = 0$. This completes the proof of exponential stability. [/guided] [/step]