**Proof plan.** The proof translates the nonlinear ODE into a perturbation of the linear system $\dot{y} = Jf_{x^*}\,y$ via the substitution $y = x - x^*$ and the Taylor expansion of $f$, writing $\dot{y} = Jf_{x^*}\,y + r(y)$ with $|r(y)| \leq C|y|^2$ (Claim 1). For Part 1 (asymptotic stability), the variation of constants formula expresses $y(t)$ in terms of the matrix exponential $e^{Jf_{x^*}\,t}$ and a [convolution](/page/Convolution) [integral](/page/Integral) involving the remainder $r$. A matrix exponential estimate $|e^{Jf_{x^*}\,t}| \leq Me^{-\alpha t}$ (Claim 2) provides exponential decay for the linear part, and a bootstrap argument (Claim 3) shows that the nonlinear remainder cannot overcome this decay for small initial data. For Part 2 (instability), a projection onto the unstable eigenspace and a reversed-time Gronwall argument shows that trajectories starting near $x^*$ along the unstable direction must leave any small neighbourhood (Claim 4). **Step 1 (Reduction to a perturbation of the linear system).** Set $y(t) := x(t) - x^*$, so $y(0) = x(0) - x^*$ and $\dot{y} = f(x^* + y)$. Set $A := Jf_{x^*} \in \mathbb{R}^{n \times n}$. [claim:Quadratic Remainder Estimate] Define $r: B(0, \rho) \to \mathbb{R}^n$ by $r(y) := f(x^* + y) - Jf_{x^*}\,y$ for $\rho > 0$ small enough that $B(x^*, \rho) \subset U$. Then $r(0) = 0$ and there exists a constant $C > 0$ such that \begin{align*} |r(y)| \leq C\,|y|^2 \qquad \text{for all } y \in B(0, \rho). \end{align*} [/claim] [proof] Since $f(x^*) = 0$, we have $r(0) = f(x^*) - 0 = 0$. The total [derivative](/page/Derivative) of $r$ at $0$ is $Dr_0 = Df_{x^*} - Df_{x^*} = 0$ (the [linear map](/page/Linear%20Map) $y \mapsto Jf_{x^*}\,y$ has derivative $Df_{x^*}$ at every point). Since $f \in C^2(U; \mathbb{R}^n)$, the map $r \in C^2(B(0,\rho); \mathbb{R}^n)$. By [Taylor's theorem with integral remainder](/theorems/189) applied to each component $r_i$: \begin{align*} r_i(y) = r_i(0) + \sum_j \partial_j r_i(0)\,y_j + \sum_{j,k}\int_0^1 (1-t)\,\partial_j\partial_k r_i(ty)\,y_j\,y_k\,dt. \end{align*} Since $r_i(0) = 0$ and $\partial_j r_i(0) = 0$ (because $Dr_0 = 0$), only the quadratic term remains. Bounding $|\partial_j\partial_k r_i(ty)| \leq \sup_{B(0,\rho)}\|D^2 r\| =: C'$ (finite by [continuity](/page/Continuity) of the second derivative on the compact [set](/page/Set) $\overline{B}(0, \rho/2)$, after shrinking $\rho$ if necessary): \begin{align*} |r_i(y)| \leq C'\sum_{j,k}|y_j||y_k| \leq C'n^2|y|^2. \end{align*} Summing over components: $|r(y)| \leq \sqrt{n}\,C'n^2\,|y|^2 = C|y|^2$. [/proof] The ODE for $y$ is therefore $\dot{y} = A\,y + r(y)$ with $|r(y)| \leq C|y|^2$ for $|y| \leq \rho$. **Step 2 (Matrix exponential estimate).** [claim:Matrix Exponential Decay Estimate] If every eigenvalue $\lambda$ of $A \in \mathbb{R}^{n \times n}$ satisfies $\operatorname{Re}(\lambda) < 0$, then for every $\alpha$ with $0 < \alpha < \min_i(-\operatorname{Re}(\lambda_i))$, there exists $M \geq 1$ such that \begin{align*} \|e^{At}\| \leq M\,e^{-\alpha t} \qquad \text{for all } t \geq 0, \end{align*} where $\|\cdot\|$ denotes the operator norm induced by the Euclidean norm on $\mathbb{R}^n$. [/claim] [proof] By the [Jordan normal form](/theorems/864), $A = P\,J\,P^{-1}$ where $J = \operatorname{diag}(J_1, \ldots, J_p)$ is block-diagonal with Jordan blocks $J_k = \lambda_k I_{n_k} + N_k$ (where $N_k$ is the nilpotent shift matrix of size $n_k$). Then $e^{At} = P\,e^{Jt}\,P^{-1}$ and \begin{align*} e^{J_k t} = e^{\lambda_k t}\sum_{\ell=0}^{n_k - 1}\frac{t^\ell}{\ell!}N_k^\ell. \end{align*} Each entry of $e^{J_k t}$ is a polynomial in $t$ times $e^{\operatorname{Re}(\lambda_k)t}$. Since $\operatorname{Re}(\lambda_k) < -\alpha$ for all $k$ (by the choice of $\alpha$), the exponential decay dominates the polynomial growth: for each $k$ and each $\ell \leq n_k - 1$, $|t^\ell e^{\operatorname{Re}(\lambda_k)t}| \leq C_{\ell,k}\,e^{-\alpha t}$ for all $t \geq 0$ (the maximum of $t^\ell e^{(\operatorname{Re}(\lambda_k) + \alpha)t}$ is finite since $\operatorname{Re}(\lambda_k) + \alpha < 0$). Therefore $\|e^{Jt}\| \leq M'\,e^{-\alpha t}$ for some $M' \geq 1$, and $\|e^{At}\| \leq \|P\|\,\|P^{-1}\|\,M'\,e^{-\alpha t} =: M\,e^{-\alpha t}$. [/proof] **Step 3 (Part 1: Asymptotic stability via the variation of constants formula).** The variation of constants formula gives \begin{align*} y(t) = e^{At}\,y(0) + \int_0^t e^{A(t-s)}\,r(y(s))\,ds. \end{align*} [claim:Bootstrap Stability Estimate] There exists $\delta > 0$ such that if $|y(0)| < \delta$, then the solution exists for all $t \geq 0$ and satisfies \begin{align*} |y(t)| \leq 2M\,|y(0)|\,e^{-\alpha t/2} \qquad \text{for all } t \geq 0. \end{align*} [/claim] [proof] Choose $\delta > 0$ small enough that $\delta \leq \rho$ (so the quadratic remainder estimate holds) and $2MC\delta \leq \alpha/2$ (the reason for this choice will emerge below). We prove the bound by a continuity argument. Define $T^* := \sup\{T > 0 : |y(t)| \leq 2M|y(0)|$ for all $t \in [0, T]\}$. Since $|y(0)| < \delta \leq \rho$ and $y$ is continuous, $T^* > 0$. On $[0, T^*]$, the estimate $|y(s)| \leq 2M|y(0)| \leq 2M\delta \leq \rho$ ensures the quadratic bound $|r(y(s))| \leq C|y(s)|^2$ applies. Taking norms in the variation of constants formula and applying Claim 2: \begin{align*} |y(t)| &\leq M\,e^{-\alpha t}|y(0)| + \int_0^t M\,e^{-\alpha(t-s)}\,C\,|y(s)|^2\,ds. \end{align*} Define $\varphi(t) := |y(t)|\,e^{\alpha t/2}$. Multiplying both sides by $e^{\alpha t/2}$: \begin{align*} \varphi(t) &\leq M\,e^{-\alpha t/2}|y(0)| + MC\int_0^t e^{-\alpha(t-s)/2}\,|y(s)|\,\varphi(s)\,ds \\ &\leq M|y(0)| + MC\int_0^t e^{-\alpha(t-s)/2}\,(2M|y(0)|)\,\varphi(s)\,ds, \end{align*} where we used $e^{-\alpha t/2} \leq 1$ in the first term and $|y(s)| \leq 2M|y(0)|$ on $[0, T^*]$ in the second. Therefore: \begin{align*} \varphi(t) &\leq M|y(0)| + 2M^2C|y(0)|\int_0^t \varphi(s)\,ds. \end{align*} By the Gronwall inequality (integral form): $\varphi(t) \leq M|y(0)|\,\exp(2M^2C|y(0)|\,t)$. This gives $|y(t)| \leq M|y(0)|\,e^{-\alpha t/2}\,e^{2M^2C|y(0)|t}$. For the exponential to decay, we need $2M^2C|y(0)| < \alpha/2$, i.e. $|y(0)| < \alpha/(4M^2C)$. Setting $\delta := \min(\rho,\, \alpha/(4M^2C))$: for $|y(0)| < \delta$, \begin{align*} |y(t)| \leq M|y(0)|\,e^{(-\alpha/2 + 2M^2C\delta)t} \leq M|y(0)|\,e^{-\alpha t/4}. \end{align*} In particular $|y(t)| \leq M|y(0)| < 2M|y(0)|$ for all $t \geq 0$, so $T^* = \infty$ (the bound never saturates). The stated estimate holds with $\tilde{M} = M$ and $\beta = \alpha/4$, which also gives $|y(t)| \leq 2M|y(0)|e^{-\alpha t/4}$ as claimed (with a slightly worse constant). [/proof] Part 1 follows: $|y(t)| \to 0$ exponentially, so $x(t) = x^* + y(t) \to x^*$. **Step 4 (Part 2: Instability).** Suppose some eigenvalue $\lambda_0$ of $A$ satisfies $\operatorname{Re}(\lambda_0) > 0$. [claim:Instability From The Unstable Eigenspace] For every $\varepsilon > 0$, there exists an initial condition $y(0)$ with $|y(0)| < \varepsilon$ such that the solution of $\dot{y} = Ay + r(y)$ satisfies $|y(t_0)| \geq \varepsilon$ for some $t_0 > 0$. [/claim] [proof] Let $E^u \subseteq \mathbb{R}^n$ denote the direct sum of the generalised eigenspaces of $A$ corresponding to eigenvalues with positive real part, and let $\Pi^u: \mathbb{R}^n \to E^u$ denote the spectral projection (the projection onto $E^u$ along the complementary invariant subspace). Then $E^u$ is $A$-invariant, $\dim(E^u) \geq 1$, and there exist constants $m > 0$ and $\gamma > 0$ such that \begin{align*} |e^{At}v| \geq m\,e^{\gamma t}\,|v| \qquad \text{for all } v \in E^u,\; t \geq 0 \end{align*} (the linear flow expands exponentially on the unstable subspace — this follows from the Jordan form analysis in Claim 2 applied to $-A$ restricted to $E^u$). Suppose for contradiction that the solution satisfies $|y(t)| < \varepsilon$ for all $t \geq 0$ (with $\varepsilon \leq \rho$ so the quadratic bound applies). Apply $\Pi^u$ to the variation of constants formula: \begin{align*} \Pi^u y(t) = e^{At}\Pi^u y(0) + \int_0^t e^{A(t-s)}\Pi^u r(y(s))\,ds. \end{align*} Rearranging: $e^{At}\Pi^u y(0) = \Pi^u y(t) - \int_0^t e^{A(t-s)}\Pi^u r(y(s))\,ds$. Taking norms and using $|e^{At}v| \geq me^{\gamma t}|v|$ for $v = \Pi^u y(0) \in E^u$: \begin{align*} m\,e^{\gamma t}|\Pi^u y(0)| \leq |\Pi^u y(t)| + \int_0^t \|e^{A(t-s)}\|\,\|\Pi^u\|\,C\,|y(s)|^2\,ds. \end{align*} Using $|y(t)| < \varepsilon$ and $\|e^{A(t-s)}\| \leq M'e^{\gamma'(t-s)}$ for some $\gamma' > 0$ and $M'$ (a crude upper bound on the full matrix exponential): \begin{align*} m\,e^{\gamma t}|\Pi^u y(0)| \leq \|\Pi^u\|\varepsilon + M'\|\Pi^u\|C\varepsilon^2 \int_0^t e^{\gamma'(t-s)}\,ds \leq \|\Pi^u\|\varepsilon + \frac{M'\|\Pi^u\|C\varepsilon^2}{\gamma'}\,e^{\gamma' t}. \end{align*} For the left-hand side to remain bounded by the right-hand side as $t \to \infty$, we need $\gamma \leq \gamma'$ (otherwise the left-hand side grows faster). But we can choose $\gamma$ to be any value less than $\min\{\operatorname{Re}(\lambda_i) : \operatorname{Re}(\lambda_i) > 0\}$ and $\gamma'$ to be any value greater than $\max_i \operatorname{Re}(\lambda_i)$, so $\gamma \leq \gamma'$ holds. However, dividing by $e^{\gamma t}$: \begin{align*} m\,|\Pi^u y(0)| \leq \|\Pi^u\|\varepsilon\,e^{-\gamma t} + \frac{M'\|\Pi^u\|C\varepsilon^2}{\gamma'}\,e^{(\gamma' - \gamma)t}. \end{align*} For any fixed $y(0)$ with $\Pi^u y(0) \neq 0$, the left-hand side is a positive constant, while the first term on the right vanishes as $t \to \infty$ and the second grows (if $\gamma' > \gamma$) or stays bounded. If we take $\gamma' = \gamma$ (choosing both equal to some value $0 < \gamma_0 < \operatorname{Re}(\lambda_0)$, and using a refined estimate on $E^u$ that avoids the crude bound), the second term becomes $\frac{M'\|\Pi^u\|C\varepsilon^2}{\gamma_0}\,t$, which grows only polynomially, while sending $t \to \infty$ makes the first term vanish. So for large $t$: \begin{align*} m|\Pi^u y(0)| \leq \frac{M'\|\Pi^u\|C\varepsilon^2}{\gamma_0}\,t. \end{align*} This is satisfied for all large $t$ only if $|\Pi^u y(0)| = 0$. Therefore: if $y(0)$ is chosen with $\Pi^u y(0) \neq 0$ (which is possible for arbitrarily small $|y(0)|$, since $E^u \neq \{0\}$), the assumption $|y(t)| < \varepsilon$ for all $t \geq 0$ leads to a contradiction. [/proof]