An equilibrium whose linearisation has all eigenvalues in the open left half-plane is locally exponentially stable

Define the symmetric matrix $Q$ entrywise by $Q_{ij}:={\int }_0^{\mathrm{\infty }}(e^{t{A}^{\mathsf{T}}}{e}^{tA}{)}_{ij}dt\mathrm{.}$ Each integrand is continuous and, for standard basis vectors $e_i,e_j$, equals $⟨e^{tA}{e}_i,e^{tA}{e}_j{⟩}_2$; Cauchy--Schwarz and [F1], [F5] bound its absolute value by $C^2{e}^{-2ct}$. Hence every entry integral converges absolutely by [F3]. The resulting matrix is symmetric because $e^{t{A}^{\mathsf{T}}}{e}^{tA}$ is symmetric for every $t$.

The matrix $Q$ is positive definite and satisfies the Lyapunov equation $A^{\mathsf{T}}Q+QA=-I_n$. Positivity: for $z\mathrm{\ne }0$, $z^{\mathsf{T}}Qz={\int }_0^{\mathrm{\infty }}\mathrm{\parallel }{e}^{tA}z{\mathrm{\parallel }}_2^{2}dt$, the integrand is continuous, nonnegative, and strictly positive on a neighbourhood of $t=0$, so the integral is $>0$; for $z=0$ it is $0$. Lyapunov equation: put $Q_T:={\int }_0^T{e}^{t{A}^{\mathsf{T}}}{e}^{tA}dt$. By [F3] and linearity of the integral on compact intervals, $A^{\mathsf{T}}{Q}_T+Q_{T}A={\int }_0^T\frac{d}{dt}\left(e^{t{A}^{\mathsf{T}}}{e}^{tA}\right)dt=e^{T{A}^{\mathsf{T}}}{e}^{TA}-I_n$, by the product rule and $\frac{d}{dt}{e}^{tA}=A{e}^{tA}=e^{tA}A$ from [F1]. Letting $T\to \mathrm{\infty }$, the left side tends to $A^{\mathsf{T}}Q+QA$ by [F3]. Each entry of $e^{T{A}^{\mathsf{T}}}{e}^{TA}$ equals $⟨e^{TA}{e}_i,e^{TA}{e}_j{⟩}_2$ for standard basis vectors $e_i,e_j$, so [F1], [F5] bound it in absolute value by $C^2{e}^{-2cT}\to 0$. Hence $A^{\mathsf{T}}Q+QA=-I_n$.

Fix $\eta >0$ with $2\mathrm{\parallel }Q{\mathrm{\parallel }}_2\eta \le \frac{1}{2}$ and let $\rho >0$ be a radius as in [F2] for this $\eta$; since $U$ is open and $0\in U$, shrink $\rho$ if necessary so that $\bar{B}(0,\rho )\subseteq U$ (the implication of [F2] persists under shrinking $\rho$; balls as in Open ball, closed ball and sphere in a metric space). Define $V(z):=z^{\mathsf{T}}Qz$; since $Q$ is positive definite with smallest eigenvalue ${\lambda }_{\min }(Q)>0$, choose $0<\alpha <{\lambda }_{\min }(Q){\rho }^2$. Then $\mathcal{S}:=\{z\in U:V(z)\le \alpha \}$ satisfies $\mathcal{S}\subseteq B(0,\rho )\subseteq \bar{B}(0,\rho )\subseteq U$, so $\mathcal{S}=\{z\in \bar{B}(0,\rho ):V(z)\le \alpha \}$ is closed and bounded, hence compact by [F6].

Let $z$ be a solution of $z^{\mathrm{\prime }}=F(z)$. On the interval where $\mathrm{\parallel }z(t){\mathrm{\parallel }}_2<\rho$, $\frac{d}{dt}V(z(t))=2z(t{)}^{\mathsf{T}}Q{z}^{\mathrm{\prime }}(t)=z^{\mathsf{T}}(A^{\mathsf{T}}Q+QA)z+2z^{\mathsf{T}}Q(F(z)-Az)\le -\mathrm{\parallel }z{\mathrm{\parallel }}_2^2+2\eta \mathrm{\parallel }Q{\mathrm{\parallel }}_2\mathrm{\parallel }z{\mathrm{\parallel }}_2^2\le -\frac{1}{2}\mathrm{\parallel }z{\mathrm{\parallel }}_2^2,$ by step 2.1, [F2] and the choice of $\eta$; the derivative exists by the chain rule (The chain rule for total derivatives: $D(g\circ f)(a)=Dg(f(a))\circ Df(a)$, [F5]). Since $V(z)\le {\lambda }_{\max }(Q)\mathrm{\parallel }z{\mathrm{\parallel }}_2^2$, this gives $\frac{d}{dt}V(z(t))\le -\frac{1}{2{\lambda }_{\max }(Q)}V(z(t))$ while the solution stays in $B(0,\rho )$.

A maximal solution with $z(0)\in \mathcal{S}$ stays in $\mathcal{S}$ and is defined for all $t\ge 0$. Indeed, while it stays in $\mathcal{S}\subseteq B(0,\rho )$, step 4.1 shows that $t\mapsto V(z(t))$ is nonincreasing, so it cannot leave $\mathcal{S}$ through the level $V=\alpha$; it cannot leave $\mathcal{S}$ through $\mathrm{\partial }U$ either, because $\mathcal{S}$ is compact in $U$. If its maximal forward interval were $[0,b)$ with $b<\mathrm{\infty }$, the solution would be trapped in the compact set $\mathcal{S}\subseteq U$ near $b$, so [F4] would extend it, a contradiction. Hence it is defined for all $t\ge 0$ and stays in $\mathcal{S}$; this positive invariance of $\mathcal{S}$ is the property used to verify the stability definition of Equilibria, Lyapunov stability and asymptotic stability of an autonomous first-order system.

Exponential decay: by step 4.1, $\frac{d}{dt}V(z(t))\le -{\gamma }_{V}V(z(t))$ for all $t\ge 0$ whenever the solution stays in $\mathcal{S}\subseteq B(0,\rho )$, with ${\gamma }_V:=\frac{1}{2{\lambda }_{\max }(Q)}>0$. A function with nonpositive derivative of $t\mapsto e^{{\gamma }_{V}t}V(z(t))$ is nonincreasing, so $V(z(t))\le e^{-{\gamma }_{V}t}V(z(0))$. Combining with ${\lambda }_{\min }(Q)\mathrm{\parallel }z{\mathrm{\parallel }}_2^2\le V(z)\le {\lambda }_{\max }(Q)\mathrm{\parallel }z{\mathrm{\parallel }}_2^2$ gives $\mathrm{\parallel }z(t){\mathrm{\parallel }}_2\le \beta e^{-{\gamma }_{V}t\mathrm{/}2}\mathrm{\parallel }z(0){\mathrm{\parallel }}_2,\beta :=\sqrt{{\lambda }_{\max }(Q)\mathrm{/}{\lambda }_{\min }(Q)}\ge 1,$ for solutions with $z(0)\in \mathcal{S}$. Choose $\delta >0$ with $B(0,\delta )\subseteq \mathcal{S}$; then every maximal solution with $\mathrm{\parallel }z(0){\mathrm{\parallel }}_2<\delta$ satisfies the displayed estimate. With $\gamma :={\gamma }_V\mathrm{/}2>0$ this proves the exponential bound, and it implies Lyapunov stability and asymptotic stability directly: given $\epsilon >0$ small, choosing initial data with $\mathrm{\parallel }z(0){\mathrm{\parallel }}_2<{\delta }^{\mathrm{\prime }}$ where ${\delta }^{\mathrm{\prime }}\le \delta$ and $\beta {\delta }^{\mathrm{\prime }}<\epsilon$ keeps $\mathrm{\parallel }z(t){\mathrm{\parallel }}_2<\epsilon$ for all $t\ge 0$, and the bound gives convergence to $0$.