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A hyperbolic equilibrium whose linearisation has an eigenvalue with positive real part is unstable

Statement

Let $n \geq 1$ , let $U \subseteq R^{n}$ be open with $0 \in U$ , and let $F : U \to R^{n}$ be of class $C^{1}$ ( $C^{k}$ Euclidean maps and diffeomorphisms) with $F (0) = 0$ . Put $A : = D F (0)$ (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case) and suppose that

no complex eigenvalue of $A$ (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism) has real part $0$ , and
at least one eigenvalue of $A$ has positive real part.

Then the equilibrium $0$ is not Lyapunov stable (Equilibria, Lyapunov stability and asymptotic stability of an autonomous first-order system): there is a real number $ε_{0} > 0$ such that for every $δ > 0$ there exist $z_{0} \in U$ with $∥ z_{0} ∥_{2} < δ$ and a time $t_{1} > 0$ such that the maximal solution $z$ of $z^{'} = F (z)$ , $z (0) = z_{0}$ is defined on $[0, t_{1}]$ and satisfies

$∥ z (t_{1}) ∥_{2} = ε_{0} .$

Facts & Assumptions

Given: The open set $U$ with $0 \in U$ , the $C^{1}$ field $F$ with $F (0) = 0$ , and the matrix $A = D F (0)$ with no purely imaginary eigenvalue and at least one eigenvalue of positive real part.

[F1]

The complexification of $A$ is $A_{C} : = {id}_{C} \otimes A$ on $C \otimes_{R} R^{n}$ (Complexification as $C \otimes_{R} V$ with its canonical real-linear embedding, Complexification of a real-linear map). In the standard basis write a complex vector as $w = p + i q$ with $p, q \in R^{n}$ and equip it with $∥ p + i q ∥_{2, C} : = (∥ p ∥_{2}^{2} + ∥ q ∥_{2}^{2})^{1 / 2}$ . The real and imaginary coordinate identification makes this a complete normed space ( $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ , $C$ is the real coordinate plane, with coordinate arithmetic), and the finite real matrix of $A_{C}$ is bounded by Cauchy--Schwarz (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation). If $λ$ is an eigenvalue of $A_{C}$ with $Re λ > 0$ , choose $w = p + i q \neq 0$ with $A_{C} w = λ w$ (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism). The exponential series gives $e^{t A_{C}} w = e^{t λ} w$ , and $∣ e^{t λ} ∣ = e^{t Re λ} \to 0$ as $t \to - \infty$ (The exponential series of a bounded operator, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ). Since $A$ is real, the real and imaginary parts of $e^{t A_{C}} w$ are $e^{t A} p$ and $e^{t A} q$ ; each tends to $0$ in the Euclidean norm. At least one of $p, q$ is nonzero. Exponential dichotomy gives $A$ -invariant subspaces $R^{n} = E_{s} \oplus E_{u}$ and constants $C \geq 1$ , $c > 0$ such that $∥ e^{t A} v ∥_{2} \leq C e^{- c t} ∥ v ∥_{2}$ for $v \in E_{s}$ , $t \geq 0$ , $∥ e^{t A} v ∥_{2} \leq C e^{c t} ∥ v ∥_{2}$ for $v \in E_{u}$ , $t \leq 0$ , and $∥ e^{t A} v ∥_{2} \geq C^{- 1} e^{c t} ∥ v ∥_{2}$ for $v \in E_{u}$ , $t \geq 0$ ; moreover $E_{s}, E_{u}$ are characterised by convergence to $0$ at $+ \infty$ and $- \infty$ respectively (Exponential dichotomy for a hyperbolic endomorphism: invariant stable and unstable subspaces with exponential estimates). Clause 4 places a nonzero real vector among $p, q$ that tends to $0$ as $t \to - \infty$ in $E_{u}$ . Thus $E_{u} \neq {0}$ .

[F2]

Differentiability at $0$ : for every $η > 0$ there is $ρ > 0$ with $∥ F (z) - A z ∥_{2} \leq η ∥ z ∥_{2}$ for $∥ z ∥_{2} < ρ$ (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder).

[F3]

Let $Π_{u}, Π_{s}$ be the projections onto $E_{u}, E_{s}$ along the decomposition; they are linear and satisfy $Π_{u} A = A Π_{u}$ , $Π_{s} A = A Π_{s}$ , $Π_{u} + Π_{s} = I$ . For a linear map $L$ between these Euclidean spaces, write $∥ L ∥_{2} : = \sup {∥ L v ∥_{2} : ∥ v ∥_{2} \leq 1}$ for its induced operator norm. In orthonormal coordinates every linear map has a finite matrix, and Cauchy--Schwarz bounds its operator norm by the square root of the sum of squares of its entries; hence $K : = \max (∥ Π_{u} ∥_{2}, ∥ Π_{s} ∥_{2}) < \infty$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

[F4]

Improper integrals For $a > 0$ , direct integration using The exponential function is smooth and $(\exp)^{'} = \exp$ gives $\int_{0}^{T} e^{- a t} d t = (1 - e^{- a T}) / a \to 1 / a$ by The exponential tends to $+ \infty$ at $+ \infty$ and to $0$ at $- \infty$ . Improper integrals of continuous functions with exponential decay converge absolutely, by comparison (Improper integrals over unbounded intervals, Comparison tests for improper integrals); Cauchy--Schwarz bounds Euclidean inner products by the product of norms (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

[F5]

Compact continuation: an autonomous solution trapped in a compact subset of its open state domain near a finite endpoint extends past that endpoint; $F$ is locally Lipschitz because it is $C^{1}$ (Compact continuation for locally Lipschitz ODEs: a solution whose graph stays in a compact subset of the domain near a finite endpoint extends past it). Closed Euclidean balls are compact because they are closed and bounded (Open ball, closed ball and sphere in a metric space, Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[F6]

If $Q$ is a symmetric positive definite endomorphism of a finite-dimensional real inner product space, the real spectral theorem gives an orthonormal eigenbasis with positive eigenvalues; writing $λ_{\max} (Q)$ for the largest eigenvalue yields $⟨ Q v, v ⟩ \leq λ_{\max} (Q) ∥ v ∥_{2}^{2}$ (Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis).

Proof

technique · direct; build an indefinite quadratic Lyapunov function that grows in the unstable directions and use exponential dichotomy estimates

1.1F1F4algebra

Choose a nonzero $u_{0} \in E_{u}$ , which is possible because $E_{u} \neq {0}$ by [F1]. On $E_{u}$ define $Q_{u} : = \int_{0}^{\infty} e^{- t A_{u}^{T}} e^{- t A_{u}} d t, A_{u} : = A ∣_{E_{u}},$ and on $E_{s}$ define $Q_{s} : = \int_{0}^{\infty} e^{t A_{s}^{T}} e^{t A_{s}} d t$ , $A_{s} : = A ∣_{E_{s}}$ ; the transposes are with respect to the inherited Euclidean inner products. If $E_{s} = {0}$ , $Q_{s}$ is the zero endomorphism of that space and its positivity assertion is vacuous. By [F1], $∥ e^{- t A_{u}} v ∥_{2} \leq C e^{- c t} ∥ v ∥_{2}$ for $v \in E_{u}$ , $t \geq 0$ , and $∥ e^{t A_{s}} v ∥_{2} \leq C e^{- c t} ∥ v ∥_{2}$ for $v \in E_{s}$ , $t \geq 0$ . Cauchy--Schwarz [F4] bounds each matrix entry of both integrands by $C^{2} e^{- 2 c t}$ , so [F4] gives absolute convergence entrywise. For nonzero $v$ in either nonzero subspace, the corresponding quadratic form equals the integral of $∥ e^{- t A_{u}} v ∥_{2}^{2}$ or $∥ e^{t A_{s}} v ∥_{2}^{2}$ ; the integrand is continuous and positive at $t = 0$ , so $Q_{u}$ and $Q_{s}$ are positive definite on their subspaces. For $Q_{u, T} : = \int_{0}^{T} e^{- t A_{u}^{T}} e^{- t A_{u}} d t$ , differentiating the integrand and applying the fundamental theorem gives $A_{u}^{T} Q_{u, T} + Q_{u, T} A_{u} = I - e^{- T A_{u}^{T}} e^{- T A_{u}}$ ; the terminal term tends to $0$ by the decay estimate, so $A_{u}^{T} Q_{u} + Q_{u} A_{u} = I_{E_{u}}$ . The same calculation with $Q_{s, T} : = \int_{0}^{T} e^{t A_{s}^{T}} e^{t A_{s}} d t$ gives $A_{s}^{T} Q_{s} + Q_{s} A_{s} = - I_{E_{s}}$ after taking $T \to \infty$ .

2.1step 1.1F2F3F4algebra

Define the function $V$ on $U$ by $V (z) : = ⟨ Q_{u} Π_{u} z, Π_{u} z ⟩_{2} - ⟨ Q_{s} Π_{s} z, Π_{s} z ⟩_{2}$ , writing $u = Π_{u} z$ , $s = Π_{s} z$ . It is a quadratic form, hence differentiable, and for a solution $z$ of $z^{'} = F (z)$ with $z (t) = u (t) + s (t)$ one has $u^{'} = A_{u} u + Π_{u} r (z)$ , $s^{'} = A_{s} s + Π_{s} r (z)$ where $r (z) : = F (z) - A z$ , because $Π_{u}, Π_{s}$ commute with $A$ [F3]. Differentiating along the solution, $\frac{d}{d t} V (z (t)) = ∥ u ∥_{2}^{2} + ∥ s ∥_{2}^{2} + 2 ⟨ Q_{u} u, Π_{u} r (z) ⟩_{2} - 2 ⟨ Q_{s} s, Π_{s} r (z) ⟩_{2} \geq ∥ u ∥_{2}^{2} + ∥ s ∥_{2}^{2} - 2 M ∥ r (z) ∥_{2} ∥ z ∥_{2},$ with $M : = ∥ Q_{u} ∥_{2} ∥ Π_{u} ∥_{2}^{2} + ∥ Q_{s} ∥_{2} ∥ Π_{s} ∥_{2}^{2} > 0$ : the two scalar-product terms are estimated by Cauchy--Schwarz and $∥ u ∥_{2} \leq ∥ Π_{u} ∥_{2} ∥ z ∥_{2}$ , $∥ s ∥_{2} \leq ∥ Π_{s} ∥_{2} ∥ z ∥_{2}$ ; no orthogonality of the splitting is assumed, and $∥ z ∥_{2} = ∥ u + s ∥_{2} \leq ∥ u ∥_{2} + ∥ s ∥_{2}$ gives $∥ z ∥_{2}^{2} \leq 2 (∥ u ∥_{2}^{2} + ∥ s ∥_{2}^{2})$ .

3.1step 1.12.1F2F3F6algebra

Choose $η > 0$ with $8 M η \leq 1$ and let $ρ > 0$ be as in [F2] for this $η$ ; shrink $ρ$ so that $\overline{B} (0, ρ) \subseteq U$ . For $∥ z ∥_{2} < ρ$ , step 2.1 and [F2] give, since $∥ z ∥_{2}^{2} \leq 2 (∥ u ∥_{2}^{2} + ∥ s ∥_{2}^{2})$ and $8 M η \leq 1$ , $\frac{d}{d t} V (z (t)) \geq ∥ u ∥_{2}^{2} + ∥ s ∥_{2}^{2} - 2 M η ∥ z ∥_{2}^{2} \geq \frac{1}{2} (∥ u ∥_{2}^{2} + ∥ s ∥_{2}^{2}) \geq 0.$ In particular, whenever $V (z) > 0$ then $u \neq 0$ and $∥ u ∥_{2}^{2} \geq V (z) / λ_{\max} (Q_{u})$ by [F6], so $\frac{d}{d t} V (z (t)) \geq \frac{1}{2} ∥ u ∥_{2}^{2} \geq γ V (z (t)), γ : = \frac{1}{2 λ_{\max} (Q_{u})} > 0.$ Set $ε_{0} : = ρ$ .

4.1step 1.13.1

Let $0 < δ$ be arbitrary and choose $z_{0} : = u_{0} \in E_{u}$ with $0 < ∥ u_{0} ∥_{2} < \min (δ, ε_{0})$ (step 1.1). Then $V (z_{0}) = ⟨ Q_{u} u_{0}, u_{0} ⟩_{2} > 0$ . Let $z : [0, b) \to U$ be the maximal forward solution through $z_{0}$ . As long as $∥ z (t) ∥_{2} < ε_{0}$ , step 3.1 gives $\frac{d}{d t} V (z (t)) \geq 0$ whenever $V (z (t)) \geq 0$ ; since $V (z (0)) > 0$ , the function $V \circ z$ stays positive and satisfies $V (z (t)) \geq e^{γ t} V (z_{0})$ on that interval.

5.1step 4.1F5F6algebra∎

The solution must reach the sphere $∥ z ∥_{2} = ε_{0}$ at some finite time. Suppose first that the solution is defined for all $t \geq 0$ and never reaches that sphere. Then $∥ z (t) ∥_{2} < ε_{0}$ for all $t$ , so step 4.1 gives $V (z (t)) \geq e^{γ t} V (z_{0}) \to + \infty$ , while $V (z) \leq λ_{\max} (Q_{u}) ∥ Π_{u} ∥_{2}^{2} ε_{0}^{2}$ on $\overline{B} (0, ε_{0})$ by [F6]; this is a contradiction. Suppose next that the maximal forward interval is $[0, b)$ with $b < \infty$ and the solution never reaches the sphere. Then its values lie in the compact set $\overline{B} (0, ε_{0}) \subseteq U$ near the endpoint $b$ by [F5], so [F5] extends the solution past $b$ , contradicting maximality. Hence in every case there is a first time $t_{1} \in (0, \infty)$ with $∥ z (t_{1}) ∥_{2} = ε_{0}$ , and the solution is defined on $[0, t_{1}]$ . Since $δ > 0$ was arbitrary, the equilibrium $0$ is not Lyapunov stable in the sense of Equilibria, Lyapunov stability and asymptotic stability of an autonomous first-order system.

Remarks

The proof uses the dichotomy hypothesis to obtain the invariant splitting with an exponential gap. If $A$ has an eigenvalue with positive real part and also a purely imaginary eigenvalue, the conclusion remains true in the classical linearisation theorem, but a different argument is needed (the centre directions require separate treatment); that case is not covered here.
The indefinite quadratic form $V$ is unbounded above, so $V$ growing is not by itself a contradiction; the contradiction in step 5.1 comes from $V$ growing while the solution stays in the bounded ball. The exit time $t_{1}$ depends on $z_{0}$ and $δ$ ; the theorem asserts instability, not a uniform escape rate for all small data.

A hyperbolic equilibrium whose linearisation has an eigenvalue with positive real part is unstable

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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