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A non-hyperbolic equilibrium with no positive eigenvalue can still be unstable: x'=x^3

Statement refuted

If $F : R \to R$ is $C^{1}$ with $F (0) = 0$ and no eigenvalue of $A = D F (0)$ has positive real part, then $0$ is a Lyapunov stable equilibrium of $x^{'} = F (x)$ .

Facts & Assumptions

Given: The function $F : R \to R$ , $F (x) = x^{3}$ , and its initial value problem $x^{'} = x^{3}$ .

[L1]

The equilibrium notion and Lyapunov stability of an equilibrium of $z^{'} = F (z)$ are defined in The local flow of an autonomous first-order system-based terms by Equilibria, Lyapunov stability and asymptotic stability of an autonomous first-order system: $0$ is stable when for every $ε > 0$ there is $δ > 0$ such that every maximal solution with $∣ x (0) ∣ < δ$ is defined for all $t \geq 0$ and satisfies $∣ x (t) ∣ < ε$ for all $t \geq 0$ .

[L2]

Every initial value problem satisfying the Picard-Lindelöf hypotheses has a unique maximal solution, and every solution through the same data is its restriction (Every Picard–Lindelöf initial value problem has one maximal solution on an open interval, Extensions, maximal solutions, maximal intervals, and global solutions, First-order systems, initial value problems, and solutions on intervals).

[L3]

For real $α$ , $u^{α}$ on $u > 0$ is continuous and has derivative $α u^{α - 1}$ (Continuity and derivatives of positive-base real powers). The polynomial $x \mapsto x^{3}$ is $C^{1}$ with derivative $3 x^{2}$ , so $A = D F (0) = 0$ ( $C^{k}$ Euclidean maps and diffeomorphisms, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ); the eigenvalue $0$ of $A$ has real part $0$ , and no eigenvalue has positive real part.

Counterexample

technique · direct; exhibit the explicit blowing-up solution through each small positive initial value

1.1givenL3

$F (x) = x^{3}$ satisfies $F (0) = 0$ and $D F (0) = 0$ by [L3], so the hypothesis of the refuted statement is satisfied with $A = 0$ .

1.2givenL3algebra

Let $x_{0} > 0$ and define $x (t) : = x_{0} (1 - 2 x_{0}^{2} t)^{- 1 / 2}$ on $J : = [0, 1 / (2 x_{0}^{2}))$ . Differentiating the power using the chain rule and the algebra of derivatives, $x^{'} (t) = x_{0} \cdot (- \frac{1}{2}) (1 - 2 x_{0}^{2} t)^{- 3 / 2} \cdot (- 2 x_{0}^{2}) = x_{0}^{3} (1 - 2 x_{0}^{2} t)^{- 3 / 2} = x (t)^{3}$ for $t \in J$ , and $x (0) = x_{0}$ ; hence $x$ is a solution on $J$ of the initial value problem.

2.1step 1.2L2

By [L2] the solution of step 1.2 is the restriction of the unique maximal solution through $(0, x_{0})$ . As $t ↑ 1 / (2 x_{0}^{2})$ the quantity $1 - 2 x_{0}^{2} t$ tends to $0$ from above, so $x (t) \to + \infty$ ; in particular the maximal solution is not defined at $t = 1 / (2 x_{0}^{2})$ and therefore is not defined for all $t \geq 0$ .

3.1step 1.12.1L1algebra∎

Instability: given any $δ > 0$ , choose $x_{0} \in (0, δ)$ . The maximal solution through $x_{0}$ is not defined for all $t \geq 0$ by step 2.1, so the defining condition of stability in [L1] fails for every $ε > 0$ : there is no $δ$ that keeps all maximal solutions with $∣ x (0) ∣ < δ$ defined for all time. Hence $0$ is not Lyapunov stable, refuting the statement.

Remarks

The example shows that the hypothesis "no eigenvalue with positive real part" alone cannot replace hyperbolicity in An equilibrium whose linearisation has all eigenvalues in the open left half-plane is locally exponentially stable or A hyperbolic equilibrium whose linearisation has an eigenvalue with positive real part is unstable, both of which assume a spectral gap at the imaginary axis. Here the derivative vanishes identically at the equilibrium, so the linearisation carries no information about the nonlinear behaviour.
With the opposite sign, $x^{'} = - x^{3}$ has $D F (0) = 0$ and the solutions $x (t) = x_{0} (1 + 2 x_{0}^{2} t)^{- 1 / 2}$ are defined for all $t \geq 0$ and decay to $0$ like $t^{- 1 / 2}$ ; thus an equilibrium with zero linear part can also be asymptotically stable. The decay is not exponential, so even asymptotic stability does not upgrade to the exponential conclusion of the hyperbolic theorem; no statement about any physical system is made by this example.

A non-hyperbolic equilibrium with no positive eigenvalue can still be unstable: x'=x^3

Statement refuted

Facts & Assumptions

Counterexample

Remarks

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