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Bootstrap closure and nonlinear stability boundaries — model 2

Example

Let $E : [0, T_{\max}) \to [0, \infty)$ be continuous, let $E_{0} \geq 0$ , $C > 0$ and $ϵ > 0$ be dimensionless, and suppose for every finite $t < T_{\max}$ that

$E (t) \leq E_{0} + C ϵ \int_{0}^{t} (1 + s)^{- 1} E (s) d s .$

Use the dimensionless thresholds $A, ϵ, η$ in the cited continuation contract, with $2 A ϵ^{2} < η$ . Gronwall gives the polynomial upper bound $E (t) \leq E_{0} (1 + t)^{C ϵ}$ . This inequality alone does not supply a uniform improvement: with $C ϵ = 1$ , $E_{0} = 1$ and $E (t) = 1 + t$ , equality holds. Let $B = A ϵ^{2} > 1$ for the thresholds in the continuation contract below; then $2 B < η$ . On the finite slab $0 \leq t \leq B$ this example has $E (0) < B$ , $E (t) \leq 2 B$ , but $E (B) > B$ . Thus the bootstrap implication from $2 B$ to $B$ fails on that slab.

Example references

A quantitative continuation and bootstrap contract

Open-closed bootstrap continuation

Gronwall's integral inequality with variable and constant coefficients

Real powers for positive bases, with the zero-base positive-exponent convention

The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$

The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t

Metivier, Chapter 3, §§3.3–3.5

Wang, §6.3

Klainerman and Szeftel, abstract and §1.2.1

Jaramillo, Valiente Kroon and Gourgoulhon, §§1.2.1 and 2.2

Facts & Assumptions

Given: The inequality in the Example on every compact interval; $E_{0} \geq 0$ ; $C, ϵ > 0$ ; and dimensionless time $t \geq 0$ .

[F1]

Gronwall's integral inequality with variable and constant coefficients: if continuous $u, a, b$ on $[0, T]$ satisfy $u, b \geq 0$ and $u (t) \leq a (t) + \int_{0}^{t} b (s) u (s) d s$ , then for nondecreasing $a$ , $u (t) \leq a (t) \exp (\int_{0}^{t} b)$ .

[F2]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: $\log^{'} (x) = 1 / x$ for $x > 0$ and $\log x = \int_{1}^{x} d u / u$ .

[F3]

Real powers for positive bases, with the zero-base positive-exponent convention: for positive $x$ , $x^{α} = \exp (α \log x)$ .

[F4]

[F5]

The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ : for differentiable $G$ on a compact interval with integrable derivative, $\int_{a}^{b} G^{'} (s) d s = G (b) - G (a)$ .

[F6]

A quantitative continuation and bootstrap contract: the bootstrap input and strict output are $E \leq 2 A ϵ^{2}$ and $E \leq A ϵ^{2}$ , respectively, with $2 A ϵ^{2} < η$ .

[F7]

Open-closed bootstrap continuation: the required contract implication must hold on every compact subinterval before the continuation conclusion applies.

Verification

1.1givenF1F2F4F5algebra

The case $t = 0$ is immediate. For finite $t > 0$ , set $u = E$ , $a (s) = E_{0}$ and $b (s) = C ϵ (1 + s)^{- 1}$ on $[0, t]$ . The given inequality holds there; $u$ is continuous and nonnegative, $a$ is a nonnegative constant and hence nondecreasing, and $b$ is continuous and nonnegative, so [F1] applies and gives $E (t) \leq E_{0} \exp ⁣ (C ϵ \int_{0}^{t} (1 + s)^{- 1} d s)$ . Since $1 + s > 0$ , [F2] and [F4] give $G^{'} (s) = (1 + s)^{- 1}$ for $G (s) = \log (1 + s)$ ; this derivative is continuous. By [F5], $\int_{0}^{t} (1 + s)^{- 1} d s = G (t) - G (0) = \log (1 + t)$ .

1.2givenF5algebra

For the stated boundary example take $T_{\max} = \infty$ , set $C ϵ = 1$ , $E_{0} = 1$ and $E (t) = 1 + t$ . Then $C ϵ (1 + s)^{- 1} E (s) = 1$ . The function $H (s) = s$ has continuous, hence integrable, derivative $1$ , so [F5] gives $\int_{0}^{t} 1 d s = H (t) - H (0) = t$ for $t > 0$ ; for $t = 0$ , both sides equal $1$ . Thus $E_{0} + C ϵ \int_{0}^{t} (1 + s)^{- 1} E (s) d s = 1 + t = E (t)$ .

2.1step 1.1F3algebra

By [F3], $\exp (C ϵ \log (1 + t)) = (1 + t)^{C ϵ}$ , which proves the stated polynomial upper bound on every finite subinterval.

3.1givenF6F7step 1.2algebra∎

Put $T = B$ . Because $B > 1$ , for $0 \leq t \leq B$ we have $E (0) = 1 < B$ and $E (t) \leq 1 + B \leq 2 B$ , but $E (B) = 1 + B > B$ . Thus the contract's input condition holds on $[0, B]$ while its claimed improvement fails there; [F7] cannot be applied from this integral inequality alone.

Bootstrap closure and nonlinear stability boundaries — model 2

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources