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Finite-slab dependence need not have a uniform exponential constant

Statement

For the scalar symmetric-hyperbolic system on $T_{2 π}^{3}$ $\partial_{t} z - 2 t z = 0, A^{0} = 1, A^{1} = A^{2} = A^{3} = 0, B (t, x, z) = - 2 t z,$ the spatially constant solution with initial value $z (0, x) = δ$ is $z_{δ} (t, x) = δ \exp (t^{2})$ . For every fixed finite slab $[0, T]$ , this solution depends continuously on $δ$ and has a finite Sobolev bound. But no constants $C > 0$ independent of $T$ can give $∥ z_{δ} (T) - z_{0} (T) ∥_{H^{s - 1}} \leq C e^{C T} ∥ z_{δ} (0) - z_{0} (0) ∥_{H^{s - 1}}$ for all $T > 0$ and all $δ \neq 0$ . For each fixed $T$ the state and time-derivative ranges are bounded, but the compact bound may depend on $T$ . This is a mathematical symmetric-hyperbolic illustration of the horizon qualification; the system is not an Einstein reduction and the example asserts no behavior of an Einstein solution.

Statement references

Finite-slab stability in a fixed hyperbolic gauge

Continuity of finite-domain geometric observables

Periodic first-order symmetric hyperbolic systems: coefficients, admissible values, data classes and solutions

The real exponential function and the number $e$ by a power series

The exponential function is smooth and $(\exp)^{'} = \exp$

The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$

The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$

Canonical naturals are positive and strictly increasing

Facts & Assumptions

Given: A dimensionless time parameter $t \geq 0$ , an integer $s \geq 5$ , a fixed finite $T > 0$ , and a real initial amplitude $δ$ .

[F1]

The symmetric-hyperbolic system is $A^{0} \partial_{t} z + \sum_{i} A^{i} \partial_{i} z + B = 0$ with symmetric matrices and positive definite $A^{0}$ (Periodic first-order symmetric hyperbolic systems: coefficients, admissible values, data classes and solutions).

[F2]

The real exponential is defined by its power series. Every factorial is a nonzero natural, and positive canonical naturals have positive real image; thus for $y \geq 0$ each series term is nonnegative and $\exp (y) \geq 1 + y > 0$ (The real exponential function and the number $e$ by a power series, The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ , Canonical naturals are positive and strictly increasing).

[F3]

The real exponential is smooth with derivative $(\exp)^{'} = \exp$ (The exponential function is smooth and $(\exp)^{'} = \exp$ ).

[F4]

The exponential satisfies $\exp (x + y) = \exp (x) \exp (y)$ (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ ).

[F5]

The finite-slab estimate permits its constant to depend on the selected compact slab and coefficient bounds (Finite-slab stability in a fixed hyperbolic gauge).

[F6]

The finite-domain geometric continuity conclusion is stated on a fixed slab and asserts no uniformity as $T \to \infty$ (Continuity of finite-domain geometric observables).

Proof technique: explicit solution and growth comparison.

Proof

1.1givenF1algebra

The coefficients in the Statement are smooth in $(t, x, z)$ , each $A^{μ}$ is symmetric, and $A^{0} = 1$ is strictly positive. Thus [F1] makes this a symmetric-hyperbolic system. Its equation reduces to $\partial_{t} z = 2 t z$ and contains no spatial derivatives.

2.1givenstep 1.1F3F5algebra

By [F3] and the chain rule, $z_{δ} (t, x) = δ \exp (t^{2})$ satisfies $\partial_{t} z_{δ} = 2 t δ \exp (t^{2}) = 2 t z_{δ}$ and $z_{δ} (0, x) = δ$ . The zero solution corresponds to $δ = 0$ . For each finite $T$ , the function $δ \exp (t^{2})$ and its time derivative are bounded on $[0, T]$ ; its values lie between the endpoint values $δ$ and $δ \exp (T^{2})$ , hence in a compact interval. Since it is spatially constant, it has finite $H^{s}$ norm for every integer $s$ . Its state and time-derivative bounds may depend on the selected finite $T$ , as permitted by the finite-slab estimate [F5].

3.1step 2.1F2F4algebra

The difference from zero is exactly $δ \exp (t^{2})$ . Homogeneity of the Sobolev norm on the spatially constant profile gives $\frac{∥ z_{δ} (T) - z_{0} (T) ∥_{H^{s - 1}}}{∥ z_{δ} (0) - z_{0} (0) ∥_{H^{s - 1}}} = \frac{∣ δ ∣ \exp (T^{2}) ∥ 1 ∥_{H^{s - 1}}}{∣ δ ∣ ∥ 1 ∥_{H^{s - 1}}} = \exp (T^{2}) .$ Suppose one constant $C > 0$ worked for all slab lengths in the displayed exponential estimate. Choose $T$ large enough that $y : = T^{2} - C T > C$ . Then $y > 0$ , and [F2] gives $\exp (y) \geq 1 + y > C$ . By [F4], $\exp (T^{2}) = \exp (C T + y) = \exp (C T) \exp (y) > C \exp (C T),$ since [F2] also gives $\exp (C T) > 0$ . This contradicts that uniform estimate.

4.1step 3.1F5F6algebra∎

For each fixed $T$ , the exact solution map is multiplication by $\exp (t^{2})$ and is Lipschitz on $[0, T]$ with finite constant $\exp (T^{2})$ ; this is consistent with [F5] and [F6], whose constants and conclusions are attached to a fixed finite slab. The calculation establishes only that finite-slab continuity alone does not supply a $T$ -independent exponential constant for general symmetric-hyperbolic systems.

Finite-slab dependence need not have a uniform exponential constant

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources