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An exactly solvable adiabatic branch: $K = I + ε I \cos θ$ and comparison with the drift bound

Example

Take $J = R$ , $S = R$ , $h (I, s) = I$ and $k (I, θ, s) = I \cos θ$ in the datum of A one-frequency parameter-dependent action-angle branch, its transformed Hamiltonian and its drift generator, with $ε = \frac{1}{10}$ , $A = 2$ , $ω_{0} = 1$ , $K = [6 / 5, 19 / 10]$ , $S_{0} = [0, 2]$ , $s_{0} = 0$ and initial datum $I (0) = 3 / 2$ , $θ (0) = 0$ . The branch system is

$I^{'} = ε I \sin θ, θ^{'} = 1 + ε \cos θ, s^{'} = ε,$

so $s (t) = ε t$ stays in $S_{0} = [0, 2]$ on $[0, A / ε] = [0, 20]$ , and $I (t)$ stays in $K$ for all $t \geq 0$ . The quantity $J : = I (1 + ε \cos θ)$ is exactly invariant, the solution stays in $K \times S_{0}$ on $[0, A / ε]$ , and

$∣ I (t) - I (0) ∣ \leq \frac{1}{3} (t \geq 0),$

while the drift bound of Adiabatic action drift bound: order- $ε$ change of the action on the $O (1 / ε)$ time scale evaluates in this instance to $∣ I (t) - I (0) ∣ \leq \frac{19}{50}$ .

Facts & Assumptions

Given: The datum $h (I, s) = I$ , $k (I, θ, s) = I \cos θ$ , $ε = 1 / 10$ , and the sets and data above.

[L1]

The branch vector field is $I^{'} = - ε \partial_{θ} k$ , $θ^{'} = \partial_{I} h + ε \partial_{I} k$ , $s^{'} = ε$ ; here $\partial_{I} h = 1$ , $\partial_{θ} k = - I \sin θ$ and $\partial_{I} k = \cos θ$ , so the field is as displayed (A one-frequency parameter-dependent action-angle branch, its transformed Hamiltonian and its drift generator). [given]

[L2]

The drift bound (Adiabatic action drift bound: order- $ε$ change of the action on the $O (1 / ε)$ time scale): if $∣ \partial_{I} h ∣ \geq ω_{0}$ and the branch solution stays in the compact sets $K \times S_{0}$ , then $∣ I (t) - I (0) ∣ \leq ε [2 \sup_{K \times [0, 2 π] \times S_{0}} ∣ g ∣ + A \sup_{K \times [0, 2 π] \times S_{0}} ∣ \partial_{I} g \partial_{θ} k - \partial_{θ} g \partial_{I} k - \partial_{s} g ∣],$ where $g = (k - ⟨ k ⟩) / \partial_{I} h$ and $⟨ k ⟩$ is the $θ$ -average of $k$ . [given]

[L3]

$(\sin θ)^{'} = \cos θ$ and $(\cos θ)^{'} = - \sin θ$ , and derivatives of sums and products follow the algebra of derivatives and the chain rule. (The derivatives of sine and cosine are cosine and minus sine, 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 $g (c) \neq 0$ )

[L4]

$∣ \sin θ ∣ \leq 1$ and $∣ \cos θ ∣ \leq 1$ for all real $θ$ (Signs, monotonicity intervals, and ranges of sine and cosine); the mean value theorem then gives $∣ \sin θ - \sin ϑ ∣ \leq ∣ θ - ϑ ∣$ and $∣ \cos θ - \cos ϑ ∣ \leq ∣ θ - ϑ ∣$ (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[L5]

Local Lipschitz continuity in the state variable is the neighborhood-uniform state-Lipschitz property of Local Lipschitz continuity in the state variable, locally uniform in time and parameters. For a continuous field with this property on an open ODE domain, there is a unique maximal solution through each datum, and a solution of an autonomous field whose state remains in a compact subset of its open state domain near a finite endpoint extends past that endpoint (First-order systems, initial value problems, and solutions on intervals, Every Picard–Lindelöf initial value problem has one maximal solution on an open interval, Extensions, maximal solutions, maximal intervals, and global solutions, 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).

[L6]

If a differentiable vector-valued curve on a compact interval has derivative norm at most $C$ , then its endpoint difference has norm at most $C$ times the interval length (If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz).

[L7]

The zero-derivative constancy fact used below is A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant: a differentiable real function on an interval whose derivative vanishes is constant.

Verification

technique · direct; exhibit an exact invariant of the branch system, check the trapping, and evaluate the theorem's constant

1.1L1L4L5algebra

Write the branch field as $V (I, θ, s) = (\frac{1}{10} I \sin θ, 1 + \frac{1}{10} \cos θ, \frac{1}{10})$ . On each cube $C_{N} = [- N, N]^{3}$ with $N > 0$ , [L4] gives, for $p = (I, θ, s)$ and $q = (I^{'}, ϑ, s^{'})$ , $∣ I \sin θ - I^{'} \sin ϑ ∣ \leq ∣ I - I^{'} ∣ + N ∣ θ - ϑ ∣, ∣ \cos θ - \cos ϑ ∣ \leq ∣ θ - ϑ ∣ .$ Hence $∥ V (p) - V (q) ∥_{2} \leq \frac{1}{10} (∣ I - I^{'} ∣ + (N + 1) ∣ θ - ϑ ∣) \leq \frac{N + 2}{10} ∥ p - q ∥_{2}$ . Thus $V$ is continuous and locally Lipschitz on its open domain $R^{3}$ ; by [L5] there is a unique maximal solution $(I, θ, s) : (α, β) \to R^{3}$ through the stated datum.

2.1step 1.1L1L3L7algebra

Along the maximal solution of step 1.1, [L1] gives $I^{'} = \frac{1}{10} I \sin θ$ , $θ^{'} = 1 + \frac{1}{10} \cos θ$ , $s^{'} = \frac{1}{10}$ . Put $J (t) : = I (t) (1 + \frac{1}{10} \cos θ (t))$ . By [L3], $J^{'} = \frac{1}{10} I \sin θ (1 + \frac{1}{10} \cos θ) + I (- \frac{1}{10} \sin θ) (1 + \frac{1}{10} \cos θ) = 0,$ so [L7] makes $J$ constant; with $I (0) = 3 / 2$ , $θ (0) = 0$ we get $J \equiv \frac{33}{20}$ .

3.1step 1.1step 2.1L2L4L5L6algebra

Trapping and global forward existence: on $[0, β)$ , step 2.1 and [L4] give $9 / 10 \leq 1 + \frac{1}{10} \cos θ (t) \leq 11 / 10$ and hence $\frac{3}{2} = \frac{33}{20} \cdot \frac{10}{11} \leq I (t) = \frac{33 / 20}{1 + \frac{1}{10} \cos θ (t)} \leq \frac{33}{20} \cdot \frac{10}{9} = \frac{11}{6} .$ Also $∣ θ^{'} (t) ∣ \leq 11 / 10$ and $∣ s^{'} (t) ∣ = 1 / 10$ , so [L6] gives $∣ θ (t) ∣ \leq 11 t / 10$ and $s (t) = t / 10$ . If $β < + \infty$ , the state curve near $β$ stays in the compact subset $[\frac{3}{2}, \frac{11}{6}] \times [- \frac{11 β}{10}, \frac{11 β}{10}] \times [0, \frac{β}{10}]$ of $R^{3}$ ; the autonomous compact-continuation clause of [L5] extends the maximal solution past $β$ , a contradiction. Hence $β = + \infty$ . Since $s (t) = t / 10$ , on $[0, 20] = [0, A / ε]$ we have $s (t) \in S_{0}$ , and the bounds above place $I (t)$ strictly inside $K = [6 / 5, 19 / 10]$ . Thus the solution stays in $K \times S_{0}$ on the required horizon, and [L2] applies.

4.1step 2.1step 3.1L2algebra∎

Evaluate the theorem's constant: $⟨ k ⟩ (I, s) = \frac{1}{2 π} \int_{0}^{2 π} I \cos θ d θ = 0$ , so $g = I \cos θ$ and $\sup_{K \times [0, 2 π] \times S_{0}} ∣ g ∣ = \frac{19}{10}$ . Moreover $\partial_{I} g = \cos θ$ , $\partial_{θ} g = - I \sin θ$ , $\partial_{s} g = 0$ , and with $\partial_{θ} k = - I \sin θ$ , $\partial_{I} k = \cos θ$ one has $\partial_{I} g \partial_{θ} k - \partial_{θ} g \partial_{I} k - \partial_{s} g = - I \cos θ \sin θ + I \sin θ \cos θ = 0$ . The theorem thus gives $∣ I (t) - I (0) ∣ \leq \frac{1}{10} \cdot 2 \cdot \frac{19}{10} = \frac{19}{50}$ , while the exact invariant gives $∣ I (t) - I (0) ∣ = \frac{1}{10} \cdot \frac{3}{2} \frac{∣ \cos θ (0) - \cos θ (t) ∣}{1 + \frac{1}{10} \cos θ (t)} \leq \frac{1}{10} \cdot \frac{3}{2} \cdot \frac{2}{9 / 10} = \frac{1}{3},$ consistent with the bound.

An exactly solvable adiabatic branch: $K = I + ε I \cos θ$ and comparison with the drift bound

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources

An exactly solvable adiabatic branch: K=I+εIcos⁡θK=I+εIcosθ and comparison with the drift bound

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources

An exactly solvable adiabatic branch: $K = I + ε I \cos θ$ and comparison with the drift bound