Polynomial decay rate of y' = -y^3

cubic_decay_asymptotic

Submitter: Kim Morrison.

Notes: Asymptotic rate y t * sqrt t -> 1/sqrt 2 for the unique solution of y' = -y^3 on (0, infty) with y continuous at 0 and y 0 = 1.

Source: Standard ODE textbook exercise.

Informal solution: The closed form is y(t) = (1 + 2t)^{-1/2}, so y(t) sqrt(t) = sqrt(t / (1 + 2t)) -> 1/sqrt(2). Uniqueness on (0, infty) follows by Grönwall: any two solutions u, v of u' = -u^3 with the same right-limit 1 at 0 satisfy |u - v|' bounded by a Lipschitz constant on bounded subintervals, and continuity at 0 forces |u(t) - v(t)| -> 0 as t -> 0+, hence u = v.

theorem declaration uses `sorry`cubic_decay_asymptotic (y : ℝ → ℝ) (hy_diff : ∀ t : ℝ, 0 < t → HasDerivAt y (-(y t) ^ 3) t)
    (hy_cont : ContinuousWithinAt y (Set.Ici 0) 0)
    (hy0 : y 0 = 1) :
    Tendsto (fun t : ℝ => y t * Real.sqrt t) atTop (𝓝 (1 / Real.sqrt 2)) := byy:ℝ → ℝhy_diff:∀ (t : ℝ), 0 < t → HasDerivAt y (-y t ^ 3) thy_cont:ContinuousWithinAt y (Set.Ici 0) 0hy0:y 0 = 1⊢ Tendsto (fun t => y t * √t) atTop (𝓝 (1 / √2))
  sorryAll goals completed! 🐙

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