Manuscript Em01-C@3 · Em01-C-v3.tex · theorem · f0953702

Theorem: Scalar singular-plus-finite expansion

Explicit scalar cone asymptotics › Response-normal derivative · lines 1601–1706 · proof included or linked

Em01-C@3#thm:scalar-derivative, Em01-C@3#eq:scalar-expansion, Em01-C@3#eq:c1, Em01-C@3#eq:scalar-finite, Em01-C@3#eq:fixed-test-exact, Em01-C@3#eq:scalar-derivative-action

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The family satisfies

U_{δ} = - δ^{- 1} c_{1} (α, h) δ_{o} + T_{sc} + O_{loc} (δ),

with

c_{1} (α, h) = \frac{π h}{4 β ^{2}} \int_{R} f_{0} (U) g_{0} (- \frac{h}{α} U) d U,

T_{sc} = c_{0} \partial_{q} δ_{o} + κ s_{1} \partial_{v} δ_{o} - \frac{κ q _{1}}{4} Δ_{ξ} δ_{o} .

The remainder satisfies

∣ O_{loc} (δ) (Ψ) ∣ \leq C δ ∥Ψ ∥_{4}

on compact test supports, uniformly on compact positive parameter sets.

For any fixed compact test equal to one near the origin,

U_{δ} (Ψ) = - c_{1} δ^{- 1}

for every sufficiently small $δ$ . Consequently the raw scalar family converges exactly when $c_{1} = 0$ . If $f_{0}, g_{0}$ are strictly positive near zero, then $c_{1} > 0$ , and a single fixed test detects divergence.

Proof. The derivative-delta action gives

U_{δ} (Ψ) = - \int P_{δ} (0, v, ξ) \partial_{q} Ψ (0, v, ξ) d v d^{2} ξ - κ δ^{- 3} \int f_{0} (\frac{4 α v - ∣ ξ ∣ ^{2}}{δ}) g_{0}^{'} (\frac{- 4 h v - ∣ ξ ∣ ^{2}}{δ}) Ψ (0, v, ξ) d v d^{2} ξ .

The second integral becomes

- κ δ^{- 1} \int F_{1} (s, X) Ψ (0, δ s, δ X) d s d^{2} X .

Its zeroth moment is

m_{1} = \frac{π}{4 β} \int_{h U + α V \leq 0} f_{0} (U) g_{0}^{'} (V) d U d V = \frac{π}{4 β} \int f_{0} (U) g_{0} (- h U / α) d U .

Therefore $κ m_{1} = c_{1}$ .

Lemma lem:scalar-moments yields

- κ δ^{- 1} \int F_{1} Ψ (0, δ s, δ X) = - c_{1} δ^{- 1} Ψ (0, 0, 0) - κ s_{1} \partial_{v} Ψ (0, 0, 0) - \frac{κ q _{1}}{4} Δ_{ξ} Ψ (0, 0, 0) + O (δ ∥Ψ ∥_{4}) .

The first integral in (eq:scalar-derivative-action) is

- c_{0} \partial_{q} Ψ (0, 0, 0) + O (δ ∥Ψ ∥_{3}) .

First derivatives of a Dirac mass act with a minus sign; the transverse Laplacian acts with a plus sign. This gives (eq:scalar-finite).

For a fixed test constant one near the origin, the shrinking rescaled supports lie in its constant region and its $q$ -derivative is zero. This proves (eq:fixed-test-exact). If $c_{1} = 0$ , (eq:scalar-expansion) gives strong convergence to $T_{sc}$ ; otherwise the fixed pairing diverges. Strict positivity near zero makes the integral defining $c_{1}$ strictly positive.

Displayed formulas in this passage

Em01-C@3#eq:scalar-expansion · line 1604
$U_{δ} = - δ^{- 1} c_{1} (α, h) δ_{o} + T_{sc} + O_{loc} (δ),$
Em01-C@3#eq:c1 · line 1613
$c_{1} (α, h) = \frac{π h}{4 β ^{2}} \int_{R} f_{0} (U) g_{0} (- \frac{h}{α} U) d U,$
Em01-C@3#eq:scalar-finite · line 1620
$T_{sc} = c_{0} \partial_{q} δ_{o} + κ s_{1} \partial_{v} δ_{o} - \frac{κ q _{1}}{4} Δ_{ξ} δ_{o} .$
Em01-C@3#eq:fixed-test-exact · line 1637
$U_{δ} (Ψ) = - c_{1} δ^{- 1}$
Em01-C@3#eq:scalar-derivative-action · line 1649
$U_{δ} (Ψ) = - \int P_{δ} (0, v, ξ) \partial_{q} Ψ (0, v, ξ) d v d^{2} ξ - κ δ^{- 3} \int f_{0} (\frac{4 α v - ∣ ξ ∣ ^{2}}{δ}) g_{0}^{'} (\frac{- 4 h v - ∣ ξ ∣ ^{2}}{δ}) Ψ (0, v, ξ) d v d^{2} ξ .$
unlabeled · line 1629
$∣ O_{loc} (δ) (Ψ) ∣ \leq C δ ∥Ψ ∥_{4}$
unlabeled · line 1663
$- κ δ^{- 1} \int F_{1} (s, X) Ψ (0, δ s, δ X) d s d^{2} X .$
unlabeled · line 1668
$m_{1} = \frac{π}{4 β} \int_{h U + α V \leq 0} f_{0} (U) g_{0}^{'} (V) d U d V = \frac{π}{4 β} \int f_{0} (U) g_{0} (- h U / α) d U .$
unlabeled · line 1680
$- κ δ^{- 1} \int F_{1} Ψ (0, δ s, δ X) = - c_{1} δ^{- 1} Ψ (0, 0, 0) - κ s_{1} \partial_{v} Ψ (0, 0, 0) - \frac{κ q _{1}}{4} Δ_{ξ} Ψ (0, 0, 0) + O (δ ∥Ψ ∥_{4}) .$
unlabeled · line 1691
$- c_{0} \partial_{q} Ψ (0, 0, 0) + O (δ ∥Ψ ∥_{3}) .$

Theorem: Scalar singular-plus-finite expansion

Displayed formulas in this passage

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