Manuscript XVI-K@1 · XVI-K.tex · lemma · aa7553bb

Lemma: Pure-trace full Hessian

The declared density and its full jet Hessian › An analytic differentiated-slot check · lines 656–691 · proof included or linked

XVI-K@1#lem:tracecheck, XVI-K@1#eq:tracehessian

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For functions depending only on $s = w^{1}$ ,

ℓ_{3} (ψ η) = - 3 ψ (ψ^{'})^{2} .

For $h = ψ η$ , $A = u η$ , and $C = v η$ ,

D^{2} ℓ_{3} (h) [A, C] = - 6 (u ψ^{'} v^{'} + v ψ^{'} u^{'} + ψ u^{'} v^{'}) .

On $ψ = s^{2}$ , this becomes

- 12 s (u v^{'} + v u^{'}) - 6 s^{2} u^{'} v^{'} .

Proof. Substitution in the finite atoms gives

r_{1} = 3 ψ^{''}, r_{2} = - 6 ψ ψ^{''} - \frac{3}{2} (ψ^{'})^{2}, r_{3} = 9 ψ^{2} ψ^{''} + \frac{9}{2} ψ (ψ^{'})^{2},

with $b_{1} = 2 ψ$ and $b_{2} = ψ^{2}$ . The $ψ^{2} ψ^{''}$ terms cancel in $r_{3} + b_{1} r_{2} + b_{2} r_{1}$ , leaving $\frac{3}{2} ψ (ψ^{'})^{2}$ . This proves the density formula. Taking the mixed variation of $- 3 (ψ + ϵ u + δ v) (ψ^{'} + ϵ u^{'} + δ v^{'})^{2}$ proves (eq:tracehessian).

Displayed formulas in this passage

XVI-K@1#eq:tracehessian · line 663
$D^{2} ℓ_{3} (h) [A, C] = - 6 (u ψ^{'} v^{'} + v ψ^{'} u^{'} + ψ u^{'} v^{'}) .$
unlabeled · line 659
$ℓ_{3} (ψ η) = - 3 ψ (ψ^{'})^{2} .$
unlabeled · line 670
$- 12 s (u v^{'} + v u^{'}) - 6 s^{2} u^{'} v^{'} .$
unlabeled · line 677
$r_{1} = 3 ψ^{''}, r_{2} = - 6 ψ ψ^{''} - \frac{3}{2} (ψ^{'})^{2}, r_{3} = 9 ψ^{2} ψ^{''} + \frac{9}{2} ψ (ψ^{'})^{2},$

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section: An analytic differentiated-slot check