Manuscript X-F-ST@1 · X-F-ST.tex · section · e71f03ef

Explicit distributional kernels

F3: three response kernels, support, and admission › Explicit distributional kernels · lines 1520–1565

X-F-ST@1#eq:treekernels

Choose finite component bases for tensors. Let $T$ be the constant-coefficient bidifferential expression for the polarized flat stress, and $B$ the expression fixed by (eq:Boperator). Define their transposes by

⟨ T^{T} Ψ, u \otimes v ⟩ = ⟨ Ψ, T (u, v) ⟩, ⟨ B^{T} Ψ, r_{1} \otimes r_{2} ⟩ = ⟨ Ψ, G_{2} [r_{1}, r_{2}] ⟩ .

The notation includes the indicated repeated source point. All component contractions are finite.

The tree kernels are

t_{mat} M (x, y) t_{self} = Sym_{3} {\frac{1}{2} T^{T} [g (y) N_{0} (x, y) δ (y - y^{'})]}, = - (Q^{(1)} [h]_{y})^{T} [σ (y) N_{0} (x, y)], = - \frac{1}{2} Sym_{y, y^{'}} {B^{T} [σ (y) N_{0} (x, y) δ (y - y^{'})]} .

Thus

T_{mat} = : ϕ^{\otimes 3} (t_{mat}) :, T_{mix} = (ϕ \otimes r) (M), T_{self} = : (ϕ \otimes r \otimes r) (t_{self}) : .

Here $Sym_{3}$ is normalized symmetrization over the identical scalar slots. It changes no cubic normal-symbol value. Only the two graviton slots are symmetrized in the self kernel.

Displayed formulas in this passage

X-F-ST@1#eq:treekernels · line 1537
$t_{mat} M (x, y) t_{self} = Sym_{3} {\frac{1}{2} T^{T} [g (y) N_{0} (x, y) δ (y - y^{'})]}, = - (Q^{(1)} [h]_{y})^{T} [σ (y) N_{0} (x, y)], = - \frac{1}{2} Sym_{y, y^{'}} {B^{T} [σ (y) N_{0} (x, y) δ (y - y^{'})]} .$
unlabeled · line 1527
$⟨ T^{T} Ψ, u \otimes v ⟩ = ⟨ Ψ, T (u, v) ⟩, ⟨ B^{T} Ψ, r_{1} \otimes r_{2} ⟩ = ⟨ Ψ, G_{2} [r_{1}, r_{2}] ⟩ .$
unlabeled · line 1554
$T_{mat} = : ϕ^{\otimes 3} (t_{mat}) :, T_{mix} = (ϕ \otimes r) (M), T_{self} = : (ϕ \otimes r \otimes r) (t_{self}) : .$

Explicit distributional kernels

Displayed formulas in this passage

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