Manuscript III@2 · III.tex · section · 82dad3e5

Zero-data response and homogeneous subtraction

The C1 triangular Cauchy quotient › Zero-data response and homogeneous subtraction · lines 1667–1689

III@2#eq:response

For scalar data $q_{0}, q_{1}$ , put

H_{t_{0}} (q_{0}, q_{1}) (t) = cos (τ Ω_{m}) q_{0} + \frac{sin ( τ Ω _{m} )}{Ω _{m}} q_{1} .

For a smooth scalar $q$ , write

H_{t_{0}} q = H_{t_{0}} (q ∣_{t_{0}}, \partial_{t} q ∣_{t_{0}}) .

Define

S_{t_{0}} (h) Φ (t) = \int_{t_{0}}^{t} \frac{sin (( t - s ) Ω _{m} )}{Ω _{m}} L_{h} Φ (s) d s .

It has zero data at $t_{0}$ and solves $P S_{t_{0}} (h) Φ = - L_{h} Φ$ . For $t < t_{0}$ , use the oriented integral.

Displayed formulas in this passage

III@2#eq:response · line 1680
$S_{t_{0}} (h) Φ (t) = \int_{t_{0}}^{t} \frac{sin (( t - s ) Ω _{m} )}{Ω _{m}} L_{h} Φ (s) d s .$
unlabeled · line 1670
$H_{t_{0}} (q_{0}, q_{1}) (t) = cos (τ Ω_{m}) q_{0} + \frac{sin ( τ Ω _{m} )}{Ω _{m}} q_{1} .$
unlabeled · line 1675
$H_{t_{0}} q = H_{t_{0}} (q ∣_{t_{0}}, \partial_{t} q ∣_{t_{0}}) .$