Manuscript Psy02-A@1 · Psy02-A.tex · section · d92ce402

An exactly evaluated invariant

A separate normalization-independent response benchmark › An exactly evaluated invariant · lines 1749–1778

Psy02-A@1#eq:Fstar

Choose $t_{0} > a > 0$ , and define

ρ_{b} (u) = {e^{- 1/ (1 - u^{2})}, 0, \abs u < 1, \abs u \geq 1,

b (t) = \frac{ρ _{b} (( t - t _{0} ) / a )}{a \int _{- 1} ^{1} ρ _{b} ( u ) ^{2} d u} .

Then

\int b (t)^{2} d t = 1, \int t b (t)^{2} d t = t_{0} .

Set

F_{*} (t, x) = (t^{2} - \abs x^{2}) b (t) b (\abs x) .

The radial factor vanishes near $x = 0$ , so this is a smooth compact test.

Displayed formulas in this passage

Psy02-A@1#eq:Fstar · line 1770
$F_{*} (t, x) = (t^{2} - \abs x^{2}) b (t) b (\abs x) .$
unlabeled · line 1752
$ρ_{b} (u) = {e^{- 1/ (1 - u^{2})}, 0, \abs u < 1, \abs u \geq 1,$
unlabeled · line 1759
$b (t) = \frac{ρ _{b} (( t - t _{0} ) / a )}{a \int _{- 1} ^{1} ρ _{b} ( u ) ^{2} d u} .$
unlabeled · line 1765
$\int b (t)^{2} d t = 1, \int t b (t)^{2} d t = t_{0} .$