Manuscript XVI-B@1 · XVI-B.tex · lemma · 6a0abffd

Lemma: Sharp-onset scalar solution

Integral and scalar diagnostics › A sharp-switch scalar solution · lines 1088–1121 · proof included or linked

XVI-B@1#lem:sharp-ode, XVI-B@1#eq:sharp-ode

The function

u_{H} (t) = H (t) (- \frac{1}{4} e^{2 i λ t} + \frac{1}{4} + \frac{i λ t}{2})

is a distributional solution of

u_{H}^{''} = λ^{2} H (t) e^{2 i λ t},

vanishes for $t < 0$ , and has zero initial position and velocity.

Proof. Let $b (t) = - e^{2 i λ t} /4 + 1/4 + i λ t /2$ . Then

b (0) = 0, b^{'} (0) = 0, b^{''} (t) = λ^{2} e^{2 i λ t} .

Distributional differentiation of $H b$ therefore produces neither a delta term nor a derivative of a delta:

(H b)^{''} = H b^{''} + δ b^{'} (0) + δ^{'} b (0) = H b^{''} .

This proves the equation. The traces at $0$ and the negative-time vanishing follow from the same formulas.

Displayed formulas in this passage

XVI-B@1#eq:sharp-ode · line 1091
$u_{H} (t) = H (t) (- \frac{1}{4} e^{2 i λ t} + \frac{1}{4} + \frac{i λ t}{2})$
unlabeled · line 1100
$u_{H}^{''} = λ^{2} H (t) e^{2 i λ t},$
unlabeled · line 1110
$b (0) = 0, b^{'} (0) = 0, b^{''} (t) = λ^{2} e^{2 i λ t} .$
unlabeled · line 1116
$(H b)^{''} = H b^{''} + δ b^{'} (0) + δ^{'} b (0) = H b^{''} .$