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

Lemma: Residual of the local oscillatory candidate

Integral and scalar diagnostics › A smooth-switch scalar residual · lines 1043–1080 · proof included or linked

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

For the real switch $σ$ and $λ > 0$ , set

v_{λ} (t) = - \frac{σ ( t )}{4} e^{2 i λ t} .

Then

v_{λ}^{''} - λ^{2} σ e^{2 i λ t} = (- i λ σ^{'} - \frac{σ ^{''}}{4}) e^{2 i λ t} .

On an interval, $v_{λ}$ solves $v_{λ}^{''} = λ^{2} σ e^{2 i λ t}$ if and only if $σ$ is constant there.

Proof. Direct differentiation gives

v_{λ}^{'} = - (\frac{σ ^{'}}{4} + \frac{i λσ}{2}) e^{2 i λ t}

and

v_{λ}^{''} = (λ^{2} σ - i λ σ^{'} - \frac{σ ^{''}}{4}) e^{2 i λ t} .

This proves (eq:ode-residual).

If the residual vanishes on an interval, its imaginary part is $- λ σ^{'} = 0$ , because $σ$ is real and $λ > 0$ . Thus $σ$ is constant there. Conversely, a constant $σ$ makes both derivative terms zero.

Displayed formulas in this passage

XVI-B@1#eq:ode-residual · line 1050
$v_{λ}^{''} - λ^{2} σ e^{2 i λ t} = (- i λ σ^{'} - \frac{σ ^{''}}{4}) e^{2 i λ t} .$
unlabeled · line 1046
$v_{λ} (t) = - \frac{σ ( t )}{4} e^{2 i λ t} .$
unlabeled · line 1063
$v_{λ}^{'} = - (\frac{σ ^{'}}{4} + \frac{i λσ}{2}) e^{2 i λ t}$
unlabeled · line 1069
$v_{λ}^{''} = (λ^{2} σ - i λ σ^{'} - \frac{σ ^{''}}{4}) e^{2 i λ t} .$

Nearby in this section

section: A smooth-switch scalar residual