Laplace Transform of Sine of Root/Proof 2
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Theorem
- $\laptrans {\sin \sqrt t} = \dfrac {\sqrt \pi} {2 s^{3/2} } \map \exp {-\dfrac 1 {4 s} }$
where $\laptrans f$ denotes the Laplace transform of the function $f$.
Proof
Let $\map y t := \sin \sqrt t$.
Differentiating twice with respect to $t$, we get:
- $(1): \quad 4 t y'' + 2 y'' + y = 0$
Let $\map Y s = \laptrans {\map t y}$ be the Laplace transform of $y$.
Then taking the Laplace transform of $(1)$:
\(\ds -4 \map {\dfrac \d {\d s} } {\laptrans {\map {y''} t} } + 2 \laptrans {\map {y'} t} + \laptrans {\map y t}\) | \(=\) | \(\ds 0\) | Derivative of Laplace Transform | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -4 \map {\dfrac \d {\d s} } {s^2 \, \map Y s - s \, \map y 0 - \map {y'} 0} + 2 \paren {s \, \map Y s - \map y 0} + \map Y s\) | \(=\) | \(\ds 0\) | Laplace Transform of Derivative, Laplace Transform of Second Derivative | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 4 s^2 \, \map {Y'} s - \paren {6 s - 1} \map Y s\) | \(=\) | \(\ds 0\) | simplifying | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map Y s\) | \(=\) | \(\ds \dfrac c {s^{3/2} } \, \map \exp {\dfrac 1 {4 s} }\) | solving the differential equation |
For small $t$, we have:
- $\sin \sqrt t \sim \sqrt t$
and:
- $\laptrans {\sqrt t} = \dfrac {\sqrt \pi} {2 s^{3/2} }$
For large $t$, we have:
- $Y \sim \dfrac c {2 s^{3/2} }$
Hence by comparison:
- $c = \dfrac {\sqrt \pi} 2$
Hence:
- $\laptrans {\sin \sqrt t} = \dfrac {\sqrt \pi} {2 s^{3/2} } \map \exp {-\dfrac 1 {4 s} }$
$\blacksquare$
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Miscellaneous Problems: $48$