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---
title: "Laplace transform"
firstLetter: "L"
publishDate: 2021-07-02
categories:
- Mathematics
- Physics
date: 2021-07-02T15:48:30+02:00
draft: false
markup: pandoc
---
# Laplace transform
The **Laplace transform** is an integral transform
that losslessly converts a function $f(t)$ of a real variable $t$,
into a function $\tilde{f}(s)$ of a complex variable $s$,
where $s$ is sometimes called the **complex frequency**,
analogously to the [Fourier transform](/know/concept/fourier-transform/).
The transform is defined as follows:
$$\begin{aligned}
\boxed{
\tilde{f}(s)
\equiv \hat{\mathcal{L}}\{f(t)\}
\equiv \int_0^\infty f(t) \exp\!(- s t) \dd{t}
}
\end{aligned}$$
Depending on $f(t)$, this integral may diverge.
This is solved by restricting the domain of $\tilde{f}(s)$
to $s$ where $\mathrm{Re}\{s\} > s_0$,
for an $s_0$ large enough to compensate for the growth of $f(t)$.
## Derivatives
The derivative of a transformed function is the transform
of the original mutliplied by its variable.
This is especially useful for transforming ODEs with variable coefficients:
$$\begin{aligned}
\boxed{
\tilde{f}'(s) = - \hat{\mathcal{L}}\{t f(t)\}
}
\end{aligned}$$
This property generalizes nicely to higher-order derivatives of $s$, so:
$$\begin{aligned}
\boxed{
\dv[n]{\tilde{f}}{s} = (-1)^n \hat{\mathcal{L}}\{t^n f(t)\}
}
\end{aligned}$$
<div class="accordion">
<input type="checkbox" id="proof-dv-s"/>
<label for="proof-dv-s">Proof</label>
<div class="hidden">
<label for="proof-dv-s">Proof.</label>
The exponential $\exp\!(- s t)$ is the only thing that depends on $s$ here:
$$\begin{aligned}
\dv[n]{\tilde{f}}{s}
&= \dv[n]{s} \int_0^\infty f(t) \exp\!(- s t) \dd{t}
\\
&= \int_0^\infty (-t)^n f(t) \exp\!(- s t) \dd{t}
= (-1)^n \hat{\mathcal{L}}\{t^n f(t)\}
\end{aligned}$$
</div>
</div>
The Laplace transform of a derivative introduces the initial conditions into the result.
Notice that $f(0)$ is the initial value in the original $t$-domain:
$$\begin{aligned}
\boxed{
\hat{\mathcal{L}}\{ f'(t) \} = - f(0) + s \tilde{f}(s)
}
\end{aligned}$$
This property generalizes to higher-order derivatives,
although it gets messy quickly.
Once again, the initial values of the lower derivatives appear:
$$\begin{aligned}
\boxed{
\hat{\mathcal{L}} \big\{ f^{(n)}(t) \big\}
= - \sum_{j = 0}^{n - 1} s^j f^{(n - 1 - j)}(0) + s^n \tilde{f}(s)
}
\end{aligned}$$
Where $f^{(n)}(t)$ is shorthand for the $n$th derivative of $f(t)$,
and $f^{(0)}(t) = f(t)$.
As an example, $\hat{\mathcal{L}}\{f'''(t)\}$ becomes
$- f''(0) - s f'(0) - s^2 f(0) + s^3 \tilde{f}(s)$.
<div class="accordion">
<input type="checkbox" id="proof-dv-t"/>
<label for="proof-dv-t">Proof</label>
<div class="hidden">
<label for="proof-dv-t">Proof.</label>
We integrate by parts and use the fact that $\lim_{x \to \infty} \exp\!(-x) = 0$:
$$\begin{aligned}
\hat{\mathcal{L}} \big\{ f^{(n)}(t) \big\}
&= \int_0^\infty f^{(n)}(t) \exp\!(- s t) \dd{t}
\\
&= \big[ f^{(n - 1)}(t) \exp\!(- s t) \big]_0^\infty + s \int_0^\infty f^{(n-1)}(t) \exp\!(- s t) \dd{t}
\\
&= - f^{(n - 1)}(0) + s \big[ f^{(n - 2)}(t) \exp\!(- s t) \big]_0^\infty + s^2 \int_0^\infty f^{(n-2)}(t) \exp\!(- s t) \dd{t}
\end{aligned}$$
And so on.
By partially integrating $n$ times in total we arrive at the conclusion.
</div>
</div>
## References
1. O. Bang,
*Applied mathematics for physicists: lecture notes*, 2019,
unpublished.
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