Categories: Mathematics, Physics.

# 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.
The transform is defined as follows:

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 $f(t)$.

The **inverse Laplace transform** $\hat{\mathcal{L}}{}^{-1}$ involves complex integration,
and is therefore a lot more difficult to calculate.
Fortunately, it is usually avoidable by rewriting a given $s$-space expression
using partial fraction decomposition,
and then looking up the individual terms.

## 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:

$\begin{aligned} \boxed{ \dvn{n}{\tilde{f}}{s} = (-1)^n \hat{\mathcal{L}}\{t^n f(t)\} } \end{aligned}$The exponential $\exp(- s t)$ is the only thing that depends on $s$ here:

$\begin{aligned} \dvn{n}{\tilde{f}}{s} &= \dvn{n}{}{s}\int_0^\infty f(t) \exp(- s t) \dd{t} \\ &= \int_0^\infty (-t)^n f(t) \exp(- s t) \dd{t} = (-1)^n \hat{\mathcal{L}}\{t^n f(t)\} \end{aligned}$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 $- f''(0) - s f'(0) - s^2 f(0) + s^3 \tilde{f}(s)$.

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.

## References

- O. Bang,
*Applied mathematics for physicists: lecture notes*, 2019, unpublished.