---
title: "Laplace transform"
sort_title: "Laplace transform"
date: 2021-07-02
categories:
- Mathematics
- Physics
layout: "concept"
---

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 $$f(t)$$.

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](/know/concept/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:

$$\begin{aligned}
    \boxed{
        \dvn{n}{\tilde{f}}{s} = (-1)^n \hat{\mathcal{L}}\{t^n f(t)\}
    }
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-dv-s"/>
<label for="proof-dv-s">Proof</label>
<div class="hidden" markdown="1">
<label for="proof-dv-s">Proof.</label>
The exponential $$\exp(- s t)$$ is the only thing that depends on $$s$$ here:

$$\begin{aligned}
    \dvn{n}{\tilde{f}}{s}
    &= \dvn{n}{}{s}\int_0^\infty f(t) \exp(- s t) \dd{t}
    \\
    &= \int_0^\infty (-t)^n f(t) \exp(- s t) \dd{t}
    = (-1)^n \hat{\mathcal{L}}\{t^n f(t)\}
\end{aligned}$$

</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" markdown="1">
<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.