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+---
+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">
+<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">
+<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.