Categories: Mathematics, Physics.

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:

\[\begin{aligned} \boxed{ \tilde{f}(s) \equiv \hat{\mathcal{L}}\{f(t)\} \equiv \int_0^\infty f(t) \exp\!(- s t) \dd{t} } \end{aligned}\]

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 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:

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

The exponential \(\exp\!(- s t)\) is the only thing that depends on \(s\) here:

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

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.

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*Applied mathematics for physicists: lecture notes*, 2019, unpublished.

© Marcus R.A. Newman, a.k.a. "Prefetch".
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