Categories: Mathematics, Physics.

Laplace transform

The Laplace transform is an integral transform that losslessly converts a function f(t)f(t) of a real variable tt, into a function f~(s)\tilde{f}(s) of a complex variable ss, where ss is sometimes called the complex frequency, analogously to the Fourier transform. The transform is defined as follows:

f~(s)L^{f(t)}0f(t)exp(st)dt\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)f(t), this integral may diverge. This is solved by restricting the domain of f~(s)\tilde{f}(s) to ss where Re{s}>s0\mathrm{Re}\{s\} > s_0, for an s0s_0 large enough to compensate for the growth of f(t)f(t).

The inverse Laplace transform L^1\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 ss-space expression using partial fraction decomposition, and then looking up the individual terms.


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:

f~(s)=L^{tf(t)}\begin{aligned} \boxed{ \tilde{f}{}'(s) = - \hat{\mathcal{L}}\{t f(t)\} } \end{aligned}

This property generalizes nicely to higher-order derivatives of ss, so:

dnf~dsn=(1)nL^{tnf(t)}\begin{aligned} \boxed{ \dvn{n}{\tilde{f}}{s} = (-1)^n \hat{\mathcal{L}}\{t^n f(t)\} } \end{aligned}

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

dnf~dsn=dndsn0f(t)exp(st)dt=0(t)nf(t)exp(st)dt=(1)nL^{tnf(t)}\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)f(0) is the initial value in the original tt-domain:

L^{f(t)}=f(0)+sf~(s)\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:

L^{f(n)(t)}=j=0n1sjf(n1j)(0)+snf~(s)\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)f^{(n)}(t) is shorthand for the nnth derivative of f(t)f(t), and f(0)(t)=f(t)f^{(0)}(t) = f(t). As an example, L^{f(t)}\hat{\mathcal{L}}\{f'''(t)\} becomes f(0)sf(0)s2f(0)+s3f~(s)- f''(0) - s f'(0) - s^2 f(0) + s^3 \tilde{f}(s).

We integrate by parts and use the fact that limxexp(x)=0\lim_{x \to \infty} \exp(-x) = 0:

L^{f(n)(t)}=0f(n)(t)exp(st)dt=[f(n1)(t)exp(st)]0+s0f(n1)(t)exp(st)dt=f(n1)(0)+s[f(n2)(t)exp(st)]0+s20f(n2)(t)exp(st)dt\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 nn times in total we arrive at the conclusion.


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 O. Bang, Applied mathematics for physicists: lecture notes, 2019, unpublished.