Logarithms turn multiplications into sums: to multiply two numbers, one adds their logarithms and comes back. The Laplace transform does something similar with differential equations: it swaps derivatives for multiplications. One solves an algebraic equation and then comes back.

The origin

Laplace came to this idea through probability. In the Analytic Theory of Probabilities, of 1812, he organized his calculations with generating functions: instead of working with a sequence of probabilities, he worked with a single function that stored them all. The continuous version of this idea is an integral that Euler had used earlier, but which Laplace explored systematically:

$$F(s) = \int_0^{\infty} e^{-st} f(t)\,dt.$$

Each function $f(t)$ becomes a new function $F(s)$, its Laplace transform, which we write $\mathcal{L}\{f\}$.

Some examples

For the constant $f(t) = 1$:

$$\mathcal{L}\{1\} = \int_0^{\infty} e^{-st}\,dt = \left[-\frac{e^{-st}}{s}\right]_0^{\infty} = \frac{1}{s} \quad (s > 0).$$

For the exponential $f(t) = e^{at}$, the exponentials combine:

$$\mathcal{L}\{e^{at}\} = \int_0^{\infty} e^{-(s-a)t}\,dt = \frac{1}{s-a} \quad (s > a).$$

Derivatives become multiplications

The property that makes the transform useful comes from an integration by parts:

$$\mathcal{L}\{f'\} = \int_0^{\infty} e^{-st} f'(t)\,dt = \Big[e^{-st}f(t)\Big]_0^{\infty} + s\int_0^{\infty} e^{-st} f(t)\,dt = sF(s) - f(0).$$

Differentiating $f$ corresponds to multiplying $F$ by $s$ (and subtracting the initial value). A differential equation becomes an algebraic equation, and the initial value is already built into the calculation.

A circuit

In a circuit with a resistor and a capacitor connected to a battery, the voltage $y(t)$ across the capacitor obeys, in convenient units,

$$y' + y = 1, \qquad y(0) = 0.$$

Applying the transform to both sides: $sY - 0 + Y = \frac{1}{s}$, that is,

$$Y(s) = \frac{1}{s(s+1)} = \frac{1}{s} - \frac{1}{s+1}.$$

Now it is enough to recognize the transforms from the table: $\frac{1}{s}$ comes from $1$ and $\frac{1}{s+1}$ comes from $e^{-t}$. Hence

$$y(t) = 1 - e^{-t}.$$

The capacitor charges up and approaches the battery voltage, as observed in the laboratory.

The return through engineering

For a long time the transform was forgotten as a probability trick. At the end of the nineteenth century the English engineer Oliver Heaviside invented an “operational calculus” for electrical circuits, in which he treated the derivative as a number that can be multiplied and divided. The results were right, but mathematicians did not understand why. In the 1910s Thomas Bromwich and others showed that Heaviside’s method was, at bottom, the Laplace transform, and in the 1930s Gustav Doetsch put it on rigorous foundations. Since then it has been in every engineering course, in control, signals and circuits.

Exercise: the transform of t

Compute $\mathcal{L}\{t\}$.

Integrating by parts, with $u = t$ and $dv = e^{-st}dt$:

$$\int_0^{\infty} t\,e^{-st}\,dt = \left[-\frac{t\,e^{-st}}{s}\right]_0^{\infty} + \frac{1}{s}\int_0^{\infty} e^{-st}\,dt = 0 + \frac{1}{s}\cdot\frac{1}{s} = \frac{1}{s^2}.$$

One can also use the derivative rule: $t$ has derivative 1 and equals 0 at $t = 0$, so $s\,\mathcal{L}\{t\} = \mathcal{L}\{1\} = \frac{1}{s}$.

Exercise: exponential growth

Solve $y' = 3y$ with $y(0) = 2$ using the transform.

Transforming: $sY - 2 = 3Y$, so $Y = \frac{2}{s-3}$. From the table, $\frac{1}{s-3}$ comes from $e^{3t}$, and therefore $y(t) = 2e^{3t}$.

Sources: Pierre-Simon Laplace, Théorie analytique des probabilités (1812), book I · Michael A. B. Deakin, “The Development of the Laplace Transform, 1737–1937”, Archive for History of Exact Sciences 25 (1981) and 26 (1982) · Gustav Doetsch, Theorie und Anwendung der Laplace-Transformation (1937) · Paul J. Nahin, Oliver Heaviside: Sage in Solitude (1988).

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