The Laplace Transform

The Laplace transform converts differential equations into algebraic equations, providing a systematic method for solving linear ODEs with constant coefficients, especially those with discontinuous or impulsive forcing functions.

Definition

Definition4.1Laplace transform

The Laplace transform of a function $f: [0, \infty) \to \mathbb{R}$ is

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

defined for all $s$ for which the integral converges. The function $F(s)$ is the Laplace transform and $f(t)$ is the inverse Laplace transform of $F$ .

Theorem4.1Existence of the Laplace transform

If $f$ is piecewise continuous on $[0, \infty)$ and of exponential order $\alpha$ (i.e., $|f(t)| \leq Me^{\alpha t}$ for $t \geq T$ ), then $\mathcal{L}\{f\}(s)$ exists for $s > \alpha$ .

Basic Transforms

ExampleStandard Laplace transforms

| $f(t)$ | $F(s) = \mathcal{L}\{f\}(s)$ | Domain | |--------|---------------------------|--------| | $1$ | $1/s$ | $s > 0$ | | $t^n$ | $n!/s^{n+1}$ | $s > 0$ | | $e^{at}$ | $1/(s-a)$ | $s > a$ | | $\sin(bt)$ | $b/(s^2+b^2)$ | $s > 0$ | | $\cos(bt)$ | $s/(s^2+b^2)$ | $s > 0$ | | $e^{at}\sin(bt)$ | $b/((s-a)^2+b^2)$ | $s > a$ |

Properties

Theorem4.2Linearity and shifting

The Laplace transform satisfies:

Linearity: $\mathcal{L}\{af + bg\} = a\mathcal{L}\{f\} + b\mathcal{L}\{g\}$ .
First shifting theorem: $\mathcal{L}\{e^{at}f(t)\}(s) = F(s-a)$ .
Transform of derivatives: $\mathcal{L}\{f'\}(s) = sF(s) - f(0)$ .
$n$ -th derivative: $\mathcal{L}\{f^{(n)}\}(s) = s^nF(s) - s^{n-1}f(0) - \cdots - f^{(n-1)}(0)$ .
Transform of integrals: $\mathcal{L}\left\{\int_0^t f(\tau)\,d\tau\right\} = F(s)/s$ .

ExampleSolving an IVP via Laplace transform

Solve $y'' + 4y = \sin t$ , $y(0) = 0$ , $y'(0) = 1$ .

Taking the Laplace transform: $s^2Y - 0 - 1 + 4Y = \frac{1}{s^2+1}$ .

$(s^2+4)Y = 1 + \frac{1}{s^2+1} = \frac{s^2+2}{s^2+1}.$

$Y = \frac{s^2+2}{(s^2+1)(s^2+4)} = \frac{1}{3}\cdot\frac{1}{s^2+1} + \frac{2}{3}\cdot\frac{1}{s^2+4}.$

Inverting: $y(t) = \frac{1}{3}\sin t + \frac{1}{3}\sin 2t$ .

Convolution

Definition4.2Convolution

The convolution of $f$ and $g$ is $(f * g)(t) = \int_0^t f(\tau)g(t-\tau)\,d\tau$ .

Theorem4.3Convolution theorem

$\mathcal{L}\{f * g\}(s) = F(s) \cdot G(s)$ . That is, convolution in the time domain corresponds to multiplication in the $s$ -domain.

RemarkInverse transforms via convolution

The convolution theorem provides a method for computing inverse Laplace transforms: $\mathcal{L}^{-1}\{F(s)G(s)\} = f * g$ . This is particularly useful when partial fractions are cumbersome.