The Convolution Theorem

The convolution theorem establishes a fundamental connection between multiplication in the transform domain and convolution in the time domain.

Statement

Theorem4.11Convolution theorem

If $\mathcal{L}\{f\} = F$ and $\mathcal{L}\{g\} = G$ , then

$\mathcal{L}\{f * g\}(s) = F(s)G(s)$

where the convolution is $(f * g)(t) = \int_0^t f(\tau)g(t - \tau)\,d\tau$ .

Proof

$F(s)G(s) = \int_0^\infty e^{-s\tau}f(\tau)\,d\tau \cdot \int_0^\infty e^{-s\sigma}g(\sigma)\,d\sigma = \int_0^\infty\int_0^\infty e^{-s(\tau+\sigma)}f(\tau)g(\sigma)\,d\tau\,d\sigma.$

Substitute $t = \tau + \sigma$ (so $\sigma = t - \tau$ , $d\sigma = dt$ ). The region of integration $\tau \geq 0, \sigma \geq 0$ becomes $0 \leq \tau \leq t$ , $t \geq 0$ :

$F(s)G(s) = \int_0^\infty e^{-st}\left(\int_0^t f(\tau)g(t-\tau)\,d\tau\right)dt = \int_0^\infty e^{-st}(f*g)(t)\,dt = \mathcal{L}\{f*g\}(s).$

The interchange of order of integration is justified by absolute convergence (exponential order condition). $\blacksquare$

■

Applications

ExampleSolving integral equations

Solve the integral equation $y(t) = t + \int_0^t y(\tau)\sin(t-\tau)\,d\tau$ .

This is $y = t + y * \sin t$ . Taking transforms: $Y = \frac{1}{s^2} + Y \cdot \frac{1}{s^2+1}$ .

$Y\left(1 - \frac{1}{s^2+1}\right) = \frac{1}{s^2} \implies Y\cdot\frac{s^2}{s^2+1} = \frac{1}{s^2} \implies Y = \frac{s^2+1}{s^4} = \frac{1}{s^2} + \frac{1}{s^4}.$

Inverting: $y(t) = t + \frac{t^3}{6}$ .

ExampleVariation of parameters via convolution

The solution to $y'' + \omega^2 y = g(t)$ , $y(0) = y'(0) = 0$ is:

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

By the convolution theorem: $y(t) = \frac{1}{\omega}\int_0^t g(\tau)\sin[\omega(t-\tau)]\,d\tau$ .

This is precisely the variation of parameters formula, derived algebraically.

Properties of Convolution

RemarkAlgebraic properties

Convolution satisfies:

Commutativity: $f * g = g * f$ .
Associativity: $(f * g) * h = f * (g * h)$ .
Distributivity: $f * (g + h) = f * g + f * h$ .
Zero element: $f * 0 = 0$ .
No identity: There is no ordinary function $e$ with $f * e = f$ for all $f$ . However, the Dirac delta $\delta$ serves as a generalized identity: $f * \delta = f$ .

ExampleDirect computation

Compute $(e^t * e^{2t})(t) = \int_0^t e^\tau e^{2(t-\tau)}\,d\tau = e^{2t}\int_0^t e^{-\tau}\,d\tau = e^{2t}(1-e^{-t}) = e^{2t} - e^t$ .

Verification: $\mathcal{L}\{e^{2t}-e^t\} = \frac{1}{s-2}-\frac{1}{s-1} = \frac{1}{(s-1)(s-2)} = \frac{1}{s-1}\cdot\frac{1}{s-2} = F(s)G(s)$ .