Recall that the Laplace transform is linear: $\alpha ,\beta \in ℝ$, $f,g:[0,\infty )\to ℝ$

• The linearity property allows solution of constant-coefficient ODEs, e.g. $y^{\text{'}\text{'}}+a{y}^{\text{'}}+by=r\left(t\right)$

• What about variable-coefficient ODEs, e.g., $y^{\text{'}\text{'}}+p\left(t\right)y^{\text{'}}+q\left(t\right)y=r$?

• This question leads to consideration of $ℒ(fg)$, and ${ℒ}^{-1}(FG)$.

Theorem. Consider $f,g:[0,\infty )\to ℝ$, with Laplace transforms $F\left(s\right)=ℒ\left(f\right)\left(s\right),G\left(s\right)=ℒ\left(g\right)\left(s\right).$ The product $H\left(s\right)=F\left(s\right)G\left(s\right)=ℒ\left(h\right)$ is the Laplace transform of $h\left(t\right)=(f\ast g)\left(t\right)$, where the convolution product $f\ast g$ is

Proof.

• Write Laplace transforms as $F\left(s\right)={\int }_0^{\infty }{e}^{-s\tau }f\left(\tau \right)d\tau$, $G\left(s\right)={\int }_0^{\infty }{e}^{-sp}g\left(p\right)dp$

• Change integration variable $p=t-\tau$, $G\left(s\right)={\int }_{\tau }^{\infty }{e}^{-s(t-\tau )}g(t-\tau )dt=e^{s\tau }{\int }_{\tau }^{\infty }{e}^{-st}g(t-\tau )dt$

• $(t,\tau )$ are independent hence

  ${\displaystyle F\left(s\right)G\left(s\right)=F\left(s\right)\underset{\tau }{\overset{\infty }{\int }}{e}^{-s(t-\tau )}g(t-\tau )dt=\underset{\tau }{\overset{\infty }{\int }}F\left(s\right)e^{-s(t-\tau )}g(t-\tau )dt=}$ ${\displaystyle =\underset{\tau }{\overset{\infty }{\int }}(\underset{0}{\overset{\infty }{\int }}{e}^{-s\tau }f\left(\tau \right)d\tau )e^{-s(t-\tau )}g(t-\tau )dt=\underset{0}{\overset{\infty }{\int }}{e}^{-st}[\underset{0}{\overset{t}{\int }}f\left(\tau \right)g(t-\tau )d\tau ]dt}$