The Laplace Transform: Convolution

Definition. The convolution $F\ast G$ of $F(t)$ and $G(t)$ is defined by
$$F(t)\ast G(t)=\int_0^t F(\tau)G(t-\tau)d\tau$$

We introduce Convolution Theorem without a proof. For those who are interested a proof can be found in  of the References below.

Theorem. Let $\mathcal{L}\{F(t)\}=f(s)$ and $\mathcal{L}\{G(t)\}=g(s)$. Suppose that $F(t)$ and $G(t)$ are piecewise continuous and are of order $e^{\alpha t}$ as $t\to\infty$. Then $\mathcal{L}\{F(t)\ast G(t)\}$ exists when $s>\alpha$ and it is $f(s)g(s)$. Equivalently,
$$\mathcal{L}^{-1}\{f(s)g(s)\}=F(t)\ast G(t)$$

Example. Let $F(t)=t$ and $G(t)=e^{at}$. Then $\mathcal{L}\{F(t)\}=\frac{1}{s^2}$ and $\mathcal{L}\{G(t)\}=\frac{1}{s-a}$. By Convolution Theorem
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{1}{s^2}\frac{1}{s-a}\right\}&=t\ast e^{at}\\
&=\int_0^t\tau e^{a(t-\tau)}d\tau\\
&=\frac{1}{a^2}(e^{at}-at-1)
\end{align*}
Note that partial fractions can be also used to obtain the result.

When $G(t)=F(t)$, we have the formula
$$[f(s)]^2=\mathcal{L}\{F(t)\ast F(t)\}$$

Example.
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{1}{(s^2+k^2)^2}\right\}&=\frac{1}{k^2}\mathcal{L}^{-1}\left\{\frac{k^2}{(s^2+k^2)^2}\right\}\\
&=\frac{1}{k^2}\int_0^t\sin kt\ast \sin kt\\
&=\frac{1}{k^2}\int_0^t\sin k\tau\sin k(t-\tau)d\tau\\
&=\frac{1}{k^3}\{\sin kt-kt\cos kt\}
\end{align*}

Theorem (Properties of Convolution).

1. $F(t)\ast G(t)=G(t)\ast F(t)$
2. $F(t)\ast[G(t)+H(t)]=F(t)\ast G(t)+F(t)\ast H(t)$
3. $F(t)\ast [kG(t)]=k[F(t)\ast G(t)]$ where $k$ is a constant.
4. $F(t)\ast[G(t)\ast H(t)]=[F(t)\ast G(t)]\ast H(t)$

Proof. We prove only the part 1.
\begin{align*}
F(t)\ast G(t)&=\int_0^t F(\tau)G(t-\tau)d\tau\\
&=\int_0^tF(t-\lambda)G(\lambda)d\lambda\ (\lambda=t-\tau)\\
&=G(t)\ast F(t)
\end{align*}

\begin{align*}
\frac{1}{s}f(s)&=\mathcal{L}\{F(t)\ast 1\}\\
&=\mathcal{L}\left\{\int_0^t F(\tau)d\tau\right\}\\
\frac{1}{s^2}f(s)&=\mathcal{L}\{F(t)\ast 1\ast 1\}\\
&=\mathcal{L}\{(F(t)\ast 1)\ast 1\}\\
&=\mathcal{L}\left\{\int_0^t(F(\tau)\ast 1)d\tau\right\}\\
&=\mathcal{L}\left\{\int_0^t\int_0^{\tau}F(\lambda)d\lambda d\tau\right\}
\end{align*}
Therefore we have the following theorem.

Theorem.
\begin{align}\label{eq:conv}\mathcal{L}^{-1}\left\{\frac{1}{s}f(s)\right\}&=\int_0^tF(\tau)d\tau\\
\label{eq:conv2}\mathcal{L}^{-1}\left\{\frac{1}{s^2}f(s)\right\}&=\int_0^t\int_0^{\tau}F(\lambda)d\lambda d\tau
\end{align}

Convolution Theorem can be used for solving non-homogeneous linear differential equations.

Example. Find the general solution of the differential equations
$$\ddot{X}(t)+k^2X(t)=F(t)$$

Solution. Let $\mathcal{L}\{X(t)\}=x(s)$. Then the transformed equation is
$$s^2x(s)-sX(0)-\dot{X}(0)+k^2x(s)=f(s)$$
So we have
$$x(s)=\frac{1}{k}\frac{k}{s^2+k^2}f(s)+\frac{s}{s^2+k^2}X(0)+\frac{1}{k}\frac{k}{s^2+k^2}\dot{X}(0)$$
Therefore by Convolution Theorem the general solution is given by
\begin{equation}
\begin{aligned}
X(t)&=\frac{1}{k}(\sin kt)\ast F(t)+X(0)\cos kt+\frac{\dot{X}(0)}{k}\sin kt\\
&=\frac{1}{k}\int_0^t\sin k(t-\tau)F(\tau)d\tau+C_1\cos kt+C_2\sin kt
\end{aligned}\label{eq:conv3}
\end{equation}

Let us redo the problem in the last example in here using \eqref{eq:conv3}.

Example. Find the general solution of the differential equation
$$\ddot{x}+4x=8\tan t,\ -\frac{\pi}{2}<x<\frac{\pi}{2}$$

Solution. Using the formula \eqref{eq:conv3} with $k=2$
\begin{align*}
x(t)&=\frac{1}{2}\int_0^t\sin 2(t-\tau)(8\tan\tau)d\tau+C_1\cos 2t+C_2\sin 2t\\
&=4\int_0^t\sin 2(t-\tau)\tan\tau d\tau+C_1\cos 2t+C_2\sin 2t\\
&=C_1\cos 2t+C_2\sin 2t-4t\cos 2t+4\sin 2t\ln(\cos t)
\end{align*}

Convolution Theorem can be also used for solving integral equations as shown in the following example.

Example. Solve the integral equation
$$X(t)=at+\int_0^tX(\tau)\sin(t-\tau)d\tau$$

Solution. $X(t)=at+X(t)\ast \sin t$ so its transformed equation is
$$x(s)=\frac{a}{s^2}+x(s)\frac{1}{s^2+1}$$
Hence we have
\begin{align*}
x(s)&=\frac{a}{s^2}\frac{s^2+1}{s^2}\\
&=a\left(\frac{1}{s^2}+\frac{1}{s^4}\right)
\end{align*}
Therefore the solution is given by
$$X(t)=a\left(t+\frac{1}{6}t^3\right)$$

References:

 Ruel V. Churchill, Operational Mathematics, McGraw-Hill, 1958