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## Section6.4Convolution

### Subsection6.4.1Convolution

If $f$ and $g$ are two piecewise continuous exponentially bounded functions, then we define the convolution product of $f$ and $g$ to be

\begin{equation*} (f*g)(t) = \int_0^t f(t - \tau) g(\tau) \, d\tau = \int_0^t f(\tau) g(t - \tau) \, d\tau. \end{equation*}

The convolution product has many properties similar to those of ordinary multiplication.

• Commutivity: $f*g = g*f\text{.}$
• Distribution: $f*(g + h) = f*g + f*h\text{.}$
• Associativity: $f*(g*h) = (f*g)*h\text{.}$
• $0 *f = f*0 = 0\text{.}$

All of these properties can be proven using the definition of convolution and the properties of integration.

There is, however, no multiplicative identity. In other words, $f*1 \neq f\text{.}$ For example, suppose that $f(t) = \cos t\text{.}$ Then

\begin{equation*} (f*1)(t) = \int_0^t \cos(t - \tau) \cdot 1 \, d\tau = \sin t. \end{equation*}

Also, it may not be the case that $f*f$ is a nonnegative function.

One important property of the Laplace transform is how convolution products behave.

If

\begin{align*} F(s) & = \int_0^\infty e^{-s\xi} f(\xi) \, d\xi\\ G(s) & = \int_0^\infty e^{-s\tau} g(\tau) \, d\tau, \end{align*}

then

\begin{align*} F(s) G(s) & = \int_0^\infty e^{-s\xi} f(\xi) \, d\xi \int_0^\infty e^{-s\tau} g(\tau) \, d\tau\\ & = \int_0^\infty g(\tau) \left( \int_\tau^\infty e^{-st} f(t - \tau) \, dt \right) \, d\tau, \end{align*}

where $\xi = t - \tau$ is the change of variable. Reversing the order of integration, we have

\begin{equation*} \int_0^\infty g(\tau) \left( \int_\tau^\infty e^{-st} f(t - \tau) \, dt \right) \, d\tau = \int_0^\infty e^{-st} \left( \int_0^t f(t - \tau) g(\tau) \, d\tau \right) \, dt. \end{equation*}

However, this last expression is just the Laplace transform of $f*g\text{.}$

### Subsection6.4.2Applying the Convolution Theorem

Let us use the Convolution Theorem (Theorem 6.4.1) to calculate the inverse Laplace transform of

\begin{equation*} H(s)= \frac{a}{s^2(s^2 + a^2)} = \frac{1}{s^2} \cdot \frac{a}{s^2 + a^2}. \end{equation*}

instead of using partial fractions. The inverse Laplace transform of $1/s^2$ is $t\text{,}$ and the inverse Laplace transform of $a/(s^2 + a^2)\text{.}$ is $\sin at\text{.}$ Using the Convolution Theorem, the inverse Laplace transform of $H(s)$ is

\begin{equation*} h(t) = \int_0^t (t - \tau) \sin a \tau \, d \tau = \frac{at - \sin at}{a^2}. \end{equation*}

We can also use the Convolution Theorem to solve initial value problems. Consider the IVP

\begin{align*} y'' + 4y & = g(t)\\ y(0) & = 3\\ y'(0) & = -1. \end{align*}

Taking the Laplace transform of both sides of the differential equation and applying the initial conditions, we obtain

\begin{equation*} s^2 Y(s) - 3s + 1 + 4Y(s) = G(s), \end{equation*}

where $G(s)$ is the Laplace transform of $g(t)\text{.}$ Solving for $Y(s)\text{,}$ we have

\begin{equation*} Y(s) = \frac{3s-1}{s^2 + 4} + \frac{G(s)}{s^2 + 4} = 3\frac{s}{s^2 + 4} - \frac{1}{2} \frac{2}{s^2 + 4} + \frac{1}{2} \frac{2}{s^2 + 4} G(s). \end{equation*}

The last term corresponds to the forcing term of our differential equation. Taking the inverse Laplace transform of both sides and applying the Convolution Theorem, we get

\begin{equation*} y = 3 \cos 2t - \frac{2}{2} \sin 2t + \frac{1}{2} \int_0^t \sin 2(t - \tau) g(\tau) \, d \tau. \end{equation*}

It is possible to write a solution for the initial value problem

\begin{align*} ay'' + by' + cy & = g(t)\\ y(0) & = y_0\\ y'(0) & = y_1 \end{align*}

using the Convolution Theorem. Taking the Laplace transform of both sides of the differential equation and using the initial conditions, we have

\begin{equation*} (as^2 + bs + c)Y(s) - (as + b)y_0 - ay_1 = G(s) \end{equation*}

or

\begin{equation*} Y(s) = \Phi(s) + \Psi(s), \end{equation*}

where

\begin{align*} \Phi(s) & = \frac{(as + b)y_0 + ay_1 }{as^2 + bs + c}\\ \Psi(s) & = \frac{G(s)}{as^2 + bs + c}. \end{align*}

Therefore,

\begin{equation*} y = \phi(t) + \psi(t), \end{equation*}

where $\phi(t) = {\mathcal L}^{-1}( \Phi(s))$ and $\psi(t) = {\mathcal L}^{-1}( \Psi(s))\text{.}$ Observe that $\phi(t)$ is the solution to the initial value problem

\begin{align*} ay'' + by' + cy & = 0\\ y(0) & = y_0\\ y'(0) & = y_1. \end{align*}

Once we have values for $a\text{,}$ $b\text{,}$ and $c\text{,}$ the function $\phi(t)$ is easy to find. To find $\psi(t)\text{,}$ we first write

\begin{equation*} \Psi(s) = \frac{1}{as^2 + bs + c} G(s) = H(s) G(s), \end{equation*}

where $H(s) = 1/(as^2 + bs + c)\text{.}$ If $h(t)$ is the inverse Laplace transform of $H(s)\text{,}$ then

\begin{equation*} \psi(t) = {\mathcal L}^{-1}(H(s)G(s)) = \int_0^t h(t - \tau) g(\tau) \, d\tau. \end{equation*}

Let us consider the case where $G(s) = 1\text{.}$ Consequently, $g(t) = \delta(t)$ and $\psi(s) = H(s)\text{.}$ This means that $h(t)$ is a solution to the initial value problem

\begin{align*} ay'' + by' + cy & = \delta(t)\\ y(0) & = y_0\\ y'(0) & = y_1. \end{align*}

For this reason, $h(t)$ is sometimes called the impulse response of the system.

### Subsection6.4.3Important Lessons

• If $f$ and $g$ are two piecewise continuous exponentially bounded functions, then we define the convolution production of $f$ and $g$ to be
\begin{equation*} (f*g)(t) = \int_0^t f(t - \tau) g(\tau) \, d\tau = \int_0^t f(\tau) g(t - \tau) \, d\tau. \end{equation*}
• The convolution product has many properties similar to those of ordinary multiplication.

• Commutivity: $f*g = g*f$
• Distribution: $f*(g + h) = f*g + f*h$
• Associativity: $f*(g*h) = (f*g)*h$
• $0 *f = f*0 = 0$

There is, however, no multiplicative identity. In other words, $f*1 \neq f\text{.}$ Also, it may not be the case that $f*f$ is a nonnegative function.

• Let $f$ and $g$ be two piecewise continuous exponentially bounded functions, and suppose that ${\mathcal L}(f)(s) = F(s)$ and ${\mathcal L}(g)(s) = G(s)$ for $s \geq a \gt 0\text{.}$ Then
\begin{equation*} F(s) G(s) = {\mathcal L}(f*g)(s) \end{equation*}
for $s \gt a\text{.}$
• It is possible to write a solution for the initial value problem
\begin{align*} ay'' + by' + cy & = g(t)\\ y(0) & = y_0\\ y'(0) & = y_1. \end{align*}
using the Convolution Theorem.

### SubsectionExercises

###### 1

Using convolution of functions, find the solution to the initial value problem

\begin{gather*} y'' + 4y' + 13y = g(t)\\ y(0) = -5, \qquad y'(0) = 2, \end{gather*}

and $g(t)$ is a piecewise continuous function.