
Section6.3Delta Functions and Forcing

Subsection6.3.1Impulse Forcing

Impulse forcing is the term used to describe a very quick push or pull on a system, such as the blow of a hammer or the force of an explosion. For example, consider the equation for a damped harmonic oscillator

\begin{equation*} \frac{d^2y}{dt^2} + 2 \frac{dy}{dt} + 26 y = g(t), \end{equation*}

where $g(t)$ is a function that is very large in a very short time interval, say $|t - t_0| \lt \tau$ and zero otherwise. The integral

\begin{equation*} I(\tau) = \int_{t_0 - \tau}^{t_0 + \tau} g(t) \, dt \end{equation*}

or since $g(t)$ is zero outside of the interval $|t - t_0| \lt \tau$

\begin{equation*} I(\tau) = \int_{-\infty}^{\infty} g(t) \, dt \end{equation*}

measures the strength or impulse of the forcing function $g(t)\text{.}$ In particular, assume that $t_0 = 0$ and

\begin{equation*} g(t) = d_{\tau}(t) = \begin{cases} 1/ 2\tau, & -\tau \lt t \lt \tau \\ 0, & \text{otherwise.} \end{cases} \end{equation*}

It is easy to see that $I(\tau) = 1$ in this case.

Now let us realize the forcing function over shorter and shorter time intervals with $\tau$ getting closer and closer to zero, we find that $I(\tau) = 1$ in all cases. Thus,

\begin{equation*} \lim_{\tau \to 0} d_\tau(t) = 0 \end{equation*}

for $t \neq 0\text{;}$ however,

\begin{equation*} \lim_{\tau \to 0} I(\tau ) = 1. \end{equation*}

We can use this information to define the unit impulse function, $\delta(t)\text{,}$ to be the function'' that imparts an impulse of magnitude one at $t = 0\text{,}$ but is zero for all values of $t$ other than zero. In other words, $\delta(t)$ has the properties

\begin{gather*} \delta(t) = 0, \qquad t \neq 0;\\ \int_{-\infty}^{\infty} \delta(t) \, dt = 1. \end{gather*}

Of course, we study no such function in calculus. The function'' $\delta$ is an example of what is known as a generalized function. We call $\delta\text{,}$ the Dirac delta function.

We can define a unit impulse at a point $t_0$ by considering the function $\delta( t - t_0)\text{.}$ In this case,

\begin{gather*} \delta(t - t_0) = 0, \qquad t \neq t_0;\\ \int_{-\infty}^{\infty} \delta(t - t_0) \, dt = 1. \end{gather*}

Subsection6.3.2The Laplace Transform of the Dirac Delta Function

Even though the Dirac delta function is not a piecewise continuous, exponentially bounded function, we can define its Laplace transform as the limit of the Laplace transform of $d_\tau(t)$ as $\tau \to 0\text{.}$ More specifically, assume that $t_0 \gt 0$ and

\begin{equation*} {\mathcal L}(\delta(t - t_0)) = \lim_{\tau \to 0} {\mathcal L}(d_\tau(t - t_0)). \end{equation*}

Assuming that $t_0 - \tau \gt 0\text{,}$ the Laplace transform of $d_\tau(t - t_0)$ is

\begin{align*} {\mathcal L}(d_\tau(t - t_0)) & = \int_0^\infty e^{-st} d_\tau(t - t_0) \, dt\\ & = \int_{t_0 - \tau}^{t_0 + \tau} e^{-st} d_\tau(t - t_0) \, dt\\ & = \frac{1}{2 \tau} \int_{t_0 - \tau}^{t_0 + \tau} e^{-st} \, dt\\ & = \frac{1}{2 s \tau} e^{-st} \bigg|_{t = t_0 - \tau}^{t = t_0 + \tau}\\ & = \frac{1}{2 s \tau} e^{-st_0} (e^{s\tau} - e^{-s \tau})\\ & = \frac{\sinh s \tau}{s \tau} e^{-st_0}. \end{align*}

We can use l'Hospital's rule to evaluate $(\sinh s \tau)/ s \tau$ as $\tau \to 0\text{,}$

\begin{equation*} \lim_{\tau \to 0} \frac{\sinh s \tau}{s \tau} = \lim_{\tau \to 0} \frac{s \cosh s \tau}{s} = 1. \end{equation*}

Thus,

\begin{equation*} {\mathcal L}(\delta(t - t_0)) = e^{-st_0}. \end{equation*}

We can extend this result to allow $t_0 = 0\text{,}$ by

\begin{equation*} \lim_{t_0 \to 0} {\mathcal L}(\delta(t - t_0)) =\lim_{t_0 \to 0} e^{-st_0} = 1. \end{equation*}

Let us now solve the initial value problem

\begin{align*} \frac{d^2y}{dt^2} + 2 \frac{dy}{dt} + 26 y & = \delta_4(t)\\ y(0) & = 1\\ y'(0) & = 0. \end{align*}

We can think of this as a damped harmonic oscillator that is struck by a hammer at time $t = 4\text{.}$ Let $Y(s) = {\mathcal L}(y)(s)$ and take the Laplace transform of both sides of the differential equation to obtain

\begin{equation*} s^2 Y(s) - sy(0) - y'(0) + 2(sY(s) - y(0)) + 26 Y(s) = {\mathcal L}(\delta_4)(s) \end{equation*}

or

\begin{equation*} s^2 Y(s) - s + 2sY(s) - 2 + 26 Y(s) = e^{-4s}. \end{equation*}

Solving for $Y(s)\text{,}$ we have

\begin{equation*} Y(s) = \frac{s + 2}{s^2 + 2s + 26} + \frac{e^{-4s}}{s^2 + 2s + 26}. \end{equation*}

The inverse Laplace transform of $Y(s)$ is

\begin{align*} y & = {\mathcal L}\left( \frac{s + 2}{s^2 + 2s + 26} \right) + {\mathcal L}\left(\frac{e^{-4s}}{s^2 + 2s + 26}\right)\\ & = {\mathcal L}\left( \frac{s + 2}{(s+ 1)^2 + 25} \right) + \frac{1}{5} {\mathcal L}\left(\frac{5e^{-4s}}{(s + 1)^2 + 25}\right)\\ & = e^{-t} \cos 5t + \frac{1}{5} e^{-t} \sin 5t + \frac{1}{5} u_4(t) e^{-(t - 4)} \sin(5(t-4)). \end{align*}

It is important to notice that we are using the Dirac delta function like an ordinary function. This requires some rigorous mathematics to justify that we can actually do this.

Subsection6.3.3Important Lessons

• Impulse forcing is the term used to describe a very quick push or pull on a system, such as the blow of a hammer or the force of an explosion. For example, consider the equation for a damped harmonic oscillator
\begin{equation*} \frac{d^2y}{dt^2} + p \frac{dy}{dt} + q y = g(t), \end{equation*}
where $g(t)$ is a function that is very large in a very short time interval, say $|t - t_0| \lt \tau$ and zero otherwise. The integral
\begin{equation*} I(\tau) = \int_{t_0 - \tau}^{t_0 + \tau} g(t) \, dt \end{equation*}
or since $g(t)$ is zero outside of the interval $|t - t_0| \lt \tau$
\begin{equation*} I(\tau) = \int_{-\infty}^{\infty} g(t) \, dt \end{equation*}
measures the strength or impulse of the forcing function $g(t)\text{.}$
• We define the unit impulse function, $\delta(t)\text{,}$ to be the function'' that imparts an impulse of magnitude one at $t = 0\text{,}$ but is zero for all values of $t$ other than zero. In other words, $\delta(t)$ has the properties
\begin{gather*} \delta(t) = 0, \qquad t \neq 0;\\ \int_{-\infty}^{\infty} \delta(t) \, dt = 1. \end{gather*}
The “function” $\delta$ is an example of what is known as a generalized function. We call $\delta\text{,}$ the Dirac delta function.
• Similarly, we can define a unit impulse at a point $t_0$ by considering the function $\delta( t - t_0)\text{.}$ In this case,
\begin{gather*} \delta(t - t_0) = 0, \qquad t \neq 0;\\ \int_{-\infty}^{\infty} \delta(t - t_0) \, dt = 1. \end{gather*}
• The Laplace transform of the Dirac delta function is
\begin{equation*} {\mathcal L}(\delta(t - t_0)) = e^{-st_0}. \end{equation*}
We can extend this result to allow $t_0 = 0\text{,}$ by
\begin{equation*} \lim_{t_0 \to 0} {\mathcal L}(\delta(t - t_0)) =\lim_{t_0 \to 0} e^{-st_0} = 1. \end{equation*}
• We can use the Dirac delta function to solve initial value problems such as
\begin{align*} \frac{d^2y}{dt^2} + 2 \frac{dy}{dt} + 26 y & = \delta_4(t)\\ y(0) & = 1\\ y'(0) & = 0, \end{align*}
or
\begin{equation*} \frac{d^2y}{dt^2} + p \frac{dy}{dt} + q y = g(t), \end{equation*}
where $g(t)$ is a function that is very large in a very short time interval.

Subsection6.3.4Exercises

1

Find the solution of the initial value problem

\begin{gather*} 2 y'' + y' + 2y = \delta(t - 5)\\ y(0) = y'(0) = 0, \end{gather*}

where $\delta(t)$ is the unit impulse function.