# The Ordinary Differential Equations Project

## 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.
Examining this 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.

#### 1.

What is impulse forcing? Give an example of a physical system where impulse forcing might be useful.

#### 2.

Is the Dirac delta function an actual function? Why or why not.

### Exercises6.3.5Exercises

#### Solving Initial Value Problems.

Solve the initial problems in Exercise Group 6.3.5.1–6 using the Laplace transform, $$\delta(t)$$ is the unit impulse function.
##### 1.
$$2 y'' + y' + 2y = \delta(t)\text{,}$$ $$y(0) = 0\text{,}$$ $$y'(0) = 0$$
##### 2.
$$y'' - y' - 2y = \delta(t - 5)\text{,}$$ $$y(0) = 0\text{,}$$ $$y'(0) = 0$$
##### 3.
$$\dfrac{d^2x}{dx^2} - 6 \dfrac{dx}{dt} + 25 x = \delta(t)\text{,}$$ $$x(0) = 1\text{,}$$ $$x'(0) = -2$$
##### 4.
$$y'' + 16y = \delta(t)\text{,}$$ $$y(0) = 1\text{,}$$ $$y'(0) = 0$$
##### 5.
$$y'' + 16y = \delta(t - 4)\text{,}$$ $$y(0) = 1\text{,}$$ $$y'(0) = 0$$
##### 6.
$$y'' + 2y' + y = \delta(t + 4)\text{,}$$ $$y(0) = -1\text{,}$$ $$y'(0) = 3$$