Section 6.3 Delta Functions and Forcing
Objectives

To understand Impulse forcing, a 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, and that an impulse function can be described by Dirac delta function, \(\delta(t)\text{,}\) which has the properties
\begin{gather*} \delta(t) = 0, \qquad t \neq 0;\\ \int_{\infty}^{\infty} \delta(t) \, dt = 1. \end{gather*} 
To understand that 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.
Subsection 6.3.1 Impulse 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
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
or since \(g(t)\) is zero outside of the interval \(t  t_0 \lt \tau\)
measures the strength or impulse of the forcing function \(g(t)\text{.}\) In particular, assume that \(t_0 = 0\) and
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,
for \(t \neq 0\text{;}\) however,
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
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,
Subsection 6.3.2 The 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
Assuming that \(t_0  \tau \gt 0\text{,}\) the Laplace transform of \(d_\tau(t  t_0)\) is
We can use l'Hospital's rule to evaluate \((\sinh s \tau)/ s \tau\) as \(\tau \to 0\text{,}\)
Thus,
We can extend this result to allow \(t_0 = 0\text{,}\) by
Example 6.3.1.
Let us now solve the initial value problem
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
or
Solving for \(Y(s)\text{,}\) we have
The inverse Laplace transform of \(Y(s)\) is
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.
Subsection 6.3.3 Important 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 â€śfunctionâ€ť \(\delta\) 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.
Reading Questions 6.3.4 Reading Questions
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.
Exercises 6.3.5 Exercises
Solving Initial Value Problems.
Solve the initial problems in Exercise GroupÂ 6.3.5.1â€“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.
\(y'' + 2y' + y = \delta(t + 4)\text{,}\) \(y(0) = 1\text{,}\) \(y'(0) = 3\)