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Coordinated Differential Equations

Thomas W. Judson, Austin Eide, Adam Larios

Section 3.3 Delta Functions and Forcing

Subsection 3.3.1 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
\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*}

Subsection 3.3.2 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
\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*}

Example 3.3.1.

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*}
described in detail following the image
a piecewise smooth solution curve for a harmonic oscillator plotted against time
Figure 3.3.2. Solution to \(y'' + 2y' + 26 y -\delta_4(t)\)

Example 3.3.3.

Solve the initial value problem
\begin{equation*} y' + y = t^2 + 2t + u_1(t)(t + 1 + \delta_1(t)), y(0) = 0 \end{equation*}
We take the Laplace transformation of both sides. To take the Laplace transformation of \(u_1(t)(t + 1 + \delta_1(t))\text{,}\) we rewrite the factor being multiplied by \(u_1(t)\) replacing \(t\) with \(t+1\text{,}\) giving \(t+2 + \delta_1(t+1) = t + 2 + \delta_0(t+1-1) = t + 2 + \delta_0(t)\text{;}\) take the Laplace of this and multiply by \(e^{-1s}\text{,}\) giving \(e^{-s}(\frac{1}{s^2} + \frac{2}{s} + 1)\text{.}\) We arrive at the equation
\begin{equation*} sY(s) + Y(s) = \frac{2}{s^3} + \frac{2}{s^2} + e^{-s}(\frac{1}{s^2} + \frac{2}{s} + 1) \end{equation*}
Therefore,
\begin{equation*} Y(s) = \frac{2}{s^3(s+1)} + \frac{2}{s^2(s+1)} + e^{-s}(\frac{1}{s^2(s+1)} + \frac{2}{s(s+1)} + \frac{1}{s+1}) \end{equation*}
It looks like we have to do extensive partial fractions, but we don’t; observe that
\begin{equation*} \frac{2}{s^3(s+1)} + \frac{2}{s^2(s+1)} = \frac{2+2s}{s^3(s+1)} = \frac{2}{s^3} \end{equation*}
and \(\frac{1}{s^2(s+1)} + \frac{2}{s(s+1)} + \frac{1}{s+1} = \frac{1+2s+s^2}{s^2(s+1)} = \frac{s+1}{s^2}=\frac1s + \frac{1}{s^2}\) So we end up at the equation
\begin{equation*} Y(s) = \frac{2}{s^3} + e^{-s}(\frac1s + \frac{1}{s^2}) \end{equation*}
We now have to take the inverse Laplace to solve for \(y(t)\text{.}\) The first term on the right gives us \(t^2\text{.}\) The inverse Laplace of \(\frac1s + \frac{1}{s^2}\) is \(1 + t\text{.}\) Since \(\frac1s + \frac{1}{s^2}\) is multiplied by \(e^{-s}\text{,}\) we subtract 1 to the arguments and multiply by \(\delta_1(t)\) to get the inverse Laplace. The final answer is
\begin{equation*} y(t) = t^2 + \delta_0(t)\cdot t \end{equation*}
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 3.3.3 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 “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.

Reading Questions 3.3.4 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 3.3.5 Exercises

Solving Initial Value Problems.

Solve the initial problems in Exercise Group 3.3.5.1–6 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.
\(y'' + 2y' + y = \delta(t + 4)\text{,}\) \(y(0) = -1\text{,}\) \(y'(0) = 3\)
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