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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*}

Figure6.6Solution to \(y'' + 2y' + 26 y -\delta_4(t)\)

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.