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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.
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*}
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.
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.