Math 555: Differential Equations

6.5 Impulse Functions

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6.5.1 Impulse Functions

In previous sections, we have seen the utility of the unit step or Heaviside function , using them to represent turning a forcing function on or off and to "ramp" from one input value to another. Other applied problems involve a force that applied to the system over a small interval of time, an impulse .

For example, imagine a car that hits a bump while driving on a highway. The effect of driving over that bump is essentially instantaneous, and we could model the shock absorbers response using that impulse added to normal response of driving over the road. Likewise, an impulse could represent a spark or a quick capacitor discharge in an electrical circuit.

To understand how to work with this function, it is necessary to introduce its mathematical construction.

6.5.2 Modeling an Impulse

We want to model an impulse , which is a large force applied over a small time interval. Suppose that

$$ d_\tau(t) = \left\{\begin{matrix} \frac{1}{2\tau}, & t\lt \vert\tau\vert \\ \ 0, & t\ge \vert\tau\vert, \end{matrix}\right. $$
where $d_\tau(t)$ is representing the force, and it is being applied over the interval $t\in (-\tau,\tau)$. The impulse itself will be represented by the integral

$$ \mathcal{l}(\tau) = \int_{-\tau}^\tau d_\tau (t)\, dt = \int_{-\infty}^\infty d_\tau(t)\, dt, $$
where these are equal because $d_\tau(t) = 0$ outside of the specified interval.
Looking at our integral $l(\tau)$, we see that regardless of the choice of $\tau\gt 0$, its value is $1$. This is because the area under the function is always a rectangle with width $2\tau$ and height $\frac{1}{2\tau}$.

The next step is to look at what happens when we take the limit as $\tau\rightarrow 0^+$. We see that it is still the case that the integral $l(\tau) = 1$ for each nonzero $\tau$, so we have

$$ \lim_{t\rightarrow 0^+} l(\tau) = 1. $$

Mathematically, we want to be able to talk about the limiting function, the one where all of the force is applied in an instant, a time interval with zero width, and the height of the function equal to "infinity" (the quotes are very important here, as infinity is not a value). This generalized function or distribution is called the unit impulse function or the Dirac delta function , after Paul Dirac (Nobel Prize winner for his work in quantum mechanics).

The idea of distributions opens up a very broad set of possibilities for studying mathematical physics and statistics, having deep connections to the fields of functional analysis and measure theory that defined a lot of the mathematical developments of the 20th century. We do not have time to explore more than the basic uses of the Dirac distribution, but this is a significant point of contact with the study of graduate level mathematics.

Integral Formulation of the Dirac Delta Function

The Dirac delta function (commonly shortened to the Dirac function, delta function, or impulse function) is not a function in the usual sense. It is defined by the following properties

1. $\delta(t) = 0,\quad t\neq 0$
   
   
2. $\displaystyle\int_{-\infty}^\infty \delta(t)\, dt = 1.$

As mentioned above, it behaves like $\delta(0) = \infty$, but this is a heuristic at best. Generalized functions such as this only truly make sense while being operated on by an integral. For instance, if we have an interval $[a,b]$, then the integral of $\delta(t)$ will be $1$ if $0\in [a,b]$ and $0$ otherwise. This can also be shifted horizontally

$$ \int_a^b \delta(t-t_0)\, dt = \left\{\begin{matrix} 1, & t_0\in [a, b]\, \\ 0, & t_0\notin [a,b].\end{matrix}\right. $$

6.5.3 Computing the Laplace Transform

Technically, the delta function does not satisfy our existence theorem for Laplace transforms, but we aren't going to let that stop us. It can be defined formally (which means the notation all works and produces a meaningful result). We'll use the fact that $\delta(t-t_0) = \lim_{\tau\rightarrow 0^+} d_\tau(t-t_0)$ to find it.

$$ \begin{aligned} \mathcal{L}\{\delta(t-t_0)\} &= \int_0^\infty e^{-st}\delta(t-t_0)\, dt \\ \\ &= \lim_{\tau\rightarrow 0^+} \int_0^\infty e^{-st}d_\tau(t-t_0)\, dt \\ \\ &= \lim_{\tau\rightarrow 0^+} \int_{t_0-\tau}^{t_0+\tau} e^{-st}d_\tau(t-t_0)\, dt \\ \\ &= \lim_{\tau\rightarrow 0^+} \frac{1}{2\tau}\int_{t_0-\tau}^{t_0+\tau} e^{-st}\, dt \\ \\ &= \lim_{\tau\rightarrow 0^+} \left.\frac{-1}{2\tau s}e^{-st}\right\vert_{\ t_0-\tau}^{\ t_0+\tau} \\ \\ &= \lim_{\tau\rightarrow 0^+} \frac{-1}{2\tau s}\left( e^{-s(t_0+\tau)} - e^{-s(t_0-\tau)}\right) \\ \\ &= \lim_{\tau\rightarrow 0^+} \frac{e^{-st_0}}{2\tau s}\left( e^{s\tau} - e^{-s\tau}\right) \\ \\ &= \lim_{\tau\rightarrow 0^+} \frac{e^{-st_0}}{s\tau}\sinh(s\tau) \\ \\ &= e^{-st_0} \end{aligned} $$
The last line comes from l'Hospital's rule, since $ \displaystyle\lim_{\tau\rightarrow 0^+} \frac{\sinh(s\tau) }{s\tau} = \displaystyle\lim_{\tau\rightarrow 0^+} \frac{s\cosh(s\tau) }{s} = 1. $

This gives us the formulas

$$ \mathcal{L}\{\delta(t-t_0)\} = e^{-st_0} $$
and

$$ \mathcal{L}\{\delta(t)\} = \lim_{t_0\rightarrow 0^+} e^{-st_0} = 1. $$

6.5.4 Properties of the Dirac Delta Function

Since we are going to be encountering Dirac delta functions in the context of Laplace transforms, it is necessary to have a way to interpret how it interacts with other functions. In particular, we are interested in the integral of the product of $\delta(t-t_0)$ with some other function

$$ \begin{aligned} \int_{-\infty}^\infty \delta(t-t_0)f(t)\, dt &= \lim_{\tau\rightarrow 0^+} \int_{-\infty}^\infty d_\tau(t-t_0)f(t)\, dt \\ \\ &= \lim_{\tau\rightarrow 0^+} \frac{1}{2\tau} \int_{t_0 - \tau}^{t_0 + \tau} f(t)\, dt \end{aligned} $$
To evaluate the integral in this expression, we utilize the Mean Value Theorem for Integration, which states that for any integrable function $f$ on $[a,b]$, there is a value $t^*\in [a,b]$ so that $\int_a^b f(t)\, dt = (b-a)f(t^*)$. Hence,

$$ \begin{aligned} \int_{-\infty}^\infty \delta(t-t_0)f(t)\, dt &= \lim_{\tau\rightarrow 0^+} \frac{1}{2\tau} \int_{t_0 - \tau}^{t_0 + \tau} f(t)\, dt \\ \\ &= \lim_{\tau\rightarrow 0^+} \frac{1}{2\tau} 2\tau f(t^*) \\ \\ &= \lim_{\tau\rightarrow 0^+} f(t^*),\quad t^*\in [t_0-\tau, t_0+\tau]. \end{aligned} $$
In the limit $\tau\rightarrow 0^+$, this gives us that

$$ \int_{-\infty}^\infty \delta(t-t_0)f(t)\, dt = f(t_0). $$
This is the primary interpretation of the Dirac delta function, which concentrates the entire effect of applying the function $f(t)$ at a single point $t = t_0$. It is an impulse of the function $f(t)$ at time $t_0$.

With this understanding, we are now ready for an example featuring an impulsive forcing function.

6.5.5 Examples with Impulsive Forcing Functions

Example 6.5.1

Solve the initial value problem

$$ y'' + 4y = 2\delta(t-\pi) - 2\delta(t-2\pi);\qquad y(0) = 0,\ \ y'(0) = 0. $$

Solution

We begin by referencing the Laplace Transform Table and converting our ODE to the $s$-domain.

$$ \begin{aligned} \mathcal{L}\left\{y'' + 4y\right\} &= \mathcal{L}\left\{2\delta(t-\pi) - 2\delta(t-2\pi)\right\} \\ \\ s^2 Y(s) - sy(0) - y'(0) + 4Y(s) &= 2e^{-\pi s} - 2e^{-2\pi s} \\ \\ (s^2 + 4)Y(s) &= 2e^{-\pi s} - 2e^{-2\pi s} \\ \\ Y(s) &= \left( e^{-\pi s} - e^{-2\pi s} \right) \underbrace{\dfrac{2}{s^2 + 4}}_{H(s)} \end{aligned} $$

We see that the terms are of the form $e^{cs}H(s)$, so the inverse transform will be a product of a Heaviside function and a shifted function. The solution is

$$ y(t) = u_\pi(t)h(t-\pi) - u_{2\pi}(t)h(t-2\pi), $$
where

$$ h(t) = \mathcal{L}^{-1}\{H(s)\} = \mathcal{L}^{-1}\left\{\frac{2}{s^2+4}\right\} = \sin(2t). $$

The plot of this function is
We see that this allows us to create a really unique effect of starting a function's typical harmonic motion (even though the initial conditions are both zero) and then stopping it. The "starting" and "stopping" are instantaneous, which utilizes the special nature of the delta function.

Example 6.5.2

In Example 1, we started motion an then turned it off using impulses. What happens if the impulse is applied in the positive direction each time, instead of alternating from positive to negative? The initial value problem in this case, for $N$ impulses at an interval of $\pi$ each, is

$$ y'' + 4y = 2\sum_{k=1}^N \delta(t-k\pi);\qquad y(0) = 0,\ \ y'(0) = 0. $$

Solution

The details here are very similar to Example 6.5.1. Work them out for yourself. The solution has the form

$$ y(t) = \sum_{k=1}^N u_{k\pi}(t)\sin(2(t-k\pi)), $$
with plot

At each time $t = k\pi$, the amplitude of the oscillation increases by $1$. Visually, this is almost indistinguishable from resonance. The effect is what happens when an impulse is applied at exactly the natural frequency of the oscillator.

Example 6.5.3

Solve the initial value problem

$$ y'' + y = \delta(t - 2\pi)\cos t;\qquad y(0) = 0,\ \ y'(0) = 1 $$
and plot the graph of the solution.

Solution

We want to use Laplace transforms to solve this problem. On the left hand side, we appeal to the Laplace Transform Table . On the right hand side, we use the definition of the Laplace transform and the property shown above for the Dirac delta function

$$ \begin{aligned} s^2Y(s) - sy(0) - y'(0) + Y(s) &= \mathcal{L}\left\{\delta(t - 2\pi)\cos t\right\} \\ \\ (s^2 + 1)Y(s) - 1 &= \int_0^\infty e^{-st}\delta(t - 2\pi)\cos t\, dt \\ \\ (s^2 + 1)Y(s) - 1 &= e^{-2\pi s}\cos(2\pi) \\ &\Downarrow \\ Y(s) &= \left(1 + e^{-2\pi s}\right)\underbrace{\frac{1}{s^2 + 1}}_{H(s)} \end{aligned} $$

The inverse transform of $H(s)$ is $\sin t$, so we have

$$ y(t) = \sin t + u_{2\pi}(t)\sin(t-2\pi) $$
as the solution.

6.5.6 Creating Plots with Technology

Most of the function plots this chapter were created in Python using the numpy and matplotlib libraries. However, while these tools are very powerful they require some set up and are best used in situations where you need a lot of control over how the final plot looks. For quicker plots, Desmos provides a versatile and intuitive interface.

The Desmos interface allows for many things, such as the definition of custom functions, plots of equations (as opposed to explicit functions), and numerical operations such as sums, derivatives, and integrals. Defining custom functions such as the Heaviside function is very valuable to us, since our solutions to the differential equations of this and the previous sections involve them. In the video below, the previous two examples are plotted using Desmos to demonstrate how to use it.

Since the answer is simpler to plot, we begin with Example 6.5.3 and then do Example 6.5.2.

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