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Ordinary Differential Equations

Section 3.4 Dirac delta

In this section, we discuss how we can model an instantaneous force acting at a single moment in time. First, we discuss the unit impulse, which is a force acting over a time interval.

Definition 3.4.1. Unit impulse.

The piecewise function
\begin{equation*} \delta_{\epsilon}(t-a)=\left\{ \begin{array}{ll} 0, & \quad 0 \leq t < a-\epsilon \\ \dfrac{1}{2\epsilon}, & \quad a-\epsilon \leq t < a+\epsilon \\ 0, & \quad t \geq a+\epsilon \end{array}\right. \end{equation*}
is called a unit impulse since
\begin{equation*} \displaystyle\int_0^{\infty} \delta_{\epsilon}(t-a)\mathrm{d}t = 1. \end{equation*}
The Dirac delta is vaguely the limit as \(\epsilon\) tends to zero in the unit impulse. But this is not actually a function! However, we will not worry too much about that fact and treat it in our calculus as if it were a function.

Definition 3.4.2. Dirac delta.

The Dirac delta can be thought of as
\begin{equation*} \delta(t-a)=\displaystyle\lim_{\epsilon \rightarrow 0} \delta_{\epsilon}(t-a) = \left\{ \begin{array}{ll} 0, & \quad t \neq a \\ \infty, & \quad t=a. \end{array}\right. \end{equation*}
Although DefinitionΒ 3.4.2 does not actually define a real-valued function (since \(\infty\) is not a real number!), this expression captures the idea of an instantaneous impulse. A "true" definition of it can come out of the theory of distributions, which is beyond the scope of our course in ordinary differential equations. Nonetheless, it does possess many properties useful in differential equations.

Example 3.4.4.

Compute the following integrals.
  1. \(\displaystyle \displaystyle\int_0^{\infty} e^t \delta(t-2)\mathrm{d}t\)
  2. \(\displaystyle \displaystyle\int_0^{\infty} \sin(t)\delta(t-\frac{3\pi}{2}) \mathrm{d}t\)

Example 3.4.5.

Consider the initial value problem
\begin{equation*} y''+y=\delta(t-\frac{\pi}{2})+\delta(t-2\pi), y(0)=1, y'(0)=0. \end{equation*}
Solve this IVP and graph the solution.

Example 3.4.6.

Consider the initial value problem
\begin{equation*} y''-6y'+9y=\delta(t-2), y(0)=0, y'(0)=1. \end{equation*}
Solve this IVP and find the exact value of \(y(1)\) and \(y(4)\text{.}\)