For any positive integer \(n\), define the function \(f_n\) by \(\displaystyle f_n (t)=\left \{\begin{array}{lr} n(1-n|t|) & \mbox { if }|t|<\frac {1}{n}\\ \\ 0 & \mbox { if }|t|\geq \frac {1}{n} \end{array}\right .\).

Here are the first few terms of the sequence \(f_1,f_2,f_3,...\)

For any positive integer \(n\) and any closed interval \(I\) containing \(\left [-\frac {1}{n},\frac {1}{n}\right ]\),

\(\displaystyle \int _{I}f_{n}\;dt=\int _{-1/n}^{1/n}f_n(t)\;dt=\) \(\displaystyle 2\int _0^{1/n} n(1-nt)\,dt=\) \(2n\left .\left (t-\frac {1}{2}nt^2\right )\right |_0^{1/n}=1\).

The Dirac delta function \(\delta \) is the limit of the above sequence of functions, that is, \(\delta =\displaystyle \lim _{n\rightarrow \infty }f_{n}\).

For all positive integers \(n\), \(f_{n}(0)=n\), so we have that
\(\displaystyle \delta (0)=\lim _{n\rightarrow \infty }f_{n}(0)=\lim _{n\rightarrow \infty }n=\infty \). Since for any \(t\neq 0\), there exists a sufficiently large positive integer \(n\) such that \(t\) is outside \(\left [-\frac {1}{n},\frac {1}{n}\right ]\) it follows that \(f_m(t)=0\) for all integers \(m\geq n\). So \(\delta (t)=0\) if \(t\neq 0\).

Let \(I\) be a closed interval not containing \(0\). Then there exists a sufficiently large positive integer \(n\) such that all of \(I\) is outside \(\left [-\frac {1}{n},\frac {1}{n}\right ]\). Hence for all integers \(m\geq n\), \(f_m=0\) on all of \(I\), and hence \(\displaystyle \int _{I}f_{m}\;dt=0\). So \(\displaystyle \int _{I}\delta \;dt=0\) for any closed interval \(I\) not containing \(0\).

Let \(I\) be a closed interval containing \(0\) (in the interior). Then there exists a sufficiently large positive integer \(n\) such that \(I\) contains all of \(\left [-\frac {1}{n},\frac {1}{n}\right ]\). Hence for all integers \(m\geq n\), \(\displaystyle \int _{I}f_{m}\;dt=1\). So \(\displaystyle \int _{I}\delta \;dt=1\) for any closed interval \(I\) containing \(0\) (in the interior).

Note that if \(0\) is an endpoint of \(I\), then a different set of functions approaching \(\delta \) (e.g. \(f_n(x)=n\) if \(x\in [0,1/n]\) and \(0\) elsewhere, or \(f_n(x)=n\) if \(x\in [-1/n,0]\) and \(0\) elsewhere) are used to have \(\displaystyle \int _I \delta \, dt=1\) rather than \(1/2\).

The Dirac delta function models an impulse, where a force is applied over a very short period of time where it makes sense to model it as happening at an instant.

Let’s find the Laplace Transform of \(\delta (t-a)\).

For any real number \(s\), \(f(t)=e^{-st}\) is a continuous function on \([a-\epsilon ,a+\epsilon ]\) for any positive \(\epsilon \) and on \([a-\epsilon ,a+\epsilon ]\), \(f(t)\) attains extreme values of \(e^{-(a-\epsilon )s}\) and \(e^{-(a+\epsilon )s}\) (one is a max. and one is a min. – which is which depends on the value of \(s\)). Then

\(\displaystyle \mathcal {L}(\delta (t-a))=\int _{0}^{\infty }e^{-st}\delta (t-a)\,dt=\) \(\displaystyle \lim _{\epsilon \rightarrow 0^{+}}\int _{a-\epsilon }^{a+\epsilon }e^{-st}\delta (t-a)\;dt\) and the integral \(\displaystyle \int _{a-\epsilon }^{a+\epsilon }e^{-st}\delta (t-a)\;dt\) is between \(e^{-(a-\epsilon )s}\) and \(e^{-(a+\epsilon )s}\), which both approach \(e^{-as}\) as \(\epsilon \rightarrow 0^{+}\).

Hence \(\mathcal {L}(\delta (t-a))=e^{-as}\) by the Squeeze Theorem.

Let’s solve the IVP \(y''+2y'+5y=\delta (t-\pi )\), \(y(0)=1\), \(y'(0)=0\) (which could represent an underdamped spring-mass system where a total upward impulse of \(1\) Ns occurs at time \(t=\pi \)).

Take the Laplace transform of both sides, letting \(Y=\mathcal {L}(y)\): \(s^2 Y-s+2(sY-1)+5Y=e^{-\pi s}\).

So \((s^2+2s+5)Y=s+2+e^{-\pi s}\rightarrow \) \(\displaystyle Y=\frac {s+2}{(s+1)^2+4}+e^{-\pi s} \frac {1}{(s+1)^2 +4}\rightarrow \)

\(\displaystyle Y=\frac {s+1}{(s+1)^2+4}+\frac {1}{2} \,\frac {2}{(s+1)^2+4}+\frac {1}{2}e^{-\pi s} \frac {2}{(s+1)^2 +4}\).

Then \(y=\mathcal {L}^{-1} (Y)=e^{-t}\cos {(2t)}+\frac {1}{2}e^{-t}\sin {(2t)}+\frac {1}{2}\mathcal {U}(t-\pi )\; [e^{-(t-\pi )}\sin {(2(t-\pi ))}]\).

Equivalently, \(y=e^{-t}\left (\cos {(2t)}+\frac {1}{2}\sin {(2t)}\right )+\frac {e^{\pi }}{2}\mathcal {U}(t-\pi )\; [e^{-t}\sin {(2t)}]\).

In a more familiar form, \(\displaystyle y(t)=\left \{\begin{array}{lr} e^{-t}\left (\cos {(2t)}+\frac {1}{2}\sin {(2t)}\right ) & \mbox { if }0\leq t<\pi \\ \\ e^{-t}\left (\cos {(2t)}+\frac {1+e^{\pi }}{2}\sin {(2t)}\right ) & \mbox { if }t\geq \pi \end{array}\right .\).

Let’s solve the IVP \(y''+\pi ^2 y=2\delta (t-1)+\delta {(t-2)}\), \(y(0)=0\), \(y'(0)=0\) (which could represent an undamped spring-mass system where total upward impulses of \(2\) Ns and \(1\) Ns occur at times \(t=1\) and \(t=2\) respectively).

We let \(Y=\mathcal {L}(y)\) and take the Laplace transform of both sides.

\(s^2 Y+\pi ^2 Y=2e^{-s}+e^{-2s}\rightarrow \)

\((s^2 +\pi ^2) Y=2e^{-s}+e^{-2s}\rightarrow \)

\( Y=\frac {2}{\pi }e^{-s}\frac {\pi }{(s^2 +\pi ^2)}+\frac {1}{\pi }e^{-2s}\frac {\pi }{(s^2 +\pi ^2)}\rightarrow \)

\( y=\mathcal {L}^{-1}(Y)=\frac {2}{\pi }\mathcal {U}(t-1)\sin {(\pi (t-1))}+\frac {1}{\pi }\mathcal {U}(t-2)\sin {(\pi (t-2))}\rightarrow \)

\( y=\mathcal {L}^{-1}(Y)=-\frac {2}{\pi }\mathcal {U}(t-1)\sin {(\pi t)}+\frac {1}{\pi }\mathcal {U}(t-2)\sin {(\pi t)}\).

In a more familiar form, \(\displaystyle y(t)=\left \{\begin{array}{lr} 0 & \mbox { if }0\leq t<1\\ \\ -\frac {2}{\pi }\sin {(\pi t)} & \mbox { if }1\leq t< 2\\ \\ -\frac {1}{\pi }\sin {(\pi t)} & \mbox { if } t\geq 2 \end{array}\right .\).