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Home / Differential Equations / Laplace Transforms / Dirac Delta Function
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### Section 4-8 : Dirac Delta Function

When we first introduced Heaviside functions we noted that we could think of them as switches changing the forcing function, $$g(t)$$, at specified times. However, Heaviside functions are really not suited to forcing functions that exert a “large” force over a “small” time frame.

Examples of this kind of forcing function would be a hammer striking an object or a short in an electrical system. In both of these cases a large force (or voltage) would be exerted on the system over a very short time frame. The Dirac Delta function is used to deal with these kinds of forcing functions.

#### Dirac Delta Function

There are many ways to actually define the Dirac Delta function. To see some of these definitions visit Wolframs MathWorld. There are three main properties of the Dirac Delta function that we need to be aware of. These are,

1. $$\delta \left( {t - a} \right) = 0,\,\,\,\,t \ne a$$

2. $$\displaystyle \int_{{\,a - \varepsilon }}^{{\,a + \varepsilon }}{{\delta \left( {t - a} \right)\,dt}} = 1,\hspace{0.25in}\varepsilon > 0$$

3. $$\displaystyle \int_{{\,\,a - \varepsilon }}^{{\,\,a + \varepsilon }}{{f\left( t \right)\delta \left( {t - a} \right)\,dt}} = f\left( a \right),\hspace{0.25in}\,\,\,\,\,\,\,\,\,\varepsilon > 0$$

At $$t = a$$ the Dirac Delta function is sometimes thought of has having an “infinite” value. So, the Dirac Delta function is a function that is zero everywhere except one point and at that point it can be thought of as either undefined or as having an “infinite” value.

Note that the integrals in the second and third property are actually true for any interval containing $$t = a$$, provided it’s not one of the endpoints. The limits given here are needed to prove the properties and so they are also given in the properties. We will however use the fact that they are true provided we are integrating over an interval containing $$t = a$$.

This is a very strange function. It is zero everywhere except one point and yet the integral of any interval containing that one point has a value of 1. The Dirac Delta function is not a real function as we think of them. It is instead an example of something called a generalized function or distribution.

Despite the strangeness of this “function” it does a very nice job of modeling sudden shocks or large forces to a system.

Before solving an IVP we will need the transform of the Dirac Delta function. We can use the third property above to get this.

$\mathcal{L}\left\{ {\delta \left( {t - a} \right)} \right\} = \int_{{\,0}}^{\infty }{{{{\bf{e}}^{ - s\,t}}\delta \left( {t - a} \right)\,dt}} = {{\bf{e}}^{ - a\,s}}\hspace{0.25in}\hspace{0.25in}{\mbox{provided }}a > 0$

Note that often the second and third properties are given with limits of infinity and negative infinity, but they are valid for any interval in which $$t = a$$ is in the interior of the interval.

With this we can now solve an IVP that involves a Dirac Delta function.

Example 1 Solve the following IVP. $y'' + 2y' - 15y = 6\delta \left( {t - 9} \right),\hspace{0.25in}\hspace{0.25in}y\left( 0 \right) = - 5\,\,\,\,\,\,y'\left( 0 \right) = 7$
Show Solution

As with all previous problems we’ll first take the Laplace transform of everything in the differential equation and apply the initial conditions.

\begin{align*}{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) + 2\left( {sY\left( s \right) - y\left( 0 \right)} \right) - 15Y\left( s \right) & = 6{{\bf{e}}^{ - 9s}}\\ \left( {{s^2} + 2s - 15} \right)Y\left( s \right) + 5s + 3 & = 6{{\bf{e}}^{ - 9s}}\end{align*}

Now solve for $$Y(s)$$.

\begin{align*}Y\left( s \right) & = \frac{{6{{\bf{e}}^{ - 9s}}}}{{\left( {s + 5} \right)\left( {s - 3} \right)}} - \frac{{5s + 3}}{{\left( {s + 5} \right)\left( {s - 3} \right)}}\\ & = 6{{\bf{e}}^{ - 9s}}F\left( s \right) - G\left( s \right)\end{align*}

We’ll leave it to you to verify the partial fractions and their inverse transforms are,

\begin{align*}F\left( s \right) & = \frac{1}{{\left( {s + 5} \right)\left( {s - 3} \right)}} = \frac{{\frac{1}{8}}}{{s - 3}} - \frac{{\frac{1}{8}}}{{s + 5}}\\ f\left( t \right) & = \frac{1}{8}{{\bf{e}}^{3t}} - \frac{1}{8}{{\bf{e}}^{ - 5t}}\end{align*} \begin{align*}G\left( s \right) & = \frac{{5s + 3}}{{\left( {s + 5} \right)\left( {s - 3} \right)}} = \frac{{\frac{9}{4}}}{{s - 3}} + \frac{{\frac{{11}}{4}}}{{s + 5}}\\ g\left( t \right) & = \frac{9}{4}{{\bf{e}}^{3t}} + \frac{{11}}{4}{{\bf{e}}^{ - 5t}}\end{align*}

The solution is then,

\begin{align*}Y\left( s \right) & = 6{{\bf{e}}^{ - 9s}}F\left( s \right) - G\left( s \right)\\ y\left( t \right) & = 6{u_9}\left( t \right)f\left( {t - 9} \right) - g\left( t \right)\end{align*}

where, $$f(t)$$ and $$g(t)$$ are defined above.

Example 2 Solve the following IVP. $2y'' + 10y = 3{u_{12}}\left( t \right) - 5\delta \left( {t - 4} \right),\hspace{0.25in}\hspace{0.25in}y\left( 0 \right) = - 1\,\,\,\,\,\,y'\left( 0 \right) = - 2$
Show Solution

Take the Laplace transform of everything in the differential equation and apply the initial conditions.

\begin{align*}2\left( {{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right)} \right) + 10Y\left( s \right) & = \frac{{3{{\bf{e}}^{ - 12s}}}}{s} - 5{{\bf{e}}^{ - 4s}}\\ \left( {2{s^2} + 10} \right)Y\left( s \right) + 2s + 4 & = \frac{{3{{\bf{e}}^{ - 12s}}}}{s} - 5{{\bf{e}}^{ - 4s}}\end{align*}

Now solve for $$Y(s)$$.

\begin{align*}Y\left( s \right) & = \frac{{3{{\bf{e}}^{ - 12s}}}}{{s\left( {2{s^2} + 10} \right)}} - \frac{{5{{\bf{e}}^{ - 4s}}}}{{2{s^2} + 10}} - \frac{{2s + 4}}{{2{s^2} + 10}}\\ & = 3{{\bf{e}}^{ - 12s}}F\left( s \right) - 5{{\bf{e}}^{ - 4s}}G\left( s \right) - H\left( s \right)\end{align*}

We’ll need to partial fraction the first function. The remaining two will just need a little work and they’ll be ready. I’ll leave the details to you to check.

\begin{align*}F\left( s \right) & = \frac{1}{{s\left( {2{s^2} + 10} \right)}} = \frac{1}{{10}}\frac{1}{s} - \frac{1}{{10}}\frac{s}{{{s^2} + 5}}\\ f\left( t \right) & = \frac{1}{{10}} - \frac{1}{{10}}\cos \left( {\sqrt 5 \,t} \right)\end{align*} $g\left( t \right) = \frac{1}{{2\sqrt 5 }}\sin \left( {\sqrt 5 \,t} \right)$ $h\left( t \right) = \cos \left( {\sqrt 5 \,t} \right) + \frac{2}{{\sqrt 5 }}\sin \left( {\sqrt 5 \,t} \right)$

The solution is then,

\begin{align*}Y\left( s \right) & = 3{{\bf{e}}^{ - 12s}}F\left( s \right) - 5{{\bf{e}}^{ - 4s}}G\left( s \right) - H\left( s \right)\\ y\left( t \right) & = 3{u_{12}}\left( t \right)f\left( {t - 12} \right) - 5{u_4}\left( t \right)g\left( {t - 4} \right) - h\left( t \right)\end{align*}

where, $$f(t)$$, $$g(t)$$ and $$h(t)$$ are defined above.

So, with the exception of the new function these work the same way that all the problems that we’ve seen to this point work. Note as well that the exponential was introduced into the transform by the Dirac Delta function, but once in the transform it doesn’t matter where it came from. In other words, when we went to the inverse transforms it came back out as a Heaviside function.

Before proceeding to the next section let’s take a quick side trip and note that we can relate the Heaviside function and the Dirac Delta function. Start with the following integral.

$\int_{{\, - \infty }}^{t}{{\delta \left( {u - a} \right)\,du}} = \left\{ {\begin{array}{*{20}{l}}0&{\hspace{0.25in}{\mbox{if }}t < a}\\1&{\hspace{0.25in}{\mbox{if }}t > a}\end{array}} \right.$

However, this is precisely the definition of the Heaviside function. So,

$\int_{{\, - \infty }}^{t}{{\delta \left( {u - a} \right)\,du}} = {u_a}\left( t \right)$

Now, recalling the Fundamental Theorem of Calculus, we get,

${u'_a}\left( t \right) = \frac{d}{{dt}}\left( {\int_{{\, - \infty }}^{t}{{\delta \left( {u - a} \right)\,du}}} \right) = \delta \left( {t - a} \right)$

So, the derivative of the Heaviside function is the Dirac Delta function.