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.
2.
3.
At 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 ,
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 .
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.
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 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.
Solution
As with all previous problems we’ll first take the Laplace transform of everything in the differential
equation and apply the initial conditions.
Now solve for Y(s).
We’ll leave it to you to verify the partial fractions and
their inverse transforms are,
The solution is then,
where, f(t) and g(t) are defined above.

Example 2 Solve
the following IVP.
Solution
Take the Laplace
transform of everything in the differential equation and apply the initial
conditions.
Now solve for Y(s).
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.
The solution is then,
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.
However, this is precisely the definition of the Heaviside
function. So,
Now, recalling the Fundamental Theorem of Calculus, we get,