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Note: 1 or 1.5 lecture, §7.6 in [EP], §6.5 in [BD]

Often in applications we study a physical system by putting in a short pulse and then seeing what the system does. The resulting behavior is often called impulse response. Let us see what we mean by a pulse. The simplest kind of a pulse is a simple rectangular pulse deﬁned by

See Figure 6.3 for a graph. Notice that

where is the unit step function.

Let us take the Laplace transform of a square pulse,

For simplicity we let , and it is convenient to set to have

That is, to have the pulse have “unit mass.” For such a pulse we compute

We generally want to be very small. That is, we wish to have the pulse be very short and very tall. By letting go to zero we arrive at the concept of the Dirac delta function.

The Dirac delta function^{5} is not exactly a function; it is sometimes called a generalized function. We avoid unnecessary details and simply say that it is an object that does not really make sense unless we integrate it. The motivation is that we would like a “function” such that for any continuous function we have

The formula should hold if we integrate over any interval that contains 0, not just . So is a “function” with all its “mass” at the single point . In other words, for any interval

Unfortunately there is no such function in the classical sense. You could informally think that is zero for and somehow inﬁnite at .

A good way to think about is as a limit of short pulses whose integral is . For example, suppose that we have a square pulse as above with , , that is . Compute

If is continuous at , then for very small , the function is approximately equal to on the interval . We approximate the integral

Therefore,

Let us therefore accept as an object that is possible to integrate. We often want to shift to another point, for example . In that case we have

Note that is the same object as . In other words, the convolution of with is again ,

As we can integrate , let us compute its Laplace transform.

In particular,

Remark 6.4.1: Notice that the Laplace transform of looks like the Laplace transform of the derivative of the Heaviside function , if we could diﬀerentiate the Heaviside function. First notice

To obtain what the Laplace transform of the derivative would be we multiply by , to obtain , which is the Laplace transform of . We see the same thing using integration,

So in a certain sense

This line of reasoning allows us to talk about derivatives of functions with jump discontinuities. We can think of the derivative of the Heaviside function as being somehow inﬁnite at , which is precisely our intuitive understanding of the delta function.

Example 6.4.1: Let us compute . So far we have always looked at proper rational functions in the variable. That is, the numerator was always of lower degree than the denominator. Not so with . We write,

The resulting object is a generalized function and only makes sense when put underneath an integral.

As we said before, in the diﬀerential equation , we think of as input, and as the output. Often it is important to ﬁnd the response to an impulse, and then we use the delta function in place of . The solution to

is called the impulse response.

Example 6.4.2: Solve (ﬁnd the impulse response)

(6.3) |

We ﬁrst apply the Laplace transform to the equation. Denote the transform of by .

Taking the inverse Laplace transform we obtain

Let us notice something about the above example. We have proved before that when the input was , then the solution to was given by

Notice that the solution for an arbitrary input is given as convolution with the impulse response. Let us see why. The key is to notice that for functions and ,

We simply diﬀerentiate twice under the integral^{6}, the details are left as an exercise. And so if we convolve the entire equation (6.3), the left hand side becomes

The right hand side becomes

Therefore is the solution to

This procedure works in general for other linear equations . If you determine the impulse response, you also know how to obtain the output for any input by simply convolving the impulse response and the input .

Let us give another quite diﬀerent example where delta functions turn up. In this case representing point loads on a steel beam. Suppose we have a beam of length , resting on two simple supports at the ends. Let denote the position on the beam, and let denote the deﬂection of the beam in the vertical direction. The deﬂection satisﬁes the Euler-Bernoulli equation^{7},

where and are constants^{8} and is the force applied per unit length at position . The situation we are interested in is when the force is applied at a single point as in Figure 6.4.

In this case the equation becomes

where is the point where the mass is applied. is the force applied and the minus sign indicates that the force is downward, that is, in the negative direction. The end points of the beam satisfy the conditions,

See § 5.2, for further information about endpoint conditions applied to beams.Example 6.4.3: Suppose that length of the beam is 2, and suppose that for simplicity. Further suppose that the force is applied at . That is, we have the equation

and the endpoint conditions are

We could integrate, but using the Laplace transform is even easier. We apply the transform in the variable rather than the variable. Let us again denote the transform of as .

We notice that and . Let us call and . We solve for ,

We take the inverse Laplace transform utilizing the second shifting property (6.1) to take the inverse of the ﬁrst term.

We still need to apply two of the endpoint conditions. As the conditions are at we can simply replace when taking the derivatives. Therefore,

and

Hence and solving for using the ﬁrst equation we obtain . Our solution for the beam deﬂection is

Exercise 6.4.7 (challenging): Solve Example 6.4.3 via integrating 4 times in the variable.

Exercise 6.4.8: Suppose we have a beam of length simply supported at the ends and suppose that force is applied at in the downward direction. Suppose that for simplicity. Find the beam deﬂection .

Exercise 6.4.103: Suppose that , , , has the solution for . Find (in closed form) the solution to , , for .

^{5}Named after the English physicist and mathematician Paul Adrien Maurice Dirac (1902–1984).

^{6}You should really think of the integral going over rather than over and simply assume that and are continuous and zero for negative .

^{7}Named for the Swiss mathematicians Jacob Bernoulli (1654–1705), Daniel Bernoulli —nephew of Jacob— (1700–1782), and Leonhard Paul Euler (1707–1783).

^{8} is the elastic modulus and is the second moment of area. Let us not worry about the details and simply think of these as some given constants.