4.5 Dirac Delta Function

INTRODUCTION

Just before the Remarks on page 229, we indicated that as an immediate consequence of Theorem 4.2.3, F(s) = 1 cannot be the Laplace transform of a function f that is piecewise continuous on [0, ∞) and of exponential order. In the discussion that follows we are going to introduce a function that is very different from the kinds that you have studied in previous courses. We shall see that there does indeed exist a function, or more precisely a generalized function, whose Laplace transform is F(s) = 1.

Unit Impulse

Mechanical systems are often acted on by an external force (or emf in an electrical circuit) of large magnitude that acts only for a very short period of time. For example, a vibrating airplane wing could be struck by lightning, a mass on a spring could be given a sharp blow by a ball peen hammer, a ball (baseball, golf ball, tennis ball) could be sent soaring when struck violently by some kind of club (baseball bat, golf club, tennis racket). See FIGURE 4.5.1. The graph of the piecewise-defined function

(1)

An illustration of a baseball player hitting the ball with his bat.

FIGURE 4.5.1 A baseball bat applies a force of large magnitude on the ball for a very short period of time

a > 0, t0 > 0, shown in FIGURE 4.5.2(a), could serve as a model for such a force. For a small value of a, δa(t − t0) is essentially a constant function of large magnitude that is “on” for just a very short period of time, around t0. The behavior of δa(t − t0) as a → 0 is illustrated in Figure 4.5.2(b). The function δa(t − t0) is called a unit impulse since it possesses the integration property (t − t0)dt = 1.

Two graphs a and b. The graph (a) represents a piecewise-defined function. Three pieces of the function are graphed as straight lines on the t y plane. The first piece starts at the point (0, 0), goes to the right, and ends at the point (t subscript 0 minus a, 0). The second piece starts at the point (t subscript 0 minus a, 1 over 2a), goes to the right, and ends at the point (t subscript 0 + a, 1 over 2a). The length of this piece is labeled 2a. The midpoint of this piece corresponds to the t value of t subscript 0. The third piece starts at the point (t subscript 0 + a, 0), goes to the right, and exits the right on the positive t axis. The graph (b) represents the behavior of a unit impulse function. Six pieces of the function are graphed as straight lines on the t y plane. The first piece starts at the point (0, 0), goes to the right, and ends at a point on the positive t axis. The second piece starts at a point in the first quadrant that corresponds to the endpoint of the first piece, goes to the right, and ends at a point in the first quadrant. The midpoint of this piece corresponds to the t value of t subscript 0. The third piece starts at a point on the positive t axis that corresponds to the endpoint of the second piece, goes to the right, and exits the right of the first quadrant. The fourth piece is above and parallel to the second piece. It starts at a point in the first quadrant, goes to the right, and ends at a point in the first quadrant. The length of this piece is less than the length of the second piece and it also has the midpoint corresponding to the t value of t subscript 0. The fifth piece is above and parallel to the fourth piece. It starts at a point in the first quadrant, goes to the right, and ends at a point in the first quadrant. The length of this piece is less than the length of the fourth piece and it also has the midpoint corresponding to the t value of t subscript 0. The sixth piece is above and parallel to the fourth piece. It starts at a point in the first quadrant, goes to the right, and ends at a point in the first quadrant. The length of this piece is less than the length of the fifth piece and it also has the midpoint corresponding to the t value of t subscript 0.

FIGURE 4.5.2 Unit impulse

The Dirac Delta Function

In practice it is convenient to work with another type of unit impulse, a “function” that approximates δa(t − t0) and is defined by the limit

(2)

The latter expression, which is not a function at all, can be characterized by the two properties

The unit impulse is called the Dirac delta function after its inventor, the English theoretical physicist Paul Adrien Maurice Dirac (1902–1984). Dirac shared the Nobel Prize in Physics in 1933.

It is possible to obtain the Laplace transform of the Dirac delta function by the formal assumption that {δ(t − t0)} = lima→0 {δa(t − t0)}.

THEOREM 4.5.1 Transform of the Dirac Delta Function

For ,

.(3)

PROOF:

To begin, we can write δa(t − t0) in terms of the unit step function by virtue of (11) and (12) of Section 4.3:

By linearity and (14) of Section 4.3, the Laplace transform of this last expression is

(4)

Since (4) has the indeterminate form 0/0 as a → 0, we apply L’Hôpital’s rule:

≡

Now when t0 = 0, it seems plausible to conclude from (3) that

The last result emphasizes the fact that δ(t) is not the usual type of function that we have been considering since we expect from Theorem 4.2.3 that {f(t)} → 0 as s → ∞.

EXAMPLE 1 Two Initial-Value Problems

Solve y″ + y = 4 δ (t − 2π) subject to

(a) y(0) = 1, y′(0) = 0

(b) y(0) = 0, y′(0) = 0.

The two initial-value problems could serve as models for describing the motion of a mass on a spring moving in a medium in which damping is negligible. At t = 2π the mass is given a sharp blow. In part (a) the mass is released from rest 1 unit below the equilibrium position. In part (b) the mass is at rest in the equilibrium position.

SOLUTION

(a) From (3) the Laplace transform of the differential equation is

s2Y(s) − s + Y(s) = 4e−2πs or Y(s) = .

Using the inverse form of the second translation theorem, (15) of Section 4.3, we find

y(t) = cos t + 4 sin(t − 2π) (t − 2π).

Since sin(t − 2π) = sin t, the foregoing solution can be written as

(5)

In FIGURE 4.5.3 we see from the graph of (5) that the mass is exhibiting simple harmonic motion until it is struck at t = 2π. The influence of the unit impulse is to increase the amplitude of vibration to for t > 2π.

Two pieces of a function are graphed on the t y plane. The first piece is a curve that starts at the point (0, 1), goes down and to the right through the point (pi over 2, 0), and reaches a low point (pi, negative 1). Then, it goes up and to the right through the point (3 pi over 2, 0), and ends at the point (2 pi, 1). The continuity of this curve is shown by a red dashed curve that follows the same oscillatory pattern. The second piece is also a curve that starts at the point (2 pi, 1), goes up and to the right, reaches a high point, then goes down and to the right through a point on the positive t axis just to the left of the point (3 pi, 0), and reaches a low point in the fourth quadrant. Again, it goes up and to the right through a point on the positive t axis just to the left of the point (4 pi, 0), reaches a high point, goes down and to the right, and ends at the right of the first quadrant.

FIGURE 4.5.3 In Example 1(a), moving mass is struck at t = 2π

(b) In this case the transform of the equation is simply

and so

(6)

The graph of (6) in FIGURE 4.5.4 shows, as we would expect from the initial conditions, that the mass exhibits no motion until it is struck at t = 2π. ≡

Two pieces of a function are graphed on the t y plane. The first piece is a straight line that starts at the point (0, 0), goes to the right, and ends at the point (2 pi, 0). The second piece is a curve that follows an oscillatory pattern. It starts at the point (2 pi, 0), goes up and to the right, reaches a high point, then goes down and to the right through the point (3 pi, 0), and reaches a low point in the fourth quadrant. Again, it goes up and to the right through the point (4 pi, 0), reaches a high point, goes down and to the right, and ends at the right of the first quadrant.

FIGURE 4.5.4 In Example 1(b), mass is at rest until struck at t = 2π

Alternative Definition

If were a function in the traditional sense, then property (i) on page 257 would imply rather than

Because the Dirac delta function did not “behave” like an ordinary function, even though its users produced correct results, it was initially met with great scorn by mathematicians. However, in the 1940s Dirac’s controversial function was put on a rigorous footing by the French mathematician Laurent Schwartz (1915–2002) in his book Théorie des distributions, and this, in turn, led to an entirely new branch of mathematics known as the theory of distributions or generalized functions. In this theory (2) is not the accepted definition of nor does one speak of a function whose values are either or 0. Although we shall not pursue this topic any further, suffice it to say that the Dirac delta function is best characterized by its effect on other functions. If f is a continuous function, then

(7)

can be taken as the definition of This result is known as the sifting property since has the effect of sifting the value out of the set of values of f on Note that with then (7) gives

(8)

Similarly, if then (7) gives immediately

(9)

The results in (8) and (9) are consistent with property (ii) and (3) on page 257. See Problems 19 and 20 in Exercises 4.5.

REMARKS

In the Remarks in Section 4.2 we indicated that the transfer function of a general linear nth-order differential equation with constant coefficients is W(s) = 1/P(s), where P(s) = ansn + an −1sn − 1 + + a0. The transfer function is the Laplace transform of function w(t), called the weight function of a linear system. But w(t) can be characterized in terms of the discussion at hand. For simplicity let us consider a second-order linear system in which the input is a unit impulse at t = 0:

a2 y″ + a1 y′ + a0 y = δ(t), y(0) = 0, y′(0) = 0.

Applying the Laplace transform and using {δ(t)} = 1 shows that the transform of the response y in this case is the transfer function

From this we can see, in general, that the weight function y = w(t) of an nth-order linear system is the zero-state response of the system to a unit impulse. For this reason w(t) is called as well the impulse response of the system.

4.5 Exercises Answers to selected odd-numbered problems begin on page ANS-11.

In Problems 1–14, use the Laplace transform to solve the given differential equation subject to the indicated initial conditions.

  1. y′ − 3y = δ(t − 2), y(0) = 0
  2. y′ + y = δ(t − 1), y(0) = 2
  3. y″ + y = δ(t − 2π), y(0) = 0, y′(0) = 1
  4. y″ + 16y = δ(t − 2π), y(0) = 0, y′(0) = 0
  5. y″ + y = δ(t − π/2) + δ(t − 3π/2), y(0) = 0, y′(0) = 0
  6. y″ + y = δ(t − 2π) + δ(t − 4π), y(0) = 1, y′(0) = 0
  7. y″ + 2y′ = δ(t − 1), y(0) = 0, y′(0) = 1
  8. y″ − 2y′ = 1 + δ(t − 2), y(0) = 0, y′(0) = 1
  9. y″ + 4y′ + 5y = δ(t − 2π), y(0) = 0, y′(0) = 0
  10. y″ + 2y′ + y = δ(t − 1), y(0) = 0, y′(0) = 0
  11. y″ + 4y′ + 13y = δ(t− π) + δ(t− 3π), y(0) = 1, y′(0) = 0
  12. y″ − 7y′ + 6y = et + δ(t− 2) + δ(t − 4), y(0) = 0, y′(0) = 0
  13. y″ + y = (t − 1) + δ(t − 2) − (t − 3), y(0) = 0, y′(0) = 0
  14. y″ + 2y′ + y = (t − 1)(t − 1) + δ(t − 5), y(0) = 0, y′(0) = 0

In Problems 15 and 16, use the Laplace transform to solve the given initial-value problem. Graph your solution on the interval

In Problems 17 and 18, a uniform beam of length L carries a concentrated load w0 at Solve the differential equation

subject to the given boundary conditions.

  1. y(0) = 0, y′(0) = 0, y″(L) = 0, y‴(L) = 0
    A beam of length L is embedded at its left end and free at its right end. A vertical line along the embedded end is labeled y. A horizontal line along the axis of the beam is labeled x. A concentrated load w subscript 0 acts on the beam at a distance L over 2 from both ends.

    FIGURE 4.5.5 Beam embedded at its left end and free at its right end

  2. y(0) = 0, y′(0) = 0, y(L) = 0, y′(L) = 0
    A beam of length L is embedded at both ends. A vertical line along the left end is labeled y. A horizontal line along the axis of the beam is labeled x. A concentrated load w subscript 0 acts on the beam at a distance L over 2 from both ends.

    FIGURE 4.5.6 Beam embedded at both ends

In Problems 19 and 20, use the Laplace transform and (7) to solve the given initial-value problem. Use a graphing utility to graph the solution.

  1. y″ + y = 2 cos t δ(t − π) + 6 cos t δ(t − 2π), y(0) = 1, y′(0) = 0
  2. y″ + 4y′ + 3y = et δ(t − 1), y(0) = 0, y′(0) = 2