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The Dirac Delta Function Appendix A The Dirac Delta Function A.1 The One-Dimensional Dirac Delta Function The Dirac delta function [1] in one-dimensional space may be defined by the pair of equations δ(x) = 0; x = 0, (A.1) ∞ δ(x)dx = 1. (A.2) −∞ It is clear from this definition that δ(x) is not a function in the ordinary mathematical sense, because if a function is zero everywhere except at a single point and the integral of this function over its entire domain of definition exists, then the value of this integral is necessarily also equal to zero. Because of this, it is more appropriate to regard δ(x) as a functional quantity with a certain well-defined symbolic meaning. For example, one can consider a sequence of functions δ(x,ε) that, with increasing values of the parameter ε, differ appreciably from zero only over a decreasing x-interval about the origin and which are such that ∞ δ(x,ε)dx = 1(A.3) −∞ for all values of ε. Although it may be tempting to try to interpret the Dirac delta function as the limit of such a sequence of well-defined functions δ(x,ε) as ε →∞, it must be recognized that this limit need not exist for all values of the independent variable x. However, the limit ∞ lim δ(x,ε)dx = 1(A.4) ε→∞ −∞ © Springer Nature Switzerland AG 2019 641 K. E. Oughstun, Electromagnetic and Optical Pulse Propagation, Springer Series in Optical Sciences 224, https://doi.org/10.1007/978-3-030-20835-6 642 A The Dirac Delta Function must exist. As a consequence, one may interpret any operation that involves the delta function δ(x) as implying that this operation is to be performed with a function δ(x,ε) of a suitable sequence and that the limit as ε →∞is to be taken at the conclusion of the calculation. The particular choice of the sequence of functions δ(x,ε) is immaterial, provided that any oscillations near the origin x = 0 are not too violent [2]. For example, each of the following functions forms a sequence with respect to the parameter ε that satisfies the required properties: ε −ε2x2 ε δ(x,ε) = √ e & δ(x,ε) = rect /ε(x) & δ(x,ε) = sinc(εx) π 1 π in Cartesian coordinates, and ε δ(r,ε) = circ (r) & δ(r,ε) = J (2πεr) 1/ε r 1 in polar coordinates, where rect1/ε(x) ≡ ε/2 when |x| < 1/ε and is zero otherwise, 2 circ1/ε(r) ≡ ε /π when r<1/ε and is zero otherwise, and sinc(x) ≡ sin (x)/x when x = 0 and sinc(0) ≡ limx→0 sin (x)/x = 1. Let f(x) be a continuous and sufficiently well-behaved function of x ∈ (−∞, ∞) and consider the value of the definite integral ∞ ∞ f(x)δ(x− a)dx = lim f(x)δ(x− a,ε)dx. −∞ ε→∞ −∞ When the parameter ε is large, the value of the integral appearing on the right- hand side of this equation depends essentially on the behavior of f(x) in the immediate neighborhood of the point x = a alone, and the error that results from the replacement of f(x) by f(a) may be made as small as desired by taking ε sufficiently large. Hence ∞ ∞ lim f(x)δ(x− a,ε)dx = f(a) lim δ(x − a,ε)dx, ε→∞ −∞ ε→∞ −∞ so that ∞ f(x)δ(x− a)dx = f(a). (A.5) −∞ This result is referred to as the sifting property of the delta function. Notice that, for this result to hold, the domain of integration need not be extended over all x ∈ (−∞, ∞); it is only necessary that the domain of integration contain the point x = a A.1 The One-Dimensional Dirac Delta Function 643 in its interior, so that a+Δ2 f(x)δ(x− a)dx = f(a), (A.6) a−Δ1 where Δ1,2 > 0. It is then seen that f(x)need only be continuous at x = a. The above results may be written symbolically as f(x)δ(x− a) = f(a)δ(x− a), (A.7) the meaning of such a statement being that the two sides yield the same result when integrated over any domain containing the point x = a. For the special case when f(x)= xk with k>0 and a = 0, Eq. (A.7) yields xkδ(x) = 0, ∀k>0. (A.8) Theorem A.1 (Similarity Relationship (Scaling Law)) For all a = 0 1 δ(ax) = δ(x). (A.9) |a| Proof In order to prove this relationship one need only compare the integrals of f(x)δ(ax)and f(x)δ(x)/|a| for any sufficiently well-behaved continuous function f(x). For the first integral one has (for any a = 0) ∞ 1 ∞ 1 f(x)δ(ax)dx =± f(y/a)δ(y)dy = f(0), −∞ a −∞ |a| where the upper or lower sign choice is taken accordingly as a>0ora<0, respectively, and for the second integral one obtains ∞ 1 1 f(x) δ(x)dx = f(0). −∞ |a| |a| Comparison of these two results then shows that δ(ax) = δ(x)/|a|, as was to be proved. For the special case a =−1, Eq. (A.9) yields δ(−x) = δ(x), (A.10) so that the delta function is an even function of its argument. 644 A The Dirac Delta Function Theorem A.2 (Composite Function Theorem) If y = f(x) is any continuous function of x with simple zeroes at the points xi [i.e., y = 0 at x = xi and f (xi) = 0] and no other zeroes, then 1 δ(f(x)) = δ(x − x ). (A.11) |f (x )| i i i Proof Let g(x) be any sufficiently well-behaved continuous function and let {xi} denote the set of points at which y = 0. Under the change of variable x = f −1(y) one has that ∞ − 1 g(x)δ(f(x))dx = g f 1(y) δ(y) dy −1 −∞ R |f f (y) | − 1 1 = g f 1(0) = g(x ) , −1 i |f f (0) | |f (xi| xi i where R denotes the range of f(x). In addition, ∞ ∞ 1 − = 1 − g(x) δ(x xi) dx g(x)δ(x xi)dx −∞ |f (x )| |f (x )| −∞ i i i i 1 = g(x ) . i |f (x | i i Comparison of these two expressions then proves the theorem. As an example, consider the function f(x) = x2 − a2 which has simple zeroes at x =±a. Then |f (±a)|=2|a| so that, for a = 0, 1 δ(x2 − a2) = (δ(x − a) + δ(x + a)) . 2|a| An additional relationship of interest that employs the Dirac delta function is ∞ δ(ξ − x)δ(x − η)dx = δ(ξ − η), (A.12) −∞ which is seen to be an extension of the sifting property to the delta function itself. This equation then implies that if both sides are multiplied by a continuous function of either ξ or η and the result integrated over all values of either ξ or η, respectively, an identity is obtained. That is, because ∞ f(ξ)δ(ξ − η)dξ = f(η), −∞ A.1 The One-Dimensional Dirac Delta Function 645 and ∞ ∞ f(ξ) δ(ξ − x)δ(x − η)dx dξ −∞ −∞ ∞ ∞ = f(ξ)δ(ξ − x)dξ δ(x − η)dx −∞ −∞ ∞ = f(x)δ(x− η)dx = f(η) −∞ then the expression in Eq. (A.12) follows. In a similar manner, because ∞ ∞ g(η) δ(ξ − x)δ(x − η)dx dη −∞ −∞ ∞ ∞ = g(η)δ(x − η)dη δ(ξ − x)dx −∞ −∞ ∞ = g(x)δ(ξ − x)dx = g(ξ) −∞ and ∞ g(η)δ(ξ − η)dη = g(ξ), −∞ then the expression in Eq. (A.12) is again obtained. Consider next what interpretation may be given to the derivatives of the delta function, accomplished through use of the function sequence δ(x,ε). Consider then ∞ the ordinary integral −∞ f(x)δ(x, ε)dx which may be evaluated by application of the method of integration by parts with u = f(x)and dv = δ (x, ε)dx, so that ∞ ∞ f(x)δ(x, ε)dx = f(∞)δ(∞,ε)− f(−∞)δ(−∞,ε)− f (x)δ(x, ε)dx. −∞ −∞ Upon proceeding to the limit as ε →∞, the first two terms appearing on the right- hand side of this equation both vanish because lim δ(±∞,ε)= 0, (A.13) ε→∞ with the result ∞ f(x)δ(x)dx =−f (0). (A.14) −∞ 646 A The Dirac Delta Function Upon repeating this procedure n times for the nth-order derivative of the delta function, one obtains the general result ∞ f(x)δ(n)(x)dx = (−1)nf (n)(0). (A.15) −∞ As a special case of Eq. (A.14), let f(x)= x so that ∞ ∞ xδ (x)dx =−1 =− δ(x)dx, −∞ −∞ and one then has the equivalence xδ (x) =−δ(x). (A.16) Because δ(x) is an even function and x is an odd function, it then follows that δ (x) is an odd function of its argument; that is δ (−x) =−δ (x). (A.17) The generalization of Eq. (A.16) may be directly obtained from Eq. (A.15) by letting f(x)= xn. In that case, f (n)(x) = n! and this relation gives ∞ ∞ xnδ (x)dx = (−1)nn!=(−1)nn! δ(x)dx, −∞ −∞ and one then has the general equivalence xnδ(n)(x) = (−1)nn!δ(x). (A.18) The even-order derivatives of the delta function are even functions and the odd-order derivatives are odd functions of the argument.
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