AMATH 731: Applied Functional Analysis Fall 2008 The Dirac “delta function” distribution (The material on pages 1-8 is taken from ERV’s Fall 2005 lecture notes for AMATH 351, “Differential Equations II.”) Introduction Suppose you have a chemical (or radioactive, for that matter) species “X” in a beaker that decays according to the rate law dx = kx. (1) dt − where x(t) is the concentration at time t. Suppose that at time t = 0, there is x0 amount of X present. Of course, if the beaker is left alone, then the amount of X at time t 0 will be given by ≥ −kt x(t)= x0e . (2) Now suppose that at time a> 0, you quickly (i.e., instantaneously) add an amount A> 0 of X to the beaker. Then what is x(t), the amount of X at time t 0? ≥ Well, for 0 t<a, i.e., for all times before you add A units of X to the beaker, x(t) is given by Eq. (2) ≤ −ka −ka above. Then at t = a, there would have been x0e in the beaker, but you added A, to give x(a)= x0e +A. Then starting at t = a, the system will evolve according to the rate law. We can consider x(a) as the initial condition and measure time from t = a. The amount of X in the beaker will be x(t) = (x e−ka + A)e−kt, for t a. (3) 0 ≥ Let’s summarize our result compactly: For the above experiment, the amount of X in the beaker will be given by x e−kt, 0 t<a, x(t)= 0 ≤ (4) (x + A)e−k(t−a), t a. 0 ≥ A qualitative sketch of the solution is given below. Clearly, the solution x(t) has a discontinuity at t = a, the time of instantaneous addition of the amount A. Otherwise, it is differentiable at all other points. x0 x(t) vs. t x(a)+ A A x(a) 0 a t Note that we can write the solution in Eq. (4) even more compactly as follows, x(t)= x e−kt + Ae−k(t−a)H(t a), t 0, (5) 0 − ≥ where H(t) is the Heaviside function reviewed earlier. We shall return to this solution a little later. We now ask whether the above operation – the instanteous addition of an amount A of X to the beaker – can be represented mathematically, perhaps with a function f(t) so that the evolution of x(t) can be expressed as dx = kx + f(t), (6) dt − 1 where f(t) models the instantaneous addition of an amount A of X at time t = a. The answer is “Yes,” and f(t) will be the so-called “Dirac delta function” f(t)= Aδ(t a). − But in order to appreciate this result, let us now consider the case of a less brutal addition of X to the beaker. Suppose that we add an amount of A units of X but over a time interval of length ∆. We’ll also add X to the beaker at a a constant rate of A/∆ units per unit time. This means that our evolution equation for x(t) will take the form kx, 0 t<a, dx − ≤ = kx + A , a t a + ∆, (7) dt − ∆ ≤ ≤ kx, t a + ∆ − ≥ We can express this in compact form as dx = kx + f (t), (8) dt − ∆ where A f (t)= [H(t a) H(t (a + ∆))] . (9) ∆ ∆ − − − A graph of the function f∆(t) is sketched below. Note that the area under the curve and above the x-axis is A, as it should be – when you integrate a rate function over a time interval [a,b], you obtain the amount added over that time interval. f∆(t) vs. t A ∆ 0 a a +∆ t We can solve the DE in (8) in two ways: (1) as a linear first-order inhomogeneous DE, (2) using Laplace Transforms. Here, we’ll use Method (2). Taking LTs of both sides of (8) gives sX(s) x + kX(s)= F (s). (10) − 0 ∆ Solving for X(s): x 1 X(s)= 0 + F (s). (11) s + k s + k ∆ 1 Noting that the inverse LT of is e−kt, we have, after taking inverse LTs: s + k x(t)= x e−kt + f (t) e−kt. (12) 0 ∆ ∗ −kt Note that the first term x0e is the solution to the homogeneous DE associated with (8). −kt Now compute the convolution of f∆ with e as follows: t −kt −k(t−τ) f∆(t) e = f∆(τ)e dτ (13) ∗ Z0 A t A t = e−k(t−τ)H(τ a) dτ e−k(t−τ)H(τ (a + ∆)) dτ ∆ Z0 − − ∆ Z0 − = I1(t)+ I2(t) Because of the Heaviside function H(τ a), the integrand in the first integral I (t) – hence the integral itself – − 1 will be zero for 0 t<a. For t a, we can compute I (t) as follows: ≤ ≥ 1 t t A −k(t−τ) A −kt I1(t)= e dτ = e dτ (14) ∆ Za ∆ Za · · · 2 A = 1 e−k(t−a) , t a. k∆ h − i ≥ Likewise, we can determine the second integral to be A −k(t−(a+∆)) I2(t)= 1 e , t a + ∆. (15) −k∆ h − i ≥ The final result for x(t) is x(t)= x e−kt + I (t)H(t a)+ I (t)H(t (a + ∆)). (16) 0 1 − 2 − The qualitative behaviour of the graph of x(t) is sketched below. It is possible that x(t) is decreasing over the interval [a,a + ∆], during which time the amount A is added. But if ∆ is sufficiently small, i.e., the rate A/∆ is large enough, x(t) will be increasing over this interval. x0 x(t) vs. t x(a)+ A A x(a) 0 a a +∆ t ∆ But these points are rather secondary. What is of prime importance is the difference in concentrations between time t = a, the time at which we began to add X to the beaker, and time t = a + ∆, the time at which we stopped adding X. Remember that regardless of ∆, we are always adding a total amount of A to the beaker. This difference in concentrations is given by A x(a + ∆) x(a) = x e−k(a+∆) + 1 e−k∆ x e−ka (17) − 0 k∆ − − 0 A = 1 e−k∆ x e−ka . − k∆ − 0 In particular, we are interested in what happens to this difference as ∆ 0, i.e., the time interval over which → we add A units of X goes to zero. To find this limit, if it exists, we expand the exponential involving ∆ to give e−k∆ =1 k∆+ O(∆2) as ∆ 0, (18) − → so that 1 e−k∆ = k∆+ O(∆2) as ∆ 0. (19) − → Substituting this result into (17) gives the result x(a + ∆) x(a) A as ∆ 0. (20) − → → In other words, in the limit ∆ 0, the graph of x(t) will exhibit a discontinuity at t = a. The magnitude of → this “jump” is A, precisely was found with the earlier method. We now return to the inhomogeneous DE in (8) and examine the behaviour of the inhomogeneous term f (t) as ∆ 0. Recall that this is the “driving” term, the function that models the addition of a total of A ∆ → 3 units of X to the beaker over the time interval [a,a + ∆]. The width of the box that makes up the graph of A f (t) is ∆. The height of the box is . In this way, the area of the box – the total amount of X delivered ∆ ∆ – is A. As ∆ decreases, the box gets thinner and higher. In the limit ∆ 0, we have produced a function → f0(t) that is zero everywhere except at t = a, where it is undefined. This is the idea behind the “Dirac delta function,” which we explore in more detail below. The Dirac delta function Let us define the following function Iǫ(t) for an ǫ> 0: 1 ǫ , 0 t ǫ, Iǫ(t)= ≤ ≤ (21) 0, t>ǫ The graph of Iǫ(t) is sketched below. Iǫ(t) vs. t 1 ǫ 0 ǫ t Clearly ∞ Iǫ(t) dt =1, for all ǫ> 0. (22) Z−∞ Now let f(t) be a continuous function on [0, ) and consider the integrals ∞ ∞ 1 ǫ f(t)Iǫ(t) dt = f(t) dt, (23) Z−∞ ǫ Z0 in particular for small ǫ and limit ǫ 0. For any ǫ > 0, because of the continuity of f(t), there exists, by the → Mean Value Theorem for Integrals, a cǫ [0,ǫ] such that ∈ ǫ f(t) dt = f(cǫ) ǫ. (24) Z0 · Therefore, ǫ f(t)Iǫ(t) dt = f(cǫ). (25) Z0 As ǫ 0, cǫ 0 since the interval [0,ǫ] 0 . Therefore → → →{ } ǫ lim f(t)Iǫ(t) dt = f(0). (26) ǫ→0 Z0 We may, of course, translate the function Iǫ(t) to produce the result ∞ lim f(t)Iǫ(t a) dt = f(a). (27) ǫ→0 Z−∞ − 4 This is essentially a definition of the “Dirac delta function,” which is not a function but rather a “generalized function” that is defined in terms of integrals over continuous functions. In proper mathematical parlance, the Dirac delta function is a distribution. The Dirac delta function δ(t) is defined by the integral ∞ f(t)δ(t) dt = f(0). (28) Z−∞ Moreover d f(t)δ(t) dt =0, if c> 0. (29) Zc As well, we have the translation result ∞ f(t)δ(t a) dt = f(a).