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 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 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 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). (30) Z−∞ − Finally, we compute the Laplace transform of the Dirac delta function. In this case f(t)= e−st so that ∞ [δ(t a)] = e−stδ(t a) dt = e−as, a 0. (31) L − Z0 − ≥ Let us now return to the substance X problem examined earlier where an amount A of substance X is added to a beaker over a time interval of length ∆. Comparing the function f∆(t) used for that problem and the function Iǫ(t) defined above, we see that ∆ = ǫ and f∆(t)= AI∆(t). (32) From our discussion above on the Dirac delta function, in the limit ∆ 0 the function f (t) becomes → ∆ f(t)= Aδ(t a). (33) − Therefore, the differential equation for x(t) modelling the instantaneous addition of A to the beaker is given by dx = kx + Aδ(t a). (34) dt − − We now compute x(t) using the result in Eq. (12) obtained from Laplace transforms. Using f(t)= Aδ(t a), − the convolution in this equation becomes (g f)(t) = e−kt Aδ(t a) (35) ∗ t∗ − = A e−k(t−τ)δ(τ a) dτ Z0 − The above integral is zero for 0 t x(t)= x e−kt + Ae−k(t−a)H(t a), (37) 0 − in agreement with the result obtained in Eq. (5). There are alternate ways to construct the Dirac delta function. For example, one could make the function Iǫ(t) symmetric about the point 0 by defining it as follows: 1 ǫ ǫ ǫ , 2 t 2 , Iǫ(t)= − ≤ ≤ (38) 0, t > ǫ | | 2 Smoother functions may also be considered. For example, consider the following function, called a “Gaus- sian,” t2 1 − 2 Gσ(t)= e 2σ . (39) σ√2π 5 Gaussian functions 2 1.5 sigma = 0.25 1 G(t) 0.5 sigma = 0.5 sigma = 1 0 -3 -2 -1 0 1 2 3 t The graphs of some Gaussian functions for various σ values are sketched below. The Gaussian function defined above is normalized, i.e. ∞ Gσ(t) dt =1. (40) Z−∞ You may have encountered this function in statistics courses – it is is the so-called “normal” distribution function for random variables (in this case, with mean zero). The quantity σ > 0 is known as the standard deviation and characterizes the width or spread of the curve. As σ 0, the width of the curve decreases and the height → increases, so as to preserve the area under the curve. In the limit σ 0, the Gaussian Gσ(t) approaches the → Dirac delta function in the context of integration – for any continuous function f(t), ∞ lim f(t)Gσ(t) dt = f(0). (41) σ→0 Z−∞ Application of the Dirac delta function to Newtonian mechanics In what follows, we consider the motion of a particle of mass m in one dimension. Let x(t) denote its position on the x-axis. If a force F (x) acts on the mass, then it will move according to Newton’s second law dv ma = m = F (x(t)). (42) dt If we integrate the above equation from t = t1 to t = t2 = t1 + ∆t, then t2 mv(t2) mv(t1)= m∆v = F (x(t)) dt. (43) − Zt1 Recall that the momentum of the particle is given by p = mv. Since the mass of the object is assumed to remain constant, the left side of the above equation is the change in momentum of the mass over the time interval [t1,t2]. The right side is referred to as the impulse. The above equation states that change in momentum = impulse. If the force F is constant over the time interval, then we have ∆p = F ∆t = I, (44) where I = F ∆t will denote the impulse associated with the interaction. Now consider the following situation. We assume that a nonzero force of constant magnitude F acts on the particle over the time interval [a,a + ∆t], where a> 0 and ∆t> 0. At all other times, the force acting on the particle is assumed to be zero. A sketch of the graph of F is shown above. Let us now examine the velocity of the particle. Suppose that its initial velocity is v(0) = v0 > 0. Because no force acts on the particle over the time interval [0,a], its velocity will remain unchanged, i.e., v(t) = v0 for 0 t a. ≤ ≤ 6 F (t) vs. t F t 0 a a +∆t Over the time interval [a,a + ∆t], Newton’s equation becomes dv dv F m = F or = . (45) dt dt m Integrating this DE from t = a to a t [a,a + ∆t] gives ∈ F F v(t) v = (t a) or v(t)= v + (t a). (46) − 0 m − 0 m − As expected, v(t) increases linearly in time over this interval. At the end of this interval, t = a+∆t, the velocity is F v = v + ∆t. (47) 1 0 m And for t>a + ∆t, v(t)= v1 since there is no force acting on the mass. A graph of v(t) vs. t is sketched below. v(t) vs. t v1 v0 t 0 a a +∆t The important feature of this graph is that the net change in the velocity due to the action of the force F over the time interval ∆t is F I ∆v = v v = ∆t = . (48) 1 − 0 m m Now suppose that we decrease the time interval ∆t but simultaneously increase the magnitude F of the force so that the impulse I = F ∆t – the area under the nonzero part of the graph of F – remains constant. (This is analogous to the function f∆(t) used to model the addition of A units of substance X over a time interval of length ∆.) This implies that the same change of velocity ∆v = v v would be produced over a 1 − 0 smaller time intervals. In the limit ∆t 0, the velocity function v(t) will become discontinuous at t = a, with → a jump of ∆v = v v = I/m. The result is sketched below. 1 − 0 Once again, the discontinuous jump in velocity at t = a is due to an idealized impulse of strength I = m(v v )= m∆v applied at t = a. Mathematically, this can be expressed by the DE, 1 − 0 dv m = Iδ(t a), (49) dt − or dv I = δ(t a) = ∆vδ(t a). (50) dt m − − 7 v(t) vs. t v1 v0 t 0 a If we integrate this DE from time t = 0 to a time t> 0, then we obtain t v(t) = v0 + (v1 v0) δ(τ a) dτ (51) − Z0 − = v + (v v )H(t a), t 0 0 1 − 0 − ≥ v , 0 t x(t) vs. t slope v1 x0 slope v0 t 0 a The most noteworthy feature of this graph is that it is not discontinuous – there is no jump in the position x(t) at t = a. Of course, there cannot be such a jump for it would imply that the velocity at that time would be infinite, which it is not. Newton’s equation in (49) can also be expressed in terms of the function x(t) as d2x m = Iδ(t a)= m∆vδ(t a). (52) dt2 − − The Dirac delta function as a generalized function Equations (29 and (30) essentially provide the definition of the Dirac delta function as a distribution or gener- alized function. Let G denote a nonempty open set in RN , N 1. Also let (G) = C∞(G), the space of test ≥ D 0 functions with compact support in G. Definition 1 We let ′(G) denote the space of generalized functions or distributions on G: An element U D ∈ ′(G) is a linear continuous functional, D U : (G) R, (53) D → such that U(aφ + bψ)= aU(φ)+ bU(ψ), for all φ, ψ (G), a,b R, (54) ∈ D ∈ and, as n , → ∞ φn φ implies U(φn) U(φ). (55) → → 8 Examples: 1. For a u L2(G), we define ∈ U(φ)= u(x)φ(x) dx, for all φ (G). (56) ZG ∈ D Then U ′(G). Moreover, if u = v in L2(G), then U = V in ′(G). ∈ D D ′ 2. The Dirac “delta function” distribution. For any a G, define δa (G) as follows, ∈ ∈ D δa(φ)= φ(a). (57) In the context of our previous discussion δa(x)= δ(x a), which was defined in terms of the integration − δa(φ)= φ(x)δ(x a) dx = φ(a). (58) ZG − Derivatives of generalized functions The derivatives of generalized functions are defined in terms of generalized derivatives, studied earlier. In what follows, let x = (ξ , , ξN ) and ∂i = ∂/∂ξi. Recall that generalized derivatives are defined by way of the 1 · · · following basic integration by parts formula: For u, v (G), ∈ D u∂iv dx = v∂iu dx. (59) ZG ZG This formula extends in a straightforward way to multiple derivatives. First define the multiindex α by an N-tuple of nonnegative integers (α , , αN ) and let α = α + + αN . We then define 1 · · · k k 1 · · · ∂kαk ∂αu = ∂α1 ∂αN u = (60) 1 · · · N ∂ξα1 ∂ξαN 1 · · · N Repeated integration by parts yields the following result: For u, v (G), ∈ D u∂αv = ( 1)kαk v∂αu dx. (61) ZG − ZG Let us now illustrate the application of integration by parts to the Dirac distribution on G = (a,b) R. ∈ For a φ (G), we shall define the derivative of δc(x) for a c G as follows: ∈ D ∈ ′ ′ ′ δc(x)φ(x) dx = δc(x)φ (x) dx = φ (c). (62) ZG − ZG − Repeated application of this integration by parts procedure yields (n) n (n) n (n) δc (x)φ(x) dx = ( 1) δc(x)φ (x) dx = ( 1) φ (c). (63) ZG − ZG − This sets up the following formula for the derivative of a generalized function U ′(G): ∈ D Definition 2 For a U ′(G), the derivative ∂αU is defined as follows ∈ D (∂αU(φ) = ( 1)kαkU(∂αφ) for all φ (G). (64) − ∈ D Some applications The above formalism now makes it possible to define the Green’s function associated with a boundary-value problem in terms of Dirac distributions. Recall the boundary-value problem on [0, 1] R studied earlier ∈ u′′(x)= f(x), u(0) = u(1) = 0. (65) − 9 When is sufficiently “nice,” e.g., f C[a,b], then the classical solution is given by ∈ 1 u(x)= g(x, y)f(y) dy, (66) Z0 where y(1 x), 0 y x 1, g(x, y)= − ≤ ≤ ≤ (67) x(1 y), 0 x y 1, − ≤ ≤ ≤ Now consider the case that f(x)= δa(x) for a (0, 1). Then the solution u(x) is given by ∈ 1 u(x)= g(x, y)δa(y) dy = g(x, a). (68) Z0 This leads to the formal statement ∂2g(x, y) = δy(x), y (0, 1). (69) ∂x2 ∈ The justification of this statement is provided in terms of generalized functions and their derivatives: Proposition 1 Let g = g(x, y) denote the Green’s function defined in (67). Then for each y (0, 1), ∈ ′′ U = δy on (0, 1), (70) D where U denotes the generalized function that corresponds to the classical function u(x) = g(x, y) for all x ∈ (0, 1). The electrostatic “point charge” One of the fundamental equations of electrostatics is Maxwell’s first equation. Suppose that there is a distri- bution of electrostatic charge in a region of D R3: Let ρ(x) denote the charge density at a point x R3, ⊂ ∈ normally assumed to be continuous over D. Then from Maxwell’s first equation, the electrostatic field at x, denoted by E(x), is given by ρ(x) div E(x)= , (71) ǫ0 where ǫ0 denotes the permittivity of the vacuum. Associated with the field E is the electrostatic potential function u, E(x)= u(x), (72) −∇ so that Eq. (71) becomes Poisson’s equation, ρ(x) 2u(x)= . (73) ∇ − ǫ0 In the absence of charge, i.e., ρ(x) = 0 in D, then we have 2u(x)=0, x D, (74) ∇ ∈ i.e., Laplace’s equation. Now consider the (idealized) case of an amount of charge Q located at a point – without loss of generality, the point x = 0. In this case, the density function ρ is obviously infinite at the origin: , x =0, ρ(x)= ∞ (75) 0, x =0, 6 But since the total amount of charge present is Q, we must have ρ(x) dx = Q. (76) ZR3 Thus, ρ does not correspond to a classical function. Instead, it will be represented by a Dirac “delta function” distribution, i.e., ρ(x)= Qδ0(x). (77) 10 The electrostatic field E(x) and associated electrostatic potential u(x) for this situation is given in elemen- tary courses in Physics: Q E(x)= x, x =0, (78) 4πǫ x 3 6 0 k k and Q u(x)= , x =0. (79) 4πǫ x 6 0 k k It is well known that u satisfies Laplace’s equation, as it should since the density function ρ(x) is zero for x = 0. 6 But what about the situation at x = 0? Can we say that u(x) represents, in some way, the solution to (73) at x = 0 as well? The answer is yes, in the context of generalized functions. It will be shown that a generalized function counterpart to u(x) may be defined so that a solution to Poisson’s equation is valid for all x R3. It ∈ then follows E(x) will be a solution to Maxwell’s equation (71) for a point charge. First of all, we write Poisson’s equation for a point charge Q in the form 2 Q u = δ0, (80) ∇ −ǫ0 Now multiply by a test function φ C∞(R3) and integrate over R3 to give ∈ 0 Q ( 2u)φ dx = φ(0). (81) ZR3 ∇ −ǫ0 We now switch from the classical function u to generalized function U: Q 2U(φ)= φ(0). (82) ∇ −ǫ0 Now consider the generalized function U which correponds to u in (79), i.e., Q φ(x) ∞ 3 U(φ)= dx, for all C0 (R ). (83) 4πǫ ZR3 x 0 k k From our earlier discussion on derivatives of generalized functions, we have 2 2 2 Q φ(x) ∞ 3 U(φ)= U( φ)= ∇ dx, for all C0 (R ). (84) ∇ ∇ 4πǫ ZR3 x 0 k k This implies that 2 φ(x) ∞ 3 ∇ dx = 4πφ(0), for all C0 (R ). (85) ZR3 x − k k We must show that this equation holds for all test functions φ with compact support over R3, including the point x = 0. To do this, define the set G = x R3 : 0 2φ(x) 2φ(x) ∇ dx = lim ∇ dx. (87) ZR3 x r→0 Z x k k G k k We now use the following Green’s identity: (f g g f) n dS = (f 2g g 2f) dV, (88) Z∂Ω ∇ − ∇ · ZΩ ∇ − ∇ 11 for Ω R3 with (sufficiently smooth) boundary ∂Ω. (To derive this identity, apply the divergence theorem ⊂ to the integral on the left.) Here Ω = G, which has an outer boundary ∂GR, with x = R, and an inner −1 k k boundary ∂Gr, with x = r. We let f = x and g = φ to give k k k k 1 1 1 1 2φ dx = φ 2 dx + φ n φ n dS + Z x ∇ Z ∇ x Z x ∇ · − ∇ x · G k k G k k ∂GR k k k k 1 1 φ n φ n dS (89) Z x ∇ · − ∇ x · ∂Gr k k k k 2 1 The first integral on the right hand side is zero since = 0 on G. The second integral over ∂GR is ∇ kxk zero because φ vanishes in a neighbourhood of the set x : x = R . Thus we are left with { k k } 1 1 1 2φ dx = φ n φ n dS (90) Z x ∇ Z r ∇ · − ∇ r · G k k ∂Gr On the boundary Gr, the unit outward normal is the vector eˆr = rˆ. As a result, only the radial derivatives − − contribute to the integrand, so that 1 1 ∂φ ∂ 1 2φ dx = + φ r2 sin φ dr dθ dφ Z x ∇ Z − r ∂r ∂r r G k k kxk=r ∂φ = r φ sin φ dr dθ dφ Zkxk=r − ∂r − 4πφ(0) as r 0. (91) → − → This establishes Eq. (85). The expression for the electrostatic field E in (78) may now be determined in terms of generalized functions as follows: E(x)φ(x) dx = U(φ)= U( φ) ZR3 −∇ ∇ φ(x) ∞ 3 = ∇ dx, for all φ C0 (R ). (92) ZR3 4πǫ x ∈ 0 k k Integration by parts will yield the desired result. 12