AMATH  Applied Functional Analysis Fall  The Dirac delta function distribution The material on pages  is taken from ERVs Fall  lectu re notes for AMATH  Dierential Equations II
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AMATH Applied Functional Analysis Fall The Dirac delta function distribution The material on pages is taken from ERVs Fall lectu re notes for AMATH Dierential Equations II

Introduction Suppose you have a chemical or radioactive for that matter species in a beaker that decays according to the rate law dx dt kx 1 where is the concentration at time Suppose that at time 0 there is amount of X present Of course if the

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AMATH Applied Functional Analysis Fall The Dirac delta function distribution The material on pages is taken from ERVs Fall lectu re notes for AMATH Dierential Equations II




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Presentation on theme: "AMATH Applied Functional Analysis Fall The Dirac delta function distribution The material on pages is taken from ERVs Fall lectu re notes for AMATH Dierential Equations II"— Presentation transcript:


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AMATH 731: Applied Functional Analysis Fall 2008 The Dirac delta function distribution (The material on pages 1-8 is taken from ERVs Fall 2005 lectu re notes for AMATH 351, Differential Equations II.) Introduction Suppose you have a chemical (or radioactive, for that matter ) species in a beaker that decays according to the rate law dx dt kx. (1) where ) is the concentration at time . Suppose that at time = 0, there is amount of X present. Of course, if the beaker is left alone, then the amount of at time 0 will be given by ) = kt (2) Now suppose that at time a>

0, you quickly (i.e., instantaneously) add an amount A> 0 of to the beaker. Then what is ), the amount of at time 0? Well, for 0 t , i.e., for all times before you add units of to the beaker, ) is given by Eq. (2) above. Then at , there would have been ka in the beaker, but you added , to give ) = ka Then starting at , the system will evolve according to the rate law. We can cons ider ) as the initial condition and measure time from . The amount of in the beaker will be ) = ( ka kt , for t a. (3) Lets summarize our result compactly: For the above experim ent, the amount of in the beaker will

be given by ) = kt t , t a. (4) A qualitative sketch of the solution is given below. Clearly , the solution ) has a discontinuity at , the time of instantaneous addition of the amount . Otherwise, it is differentiable at all other points. ) + ) vs. Note that we can write the solution in Eq. (4) even more compac tly as follows, ) = kt Ae , t (5) where ) is the Heaviside function reviewed earlier. We shall retur n to this solution a little later. We now ask whether the above operation the instanteous addi tion of an amount of to the beaker can be represented mathematically, perhaps with

a function ) so that the evolution of ) can be expressed as dx dt kx (6)
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where ) models the instantaneous addition of an amount of at time . The answer is Yes, and ) will be the so-called Dirac delta function ) = A ). But in order to appreciate this result, let us now consider th e case of a less brutal addition of to the beaker. Suppose that we add an amount of units of but over a time interval of length ∆. Well also add to the beaker at a a constant rate of A/ ∆ units per unit time. This means that our evolution equation for ) will take the form dx dt kx, t

kx , a + kx, t + (7) We can express this in compact form as dx dt kx (8) where ) = + ∆))] (9) A graph of the function ) is sketched below. Note that the area under the curve and abo ve the -axis is 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. + ) vs. We can solve the DE in (8) in two ways: (1) as a linear first-orde r inhomogeneous DE, (2) using Laplace Transforms. Here, well use Method (2). Taking LTs of both si des of (8) gives sX kX ) = (10) Solving for ): ) = (11) Noting that the inverse

LT of is kt , we have, after taking inverse LTs: ) = kt kt (12) Note that the first term kt is the solution to the homogeneous DE associated with (8). Now compute the convolution of with kt as follows: kt d (13) d + ∆)) d ) + Because of the Heaviside function ), the integrand in the first integral ) hence the integral itself will be zero for 0 t . For , we can compute ) as follows: ) = d kt d (14)
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, t a. Likewise, we can determine the second integral to be ) = +∆)) , t + (15) The final result for ) is ) = kt ) + + ∆)) (16) The qualitative

behaviour of the graph of ) is sketched below. It is possible that ) is decreasing over the interval [ a,a + ∆], during which time the amount is added. But if ∆ is sufficiently small, i.e., the rate A/ is large enough, ) will be increasing over this interval. ) vs. ) + + But these points are rather secondary. What is of prime impor tance is the difference in concentrations between time , the time at which we began to add to the beaker, and time + ∆, the time at which we stopped adding . Remember that regardless of ∆, we are always adding a total a mount of to

the beaker. This difference in concentrations is given by + ∆) ) = +∆) ka (17) ka In particular, we are interested in what happens to this diffe rence as 0, i.e., the time interval over which we add units of goes to zero. To find this limit, if it exists, we expand the exp onential involving ∆ to give = 1 ∆ + ( as (18) so that ∆ + ( as (19) Substituting this result into (17) gives the result + ∆) A as (20) In other words, in the limit 0, the graph of ) will exhibit a discontinuity at . The magnitude of this jump is , precisely was found

with the earlier method. We now return to the inhomogeneous DE in (8) and examine the be haviour of the inhomogeneous term ) as 0. Recall that this is the driving term, the function that m odels the addition of a total of
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units of to the beaker over the time interval [ a,a + ∆]. The width of the box that makes up the graph of ) is ∆. The height of the box is . In this way, the area of the box the total amount of delivered is . As ∆ decreases, the box gets thinner and higher. In the limit 0, we have produced a function ) that is zero everywhere except at

, where it is undefined. This is the idea behind the Dirac delt function, which we explore in more detail below. The Dirac delta function Let us define the following function ) for an > 0: ) = , , t> (21) The graph of ) is sketched below. ) vs. t Clearly dt = 1 for all > (22) Now let ) be a continuous function on [0 ) and consider the integrals dt dt, (23) in particular for small and limit 0. For any > 0, because of the continuity of ), there exists, by the Mean Value Theorem for Integrals, a [0 , ] such that dt . (24) Therefore, dt (25) As

0, 0 since the interval [0 , →{ . Therefore lim dt (0) (26) We may, of course, translate the function ) to produce the result lim dt (27)
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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 continuo us functions. In proper mathematical parlance, the Dirac delta function is a distribution . The Dirac delta function ) is defined by the integral dt (0) (28) Moreover dt = 0 if c> (29) As well, we have the translation result dt (30) Finally, we

compute the Laplace transform of the Dirac delta function. In this case ) = st so that )] = st dt as , a (31) Let us now return to the substance problem examined earlier where an amount of substance is added to a beaker over a time interval of length ∆. Comparing t he function ) used for that problem and the function ) defined above, we see that ∆ = and ) = AI (32) From our discussion above on the Dirac delta function, in the limit 0 the function ) becomes ) = A (33) Therefore, the differential equation for ) modelling the instantaneous addition of to the beaker is

given by dx dt kx A (34) We now compute ) using the result in Eq. (12) obtained from Laplace transfor ms. Using ) = A ), the convolution in this equation becomes )( ) = kt A (35) d The above integral is zero for 0 t . For , this integral is nonzero and becomes d Ae , t a. (36) Therefore, the solution ) becomes ) = kt Ae (37) in agreement with the result obtained in Eq. (5). There are alternate ways to construct the Dirac delta functi on. For example, one could make the function ) symmetric about the point 0 by defining it as follows: ) = (38) Smoother functions may also be considered.

For example, con sider the following function, called a Gaus- sian, ) = (39)
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0 0.5 1 1.5 2 -3 -2 -1 0 1 2 3 G(t) Gaussian functions sigma = 1 sigma = 0.5 sigma = 0.25 The graphs of some Gaussian functions for various values are sketched below. The Gaussian function defined above is normalized, i.e. dt = 1 (40) You may have encountered this function in statistics course s it is is the so-called normal distribution function for random variables (in this case, with mean zero). The quan tity σ > 0 is known as the standard deviation and characterizes the width or

spread of the curve. As 0, the width of the curve decreases and the height increases, so as to preserve the area under the curve. In the l imit 0, the Gaussian ) approaches the Dirac delta function in the context of integration for any c ontinuous function ), lim dt (0) (41) Application of the Dirac delta function to Newtonian mechan ics In what follows, we consider the motion of a particle of mass in one dimension. Let ) denote its position on the -axis. If a force ) acts on the mass, then it will move according to Newtons sec ond law ma dv dt )) (42) If we integrate the above equation

from to + , then mv mv ) = )) dt. (43) Recall that the momentum of the particle is given by mv . Since the mass of the object is assumed to remain constant, the left side of the above equation is the ch ange in momentum of the mass over the time interval [ ,t ]. The right side is referred to as the impulse . The above equation states that change in momentum = impulse. If the force is constant over the time interval, then we have I, (44) where will denote the impulse associated with the interaction. Now consider the following situation. We assume that a nonze ro force of constant magnitude acts

on the particle over the time interval [ a,a + ], 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 is shown above. Let us now examine the velocity of the particle. Suppose that its initial velocity is (0) = 0. Because no force acts on the particle over the time interval [0 ,a ], its velocity will remain unchanged, i.e., ) = for
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a a + ) vs. Over the time interval [ a,a + ], Newtons equation becomes dv dt or dv dt (45) Integrating this DE from to a a,a + ] gives ) or ) = (46) As expected, ) increases

linearly in time over this interval. At the end of this interval, + , the velocity is t. (47) And for t>a + ) = since there is no force acting on the mass. A graph of ) vs. is sketched below. a a + ) vs. The important feature of this graph is that the net change in t he velocity due to the action of the force over the time interval is (48) Now suppose that we decrease the time interval but simultaneously increase the magnitude of the force so that the impulse the area under the nonzero part of the graph of remains constant. (This is analogous to the function ) used to model the addition of

units of substance over a time interval of length ∆.) This implies that the same change of ve locity would be produced over a smaller time intervals. In the limit 0, the velocity function ) will become discontinuous at , with a jump of I/m . The result is sketched below. Once again, the discontinuous jump in velocity at is due to an idealized impulse of strength ) = applied at . Mathematically, this can be expressed by the DE, dv dt I (49) or dv dt ) = v (50)
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) vs. If we integrate this DE from time = 0 to a time t> 0, then we obtain ) = + ( d (51) + ( , t t , t a. The

graph of ) coincides with the graph sketched above. We can integrate this discontinuous velocity-time graph to obtain the position ). The graph of ) will have two components: (1) a straight line of slope for 0 , (2) a straight line of slope for t>a . The qualitative behaviour of this graph is sketched below. ) vs. slope slope The most noteworthy feature of this graph is that it is not discontinuous there is no jump in the position ) at . 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. Newtons equation in (49)

can also be expressed in terms of th e function ) as dt I ) = v (52) 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 denote a nonempty open set in 1. Also let ) = ), the space of test functions with compact support in Definition 1 We let denote the space of generalized functions or distributions on : An element is a linear continuous functional, (53) such that a b ) = aU ) + bU for all φ, ∈D , a,b (54) and, as implies (55)
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Examples: 1. For a ), we define ) = dx, for all ∈D (56) Then ∈D ). Moreover, if in ), then in ). 2. The Dirac delta function distribution. For any , define ∈D ) as follows, ) = (57) In the context of our previous discussion ) = ), which was defined in terms of the integration ) = dx (58) Derivatives of generalized functions The derivatives of generalized functions are defined in term s of generalized derivatives, studied earlier. In what follows, let = ( , ) and ∂/ . Recall that generalized derivatives are defined by way of th

following basic integration by parts formula: For u,v ∈D ), u vdx v udx. (59) This formula extends in a straightforward way to multiple de rivatives. First define the multiindex by an -tuple of nonnegative integers ( , ) and let . We then define (60) Repeated integration by parts yields the following result: For u,v ∈D ), u = ( 1) v udx. (61) Let us now illustrate the application of integration by part s to the Dirac distribution on = ( a,b For a ∈D ), we shall define the derivative of ) for a as follows: dx dx (62) Repeated application of this integration

by parts procedur e yields dx = ( 1) dx = ( 1) (63) This sets up the following formula for the derivative of a gen eralized function ∈D ): Definition 2 For a ∈D , the derivative is defined as follows ) = ( 1) for all ∈D (64) 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] studied earlier 00 ) = , u (0) = (1) = 0 (65)
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When is sufficiently nice, e.g., a,b ], then the classical

solution is given by ) = x,y dy, (66) where x,y ) = (1 (1 (67) Now consider the case that ) = ) for (0 1). Then the solution ) is given by ) = x,y dy x,a (68) This leads to the formal statement x,y ∂x , y (0 1) (69) The justification of this statement is provided in terms of ge neralized functions and their derivatives: Proposition 1 Let x,y denote the Greens function defined in (67). Then for each (0 1) 00 on (0 1) (70) where denotes the generalized function that corresponds to the cl assical function ) = x,y for all (0 1) The electrostatic point charge One of the

fundamental equations of electrostatics is Maxwe lls first equation. Suppose that there is a distri- bution of electrostatic charge in a region of : Let ) denote the charge density at a point normally assumed to be continuous over . Then from Maxwells first equation, the electrostatic field at denoted by ), is given by div ) = (71) where denotes the permittivity of the vacuum . Associated with the field is the electrostatic potential function ) = (72) so that Eq. (71) becomes Poissons equation ) = (73) In the absence of charge, i.e., ) = 0 in , then we have ) = 0 , x

D, (74) i.e., Laplaces equation. Now consider the (idealized) case of an amount of charge located at a point without loss of generality, the point = 0. In this case, the density function is obviously infinite at the origin: ) = , x = 0 , x = 0 (75) But since the total amount of charge present is , we must have dx Q. (76) Thus, does not correspond to a classical function. Instead, it wil l be represented by a Dirac delta function distribution, i.e., ) = Q (77) 10
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The electrostatic field ) and associated electrostatic potential ) for this situation is given in

elemen- tary courses in Physics: ) = , x = 0 (78) and ) = , x = 0 (79) It is well known that satisfies Laplaces equation, as it should since the density function ) is zero for = 0. But what about the situation at = 0? Can we say that ) represents, in some way, the solution to (73) at = 0 as well? The answer is yes, in the context of generalized fu nctions. It will be shown that a generalized function counterpart to ) may be defined so that a solution to Poissons equation is val id for all . It then follows ) will be a solution to Maxwells equation (71) for a point cha rge. First

of all, we write Poissons equation for a point charge in the form (80) Now multiply by a test function ) and integrate over to give φdx (0) (81) We now switch from the classical function to generalized function ) = (0) (82) Now consider the generalized function which correponds to in (79), i.e., ) = dx, for all (83) From our earlier discussion on derivatives of generalized f unctions, we have ) = ) = dx, for all (84) This implies that dx (0) for all (85) We must show that this equation holds for all test functions with compact support over , including the point = 0. To do this,

define the set : 0 . Since the only point of concern is = 0, it is sufficient to consider an R> 0 and test functions that vanish for | R/ 2. Note that dx sin φdrdφdθ, (86) which implies that the integral on the left is finite. Therefo re, dx = lim dx. (87) We now use the following Greens identity dS dV, (88) 11
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for with (sufficiently smooth) boundary Ω. (To derive this identity, apply the divergence theorem to the integral on the left.) Here Ω = , which has an outer boundary ∂G , with , and an inner boundary ∂G ,

with . We let and to give φdx dx ∂G dS ∂G dS (89) The first integral on the right hand side is zero since = 0 on . The second integral over ∂G is zero because vanishes in a neighbourhood of the set . Thus we are left with φdx ∂G dS (90) On the boundary , the unit outward normal is the vector . As a result, only the radial derivatives contribute to the integrand, so that φdx ∂r ∂r sin φdrdθd ∂r sin φdrdθd (0) as (91) This establishes Eq. (85). The expression for the electrostatic field in (78) may now be

determined in terms of generalized functio ns as follows: dx ) = dx, for all (92) Integration by parts will yield the desired result. 12