New Physics: Sae Mulli, Vol. 70, No. 9, September 2020, pp. 759∼765 http://dx.doi.org/10.3938/NPSM.70.759

General Definitions of Transforms for

Dongseung Kang Department of Education, Dankook University, Gyeonggi 16890, Korea

Hoewoon Kim Department of Mathematics, Oregon State University, Corvallis, Oregon 97331, USA

Bongwoo Lee∗ Department of Science Education, Dankook University, Gyeonggi 16890, Korea

(Received 10 July 2020 : revised 12 August 2020 : accepted 18 August 2020)

The Laplace and the Fourier transforms are famous integral transform methods in mathematical physics for solving differential equations. However, most undergraduate textbooks on differential equations only contain a few chapters covering Laplace and Fourier transforms, begin with their definitions as improper on a half line, (0, ∞), or on an entire line, (−∞, ∞), respectively, proceed with their properties, and then apply them to solve differential equations. Many students just follow these procedures while wondering how those integral transform methods arise naturally for physically reasonable situations. This present article presents new perspectives on derive the general definitions of the Laplace and the Fourier transforms and presents examples in physicsto help students discover the Laplace and the Fourier transforms on various domains, in contrast to only a half line and a whole line in most textbooks.

Keywords: , , Integral transforms, Mathematical physics, Differential equations, Undergraduate mathematics

I. Introduction physics courses, is in undergraduate physics curriculum to learn and study mathematics also shows that math- Paul A. M. Dirac, an English theoretical physicist, ematics plays an important role in the investigation of communicated his idea of the relation between physics physical sciences. In the course of mathematical physics, and mathematics on presentation of the James Scott integral transforms such as Laplace and Fourier trans- Prize in 1939 saying that there should be two meth- forms are dealt with and provide the students with an ods in the study of natural phenomena making a success opportunity to think about the modern approaches to of research: (1) the method of experiment and observa- solve differential equations as other topics in linear alge- tion and (2) the one of mathematical reasoning [1]. As bra and differential equations. he described the connection between mathematics and In mathematical physics, we use pairs of functions re- physics goes far deeper than the simple perspective of lated by an expression of the following form, “mathematical quality in Nature” and one would agree ∫ ∞ that physics is closely related to mathematics in even g(α) = f(t)K(α, t) dt. −∞ education, not only in research. In particular, the rea- son that mathematical physics, as one of requirements in The g(α) is called the integral transform of f(t) by the kernel K(α, t) like below: (a) the Laplace integral − ∗E-mail: [email protected] transform with the kernel function K(α, t) = e αt if t >

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0 and K(α, t) = 0 if t ≤ 0, (b) the Fourier integral From a pedagogical point of view, this approach might transform with the kernel function K(α, t) = e−iαt for make the students confused by the definition because it all t. looks rather strange and demotivating as starting with Integral transforms, especially Laplace transform and an out of the blue [6, 7]. They are Fourier transform, have many special physical applica- even clueless about what H(s) is, not to mention the tions and interpretations. If an original problem can be applications of it. Students often ask questions like ‘Why solved only with difficulty in the original space, it often do I have to use the Laplace transform to solve an electric happen that integral transform of the problem can be circuit?’ and they might find it difficult to understand solved relatively easily. Then we can get an answer with what they are doing when they use the Laplace transform inverse integral transform. For example, Laplace trans- [8]. From even teacher’s perspectives, not only students’ form can be used to solve ordinary differential equations views, it is one of the most difficult topics for students (ODEs) by reducing a linear to an to grasp when applying it in applications such as the algebraic equation, which can be easily solved [2]. The electric circuit theory at undergraduate level [8,9]. Fourier transform can also be used to solve partial dif- The integral transform is very important methods be- ferential equations (PDEs). Moreover, because Fourier cause it provides the students with an opportunity to transform decomposes a function (often a function of think about the modern approaches to solve differential time, or a signal) into its constituent frequencies, this equations in terms of operators and functionals. In spite transform enables us to consider physical phenomena of of its importance there seems a lack of researches that time and space in frequency domains and vice versa and could give pedagogical approaches to teach and learn the hence it in turn gives us an interesting insight into the integral transform methods. interpretation of problems in physics. As Joseph Fourier Ngo and Ouzomgi provided a visual way of evaluating introduced the transform in his study of heat transfer, the Laplace transform and its inverse by commutative Fourier transform can be applied to thermal physics, diagrams [10]. However, the article made no considera- acoustics, optics, electrodynamics and quantum mechan- tion of motivation of the transform or any approaches to ics. Of course, integral transforms are not only used for discover and develop it. More recently Randall started the solutions of ODEs but also play an important role with a simple differential equation, solved it by the in- in the analysis of experimental data in physics [3]. The tegrating factor method, and extended the method by use of integral transforms in physics is still one of signifi- applying the concept of the area function for antideriva- cant research areas (e.g. electromagnetic propagation [4] tives to define the Laplace transform [7]. Although he and integral imaging [5].) In particular, the undergradu- tried to convince that his “evolutionary” approach can ate students in scientific fields including physics have to give the understanding of the parameter α in the def- spend considerable time in studying the integral trans- inition of the Laplace transform, there seems no clear forms on the subject of mathematical physics for their explanation of the minus sign just before the parameter majors. α unless we use the same kind of the differential equa- If we describe the process of learning integral trans- tion in his paper. For this matter in 2008 Quinn and forms in mathematical physics by taking an example of Rai presented an approach similar to Randall’s by using Laplace transform, then we are going to: a second order linear differential equation with constant 1. Define the Laplace transform of a piece-wise andof coefficients where they introduced the kernel e−st so that exponential order function f(t) as a function of s, H(s) the involving improper integrals in the process of solving 2. Do some simple examples of Laplace transforms such the equation should be able to converge for the solution as step functions and partial fraction expansions to exist [11]. In 2011 Dwyer offered another way to de- 3. Derive of Laplace transform fine the Laplace transform by using the inner product of 4. Apply the Laplace transform to simple physics prob- functions in a as a vector space with a lems such as simple harmonic oscillator, RLC circuits kernel K(s, t) and deriving a differential equation of the and electromagnetic waves. kernel from the [6]. General Definitions of Integral Transforms for Mathematical Physics – Dongseung Kang · Hoewoon Kim 761

∂ρ Even though advanced textbooks dealing with the in- (incompressible flow), the equation + ∇ · (ρv) = 0 ∂t tegral transform such as complex analysis give rigorous reduces to ∇ · v = 0. If we suppose, in addition, that investigations of the formula, they have no motivation the flow is irrotational -that is, ∇ × v = 0 or v = ∇u for for students. some potential function u, then we obtain a differential The goal of this paper, therefore, is to present an equation (the Laplace equation) alternative approach to the general definitions of the “common” integral transforms. We will present alterna- ▽2 tive approach to derive the “general” definitions of inte- u = 0 gral transforms on the complex plane with an interesting ∂2 ∂2 ∂2 for a scalar function u and ▽2 = + + . We example of the Laplace equation in potential theory. ∂x2 ∂y2 ∂z2 shall illustrate how the Laplace transform arises gener- Our paper is organized as follows. The second section ally in this practical application of the Laplace equation of this paper will start with an interesting problem from on a half-plane. As we will realize later we don’t have to physics, solve it, and define the general definitions of take the half space as the domain on which the Laplace Laplace and Fourier transforms on the complex plane equation is considered. However, in order to derive the by analyzing the solution of it. The third section will “general” definition of Laplace transform integrated over explore some examples of Laplace and Fourier transforms the half line we just need to take one of two , say on various domains. The last section as a conclusion x, to be nonnegative. makes some discussions and remarks about the integral We consider the Laplace equation of a function u(x, y) transform methods for differential equations. ∈ R2 ≥ ∈ R in the half-plane (x, y) + where x 0 and y (i.e., y is a real√ number) assuming that the function u II. General Definitions of Laplace and tends to 0 as x2 + y2 → ∞: Fourier Transforms ∂2u ∂2u ▽2u = + = 0. We begin this section for the general defini- ∂x2 ∂y2 tion of Laplace transform by considering a region Now we’ll use the method that 3 D ⊂ R filled with a fluid having variable den- is a generalization of method in 3 sity ρ(x, y, z, t)(g/cm ) and the flow of the fluid ordinary differential equations by assuming that the so- is defined by the vector velocity field v(x, y, z, t) = lutions of the Laplace equation are of the form ⟨v1(x, y, z, t), v2(x, y, z, t), v3(x, y, z, t)⟩ (cm/s). By the principle of conservation of mass the rate of increase of u(x, y) = X(x)Y (y) mass in a fixed subregion W of D (W does not change where X is a function of x alone and Y is a function of with time) equals the rate at which mass is crossing the y only. It is not hard to show that ▽2u = 0 will have boundary of W , ∂W in the inward direction; i.e., solutions of this form if we can find functions X and Y ∫ ∫ d such that ρ dV = − ρv · n dA (1) ( ) ( ) 2 2 dt W ∂W 1 d X 1 d Y − = (3) where n is the outward normal vector to the boundary of X dx2 Y dy2 W , ∂W , assuming there is no distribution of mass source. R2 for all (x, y) in the half-plane +. Noting that the left- This is the integral form of the law of conservation of hand side of the equation (3) is a function of x only and mass. Using the (or Gauss) theorem we show the right-hand side is a function of y alone, we see that that the integral form (1) above is equivalent to both sides must be equal to a constant, say −t2 with ∫ { } ∂ρ t ≥ 0. Therefore, X and Y will satisfy the following + ∇ · (ρv) dV = 0 (2) W ∂t ordinary differential equations ∂ρ and hence + ∇ · (ρv) = 0 since the integral form d2X d2Y ∂t = t2X, = −t2Y. (4) (2) is to hold for all W . If the density is a constant dx2 dy2 762 New Physics: Sae Mulli, Vol. 70, No. 9, September 2020

By solving the differential equations of X and Y in (4) Now let us extend t in (4) to be any real since we just above we see that the non-trivial solution had said that t is a constant. Then we take X(x) = e−|t|x and Y (y) = eity, which shows us that u(x, y) = u(x, y) = e−txe−ity = e−tx−ity, t > 0 e−|t|x−ity for any real number t and we, therefore, have ∫ where eiθ = cos(θ) + i sin(θ). Here we imposed the end ∞ u(x, y) = f(t) e−|t|x−ity dt (6) √behavior of the solution u, that is, u tends to zero as −∞ x2 + y2 → ∞ and we also took e−ity as the solution to for some certain function f(t). Evidently, we should the equation of Y above. For the solution to the differen- specify the nature of the flowing at the boundary of the tial equation of Y in (4) we are able to think of eity as the domain to obtain a complete description of the physi- solution for it. Then we will have an alternative Fourier cal situation and so we impose the boundary condition transform and its inverse Fourier transform should be u(0, y) = F (y) for a given function F (y) where y is any the one we are going to define for the Fourier transform real. Simply applying the boundary data or x = 0 in (6) in this paper and vice versa. From the theory of Fourier shows us that analysis they are just a couple with the transform and ∫ ∞ inverse transform relationship. u(0, y) = F (y) = f(t) e−ity dt (7) By direct computations we show that for a certain −∞ function f(t) the function or ∫ ∫ ∞ ∫ ∞ ∞ − u(x, y) = f(t) e−tx−ity dt = f(t) e−t(x+iy) dt F (ξ) = f(t) e iξt dt, ξ is any real, (8) 0 0 −∞ also satisfies the Laplace equation, ▽2u = 0 in the half- and this is the exact definition of Fourier transform of plane. This is the concept of the superposition princi- f(t) as we have seen it in many of the textbooks for ple of linear differential equations. Therefore, the solu- mathematical physics. For example, if f(t) = e−|t| then 2 tion u to the Laplace equation in the half plane R2 is we are able to find its Fourier transform F (ξ) = as + 1 + ξ2 just the Laplace transform of a certain function f(t) as one of exercise problems on this topic. Definitely we also the complex-valued function u(x, y) = u(z) where z is think of the Fourier transform of f(t) as the resulting a complex number consisting of x > 0 and y as its real integral in (6) by choosing the imaginary part of the and imaginary parts, respectively. Therefore, we finally complex variable z = x + iy or (x, y), i.e., x = 0 as we define the “general” Laplace transform of a function of did for the definition of Laplace transform where we took ∞ → R C t, f(t) : [0, ) (or ), as a complex-valued func- y = 0. ∈ R2 tion u(z) with z = (x, y) + (or on the half complex To summarize both Laplace and Fourier transforms plane), by an improper integral are just the resulting integrals by taking the real and ∫ ∞ imaginary parts of the complex variable in the solution u(z) = u(x + iy) = f(t) e−t(x+iy) dt. (5) 0 u(z) = u(x + iy) = u(x, y) to the Laplace equation on ∈ R2 ≥ ∈ R Let f(t) = e−ct in (5), for example. Then if you think of the half space (x, y) + where x 0 and y , u as a real-valued function, i.e., y = 0 then we will have respectively. 1 u(x) = . x + c III. Laplace Transforms and Fourier It might be very familiar to us because this shows that Transforms on Various Domains the “common” Laplace transform of eat with a constant 1 a is from a table of Elementary Laplace Trans- In this section we will present some examples of Laplace s − a forms in the class of differential equations you take where transforms of a function f(x) defined on both a finite the variable s represents the independent variable of the interval, say (0, 1) for simplicity, and an infinite interval Laplace transformed function. (1, ∞). General Definitions of Integral Transforms for Mathematical Physics – Dongseung Kang · Hoewoon Kim 763

We consider the Laplace equation of a function u(x, y) where a˜ and ˜b are constants, and this function is 2π in the unit disk centered at the origin on the plane R2. periodic only when c = n2 for n = 0, 1, 2. ··· . For such In general finding the temperature u in a solid which is n, assumed that it is rigid so that the only energy present is Θ =a ˜ cos(nθ) + b˜ sin(nθ) (15) thermal energy involves the law of conservation of energy n n n and the Fourier’s law to formulate the Poisson equation is the solution to (13). If c = n2 the equation (12) is an or inhomogeneous Laplace equation. When there is no Euler equation which has the general solution sink or source in the medium the resulting one is called 2 −n n the Laplace equation, ▽ u = 0 we actually solved on the Rn = αnr + βnr . (16) half space in the previous section earlier. Thus consid- We must choose α = 0 to eliminate the singularity at ering the Laplace equation in the unit disk centered at n r = 0 and thus we have the origin on the plane R2 is equivalent to finding the steady-state temperature in the unit circular plate lying n Rn = βnr . (17) on the xy plane. Since we have a circular region it is natural to use polar Combining (15) and (17) gives the separated solutions coordinates and express the Laplace equation as n un(r, θ) = r (an cos(nθ) + bn sin(nθ)), (18) ∂2u 1 ∂u ∂2u 2 + + 2 = 0, 0 < r < 1 (9) ∂r r ∂r ∂θ where an and bn are arbitrary constants. Superposing where r is the distance from the origin and θ is the angle the solutions we just found we obtain measured counterclockwise from the positive x-axis in ∑∞ n the polar coordinates. As we did in the previous section u(r, θ) = r (an cos(nθ) + bn sin(nθ)) (19) n=0 we seek separated solutions As we defined the Laplace transform of a function f in u(r, θ) = R(r)Θ(θ). (10) the previous section we can define the Laplace transform

of a function f : Z+ → R with its domain of the set of Since (r, θ) and (r, θ + 2π) represent the same point on all the non-negative integers, Z+, as the plane, it is useful to regard u as 2π periodic in θ; ∑∞ equivalently, Θ is 2π periodic. For (10) to satisfy the n u(r) = L(0,1)(f(n))(r) = f(n) r (20) equation (9) we should have n=0 ( ) 1 1 for 0 < r < 1 by taking θ = 0 or more generally θ = 2kπ R′′ + R′ Θ + RΘ′′ = 0. (11) r r2 for an integer k in u(r, θ) in (19). In other words, the Laplace transform of f : Z → R is just a power of Dividing the equation by RΘ/r2 we obtain the following + r on the interval (0, 1) such that its coefficients are the equations function values f(n) for all integers n ≥ 0. ′′ 1 ′ − c Similarly, if we go through all the computation for the R + R 2 R = 0, (12) r r Laplace equation in the exterior domain of the unit circle centered at the origin on the plane R2, i.e., r > 1, as we did in previous example then we arrive the following Θ′′ + cΘ = 0, Θ(θ) = Θ(θ + 2π) (13) separated solution ( ) ≥ ∞ n where c 0 is the separation constant (If c < 0, then ∑ 1 u(r, θ) = (a cos(nθ) + b sin(nθ)) (21) the differential equation has real exponential solutions, r n n n=0 not periodic ones). Thus we have the solutions for 1 < r < ∞ by noting that we have to take β = √ √ n Θ =a ˜ cos( cθ) + ˜b sin( cθ) (14) 0 in (16) for the decay of the solution at the infinity. 764 New Physics: Sae Mulli, Vol. 70, No. 9, September 2020

Therefore the Laplace transform of a function f : Z+ → the resulting integrals by taking the real and imaginary R is defined on (1, ∞) as parts of the complex variable in the solution u(z) = u(x+ ( ) ∑∞ n iy) = u(x, y) to the Laplace equation on the half space 1 2 u(r) = L ∞ (f(n))(r) = f(n) . (22) (x, y) ∈ R where x ≥ 0 and y ∈ R, respectively. Then (1, ) r + n=0 we help students understand the Laplace and Fourier Surprisingly, its’ just the Laurent expansion of a real- transforms on various domains such as the unit interval, valued function on (1, ∞) with the coefficients f(n) for (0, 1) and (1, ∞) by solving a boundary value of problem each integer n ≥ 0. of Laplace equation on the unit disk. In order to define the Fourier transform in this casewe Specifically, motivated by the general definition ofthe assume that the top and bottom of the plate are insulted Laplace transform on the half-complex plane we can de- L L and that its circumference is kept at a prescribed temper- fine two different Laplace transforms, (0,1) and (1,∞), ature f(θ) and now let us define the Fourier transform of on both a finite length of an interval, (0, 1), and an infi- f(θ) by taking r = 1 or imposing a Dirichlet boundary nite interval, (1, ∞). As we already noticed in the pre- condition u(1, θ) = f(θ) in either the interior or exterior vious examples these are just the power and Laurent ex- domain as pansions of a variable r, respectively. One of applications ∞ ∑∞ of the Laplace transform on (0, ) we usually teach in u(1, θ) = f(θ) = (an cos(nθ) + bn sin(nθ)) (23) the courses of differential equations for the undergrad- n=0 uates is to solve initial value problems by transforming and this trigonometric series is the exact definition of differential equations into corresponding algebraic equa- of a periodic function f(θ) on the unit circle tions where we apply the integration by parts for terms where an and bn are the Fourier coefficients of f(θ) as involving the derivatives. Hence a natural question now we might expect in the first place. Very interestingly, the arises. How do these Laplace transforms we defined dif- Fourier transform of a periodic function in the interval ferently provide solutions to differential equations that (0, 2π) is just the Fourier series expansion of the function can not be solved by more elementary means? The an- that is one of the most important topics dealing with swer should be simple as we might already know, that problems arising from mathematical physics. is, it’s the method. In other words, we seek to find the solution in the form of the power series by identifying all the coefficients in it. Also it should be IV. Conclusion noted that there are infinitely many ways to choose the angle θ, not only θ = 0, to define the Laplace trans- As pointed out the approach taken and adopted by forms on those domains. If you take the angle θ to be many textbooks in both mathematical physics and even different angle from the x-axis then we will have a new mathematics to deal with integral transforms such as Laplace transform containing Fourier series expansion of Laplace and Fourier transforms might make the students a periodic function. confused by their definitions because they look rather For the Fourier transforms of functions on the unit strange and demotivating as starting with improper in- circle or periodic functions in the interval (0, 2π) we re- tegrals out of nowhere. Students also might find it dif- alize from the examples in the previous section that it’s ficult to understand what they are doing when they use just the Fourier series expansion of the periodic function those integral transforms to solve application problems and hence the Fourier transform on the entire domain in their majors. can be thought of as the extension of the Fourier series Therefore, in this article, we present an alternative of periodic functions on a finite interval. Even though approach to derive the “general” definitions of integral we have the case of an infinite interval with a boundary transforms on the complex plane with the example of the point, say, (1, ∞) we still have the same Fourier series Laplace equation on the half space in potential theory. In expansion due to the boundary condition at r = 1 as a summary both Laplace and Fourier transforms are just expected. General Definitions of Integral Transforms for Mathematical Physics – Dongseung Kang · Hoewoon Kim 765

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