Chapter 12 Integral Transforms 12.1 Goal In this chapter, we introduce the idea of integral transforms and discuss some of their applications. More specifically, we will address the following: 1. What are integral transforms? 2. How many integral transforms are there? 3. What do they represent? 4. What are their applications? 5. What are their properties? 6. How do we compute them? The last item will not be addressed. We will simply use tables of known integral transforms. We will look deeper into applications in the next chapters as we will focus on one integral transform: the Fourier transform. 12.2 Integral Transforms: Definitions and Prop- erties We begin by giving a general idea of what integrals transforms are, and how they are used. There is a whole family of integral transforms which include the Fourier transform, the Laplace transform, the Mellin transform, the Hankel transform, ... In addition, as we will see, each transform goes with its inverse. Definition 96 An integral transform is an operator of the form: B F [f]=F (s)= f (t) K (s, t) dt (12.1) A 93 94 CHAPTER 12. INTEGRAL TRANSFORMS The function K is called the kernel of the transformation.Itisthemajor part of the transform. What makes the difference between the various transforms mentioned is mostly their kernel. It is also the interval of integration, [A, B] in the general definition. Remark 97 You will note that we used two notations: F [f] and F (s).Asan operator denoted F, an integral transform acts on a function. This is why we used the notation F[f]. Some texts also use F[f](s) because F[f] is a function of a new variable we call s in the definition. For that reason, we also use the notation F (s). In other words, we apply an integral transform to a function f (t) and we obtain a new function F (s). One of the features of integral transforms, one we will use in this chapters and the next two, is that integral transforms eliminate partial derivatives with respect to one of the variables. The new equation has one less variable. So, if we start with a PDE having two variables and partial derivatives with respect to both, then after applying an integral transform, we would have an ODE. For example, if we apply an integral transform to the heat equation 2 ut = α uxx to eliminate the time derivative, then we would be left with an ODE in x.Ifwe applied an integral transform to the equation uxx + uyy + uzz =0 to eliminate the derivative in x, we would be left with a PDE only involving y and z. We could apply an integral transform again to eliminate the derivative in y. We would be left with an ODE. Once the new equation is solved, we need to express the solution in terms of the original variables and functions. This is done using the inverse integral transform. This is why each transform goes in pair with its inverse. A transform and its inverse form what is called a transform pair. Both are needed when solving an equation. The following flow chart illustrates better than words how integral transforms can be applied. 12.2. INTEGRAL TRANSFORMS: DEFINITIONS AND PROPERTIES 95 Figure 12.1: Using an Integral Transform to Solve a PDE We finish this section by listing some common transforms and their inverse. Transform Interval Transform Inverse Transform 2 [0 ∞) F [ ]= ( )= ∞ ( )cos F −1 [ ]= ( )= ∞ ( )sin Fourier Sine , s f F ω 0 f x ωxdx s F f x 0 F ω ωxdω π 2 ∞ −1 ∞ Fourier Cosine [0, ∞) Fc [f]=F (ω)= f (x)cosωxdx F [F ]=f (x)= F (ω)cosωxdω π 0 c 0 1 ∞ 1 ∞ Fourier (−∞, ∞) F [f]=F (ω)=√ f (x) e−iωxdx F −1 [F ]=f (x)=√ F (ω) eiωxdω 2π −∞ 2π −∞ ∞ 2 L nπx −1 nπx Finite Sine [0,L] Fs [f]=Sn = 0 f (x)sin dx F s [Sn]=f (x)= Sn sin L L n=1 L 2 ∞ [0 ] F [ ]= = L ( )cosnπx F −1 [ ]= ( )= cos nπx Finite Cosine ,L c f Cn 0 f x dx c F f x Cn L L n=1 L ∞ 1 c+i∞ Laplace [0, ∞) L[f]=F (s)= f (x) e−sxdx L−1 [F ]=f (x)= F (s) esxds 0 2πi c−i∞ [0 ∞) [ ]= ( )= ∞ ( ) ( ) −1 [ ]= ( )= ∞ ( ) ( ) Hankel , H f Fn ξ 0 rJn ξr f r dr H Fn f r 0 Jn ξr Fn ξ dξ ∞ 1 c+i∞ Mellin [0, ∞) M [f]=F (s)= xs−1f (x) dx M −1 [F ]=f (x)= x−sf (s) ds 0 2πi c−i∞ Remark 98 For the inverse Laplace and Mellin transforms, c is a real number. Remark 99 For the Hankel transform, Jn is the Bessel function of the first −1 ≥ kind,orordern 2 . We are not going to study all these transforms in detail. In this chapter, we will only study the sine and cosine transforms. In the next two, we will study the Fourier transform. 96 CHAPTER 12. INTEGRAL TRANSFORMS 12.3 Some Properties of the Sine and Cosine Transforms We begin with general properties the Fourier sine and cosine transforms have. We will then give the transform of some standard functions. These properties are very important. They will be used to transform a PDE problem into an easier problem. Proposition 100 All the integral transforms and their inverses in the table above are linear operators. Proof. We prove the result for the Fourier sine transform. The proof for the remaining transforms is left as an exercise. If f and g are two functions and α isarealnumber,weneedtoprovethatFs [f + g]=Fs [f]+Fs [g] and Fs [αf]=αFs [f]. 2 ∞ Fs [f + g]= (f (x)+g (x)) cos ωxdx π 0 2 ∞ = [f (x)cosωx + g (x)cosωx] dx π 0 2 ∞ 2 ∞ = f (x)cosωxdx + g (x)cosωxdx π 0 π 0 = Fs [f]+Fs [g] And 2 ∞ Fs [αf]= (αf (x)) cos ωxdx π 0 2 ∞ = α f (x)cosωxdx π 0 = αFs [f] Proposition 101 (Transforms of derivatives) Let f be a function such that f (x) → 0 and f (x) → 0 as x →∞∞. The following are true: 1. Fs [f ]=−ωFc [f] 2 2 2. Fs [f ]= ωf (0) − ω Fs [f] π −2 3. Fc [f ]= f (0) + ωFs [f] π −2 2 4. Fc [f ]= f (0) − ω Fc [f] π 12.3. SOME PROPERTIES OF THE SINE AND COSINE TRANSFORMS97 Proof. We prove part 1. The remaining parts are left as exercises. For these proofs, we will use integration by parts. Also, recall that to compute an integral ∞ u of the form f (x) dx, we must compute lim f (x) dx.Withthisinmind, a u→∞ a we have ∞ 2 Fs [f ]= (f (x)) cos ωxdx π 0 2 u = lim (f (x)) cos ωxdx π u→∞ 0 u 2 u = lim f (x)sinωx| − ω f (x)cosωxdx (integration by parts) π u→∞ 0 0 2 ∞ = lim f (u)sinωu − f (0) sin 0 − ω f (x)cosωxdx π u→∞ 0 2 ∞ = −ω f (x)cosωxdx π 0 = −ωFc [f] For this computation we also used the fact that lim f (u)sinωu =0since by u→∞ assumption lim f (u)=0and we know that sin ωu is bounded. u→∞ Proposition 102 (Derivatives of transforms) Let f be a function. The fol- lowing are true: d 1. Fc [xf (x)] = Fs [f (x)] dw d 2. Fs [xf (x)] = − Fc [f (x)] dw Proof. We start from the expression on the right and we use Leibniz rule, which we reviewed several chapters earlier (see theorem 31). We have d d 2 ∞ Fs [f (x)] = f (x)sinωxdx dw dw π 0 2 ∞ d = (f (x)sinωx) dx (by Leibniz rule) π 0 dw 2 ∞ = xf (x)cosωxdx π 0 = Fc [xf (x)] We now list the sine and cosine transforms of some standard functions. Be- fore we do so, we introduce some new functions which play a role in the tables we are about to give. 98 CHAPTER 12. INTEGRAL TRANSFORMS Definition 103 We define four important functions. 1. The Heaviside function,denotedH (x − a),isdefinedtobe 0 if x<a H (x − a)= 1 if x ≥ a 2. The Reflected Heaviside function,denotedH (a − x),isdefinedtobe 0 if x>a H (a − x)= 1 if x ≤ a 3. The error function, also called the Gauss error function,denoted erf (x),isdefinedtobe x 2 2 erf (x)=√ e−t dt π 0 4. The complementary error function,denotederf c (x) is defined to be erf c (x)=1− erf (x) ∞ 2 2 = √ e−t dt π x Proposition 104 The following table gives the sine transform pair for some standard functions. ∞ 2 ∞ f (x)= F (ω)sinωxdω F (ω)= f (x)sinωxdx 0 π 0 0 0 1 ω f (ax) F a a 2ω e−ax π (a2 + ω2) 1 2 − 1 2 x 2 πω 2 H (a − x) [1 − cos ωa] πω x−1 1 x e−aω a2 + x2 x 1 e−ω sin ω x4 +4 2 1 − −aω tan−1 a e x 2 ω −x2f (x) F (ω) π 2 x√ 2 1 − e−aω erf c a 2 π ω 12.3. SOME PROPERTIES OF THE SINE AND COSINE TRANSFORMS99 Proposition 105 The following table gives the cosine transform pair for some standard functions. ∞ 2 ∞ f (x)= F (ω)sinωxdω F (ω)= f (x)sinωxdx 0 π 0 0 0 1 ω f (ax) F a a 2a e−ax π (a2 + ω2) 1 2 − 1 2 x 2 πω 2 H (a − x) sin aω πω ω2 2 1 − e−ax √ e 4a πa sin ax H (a − ω) xa e−aω x2 + a2 2 −x2f (x) F (ω) π Finding the Fourier transform of a function is a matter of computing the corresponding integral, which can be hard.