Integral Transforms

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 following 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

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. We illustrate this with an example.

Example 106 Find the Fourier cosine transform of f (x)=e−ax, (a>0, x>0). Note these conditions mean that f (x) → 0 as x →∞.

∞ 2 −ax Fc [f]= e cos ωxdx π 0

From a table on integrals,

−ax −ax cos = e [ sin − cos ] e ωxdx 2 + 2 ω ωx a ωx a ω a ω = e−ax sin ωx − cos ωx a2 + ω2 a

Thus 2 ∞ F [ ]= a −ax ω sin − cos c f 2 2 e ωx ωx π a + ω a 0 2 = a π a2 + ω2 100 CHAPTER 12. INTEGRAL TRANSFORMS 12.4 The Sine Transform: An Application

We show how the Fourier sine transform can be used to solve a PDE. Consider the following IBVP ⎧ 2 ⎨ PDE ut = α uxx 0

We apply Fs, the Fourier transform with respect to x to each side to get 2 Fs ut − α uxx = Fs [0]

Using linearity of F s and the fact that F x [0] = 0,weget 2 Fs [ut] − α Fs [uxx]=0 which can be written as 2 Fs [ut]=α Fs [uxx] Now, we evaluate the Fourier sine transform of each quantity separately. Remember that we are taking the sine transform with respect to x.So, we have a dx integral. 2 ∞ Fs [ut]= ut (x, t)sinωxdx π 0 ∂ 2 ∞ = u (x, t)sinωxdx (Leibniz rule) ∂t π 0 d = Fs [u] dt d = U (t) dt

Note that while u was a function of x and t, its transform Fs [u] is a function of ω and t. The new variable ω will be treated as a parameter. Hence our notation Fs [u]=U (t). Forthesecondtransform,wegetusing proposition 101

2 2 Fs [uxx]= ωu(0,t) − ω Fs [u] π 2Aω = − ω2U (t) π 12.4. THE SINE TRANSFORM: AN APPLICATION 101

Replacing in the PDE, we get d 2Aω U (t)=α2 − ω2U (t) dt π d 2Aωα2 U (t)+ω2α2U (t)= dt π We need to ﬁnd the IC for this new equation, by applying the same transform to the original IC. So, we get

Fs [u (x, 0)] = 0 U (0) = 0

So, our ODE problem is ⎧ ⎨ d 2Aωα2 U (t)+ω2α2U (t)= 0

Step 2: Solve the ODE This is an ODE we have already solved using the 2 2 integrating factor technique (multiply each side by eω α t). Its solution is 2A 2 2 U (t)= 1 − e−ω α t πω

Step 3: Applying the inverse transform to the solution F s changed the −1 variable x into ω. F s will do the reverse. In this process, t is just a con- stant. Since U (t)=Fs [u], it follows that

−1 u (x, t)=Fs [U (t)]

−1 To ﬁnd u (x, t) we can either compute the integral corresponding to Fs or 2 2 1 − e−aω x√ use tables. From the tables, we see that F −1 =erfc a . s π ω 2 2A 2 2 Ourfunctionisalmostthat.WehaveU (t)= 1 − e−ω α t = 2 2 πω 2 1 − e−a ω t A . Thus, using linearity of F −1,wehave π ω s 2 2 2 1 − e−a ω t F −1 [U (t)] = F −1 A s s π ω 2 2 2 1 − e−a ω t = AF −1 s π ω √ = erf x A c 2 α t 102 CHAPTER 12. INTEGRAL TRANSFORMS 12.5 Problems

1. Finish proving proposition 100. 2. Finish proving proposition 101. 3. Finish proving proposition 102. 4. Solve the ODE: ⎧ ⎨ d 2Aωα2 U (t)+ω2α2U (t)= 0

5.Solveusingthesineorcosinetransform ⎧ 2 ⎨ PDE ut = α uxx 0