# Solving Generalized Wave and Heat Equations Using Linear Canonical Transform and Sampling Formulae

Authors: Mawardi Bahri; Ryuichi Ashino
Journal: Abstract and Applied Analysis
 (2020)

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## Abstract

Several essential properties of the linear canonical transform (LCT) are provided. Some results related to the sampling theorem in the LCT domain are investigated. Generalized wave and heat equations on the real line are introduced, and their solutions are constructed using the sampling formulae. Some examples are presented to illustrate our results.


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## Body

## 1. Introduction

The linear canonical transform (LCT) is one of nontrivial generalizations of the Fourier transform (FT). Recently, the LCT has been investigated extensively both in theory and applications. In the connection with the LCT theory, many interesting papers on this topic have been already available in the literature (see [1–5]). From the applied point of view, the LCT also has been widely used in various areas such as optics, radar system analysis, phase retrieval, pattern recognition, and many other applications (see [6–8]).It is known that the wave and heat equations are important in mathematical physics, applied mathematics, and engineering. Common methods to solve these equations are the Fourier and Laplace transforms, which give their exact solutions. The other techniques to have the solutions of these equations are the variation iteration method and the domain decomposition method, which coresponds to the adomian decomposition method [9]. Recently, the fundamental solution of the generalized wave equation was obtained using the fractional Fourier transform (FrFT) by [10]. Since the LCT is a general form of the FrFT and also it is closely related to the FT, it is possible to extend wave and heat equations into the LCT domain. However, as far as we observe, up to now, the solution of generalized heat equation using the sampling formulae and the LCT has not yet been published in the literature.Therefore, in the present article, we provide useful properties and examples of the LCT. We will consider the wave equation in the LCT domain, and we will explore the solution of this equation using the LCT as exact analytic tool for solutions. We find a relation that the solution of the wave equation using the FT and the FrFT is a special case of the solution of these generalized wave equation. We explicitly discuss a generalized version of the heat equation in the LCT domains. Based on the sampling formulae related to the LCT, we further derive its solution followed by using the LCT. We emphasize that our solutions are nontrivial generalizations of the solutions of the heat equation using the sampling formulae and FT methods. For implementations, we finally provide a few examples to illustrate our results.The remnants of the article are organized as follows: Section2 introduces definitions and examples of the LCT, which will be needed in the sequel. The useful properties of the LCT are presented in Section 3. Some results related to the sampling theorems are demonstrated in Section 4. A general form of wave equation in the LCT domain is provided in Section 5. A general form of heat equation in the LCT domain and its solution using the sampling formulae are presented in Section 6. Its solution using LCT is discussed in detail in Section 7. Finally, Section 8 concludes our article.


## 2. Linear Canonical Transform

Stuart A. Collin [11] introduced the linear canonical transforms (LCTs) during the early 1970s in paraxial optics. Independently, Moshinsky and Quesne [12] studied LCTs in quantum mechanics. Recently, Wolf [13] discussed in detail the development of LCTs including several complex extensions of the LCTs.Let us start with the definition of the several dimensional linear canonical transform:Definition 1. (see [12, 13]).
For an arbitrary2N×2N symplectic matrix
(1)E=abcd,EΩET=Ω,Ω=01N−1N0,the N-dimensional of the linear canonical transform (LCT) is then defined as
(2)LEfξ=∫ℝnftKEξ,tdt,where
(3)KEξ,t=12πiNdetbei2ξTb−1dξ−ξTb−1t+12tTab−1t,where i2=−1. Here, E is invertible and its inverse is given by
(4)E−1=abcd−1=ΩETΩT=dT−bT−cTaT.Secondly, let us introduce the inversion formula. Because of(5)KEξ,tå=KE−1t,ξ,then, inversion formula of (2) has the following form:
(6)ft=∫ℝnLEfξKE−1t,ξdξ.To solve the generalized wave and heat equations, we will only deal with the case thatN=1. When N=1, Definition 1 mentioned above becomes
(7)LEfξ=FEξ=∫ℝftKEξ,tdt,b≠0,deicd/2ξ2fdξ,b=0,where KEξ,t is so-called kernel of the LCT given by
(8)KEξ,t=12πbie1/2a/bt2−2/btξ+d/bξ2,and the matrix parameter E=abcd satisfies detE=ad−bc=1, that is, E∈Sl2,ℝ, which denotes the special linear group over ℝ.Based on (5), the LCT kernel (8) satisfies
(9)KE−1t,ξ=KEξ,t★=1−2πbie−i/2a/bt2−2/btξ+d/bξ2,where f★ stands for the complex conjugation of f. Equation (7) implies that for b=0, the LCT of a signal is essentially a chirp multiplication. Therefore, we only consider the case of b≠0 in this work. In the special case where E=01−10, the LCT mentioned above reduces to the Fourier transform (FT), that is,
(10)LEftξ=1iFftξ.Here, the FT and its inverse are defined by (see [14–16])
(11)Fftξ=12π∫ℝfte−itξdt,ft=12π∫ℝFftξeitξdξ,provided that the integral exists. The connection between the FT and LCT is given by
(12)LEftξ=1ibeid/2bξ2Feia/2bt2fξb.An application of (9), we get the inverse transform of the LCT, that is,
(13)LE−1LEft=ft=liml→∞∫−llLEfξKE−1t,ξdξ=∫ℝLEfξKE−1t,ξdξ.Or, equivalently,(14)ft=1−2πbi∫ℝLEfξe−i/2a/bt2−2/btξ+d/bξ2dξ.Definition 2.
For everyE=abcd∈Sl2,ℂ, the complex LCT is defined by
(15)LEfz=∫ℝfxKEz,xdx,where
(16)KEz,x=12πbiei/2a/bx2−2/bxz+d/bz2,provided Reia/b<0. For a is a real number, b must satisfy −π<argb<0. For an example, taking a=2−1/2=d,b=i2−1/2=c, we obtain (compare to [17])
(17)KEz,x=π−1/22−1/4e−1/2x2+z2+21/2xz.The next, following [13], we have.Remark 3.
WhenE=1−2it01, we obtain
(18)KEξ,x=14πte−x−ξ2/4t,t>0.This form is often so-called heat kernel or Gauss’s kernel.Definition 4.
The inversion formula of the complex LCT is defined as in [13, 17].
(19)fx=∫ℂLEfzKE−1x,zd2μz,zå,d2μz,zå=vmz,zådRezdImz,where
(20)vmz,zå=2πve1/2vuz2−2zzå+uåzå2,u=aåd−båc,v=2Imabå>0.Example 5. (Dirac delta function).
The Dirac delta functionδt is a generalized function satisfying the following:
(21)δt=0fort≠0,δ0=∞,∫ℝδtdt=1.By applying (7) to δt, we obtain
(22)LEδtξ=∫ℝδtKEξ,tdt=KEξ,0=12πbieid/2bξ2.Example 6. (Gaussian function).
The Gaussian function is defined by(23)gt=e−kt2,k>0.The LCT of the Gaussian functiongt is given by (see [1])
(24)LEe−kt2ξ=1a+2kbiec+2kdi/22kb−aiξ2.Let us introduce the convolution of two complex functions in the LCT domain and the relationship between the convolution and its LCT (see [5, 18–20]).Definition 7.
Letf, g∈L1ℝ. Then, the convolution operator associated with the LCT is given by
(25)f∗gx=∫ℝeitt−xabftgx−tdt.The LCT of the convolution off and g is given by
(26)12πbieidξ2/2bLEf∗gξ=LEfξLEgξ.


## 3. Useful Properties of LCT

The following Theorem8 describes the LCT of the derivative of a function (see [21]). It is very useful to solve partial differential equations related to the LCT.Theorem 8.
Letft∈L1ℝ. Assume that LEdn/dtnftξ exist. Then, we have
(27)LEdndtnftξ=−cξi+addξnLEftξ.Let us give a simple application of Theorem8.Example 9. (Mexican hat wavelet).
The Mexican hat wavelet is defined by (see [22])
(28)Mt=1−t2e−t2/2.The alternative form of (28) is
(29)Mt=−d2dt2e−t2/2.The LCT of (28) is given by
(30)LEMtξ=−1a+biehξ2−ica+6ha2ξ−4hica+c2ξ2+4h2a2ξ3.Let us give a sketch of the proof of (30) below.By applying Theorem8 with n=2 and Example 6, we get
(31)LEMtξ=−LEd2dt2e−t2/2ξ=−−cξi+addξ2LEe−t2/2ξ=−−cξi2−icaξddξ−icaddξξ+a2d2dξ21a+biec+di/2b−aiξ2.By algebraic simplifications, we obtain(32)LEMtξ=−1a+bi−c2ξ2−icaξddξ−icaddξξ+a2d2dξ2e1/2ac+bdi−1/a2+b2ξ2.Now, settingh=1/2ac+bdi−1/a2+b2, we get
(33)LEMtξ=−1a+bi−c2ξ2−icaξddξ−icaddξξ+a2d2dξ2ehξ2=−1a+bi−c2ξ2ehξ2−icaξddξehξ2−icaddξξehξ2+a2d2dξ2ehξ2=−1a+bi−c2ξ2ehξ2−2hicaξ2ehξ2−ica1+2hξ2ehξ2+a2ddξ1+2hξ2ehξ2=−1a+bi−c2ξ2ehξ2−2hicaξ2ehξ2−ica1+2hξ2ehξ2+a24hξ+1+2hξ22hξehξ2=−1a+biehξ2−c2ξ2−2hicaξ2−ica−2hicaξ2+4ha2ξ+2ha2ξ+4h2a2ξ3=−1a+biehξ2−ica+6ha2ξ−4hica+c2ξ2+4h2a2ξ3.Definition 10.
The Schwartz spaceSℝ of rapidly decaying functions is defined by a collection of complex valued functions such that
(34)S=ψt∈C∞ℝ:supt∈ℝ∣tmDnψt∣<∞,∀m,n∈ℕ,where D=d/dt.The generalization of the Schwartz space in the LCT domain is described in the following.Definition 11. ([21]).
The Schwartz spaceSEℝ of rapidly decaying functions associated with the LCT is defined by a collection of complex valued functions such that
(35)SE=ψt∈C∞ℝ:supt∈ℝ∣tmDEnψt∣<∞,∀m,n∈ℕ,where DE=d/dt−ia/bt.From the above definition, we obtain the following important result (see [21]).Theorem 12.
LetKEξ,t be the LCT kernel and D¯E=−d/dt−ia/bt. Then, we have
(36)DEnKEξ,t=−iξbnKEξ,t,∀n∈ℕ,LED¯Enψtξ=−iξbnLEψtξ,∀n∈ℕandψ∈SE.


## 4. Sampling Theorem in LCT Domain

The sampling theorem [23] states that a class of functions can be determined perfectly from samples taken at equal interval. In this part, we investigate some results of sampling theorem associated with the LCT. We first introduce the following result, which was investigated in [24, 25].Theorem 13.
Suppose thatf is band limited to ΩE in the LCT domain. Then, for every x and b>0, one has
(37)fx=∑n=−∞n=∞e−ia/2bx2fnTeia/2bnT2sincΩEx−nTbπ.Here,T=πb/ΩE, and it is the sampling period.This gives the following theorem which is very useful to obtain the solution of heat equation in the LCT domain.Theorem 14.
Under assumptions as in Theorem13, we have
(38)LEfξ=H0ξ+ΩE−H0ξ−ΩEπbπΩE2ieid/2bξ2∑n=−∞n=∞fnTeia/2bnT2e−inTξ/b,where the step function is defined by
(39)H0ξ+ΩE−H0ξ−ΩE=1ifΩE<ξ<ΩE0otherwise.Proof.
Substituting (37) into the LCT definition, it is not difficult to arrive at
(40)LEfξ=∫ℝfx12πbiei/2a/bx2−2/bxξ+d/bξ2dx=∫ℝ∑n=−∞n=∞fnTeia/2bnT2sincΩEx−nTbπ12πbiei/2−2/bxξ+d/bξ2dx=12πbieid/2bξ2∑n=−∞n=∞fnTeia/2bnT2∫ℝsincΩEx−nTbπe−ixξ/bdx=12πbieid/2bξ2∑n=−∞n=∞fnTeia/2bnT2∫ℝsinΩE/bπx−ΩET/bπ2nπΩE/bπx−ΩET/bπ2nπe−ixξ/bdx=H0ξ+ΩE−H0ξ−ΩEeid/2bξ2∑n=−∞n=∞fnTeia/2bnT2bπΩEe−inTξ/bπ2bi=H0ξ+ΩE−H0ξ−ΩEπbπΩE2ieid/2bξ2∑n=−∞n=∞fnTeia/2bnT2e−inTξ/b.This is the desired result.


## 5. Generalized Wave Equations in Linear Canonical Transform Domain

In the present section, we study how the LCT can be applied to solve partial differential equations. In particular, we are going to obtain the solution of generalized wave equation. In [26], the author have obtained the solution of wave equation using the LCT. Therefore, in this part, we propose a different way to obtain the solution of wave equations related to the LCT. We also provide a simple example to illustrate the result. Let us therefore introduce the generalized wave equation in the LCT domain.Consider the following generalized one-dimensional wave equation associated with the LCT:(41)∂2ϕx,t∂t2=k2D¯E2ϕx,t,−∞<x<∞,t>0,with the initial condition:
(42)ϕx,0=fx,∂ϕx,t∂tt=0=gx,−∞<x<∞,where k is a positive constant and fx, gx∈L1ℝ. Note that D¯E=−d/dx−ia/bx. The solution of the generalized one-dimensional wave equation mentioned above is given by
(43)ϕx,t=sgnbe−ia/2bx22×fx+kteia/2bx+kt2+fx−kteia/2bx−kt2+1k∫x−ktx+ktgξeia/2bξ2dξ.Note that, in the case whenE=01−10, the solution (43) becomes the well-known d’Alembert solution of the classical wave equation, that is,
(44)ϕx,t=12fx+kt+fx−kt+1k∫x−ktx+ktgξdξ.WhenE=cosαsinα−sinαcosα, (43) leads to
(45)ϕx,t=sgnsinαe−icotα/2x22fx+kteicotα/2x+kt2+fx−kteicotα/2x−kt2+1k∫x−ktx+ktgξeicotα/2ξ2dξ,which is the solution of the generalized wave equation using the FrFT (compare to [10]). Therefore, (43) can be seen as follows:For fixedt, applying the LCT on both sides of (41) and using Theorem 12, we obtain
(46)∂2ϕEξ,t∂t2=k2−iξb2ϕEξ,t,where ϕEξ,t=LEϕx,t is the LCT of ϕx,t with respect to x only. The solution of equation (46) takes the form:
(47)ϕEξ,t=PEξeik/bξt+QEξe−ik/bξt.Differentiating (47) with respect to t, we have
(48)ϕtEξ,t=kbξiPEξeik/bξt−kbξiQEξe−ik/bξt.Now, we use the notation(49)FEξ=LEfxξ=LEϕx,0ξ=ϕEξ,0,(50)GEξ=LEgxξ=LEϕtx,0ξ=ϕtEξ,0.Consequently, we obtain(51)FEξ=ϕEξ,0=PEξ+QEξ,(52)GEξ=ϕtEξ,0=kξξiPEξ−QEξ.From equations (51) and (52), we get
(53)PEξ=12FEξ+bikξGEξ,(54)QEξ=12FEξ−bikξGEξ.Substituting (54) in (47) yields
(55)ϕEξ,t=12FEξeik/bξt+e−ik/bξt+b2ikξGEξeik/bξt−e−ik/bξt.Now, applying inverse transform of the LCT defined by (14) into both sides of (55), we immediately get
(56)ϕx,t=1−2πbi∫ℝ12FEξeik/bξt+e−ik/bξt+b2ikξGEξeik/bξt−e−ik/bξt×e−i/2a/bx2−2/bxξ+d/bξ2dξ=1−2πbi∫ℝ1212πbi∫ℝfεei/2a/bε2−2/bεξ+d/bξ2dεeik/bξt+e−ik/bξt+b2ikξ12πbi∫ℝgεei/2a/bε2−2/bεξ+d/bξ2dεeik/bξt−e−ik/bξte−i/2a/bx2−2/bxξ+d/bξ2dξ=1−2πbi12πbi∫ℝ12∫ℝfεei/2a/bε2−2/bεξ+d/bξ2dεeik/bξt+e−ik/bξt+b2ikξ∫ℝgεei/2a/bε2−2/bεξ+d/bξ2dεeik/bξt−e−ik/bξte−i/2a/bx2−2/bxξ+d/bξ2dξ=12π∣b∣∫ℝ12∫ℝfεei/2a/bε2−2/bεξeik/bξt+e−ik/bξte−i/2a/bx2−2/bxξdε+b2ikξ∫ℝgεei/2a/bε2−2/bεξeik/bξt−e−ik/bξte−i/2a/bx2−2/bxξdεdξ=e−ia/2bx22π∣b∣∫ℝ12∫ℝfεei/2a/bε2−2/bεξeik/bξt+e−ik/bξte−i/2−2/bxξdε+b2ikξ∫ℝgεei/2a/bε2−2/bεξeik/bξt−e−ik/bξte−i/2−2/bxξdεdξ=e−ia/2bx22π∣b∣∫ℝ12∫ℝfεeia/2bε2e−iεξ/beik/bξt+e−ik/bξte−i/2−2/bxξdε+b2ikξ∫ℝgεeia/2bε2e−iεξ/beik/bξt−e−ik/bξte−i/2−2/bxξdεdξeia/2bε2e−iεξ/b.Setf¯¯ε=fεeia/2bε2,g¯¯ε=gεeia/2bε2, and v=ξ/b, then we have
(57)ϕx,t=be−ia/2bx222π∣b∣∫ℝ12π∫ℝf¯¯εe−ivεdεeikvt+e−ikvteixv+1ikv12π∫ℝg¯¯εe−ivεdεeikvt−e−ikvteixvdv.By using the FT definition (11), we immediately obtain
(58)ϕx,t=b2∣b∣e−ia/2bx22π∫ℝFf¯¯veivx+kt+eivx−kt+bikvFg¯¯veivx+kt−eivx−ktdv=sgnb2e−ia/2bx22π2πf¯¯x+kt+f¯¯x−kt+1k∫ℝFg¯¯v∫x−ktx+kteivαdαdv=sgnb2e−ia/2bx22π2πfx+kteia/2bx+kt2+fx−kteia/2bx−kt2+1k∫x−ktx+kt∫ℝFg¯¯veivαdvdα=sgnbe−ia/2bx22fx+kteia/2bx+kt2+fx−kteia/2bx−kt2+1k∫x−ktx+kt12π∫ℝFg¯¯veivαdvdα=sgnbe−ia/2bx22fx+kteia/2bx+kt2+fx−kteia/2bx−kt2+1k∫x−ktx+ktg¯¯αdα=sgnbe−ia/2bx22fx+kteia/2bx+kt2+fx−kteia/2bx−kt2+1k∫x−ktx+ktgαeia/2bα2dα,as claimed in (43). For different values of (43) is shown in Figures 1 and 2.Figure 1
Real and imaginary parts of solution of (43) for a=b=1,k=1/2,t=1,t=2,t=4,t=5,t=7, and gx=0.
(a)(b)Figure 2
Real and imaginary parts of solution of (43) for a=1/2,b=2,k=3/2,t=1/2,t=3/2,t=5/2,t=5,t=8, and gx=signumx.
(a)(b)For an illustration, we provide the following simple example.Example 5.1 Let us consider the following initial wave equation:(59)∂2ϕx,t∂t2=0.25D¯E2ϕx,t,−∞<x<∞,t>0ϕx,0=e−1/2x2,∂ϕx,t∂tt=0=e−1/2x2.According to (43), we get the solution in the form:
(60)ϕx,t=sgnb2e−ia/2bx2e−1/2x+0.5t2ei/2x+0.5t2+e−1/2x−0.52ei/2x−0.5t2+2∫x−0.5tx+0.5te−1/2ξ2eia/2bξ2dξ.On the other hand, fora=b=1, we get
(61)∫x−0.5tx+0.5te−1/2ξ2eia/2bξ2dξ=∫x−0.5tx+0.5te−1/21−iξ2dξ=∫x−0.5tx+0.5te−1/21−iξ2dξ.Settingu=1/21−iξ, then
(62)∫x−0.5tx+0.5te−1/2ξ2eia/2bξ2dξ=11/21−i∫1/21−ix−0.5t1/21−ix+0.5te−u2du=22−2i∫1/42−2i2x−t1/42−2i2x+te−u2du=π2−2i2π∫1/42−2i2x−t0e−u2du+2π∫01/42−2i2x+te−u2du=π2−2i2π∫01/42−2it−2xe−u2du+2π∫01/42−2i2x+te−u2du=π2−2ierf142−2i−2x+t+erf142−2i2x+t.It means that equation (60) takes the form
(63)ϕx,t=e−i/2x22e−1/2x+0.5t2ei/2x+0.5t2+e−1/2x−0.5t2ei/2x−0.5t2+2π2−2ierf142−2i−2x+t+erf142−2i2x+t,where the error function is defined by
(64)erfx=2π∫0xe−t2dt,forallx.In particular, ifgx=0, the above expression becomes
(65)ϕx,t=eit2/82e−1/2x+0.5t2ei0.5xt+e−1/2x−0.5t2e−i0.5xt.


## 6. Solution of Generalized Heat Equation Using Sampling Formulae

Let us consider the following generalized one-dimensional heat equation associated with the LCT:(66)∂φx,t∂t=k2D¯E2φx,t,−∞<x<∞,t>0,with the initial condition:
(67)φx,0=fx,where k≠0 is a constant and D¯E=−d/dx−ia/bx.Theorem 15.
Assume thatfx in (67) is band limited to ΩE in the LCT domain. Then, the solution of (66) is given by
(68)φx,t=πΩEe−ia/2bx2∑n=−∞n=∞fnTeia/2bnT2∫−ΩEΩEe−k2/b2ξ2tcosξnTb−xξbdξ.Proof.
Applying the LCT to both sides of (66) yields
(69)∂φEξ,t∂t=−k2b2ξ2φEξ,t.This gives(70)φEξ,t=FEξe−k2/b2ξ2t,where FEξ is defined by (49). Taking the inverse transform of the LCT in (70), it is easy to obtain
(71)φx,t=LE−1FEξe−k2/b2ξ2t=1−2πbi∫ℝFEξe−k2/b2ξ2tei/2a/bx2−2/bxξ+d/bξ2dξ.Becausefx is band limited to ΩE, then LEfξ=0 for ∣ξ∣>ΩE. Applying this fact and then substituting (38) in (71), we obtain
(72)φx,t=1−2πbie−ia/2bx2∫ℝe−id/2bξ2FEξe−k2/b2ξ2te−ixξ/bdξ=1−2πbie−ia/2bx2∫−ΩEΩEe−id/2bξ2eid/2bξ2∑n=−∞n=∞fnTeia/2bnT2πbπΩE2ie−inTξ/be−k2/b2ξ2te−ixξ/bdξ=1−2πbie−ia/2bx2∫−ΩEΩE∑n=−∞n=∞fnTeia/2bnT2πbπΩE2ie−inTξ/be−k2/b2ξ2te−ixξ/bdξ=1−2πbie−ia/2bx2∑n=−∞n=∞fnTeia/2bnT2πbπΩE2i∫−ΩEΩEe−k2/b2ξ2te−iξnT/b−xξ/bdξ=π2ΩEe−ia/2bx2∑n=−∞n=∞fnTeia/2bnT2∫−ΩEΩEe−k2/b2ξ2te−iξnT/b−xξ/bdξ=πΩEe−ia/2bx2∑n=−∞n=∞fnTeia/2bnT2∫−ΩEΩEe−k2/b2ξ2tcosξnTb−xξbdξ.The proof is complete.


## 7. Solution of Generalized Heat Equation Using LCT

The purpose of this part is to solve (66) using the LCT. We prove the following result.Theorem 16.
Assume thatfx in (67) belong L1ℝ. Then, the solution of (66) is given by
(73)φx,t=1−4πk2t∫ℝeiuu−xa/bfue1/2−b2c−2ak2it/2k2btx−u2du.Proof.
By using the inverse transform of the LCT (14), we have
(74)LE−1e−k2/b2ξ2t=1−2πbi∫ℝe−k2/b2ξ2te−i/2a/bx2−2/bxξ+d/bξ2dξ=e−ia/2bx2−2πbi∫ℝe−k2/b2t−di/2bξ2e−iξx/bdξ=e−ia/2bx2−2πbi∫ℝe2k2t−idb/2b2ξ2−2ibξx/2k2t−idbdξ=e−ia/2bx2−2πbi∫ℝe2k2t−idb/2b2ξ2−2ibξx/2k2t−idb+ibx/2k2t−idb2−ibx/2k2t−idb2dξ=e−ia/2bx2−2πbie2k2t−idb/2b2−ibx/2k2t−dbi2∫ℝe2k2t−dbi/2b2ξ2−2ibξx/2k2t−dbi+ibx/2k2t−idb2dξ=e−ia/2bx2−2πbiex2/4k2t−2dbi∫ℝe−dbi−2k2t/2b2ξ−ibx/2k2t−idb2dξ.Based on Gaussian integral and basic property of matrix parameter of the LCT, we obtain(75)LE−1e−k2/b2ξ2t=1−2πbie−a4k2t−2dbi/2b4k2t−2dbix2ie2bx2/2b4k2t−2dbi2b2πdbi−2k2t=b2k2ti+db2e−4k2tai−2adb+2b/2b4k2t−2dbix2=b2k2ti+db2e−4k2tai−2b2c/2b4k2t−2dbix2=b2k2ti+db2e1/2−2k2tai−b2c/2k2bt−db2ix2.Puthx,t=e1/2−b2c−2ak2it/2k2bt−ib2dx2. Then, we get
(76)LEb2k2ti+db2hx,t=e−k2/b2ξ2t.From (70), we obtain
(77)φx,t=LE−1FEξe−k2/b2ξ2t.By inserting (76) in (77), it is easy to see that
(78)φx,t=LE−1LEfxLEb2k2ti+db2hx,t=b2k2ti+db2LE−1LEfxLEhx,t=b2k2ti+db2LE−112πbieidξ2/2bLEf∗h.For simplicity, let us assumeE~=abc0. Then, the above identity takes the form:
(79)φx,t=b/2k2ti2πbiLA~−1LA~f∗h.Applying the convolution theorem for the LCT defined by (25), we obtain
(80)φx,t=1−4πk2tLA~−1LA~f∗h=1−4πk2t∫ℝeiuu−xa/bfuhx−u,tdu=1−4πk2t∫ℝeiuu−xa/bfue1/2−b2c−2ak2it/2k2btx−u2du,as claimed in (73).This solution of (73) with initial conditions fx=e−1/2x2 and the Heaviside step function
(81)fx=ux=0x<0,1x≥0,is described in Figures 3 and 4, respectively.Remark 17.
Notice first that, whenE=01−10, (73) will lead to
(82)φx,t=1ik4πt∫ℝfuex−u2/4tk2du.Figure 3
Real and imaginary parts of solution of (73) for a=b=c=1,k=1/2,t=1/2,t=2,t=3,t=5,t=10, and fx=e−1/2x2.
(a)(b)Figure 4
Real and imaginary parts of solution of (73) for a=1/3,b=1/2,c=5/2,k=1,t=1/2,t=2,t=3,t=5,t=10, and fx=ux.
(a)(b)The above identity is solution of the classical heat equation.Let us consider(83)Gx−u,t=1−4πk2teiuu−xa/be1/2−b2c−2ak2it/2k2btx−u2.Then, the solution of heat equation associated with the LCT (73) reduces to the form:
(84)φx,t=∫ℝfuGx−u,tdu.Whenfx=δx, (84) becomes
(85)φx,t=1−4πk2te1/2−b2c−2ak2it/2k2btx2.


## 8. Conclusion

We have established the solution of generalized wave equation using the lCT and obtained the solution of heat equation using the sampling formulae of band limited function related to the LCT. The solutions have been obtained by applying the properties of the LCT and the relationship between the FT and the LCT. It is found that the solutions are general form of the solutions of the classical wave and heat equations.



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