# On the Fractional Derivative of Dirac Delta Function and Its Application

Authors: Zaiyong Feng; Linghua Ye; Yi Zhang
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

The Dirac delta function and its integer-order derivative are widely used to solve integer-order differential/integral equation and integer-order system in related fields. On the other hand, the fractional-order system gets more and more attention. This paper investigates the fractional derivative of the Dirac delta function and its Laplace transform to explore the solution for fractional-order system. The paper presents the Riemann-Liouville and the Caputo fractional derivative of the Dirac delta function, and their analytic expression. The Laplace transform of the fractional derivative of the Dirac delta function is given later. The proposed fractional derivative of the Dirac delta function and its Laplace transform are effectively used to solve fractional-order integral equation and fractional-order system, the correctness of each solution is also verified.


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## Body

## 1. Introduction

The Dirac delta functionδt was introduced by the theoretical physicist Paul Dirac to develop tools for the quantum field theory [1]. Till now, it is widely used to describe the impulse phenomenon in physics, mathematics, and engineering fields [2–5]. In control theory and signal process, the Dirac delta function is a basic typical input or testing signal to study the system’s response [6, 7]. The integer-order derivative of the Dirac delta function δt, i.e., ith order derivative δiti∈ℕis also well-studied and finds great significance in applications of related fields [8, 9]. The Laplace transform of the Dirac delta function δt and its integer-order derivative δit are Lδt=s and Lδit=si,i∈ℕ, respectively [10]. They are effectively used to solve various kinds of integer-order differential/integral equation and integer-order system.On the other hand, the fractional-order system which is characterized with “process memory” and “historical heredity” gets more and more attention [11–15]. Many different definitions of integrals and fractional derivatives, such as the ψ-Hilfer integral and the ψ-Hilfer fractional derivative, are proposed and find great significance in applications [16–18]. Just as solving integer-order integral/differential equation and system, it will be significant to investigate the fractional derivative of The Dirac delta function δt, i.e., D0tαδt and its Laplace transformLD0tαδt for exploring the solution for fractional-order integral/differential equation or fractional-order system.Research on the fractional derivative of the Dirac delta functionD0tαδt is just beginning so far.From the point of view of viscoelasticity, the literature [19] showed that the memory function of complex materials is the fractional derivative of the Dirac delta function, and the Laplace transform of the fractional derivative of the Dirac delta function is derived. However, the literature [19] only discussed the Riemann-Liouville fractional derivative, and the fractional derivative order discussed in [19] is limited in ℝ+.In this paper, we explore the Riemann-Liouville fractional derivative and the Caputo fractional derivative of the Dirac delta function under the widely used sense, and the order of the fractional derivative is generalized toℝ. Comparing with [19], our research method is simpler while results are more general.This paper is organized as follows: Section2 Basic Preparation and Section 3 investigates the fractional derivative of Dirac delta function, including the Riemann-Liouville fractional derivative ofδt, i.e.,D0Rtαδt and the Caputo fractional derivative of δt, i.e., D0Ctαδt. Section 4 studies the Laplace transform of the fractional derivative of Dirac delta function; Section 5 presents application with examples to verify the correctness of the theorems proposed in the paper. Section 6 briefly summarizes the main conclusions of the paper.


## 2. Basic Preparation

The fractional derivative of functionf, i.e., Dtαaf is the basic tool used in fractional calculus and related applications. The basis for its definition is an arbitrary-order integral Itpaf, namely definition 1 given below [11].Definition 1.
Itpaft=1/Γp∫att−τp−1fτdτp>0.With Definition1 as the basis, the full definition of the function f’s Riemann-Liouville fractional derivative DaRtαfand its Caputo fractional derivative DaCtαf are as follows:Definition 2.
The Riemann-Liouville fractional derivative of the continuous functionf,DaRtαf is as follows:
(1)DaRtαft=It−αaft=1Γ−α∫att−τ−α−1fτdτ,α<0.dmdtmItm−αaft=1Γm−αdmdtm∫0tt−τm−α−1fτdτ,0≤m−1<α<m,m∈ℕ+.SupposingLft=Fs, the Laplace transform of D0Rtαf is as follows:Property 1.
LD0Rtαft=sαFs−∑k=0m−1skD0Rtα−k−1ftt=0,0≤m−1<α<m.Definition 3.
If the functionf has continuous derivatives up to the m-order, then its Caputo fractional derivative D0Ctαf is as follows:
(2)DaCtαft=It−αaft=1Γ−α∫att−τ−α−1fτdτ,α<0.Itm−αafmt=1Γm−α∫0tt−τm−α−1fmτdτ,0≤m−1<α<m,m∈ℕ+.The Laplace transform ofD0Ctαf only involves the initial value of the integer-order derivative of the function ft, which brings great convenience to its application in engineering.Property 2.
LD0Ctαft=sαFs−∑k=0m−1sα−k−1fk0,0≤m−1<α<m.The Dirac delta functionδt and its ith-order derivative δiti∈ℕbelong to a generalized function. δtis generally defined as follows [1]:Definition 4.
δt=0,t≠0∞,t=0, ∫−∞+∞δtdt=1.Main useful properties ofδt include [8, 10]:Property 3.
Supposef is continuous within the interval a,b containing 0, then we have:∫abftδtdt=f0, particularly, ∫abδtdt=1.Property 4.
Theith derivative of δt, δit=0,t≠0,i∈ℕ. If the function f is continuous and ith differentiable at t=0, within the interval a,b, then ∫abftδitdt=−1ifi0.Property 5.
Laplace transform ofδit, Lδit=si,i∈ℕ.


## 3. The Fractional Derivative of Dirac Delta Function

Corresponding to the above two definitions of fractional derivative, the Riemann-Liouville fractional derivative ofδt, i.e., D0Rtαδt and the Caputo fractional derivative of δt, i.e., D0Ctαδt could be defined as in Definition 2 and Definition 3. In fact, we can give the analytic expression for them further. Theorem 1 gives the analytic formula of D0Rtαδt.Theorem 1.
D0Rtαδt=1/Γ−αtα+1,α∉ℕ.Proof.
According to the value ofα∉ℕ:
(1)
Whenα<0, Definition 2, Definition 4, and Property 3 result in:
(3)D0Rtαδt=It−α0δt=1Γ−α∫0tt−τ−α−1δτdτ=1Γ−α∫0t−εt−τ−α−1δτdτε>0=1Γ−αt−τ−α−1τ=0=1Γ−αtα+1.(2)
When0≤m−1<α<m,m∈ℕ+, Definition 1, Definition 2, and Property 3 result in:
(4)D0Rtαδt=dmdtmItm−αaδt=1Γm−αdmdtm∫0tt−τm−α−1δτdτ=1Γm−αdmtm−α−1dtm=m−α−1Γm−αdm−1tm−α−2dtm−1=1Γm−α−1dm−1tm−α−2dtm−1=⋯⏞m−1times=1Γ−αtα+1.
Theorem2 shows us that D0Ctαδt=D0Rtαδt.Theorem 2.
D0Ctαδt=D0Rtαδt=1/Γ−αtα+1,α∉ℕ.Proof.
According to the value ofα∉ℕ:
(1)
Whenα<0, Definition 3, Definition 4, and Property 3 result in:
(5)D0Ctαδt=It−α0δt=1Γ−α∫0tt−τ−α−1δτdτ=1Γ−αt−τ−α−1τ=0=1Γ−αtα+1.(2)
When0≤m−1<α<m,m∈ℕ+, Definition 1, Definition 3, and Property 4 result in:
(6)D0Ctαδt=Itm−α0δmt=1Γm−α∫0tt−τm−α−1δmτdτ=−1mΓm−αdmt−τm−α−1dτmτ=0=−1m×−1Γm−αm−α−1dm−1t−τm−α−2dτm−1τ=0=−1m−1Γm−α−1dm−1t−τm−α−2dτm−1τ=0=⋯⏞m−1times=1Γ−αtα+1.Remark 1.
Theorem2 shows that the analytic formulas of D0Rtαδt is the same as that of D0Ctαδt. DenotingD0Ctαδt=D0Rtαδt=1/Γ−αtα+1,α∉ℕ as D0tαδt; thus, D0tαδt=1/Γ−αtα+1,α∉ℕ. We will use D0tαδt to replace D0Rtαδtor D0Ctαδt below. The different forms of Definition 2 and Definition 3 allow us to adopt suitable one as the definition of D0tαδt when dealing with specific problem related to D0tαδt.


## 4. The Laplace Transform of the Fractional Derivative of Dirac Delta Function

The Laplace transform is of great importance in solving differential/integral equation and system analysis. Theorem3 gives the Laplace transform of D0tαδt.Theorem 3.
LD0tαδt=sα,α∈ℝ.Proof.
We will prove it by classifying the different values ofα.
(1)
Whenα<0,LD0tαδt=LI0t−αδt=L1/Γ−α∫0tt−τ−α−1δτdτ=1/Γ−αLt−α−1∗δt=1/Γ−αLt−α−1∙Lδt.Because−α−1>−1, thus, Lt−α−1=Γ−α/s−α [20]. Combining Property 5, we further obtain:
(7)LD0tαδt=1Γ−α∙Γ−αs−α=sα.(2)
Whenα=m∈ℕ, it is known that Lδmt=sm,m∈ℕ from Property 5.(3)
When0≤m−1<α<m,m∈ℕ+, we adopt Definition 3 as the definition of D0tαδt; thus,
(8)LD0tαδt=LD0Ctαδt=L1Γm−α∫0tt−τm−α−1δmτdτ=1Γm−αLtm−α−1∗δmt=1Γm−αLtm−α−1∙Lδmt.
Because0≤m−1<α<m,m∈ℕ+, −1<m−α−1<0; thus, we can obtain:
(9)LD0tαδt=1Γm−αLtm−α−1∙Lδmt=1Γm−α∙Γm−αsm−α∙sm=sα.Remark 2.
Theorem3 proves LD0tαδt=sα,α∈ℝ, while Property 5 givesLδit=si, in which i∈ℕ. Theorem 3 extends the Laplace transform of the function δiti∈ℕ, to the Laplace transform of D0tαδtα∈ℝ. Obviously, the Laplace transforms of δiti∈ℕ and that of D0tαδtα∈ℝ are consistent in the form, the Laplace transform of D0tαδtα∈ℝis the generalization of the Laplace transform of the integer-order derivative of δt.Theorem4 could be derived from implementing the inverse Laplace transform to Theorem 3.Theorem 4.
L−1sα=D0tαδt,α∈ℝ.Remark 3.
The conclusion “Whenα<0,L−1sα=L−11/s−α=t−α−1/Γ−α” is well known and often used [20]. By introducing the nonclassical function δt and its fractional derivative D0tαδt, the inverse Laplace transform of sα,α∈ℝ is obtained.Remark 4.
It is well-known that the role of the Laplace transform in differential/integral equation and system analysis is significant. In classical cases, objects involved in the Laplace transform are often the rational proper fraction ofs [21]. For example, to the transfer function essentially discussed in control theory, the case that the power of the polynomial of s in the numerator is higher than the power of the polynomial of s in the denominator is nearly ignored. Therefore, Theorem 4 allows us to implement Laplace transform to general rational fraction. This provides a new way for the solution of differential equations and the analysis of control systems, especially of the singular systems.


## 5. Application with Examples

Example 1 (Solving fractional Volterra integral equation of the first kind [22]).
Investigate the following integral equation:(10)1Γp∫0tt−τp−1yτdτ=ktm,p>0,m>−1.Obviously, the left hand of (10) is the p-order integral of the function yτ, implementing the Laplace transform on the both side of the equation results in: Ys/sp=kΓm+1/sm+1. Consequently,
(11)Ys=kΓm+1sp−m−1.The solution of (10) could be derived by implementing the inverse Laplace transform on (11) with Theorem 4:
(12)yt=kΓm+1Dtp−m−10δt.We verify the correctness of (12) according to p−m−1>0 and p−m−1<0 respectively.
(a)
In case ofp−m−1>0, Dtp−m−10δt represents p−m−1 order derivative. Takingp=m+2 as an example, in this case:
(13)yt=kΓm+1Dt10δt=kΓm+1δ′t.Substituting it into the left hand of (10) yields:
(14)1Γp∫0tt−τp−1yτdτ=kΓm+1Γm+2∫0tt−τm+1δ′τdτ=km+1m+1×−1×−1×tm=ktm.(b)
In case ofp−m−1<0, Dtp−m−10δt represents p−m−1 order integration. Taking p=1,m=5/2asanexample,inthiscase:(15)yt=kΓ52+1Dt−2.50δt.Figure1 plots the curve of yt with k=2.Figure 1
Curve ofyt in Example 1.Verification: substituting (15) into the left hand of (10) yields:
(16)1Γp∫0tt−τp−1yτdτ=kΓ52+11Γ1∫0tDt−2.50δτdτ=kΓ52+1∫0t1Γ2.5τ1.5dτ=kΓ2.5+1Γ2.5∫0tτ1.5dτ=kΓ2.5+12.5×Γ2.5t2.5=kt5/2.Example 2 (Computation of convolution of power functions).
Compute the convolution of power functions:(17)ft=∫0tt−τατβdτα,β>−1.Implementing the Laplace transform on the both side of (17), we get
(18)Fs=Γα+1sα+1∙Γβ+1sβ+1=Γα+1Γβ+1sα+β+2=Γα+1Γβ+1s−α−β−2.Implementing the inverse Laplace transform on the both side of (18) with Theorem 4, we have
(19)ft=Γα+1Γβ+1Dt−α−β−20δt.Verification: for instance, takingα=1, β=0.5 results inft=∫0tt−ττdτ. Computation of ft(20)ft=∫0tt−ττdτ=∫0ttτ1/2−τ3/2dτ=t23t3/2−25t5/2=415t5/2.Investigating (19), we have the analytic form of ft:
(21)ft=Γ2Γ1.5Dt−1−0.5−20δt=Γ1.5Γ3.51t−2.5=Γ1.52.5×1.5×Γ1.5t2.5=415t5/2.Obviously, the correctness of (19) is confirmed.Note that this kind of convolution is not easy to compute for the vast major value ofα and β by traditional way, such as integration by substitution.Example 3 (Solution for fractional singular system[23]).
Find the solution for the next fractional singular system:(22)0100Dt0.5xt0C=xt+−11ut,x0=x10x20.Implementing the Laplace transform on the both side of (22) yields:
(23)0100s0.5Xs−s−0.5x0=Xs+−11Us.After transposition and sorting out:(24)0100s0.5−IXs=0100s−0.5x0+−11Us.Thus,(25)Xs=0100s0.5−I−10100s−0.5x0+−11Us.0100s0.5−I−1 could be expanded by the Neumann series [24], namely
(26)0100s0.5−I−1=−∑k=0∞0100ks0.5×k=−1001−0100s0.5.Substituting (26) into (25) yields
(27)Xs=−1001−0100s0.50100s−0.5x0+−11Us=−0100s−0.5x0+−11Us−10s0.5Us.Implementing the inverse Laplace transform on the both side of (27) with Theorem 4, we obtain
(28)xt,u,x0=x1tx2t=−Dt−0.50Cδtx20+ut−Dt0.50Cut−Dt−0.50Cδtu0−ut.Verification: system (22) is equal to
(29)Dt0.5x2t=x1t−ut0C29−10=x2t+ut29−2.The correctness ofx2t=−ut is obvious from (29-2). Takingx2t=−ut,x20=−u0 into account, substituting x1t in (28) into the right side of (29-1) brings us
(30)−Dt0.50Cut=−Dt−0.50Cδtx20+ut−Dt0.50Cut−Dt−0.50Cδtu0−ut=−Dt0.50Cut−Dt−0.50Cδtx20+u0.Sincex20=−u0, thus, x20+u0=0. The right side of (16-1) is changed into −Dt0.50Cut, which is exactly equal to the left side of (16-1). Figure 2 plots the curves of xt with x20=1.Figure 2
Curves ofxt in Example 3.


## 6. Conclusion

This paper investigated the fractional derivative of the Dirac delta functionδt, its Laplace transform and their application. It is proved that D0Ctαδt is equal to D0Rtαδt, and we further have D0Ctαδt=D0Rtαδt=1/Γ−αtα+1,α∉ℕ. Consequently, we denote them as D0tαδt. The paper generalized the Laplace transform of Lδit=si,i∈ℕ to LD0tαδt=sα,α∈ℝ. The proposed fractional derivative of the Dirac delta function δt, i.e., D0tαδtand its inverse Laplace transform thatL−1sα=D0tαδt,α∈ℝare effectively used in solving the fractional volterra integral equation, computation of convolution of power functions, and the solution for fractional singular system.



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