On a New Integral Transform and Differential Equations

Abstract

Integral transform method is widely used to solve the several differential equations with the initial values or boundary conditions which are represented by integral equations. With this purpose, the Sumudu transform was introduced as a new integral transform by Watugala to solve some ordinary differential equations in control engineering. Later, it was proved that Sumudu transform has very special and useful properties. In this paper we study this interesting integral transform and its efficiency in solving the linear ordinary differential equations with constant and nonconstant coefficients as well as system of differential equations.

1. Introduction

In the literature there are numerous integral transforms and widely used in physics, astronomy as well as in engineering. In order to solve the differential equations, the integral transform were extensively used and thus there are several works on the theory and application of integral transform such as the Laplace, Fourier, Mellin, and Hankel, to name but a few. In the sequence of these transform, in early 90's Watugala [1] introduced a new integral transform, named the Sumudu transform and further applied it to the solution of ordinary differential equation in control engineering problems. For further detail and properties about Sumudu transform (see [2–7]) and many others. The Sumudu transform is defined over the set of the functions𝐴={𝑓(𝑡)∶∃𝑀,𝜏₁,𝜏₂>0,|𝑓(𝑡)|<𝑀𝑒^{𝑡/𝜏𝑗},if𝑡∈(−1)^𝑗×[0,∞)}, (1.1) by the following formula𝐺(𝑢)=𝑆[𝑓(𝑡);𝑢]=∶^∞₀𝑓(𝑢𝑡)𝑒^−𝑡𝑑𝑡,𝑢∈(−𝜏₁,𝜏₂). (1.2) The existence and the uniqueness was discussed in [5], for further details and properties of the Sumudu transform and its derivatives we refer to [2] and applied this new transform to solve the ordinary differential equations and control engineering problems, see [1, 8, 9].

Further, in [3], some fundamental properties of the Sumudu transform were established. In [10], this new transform was applied to the one-dimensional neutron transport equation. In fact one can easily show that there is a strong relationship between double Sumudu and double Laplace transforms see, [5]. In [6], the Sumudu transform was extended to the distributions and some of their properties were studied.

In [11, 12], some further fundamental properties of the Sumudu transform were established. It turns out that the Sumudu transform has very special and useful properties and further it can help with intricate applications in sciences and engineering, see, for example, [10, 13]. Once more, Watugala's work was followed by Weerakoon in [14, 15] by introducing a complex inversion formula for the Sumudu transform. In fact in [5], the existence and the uniqueness of Sumudu transform was discussed and some of the fundamental relationship established. In [16, 17], the further properties of Sumudu transform were studied and applied to the regular system of differential equations with convolution terms, respectively.

In this study, our purpose is to show the applicability of this interesting new transform and its efficiency in solving the linear ordinary differential equations with constant and non constant coefficients.

The existence and the uniqueness of Sumudu transform was discussed in [5] we reproduce as the next theorem that we will refer to it in the development of the present paper.

Theorem 1.1. If 𝑓 is of exponential order, then its Sumudu transform 𝑆[𝑓(𝑡)]=𝐹(𝑢) is given by 𝐹(𝑢)=^∞₀𝑒^−𝑡/𝑢𝑓(𝑡)𝑑𝑡, (1.3) where 1/𝑢=1/𝜂+𝑖/𝜏. Then the integral exists in the right half plane 𝜂>𝐾 and 𝜁>𝐿. Further, if 𝑓(𝑡) and 𝑔(𝑡) are having Sumudu transforms and given by 𝐹(𝑢) and 𝐺(𝑢), respectively, and if 𝐹(𝑢)=𝐺(𝑢) then 𝑓(𝑡)=𝑔(𝑡) where 𝑢 is a complex number.

Theorem 1.2. Let 𝑓 be Sumudu transformable and satisfy 𝑓(𝑡)=0 for 𝑡<0. Then lim_𝑢→∞(𝑆[𝑓](𝑢))=0.

Proof. Let 𝑓₀(𝑡)=𝑓(𝑡)[1−𝐻(𝑡−1)],𝑓₁(𝑡)=𝑓(𝑡)𝐻(𝑡−1). Since 𝑓₀ vanishes outside [0,1], then we have 𝑆[𝑓](𝑢)=∫¹₀𝑒^−𝑡/𝑢𝑓(𝑡)𝑑𝑡 for any 1/𝑢. Moreover, 𝑓=𝑓₀+𝑓₁,dom(𝑆(𝑓₁))=dom(𝑆(𝑓)) and 𝑆[𝑓](𝑢)=∫¹₀𝑒^−𝑡/𝑢𝑓(𝑡)𝑑𝑡+𝑆(𝑓₁)(𝑢) for all 1/𝑢∈dom(𝑆(𝑓)). Let 1/𝑢₀∈dom(𝑆(𝑓)) and apply |𝑆(𝑓)(𝑢)|≤𝐴𝑒^−𝑐/𝑢 to 𝑓₁ we conclude that there is a constant 𝐴 such that |𝑆(𝑓)(𝑢)|≤¹₀𝑒^−𝑡/𝑢|𝑓(𝑡)|𝑑𝑡+𝐴𝑒^−1/𝑢∀1𝑢≥1𝑢₀ (1.4) as 1/𝑢→∞ the second term on the right clearly tends to zero. The same applies to the first term.

Next we prove the following theorem that is very useful in the rest of this study.

Theorem 1.3. Let 𝜆>−1. Then(i)If 𝑓∈loc₊ and lim_𝑡→∞[𝑓(𝑡)/𝑡^𝜆] exists, so does lim_1/𝑢→0+[𝑆[𝑓(𝑡)](𝑢)/𝑢^𝜆+1] and one has lim_𝑡→∞[𝑓(𝑡)𝑡^𝜆1]=Γ(𝜆+1)lim_1/𝑢→0+[𝑆[𝑓(𝑡)](𝑢)𝑢^𝜆+1]. (1.5)(ii)if 𝑓 is Sumudu transformable and satisfies 𝑓(𝑡)=0 for 𝑡<0 and if lim_𝑡→0+[𝑓(𝑡)/𝑡^𝜆] also lim_1/𝑢→∞[𝑆[𝑓(𝑡)](𝑢)/𝑢^𝜆+1] and one has lim_𝑡→0+[𝑓(𝑡)𝑡^𝜆1]=Γ(𝜆+1)lim_1/𝑢→∞[𝑆[𝑓(𝑡)](𝑢)𝑢^𝜆+1]. (1.6)

Proof. (i) Let 𝑓(𝑡)/𝑡→𝛼 as 𝑡→∞. This implies that there are constants 𝐴 and 𝜌>0 such that |𝑓(𝑡)|/𝑡^𝜆≤𝐴 for 𝑡>𝜌. This further implies that 𝑒^−𝑡/𝑢𝑓(𝑡) is integrable for all 1/𝑢>0 so that we may write, if 𝑓(𝑡)=0 for 𝑡<𝑐,𝑆[𝑓(𝑡)](𝑢)=^∞_𝑐𝑒^−𝑡/𝑢𝑓(𝑡)𝑑𝑡=^𝜌_𝑐𝑒^−𝑡/𝑢𝑓(𝑡)𝑑𝑡+^∞_𝜌𝑒^−𝑡/𝑢𝑓(𝑡)𝑑𝑡, (1.7) now it is easy to see that 1/𝑢^𝜆+1 time the first term on the right of (1.7) tends to zero as 1/𝑢→0⁺ and 1/𝑢^𝜆+1 times the second term on the right of (1.7) may be written for 1/𝑢>0 as follows: 1𝑢^𝜆∞_𝜌/𝑢𝑒^−𝑥𝑓(𝑢𝑥)𝑑𝑥=^∞_𝜌/𝑢𝑥^𝜆𝑒^−𝑥𝑓(𝑢𝑥)(𝑢𝑥)^𝜆𝑑𝑥 (1.8) as 1/𝑢→0⁺ and 𝑓(𝑢𝑥)/(𝑢𝑥)^𝜆 tends to 𝛼 and, since it is bounded in the range of integration by the constant 𝐴. Then by the dominated convergence theorem and we conclude that 1𝑢^𝜆+1𝑆[𝑓(𝑡)](𝑢)→^∞_𝜌/𝑢𝑥^𝜆𝑒^−𝑥1𝛼𝑑𝑥=𝛼Γ(𝜆+1)as𝑢→0⁺, (1.9) which completes the proof.
(ii) Let 𝑓(𝑡)/𝑡^𝜆→𝛽 as 𝑡→0⁺. Since this function is a bounded in the neighborhood of zero then there are constants 𝐵 and 𝜎>0 such that |𝑓(𝑡)|/𝑡^𝜆≤𝐵 for 0<𝑡<𝜎. Using a method similar to that in the proof of Theorem 1.3 we let 𝑓₀=𝑓(𝑡)[1−𝐻(𝑡−𝜎)] and 𝑓₁(𝑡)=𝑓(𝑡)𝐻(𝑡−𝜎). Then𝑆[𝑓(𝑡)](𝑢)=^𝜎₀𝑒^−𝑡/𝑢𝑓(𝑡)𝑑𝑡+𝑆[𝑓₁1(𝑡)](𝑢)for𝑢∈dom𝑆[𝑓], (1.10) and apply |𝑆(𝑓)(𝑢)|≤𝐴𝑒^−𝑐/𝑢 we have |𝑆(𝑓₁)(𝑢)|≤𝐾𝑒^−𝜎/𝑢 for some constant 𝐾 and 1/𝑢 sufficiently large, and therefore (1/𝑢^𝜆+1)𝑆[𝑓₁(𝑡)](𝑢)→0 as 1/𝑢→∞. Also, by a similar argument to that used in (i) then 1/𝑢^𝜆+1 times the first term on the right of (1.10) tends to 𝛽Γ(𝜆+1) as 1/𝑢→∞, which completes the proof.

In the next proposition we prove the existence of Sumudu transform for the derivatives. In fact similar results were also reported, for example, in [10, 12, 13] by using different methods.

Proposition 1.4 (Sumudu transform of derivative). (i) Let 𝑓 be differentiable on (0,∞) and let 𝑓(𝑡)=0 for 𝑡<0. Suppose that 𝑓^′∈𝐿_loc. Then 𝑓^′∈𝐿_loc,dom(𝑆𝑓)⊂dom(𝑓^′) and𝑆(𝑓^′1)=𝑢1𝑆(𝑓)−𝑢𝑓(0+)for𝑢∈dom(𝑆(𝑓)). (1.11)
 (ii) More generally, if 𝑓 is differentiable on (𝑐,∞), 𝑓(𝑡)=0 for 𝑡<0 and 𝑓^′∈𝐿_loc then𝑆(𝑓^′1)=𝑢1𝑆(𝑓)−𝑢𝑒^−𝑐/𝑢𝑓(𝑐+)for𝑢∈dom(𝑆(𝑓)). (1.12)

Proof. We start by (1.2) as follows, the local integrability of implies that 𝑓(𝑐+) exists, because, if 𝑥>𝑐,𝑓(𝑥)=𝑓(𝑐+1)−_𝑥^𝑐+1𝑓^′(𝑡)𝑑𝑡→𝑓(𝑐+1)−_𝑐^𝑐+1𝑓^′(𝑡)𝑑𝑡as𝑥→𝑐⁺. (1.13) Let 𝑢∈dom(𝑆(𝑓)). If 𝜔∈𝐷₀={𝑤∶𝑤differentiableand𝑤(0)=0}, integrating by part, we have 1𝑢𝑡𝜔(𝜆)𝑒^−𝑡/𝑢𝑓^′1(𝑡)𝑑𝑡=𝑢^∞_𝑐𝑡𝜔(𝜆)𝑒^−𝑡/𝑢𝑓^′(𝑡)𝑑𝑡=lim_𝑥→𝑐+1[−𝑢𝑥𝜔(𝜆)𝑒^−𝑥/𝑢−1𝑓(𝑥)]𝑢^∞_𝑐𝑒^−𝑡/𝑢[1𝜆𝜔^′(𝑡𝜆1)−𝑢𝑡𝜔(𝜆)]𝑓(𝑡)𝑑𝑡. (1.14) Then we have −1𝑢𝑐𝜔(𝜆)𝑒^−𝑐/𝑢1𝑓(𝑐+)→−𝑢𝜔(0)𝑒^−𝑐/𝑢−1𝑓(𝑐+)as𝜆→∞,𝑒𝑢𝜆^−𝑡/𝑢𝜔^′(𝑡𝜆1)𝑓(𝑡)𝑑𝑡+𝑢²𝑒^−𝑡/𝑢𝑡𝜔(𝜆1)𝑓(𝑡)𝑑𝑡,→0+𝑢𝜔(0)𝑆(𝑓) (1.15) as 𝜆→∞ thus for any 𝜔∈𝐷₀,lim_𝜆→∞[1𝑢𝑡𝜔(𝜆)𝑒^−𝑡/𝑢𝑓^′(𝑡)𝑑𝑡]=𝜔(0)𝑢[𝑆(𝑓)−𝑓(𝑐+)]. (1.16) This implies that 𝑒^−𝑡/𝑢𝑓^′(𝑡) is convergent, that is, 𝑢∈dom(𝑆(𝑓)), and that 𝑆(𝑓^′1)=𝑢1𝑆(𝑓)−𝑢𝑒^−𝑐/𝑢𝑓(𝑐+). (1.17)

In general case, if 𝑓 is differentiable on (𝑎,𝑏) with 𝑎<𝑏, and 𝑓(𝑡)=0 for 𝑡<𝑎 or 𝑡>𝑏 and 𝑓^′∈𝐿_loc then, for all 𝑢𝑆(𝑓^′1)=𝑢1𝑆(𝑓)−𝑢𝑒^−𝑎/𝑢1𝑓(𝑎+)+𝑢𝑒^−𝑏/𝑢𝑓(𝑏). (1.18) In the next example we show that Sumudu transform of a function can be obtained by using the differential equations.

Example 1.5. Let 𝑦(𝑡)=sinh(√𝑡). Then 𝑑𝑑𝑡[𝑡𝑦^′1(𝑡)]=4sinh(√1𝑡)+4√𝑡cosh(√𝑡)=𝑦(𝑡)4+𝑦^′(𝑡)2,𝑡>0. (1.19) Now by taking Sumudu transform we have 1𝑢𝑆[𝑡𝑦^′(𝑡)](𝑢)−𝑘=𝑆[𝑦(𝑡)](𝑢)4+𝑆[𝑦^′(𝑡)](𝑢)2, (1.20) where 𝑘=lim_𝑡→0+[𝑡𝑓^′(𝑡)]. Then we have 1𝑢𝑆[𝑡𝑦^′(𝑡)](𝑢)=𝑆[𝑦(𝑡)](𝑢)4+12𝑢𝑆[𝑦(𝑡)](𝑢) (1.21) on noting that 𝑓(0)=0. Then it follows that 16𝑢𝐹^′(𝑢)−(6−𝑢)𝐹(𝑢)=0, (1.22) and the solution given by 𝐹(𝑢)=𝐶^′𝑒^(1/4)𝑢√𝑢³, (1.23) replacing 𝑢 by 1/𝑠 we obtain 𝑆[sinh(√√𝑡)](𝑢)=𝜋2𝑠^3/2𝑒^1/4𝑠 (1.24) on noting that Γ(3/2)=(1/2)Γ(1/2)=√𝜋/2 and lim_𝑡→0+(𝑓(𝑡)/√𝑡)=lim_𝑡→0+(sinh√𝑡/√𝑡)=1. This shows that the solution of some differential equations with non constant coefficients can be expressed as Sumudu transform.

In the next section we consider the Sumudu transform of higher derivatives and representation in the matrix form. However, first of all we introduce the following notation. Let 𝑃(𝑥)=∑^𝑛_𝑘=0(𝑎_𝑘/𝑥^𝑘) be a polynomial in 1/𝑥, where 𝑛≥0 and 𝑎_𝑛≠0 then we define 𝑀_𝑃(𝑥) to be the 1×𝑛 matrix given by𝑀_𝑃1(𝑥)=(𝑥1𝑥²1𝑥³⋯1𝑥^𝑛−1)𝑎₁𝑎₂⋅⋅⋅𝑎_𝑛𝑎₂𝑎₃⋅⋅𝑎_𝑛0𝑎₃⋅⋅𝑎_𝑛𝑎00⋅⋅⋅⋅⋅⋅_𝑛0⋅⋅⋅0. (1.25) Thus 𝑀_𝑃(𝑥) defines a linear mapping of ℂ^𝑛 into ℂ in obvious way. We will write vectors 𝑦 in ℂ^𝑛 as the row vectors or the column vectors interchangeably, whichever is convenient although, when 𝑀_𝑃(𝑥)𝑦 is to be compute and the matrix representation by (1.25) of 𝑀_𝑃(𝑥) is used, then of course 𝑦must be written as a column vector𝑀_𝑃(𝑥)𝑦=^𝑛_𝑖=11𝑥^{𝑖𝑛−𝑖}_𝑘=0𝑎_𝑖+𝑘𝑦_𝑘 (1.26) for any 𝑦=(𝑦₀,𝑦₁,…,𝑦_𝑛−1)∈ℂ^𝑛. If 𝑛=0, then 𝑀_𝑃(𝑥) is a unique linear mapping of {0}=ℂ⁰ into ℂ (empty matrix). In general, if 𝑛>0 and 𝑓 is 𝑛−1 times differentiable on an interval (𝑎,𝑏), with 𝑎<𝑏, we will write 𝜑(𝑓;𝑎;𝑛)=(𝑓(𝑎+),𝑓^′(𝑎+),…,𝑓^(𝑛−1)(𝑎+))∈ℂ^𝑛,𝜙(𝑓;𝑏;𝑛)=(𝑓(𝑏−),𝑓^′(𝑏−),…,𝑓^(𝑛−1)(𝑏−))∈ℂ^𝑛. (1.27) If 𝑎=0, we write 𝜑(𝑓;𝑛) for 𝜑(𝑓;0;𝑛). If 𝑛=0, we define 𝜑(𝑓;𝑎;0)=𝜙(𝑓;𝑎;0)=0∈ℂ⁰. (1.28) Next we have the following proposition.

Proposition 1.6 (Sumudu transform of higher derivatives). Let 𝑓 be 𝑛 times differentiable on (0,∞) and let 𝑓(𝑡)=0 for 𝑡<0. Suppose that 𝑓^(𝑛)∈𝐿_loc. Then 𝑓^(𝑘)∈𝐿_loc for 0≤𝑘≤𝑛−1,dom(𝑆𝑓)⊂dom(𝑆𝑓^(𝑛)) and, for any polynomial 𝑃 of degree 𝑛,𝑃(𝑢)𝑆(𝑦)(𝑢)=𝑆(𝑓)(𝑢)+𝑀_𝑃(𝑢)𝜑(𝑦,𝑛) (1.29) for 𝑢∈dom(𝑆𝑓). In particular (𝑆𝑓^(𝑛)1)(𝑢)=𝑢^𝑛1(𝑆𝑓)(𝑢)−(𝑢^𝑛,1𝑢^𝑛−11,…,𝑢)𝜑(𝑓;𝑛) (1.30) (with 𝜑(𝑓;𝑛) here written as a column vector). For 𝑛=2 one has (𝑆𝑓^′′1)(𝑢)=𝑢²1(𝑆𝑓)(𝑢)−𝑢²1𝑓(0+)−𝑢𝑓^′(0+). (1.31)

Proof. We use induction on 𝑛. The result is trivially true if 𝑛=0, and the case 𝑛=1 is equivalent to the Proposition 1.4 (1.11). Now suppose that the result is true for some 𝑛≥1 and let 𝑃(𝑥)=∑^𝑛+1_𝑘=0(𝑎_𝑘/𝑥^𝑘) having degree 𝑛+1 and writing in the form 𝑃(𝑥)=𝑎₀+(1/𝑥)𝑊(𝑥), where 𝑊(𝑥)=∑^𝑛_𝑘=0(𝑎_𝑘+1/𝑥^𝑘). Then it follows that 𝑃(𝐷̇)𝑓=𝑎₀𝑓+𝑊(𝐷̇)𝑧 therefore by using Theorem 1.1 we have 𝑆[𝑃(𝐷̇)𝑓](𝑢)=𝑎₀𝑆[𝑓](𝑢)+𝑆[𝑊(𝐷̇)𝑧](𝑢)−𝑀_𝑊(𝑢)𝜑(𝑧;𝑛)=𝑎₀1𝑆[𝑓](𝑢)+𝑊(𝑢)[𝑢1𝑆[𝑓](𝑢)−𝑢−𝑓(0+)]_𝑛𝑖=11𝑢^{𝑖𝑛−𝑖}_𝑘=0𝑎_𝑖+𝑘+1𝑓^(𝑘+1)(0+) (1.32) on using (1.26) and setting 𝑧^(𝑘)=𝑓^(𝑘+1). Then the summation can be written in the form of ^𝑛_𝑖=11𝑢^{𝑖𝑛−𝑖+1}_𝑘=1𝑎_𝑖+𝑘𝑓^(𝑘)(0+)=^𝑛+1_𝑖=11𝑢^{𝑖𝑛−𝑖+1}_𝑘=0𝑎_𝑖+𝑘𝑓^(𝑘)−1(0+)𝑢[1𝑢^𝑛𝑎_𝑛+1𝑓(0+)+^𝑛_𝑖=11𝑢^𝑖−1𝑎_𝑖𝑓(0+)]=𝑀_𝑃1(𝑢)𝜑(𝑓;𝑛)−𝑢𝑊(𝑢)𝑓(0+). (1.33) Thus we have 𝑆[𝑃(𝐷̇)𝑓](𝑢)=[𝑎₀+𝑊(𝑢)1𝑢1]𝑆[𝑓](𝑢)−𝑢𝑊(𝑢)𝑓(0+)−𝑀_𝑃1(𝑢)𝜑(𝑓;𝑛)+𝑢𝑊(𝑢)𝑓(0+)=𝑃(𝑢)𝑆(𝑓)(𝑢)−𝑀_𝑃(𝑢)𝜑(𝑓;𝑛). (1.34)

In general, if 𝑓 is differentiable on the open interval (𝑎,𝑏), and 𝑓(𝑡)=0 for 𝑡<𝑎 or 𝑡>𝑏 then 𝑓^(𝑛)∈𝐿_loc and𝑆[𝑃(𝐷̇)𝑓](𝑢)=𝑃(𝑢)(𝑆𝑓)(𝑢)−𝑀_𝑃(𝑢)[𝑒^−𝑎/𝑢𝜑(𝑓;𝑎;𝑛)−𝑒^−𝑏/𝑢𝜙(𝑓;𝑏;𝑛)] (1.35) for all 𝑢.

In particular case if we consider 𝑦(𝑡)=sin(𝑡) then clearly 𝑦^′′+𝑦=0 and in the operator form we write (𝐷²+1)𝑓=0. (1.36) Since dom(𝑆𝑓) contain (0,∞) then on using (1.29) and (1.42) with 𝑛=2 and 𝑃(𝑥)=𝑥²+1, for 𝑢>0,10=(𝑢²1+1)𝑆(𝑓)−(𝑢1𝑢²)010110. (1.37) Since 𝜑(𝑦,2)=(𝑓(0),𝑓^′(0))=(0,1). Thus we can obtain the same result without using definitions or transforms table as 𝑢𝑆[sin(𝑡)𝐻(𝑡)]=𝑢².+1 (1.38) Now, in general form if we want to solve 𝑎_𝑛𝑦^(𝑛)+𝑎_𝑛−1𝑦^(𝑛−1)+𝑎_𝑛−2𝑦^(𝑛−2)⋯+𝑎₀=𝑓∗𝑔, (1.39) then we rewrite in the form of𝑃(𝐷̇)𝑦=𝑓∗𝑔, (1.40) under the initial condition 𝑦(0)=𝑦₀,𝑦^′(0)=𝑦₁,…,𝑦^(𝑛−1)(0)=𝑦_𝑛−1, (1.41) where 𝑦 is 𝑛 times differentiable on (0,∞), zero on (−∞,0). Since 𝑦^(𝑘) is locally integrable therefore Sumudu transformable for 0≤𝑘≤𝑛 and, for every such 𝑘, then on using the Sumudu transform of (1.40) we have𝑀_𝑃1(𝑢)𝜑(𝑦,𝑛)=(𝑢1𝑢²⋯1𝑢^𝑛)𝑎₁𝑎₂⋅⋅⋅𝑎_𝑛𝑎₂𝑎₃⋅⋅𝑎_𝑛0𝑎₃⋅⋅𝑎_𝑛𝑎00⋅⋅⋅⋅⋅⋅_𝑛𝑦0⋅⋅⋅0⁰𝑦₁⋅⋅⋅𝑦_𝑛−1, (1.42) where 𝑃(𝑢)=𝑎_𝑛/𝑢^𝑛+𝑎_𝑛−1/𝑢^𝑛−1+⋯+𝑎₀, and the nonhomogeneous term is single convolution. In particular, if 𝑛=2 we have (𝑎₂𝑢²+𝑎₁𝑢+𝑎₀1)𝑆(𝑦)(𝑢)=𝑆(𝑓∗𝑔)(𝑢)+(𝑢1𝑢²)𝑎₁𝑎₂𝑎₂0𝑦₀𝑦₁. (1.43) In order to get the solution of (1.40), we are taking inverse Sumudu transform for (1.29) as follows:𝑦(𝑡)=𝑆⁻¹[(𝑓∗𝑔)(𝑢)𝑃(𝑢)]+𝑆⁻¹[𝑀_𝑃(𝑢)𝑃(𝑢)𝜙(𝑦,𝑛)] (1.44) provided that the inverse exist for each terms in the right-hand side of (1.44).

Now, multiply the right-hand side of (1.40) by polynomial Ψ(𝑡)=∑^𝑛_𝑘=0𝑡^𝑘, we obtain the following equation that is having non constant coefficients:Ψ(𝑡)∗[𝑃(𝐷̇)𝑦]=𝑓∗𝑔, (1.45) under the same initial conditions as above. By taking Sumudu transform for (1.45) and inverse Sumudu transform we have𝑦=𝑆⁻¹[𝐹(𝑢)𝐺(𝑢)𝑘!𝑢^𝑘+1𝑃(𝑢)(1𝑃(𝑢)𝑢1𝑢²⋯1𝑢^𝑛+1)𝑎₁𝑎₂⋅⋅⋅𝑎_𝑛𝑎₂𝑎₃⋅⋅𝑎_𝑛0𝑎₃⋅⋅𝑎_𝑛𝑎00⋅⋅⋅⋅⋅._𝑛𝑦0⋅⋅⋅0⁰𝑦₁⋅⋅⋅𝑦_𝑛−1], (1.46) provided that the inverse transform exists. Now, if we substitute (1.46) into (1.45), we obtain the non homogeneous term of (1.45) 𝑓∗𝑔 and polynomial in the form of Φ(𝑡)=−∑^𝑛_𝑘=1(1/𝑘!)𝑡^𝑘.

In particular consider the differential equation in the form of𝑦^′′′−𝑦^′′+4𝑦^′−4𝑦=𝑒^𝑡∗sin(𝑡),𝑡>0,𝑦(0)=1,𝑦^′(0)=4,𝑦^′′(0)=1. (1.47) On using (1.29) we have(1𝑢³−1𝑢²+4𝑢−4)𝑆(𝑦)(𝑢)=𝑢𝑆[𝑒^𝑡]𝑆[sin(𝑡)]+𝑀_𝑃𝑀(𝑢)𝜑(𝑦,4),_𝑃1(𝑢)𝜑(𝑦,4)=(𝑢1𝑢²1𝑢³)141=14−11−110100𝑢+3𝑢²+1𝑢³ (1.48)(1.49) after simplifying (1.48), we have𝑢𝑌(𝑢)=⁵(1−𝑢)(4𝑢²+1)(1−𝑢+4𝑢²−4𝑢³)+(𝑢²+3𝑢+1)(1−𝑢+4𝑢²−4𝑢³). (1.50) By replacing the complex variables 𝑢 by 1/𝑠 then (1.50) becomes1𝑌(𝑠𝑠)=(𝑠²+4)(𝑠²+1)(𝑠−1)²+𝑠(𝑠²+3𝑠+1)(𝑠²+1)(𝑠−1). (1.51) To obtain the inverse Sumudu transform for (1.51) we use the following formula𝑆⁻¹1(𝑌(𝑠))=2𝜋𝑖^𝛾+𝑖∞_{𝛾−𝑖∞}𝑒^𝑠𝑡1𝑌(𝑠)𝑑𝑠𝑠=residues[𝑒^𝑠𝑡𝑌(1/𝑠)𝑠], (1.52) for more details see [18]. Then the solution of (1.47) given by 1𝑦(𝑡)=61cos(𝑡)+𝑒10^𝑡𝑡+43𝑒50^𝑡−275cos(2𝑡)+3825sin(2𝑡). (1.53) Now if we multiply the right-hand side of (1.47) by 𝑡²∗ then the equation becomes𝑡²∗(𝑦^′′′−𝑦^′′+4𝑦^′−4𝑦)=2cos(2𝑡)−sin(2𝑡),𝑡>0,𝑦(0)=1,𝑦^′(0)=4,𝑦^′′(0)=1, (1.54) by applying similar method that we used above we obtain the solution of (1.54) in the form of𝑦₁=112sin(𝑡)+27𝑒25^𝑡+1𝑒20^𝑡2𝑡−25cos(2𝑡)+209150sin(2𝑡). (1.55) Now, if we substitute the solution of nonconstant coefficient in (1.54) we obtain the solution of the non homogeneous part of (1.54) plus the following terms −(1/2)𝑡².

In fact the Sumudu transform is also applicable to the system of differential equations, see the details in [17].

Example 1.7. Solve for 𝑡>0 the system of two equations 𝑥^′′+2𝑦^′−2𝑥=−sin(𝑡),𝑥(0)=1,𝑥^′𝑦(0)=2,_′′−2𝑥^′−2𝑦=cos(𝑡)−2,𝑦(0)=0,𝑦^′(0)=1. (1.56)

The matrix1𝑃(𝑢)=𝑢²2−2𝑢−2𝑢1𝑢²−2, (1.57) and we have det[𝑃(𝑢)]=1/𝑢⁴+4 which has degree 4=𝑁(𝑃). Thus 𝑃 regular. Now by applying Sumudu transform to the above system we have𝑥𝑦−𝑢𝑃(𝑢)𝑆(𝑢)=𝑢²−+11+2𝑢²𝑢²+1+Ψ_𝑃(𝑢)𝜑(𝑦,𝑁(𝑃)), (1.58) where Ψ_𝑃(𝑢)𝜑(𝑦,𝑁(𝑃)) given by Ψ_𝑃1(𝑢)𝜑(𝑦,𝑁(𝑃))=𝑢²1𝑢1𝑢0−1𝑢01𝑢²1𝑢1201=1+2𝑢𝑢²−1𝑢. (1.59) Then we obtain 𝑃(𝑢)⁻¹=𝑢²(1−2𝑢²)1+4𝑢⁴−2𝑢³1+4𝑢⁴2𝑢³1+4𝑢⁴𝑢²(1−2𝑢²)1+4𝑢⁴. (1.60) Equation (1.58) becomes𝑆𝑥𝑦(𝑢)=𝑃⁻¹−𝑢(𝑢)𝑢²−+11+2𝑢²𝑢²+1+𝑃⁻¹(𝑢)[Ψ_𝑃(𝑢)𝜑(𝑦,𝑁(𝑃))], (1.61) finally, by taking inverse Sumudu transform (1.61) we obtain the solution of the system as follows𝑥(𝑡)=sin(𝑡)+𝑒^𝑡cos(𝑡),𝑦(𝑡)=−cos(𝑡)+𝑒^𝑡sin(𝑡)+1, (1.62) thus based on the above discussions we note that the Sumudu transform can be applied for system of differential equations thus can be used in many engineering problems. Similar applications can also be seen in [5, 10, 11, 13, 19].

Acknowledgments

The authors gratefully acknowledge that this research was partially supported by the University Putra Malaysia under the Research University Grant Scheme 05-01-09-0720RU and Fundamental Research Grant Scheme 01-11-09-723FR. The authors also thank the referee(s) for very constructive comments and suggestions that improved the paper.