Abstract

We study the relationship between Sumudu and Laplace transforms and further make some comparison on the solutions. We provide some counterexamples where if the solution of differential equations exists by Laplace transform, the solution does not necessarily exist by using the Sumudu transform; however, the examples indicate that if the solution of differential equation by Sumudu transform exists then the solution necessarily exists by Laplace transform.

1. Introduction

In order to solve the differential equations, the integral transform is extensively applied and thus there are several works on the theory and application of integral transforms. In the sequence of these transforms, Watugala introduced a new integral transform, named the Sumudu transform, and further applied it to the solution of ordinary differential equation in control engineering problems; see [1]. For further details and properties of Sumudu transform see [2–7] and many others. The Sumudu transform is defined over the set of the functions 𝐴=𝑓(𝑡)∶∃𝑀,𝜏1,𝜏2||||>0,𝑓(𝑡)<𝑀𝑒𝑡/𝜏𝑗,if𝑡∈(−1)𝑗×[0,∞)(1.1) by the following formula:[]𝑓(𝑢)=𝑆𝑓(𝑡);𝑢=∶∞0𝑓(𝑢𝑡)𝑒−𝑡𝑑𝑡,𝑢∈−𝜏1,𝜏2.(1.2)

The existence and the uniqueness was discussed in [8]; for further properties of Sumudu transform and its derivatives, we refer to [2]. In [3], some fundamental further properties of Sumudu transform were also established.

Similarly, this new transform was applied to the one-dimensional neutron transport equation in [9]. In fact, one can easily show that there is strong relationship between Sumudu and other integral transforms. In particular, the relation between Sumudu transform and Laplace transforms was proved in [8].

Further in [6], the Sumudu transform was extended to the distributions (generalized functions) and some of their properties were also studied in [10]. Recently, Kılıçman et al. applied this transform to solve the system of differential equations; see [7, 11].

Now, let us recall the following definition which is held in Estrin and Higgins; see [12]; the double Laplace Transform is defined by𝐿𝑥𝐿𝑡[]𝑓(𝑥,𝑠)=𝐹(𝑝,𝑠)=∞0𝑒−𝑝𝑥∞0𝑒−𝑠𝑡𝑓(𝑥,𝑡)𝑑𝑡𝑑𝑥,(1.3) where 𝑥,𝑡>0 and 𝑝, 𝑠  are complex numbers. The double Sumudu transform of second partial derivative with respect to 𝑥 is given by𝑆2𝜕2𝑓(𝑡,𝑥)𝜕𝑥2=1;(𝑣,𝑢)𝑢𝑣∞0∞0𝑒−(𝑡/𝑣+𝑥/𝑢)𝜕2𝑓(𝑡,𝑥)𝜕𝑥2=1𝑑𝑡𝑑𝑥𝑢∞0𝑒−𝑡/𝑣1𝑣∞0𝑒−𝑥/𝑢𝜕2𝑓(𝑡,𝑥)𝜕𝑥2𝑑𝑥𝑑𝑡.(1.4) Then, the interior integral is given by1𝑢∞0𝑒−𝑥/𝑣𝜕2𝑓(𝑡,𝑥)𝜕𝑥21𝑑𝑥=𝑢21𝐹(𝑡,𝑣)−𝑢21𝐹(𝑡,0)−𝑢𝜕𝑓(𝑡,0).𝜕𝑥(1.5)

By taking Sumudu transform with respect to 𝑡 for (1.5), we get double Sumudu transform in the form of𝑆2𝜕2𝑓(𝑡,𝑥)𝜕𝑥2=1;(𝑢,𝑣)𝑢21𝐹(𝑣,𝑢)−𝑢21𝐹(0,𝑢)−𝑢𝜕𝑓(0,𝑢)𝜕𝑥.(1.6)

Similarly, double Sumudu transform of 𝜕2𝑓(𝑡,𝑥)/𝜕𝑡2 is given by𝑆𝜕2𝑓(𝑡,𝑥)𝜕𝑡2=1;(𝑢,𝑣)𝑣21𝐹(𝑣,𝑢)−𝑣21𝐹(𝑣,0)−𝑣𝜕𝑓(𝑣,0)𝜕𝑡.(1.7) And double Laplace transform defined the first-order partial derivative as𝐿𝑥𝐿𝑡𝜕𝑓(𝑥,𝑡)𝜕𝑥=𝑝𝐹(𝑝,𝑠)−𝐹(0,𝑠).(1.8)

The double Laplace transform for second partial derivative with respect to 𝑥 is given by𝐿𝑥𝑥𝜕2𝑓(𝑥,𝑡)𝜕2𝑥=𝑝2𝐹(𝑝,𝑠)−𝑝𝐹(0,𝑠)−𝜕𝐹(0,𝑠),𝜕𝑥(1.9) and double Laplace transform for second partial derivative with respect to 𝑡 similarly as above is given by𝐿𝑡𝑡𝜕2𝑓(𝑥,𝑡)𝜕2𝑡=𝑠2𝐹(𝑝,𝑠)−𝑠𝐹(𝑝,0)−𝜕𝐹(𝑝,0).𝜕𝑡(1.10)

In a similar manner, the double Laplace transform of a mixed partial derivative can be deduced from single Laplace transform as𝐿𝑥𝐿𝑡𝜕2𝑓(𝑥,𝑡)𝜕𝑥𝜕𝑡=𝑝𝑠𝐹(𝑝,𝑠)−𝑝𝐹(𝑝,0)−𝑠𝐹(0,𝑠)−𝐹(0,0).(1.11)

Our purpose here is to show the difference between Laplace transform and Sumudu transform by solving partial differential equations. In fact, the double Sumudu transform and double Laplace transform have a strong relationship that may be expressed either as(I)𝑢𝑣𝐹(𝑢,𝑣)=ℒ21𝑓(𝑥,𝑦);𝑢,1𝑣,(II)𝑝𝑠𝐹(𝑝,𝑠)=ℒ21𝑓(𝑥,𝑦);𝑝,1𝑠,(1.12) where ℒ2 represents the operation of double Laplace transform. In particular, this relation is best illustrated by the fact that the double Sumudu and double Laplace transforms interchange the image of sin(𝑥+𝑡) and cos(𝑥+𝑡). It turns out that𝑆2[]sin(𝑥+𝑡)=ℒ2[]=cos(𝑥+𝑡)𝑢+𝑣(1+𝑢)2(1+𝑣)2,𝑆2[]cos(𝑥+𝑡)=ℒ2[]=1sin(𝑥+𝑡)(1+𝑢)2(1+𝑣)2.(1.13)

Further, the double Laplace and Sumudu transforms interchange the images of the Dirac function, 𝛿(𝑥,𝑡)=𝛿(𝑥)𝛿(𝑡) and the Heaviside function, 𝐻(𝑥,𝑡)=𝐻(𝑥)⊗𝐻(𝑡), since𝑆2[]𝐻(𝑥,𝑡)=ℒ2[]𝑆𝛿(𝑥,𝑡)=1,2[]𝛿(𝑥,𝑡)=ℒ2[]=1𝐻(𝑥,𝑡),𝑢𝑣(1.14) where the symbol ⊗ means the tensor product thus the relation between the double Sumudu and double Laplace transform of convolution was given by𝑆2[]=1(𝑓∗∗𝑔)(𝑡,𝑥);𝑣,𝑢ℒ𝑢𝑣2[](𝑓∗∗𝑔)(𝑡,𝑥)(1.15) (see [10]).

Note that since many practical engineering problems involve mechanical or electrical systems acted upon by discontinuous or impulsive forcing terms, then the Sumudu transform can be effectively used to solve ordinary differential equations as well as partial differential equations in engineering problems; see [13]. In this paper, we study the relationship between Sumudu and Laplace transforms and further make some comparison on the solutions. Provide a counterexample where if the solution of differential equation by Laplace transform exists then it does not necessarily exist by using the Sumudu transform, however, if the solution of differential equation by Sumudu transform exist, then solution necessarily exists by Laplace transform. First of all we need the following concept related to the Sumudu transform of derivatives.

Proposition 1.1 (Sumudu Transform of Derivative). Let 𝑓 be differentiable on (0,∞) and let 𝑓(𝑡)=0 for 𝑡<0. Suppose that 𝑓′∈𝐿loc. Then, 𝑓∈𝐿loc, dom(𝑆𝑓)⊂dom(𝑓), and 𝑆𝑓=1𝑢1𝑆(𝑓)−𝑢𝑓(0+)for𝑢∈dom(𝑆(𝑓)).(1.16)
More generally, if 𝑓 is differentiable on (𝑐,∞), 𝑓(𝑡)=0 for 𝑡<0, and 𝑓′∈𝐿loc, then 𝑆𝑓=1𝑢1𝑆(𝑓)−𝑢𝑒−𝑐/𝑢𝑓(𝑐+)for𝑢∈dom(𝑆(𝑓)).(1.17)

Proof. For the proof of this proposition, see [14].

In general case, if 𝑓 is a differentiable function on (𝑎,𝑏) with 𝑎<𝑏, and 𝑓(𝑡)=0 for 𝑡<𝑎 or 𝑡>𝑏 and 𝑓′∈𝐿loc, then, for all 𝑢, 𝑆𝑓=1𝑢1𝑆(𝑓)−𝑢𝑒−𝑎/𝑢1𝑓(𝑎+)+𝑢𝑒−𝑏/𝑢𝑓(𝑏−).(1.18) Proposition 1.1 can be extended to higher derivatives, before extension, we introduce the following notation as in [11]. Let ∑𝑃(𝑥)=𝑛𝑘=0𝑎𝑘𝑥𝑘 be a polynomial in 𝑥, where 𝑛≥0 and 𝑎𝑛≠0. We define 𝑀𝑃(𝑥) to be the 1×𝑛 matrix of polynomials given by the matrix product:𝑀𝑃(𝑥)=1𝑥𝑥2⋯𝑥𝑛−1⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝𝑎1𝑎2⋅⋅⋅𝑎𝑛𝑎2𝑎3⋅⋅𝑎𝑛0𝑎3⋅⋅𝑎𝑛𝑎00⋅⋅⋅⋅⋅⋅𝑛⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠0⋅⋅⋅0.(1.19) For each complex number 𝑥, the map 𝑀𝑃(𝑥) defines a linear mapping of ℂ𝑛 into ℂ in obvious way. We will write vectors 𝑦 in ℂ𝑛 as row vectors or column vectors interchangeably, whichever, is convenient, although when 𝑀𝑃(𝑥)𝑦 is to be compute and the matrix representation by (1.19) of 𝑀𝑃(𝑥) is used, then of course 𝑦must be written as a column vector:𝑀𝑃(𝑥)𝑦=𝑛𝑖=1𝑥𝑖−1𝑛−𝑖𝑘=0𝑎𝑖+𝑘𝑦𝑘,(1.20) for any 𝑦=(𝑦0,𝑦1,…,𝑦𝑛−1)∈ℂ𝑛. If 𝑛=0, then 𝑀𝑃(𝑥) defines 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, then we shall write𝜑𝑓(𝑓;𝑎;𝑛)=(𝑎+),𝑓(𝑎+),…,𝑓(𝑛−1)(𝑎+)∈ℂ𝑛,𝜙(𝑓;𝑏;𝑛)=𝑓(𝑏−),𝑓(𝑏−),…,𝑓(𝑛−1)(𝑏−)∈ℂ𝑛.(1.21) If 𝑎=0, we write 𝜑(𝑓;𝑛) for 𝜑(𝑓;0;𝑛). If 𝑛=0, then we define𝜑(𝑓;𝑎;0)=𝜙(𝑓;𝑎;0)=0∈ℂ0.(1.22)

Now, we need to consider the transform of higher derivatives as follows.

Proposition 1.2 (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)∈𝐿loc. Then, 𝑓(𝑘)∈𝐿locfor 0≤𝑘≤𝑛−1, dom(𝑆𝑓)⊂dom(𝑆𝑓(𝑛)) and, for any polynomial 𝑃 of degree 𝑛, 𝑆𝑃̇𝐷𝑓(𝑢)=𝑃(𝑢)𝑆(𝑓)(𝑢)−𝑀𝑃(𝑢)𝜑(𝑓;𝑛)(1.23) for 𝑢∈dom(𝑆𝑓). In particular, 𝑆𝑓(𝑛)1(𝑢)=𝑢𝑛1(𝑆𝑓)(𝑢)−𝑢𝑛,1𝑢𝑛−11,…,𝑢𝜑(𝑓;𝑛)(1.24) (with 𝜑(𝑓;𝑛) here written as a column vector). For 𝑛=2, one has 𝑆𝑓1(𝑢)=𝑢21(𝑆𝑓)(𝑢)−𝑢21𝑓(0+)−𝑢𝑓(0+).(1.25)

Proof. For the proof of this proposition, see [8].

In general, if 𝑓 is differentiable on (𝑎,𝑏)with 𝑎<𝑏, and 𝑓(𝑡)=0 for 𝑡<𝑎 or 𝑡>𝑏 and 𝑓(𝑛)∈𝐿loc then we have, for all 𝑢,𝑆𝑃̇𝐷𝑓(𝑢)=𝑃(𝑢)(𝑆𝑓)(𝑢)−𝑀𝑃𝑒(𝑢)−𝑎/𝑢𝜑(𝑓;𝑎;𝑛)−𝑒−𝑏/𝑢.𝜙(𝑓;𝑏;𝑛)(1.26)

2. Solution of Differential Equations by Convolution Methods

In this section, we give the solution of the following equation:𝑎𝑛𝑦(𝑛)+𝑎𝑛−1𝑦(𝑛−1)+⋯+𝑎1𝑦+𝑎0𝑦=𝑓(2.1) on (0,∞). We prove an existence and uniqueness and provide a formula for the solution.

If we define 𝑓 and 𝑦 to be zero on (−∞,0), then (2.1) is equivalent to the equation𝑃̇𝐷𝑦=𝑓,(2.2) where𝑃(𝑥)=𝑛𝑘=0𝑎𝑘𝑥𝑘.(2.3)

First of all, we establish first an important result for homogeneous equation.

Theorem 2.1 (Properties of solution of the homogeneous equation). Let 𝑛≥0 and let 𝑎𝑘(0≤𝑘≤𝑛) be complex constant such that 𝑎𝑛≠0. Let 𝑦 be 𝑛 differentiable on (0,∞) and zero (−∞,0) and satisfy 𝑃̇𝐷𝑦=0,(2.4) then one has the following.(1)𝑦 is infinitely differentiable on (0,∞).(2) For every integer 𝑘≥0, the limits 𝑦𝑘(0+) exist.(3) If 𝑛>0,𝑦 then is given (except at 0) by the formula 𝑦=𝑆−1𝑀𝑃(𝑢)𝑃(𝑢)𝜑(𝑦;𝑛)=𝑀𝑃̇𝐷𝑉𝜑(𝑦;𝑛),(2.5) where 𝑉=𝑆−11𝑃.(2.6)

Proof. To avoid trivial statements, suppose that 𝑛>0. Let 0<𝑎<𝑏. The function 𝑧=𝑦(𝐻𝑎−𝐻𝑏) is 𝑛 times differentiable on (𝑎,𝑏), and 𝑧, and 𝑧𝑛 are locally integrable and dom(𝑆(𝑧))=ℝ. By (1.26) and the relation 𝜑(𝑧;𝑎;𝑛)=𝜑(𝑦;𝑎;𝑛), 𝜙(𝑧;𝑏;𝑛)=𝜑(𝑦;𝑏;𝑛); we have, for all 𝑢𝑆𝑃̇𝐷𝑧(𝑢)=𝑃(𝑢)(𝑆𝑧)(𝑢)−𝑀𝑃𝑒(𝑢)−𝑎/𝑢𝜑(𝑦;𝑎;𝑛)−𝑒−𝑏/𝑢.𝜑(𝑦;𝑏;𝑛)(2.7) Since ̇𝑃(𝐷)𝑧=0, we obtain for large 𝑢𝑀(𝑆𝑧)(𝑢)=𝑃(𝑢)𝑒𝑃(𝑢)−𝑎/𝑢𝜑(𝑦;𝑎;𝑛)−𝑒−𝑏/𝑢.𝜑(𝑦;𝑏;𝑛)(2.8) Now, (𝑀𝑝(𝑢)/𝑃(𝑢))[𝜑(𝑦;𝑎;𝑛)] is a proper rational function of 𝑢; there is a function 𝑔𝑎, analytic on ℝ, such that, for 𝑢 sufficiently large, 𝑆𝑔𝑎𝐻𝑀(𝑢)=𝑃(𝑢)𝑃(𝑢)𝜑(𝑦;𝑎;𝑛).(2.9) With 𝑔𝑏 defined analogously, we deduce from (2.8) and the shift rule that []𝑦(𝑡)𝐻(𝑡−𝑎)−𝐻(𝑡−𝑏)=𝑔𝑎(𝑡−𝑎)𝐻(𝑡−𝑎)−𝑔𝑏(𝑡−𝑏)𝐻(𝑡−𝑏)(2.10) if 𝑎<𝑡<𝑏, this gives 𝑦(𝑡)=𝑔𝑎(𝑡−𝑎),(2.11) thus 𝑦 is analytic on every open interval (𝑎,𝑏), with 0<𝑎<𝑏; hence, 𝑦 is analytic on (0,∞). However, on the interval (1,2), we have 𝑦(𝑡)=𝑔1(𝑡−1). We conclude that this formula must hold for all 𝑎>0, which allows us to write 𝑦(𝑡)=𝑔1(𝑡−1)𝐻(𝑡)(2.12) for all 𝑡≠0; from this formula, (1.1) and (1.2) follow immediately. Now, we write, for 𝑡≠0, 𝑦(𝑡)=𝑔1[](𝑡−1)𝐻(𝑡)−𝐻(𝑡−1)+𝑔1(𝑡−1)𝐻(𝑡−1).(2.13) The first term on the right is clearly Sumudu transformable. Similarly, the second term is also transformable since it is merely the translated function (𝑔1𝐻)(𝑡−1), thus 𝑦 is Sumudu transformable. To obtain the formula (2.5), we apply (1.23) and get 𝑆[𝑦]𝑀(𝑢)=𝑃(𝑢)𝑃(𝑢)𝜑(𝑦;𝑛).(2.14) Since 𝑀𝑃(𝑢)𝜑(𝑦;𝑛) is a polynomial of degree less than 𝑛, by using Sumudu inverse transform, we obtain 𝑦=𝑆−1𝑀𝑃(𝑢)𝑃(𝑢)𝜑(𝑦;𝑛)=𝑀𝑃̇𝐷𝑉𝜑(𝑦;𝑛).(2.15)

Now, we extend the above theorem to the nonhomogeneous equation as follows.

Proposition 2.2. Let 𝑎𝑘 be as in the above theorem. Let 𝑓 be continuous on (0,∞) and zero on (−∞,0) and let 𝑓 be locally integrable. Let 𝑦 be 𝑛 times differentiable on (0,∞) and zero on (−∞,0) and satisfy 𝑃̇𝐷𝑦=𝑓,(2.16) then one has the following.(1)𝑦𝑛 is continuous on (0,∞) and locally integrable on ℝ.(2) For 0≤𝑘≤𝑛−1, the limits 𝑦(𝑘)(0+) exist.(3) If 𝑛>0, then 𝑦 is given by the formula𝑦=𝑉∗𝑓+𝑀𝑃̇D𝑉𝜑(𝑦;𝑛).(2.17)

Proof. The result is trivial if 𝑛=0. Suppose that 𝑛>0. Let 𝑧=𝑉∗𝑓. if 𝑓 is continuous on an open interval 𝐼, then 𝑉∗𝑓 is 𝑛 times differentiable on 𝐼. We have ̇𝑃(𝐷)𝑧=𝑓. Also, consider that 𝑧(𝑛) is continuous for 0≤𝑘≤𝑛−1, and hence 𝑧(𝑛) is locally integrable on ℝ and continuous on (0,∞). Let 𝑤=𝑦−𝑧. Then, ̇𝑃(𝐷)𝑤=0. By using the above theorem, 𝑤(𝑛) is locally integrable and 𝑤(𝑘)(0+) exist for all 𝑘≥0, and we have 𝑤=𝑀𝑃(̇𝐷)𝑉𝜑(𝑦;𝑛). If we now write 𝑦=𝑤+𝑧, then the properties (1) and (2) follow immediately. Equation (2.17) also follows because 𝜑(𝑧;𝑛)=0 by using the statement if 𝑓(𝑡)=0 for 𝑡<𝑐, then (𝑉∗𝑓)(𝑘)(𝑐)=0 for 0≤𝑘≤𝑛−1, where 𝑉=𝑆−1[1/𝑃(𝑢)], and hence 𝜑(𝑤;𝑛)=𝜑(𝑦;𝑛).

In the next theorem, we provide a complete solution of a non-homogeneous equation.

Theorem 2.3 (Existence and Uniqueness). Let 𝑓 be continuous function on (0,∞), zero on (−∞,0), and locally integrable. Let 𝛾∈ℂ𝑛. Then, there exists a unique (except on zero) function 𝑦 that is 𝑛 times differentiable on (0,∞) and zero on (−∞,0) satisfying 𝑃̇𝐷𝑦=𝑓,𝛾=𝜑(𝑦;𝑛).(2.18) If 𝑛>0, then 𝑦 is given (except on zero) by 𝑦=𝑉∗𝑓+𝑆−1𝑀𝑃(𝑢)𝛾𝑃(𝑢)=𝑉∗𝑓+𝑀𝑃̇𝐷𝑉𝛾.(2.19)

Proof. Uniqueness is obvious by (2.17), thus 𝑦 is given by (2.19). To establish existence, we consider 𝑦 by (2.19). Since 𝑃̇𝐷𝑀𝑃̇𝐷𝑉𝛾=𝑀𝑃̇𝐷𝑃̇𝐷𝑉𝛾=0(2.20) by taking Sumudu transform ̇𝑃(𝐷)𝑉 and using (1.23), we have 𝑆𝑃̇𝐷𝑉[𝑉]=𝑃(𝑢)𝑆−𝑀𝑃(𝑢)𝜑(𝑉;𝑛)=1−𝑀𝑃(𝑢)𝜑(𝑉;𝑛).(2.21) This is a polynomial in 𝑢 which by virtue of the statement (if 𝑓 be Sumudu transformable and satisfy 𝑓(𝑡)=0 for 𝑡<0. Then, lim𝑢→∞𝑆[𝑓](𝑢)=0) must be identically zero. Thus, ̇𝑃(𝐷)𝑉=0, then (2.20) is true. We have 𝑃̇𝐷̇𝐷𝑦=𝑃(𝑉∗𝑓)=𝑓.(2.22) To verify that 𝑦 satisfies the initial conditions, we first observe that by (2.17) we must have 𝑦=𝑉∗𝑓+𝑀𝑃(̇𝐷)𝑉𝜑(𝑦;𝑛). We deduce that 𝑀𝑃̇𝐷𝑉[]𝛾−𝜑(𝑦;𝑛)=0.(2.23) Taking Sumudu transform and using the relation, if 𝑅 is a polynomial of degree less than 𝑛, then 𝑆−1̇[𝑅(𝑢)/𝑃(𝑢)]=𝑅(𝐷)𝑉, then it follows that 𝑀𝑃[](𝑢)𝑉𝛾−𝜑(𝑦;𝑛)=0(2.24) for all 𝑢 sufficiently large (and hence for all real 𝑢 since the left side is a polynomial). From the relation 𝑦∈ℂ𝑛 and 𝑀𝑃(𝑥)𝑦=0 for all 𝑥, we have 𝑦=0. We conclude that 𝛾=𝜑(𝑦;𝑛). This establishes existence.

The last term in the right-hand side of (2.20) can be written in the form of𝑀𝑃̇𝐷𝑉𝛾=𝑉𝑉′⋯𝑉(𝑛−1)⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝𝑎1𝑎2⋅𝑎2𝑎3⋅𝑎3𝑎⋅⋅⋅⋅⋅𝑛⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝0⋅⋅⋅⋅⋅𝑎𝑛0𝑎𝑛⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠⎛⎜⎜⎜⎜⎜⎜⎝𝛾00⋅⋅⋅⋅⋅00𝛾1⋮𝛾𝑛−1⎞⎟⎟⎟⎟⎟⎟⎠(2.25) for any vector 𝛾=(𝛾0,𝛾1,…,𝛾𝑛−1)∈ℂ𝑛. In the next section, we provide an example to make a comparison.

3. A Comparison on Solutions

Consider that the steady-state temperature distribution function 𝑓(𝑥,𝑦) in a long square bar with one face held at constant temperature 𝑇0 and the other faces held at zero temperature is governed by the boundary-value problem𝜕2𝑓(𝑥,𝑦)𝜕𝑥2+𝜕2𝑓(𝑥,𝑦)𝜕𝑦2=0(3.1) under the boundary conditions𝑓(0,𝑦)=0,𝑓(𝑥,0)=0,𝑓(𝑥,𝜋)=0,𝑓(𝜋,𝑦)=𝑇0.(3.2)

3.1. Solution with Laplace Transform

If we apply the multiple Laplace transform with respect to the variables 𝑥 and 𝑦 for (3.1), and single Laplace transform for the first and second boundary condition, we obtain𝑝2+𝑞2𝜕𝐹(𝑝,𝑞)=𝑓(0,𝑞)+𝜕𝜕𝑥𝑓(𝑝,0).𝜕𝑦(3.3) If we look at (3.3) then we can easily notice that the right-hand side is transforms of the functions 𝜕𝑓(0,𝑦)/𝜕𝑥 and 𝜕𝑓(𝑥,0)/𝜕𝑦 which are not among the boundary condition of (3.1). Hence, 𝜕𝑓(0,𝑦)/𝜕𝑥 and 𝜕𝑓(𝑥,0)/𝜕𝑦 are taken to be the unknown functions ℎ(𝑦) and 𝑔(𝑥), respectively. Then, by using single Laplace transform, we have𝜕𝑓(0,𝑞)𝜕𝜕𝑥=𝐻(𝑞),𝑓(𝑝,0)𝜕𝑦=𝐺(𝑝).(3.4)

By substituting (3.4) into (3.3) and rearrangement, we obtain𝐹(𝑝,𝑞)=H(𝑞)𝑝2+𝑞2+𝐺(𝑝)𝑝2+𝑞2.(3.5)

On using single inverse Laplace transform with respect to 𝑝 for (3.3) and using the Laplace transform of convolution, we have𝑓(𝑥,𝑞)=𝐻(𝑞)sin𝑞𝑥+𝑥0𝐺(𝛽)sin𝑞(𝑥−𝛽)𝑑𝛽.(3.6) Now, we take Laplace transform for third boundary condition and substitute in (3.6), using integral property𝜋0𝑄(𝛽)𝑑𝛽=𝑥0𝑄(𝛽)𝑑𝛽+𝜋𝑥𝑄(𝛽)𝑑𝛽,(3.7) and trigonometric manipulation, then (3.6) can be written in the form of1𝑓(𝑥,𝑞)=−sin𝑞𝜋sin𝑞(𝑥−𝜋)𝑥0𝐺(𝛽)sin𝑞𝛽𝑑𝛽+sin𝑞𝑥𝜋𝑥𝐺(𝛽)sin𝑞(𝜋−𝛽)𝑑𝛽.(3.8)

In order to obtain the single inverse Laplace transform with respect to 𝑞, we use Cauchy’s residue theorem, so we have a simple pole at 𝑞=±𝑛; we use𝑓(𝑥,𝑦)=residues𝑒𝑦𝑞𝑓(𝑥,𝑞).(3.9) We compute the residues at 𝑞=±𝑛, and adding together, then we obtain𝑓(𝑥,𝑦)=∞𝑛=1odd2sin𝑛𝑥sinh𝑛𝑦𝜋0𝐺(𝛽)sin𝑛𝛽𝑑𝛽,(3.10) by using the last boundary condition and Fourier series to compute the unknown integral, then we obtain the solution of (3.1) as follows:𝑓(𝑥,𝑦)=4𝑇0𝜋∞𝑛=1oddsinh𝑛𝑦𝑛sinh𝑛𝜋sin𝑛𝑥.(3.11)

3.2. Solution with Sumudu Transform

Now, we apply multiple Sumudu transform for the same problem to check the solution whether equal or not equal or probably does not exist. By applying multiple Sumudu transform for (3.1) as follows:1𝑢21𝑓(𝑢,𝑣)−𝑢21𝑓(0,𝑣)−𝑢𝜕𝜕𝑥1𝑓(0,𝑣)+𝑣21𝑓(𝑢,𝑣)−𝑣21𝑓(𝑢,0)−𝑣𝜕𝜕𝑦𝑓(𝑢,0)=0.(3.12) and by taking single Sumudu transform for first two boundary conditions of (3.2), we have𝑓(0,𝑣)=0,𝑓(𝑢,0)=0,(3.13) where 𝑓(𝑥,0) and 𝜕𝑓(0,𝑦)/𝜕𝑥 are taken to be unknown functions ℎ(𝑦) and 𝑔(𝑥). Then, the Sumudu transform of unknown functions is given by𝜕𝜕𝑥𝜕𝑓(𝑢,0)=𝐺(𝑢),𝜕𝑥𝑓(0,𝑣)=𝐻(𝑣).(3.14) By substituting (3.13) and (3.14) into (3.12) and rearranging, we have𝑓(𝑢,𝑣)=𝑣𝑢2𝐺(𝑢)𝑢2+𝑣2+𝑢𝑣2𝐻(𝑣)𝑢2+𝑣2.(3.15) By taking inverse Sumudu transform with respect to 𝑢 for (3.15), and using convolution, we have 𝑓(𝑥,𝑣)=𝑥0𝐺(𝛽)sin𝑣(𝑥−𝛽)𝑑𝛽+𝑣2𝐻(𝑣)sin𝑣𝑥.(3.16) Now by taking single Sumudu transform for third boundary condition, we have𝑓(𝑢,𝜋)=0, and use 𝑥=𝜋, we have−∫𝐻(𝑣)=𝜋0𝐺(𝛽)sin𝑣(𝜋−𝛽)𝑑𝛽𝑣2sin𝜋𝑣.(3.17) Substituting (3.17) into (3.16), we obtain𝑓(𝑥,𝑣)=𝑥0−∫𝐺(𝛽)sin𝑣(𝑥−𝛽)𝑑𝛽+sin𝑣𝑥𝜋0𝐺(𝛽)sin𝑣(𝜋−𝛽)𝑑𝛽sin𝜋𝑣.(3.18) By using trigonometric properties, we have1𝑓(𝑥,𝑣)=−sin𝜋𝑣sin𝜋𝑣𝑥0𝐺(𝛽)sin𝑣(𝑥−𝛽)𝑑𝛽+sin𝑣𝑥𝜋0𝐺(𝛽)sin𝑣(𝜋−𝛽)𝑑𝛽.(3.19) By rearrangement the above equation, we have 1𝑓(𝑥,𝑣)=−sin𝜋𝑣sin𝑣(𝑥−𝜋)𝑥0𝐺(𝛽)sin𝑣𝛽𝑑𝛽+sin𝑣𝑥𝜋𝑥𝐺(𝛽)sin𝑣(𝜋−𝛽)𝑑𝛽.(3.20) In order to obtain inverse Sumudu transform for (3.20), we use the Cauchy’s residue theorem, and then we have𝑓(𝑥,𝑦)=residues𝑒𝑦𝑞𝑓(𝑥,1/𝑞)𝑞.(3.21) If we replace the variable 𝑣 by 1/𝑞 and divide the equation by 𝑞, then we obtain1𝑓(𝑥,𝑣)=−𝑞sin𝜋/𝑞sin(𝑥−𝜋)𝑞𝑥0𝛽𝐺(𝛽)sin𝑞𝑥𝑑𝛽+sin𝑞𝜋𝑥𝐺(𝛽)sin(𝜋−𝛽)𝑞𝑑𝛽.(3.22)

By using Cauchy’s residue theorem, we have simple poles at 𝑞=0 and 𝑞=±1/𝑛, thus it follows that the limit does not exist at 𝑞=0, then the solution does not exist. That leads us to make a remark that if the solution of differential equation by Laplace transform exists then it does not necessarily exist by using the Sumudu transform, but if the solution of differential equation by Sumudu transform exists then it necessarily exists by Laplace transform.

Acknowledgment

The authors gratefully acknowledge that this research was partially supported by University Putra Malaysia under the Research University Grant Scheme  05-01-09-0720RU.