A Numerical Method for Partial Differential Algebraic Equations Based on Differential Transform Method

Abstract

We have considered linear partial differential algebraic equations (LPDAEs) of the form 𝐴𝑢_𝑡(𝑡,𝑥)+𝐵𝑢_𝑥𝑥(𝑡,𝑥)+𝐶𝑢(𝑡,𝑥)=𝑓(𝑡,𝑥), which has at least one singular matrix of 𝐴,𝐵∈ℝ^𝑛×𝑛. We have first introduced a uniform differential time index and a differential space index. The initial conditions and boundary conditions of the given system cannot be prescribed for all components of the solution vector 𝑢 here. To overcome this, we introduced these indexes. Furthermore, differential transform method has been given to solve LPDAEs. We have applied this method to a test problem, and numerical solution of the problem has been compared with analytical solution.

1. Introduction

The partial differential algebraic equation was first studied by Marszalek. He also studied the analysis of the partial differential algebraic equations [1]. Lucht et al. [2–4] studied the numerical solution and indexes of the linear partial differential equations with constant coefficients. A study about characteristics analysis and differential index of the partial differential algebraic equations was given by Martinson and Barton [5, 6]. Debrabant and Strehmel investigated the convergence of Runge-Kutta method for linear partial differential algebraic equations [7].

There are numerous LPDAEs applications in scientific areas given, for instance, in the field of Navier-Stokes equations, in chemical engineering, in magnetohydrodynamics, and in the theory of elastic multibody systems [4, 8–12].

On the other hand, the differential transform method was used by Zhou [13] to solve linear and nonlinear initial value problems in electric circuit analysis. Analysis of nonlinear circuits by using differential Taylor transform was given by Köksal and Herdem [14]. Using one-dimensional differential transform, Abdel-Halim Hassan [15] proposed a method to solve eigenvalue problems. The two-dimensional differential transform methods have been applied to the partial differential equations [16–19]. The differential transform method extended to solve differential-difference equations by Arikoglu and Ozkol [20]. Jang et al. have used differential transform method to solve initial value problems [21]. The numerical solution of the differential-algebraic equation systems has been studied by using differential transform method [22, 23].

In this paper, we have considered linear partial differential equations with constant coefficients of the form 𝐴𝑢_𝑡(𝑡,𝑥)+𝐵𝑢_𝑥𝑥()()(),()𝑡,𝑥+𝐶𝑢𝑡,𝑥=𝑓𝑡,𝑥𝑡,𝑥∈𝐽×Ω, (1) where 𝐽=[0,∞), Ω=[−𝑙,𝑙],𝑙>0, and 𝐴,𝐵,𝐶∈ℝ^𝑛×𝑛. In (1) at least one of the matrices 𝐴,𝐵∈ℝ^𝑛×𝑛 should be singular. If 𝐴=0 or 𝐵=0, then (1) becomes ordinary differential equation or differential algebraic equation, so we assume that none of the matrices 𝐴 or 𝐵 is the zero matrix.

2. Indexes of Partial Differential Algebraic Equation

Let us consider (1), with initial values and boundary conditions given as follows: 𝑢_𝑗()𝑢𝑡,±𝑙=0for𝑡∈𝐽,_𝑖(0,𝑥)=𝑔(𝑥)for𝑥∈Ω, (2) where 𝑗∈𝔐_𝐵𝐶⊆{1,2,…,𝑛}, 𝔐_𝐵𝐶 is the set of indices of components of 𝑢 for which boundary conditions can be prescribed arbitrarily, and 𝑖∈𝔐_𝐼𝐶⊆{1,2,…,𝑛}, 𝔐_𝐼𝐶 is the set of indices of components of 𝑢 for which initial conditions can be prescribed arbitrarily. The initial boundary value problem (IBVP) (1) has only one solution where a function 𝑢 is a solution of the problem, if it is sufficiently smooth, uniquely determined by its initial values (IVs) and boundary values (BVs), and if it solves the LPDAE point wise.

Definition of the indexes can be given using the following assumptions.(i)Each component of the vectors 𝑢,𝑢_𝑡, and 𝑓 satisfy the following condition: 𝑦(𝑡,𝑥)≤𝑀𝑒^𝛼𝑡,𝛼≥0,𝑡≥0, (3) where 𝑀 and 𝛼 are independent of 𝑡 and 𝑥. (ii)(𝐵,𝜉𝐴+𝐶),Re(𝜉)>𝛼, called as the matrix pencil, is regular. (iii)(𝐴,𝜇_𝑘𝐵+𝐶) is regular for all 𝑘, where 𝜇_𝑘 is an eigenvalue of the operator 𝜕²/𝜕𝑥² together with prescribed BCs. (iv)The vector 𝑓(𝑡,𝑥) and the initial vector 𝑔(𝑥) are sufficiently smooth.

If we use Laplace transform, from assumption (ii), (1) can be transformed into 𝐵𝑢_𝜉^′′(𝑥)+(𝜉𝐴+𝐶)𝑢_𝜉(𝑥)=𝑓_𝜉(𝑥)+𝐴𝑔(𝑥),Re(𝜉)>𝛼, (4) if 𝐵 is a singular matrix, then (4) is a DAE depending on the parameter 𝜉. To characterize 𝔐_𝐵𝐶, we introduce 𝑗∈𝔐^(𝜉)_𝐵𝐶⊆{1,2,…,𝑛} as the set of indices of components of 𝑢_𝜉 for which boundary conditions can be prescribed arbitrarily.

In order to define a spatial index, we need the Kronecker normal form of the DAE (4). Assumption (iii) guarantees that there are nonsingular matrices 𝑃_𝐿,𝜉,𝑄_𝐿,𝜉∈ℂ^𝑛×𝑛 such that 𝑃_𝐿,𝜉𝐵𝑄_𝐿,𝜉=𝐼_𝑚100𝑁_𝐿,𝜉,𝑃_𝐿,𝜉(𝜉𝐴+𝐶)𝑄_𝐿,𝜉=𝑅_𝐿,𝜉00𝐼_𝑚2, (5) where 𝑅_𝐿,𝜉∈ℂ^𝑚1×𝑚2 and 𝑁_𝐿,𝜉∈ℝ^𝑚2×𝑚2 is a nilpotent Jordan chain matrix with 𝑚₁+𝑚₂=𝑛.𝐼_𝑘 is the unit matrix of order 𝑘. The Riesz index (or nilpotency) of 𝑁_𝐿,𝜉 is denoted by 𝜈_𝐿,𝜉 (i.e., 𝑁^{𝜈𝐿,𝜉}_𝐿,𝜉=0,𝑁^{𝜈𝐿,𝜉−1}_𝐿,𝜉≠0).

Here, we will assume that there is a real number 𝛼^∗≥𝛼 such that the index set 𝔐^(𝜉)_𝐵𝐶 is independent of the Laplace parameter 𝜉, provided Re(𝜉)≥𝛼^∗.

Definition 1. Let 𝛼^∗∈ℝ⁺ be a number with 𝛼^∗≥𝛼, such that for all 𝜉∈ℂ with Re(𝜉)≥𝛼^∗(1)the matrix pencil (𝐵,𝜉𝐴+𝐶) is regular, (2)𝔐^(𝜉)_𝐵𝐶 is independent of 𝜉, i.e., 𝔐^(𝜉)_𝐵𝐶=𝔐_𝐵𝐶,(3)the nilpotency of 𝑁_𝐿,𝜉 is 𝜈_𝐿≥1.
Then 𝜈_𝑑,𝑥=2𝜈_𝐿−1 is called the “differential spatial index” of the LPDAE. If 𝜈_𝐿=0, then the differential spatial index of LPDAE is defined to be zero.

If we use Fourier transform, (1) can be transformed into 𝐴𝑢^′_𝑘(𝑡)+𝜇_𝑘𝐵+𝐶𝑢_𝑘(𝑡)=𝑓_𝑘(𝑡)+𝐵𝜌_𝑘(𝑡) (6) with 𝜌_𝑘(𝑡)=(𝜌_𝑘1(𝑡),…,𝜌_𝑘𝑛(𝑡))^𝑇 and 𝜌_𝑘𝑖(𝑡)=0for𝑖∈𝔐_𝐵𝐶,𝜌_𝑘𝑗(𝑡)=1𝑙𝜙^′_𝑘(𝑥)𝑢_𝑗(𝑡,𝑥)−𝜙_𝑘(𝑥)𝑢_𝑥,𝑗(𝑡,𝑥)^𝑥=𝑙_𝑥=−𝑙 (7) for 𝑗∉𝔐_𝐵𝐶, which results from partial integration of the term ∫^𝑙_−𝑙𝑢_𝑥𝑥(𝑡,𝑥)𝜙_𝑘(𝑥)𝑑𝑥.

If 𝐴 is a singular matrix, then (6) is a DAE depending on the parameter 𝜇_𝑘 which can be solved uniquely with suitable ICs under the assumptions (iv) and (v). Analogous to the case of the Laplace transform, the above assumption (iv) implies that there exist regular matrices 𝑃_𝐹,𝑘,𝑄_𝐹,𝑘 such that 𝑃_𝐹,𝑘𝐴𝑄_𝐹,𝑘=𝐼_𝑛100𝑁_𝐹,𝑘,𝑃_𝐹,𝑘𝜇_𝑘𝐵+𝐶𝑄_𝐹,𝑘=𝑅_𝐹,𝑘00𝐼_𝑛2. (8) With 𝑅_𝐹,𝑘∈ℝ^𝑛1×𝑛1. 𝑁_𝐹,𝑘∈ℝ^𝑛2×𝑛2 is again a nilpotent Jordan chain matrix with Riesz index 𝜈_𝐹,𝑘, where 𝑛₁+𝑛₂=𝑛.

To characterize 𝔐_𝐼𝐶, we introduce 𝔐^(𝑘)_𝐼𝐶⊆{1,2,…,𝑛} as the set of indices of components of 𝑢_𝑘 for which initial conditions can be prescribed arbitrarily. Therefore, we always assume in the context of a Fourier analysis of 𝑢 that 𝔐^(𝑘)_𝐼𝐶 is independent of 𝑘∈ℕ₊, i.e., 𝔐^(𝑘)_𝐼𝐶=𝔐_𝐼𝐶.

Definition 2. Assume for 𝑘=1,2,… that(1)the matrix pencil (𝐴,𝜇_𝑘𝐵+𝐶) is regular,(2)𝔐^(𝑘)_𝐼𝐶 is independent of 𝑘, i.e., 𝔐^(𝑘)_𝐼𝐶=𝔐_𝐼𝐶,(3)the nilpotency of 𝑁_𝐹,𝑘 is 𝜈_𝐹,𝑘=𝜈_𝐹.
Then the PDAE (1) is said to have uniform differential time index 𝜈_𝑑,𝑡=𝜈_𝐹.

The differential spatial and time indexes are used to decide which initial and boundary values can be taken to solve the problem.

3. Two-Dimensional Differential Transform Method

The two-dimensional differential transform of function 𝑤(𝑥,𝑦) is defined as 𝑊()=1𝑘,ℎ𝜕𝑘!ℎ!^𝑘+ℎ𝑤𝑥,𝑦𝜕𝑥^𝑘𝜕𝑦^ℎ_𝑥=0𝑦=0, (9) where it is noted that upper case symbol 𝑊(𝑘,ℎ) is used to denote the two-dimensional differential transform of a function represented by a corresponding lower case symbol 𝑤(𝑥,𝑦). The differential inverse transform of 𝑊(𝑘,ℎ) is defined as 𝑤𝑥,𝑦=^∞∞_𝑘=0ℎ=0𝑊(𝑘,ℎ)𝑥^𝑘𝑦^ℎ. (10) From (9) and (10), we obtain 𝑤𝑥,𝑦=^∞∞_𝑘=0ℎ=0𝑥^𝑘𝑦^ℎ𝜕𝑘!ℎ!^𝑘+ℎ𝑤𝑥,𝑦𝜕𝑥^𝑘𝜕𝑦^ℎ_𝑥=0𝑦=0. (11) The concept of two-dimensional differential transform is derived from two-dimensional Taylor series expansion, but the method doesn’t evaluate the derivatives symbolically.

Theorem 3. Differential transform of the function 𝑤(𝑥,𝑦)=𝑢(𝑥,𝑦)±𝑣(𝑥,𝑦) is 𝑊(𝑘,ℎ)=𝑈(𝑘,ℎ)±𝑉(𝑘,ℎ), (12) see [17].

Theorem 4. Differential transform of the function 𝑤(𝑥,𝑦)=𝜆𝑢(𝑥,𝑦) is 𝑊(𝑘,ℎ)=𝜆𝑈(𝑘,ℎ), (13) see [17].

Theorem 5. Differential transform of the function 𝑤(𝑥,𝑦)=𝜕𝑢(𝑥,𝑦)/𝜕𝑥 is 𝑊(𝑘,ℎ)=(𝑘+1)𝑈(𝑘+1,ℎ), (14) see [17].

Theorem 6. Differential transform of the function 𝑤(𝑥,𝑦)=𝜕𝑢(𝑥,𝑦)/𝜕𝑦 is 𝑊(𝑘,ℎ)=(ℎ+1)𝑈(𝑘,ℎ+1), (15) see [17].

Theorem 7. Differential transform of the function 𝑤(𝑥,𝑦)=𝜕^𝑟+𝑠𝑢(𝑥,𝑦)/𝜕𝑥^𝑟𝜕𝑦^𝑠 is 𝑊()=()⋯()×()⋯()𝑈(),𝑘,ℎ𝑘+1)(𝑘+2𝑘+𝑟)(ℎ+1ℎ+2ℎ+𝑠𝑘+𝑟,ℎ+𝑠 (16) see [17].

Theorem 8. Differential transform of the function 𝑤(𝑥,𝑦)=𝑢(𝑥,𝑦)⋅𝑣(𝑥,𝑦) is 𝑊(𝑘,ℎ)=^𝑘ℎ_𝑟=0𝑠=0𝑈(𝑟,ℎ−𝑠)𝑉(𝑘−𝑟,𝑠), (17) see [17].

Theorem 9. Differential transform of the function 𝑤(𝑥,𝑦)=𝑥^𝑚𝑦^𝑛 is 𝑊(𝑘,ℎ)=𝛿(𝑘−𝑚,ℎ−𝑛)=𝛿(𝑘−𝑚)𝛿(ℎ−𝑛), (18) see [17], where 𝛿()=𝛿()=𝑘−𝑚1,𝑘=𝑚0,𝑘≠𝑚,ℎ−𝑛1,ℎ=𝑛0,ℎ≠𝑛. (19)

Theorem 10. Differential transform of the function 𝑤(𝑥,𝑦)=𝑔(𝑥+𝑎,𝑦) is 𝑊(𝑘,ℎ)=^𝑁_𝑝=𝑘𝑝𝑘𝑎^𝑝−𝑘𝐺𝑝,ℎ. (20)

Proof. From Definition 1, we can write 𝑤𝑥,𝑦=^∞∞_ℎ=0𝑘=0𝐺(𝑘,ℎ)(𝑥+𝑎)^𝑘𝑦^ℎ()()()()𝑥()𝑦=𝐺0,0+𝐺0,1𝑦+𝑎𝐺1,0+𝐺1,0+𝐺0,2²+𝐺(1,1)𝑥𝑦+𝑎𝐺(1,1)𝑦+𝑎²𝐺(2,0)+2𝑎𝐺(2,0)𝑥+𝐺(2,0)𝑥²+𝐺(0,3)𝑦³+𝐺(1,2)𝑥𝑦²+𝑎𝐺(1,2)𝑦²+𝐺(2,1)𝑥²𝑦+2𝑎𝐺(2,1)𝑥𝑦+𝑎²𝐺(2,1)𝑦+𝑎³𝐺(3,0)+3𝑎²𝐺(3,0)𝑥+3𝑎𝐺(3,0)𝑥²+𝐺(3,0)𝑥³𝑤=𝐺()()+⋯,𝑥,𝑦0,0+𝑎𝐺1,0+𝑎²𝐺(2,0)+𝑎³𝐺()𝐺()()3,0+⋯+𝑥1,0+2𝑎𝐺2,0+3𝑎²𝐺()𝐺()()3,0+⋯+𝑦0,1+𝑎𝐺1,1+𝑎²𝐺(2,1)+⋯+𝑥²[𝐺()()][𝐺()()]2,0+3𝑎𝐺3,0+⋯+𝑥𝑦1,1+2𝑎𝐺2,1+⋯+𝑦²[𝐺()()]=0,2+𝑎𝐺1,2+⋯+⋯_∞𝑝=0𝑎^𝑝𝐺𝑝,0+𝑥^∞_𝑝=1𝑝𝑎^𝑝−1𝐺𝑝,0+𝑦^∞_𝑝=0𝑎^𝑝𝐺𝑝,1+𝑥^2∞_𝑝=2𝑝!𝑝−2!2!×𝑎^𝑝−2𝐺𝑝,0+𝑥𝑦^∞_𝑝=1𝑝𝑎^𝑝−1𝐺𝑝,1+⋯, (21) where 𝑤𝑥,𝑦=^∞∞_ℎ=0^∞_{𝑘=0𝑝=𝑘}𝑝𝑘𝑎^𝑝−𝑘𝐺𝑝,ℎ𝑥^𝑘𝑦^ℎ (22) hence, 𝑊(𝑘,ℎ)=^𝑁_𝑝=𝑘𝑝𝑘𝑎^𝑝−𝑘𝐺𝑝,ℎ. (23)

Theorem 11. Differential transform of the function 𝑤(𝑥,𝑦)=𝑔(𝑥+𝑎,𝑦+𝑏) is 𝑊(𝑘,ℎ)=^𝑁𝑁_{𝑝=𝑘𝑞=ℎ}𝑞ℎ𝑝𝑘𝑎^𝑝−𝑘𝑏^𝑞−ℎ𝐺𝑝,𝑞. (24)

Proof. From Definition 2, we can write 𝑤𝑥,𝑦=^∞∞_ℎ=0𝑘=0𝐺(𝑘,ℎ)(𝑥+𝑎)^𝑘𝑦+𝑏^ℎ()()()()()()𝑥=𝐺0,0+𝐺1,0𝑥+𝑎𝐺1,0+𝐺0,1𝑦+𝑏𝐺0,1+𝐺2,0²+2𝑎𝐺(2,0)𝑥+𝑎²𝐺()()()()()()𝑦2,0+𝐺1,1𝑥𝑦+𝑏𝐺1,1𝑥+𝑎𝐺1,1𝑦+𝑎𝑏𝐺1,1+𝐺0,2²+2𝑏𝐺(0,2)𝑦+𝑏²𝐺()()𝑥0,2+𝐺3,0³+3𝑎𝐺(3,0)𝑥²+3𝑎²𝐺(3,0)𝑥+𝑎³𝐺(3,0)+𝐺(2,1)𝑥²𝑤=𝐺()()()𝑦+⋯𝑥,𝑦0,0+𝑎𝐺1,0+𝑏𝐺0,1+𝑎²𝐺()()+𝐺()()()2,0+𝑎𝑏𝐺1,1+⋯1,0+2𝑎𝐺2,0+𝑏𝐺1,1+3𝑎²𝐺()()𝑥+𝐺()()()3,0+2𝑎𝑏𝐺2,1+⋯0,1+𝑎𝐺1,1+2𝑏𝐺0,2+𝑎²𝐺()=2,1+⋯𝑦+⋯_{𝑁𝑁𝑝=0𝑞=0}𝑎^𝑝𝑏^𝑞𝐺𝑝,𝑞+𝑥^𝑁𝑁_𝑝=1𝑞=0𝑝𝑎^𝑝−1𝑏^𝑞𝐺𝑝,𝑞+𝑦^𝑁𝑁_𝑝=0𝑞=1𝑞𝑎^𝑝𝑏^𝑞−1𝐺𝑝,𝑞+𝑥^2𝑁𝑁_𝑝=2𝑞=0𝑝!𝑎𝑝−2!2!^𝑝−2𝑏^𝑞𝐺𝑝,𝑞+𝑥𝑦^𝑁𝑁_𝑝=1𝑞=1𝑝𝑞𝑎^𝑝−1𝑏^𝑞−1𝐺𝑝,𝑞+⋯. (25) Hence, we can write 𝑤𝑥,𝑦=^∞∞_ℎ=0^𝑁_𝑘=0^𝑁_{𝑝=𝑘𝑞=ℎ}𝑞ℎ𝑝𝑘×𝑎^𝑝−𝑘𝑏^𝑞−ℎ𝐺𝑝,𝑞𝑥^𝑘𝑦^ℎ. (26) Using Definition 2, we obtain 𝑊(𝑘,ℎ)=^𝑁𝑁_{𝑝=𝑘𝑞=ℎ}𝑞ℎ𝑝𝑘𝑎^𝑝−𝑘𝑏^𝑞−ℎ𝐺𝑝,𝑞. (27)

Theorem 12. Differential transform of the function 𝑤(𝑥,𝑦)=𝜕^𝑟+𝑠𝑔(𝑥+𝑎,𝑦+𝑏)/𝜕𝑥^𝑟𝜕𝑦^𝑠 is 𝑊()=()!𝑘,ℎ𝑘+𝑟()!𝑘!ℎ+𝑠×ℎ!_{𝑁𝑁𝑝=𝑘+𝑟𝑞=ℎ+𝑠}𝑞𝑝ℎ+𝑠𝑘+𝑟×𝑎^{𝑝−𝑘−𝑟}𝑏^{𝑞−ℎ−𝑠}𝐺𝑝,𝑞. (28)

Proof. Let 𝐶(𝑘,ℎ) be differential transform of the function 𝑔(𝑥+𝑎,𝑦+𝑏). From Theorem 7, we can write that differential transform of the function 𝑤(𝑥,𝑦) is 𝑊()=()!𝑘,ℎ𝑘+𝑟()!𝑘!ℎ+𝑠𝐶(),ℎ!𝑘+𝑟,ℎ+𝑠 (29) from Theorem 4, we can write 𝐶(𝑘+𝑟,ℎ+𝑠)=^𝑁𝑁_{𝑝=𝑘+𝑟𝑞=ℎ+𝑠}𝑞𝑝ℎ+𝑠𝑘+𝑟×𝑎^{𝑝−𝑘−𝑟}𝑏^{𝑞−ℎ−𝑠}𝐺𝑝,𝑞. (30) If we substitute (30) into (29), we find 𝑊()=()!𝑘,ℎ𝑘+𝑟()!𝑘!ℎ+𝑠×ℎ!_{𝑁𝑁𝑝=𝑘+𝑟𝑞=ℎ+𝑠}𝑞𝑝ℎ+𝑠𝑘+𝑟×𝑎^{𝑝−𝑘−𝑟}𝑏^{𝑞−ℎ−𝑠}𝐺𝑝,𝑞. (31)

4. Application

We have considered the following PDAE as a test problem: 𝑢1100^𝑡+𝑢−10−10^𝑥𝑥+[)[],11−10𝑢=𝑓,𝑡∈0,∞,𝑥∈−1,1 (32) with initial values and boundary values 𝑢₁(0,𝑥)=𝑥³−𝑥,𝑢₂(0,𝑥)=𝑥⁴𝑢−1,₁(𝑡,1)=𝑢₁(𝑡,−1)=0,𝑢₂(𝑡,1)=𝑢₂(𝑡,−1)=0. (33) The right hand side function 𝑓 is 𝑓=𝑥⁴−1(cos𝑡−sin𝑡)−6𝑥𝑒^−𝑡,−𝑒^−𝑡𝑥³+5𝑥^𝑇, (34) and the exact solutions are 𝑢₁(𝑡,𝑥)=𝑥³−𝑥𝑒^−𝑡,𝑢₂(𝑡,𝑥)=𝑥⁴−1cos𝑡. (35) If nonsingular matrices 𝑃_𝐹,𝑘, 𝑄_𝐹,𝑘, 𝑃_𝐿,𝜉, and 𝑄_𝐿,𝜉 are chosen such as 𝑃_𝐹,𝑘=1−𝜇_𝑘𝜇_𝑘01−1𝜇_𝑘−1,𝑄_𝐹,𝑘=,𝑃0−11−1_𝐿,𝜉=011𝜉+1−1𝜉+1𝜉+1,𝑄_𝐿,𝜉=,−𝜉−10𝜉1 (36) matrices 𝑃_𝐹,𝑘𝐴𝑄_𝐹,𝑘 and 𝑃_𝐿,𝜉𝐵𝑄_𝐿,𝜉 are found as 𝑃_𝐹,𝑘𝐴𝑄_𝐹,𝑘=1000,𝑃_𝐿,𝜉𝐵𝑄_𝐿,𝜉=1000. (37) From (38), we have 𝑁_𝐿,𝜉=0 and 𝑁_𝐹,𝑘=0. Then the PDAE (32) has differential spatial index 1 and differential time index 1. So, it is enough to take 𝔐^(𝜉)_𝐵𝐶={1} and 𝔐^(𝑘)_𝐼𝐶={2} to solve the problem.

Taking differential transformation of (32), we obtain (𝑘+1)𝑈₁(𝑘+1,ℎ)+(𝑘+1)𝑈₂()−()𝑈𝑘+1,ℎℎ+1)(ℎ+2₁(𝑘,ℎ+2)+𝑈₁(𝑘,ℎ)+𝑈₂(𝑘,ℎ)=𝐹₁(),−()𝑈𝑘,ℎℎ+1)(ℎ+2₁(𝑘,ℎ+2)−𝑈₁(𝑘,ℎ)=𝐹₂(𝑘,ℎ). (38)(39) The Taylor series of functions 𝑓₁ and 𝑓₂ about 𝑥=0,𝑡=0 are 𝑓₁()1𝑡,𝑥=−1+𝑡−6𝑥+2𝑡²−1+6𝑥𝑡6𝑡³−3𝑥𝑡²−1𝑡24⁴+𝑥⁴+𝑥𝑡³−𝑥⁴1𝑡+𝑡120⁵−14𝑥𝑡⁴+1𝑡720⁶−12𝑥⁴𝑡²+120𝑥𝑡⁵+16𝑥⁴𝑡³−1𝑡5040⁷−1120𝑥𝑡⁶𝑓+⋯,₂()5𝑡,𝑥=−5𝑥+5𝑥𝑡−2𝑥𝑡²−𝑥³+56𝑥𝑡³+𝑥³1𝑡−2𝑥³𝑡²−524𝑥𝑡⁴+16𝑥³𝑡³+124𝑥𝑡⁵−1144𝑥𝑡⁶−1𝑥24³𝑡⁴+⋯. (40)(41) The values 𝐹₁(𝑘,ℎ)and𝐹₂(𝑘,ℎ) in (39) and (40) are coefficients of polynomials (41) and (42). If we use Theorem 3 for boundary values, we obtain ⁷_𝑖=0𝑈₁𝑗,𝑖=0,𝑗=0,1,…,7,_7𝑖=0(−1)^𝑖𝑈₁𝑗,𝑖=0,𝑗=0,1,…,7. (42)(43) In order to write 𝑘=0 and ℎ=0,1,2,3,4,5 in (40), we have 2𝑈₁(0,2)+𝑈₁(0,0)=0,20𝑈₁(0,5)+𝑈₁(0,3)=1,6𝑈₁(0,3)+𝑈₁(0,1)=5,30𝑈₁(0,6)+𝑈₁(0,4)=0,12𝑈₁(0,4)+𝑈₁(0,2)=0,42𝑈₁(0,7)+𝑈₁(0,5)=0. (44) If we take 𝑗=0 in (43) and (44), we obtain 𝑈₁(0,0)+𝑈₁(0,1)+𝑈₁(0,2)+𝑈₁(0,3)+𝑈₁(0,4)+𝑈₁(0,5)+𝑈₁(0,6)+𝑈₁()𝑈0,7=0,₁(0,0)−𝑈₁(0,1)+𝑈₁(0,2)−𝑈₁(0,3)+𝑈₁(0,4)−𝑈₁(0,5)+𝑈₁(0,6)−𝑈₁(0,7)=0. (45) From (45) and (46), we find 𝑈₁(0,0)=0,𝑈₁(0,1)=−1,𝑈₁()𝑈0,2=0,₁(0,3)=1,𝑈₁(0,4)=0,𝑈₁()𝑈0,5=0,₁(0,6)=0,𝑈₁(0,7)=0. (46) In this manner, from (40), (44), and (45), the coefficients of the 𝑢₁ are obtained as follows: 𝑈₁(1,0)=0,𝑈₁(1,1)=1,𝑈₁()𝑈1,2=0,₁(1,3)=−1,𝑈₁(1,4)=0,𝑈₁()𝑈1,5=0,₁(1,6)=0,𝑈₁(2,0)=0,𝑈₁()12,1=−2,𝑈₁(2,2)=0,𝑈₁()=12,32,𝑈₁()𝑈2,4=0,₁(2,5)=0,𝑈₁(3,0)=0,𝑈₁()=13,16,𝑈₁(3,2)=0,𝑈₁()13,3=−6,𝑈₁()𝑈3,4=0,₁(4,0)=0,𝑈₁()14,1=−24,𝑈₁()𝑈4,2=0,₁()=14,324,𝑈₁(5,0)=0,𝑈₁()=15,1,𝑈120₁(5,2)=0,𝑈₁(6,0)=0,𝑈₁()16,1=−,𝑈720₁(7,0)=0. (47) Using the initial values for the second component, we obtain the following coefficients: 𝑈₂(0,0)=−1,𝑈₂(0,1)=0,𝑈₂()𝑈0,2=0,₂(0,3)=0,𝑈₂(0,4)=1,𝑈₂()𝑈0,5=0,₂(0,6)=0,𝑈₂(0,7)=0. (48) The coefficients of the 𝑢₂ can be found using (47), (48), (49), and taking 𝑘=0,1,2,… and ℎ=0,1,2,… in (39) as follows: 𝑈₂(1,2)=0,𝑈₂(1,3)=0,𝑈₂()𝑈1,4=0,₂(1,5)=0,𝑈₂(1,6)=0,𝑈₂()𝑈2,1=0,₂(2,2)=0,𝑈₂(2,3)=0,𝑈₂()12,4=−2,𝑈₂(2,5)=0,𝑈₂(3,0)=0,𝑈₂()𝑈3,1=0,₂(3,2)=0,𝑈₂(3,3)=0,𝑈₂()𝑈3,4=0,₂(4,1)=0,𝑈₂(4,2)=0,𝑈₂()𝑈4,3=0,₂()14,0=−24,𝑈₂(5,0)=0,𝑈₂()𝑈5,1=0,₂(5,2)=0,𝑈₂()=16,0720,𝑈₂()𝑈6,1=0,₂(7,0)=0. (49) If we write the above values in (39) and (40), then we have 𝑢^∗₁()1𝑡,𝑥=−𝑥+𝑥𝑡−2𝑥𝑡²+𝑥³+16𝑥𝑡³−𝑥³1𝑡+2𝑥³𝑡²−124𝑥𝑡⁴−16𝑥³𝑡³+1120𝑥𝑡⁵−1720𝑥𝑡⁶+1𝑥24³𝑡⁴𝑢+⋯,_∗2()1𝑡,𝑥=−1+2𝑡²−1𝑡24⁴+𝑥⁴+1𝑡720⁶−12𝑥⁴𝑡²+⋯. (50)(51) Numerical and exact solution of the given problem has been compared in Tables 1 and 2, and simulations of solutions have been depicted in Figures 1, 2, 3, and 4, respectively.

Table 1 

The numerical and exact solution of the test problem (32), where 𝑢₁(𝑡,𝑥) is the exact solution and 𝑢^∗₁(𝑡,𝑥) is the numerical solution, for 𝑥=0.1.

Table 2 

The numerical and exact solution of the test problem (32), where 𝑢₂(𝑡,𝑥) is exact solution and 𝑢^∗₂(𝑡,𝑥) is numerical solution, for 𝑥=0.1.

Figure 1 

The graphic of the function 𝑢1(𝑡,𝑥) in the test problem (32).

Figure 2 

The graphic of the function 𝑢∗1(𝑡,𝑥) in the test problem (32).

Figure 3 

The graphic of the function 𝑢2(𝑡,𝑥) in the test problem (32).

Figure 4 

The graphic of the function 𝑢∗2(𝑡,𝑥) in the test problem (32).

5. Conclusion

The computations associated with the example discussed above were performed by using Computer Algebra Techniques [24]. We show the results in Tables 1 and 2 for the solution of (32) by numerical method. The numerical values on Tables 1 and 2 obtained above are in full agreement with the exact solutions of (32). This study has shown that the differential transform method often shows superior performance over series approximants, providing a promising tool for using in applied fields.