Numerical Methods for Partial Differential Equations

Received: 12 September 2017

Revised: 30 October 2017

Accepted: 1 November 2017

DOI: 10.1002/num.22236

RESEARCH ARTICLE

Solutions of time-fractional Tricomi and Keldysh equations of Dirichlet functions types in Hilbert space Omar Abu Arqub Department of Mathematics, Faculty of Science, Al-Balqa Applied University, Salt 19117, Jordan Correspondence Omar Abu Arqub, Department of Mathematics, Faculty of Science, Al-Balqa Applied University, Salt 19117, Jordan. Email: [email protected]

The subject of the fractional calculus theory has gained considerable popularity and importance due to their attractive

applications in widespread fields of physics and engineering. The purpose of this research article is to present results on the numerical simulation for time-fractional Tricomi and Keldysh equations of Dirichlet functions types in Hilbert space that were found in the transonic flows. Those resulting mathematical models are solved using the reproducing kernel algorithm which provide appropriate solutions in term

of infinite series formula. Convergence analysis, error estimations, and error bounds under some hypotheses which

provide the theoretical basis of the proposed algorithm are also discussed. The dynamical properties of these numerical solutions are discussed and the profiles of several representative numerical solutions are illustrated. Finally, the prospects of the gained results and the algorithm are discussed through academic validations. KEYWORDS

Dirichlet functions types, Keldysh equation, reproducing kernel algorithm, time-fractional partial differential equations, Tricomi equation

1

I N T R O D U CTI ON

The field of time-fractional partial differential equations (PDEs) is a major link between mathematics and its applications. Time-fractional PDEs govern a broad spectrum of physical phenomena such as multiphase fluid flow problems, electromagnetic theory, diffusive transport, and electrical networks [1–14]. Developing analytical and numerical algorithms for the solutions of time-fractional PDEs is Numer Methods Partial Differential Differential Eq. 2018;34:1759–1780. wileyonlinelibrary.com/journal/num 2017;00:1–22. wileyonlinelibrary.com/journal/num

© 2017 Wiley Periodicals, Inc.Inc. 1759 © 2017 Wiley Periodicals, 1

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a very important task, and fast computations is essential to reach the results we seek to make the right decision in due time. Anyhow, attempts have been made to propose analytical methods that approximate the exact solutions of such time-fractional PDEs. The aim of this article is to present results on the numerical simulation for classes of time-fractional PDEs such as those found in the transonic flows. Such classes yield of the mathematical models which describe a compressible two-phase flow in the framework of the theory of thermodynamically compatible hyperbolic-elliptic systems of conservation laws. The models are independent of the type of the numerical methods used to resolve them and can be applied for different real-world applications including hyperbolic and elliptic basic type, hyperbolic, and parabolic degeneracy. More specifically, the governing equations presented in this work were solved numerically by well-defined numerical methods such as the spectral collocation method [15], the finite element method [16], and the element free Galerkin method [17]. Verification and simulations become very easy on a computer so they are more than necessary in applied and other apllied sciences. Anyhow, the reproducing kernel algorithm (RKA) has been investigated systematically for the numerical solutions of the following set of PDEs: • The time-fractional Tricomi equation: ∂tαα u(x, t) + ξ(t)∂x22 u(x, t) = f (x, t).

(1)

• The time-fractional Keldysh equation: ζ (t)∂tαα u(x, t) + ∂x22 u(x, t) = f (x, t).

(2)

Here, Equations 1 and 2 are subject to the following set of Dirichlet functions types: ⎧ ⎪ u(x, 0) = ω1 (x), ⎪ ⎪ ⎪ ⎨∂ u(x, 0) = ω (x), t 2 ⎪ u(0, t) = υ1 (t), ⎪ ⎪ ⎪ ⎩ u(L, t) = υ2 (t).

(3)

To covering this research precisely, the following assumptions for Equations 1–3 will be assumed: • (x, y) ∈ [0, L] × [0, T ] and α ∈ (1, 2],

• f is continuos real-valued function in the given domain, • u is an unknown function to be determined,

• ω1 (x), ω1 (x), υ1 (t), and υ2 (t) are continuos real-valued functions in the given domain,

• ∂x = ∂/∂x, ∂x22 = ∂ 2 /∂x 2 , and so on,

• Equations 1–3 have unique analytical solutions over the rectangle [0, L] × [0, T ],

• ∂tαα denote the Caputo time-fractional derivative of order α. In this aspect, ∂tαα = ∂ α /∂t α and ∂tαα u(x, t)

1 = �(2 − α)

�

0

t

(t − τ )1−α ∂τ22 u(x, τ )dτ ,

0 < τ < t, 1 < α < 2.

(4)

A brief survey of the RK theory and the RKA can be summarized shortly as follows: this theory was used first at 1907 as a solver for harmonic and biharmonic Dirichlet problems [18]. In 1950, it

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was established by knitting it with the RK functions [19]. This theory, which is proxy in the RKA, has been coming out well used to diverse application in applied mathematics and engineering modeling [20–23]. Newly, large scale researches has been applied the RKA for the solutions of several integral and differential operators alongside with their theories [24–44]. After this brief introduction, the organization of the article is as follows: • For clarity of presentation, in Section 2, several Hilbert spaces and several RK functions are listed, • To formulate the solution, in Section 3, two extended Hilbert spaces and two extended RK functions are fitted, • In Section 4, problem formulation and computational algorithm are presented,

• In Section 5, essential theoretical results are presented to fixtures the stable numerical solution, • To capture the behavior of the numerical solution, error estimations and error bounds are derived in Section 6, • Numerical outcomes are discussed to demonstrate the accuracy and the applicability of the presented algorithm as utilized in Section 7, • Finally, in Section 8, concluding remarks are listed.

2 R E Q U I R E M ENTS F ROM THE REPRO DUCING K ERNEL T HE O R Y The main idea in this section is to present several inner product spaces that satisfying the constraints Dirichlet functions types of the given time-fractional PDEs. To do so, some basic requirements are presented. Let K be a Hilbert space of real-valued functions defined on a nonempty set �. A function � : � × � → R is a reproducing kernel of K if it satisfies the following: • For each t ∈ �, �(·, t) ∈ K,

• For each θ ∈ K and each t ∈ �, �θ (·), �(·, t)�K = θ(t).

A Hilbert space which possesses a reproducing kernel is called a reproducing kernel Hilbert space (RKHS). Through this section, we denote �z�2 = �z(∗), z(∗)� , where z ∈ , ∗ ∈ [0, 1], and ˆ 21 , W22 , W23 }.  ∈ {W21 , W Definition 1 ([24]) Suppose that z� is in L 2 [0, T ]. The space W21 [0, T ] is defined as W21 [0, T ] = {z = z(t) : z is absolutely continuous function on [0, T ]}. Whilst, its metric system structure lie in �z1 (t), z2 (t)�W 1 = z1 (0)z2 (0) + 2



T 0

z1� (t)z2� (t)dt.

(5)

Theorem 1 ([24]) The Hilbert space W21 [0, T ] is a complete RK with structure kernel function Rs{1} (t) = 1 + min(s, t).

(6)

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ˆ 21 [0, L] can In the simiar fashion, if [0, L] is the domain in the x direction, then the space W  L space {1} � � ˆ be defined as �z1 (x), z2 (x)�Wˆ 1 = z1 (0)z2 (0) + 0 z1 (x)z2 (x)dx and Ry (x) = 1 + min(x, y). 2

Definition 2 ([25]) Suppose that z�� is in L 2 [0, T ]. The space W23 [0, T ] is defined as W23 [0, T ] = {z = z(t) : z, z� , z�� are absolutely continuous functions on [0, T ] and z(0) = z� (0) = 0}. While, its metric system structure lie in �z1 (x), z2 (x)�W 3 = 2

2  i=0

z1(i) (0)z2(i) (0)

+



T 0

z1�� (t)z2�� (t)dt.

(7)

Theorem 2 ([25]) The Hilbert space W23 [0, T ] is a complete RK with structure kernel function Rs{3} (t)

=



1 2 3 t (t − 5t 2 s + 10s2 (t + 3)), 120 1 2 3 s (s − 5s2 t + 10t 2 (s + 3)), 120

t ≤ s,

t > s.

(8)

ˆ 23 [0, L] is defined as Definition 3 ([26]) Suppose that z��� is in L 2 [0, L]. The space W ˆ 23 [0, L] = {z = z(x) : z, z� , z�� are absolutely continuous functions on [0, L] and z(0) = W z(L) = 0}. While, its metric system structure lie in �z1 (x), z2 (x)�Wˆ 3 = 2

1  i=0

z1(i) (0)z2(i) (0) + z1 (L)z2 (L) +



L 0

z1��� (x)z2��� (x)dx.

(9)

ˆ 23 [0, L] is a complete RK with structure kernel Theorem 3 ([26]) The Hilbert space W function Rˆ y{3} (x)

=



y (�1 (x, y) 120L 2 x (�1 (y, x) 120L 2

+ L 2 �2 (x, y) + 5L�3 (x, y)), x ≤ y,

+ L 2 �2 (y, x) + 5L�3 (y, x)), x > y.

(10)

in which its symmetric piecewise internal structure is given as �1 (x, y) = x 2 y(6L 3 + 120 − x 3 − y3 ), �2 (x, y) = y(y3 − 10x 3 ) − 5x(−24 + y3 ), �3 (x, y) = x(y(x 3 − 24) + x(y3 − 24)).

3

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F I TTI N G APP ROPRI ATE I NNER PRO DUCT S PACES

Here, two extended inner product spaces and two extended RK functions are fitted to formulate the solutions in the mentioned spaces. Henceforth, we denote the following markers: � = [0, L] ⊗ [0, T ], i+j ∂xi t j = (∂ i /∂x i )(∂ j /∂t j ), whenever i, j = 1, 2 and �u�2 = �u(∗, ◦), u(∗, ◦)� , where u ∈ , ∗, ◦ ∈ �,  ∈ {H, W }. Definition 4 Suppose that ∂x33 ∂t33 u is in L 2 (�). The Hilbert space W (�) can be defined as W (�) = {u = u(x, t) : ∂x22 ∂t22 u is complete continuous function in � and u(x, 0) =

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∂t u(x, 0) = u(0, t) = u(L, t) = 0}. Whilst, its metric system structure lie in �u1 (x, t), u2 (x, t)�W =

2 

j

2

j=0

+ +

j

�∂t j u1 (x, 0), ∂t j u2 (x, 0)�Wˆ 3



T

0



T

0

 

1  j=0

0

L

j j ∂t22 ∂xj u1 (0, t)∂t22 ∂xj u2 (0, t)

+



∂t22 u1 (L, t)∂t22 u2 (L, t)

dt

∂x33 ∂t22 u1 (x, t)∂x33 ∂t22 u2 (x, t)dxdt.

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Theorem 4 The Hilbert space W (�) is a complete RK with structure kernel function R(y,s) (x, t) = Rˆ y{3} (x)Rs{3} (t),

(13)

such that for any u(x, t) ∈ W (�), we have �u(x, t), R(y,s) (x, t)�W = u(y, s) and R(y,s) (x, t) = ˆ 23 [0, L] and R(x,t) (y, s), where Rˆ y{3} (x) and Rs{3} (t) are the RK functions of the spaces W 3 W2 [0, T ], respectively. ˆ 23 [0, L] and W23 [0, T ] with respect Proof Using properties of the inner product spaces W to the differentials dx and dt, respectively, one can find �u(x, t), Rˆ y{3} (x)Rs{3} (t)�W =

2 

+

0

T

T 0

L

+



dt



dt

∂t22 u(L, t)∂t22 Rˆ y{3} (L)Rs{3} (t)

∂x33 ∂t22 u(x, t)∂x33 ∂t22 Rˆ y{3} (x)Rs{3} (t)dxdt

 

1  j=0

0

L

j j ∂t22 ∂xj u(0, t)∂t22 Rs{3} (t)∂xj Rˆ y{3} (0)

+

∂t22 u(L, t)Rˆ y{3} (L)∂t22 Rs{3} (t)

∂x33 ∂t22 u(x, t)∂x33 Rˆ y{3} (x)∂t22 Rs{3} (t)dxdt

j j ∂t j �u(x, 0), Rˆ y{3} (x)�Wˆ 3 ∂t j Rs{3} (0) 2



j=0

j=0

j j ∂t22 ∂xj u(0, t)∂t22 ∂xj Rˆ y{3} (0)Rs{3} (t)

2

j=0

2 



1 

0

0





j j �∂t j u(x, 0), Rˆ y{3} (x)∂t j Rs{3} (0)�Wˆ 3



2 

+ =

T

j=0

+

T

0

2 

+

=





j

2

j=0

+

=

j

�∂t j u(x, 0), ∂t j Rˆ y{3} (x)Rs{3} (0)�Wˆ 3

T

0 j

∂t22 Rs{3} (t)∂t22 j



1  j=0

j j ∂xj u(0, t)∂xj Rˆ y{3} (0)

∂t j u(y, 0)∂t j Rs{3} (0) +



T 0

+

u(L, t)Rˆ y{3} (L)

+

∂t22 Rs{3} (t)∂t22 �u(x, t), Rˆ y{3} (x)�Wˆ 3 dt 2



L 0

∂x33 u(x, t)∂x33 Rˆ y{3} (x)dx



dt

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=

2  j=0

j

j

∂t j u(y, 0)∂t j Rs{3} (0) +



T 0

∂t22 Rs{3} (t)∂t22 u(y, t)dt

= �u(y, t), Rs{3} (t)�Wˆ 3 = u(y, s).

(14)

2

Thus, �u(x, t), R(y,s) (x, t)�W = u(y, s). While conversely, R(y,s) (x, t) = �R(y,s) (ξ , ζ ), R(x,t) (ξ , ζ )�W = �R(x,t) (ξ , ζ ), R(y,s) (ξ , ζ )�W = R(x,t) (y, s). Definition 5 Suppose that ∂x ∂t u is in L 2 (�). The Hilbert space H(�) can be defined as H(�) = {u = u(x, t) : u is complete continuous function in �}. While, its metric system structure lie in �u1 (x, t), u2 (x, t)�H = �u1 (x, 0), u2 (x, 0)�Wˆ 1 2   T ∂t u1 (0, t)∂t u2 (0, t)dt + +

T

0

0



0

L

∂xt2 u1 (x, t)∂xt2 u2 (x, t)dxdt.

(15)

Theorem 5 The Hilbert space H(�) is a complete RK with structure kernel function r(y,s) (x, t) = Rˆ y{1} (x)Rs{1} (t),

(16)

such that for any u(x, t) ∈ H(�), we have �u(x, t), r(y,s) (x, t)�H = u(y, s) and r(y,s) (x, t) = ˆ 21 [0, L] and W21 [0, T ], r(x,t) (y, s), where Rˆ y{1} (x) and Rs{1} (t) are the RK functions of spaces W respectively. Proof Similar to the proof of Theorem 4.

4 P R O B L E M F ORMULATI ON AND CO MPUTATIO NAL A L G O R I TH M To set the stage for solving the time-fractional PDEs of Equations 1–3, it is necessary first to formulate it in a manner not only reflecting the situation being modeled, but so as to be amenable to computational technique. Through the remainder sections, we will use the following markers: P = P(x,t), and Pk = P(x k ,t k ) whenever k = 1,2,3,...,∞. Let us consider the following general form of time-fractional PDE that described completely Equations 1 and 2: ζ (t)∂tαα u(x, t) + ξ(t)∂x22 u(x, t) = f (x, t).

(17)

Note that, when ζ (t) = 1, then Tricomi equation well be obtained and when ξ(t) = 1, then Keldysh equation well be obtained. To apply the RKA, we must homogenized the nonhomogeneous constraints conditions by suitable transformations. This normalizing is needed to put the mentioned conditions into the space W(�) to make an agreement feasible between the constructed spaces and the corresponding RK functions which are on �, for the convenience, we still denote the solution of the new equation by u(x,t). So, let ζ (t)∂tαα u(x, t) + ξ(t)∂x22 u(x, t) = P(x, t),

(18)

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subject to the following Dirichlet functions types: ⎧ ⎪ u(x, 0) = 0, ⎪ ⎪ ⎪ ⎨∂ u(x, 0) = 0, t ⎪ u(0, t) = 0, ⎪ ⎪ ⎪ ⎩ u(L, t) = 0.

(19)

For the conduct of proceedings, we define the fractional differential linear operator �:W (�)→H(�) such that �u(x, t) := ζ (t)∂tαα u(x, t) + ξ(t)∂x22 u(x, t).

(20)

Thus, the time-fractional PDE of Equation 18 is governed by the following equivalent functional equation: �u(x, t) = P(x, t).

(21)

To build an orthogonal function systems of the space W (�), we choose a countable dense subset ∗ ∗ ((xi , ti ))∞ i=1 in �, define . and ψi (x, t) = � ϕi (x, t), where � : H(�) → W (�) is the adjoint operator of � and is uniquely determined. � �∞ The normalized orthonormal function systems ψ i (x, t) i=1 of W (�) is usually constructed from the process of the Gram-Schmidt orthogonalization of {ψi (x, t)}∞ i=1 as ψ i (x, t) =

i �

μik ψk (x, t).

(22)

k=1

To apply the RKA, we divide the finite domain � into a p × q mesh point with the space step size �x = Lp in the x direction of [0, L] and the time step size �t = Tq in the t direction of [0, T ], respectively, in which p and q are positive integers. Anyhow the grid points (xl , tm ) in the space-time domain � are defined simultaneously as (xl , tm ) = (l�x, m�t),

l = 0, 1, 2, · · · , p, m = 0, 1, 2, · · · , q.

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The numerical solvability of Tricomi and Keldysh equations presented in this work can be done by applying the following algorithm. The steps in the algorithm are explained in more detail in the next sections. Algorithm 1 To approximate the solution un (x, t) of u(x, t) for Eq. (21) and (19) on �, do the following main steps. Step 1: Choose n = pq collocation points in the domain �; Step 2: Set ψi (xi , ti ) = �(y,s) R(x, t)|(y,s)=(xi ,ti ) ; Step 3: Obtain the orthogonalization coefficients μik ; � Step 4: Set ψ i (xi , ti ) = il=1 μik ψi (xi , ti ), i = 1, 2, . . . , n; Step 5: Choose an initial approximation u0 (x1 , t1 ); Step 6: Set i = 1; � Step 7: Set Bi = il=1 μik Pk ; � � Step 8: Set ui (xi , ti ) = ik=1 nj=1 Bk ψ k (xk , tk ); Step 9: If i < η, then set i = i + 1 and go to step 7, else stop.

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5

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F I X T U R E S OF THE STABLE NUMERICAL S O LUTIO N

 ∞ In this section, we will show that {ψi (x, t)}∞ i=1 is a complete function system, while R(xi ,ti ) (x, t) i=1 is a linearly independent in the space W (�). After that, an iterative formulas of obtaining the numerical solutions are utilized. Through the remainder sections, we suppose that the sequence {(xi , ti )}∞ i=1 is dense on �. At first, depending on the the Schwarz inequality it is easy to see that � : W (�) → H(�) is a bounded linear operator, that is ��u(x, t)�2W 1 ≤ M�u�2W with M > 0. For more details see the proof 2 of Lemma 1 in [27]. Lemma 1 The sequence {ψi (x, t)}∞ i=1 is a complete function system in W (�) with ψi (x, t) = �(y,s) R(x, t)|(y,s)=(xi ,ti ) .

(24)

Proof Here, �(y,s) indicates that the operator � applies to the function of (y, s). Indeed ψi (x, t) = �∗ ϕi (x, t)

= ��∗ ϕi (y, s), R(x,t) (y, s)�W

= �ϕi (y, s), �(y,s) R(x,t) (y, s)�H = �(y,s) R(x,t) (y, s)|(y,s)=(xi ,ti )

= �(y,s) R(y,s) (x, t)|(y,s)=(xi ,ti ) ∈ W (�).

(25)

Now, for each fixed u ∈ W (�), let �u(x, t), ψi (x, t)�W = 0, i = 1, 2, . . . . Then, �u(x, t), ψi (x, t)�W = �u(x, t), �∗ ϕi (x, t)�W = ��u(x, t), ϕi (t)�H = �u(xi , ti ) = 0. While, −1 {(xi , ti )}∞ i=1 is dense on �, we must have �u(x, t) = 0 from the existence of � , it follows that u = 0. ∞  Theorem 6 The sequence R(xi ,ti ) (x, t) i=1 is a linearly independent in W (�).

 m Proof It is adequate to show that R(xi ,ti ) (x, t) i=1 is a linearly independent for each  m ≥ 1. In fact, if {ci }mi=1 satisfies mi=1 ci R(xi ,ti ) (x, t) = 0, taking hk (x, t) ∈ W (�) such that hk (xl , tl ) = δl,k for each l = 1, 2, · · · , m. Then 

0 = hk (x, t) = =

m  i=1

m  i=1

m  i=1

 ci R(xi ,ti ) (x, t)

W

ci �hk (x, t), R(xi ,ti ) (x, t)�W ci hk (xi , ti ) = ck .

(26)

Thus ck = 0 or k = 1, 2, · · · , m. In the next theorem, μik are the orthogonalization coefficients of the orthonormal systems ψ ij (t) and these basis can be obtained directly from the Gram-Schmidt orthogonalization process of Equation 22.

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Theorem 7 Suppose that Ai = 21 and 19, then

i

k=1

μik Pk . If u ∈ W (�) is the solution of Equations ∞ 

u(x, t) =

Ai ψ i (x, t).

(27)

i=1

 u ∈ W (�), while, ∞ Proof As, �u(x, t), ϕi (x, t)�W = u(x i=1 Ai ψ i (x, t) is  i , ti ) foreach ∞ the Fourier series expansion about ψ i (x, t) i=1 , then it is a convergent in the sense of � · �W . Thus, u(x, t) = = = = = In other words,

∞

i=1

i ∞   i=1 k=1

∞  

�u(x, t), ψ i (x, t)�W ψ i (x, t)

u(x, t),

i=1

i ∞   i=1 k=1

i ∞  

i  k=1

 μik ψk (x, t) ψ i (x, t) W

μik �u(x, t), �∗ ϕk (x, t)�W ψ i (x, t) μik �u(xk , tk )ψ i (x, t)

i=1 k=1

∞ 

Ai ψ i (x, t).

(28)

i=1

Ai ψ i (x, t) is the exact solution.

For numerical computations, put (x1 , t1 ) = (0, 0), then from the constraints conditions of Equation 19, the value of u(x1 , t1 ) is known. Set u0 (x1 , t1 ) = u(x1 , t1 ) and define the n-term numerical solution of u(x, t) using the truncating version as un (x, t) =

n 

Ai ψ i (x, t).

(29)

i=1

 In order that W (�) is a Hilbert space, then the series ∞ i=1 Ai ψ i (x, t) < ∞. Thus, we can guarantee that the numerical solution un (x, t) satisfies the constraints conditions of Equation 19.

6

C O N V E R GENCE ANALYSI S AND ERRO R ES TIMATIO NS

In this part, the theoretical basis of the proposed algorithm under some hypotheses for the obtained approximation is studied in terms of convergence and error estimates in the space W (�). Because �un �W is bounded, {(xi , ti )}∞ i=1 is dense on �, and the solution of Equations 21 and 19 is unique. Then the n-term numerical solution un (x, t) must converges to the exact solution u(x, t) as n → ∞. j

Theorem 8 The partial derivatives of the numerical solution ∂xi i ∂t j un (x, t) are conj verging uniformly to the partial derivatives of the exact solution ∂xi i ∂t j u(x, t), whenever i = 0, 1, 2, j = 0, 1 as n → ∞.

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Proof As W (�) is a Hilbert space, from Equations 21 and 19, it is follows that, �u − un �W → 0 as n → ∞. Again, as j

j

j

|∂xi i ∂t j u(x, t) − ∂xi i ∂t j un (x, t)| = |�u(y, s) − un (y, s), ∂xi i ∂t j �R(x,t) (y, s)�W | j

≤ �u − un �W �∂xi i ∂t j �R(x,t) (y, s)�W

≤ Mi,j �u − un �W . j

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j

Thus, |∂xi i ∂t j u(x, t) − ∂xi i ∂t j un (x, t)| → 0 as n → ∞.

For errors estimations, the following assumptions have been made in defining the proof of requested L T two next theorems: hx = n−1 with xi = (i − 1)hx and ht = n−1 with tj = (j − 1)ht , and n is a positive integer, where i = 1, 2, . . . , n and j = 1, 2, . . . , n. In the case of two variables functions type, we may use a norm that represents the maximum absolute value at each point, that is �u(x, t)�∞ = max(x,t)∈� |u(x, t)|. Theorem 9 Let u(x, t) and un (x, t) be the exact and the numerical solutions of Equations 21 and 19, respectively. If u(x, t) − un (x, t) is the nature error at (x, t) ∈ � in the space W (�), then there is a positive constant C such that �u(x, t) − un (x, t)�∞ ≤ C(hx2 + hx ht ).

(31)

Proof In each subdomain [xi , xi+1 ] × [tj , tj+1 ] ⊂ �, one can write ∂x u(x, t) − ∂x un (x, t) = ∂x u(x, t) − ∂x u(xi , tj ) + ∂x un (xi , tj ) − ∂x un (x, t) + ∂x u(xi , tj ) − ∂x un (xi , tj ). (32) Expand ∂x u(x, t) about the point (xi , tj ) using Taylor series of function of two variables, we get ∂x u(x, t) = ∂x u(xi , tj ) + (hx ∂x + ht ∂t )∂x u(xi + ζ hx , tj + ζ ht ) + · · · , ζ ∈ [0, 1].

(33)

By means of the continuation of ∂xt u and ∂x22 u on �, one can show that �∂x u(x, t) − ∂x un (xi , tj )�∞ = O(hx + ht ).

(34)

On the other aspect as well, we know that |∂x un (xi , tj ) − ∂x un (x, t)| ≤



xi

x

|∂y22 un (y, tj )|dy

+



t tj

|∂xs un (x, s)|ds.

(35)

In the means of ∞ th norm, it follows that �∂x un (xi , tj ) − ∂x un (x, t)�∞ = O(hx + ht ).

(36)

Again, given any ε > 0, using Theorem 8, there is an n sufficiently large such that �∂x u(xi , tj ) − ∂x un (xi , tj )�∞ < ε.

(37)

ARQUB

1769 11

As ε is arbitrary and by combining Equations 32–37 for the chosen value of n, we must have �∂x u(x, t) − ∂x un (x, t)�∞ = O(hx + ht ).

(38)

Using the integral property for differentiable functions, we have u(x, t) − un (x, t) = u(xi , t) − un (xi , t) +



xi

x

(∂y u(y, t) − ∂y un (y, t))dy.

(39)

Using Equations 38 and 39 and applying Theorem 8, we can straightforwardly see that �u(x, t) − un (x, t)�∞ ≤ C(hx2 + hx ht ),

(40)

where C > 0. Theorem 10 Let u(x, t) and un (x, t) be the exact and the numerical solutions of Equations 20 and 18, respectively. If u(x, t) − un (x, t) is nature error at (x, t) ∈ � in the space W (�) with ∂x33 ∂t u(x, t), ∂x22 ∂t22 u(x, t) ∈ (�) such that �∂x33 ∂t u(x, t)�∞ and �∂x22 ∂t22 u(x, t)�∞ are bounded, then there is a positive constant D such that �u(x, t) − un (x, t)�∞ ≤ D(ht hx3 + ht2 hx2 ).

(41)

Proof In each subdomain [xi , xi+1 ] × [tj , tj+1 ] ⊂ �, one can write ∂x22 ∂t u(x, t) − ∂x22 ∂t un (x, t) = ∂x22 ∂t u(x, t) − ∂x22 ∂t u(xi , tj )

+ ∂x22 ∂t un (xi , tj ) − ∂x22 ∂t un (x, t) + ∂x22 ∂t u(xi , tj ) − ∂x22 ∂t un (xi , tj ).

(42)

Expand ∂x22 ∂t u(x, t) bout the point (xi , tj ) using Taylor series of function of two variables, we get ∂x22 ∂t u(x, t) = ∂x22 ∂t u(xi , tj ) + (hx ∂x + ht ∂t )∂x22 ∂t u(xi + ζ hx , tj + ζ ht ) + · · · , ζ ∈ [0, 1].

(43)

By means of the continuation of ∂x33 ∂t u and ∂x22 ∂t22 u on �, one can show that �∂x22 ∂t u(x, t) − ∂x22 ∂t un (xi , tj )�∞ = O(hx + ht ).

(44)

On the other aspect as well, one can write |∂x22 ∂t un (xi , tj ) − ∂x22 ∂t un (x, t)| ≤



xi

x

|∂x33 ∂t un (y, tj )|dy +



t tj

|∂x22 ∂s22 un (x, s)|ds.

(45)

In the means of ∞ th norm, it follows that �∂x22 ∂t un (xi , tj ) − ∂x22 ∂t un (x, t)�∞ = O(hx + ht ).

(46)

1770 12

ARQUB ARQUB

Again, given any ε > 0, using Theorem 8, there is an n sufficiently large such that �∂x22 ∂t u(xi , tj ) − ∂x22 ∂t un (xi , tj )�∞ < ε.

(47)

As ε is arbitrary and by combining Eqs. (42–47) for the chosen value of n, we must have �∂x22 ∂t u(x, t) − ∂x22 ∂t un (x, t)�∞ = O(hx + ht ).

(48)

Using the integral property for differentiable functions, we have ∂x ∂t u(x, t) − ∂x ∂t un (x, t) = ∂x ∂t u(xi , t) − ∂x ∂t un (xi , t) + ∂t u(x, t) − ∂t un (x, t) = ∂t u(xi , t) − ∂t un (xi , t) + u(x, t) − un (x, t) = u(x, ti ) − un (x, ti ) +



t ti



x

xi



xi

x

(∂y22 ∂t u(y, t) − ∂y22 ∂t un (y, t))dy.

(∂y ∂t u(y, t) − ∂y ∂t un (y, t))dy.

(∂s u(x, s) − ∂s un (x, s))ds.

(49) (50) (51)

Using Equations 48 and 51 and applying Theorem 8, we can straightforwardly see that �u(x, t) − un (x, t)�∞ ≤ D(ht hx3 + ht2 hx2 ),

(52)

where D > 0. Theorem 11 Let u(x, t) and un (x, t) be the exact and the numerical solutions of Equations 21 and 19, respectively. If u(x, t) − un (x, t) is the nature error at (x, t) ∈ � in the space W (�), then {u(x, t) − un (x, t)}∞ n=1 is decreasing in the sense of the norm of W (�) and �u(x, t) − un (x, t)�W → 0 as n → ∞. Proof

Using prevoius results described in this article, one can write �u(x, t) −

un (x, t)�2W

=

∞ 

i=n+1



i  k=1

μik Pk

2

.

(53)

Consequently {u(x, t) − un (x, t)}∞ n=1 is decreasing in the sense of the norm of � · �W . ∞ In addition, since i=1 Ai ψ i (x, t) is convergent. Thus, �u(x, t) − un (x, t)�2W → 0 as n → ∞.

7

N U M E R I CAL RESULTS

In numerical analysis problems there are some basic unknowns. If they are found, the behavior of the entire structure can be predicted. The basic unknowns or the field variables which are encountered are displacements in the applied mathematics and engineering problems. This section presents the numerical solutions for some of one-two- dimensional Tricomi and Keldysh equations of fractional order using the RKA. The RKA presented in this article is used to numerically solve the given Tricomi and Keldysh equations. The input data to the algorithm is divided into two parts; the RKA related parameters and the time-fractional PDEs related parameters. Anyhow, the input data to the algorithm are as follows:

ARQUB

1771 13

Parameter or concept � = [0, L] ⊗ [0, T ] p = 25 q = 25 �x = L/25 �t = T /25 (xl , tm ) = (l�x, m�t) α ∈ (0, 1] ω1 (x) ω2 (x) υ1 (t) υ2 (t) R(y,s) (x, t) = Ry{3} (x)Rs{2} (t) r(y,s) (x, t) = Rˆ y{1} (x)Rs{1} (t) μik

Description Dirct product between the x direction domain and the t direction domain The number of mishpoint in the x direction domain of [0, L] The number of mishpoint in the t direction domain of [0, T ] The space step size in the x direction domain of [0, L] The time step size in the t direction domain of [0, T ] The two-dimensional partition of the domain � in un (xl , tm ) The order of the fractional derivative in the Caputo sense The function form of initial condition at t = 0 The function form of derivative of initial condition at t = 0 The function form of Dirichlet condition at x = 0 The function form of Dirichlet condition at x = L The structure of the RK function of the Hilbert space W (�) The structure of the RK function of the Hilbert space H(�) The orthogonalization coefficients of the orthonormal systems ψ ij (t)

Next, throughout the numerical computations; some graphical results, tabulate data, and numerical comparison are presented and discussed quantitatively at some selected grid points on � to illustrate the numerical solutions for the following sets of Tricomi and Keldysh equations.

7.1 Solutions of time-fractional one-dimensional Tricomi and Keldysh equations Here, we present two examples on time-fractional Tricomi and Keldysh equations on a finite domain � = [0, 1]2 to experiment the efficiency and applicability of the RKA. Example 1

Consider the following time-fractional Tricomi equation: ∂tαα u(x, t) − et ∂x22 u(x, t) = f (x, t),

(54)

subject to the following Dirichlet functions types: ⎧ ⎪ u(x, 0) = 0, ⎪ ⎪ ⎪ ⎨∂ u(x, 0) = 0, t ⎪ u(0, t) = 0, ⎪ ⎪ ⎪ ⎩ u(1, t) = 0,

(55)

where (x, t) ∈ � and α ∈ (1, 2]. Here, the exact solution is 5

u(x, t) = t 1.5α (x − x 2 ) . Example 2

(56)

Consider the following time-fractional Keldysh equation: cos(t)∂tαα u(x, t) − ∂x22 u(x, t) = f (x, t),

(57)

1772 14

ARQUB ARQUB

T AB LE 1 Absolute errors of approximating the solution in Example 1 over the domain � x

t

.2

.25 .5 .75 1

.4

.25 .5 .75 1

.6

.25 .5 .75 1

.8

.25 .5 .75 1

α = 1.25

5.0371511342 × 10−5 6.7451574099 × 10

−5

4.8002996405 × 10

−5

7.4176077574 × 10

−5

9.5123895875 × 10

−5

5.2477342420 × 10

−5

3.5390736187 × 10−5 9.1720802309 × 10

−5

5.2407221976 × 10

−5

5.9468159080 × 10

−5

2.1915285678 × 10−5 8.5014553545 × 10

−5

7.6471343756 × 10

−5

9.3381685129 × 10

−5

2.1450234362 × 10−5 1.5664618355 × 10

−5

α = 1.5

6.1071033630 × 10−6 9.6240869104 × 10

−6

6.0195664876 × 10

−6

3.9201491091 × 10

−6

7.5571137336 × 10

−6

6.4288283759 × 10

−6

6.5333880368 × 10

−6

6.7922469209 × 10

−6

9.2438982281 × 10

−6

7.7992676696 × 10

−6

9.6265796295 × 10−6 8.1898823219 × 10

−6

7.4674961637 × 10−6 9.0891095380 × 10

−6

7.2125156323 × 10−6 6.5365381091 × 10

−6

α = 1.75

9.6390353713 × 10−7 5.2952153856 × 10

−7

6.6371942625 × 10

−6

9.2931210619 × 10

−7

1.7631848555 × 10

−6

6.7843236093 × 10

−7

3.7297224772 × 10

−6

7.7571548168 × 10

−6

7.6744655209 × 10

−6

7.4123825088 × 10

−7

8.9691068614 × 10−7 6.2530082870 × 10

−7

8.4785665056 × 10−6 8.1402491901 × 10

−7

9.5310612896 × 10−7 9.2209942638 × 10

−7

α=2

6.8097620130 × 10−7 5.3023455411 × 10−7 7.5913206807 × 10−7

3.1769964682 × 10−7

5.9946572210 × 10−7 2.3715909804 × 10−7 9.5003498558 × 10−7 3.1897299009 × 10−7 7.4447828291 × 10−7

4.9270687832 × 10−7 5.1055962776 × 10−7 3.4794895229 × 10−7

5.5687107850 × 10−7 1.8027141527 × 10−7 6.0276809859 × 10−7 7.7356131697 × 10−7

subject to the following Dirichlet functions types: ⎧ ⎪ u(x, 0) = 0, ⎪ ⎪ ⎪ ⎨∂ u(x, 0) = 0, t ⎪ u(0, t) = t 3α−1.5 , ⎪ ⎪ ⎪ ⎩ u(1, t) = 0,

(58)

where (x, t) ∈ � and α ∈ (1, 2]. Here, the exact solution is u(x, t) = t 3α−1.5 cos(2.5π x).

(59)

Remark 1 In the previous examples, the function f (x, t) is chosen, so that, the given exact solution is satisfied in the both LHS and RHS of the mentioned equation over the domain �. With a view to demonstrate the agreement between the exact and the RKA approximate solutions, Tables 1 and 2 show the absolute error of approximate solution of Examples 1 and 2, respectively, obtained at various (x, t) in � when α ∈ {1.25, 1.5, 1.75, 2}. From the tables, it can be seen that the error estimate show that the accuracy of the numerical solutions is closely related to the fill time tm = m�t, m = 0, 1, . . . , q, where �t = q1 and fill distance xl = l�x, l = 0, 1, . . . , p, where �x = p1 . So, more accurate numerical solutions can be obtained using more mesh points. The geometric behaviors of the memory and hereditary properties of the RKA solutions and their level characteristics are investigated next. Anyhow, the comparisons of between the computational values of the RKA solutions for different values of α ∈{1.25,1.5,1.75,2} for Example 1 have been depicted on the domain � as shown in Figure 1, while, Figure 2 shows the comparisons of for Example 2.

ARQUB

1773 15

T AB L E 2 x .2

t .25 .5 .75 1

.4

.25 .5 .75 1

.6

.25 .5 .75 1

.8

.25 .5 .75 1

Absolute errors of approximating the solution in Example 2 over the domain � α = 1.25

6.9084952600 × 10 3.4987946026 × 10

−5 −5

9.1870150528 × 10

−5

3.3398495324 × 10

−5

9.0317811277 × 10−5 7.3184639007 × 10

−5

5.0592778839 × 10

−5

1.4171164453 × 10

−5

8.7816023520 × 10

−5

7.0341320910 × 10

−5

7.8210524435 × 10−5 7.6701961020 × 10

−5

7.5768292263 × 10−5 8.6045091087 × 10

7.3315578340 × 10

−5 −5

5.8766689271 × 10−5

α = 1.5

2.9167127623 × 10

−6

9.0730688038 × 10

−6

4.1600063087 × 10

−6

6.9665115955 × 10

−6

6.9665554113 × 10

−6

8.6981800475 × 10

−6

8.3266675557 × 10

−6

8.1245953783 × 10

−6

7.5820146434 × 10

−6

8.1054226630 × 10−6 5.3266970061 × 10

−6

4.4191301247 × 10−6 7.9905773607 × 10

−6

2.0954711139 × 10−6 9.5668066502 × 10

−6

9.4105778789 × 10−6

α = 1.75

5.9580314543 × 10

−7

8.4387702272 × 10

−6

3.0479375480 × 10

−7

9.4059590141 × 10

−7

5.2892768970 × 10−6 6.2874183115 × 10

−7

3.8648448787 × 10

−6

9.5210421270 × 10

−7

3.4517701570 × 10−6 3.3742037925 × 10

−7

5.9462014745 × 10

−6

3.3096343993 × 10

−6

8.5346778544 × 10

−7

7.4164302486 × 10−6 9.0242036313 × 10

−7

7.3867786081 × 10−7

α=2

6.7753537408 × 10−7 7.9039799715 × 10−7 7.0823031725 × 10−7 8.9650552931 × 10−7

9.5865893520 × 10−7 6.2533184700 × 10−7 7.6917663063 × 10−7

2.6294381112 × 10−7 7.1562432205 × 10−7 7.4334871393 × 10−7

6.4249959690 × 10−7 4.0612281055 × 10−7 8.8858427609 × 10−7

7.4963957030 × 10−7 9.1770414599 × 10−7 7.8929437749 × 10−7

F IGU R E 1 Comparisons of between the computational values of the RKA solutions at various values of (x, t) ∈ � when α ∈ {1.25, 1.5, 1.75, 1} for Example 1: black α = 2; blue: α = 1.75; green: α = 1.5; red: α = 1.25 [Color figure can be viewed at wileyonlinelibrary.com]

1774 16

ARQUB ARQUB

F IGU R E 2 Comparisons of between the computational values of the RKA solutions at various values of

(x, t) ∈ � when α ∈ {1.25, 1.5, 1.75, 1} for Example 2: black α = 2; blue: α = 1.75; green: α = 1.5; red: α = 1.25 [Color figure can be viewed at wileyonlinelibrary.com]

It is clear from the Figures 1 and 2 that each of the graphs are nearly coinciding and similar in their behaviors with good agreement with the RKA solutions when the ordinary derivatives are considered. Whilst, one can note that the RKA solutions are continuously depends on the time-fractional derivatives assumed. As a fast scan, the process described by the RKA is slightly more skewed to the down and drag to the right than that modeled by the standard equations. In the meantime, from the graphs, it can be seen that, the RKA solutions are stable and convergent.

7.2 Solutions of time-fractional two-dimensional Tricomi and Keldysh equations Here, we present two examples on time-fractional Tricomi and Keldysh equations on a finite domain � = [0, 1]3 to experiment the efficiency and applicability of the RKA. Example 3 Consider the following time-fractional Tricomi equation: ∂tαα u(x, y, t) − t 2 (∂x22 u(x, y, t) + ∂y22 u(x, y, t)) = f (x, y, t),

(60)

ARQUB

1775 17

subject to the following Dirichlet functions types: ⎧ ⎪ u(x, y, 0) = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ∂t u(x, y, 0) = 0, ⎪ ⎪ ⎪ 2 ⎨u(0, y, t) = t 2 e −10 α [0.25+(y−0.5) ] , −10 2 ⎪ u(1, y, t) = t 2 e α [0.25+(y−0.5) ] , ⎪ ⎪ ⎪ −10 2 ⎪ ⎪ u(x, 0, t) = t 2 e α [0.25+(x−0.5) ] , ⎪ ⎪ ⎪ −10 ⎩ 2 u(x, 1, t) = t 2 e α [0.25+(x−0.5) ] ,

(61)

where (x, y, t) ∈ � and α ∈ (1, 2]. Here, the exact solution is u(x, y, t) = t 2 e Example 4

−10 [(x−0.5)2 +(y−0.5)2 ] α

.

(62)

Consider the following time-fractional Keldysh equation: t∂tαα u(x, y, t) − (∂x22 u(x, y, t) + ∂y22 u(x, y, t)) = f (x, y, t),

(63)

subject to the following Dirichlet functions types: ⎧ ⎪ u(x, y, 0) = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ∂t u(x, y, 0) = 0, ⎪ ⎪ ⎪ ⎨u(0, y, t) = 0, ⎪ u(1, y, t) = t 2 sin2 (απ )sin2 (απ y), ⎪ ⎪ ⎪ ⎪ ⎪ u(x, 0, t) = 0, ⎪ ⎪ ⎪ ⎩ u(x, 1, t) = t 2 sin2 (απ x)sin2 (απ ),

(64)

where (x, y, t) ∈ � and α ∈ (1, 2]. Here, the exact solution is u(x, y, t) = t 2 sin2 (απ x)sin2 (απ y). Remark 2 In the previous examples, the function f (x, y, t) is chosen, so that, the given exact solution is satisfied in the both LHS and RHS of the mentioned equation over the domain �. Again, to demonstrate the agreement between the exact and the RKA approximate solutions, Tables 3 and 5 show the absolute error of approximate solutions of Examples 3 and 4, respectively, obtained at various (x, y, 0.5) in � when α ∈ {1.25, 1.5, 1.75, 2}, whilst, Tables 4 and 6 show the comparisons of for Examples 3 and 4 at various (x, y, 1) in �. From the tables, it can be seen that the error estimate show that the accuracy of the numerical solutions is closely related to the fill time tm = m�t, m = 0, 1, . . . , h, where �t = h1 , fill x-distance xl = l�x, l = 0, 1, . . . , p, where �x = p1 , and fill y-distance yk = k�y, l = 0, 1, . . . , q, where �y = q1 . Finally, the comparisons of between the computational values of the RKA solutions for different values of α ∈{1.25,1.5,1.75,2} for Example 3 have been depicted on the domain � at t = .5 as shown in Figure 3, while, Figure 4 show the comparisons of for Example 4 on the same domain at t = .5.

(65)

1776 18

ARQUB ARQUB

T AB LE 3 Absolute errors of approximating the solution in Example 3 over the domain  at t = 0.5 x .2

y .25 .5 .75

.4

.25 .5 .75

.6

.25 .5 .75

.8

.25 .5 .75

T AB LE 4 x .2

y .25 .5 .75

.4

.25 .5 .75

.6

.25 .5 .75

.8

.25 .5 .75

α = 1.25

3.5773526583 × 10

−4

6.8702501178 × 10

−5

2.7558927334 × 10

−4

6.8345999506 × 10

−4

5.8279953239 × 10

−4

6.3452768239 × 10

−4

−4

9.6433370954 × 10

7.5077755774 × 10−4 −4

9.6045654602 × 10

8.6880998271 × 10−4 −4

8.4074350111 × 10

5.5064350404 × 10−4

α = 1.5

8.5978274601 × 10

−5

7.6070390568 × 10

−5

4.9602594260 × 10

−5

9.5255284591 × 10

−5

6.1099374815 × 10

−5

4.0946009260 × 10

−5

9.3018487112 × 10

−5

4.3716899715 × 10−5 3.6678280620 × 10

−5

6.9481140252 × 10−5 6.3670689333 × 10

−5

6.0000825814 × 10−5

α = 1.75

4.9694538134 × 10

−6

9.5594847890 × 10

−6

8.1557771493 × 10

−6

9.5514572633 × 10

−6

7.2979496282 × 10

−6

3.7168111292 × 10

−6

6.6307527399 × 10−6 3.2592747420 × 10

−6

8.0268235476 × 10−6 7.2691900501 × 10 3.8501788839 × 10

−6 −6

2.8625793971 × 10−6

α=2

5.2451779132 × 10−7 9.2213445288 × 10−7

7.9494555904 × 10−6 2.8164979166 × 10−6 5.1905236746 × 10−6 7.7222900125 × 10−6 5.7157820766 × 10−6 4.5331764758 × 10−6 6.4168109743 × 10−6

7.8857061154 × 10−6

8.7291501622 × 10−7 7.4858674797 × 10−7

Absolute errors of approximating the solution in Example 3 over the domain  at t = 1 α = 1.25

7.0658536245 × 10

−4

4.5452435209 × 10

−4

6.3106548941 × 10

−4

2.8977402407 × 10

−4

5.1025162097 × 10

−4

8.4911510873 × 10

−4

6.7978249432 × 10

−4

6.8101995751 × 10−4 8.3307912576 × 10

−4

6.1816945149 × 10−4 8.3628239416 × 10

−4

8.3317016860 × 10−4

α = 1.5

6.4317206922 × 10 9.2408012108 × 10

−5 −5

9.0687754545 × 10

−4

6.9716379596 × 10

−4

7.5795546401 × 10

−4

5.4464858234 × 10

−4

7.2475178019 × 10−4 5.2059846648 × 10

−4

7.3469483250 × 10−4 9.7184083279 × 10

−6

4.5697871930 × 10

−6

9.5185780021 × 10−6

α = 1.75

3.4269903241 × 10

−6

3.4304338881 × 10

−5

8.9616549346 × 10

−5

7.7984902605 × 10

−5

5.0661624899 × 10

−5

6.3387281012 × 10

−6

9.1992494079 × 10

−6

8.4530064960 × 10−5 4.8903819492 × 10

−5

6.0419366506 × 10−5 7.3907821393 × 10

−5

7.8127986046 × 10−6

α=2

2.1586616172 × 10−6 9.5832680194 × 10−6 3.6729919268 × 10−6 9.4417155039 × 10−6

9.2994797004 × 10−6 3.6757645631 × 10−6 4.9572419551 × 10−6 2.0367870599 × 10−6 9.0871290281 × 10−6 8.9552802086 × 10−6 9.3907124648 × 10−6 8.4018068121 × 10−6

T AB LE 5 Absolute errors of approximating the solution in Example 4 over the domain  at t = 0.5 x .2

y .25 .5 .75

.4

.25 .5 .75

.6

.25 .5 .75

.8

.25 .5 .75

α = 1.25

6.5916694548 × 10

−4

6.5237984509 × 10

−4

2.9363024016 × 10

−4

4.5723291883 × 10

−4

8.0879902480 × 10

−4

8.8405206695 × 10−4 9.8540538024 × 10

−4

8.6483987342 × 10

−4

6.5624388639 × 10

−4

2.3029079284 × 10

−4

9.8228212991 × 10−4 3.1307174600 × 10

−4

α = 1.5

3.3378591094 × 10

−5

3.1366509365 × 10

−5

7.1737678720 × 10

−5

2.1116016400 × 10

−5

2.9994509827 × 10

−5

9.2938618127 × 10

−5

5.3418171740 × 10

−5

2.8994171014 × 10

−5

3.2276249604 × 10−5 5.9010560780 × 10

−5

8.1986069832 × 10−5 7.7517335436 × 10

−5

α = 1.75

5.9612838383 × 10

−6

4.5272718104 × 10

−6

8.6385184313 × 10

−6

3.3347161804 × 10

−5

4.3239651147 × 10

−5

4.2345602024 × 10

−5

8.5152496622 × 10

−5

8.8663565293 × 10−5 9.9753715022 × 10

−5

7.6481589131 × 10−5 8.0850295017 × 10 1.1168962299 × 10

−6 −6

α=2

5.5880563478 × 10−6 9.9893072377 × 10−6 7.7212361136 × 10−6 3.0987008196 × 10−6

8.9909933542 × 10−6 6.3346461542 × 10−6 3.8278094554 × 10−6

7.8414265510 × 10−6 4.3739816879 × 10−6 6.4534266331 × 10−6 9.9982346547 × 10−6 7.5021470619 × 10−6

ARQUB

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T AB LE 6 x

y

.2

.25 .5 .75

.4

.25 .5 .75

.6

.25 .5 .75

.8

.25 .5 .75

Absolute errors of approximating the solution in Example 4 over the domain � at t = 1 α = 1.25

7.7702733224 × 10−4 5.9759406266 × 10

−4

3.1382554609 × 10

−4

8.6468998701 × 10

−4

2.9421560909 × 10

−4

4.6086895198 × 10

−4

3.8489325267 × 10

−4

6.4000536266 × 10

−4

8.8186751598 × 10−4 2.3873893864 × 10

−4

5.9810934932 × 10−4 9.4265748266 × 10

−4

α = 1.5

4.7671621772 × 10−5 5.8802996916 × 10

−5

9.5627442954 × 10

−4

5.9071369603 × 10

−4

4.5490284899 × 10

−4

4.7929596099 × 10

−4

8.0829878624 × 10

−4

7.2448733258 × 10

−5

8.5796041965 × 10−4 5.5702893734 × 10

−4

5.3265684840 × 10−5 1.1509109562 × 10

−5

α = 1.75

8.5901225613 × 10−5 4.8892644965 × 10

−5

3.7786325593 × 10

−5

9.6844369375 × 10

−5

8.4021089586 × 10

−5

2.4236310888 × 10

−5

2.5820369334 × 10

−5

4.5268873821 × 10

−5

4.4074493029 × 10−5 9.4810367003 × 10

−5

9.5837208656 × 10−5 7.6368661965 × 10

−5

α=2

8.1016129691 × 10−6 7.4309983643 × 10−6 7.6811613345 × 10−6 5.5373125966 × 10−6 8.7759289673 × 10−6 1.3957421929 × 10−6 7.6063986091 × 10−6

6.3757032040 × 10−6 7.1643452255 × 10−6 1.1382468846 × 10−6

8.3948570340 × 10−6 4.0523175497 × 10−6

F IGU R E 3 Comparisons of between the computational values of the RKA solutions at various values of (x, y, 0.5) ∈ �

when α ∈ {1.25, 1.5, 1.75, 1} for Example 3: black α = 2; blue: α = 1.75; green: α = 1.5; red: α = 1.25 [Color figure can be viewed at wileyonlinelibrary.com]

8

C O N C L U DI NG REMARKS

The mathematical models in hand have been applied successfully to investigate the relative velocity between gas flows and nearly sonic speeds as demonstrated in this paper. Simulation results are

1778 20

ARQUB ARQUB

F IGU R E 4 Comparisons of between the computational values of the RKA solutions at various values of (x, y, 0.5) ∈ �

when α ∈ {1.25, 1.5, 1.75, 1} for Example 4: black α = 2; blue: α = 1.75; green: α = 1.5; red: α = 1.25 [Color figure can be viewed at wileyonlinelibrary.com]

systematically validated through a series of numerical test cases. Strong evidence showed that the model and the method are accurate, robust and conservative. Following the success of the current research presented in this article along with its publication within the largest applied mathematics and computation community, both the model and numerical tools remain open for further modeling and applications involving multiphase flow problems. The model currently is used to simulate future systems performance for a specific problem related to hyperbolic and elliptic basic type, hyperbolic and parabolic degeneracy. These new simulation capabilities will increase the international community confidence and regulatory acceptance of scientific predictions generated through mathematical modelling and numerical simulations presented in this article.

ACKNOWLEDGME N T The author would like to express his gratitude to the unknown referees for carefully reading the paper and their helpful comments.

ORCID Omar Abu Arqub

http://orcid.org/0000-0001-9526-6095

ARQUB

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How to Tricomi and and Keldysh Keldysh equations equationsof of to cite citethis thisarticle: article: Arqub Arqub OA. OA.Solutions Solutions of oftime-fractional time-fractional Tricomi Dirichlet functions inin Hilbert space. Numer. Methods Partial Differential Eq. 2018;34:1759–1780. Dirichlet functionstypes types Hilbert space. Numer. Methods Partial Differential Eq. 2017;00:1–22. https://doi.org/10.1002/num.22236 https://doi.org/10.1002/num.22236