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Pakchin et al. / IJOCTA, Vol.15, No.3, pp.535-548 (2025)
                                                              following formula Equation (41) was used to cal-
                              L                               culate the convergence order:
                           1  X
                 k+1                   µ p              2
               |R q  | ≤ P 1    b 1 (µ p )b (∆t) 2−µ p  + P 2 h
                                       1
                          L                                                            e(h, ∆t)
                             p=0                                         Rate h = log (        ),
                                                                                     2
                                                                                         h
                       L                                                               e( , ∆t)
                                                                                         2
                    1  X        ν p              2                                                       (41)
               + P 3      b 2 (ν p )b (∆t) 4−ν p  + P 4 h                               e(h, ∆t)
                                2
                    L                                                    Rate ∆t = log (        ).
                      p=0                                                             2  e(h,  ∆t )
                                                                                             2
                     α
               + P 5 b (∆t) 4−α  + P 6 ∆t
                                                              5.1. Example 1
                ≤ b 1 P 1 max {b 1 (µ)}(∆t) 2−µ L  + P 2 h 2
                      µ∈(0,1)                                 Consider Equation (42):
               + b 2 P 3 max {b 2 (ν)}(∆t) 4−ν L  + P 4 h 2
                      ν∈(1,2)                                      Z  1
                     α    4−α                                                  C  µ
               + P 5 b (∆t)   + P 6 ∆t                                 Γ(5 − µ) D u(x, t)dµ
                                                                                  t
                                                                    0
                                                       (37)          Z  2
            in which it can be defined as in Equation (38):        +     Γ(5 − ν) D u(x, t)dν =          (42)
                                                                                 C
                                                                                    ν
                                                                                    t
                                                                       1
                                                                           α          2
                          2
                        ∂ u(x q , t)                               − (−∆) 2 u(x, t) + u (x, t) + g(x, t)
                                                  ′′
             P 1 = max |          |, P 2 = max |F 1 |,
                  t∈[0,T]  ∂t 2           µ∈(0,1)             where in Equation (43):
                            4
                         1 ∂ U(x q , t)
                                                     ′′
             P 3 = max |             |, P 4 = max |F 2 |,                             4    2
                  t∈[0,T] 24  ∂t 4           ν∈(1,2)           g(x, t) = 24(x(1 − x)) 3 (t − t )  − t (x(1 − x)) 6
                                                                                                8
                             4
                   1        ∂ U                                                        ln t
                                                                                        απ
             P 5 =    max |     (x, t k )|,                    + g 1 (x, t)g 1 (x, t) = (2cos(  )) −1 4
                                                                                               t
                  24 x∈[0,L] ∂x 4                                                        2
                               ∂f                                   Γ(4)
             P 6 =   max     |{   (x q , ζ k+1 , U(x q , ζ k+1 ))  ×       (x 3−α  + (1 − x) 3−α )
                  ζ k+1 ∈[t k ,t k+1 ]  ∂t                         Γ(4 − α)
                    ∂f                      ∂U                     3Γ(5)
                  +    (x q , ζ k+1 , U(x q , ζ k+1 ))  (x q , ζ k+1 )}|,  −  (x 4−α  + (1 − x) 4−α )
                    ∂U                       ∂t                   Γ(5 − α)
                                        ν p
                        µ p
             b 1 = max{b }, b 2 = max{b }                          3Γ(6)
                   p    1           p   2                      +          (x 5−α  + (1 − x) 5−α )
                                                       (38)       Γ(6 − α)
                             α
            and b α = max α {b }. Then, we get Equation (39):       Γ(7)
                                                                −         (x 6−α  + (1 − x) 6−α )
                                                                 Γ(7 − α)
                  k+1            2−µ L    2       4−ν L                                                  (43)
              ∥ R    ∥ ∞ ≤ C((∆t)     + h + (∆t)
                                                       (39)   with conditions in Equation (44):
              + (∆t) 4−α  + (∆t))
                                              α
            where C = max{b 1 P 1 , P 2 , b 2 P 3 , P 4 , b P 5 , P 6 , }.  u(x, 0) = 0, x ∈ (0, 1),
                                                                                                         (44)
                                                                      u(0, t) = 0, u(1, t) = 0, t ∈ (0, 1]
            5. Illustrative examples                              The exact solution is u(x, t) = t (x(1 −
                                                                                                     4
                                                                 3
            Some numerical examples are demonstrated in       x)) . We solved this problem using the numer-
            this part to show the effectiveness of the studied  ical method developed in this study. The method
            numerical approach. The absolute errors for these  was applied with L = 20, M = N = 30, and
            illustrative examples at different points (x, t) are  various values of α. Figure 1 presents the ap-
            computed as follows in Equation (40):             proximate solutions obtained by our method for
                                                              these parameter values and different choices of α.
                                               k
                       |e(x q , t k )| = |u(x q , t k ) − u |,  (40)  This figure provides a visual representation of how
                                               q
                       k = 1, . . . , N, q = 1, . . . , M,    the solution evolves across the spatial domain and
                                                              time for different choices of α, offering insight into
                                   k
            in which u(x q , t k ) and u show the exact and ap-  the influence of the fractional order on the over-
                                   q
            proximate solutions, respectively. Additionally,  all behavior of the solution. Figure 2 compares
            all numerical simulations presented in this study  the exact solution with the approximate solution
            were conducted using Mathematica software (ver-   at L = 20, M = N = 30, and various choices of
            sion 11). The computations were performed on a    α in u(x, 0.5). Figure 3 illustrates the absolute
            laptop equipped with an Intel Core i5 processor   error functions for L = 20, M = N = 30, and
            (2.40 GHz) and 16.00 GB of RAM. Moreover, the     various choices of α. These graphs illustrate the
                                                           542
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