IHJPAS. 36(1)2023 

345 
 This work is licensed under a Creative Commons Attribution 4.0 International License 

 

   

 

 

Improved Runge-Kutta Method for Oscillatory Problem Solution Using 

Trigonometric Fitting Approach 

 

 

 
 

 

 

Abstract 

  This paper provides a four-stage Trigonometrically Fitted Improved Runge-Kutta (TFIRK4) 

method of four orders to solve oscillatory problems, which contains an oscillatory character in the 

solutions. Compared to the traditional Runge-Kutta method, the Improved Runge-Kutta (IRK) 

method is a natural two-step method requiring fewer steps. The suggested method extends the 

fourth-order Improved Runge-Kutta (IRK4) method with trigonometric calculations. This 

approach is intended to integrate problems with particular initial value problems (IVPs) using the 

set functions 𝑒𝜔𝑥 and 𝑒−𝜔𝑥  for trigonometrically fitted. To improve the method's accuracy, the 

problem primary frequency 𝜔 ∈ 𝑅 is used. The novel method is more accurate than the 

conventional Runge-Kutta method and IRK4. Several test problems for the system of first-order 

ordinary differential equations carry out numerically to demonstrate the effectiveness of this 

approach. The computational studies show that the TFIRK4 approach is more efficient than the 

existing Runge-Kutta methods. 

Keywords: Improved Runge-Kutta Method, Trigonometrically-Fitted, Initial Value Problem, 

Oscillating Solution. 

1. Introduction  

  This paper focuses on solving the system of the first-order ordinary differential equations (ODEs) 

of the form: 

                        𝑢′(𝑡) = 𝑓(𝑡, 𝑢), 𝑢(𝑡0) = 𝑢0.                                                                         (1) 

  These issues are frequently seen in various applied sciences, including quantum mechanics, 

electronics, chemical physics, and astronomy (see [1] and [2]). Traditionally, Runge-Kutta (RK) 

doi.org/10.30526/36.1.2963 

Article history: Received 29 July 2022, Accepted 1 September 2022, Published in January 2023. 

 

 

 

Ibn Al-Haitham Journal for Pure and Applied Sciences 

Journal homepage: jih.uobaghdad.edu.iq 

 

Kasim Abbas Hussain 

Department of Mathematics, College of Sciences, 

Mustansiriyah University, Baghdad, Iraq 

kasimabbas@uomustansiriyah.edu.iq 
 

Waleed Jamal Hasan 

Department of Mathematics, College of Sciences, 

Mustansiriyah University, Baghdad, Iraq 

waleedjamalhasan1979@gmail.com 

 
 

https://creativecommons.org/licenses/by/4.0/
mailto:kasimabbas@uomustansiriyah.edu.iq
mailto:waleedjamalhasan1979@gmail.com


IHJPAS. 36(1)2023 
 

346 
 

methods or two-step methods are used to solve equation (1) [3]. In 2013, [4] developed the 

Improved Runge-Kutta technique for solving first ODEs using new terms 𝑘−𝑖  taking the values 

of 𝑘𝑖 , 𝑖 ≥ 2 from earlier stages.. Another choice for coming up with approaches to solve the 

oscillatory ODEs is the trigonometrically-fitted approach. These techniques are an improved 

version of any earlier techniques. In 2019, [5] developed trigonometrically-fitted sixth-order two-

derivative Runge-Kutta method for solving oscillatory problems. In 2021, [6] derived the fourth 

and fifth-order modified Runge-Kutta method to resolve oscillatory problems using phase-Lag 

properties. Recently, [7] derived the trigonometrically-fitted third-order Improved Runge-Kutta 

method for solving oscillatory problems.    

    This paper aims to develop the fourth-order IRK method called Trigonometrically Fitted 

Improved Runge-Kutta (TFIRK4) to solve a system of first-order ODEs with oscillatory solutions. 

Numerical experiments demonstrate the accuracy of the newly proposed method over other 

methods. A proposed TFIRK technique derivation is presented in Section 2. A numerical test and 

comparison with different approaches are discussed in Section 3 to show the efficiency of the 

TFIRK4 method. Section 4 contains the discussion and conclusion of this paper. 

2. The Derivation of TFIRK4 Method 

The Improved Runge-Kutta (IRK) method for solving equation (1) has the following form: [4]         

𝑢𝑛+1
′ = 𝑢𝑛 + ℎ(𝑏1𝑘1 − 𝑏−1𝑘−1 + ∑ 𝑏𝑖 (𝑘𝑖 − 𝑘−𝑖 ),

𝑠
𝑖=2                                                 (2) 

      𝑘1 = 𝑓(𝑡𝑛, 𝑢𝑛 ),                                                                                                       (3) 

      𝑘−1 = 𝑓(𝑡𝑛−1, 𝑢𝑛−1),                                                                                             (4) 

      𝑘𝑖 = 𝑓(𝑡𝑛 + 𝑐𝑖 ℎ, 𝑢𝑛 + ℎ ∑ 𝑎𝑖𝑗 𝑘𝑗 ),
𝑖−1
𝑗=1                                                                      (5) 

      𝑘−𝑖 = 𝑓(𝑡𝑛−1 + 𝑐𝑖 ℎ, 𝑢𝑛−1 + ℎ ∑ 𝑎𝑖𝑗 𝑘−𝑗 ).
𝑖−1
𝑗=1                                                          (6) 

Where  𝑐𝑖 , 𝑏𝑖 , 𝑏−1,  and  𝑎𝑖𝑗  are real numbers and  𝑖, 𝑗 = 1,2, … , 𝑠. IRK method (2)-(6) can be 

expressed using the following Butcher Tableau (see Table 1): 

                                             Table 1. 𝑠 − Stages IRK method  

 

 

 

 

 

   0  

𝑐2 𝑎21 
𝑐3 𝑎31       𝑎32 
.   .            .         .  
.   .            .               . 
.   .            .                    . 

𝑐𝑠 𝑎𝑠1       𝑎𝑠2          …         𝑎𝑠𝑠−1 
  𝑏−1  𝑏1         𝑏2           …         𝑏𝑠−1       𝑏𝑠 



IHJPAS. 36(1)2023 
 

347 
 

 

Following are the four stages of the IRK4 method in its general form: 

𝑢𝑛+1
′ = 𝑢𝑛 + ℎ(𝑏1𝑘1 − 𝑏−1𝑘−1 + 𝑏2(𝑘2 − 𝑘−2) + 𝑏3(𝑘3 − 𝑘−3) + 𝑏4(𝑘4 − 𝑘−4),  (7) 

𝑘1 = 𝑓(𝑡𝑛, 𝑢𝑛 ),                                                                                                              (8)                                                           

𝑘−1 = 𝑓(𝑡𝑛−1, 𝑢𝑛−1),                                                                                                    (9) 

𝑘2 = 𝑓(𝑡𝑛 + 𝑐2ℎ, 𝑢𝑛 + ℎ 𝑎21𝑘1),                                                                                 (10) 

𝑘−2 = 𝑓(𝑡𝑛−1 + 𝑐2ℎ, 𝑢𝑛−1 + ℎ 𝑎21𝑘−1),                                                                     (11) 

𝑘3 = 𝑓(𝑡𝑛 + 𝑐3ℎ, 𝑢𝑛 + ℎ (𝑎31𝑘1 + 𝑎32𝑘2)),                                                               (12) 

𝑘−3 = 𝑓(𝑡𝑛−1 + 𝑐3ℎ, 𝑢𝑛−1 + ℎ (𝑎31𝑘−1 + 𝑎32𝑘−2)),                                                  (13) 

𝑘4 = 𝑓(𝑡𝑛 + 𝑐4ℎ, 𝑢𝑛 + ℎ (𝑎41𝑘1 + 𝑎42𝑘2 + 𝑎43𝑘3)),                                                (14)  

𝑘−4 = 𝑓(𝑡𝑛 + 𝑐4ℎ, 𝑢𝑛−1 + ℎ (𝑎41𝑘−1 + 𝑎42𝑘−2 + 𝑎43𝑘−3)).                                    (15) 

The coefficients of the IRK4 method in [4] are offered in Table 2: 

 

                                                Table 2. IRK4 method 

 

Applying the exponential function into the IRK4 method (7)-(15), the trigonometrically fitting 

approach is implemented by allow: 

   𝑢n = 𝑢(𝑡n) = e
iω𝑡n ,                                                                                               (16)                                                                                             

   𝑢n+1 = 𝑢(𝑡n+1) = e
iω(𝑡n+ℎ),                                                                                  (17) 

  𝑢n−1 = 𝑢(𝑡n−1) = e
iω(𝑡n−ℎ).                                                                                   (18) 

Using Euler formula   𝑒 𝑖𝑣 = cos(𝑣) + 𝑖𝑠𝑖𝑛(𝑣), and substituting the equations (16)-(18) into 

equation IRK4 method (7)-(15), we get:  

  𝑒𝑖𝑣 = cos(𝑣) + 𝑖𝑠𝑖𝑛(𝑣) = 1 − 𝑣𝑏−1 sin(𝑣) − 𝑣
2𝑏2 𝑎21 − 𝑣𝑏2 sin(𝑣) − 𝑣

2𝑏3 𝑎31 

         +𝑣2 𝑏2𝑎21 cos (𝑣) − 𝑣
2𝑏3 𝑎32 − 𝑣𝑏3 sin(𝑣) + 𝑣

2 𝑏3𝑎31 cos (𝑣) − 𝑣
2𝑏4 𝑎41 

          +𝑣2 𝑏3𝑎32 cos(𝑣) + 𝑣
3 𝑏3𝑎32𝑎21 sin(𝑣) − 𝑣

2 𝑏4𝑎42 − 𝑣
2 𝑏4𝑎43 − 𝑣𝑏4 sin(𝑣) 

          +𝑣4 𝑏4𝑎43𝑎32𝑎21 + 𝑣
2 𝑏4𝑎41 cos(𝑣) + 𝑣

2 𝑏4𝑎42 cos(𝑣) + 𝑣
3 𝑏4𝑎42𝑎21 sin(𝑣) 

   0  
1

5
 

1

5
 

3

5
  0                 

3

5
 

4

5
 

2

15
              

4

25
            

38

75
 

19

288
 

307

288
      −  

25

144
         

25

144
          

125

288
 



IHJPAS. 36(1)2023 
 

348 
 

           +𝑣2 𝑏4𝑎43 cos(𝑣) + 𝑣
3 𝑏4𝑎43𝑎31 sin(𝑣) + 𝑣

3 𝑏4𝑎43𝑎32 sin(𝑣) 

            −𝑣3 𝑏4𝑎43𝑎32 𝑎21 cos(𝑣) + 𝑖𝑣𝑏1 − 𝑖𝑣𝑏−1 cos(𝑣) + 𝑖𝑣𝑏2 − 𝑖𝑣𝑏2 cos(𝑣) 

           +𝑖𝑣𝑏3 − 𝑖𝑣
2𝑏2𝑎21 sin(𝑣) − 𝑖𝑣

3 𝑏3𝑎32𝑎21 − 𝑖𝑣𝑏3 cos(𝑣) − 𝑖𝑣
2𝑏3𝑎31 sin(𝑣) 

      +𝑖𝑣𝑏4 + 𝑖𝑣
3𝑏3𝑎32𝑎21 cos(𝑣) − 𝑖𝑣

3𝑏4𝑎42𝑎21 − 𝑖𝑣
3𝑏4𝑎43𝑎31 − 𝑖𝑣

3𝑏4𝑎43𝑎32      

       −𝑖𝑣𝑏4 cos(𝑣) − 𝑖𝑣
2𝑏4𝑎41 sin(𝑣) − 𝑖𝑣

2𝑏4𝑎42 sin(𝑣) − 𝑖𝑣
2𝑏4𝑎43 sin(𝑣) 

       +𝑖𝑣3𝑏4𝑎42𝑎21cos (𝑣) + 𝑖𝑣
3𝑏4𝑎43𝑎31cos (𝑣) + 𝑖𝑣

3𝑏4𝑎43𝑎32cos (𝑣) 

                                               +𝑖𝑣4𝑏4𝑎43𝑎32𝑎21 sin(𝑣) − 𝑖𝑣
2𝑏3𝑎32 sin(𝑣).                 (19) 

 Where  𝑣 = 𝜔ℎ. We obtain the trigonometrically fitting order conditions by equating the real and 

imaginary parts: 

 cos(𝑣) = 1 − 𝑣𝑏−1 sin(𝑣) − 𝑣
2𝑏2 𝑎21 − 𝑣𝑏2 sin(𝑣) − 𝑣

2𝑏3 𝑎31+𝑣
2 𝑏2𝑎21 cos (𝑣) 

            −𝑣2𝑏3 𝑎32 − 𝑣𝑏3 sin(𝑣) + 𝑣
2 𝑏3𝑎31 cos (𝑣) − 𝑣

2𝑏4 𝑎41 + 𝑣
2 𝑏3𝑎32 cos(𝑣) 

            +𝑣3 𝑏3𝑎32𝑎21 sin(𝑣) − 𝑣
2 𝑏4𝑎42 − 𝑣

2 𝑏4𝑎43 − 𝑣𝑏4 sin(𝑣) + 𝑣
4 𝑏4𝑎43𝑎32𝑎21 

          +𝑣2 𝑏4𝑎41 cos(𝑣) + 𝑣
2 𝑏4𝑎42 cos(𝑣) + 𝑣

3 𝑏4𝑎42𝑎21 sin(𝑣) + 𝑣
2 𝑏4𝑎43 cos(𝑣) 

             +𝑣3 𝑏4𝑎43𝑎31 sin(𝑣) + 𝑣
3 𝑏4𝑎43𝑎32 sin(𝑣) − 𝑣

3 𝑏4𝑎43𝑎32 𝑎21 cos(𝑣) ,            (20) 

 sin(𝑣) = 𝑣𝑏1 − 𝑣𝑏−1 cos(𝑣) + 𝑣𝑏2 − 𝑣𝑏2 cos(𝑣) − 𝑣𝑏3 cos(𝑣) − 𝑣
2𝑏3𝑎31 sin(𝑣) 

             +𝑣𝑏3 − 𝑣
2𝑏2𝑎21 sin(𝑣) − 𝑣

3 𝑏3𝑎32𝑎21 − 𝑣
3𝑏4𝑎43𝑎31 − 𝑣

3𝑏4𝑎43𝑎32 

         +𝑣𝑏4 + 𝑣
3𝑏3𝑎32𝑎21 cos(𝑣) − 𝑣

3𝑏4𝑎42𝑎21 − 𝑣
2𝑏4𝑎42 sin(𝑣) − 𝑣

2𝑏4𝑎43 sin(𝑣)      

         −𝑣𝑏4 cos(𝑣) − 𝑣
2𝑏4𝑎41 sin(𝑣) + 𝑣

3𝑏4𝑎43𝑎31cos (𝑣) + 𝑣
3𝑏4𝑎43𝑎32cos (𝑣) 

                    +𝑣3𝑏4𝑎42𝑎21cos (𝑣) + 𝑣
4𝑏4𝑎43𝑎32𝑎21 sin(𝑣) − 𝑣

2𝑏3𝑎32 sin(𝑣).              (21)    

Solving equations (20) and (21) using the coefficients of the IRK4 method in Table2 for unknowns 

parameters  𝑎31  and  𝑎41 we obtain the following solution: 

𝑎31 =
−72

95 𝑣3(sin2(𝑣) + cos2(𝑣) − 2 cos(𝑣) + 1)
 (−12 cos(𝑣) sin(𝑣) + 12 sin (𝑣)  

          −3 𝑣 sin2(𝑣) + 𝑣3 sin2(𝑣) + 12 𝑣 𝑐𝑜𝑠(𝑣) − 3 𝑣 cos2(𝑣) − 2 𝑣3 cos(𝑣) 

                                                                                      −9𝑣 + 𝑣3 cos2(𝑣) + 𝑣3),            (22) 

 𝑎41 =
2

35625 𝑣3(sin2(𝑣)+cos2(𝑣)−2 cos(𝑣)+1)
(−7560 𝑣 − 9325 𝑣3  

                 −17280𝑣 𝑐𝑜𝑠(𝑣) + 24840 𝑣 cos2(𝑣) + 18650 𝑣3 cos(𝑣) 

                  +1083 𝑣5 cos2(𝑣) + 64800 sin(𝑣) − 57240 𝑣 sin2(𝑣) + 1083 𝑣5 

                 −64800 cos(𝑣) sin(𝑣) − 9325 𝑣3sin2(𝑣) − 2166 𝑣5 cos(𝑣)   

                              −9325 𝑣3 cos2(𝑣) + 41040𝑣2 sin(𝑣) + 1083 𝑣5 sin2(𝑣)).           (23) 



IHJPAS. 36(1)2023 
 

349 
 

It can be noticed that for  𝑣 → 0 using Maple package, the series expansions are given in following 

form: 

    𝑎31 = −
18

475
 𝑣2 +

3

3325
 𝑣4 −

1

79800
 𝑣6 +

1

8778000
 𝑣8 −

1

1369368000
 𝑣10 

                                                         +
1

28756728000
 𝑣12 −

1

78218300160000
 𝑣14 + ⋯ ,        (24) 

    𝑎41 =
2

15
−

276

11875
 𝑣2 +

689

249375
 𝑣4 −

77

1425000
 𝑣6 +

1069

1975050000
 𝑣8 

              −
61849

10783773000000
 𝑣10 −

1

172540368000
 𝑣12 −

77963

87995587680000000
 𝑣14 + ⋯ ,           (25) 

which lead to the new TFIRK4 method. As  𝑣 → 0, the newly obtained parameters   𝑎31    and   

𝑎41  turn into the parameters of the original method. 

3. Numerical Results 

  To evaluate the efficacy of the TFIRK4 method proposed, we apply them to oscillatory test 

problems and their numerical results are compared with the existing effective methods. The 

following numerical methods are applied in the comparison: 

▪ Step size. 

▪ TFIRK4: the Trigonometrically Fitted four-stage fourth-order IRK method presented here. 

▪ IRK4: fourth-order IRK method derive in [4]. 

▪ TFRK4: Trigonometrically-Fitted RK method developed in [8]. 

▪ RK4Z: fifth-order RK method given in [1]. 

▪ MRK4: modified RK method of order four proposed in [9]. 

▪ Max Error: max (|𝑢(𝑥𝑛) − 𝑢𝑛|) This is the maximum between absolute errors of the 
exact solutions and the computed solutions. 

Problem 1: [10] Inhomogeneous problem: 

                       𝑢1
′ (𝑡) = 𝑢2(𝑡),    𝑢1(0) = 1,   

                          𝑢2
′ (𝑡) = −100 𝑢1(𝑡) + 99 sin(𝑡),   𝑢2(0) = 11. 

Exact solution is: 

                       𝑢1(𝑡) = cos(10𝑡) + sin(10𝑡) + sin(𝑡), 

                       𝑢2(𝑡) = −10 sin(10𝑡) + 10 cos(10𝑡) + cos(𝑡). 

 

Problem 2: [7] Inhomogeneous problem: 

                       𝑢1
′ (𝑡) = 𝑢2(𝑡),    𝑢1(0) = 1, 

                          𝑢2
′ (𝑡) = −𝑢1(𝑡) + 𝑡, 𝑢2(0) = −2. 

Exact solution is: 

                        𝑢1(𝑡) = sin(𝑡) + cos(𝑡) + 𝑡, 

                        𝑢2(𝑡) = cos(𝑡) − sin(𝑡) + 1 

 

Problem 3: [11] Duffing problem: 

                𝑢1
′ (𝑡) = 𝑢2(𝑡),   𝑢1(0) =  0.200426728067, 

                𝑢3
′ (𝑡) = −𝑢1(𝑡) − (𝑢1(𝑡))

3
+ 0.002 cos(1.01 𝑡),  𝑢2(0) = 0, 

Exact solution is: 

   𝑢1(𝑡) =  0.200179477536 cos(1.01 𝑡) +  2.46946143 × 10
−4 cos(3.03 𝑡) 



IHJPAS. 36(1)2023 
 

350 
 

                +3.04014 × 10−7 cos(5.05 𝑡) +  3.74 10−10 cos(7.07 𝑡), 

   𝑢2(𝑡) =  −0.2021812723 sin(1.01 𝑡)  −  7.482468133 × 10
−4  sin(3.03 𝑡) 

                −1.53527070 × 10−6 sin(5.05 𝑡)   −  2.64418 ×  10−9 sin(7.07 𝑡). 

 

Problem 4: [8]  

     𝑢1
′ (𝑡) = 𝑢3(𝑡),   𝑢1(0) = 1, 

     𝑢3
′ (𝑡) =

−𝑢1(𝑡)

(√(𝑢1(𝑡))
2+(𝑢2(𝑡))

2)
3  𝑢3(0) = 0, 

    𝑢2
′ (𝑡) = 𝑢4(𝑡),   𝑢2(0) = 0, 

       𝑢4
′ (𝑡) =

−𝑢2(𝑡)

(√(𝑢1(𝑡))
2+(𝑢2(𝑡))

2)
3  𝑢4(0) = 1. 

Exact solution is: 

     𝑢1(𝑡) = cos(𝑡), 

     𝑢1(𝑡) = sin(𝑡), 

     𝑢3(𝑡) = −sin(𝑡), 

     𝑢4(𝑡) = cos(𝑡), 

 

Problem 5: [11] 

              𝑢1
′ (𝑡) = 𝑢3(𝑡),   𝑢1(0) = 1, 

                𝑢3
′ (𝑡) = −𝑢1(𝑡) + 0.001 cos(𝑡),  𝑢3(0) = 0, 

                𝑢2
′ (𝑡) = 𝑢4(𝑡),   𝑢2(0) = 0, 

                   𝑢4
′ (𝑡) = −𝑢2(𝑡) + 0.001 sin(𝑡),   𝑢4(0) = 0.9995. 

Exact solution is: 

                 𝑢1(𝑡) = cos(𝑡) + 0.0005 𝑡 sin(𝑡), 

                 𝑢2(𝑡) = −0.9995 sin(𝑡) + 0.0005 𝑡 cos(𝑡), 

                 𝑢3(𝑡) = sin(𝑡) − 0.0005 𝑡 cos(𝑡), 

                 𝑢4(𝑡) = 0.9995 cos(𝑡) + 0.0005 𝑡 sin(𝑡). 

      

                



IHJPAS. 36(1)2023 
 

351 
 

        

             Figure 1. The curves comparisons for Problem 1 with step size  ℎ =
1

2𝑟
, 𝑟 = 0,1,2,4.   

    

        

             Figure 2. The curves comparisons for Problem 2 with step size  ℎ =
1

2𝑟
, 𝑟 = 0,1,2,4.   

 



IHJPAS. 36(1)2023 
 

352 
 

          

                Figure 3. The curves comparisons for Problem 3 with step size  ℎ =
1

2𝑟
, 𝑟 = 0,1,2,4. 

 

 

             

             Figure 4. The curves comparisons for Problem 4 with step size  ℎ =
1

2𝑟
, 𝑟 = 0,1,2,4. 

 

 



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                          Figure 5. The curves comparisons for Problem 5 with step size  ℎ =
1

2𝑟
, 𝑟 = 0,1,2,4.  

 

To assess the method's accuracy, we utilize the absolute error criterion. The step size is ℎ =
1

2𝑟
, 𝑟 = 0,1,2,4.  and integration interval is [0, 1000] for all problems. The accuracy of the new 

TFIRK4 approach is depicted in Figures 1–5 in terms of the greatest global absolute error versus 

the step sizes required by each method. Compared to other RK methods of the same order, the 

TFIRK4 approaches, as shown in Figures 1–5, have the smallest maximum global error per step. 

The TFIRK4 produces results that are more accurate than those of other research in the literature, 

as seen in Figures 1–5. 

 

4. Conclusions 

  We derived the conditions of the Trigonometrically-Fitted IRK approach to solving oscillatory 

problems in this paper. As a result, we developed the TFIRK4 method, a four-stage, fourth-order 

IRK method that is trigonometrically fitted. The Figures show how the step size was used to 

calculate the common logarithm of the most significant global error during integration and 

computing cost. The numerical results made it clear that the TFIRK4 approach using a 

trigonometrically fitted strategy, had less global error than the existing methods used to solve 

oscillatory problems. 

 

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