volume 120 no. 1 2018,27-50 - ijpam · the order state about rktg method up to order six were...

International Journal of Pure and Applied Mathematics

Volume 120 No. 1 2018, 27-50ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version)url: http://www.ijpam.eudoi: 10.12732/ijpam.v120i1.3

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AN EFFICIENT OF DIRECT INTEGRATOR OF

RUNGE-KUTTA TYPE METHOD FOR SOLVING

y′′′ = f(x, y, y′) WITH APPLICATION TO THIN

FILM FLOW PROBLEM

Firas Adel Fawzi1 §, Norazak Senu2, Fudziah Ismail2

and Zanariah Abd. Majid2

1Department of MathematicsFaculty of Computer Science and Mathematics

Tikrit University, Sallah AL-Deen, IRAQ2Institute for Mathematical Research

Universiti Putra Malaysia43400 UPM, Serdang, Selangor, MALAYSIA

Abstract: In this paper, we proposed a fifth-order Runge-Kutta (RK) technique for regu-

lating coordination about third-order ordinary differential equations (ODEs) of the structure

y′′′ = f(x, y, y′) indicated similarly as RKTG method is constructed. The order state about

RKTG method up to order six were proved and verified. In view of those order conditions de-

veloped, four-stage fifth-order express Runge-Kutta methods of techniques were constructed.

The zero Strength of the new system was indicated. The Different types for third-order ODEs

need been derived utilizing the new system and also some numerical comparison were con-

ducted when the same issue will be decreased of the first-order framework of equations which

are solved using existing Runge-Kutta techniques. The numerical investigation of a third-

order tribute on thin film flow for viscous liquid in applied mathematical physics. Numerical

outcomes indicated that those new proposed method is more efficient in terms of accuracy

and number of function evaluations of capacity assessments.

Received: May 15, 2017

Revised: June 19, 2018

Published: August 13, 2018

c© 2018 Academic Publications, Ltd.

url: www.acadpubl.eu

§Correspondence author

28 F.A. Fawzi, N. Senu, F. Ismail

AMS Subject Classification: 65L05, 65L06

Key Words: Runge-Kutta type methods, general third-order ODEs, order conditions, thin

film flow

1. Introduction

We are dealing in this work with numerical integration of third-order ODE ofthe form:

y′′′(x) = f(

x, y, y′)

, (1)

with initial conditions

y(x0) = δ, y′(x0) = σ, y′′(x0) = ξ, x ≥ x0.

where y, y′, y′′ ∈ Rd, f : R × R

d → Rd is a continuous vector-valued function.

This sort of problems often found in numerous physical problems like thin filmflow, gravity-driven flows and electromagnetic waves. The general solution of(1) is by decreasing it to an equal series 1st-order framework which is threetimes extent and can be solved utilizing standard Runge-Kutta or multi-stepmethod. A lot of researches have been solved problem (1) by converting (1)to a system of 1st-order equations. Furthermore, there were several authorsstudied different numerical techniques which solved the problem (1) straight-forwardly for instance Jator [1], Awoyemi and Idowu [2], introduced generalfamily of hybrid methods for solving of higher-order ODEs. You and Chen[3], constructed direct integrations of RK type for special third-order ODEs.Waeleh et al. [4], proposed a new algorithm for solving higher-order IVPs ofODEs. Jator [5], constructed by hybrid multi-step method solving second orderIVPs without predictors. Samat and Ismail [6], developed a block multi-stepmethod which could straightforwardly solve general third-order equations, fur-thermore, Ibrahim et al. [7], found a process by using multi-step techniquewhich could solve stiff 3rd-order differential equations. Mechee et al. [8], con-structed a three-stage 5th-order RK type method for directly solving special3rd-order ODEs. Kasim et al. [9], proposed integration of 3rd-order ODEs us-ing improved RK technique. Subsequently, Senu et al. [10] constructed a newembedded explicit RK technique for solving special 3rd-order ODEs. In thispaper, the main aim is to proposed a one-step technique of order five to solvethird-order ODEs easily. The derivation of order conditions are given in Section2. In Section 3, the zero-stability of the new method is given. Four-stage fifthorder is constructed in Section 4. The effectiveness of the new technique, when

AN EFFICIENT OF DIRECT INTEGRATOR OF... 29

compared with existing method is given in Section 5. The Thin Film Flowproblem discussed in Section 6.

2. Derivation of the New Method

The general type of RKTG technique with m-stage for solving the IVPs (1) canbe indited as follows:

yn+1 = yn + h y′n +h2

2y′′n + h3

m∑

i=1

biki, (2)

y′n+1 = y′n + h y′′n + h2m∑

i=1

b′iki, (3)

y′′n+1 = y′′n + h

m∑

i=1

b′′i ki, (4)

where

k1 = f(

xn, yn, y′n

)

,

ki = f

(

xn + cih, yn + ci h y′n +

h2

2c2i y

′′n + h3

i−1∑

j=1

aijkj ,

y′n + ci h y′′n + h2

i−1∑

j=1

aijkj

)

(5)

for i = 2, 3, ... ,m.

The new parameters bi, b′i, b

′′i , aij , aij and ci of the RKTG method assumed to

be real and used for i, j = 1, 2, ...,m. The technique is explicitly if aij = aij = 0for i ≤ j and it is implicitly if aij 6= 0 and aij 6= 0 for i ≤ j. The new proposedtechnique is presented by the tableau below:

c A A

bT b′T b′′T(6)

Expressing the new proposed method of technique parameters provided by (2)-(5), the RKTG technique is expanded utilizing Taylor’s series expansion. After


performing a few algebraic manipulations, this expansion is equated to the truesolution that is given by Taylor’s series expansion. The direct expansion ofthe local truncation error utilized to derive the general order conditions forthe RKTG method. This conception predicated on the derivation of orderconditions for the RK method proposed by [11] and [12]. The new methodRKTG can be shown as follows:

yn+1 =yn + hϕ(xn, yn, y′n),

y′n+1 =y′n + hϕ′(xn, yn, y′n),

y′′n+1 =y′′n + hϕ′′(xn, yn, y′n). (7)

where the increment functions are

ϕ(xn, yn, y′n) = y′n +

h

2y′′n + h2

m∑

i=1

biki,

ϕ′(xn, yn, y′n) = y′′n + h

m∑

i=1

b′iki,

ϕ′′(xn, yn, y′n) =

m∑

i=1

b′′i ki. (8)

where ki is given in (5). If we postulate that ∆, ∆′ and ∆′′ are the Taylor seriesincrement function. Thus, the local truncation errors of y(x), y′(x) and y′′(x)can be acquired by superseding the precise solution of (1) into (8) as follows:

tn+1 = h[ϕ−∆],

t′n+1 = h[ϕ′ −∆′],

t′′n+1 = h[ϕ′′ −∆′′]. (9)

In the terms of elementary differentials, these expressions are best given andthe Taylor series might be expressed as follows:

∆ = y′ +1

2h y′′ +

1

6h2 F

(3)1 +

1

24h3 F

(4)1 +O(h4),

∆′ = y′′ +1

2hF

(3)1 +

1

6h2 F

(4)1 +

1

24h3 F

(5)1 +O(h4),

∆′′ = F(3)1 +

1

2hF

(4)1 +

1

6h2 F

(5)1 +O(h3). (10)

The first few elementary differentials for the scalar case are

F(3)1 = f,


F(4)1 = fx + fyyx + fy′yxx,

F(5)1 = fxx + yxfxy + fxy′yxx + y2xfyy + fyy′yxyxx + fyyxx + fy′y′y

2xx

+ fy′f. (11)

Substituting (11) into (8), the increment functions ϕ,ϕ′ and ϕ′′ for new methodwill become

m∑

i=1

biki =m∑

i=1

bif +m∑

i=1

bici(

fx + fyyx + fy′yxx)

h+1

2

m∑

i=1

bic2i

(

fxx

+ yxfxy + fxzyxx + y2xfyy + fyy′yxyxx + fyyxx + fy′y′y2xx + fy′f

)

h2

+O(h3),m∑

i=1

b′iki =

m∑

i=1

b′if +

m∑

i=1

b′ici(

fx + fyyx)

h+1

2

m∑

i=1

b′ic2i

(

fxx + yxfxy

+ fxy′yxx + fy′yxx + y2xfyy + fyy′yxyxx + fyyxx + fy′y′y2xx + fy′f

)

h2

+O(h3)

,m∑

i=1

b′′i ki =m∑

i=1

b′′i f +m∑

i=1

b′′i ci(

fx + fyyx + fy′yxx)

h+1

2

m∑

i=1

b′′i c2i

(

fxx

+ yxfxy + y2xfyy + Fyy′yxyxx + fyyxx + Fy′y′y2xx + fy′f

)

h2 +O(h3). (12)

From (10) and (12), the local truncation error (9) can be expressed as follows:

tn+1 = h3

[

m∑

i=1

biki −(

1

6F

(3)1 +

1

24hF

(4)1 + ...

)

]

,

t′n+1 = h2

[

m∑

i=1

b′iki −(

1

2F

(3)1 +

1

6hF

(4)1 + ...

)

]

,

t′′n+1 = h

[

m∑

i=1

b′′i ki −(

F(3)1 +

1

2hF

(4)1 +

1

6h2F

(5)1 + ...

)

]

. (13)

Substituting (12) into (13) by Taylors expansion employing the Maple softwareobtaining the truncated errors for m-stages up to order six for the proposedtechnique can be expressed as follows:

The order terms for y:


3rd-order∑

bi =1

6(14)

4th-order∑

bici =1

24(15)

5th-order∑

bic2i =

1

60,

∑

biaij =1

120(16)

6th-order∑

biaijcj =1

720,

∑

bic3i =

1

120(17)

∑

biaijci =1

240,

∑

biaij =1

720(18)

The order terms for y′:

2nd-order∑

b′i =1

2(19)

3rd-order∑

b′ici =1

6(20)

4th-order∑

b′ic2i =

1

12,

∑

b′iaij =1

24(21)

5th-order∑

b′ic3i =

1

20,

∑

b′iaijcj =1

120(22)

∑

b′iaijci =1

40,

∑

b′iaij =1

120(23)

6th-order

∑

b′ic2i aij =

1

60,

∑

b′iciaijcj =1

180(24)

∑

b′ic4i =

1

30,∑

b′ic2j aij +

∑

b′iciaijcj =1

120(25)

1

2

∑

b′ic2j aij =

1

720,1

2

∑

b′ic2j aij +

∑

b′iaijcj =1

360(26)

∑

b′iaijci =1

180,1

2

∑

b′ic2i aij +

∑

b′iaijci =1

72(27)


1

2

∑

b′ic2j aij +

∑

b′iciaijcj =1

144,∑

b′iaijcj =1

720(28)

∑

b′iaij ajk =1

720,1

2

∑

b′i aik aij =1

240(29)

The order terms for y′′:

1st-order∑

b′′i = 1 (30)

2nd-order∑

b′′i ci =1

2(31)

3rd-order∑

b′′i c2i =

1

3,

∑

b′′i aij =1

6(32)

4th-order∑

b′′i c3i =

1

4,

∑

b′′i aijcj =1

24(33)

∑

b′′i aij =1

24,

∑

b′′i ciaij =1

8. (34)

5th-order

∑

b′′i c4i =

1

5,

∑

b′′i aijc2j +

∑

b′′i aijcicj =1

20, (35)

∑

b′′i aijc2j =

1

60,

1

2

∑

b′′i aijc2j +

∑

b′′i aijcj =1

60, (36)

∑

b′′i aijci =1

30,

1

2

∑

b′′i c2i aij +

∑

b′′i aijci =1

12, (37)

1

2

∑

b′′i aijc2j +

∑

b′′i aijcicj =1

24,

∑

b′′i aij ajk =1

120, (38)

1

2

∑

b′′i aikaij =1

40,

∑

b′′i aijcj =1

120,

∑

b′′i aijcicj =1

30(39)

6th-order

∑

b′′i c5i =

1

6,

1

2

∑

b′′i aijc3i +

1

2

∑

b′′i aijc2i =

1

18(40)

1

2

∑

b′′i aijc3j +

∑

b′′i aijc2i cj =

23

720,

1

2

∑


1

72(41)

1

6

∑

b′′i aijc3i =

1

72,

1

2

∑

b′′i aijcic2j +

∑

b′′i aijc2j =

7

720(42)


∑

b′′i aijcicj =1

144,

1

6

∑

b′′i aijc3j +

1

2

∑


11

720(43)

∑

b′′i aijcicj =1

144,

1

6

∑

b′′i aijc3j +

1

2

∑


11

720(44)

∑

b′′i ciaijc2j =

1

72,

1

2

∑

b′′i c2i aijcj +

∑

b′′i ciaijcj =1

48(45)

1

2

∑

b′′i aijc3j +

∑

b′′i ci aijcj =1

90,

1

2

∑

b′′i c2i aij =

1

72(46)

1

2

∑

b′′i aijc2j +

∑

b′′i ciaijcj =1

120,

∑

b′′i aij ajkck =1

720(47)

1

2

∑


1

2

∑

b′′i aijc3j =

13

720,

∑

b′′i aij aik =1

72(48)

1

2

∑

b′′i c3i aij +

∑

b′′i c2i aij =

5

75,

1

2

∑

b′′i aijc2j =

1

720(49)

1

2

∑

b′′i aijc2j +

1

2

∑

b′′i ciaijc2j =

1

120(50)

1

2

∑


1

2

∑

b′′i aijc3j +

∑

b′′i ciaijcj =1

40(51)

∑

b′′i aijajk +∑

b′′i aij ajk =1

360(52)

∑

b′′i aijc2j +

∑

b′′i ciaijcj =7

720(53)

1

2

∑

b′′i ciaijc2j +

1

2

∑


∑

b′′i aijc2j +

∑

b′′i ciaijcj =11

360(54)

1

2

∑

b′′i ciaij aik =1

48,∑

b′′i aij aikcj =1

72(55)

∑

b′′i aij aikck =1

72(56)

∑

b′′i aij ajkci +∑

b′′i aij ajkcj +∑

b′′i aij aikcj =1

40(57)

All indexes are run from one to m. To obtain the higher-order RKTG techniquewe assume the following equation to eliminate and reduce the problem as statedbelow:

∑

aij =c2i2,

b′i =b′′i(

1− ci)

,

bi =b′′i

(

1− ci)2

2. i = 1, ... ,m. (58)


3. Zero-Stability of the New Model

Here, we will discuss the zero-stability of new technique of convergence. It isstable at zero significance to prove the convergence of multi-step techniquesand stability (see [13], [14]). Hairer et al. [15], also discussed on the zero-stability to obtained the upper boundedness of the multi-steps methods. Now,the first characteristic polynomial for the RKTG method (2)-(5) is based onthe following equation:

1 0 00 1 00 0 1

yn+1

hy′n+1

h2y′′n+1

=

1 1 12

0 1 10 0 1

ynhy′nh2y′′n

,

where I =

1 0 00 1 00 0 1

is the identity matrix coefficient of yn+1, h y′n+1 and

h2y′′n+1

and A =

1 1 12

0 1 10 0 1

is matrix coefficient of yn, h y′n and h2y′′n, respectively.

Then, the first characteristic polynomial of new method is

ρ(ζ) = det[Iζ −A] =

∣

∣

∣

∣

∣

∣

ζ − 1 −1 −12

0 ζ − 1 −10 0 ζ − 1

∣

∣

∣

∣

∣

∣

.

thus,

ρ(ζ) = (ζ − 1)3.

Therefore, this technique is stable at zero interval whereby the roots, ζ1, 2, 3are all equivalence to unity.

4. Construction of the RKTG Methods

By the order conditions stated in Section 2 above we proceed to constructexplicit RKTG methods. The local truncated error for the p order RKTG


technique is defined as follows:

‖ t(p+1)g ‖2=

np+1∑

i=1

(

t(p+1)i

)2+

n′

p+1

∑

i=1

(

t′i(p+1)

)2+

n′′

p+1

∑

i=1

(

t′′i(p+1)

)2

1

2

(59)

where t(p+1), t′(p+1) and t′′(p+1) are the local truncation error terms for y, y′ andy′′ respectively, tg

(p+1) is the global local truncation error.

4.1. A Four-stage Fifth-order RKTG Method

In this part, we discuss mainly on the deriving the four-stage RKTG techniqueof 5th-order and the algebraic conditions

(

(14)–(16), (19)–(23), (30)–(39))

willbe use because of the high number consisting of some non-linear equations,therefore we use the simplifying assumption (58) to reduce the system of equa-tions to 19 equations with 16 unknowns and left with 3 degree of freedom.Solving the system simultaneously and the family of solution in term of a42, a43and c3 are given as follows:

a21 =(5c3 − 3)2

50(2c3 − 1)2, a31 = −c3(−35 c23 + 19 c3 − 3 + 20 c33)

10(c3 − 3),

a32 =(−10 c3 + 3 + 10 c23) c3 (2 c3 − 1)

(10 c3 − 3),

a41 = − 10 c33 − 17 c23 + 10 c3 − 2

2 c3 (3− 12 c3 + 10 c23) (5 c3 − 3),

a42 =(−85 c23 + 45 c3 + 50 c33 − 7) (2 c3 − 1) (−2 + 5 c3)

2 (−10 c3 + 3 + 10 c23) (5 c3 − 3) (3 − 12 c3 + 10 c23),

a43 = − (−2 + 5 c3) (−1 + c3)2

2 (−10 c3 + 3 + 10 c23) (3 − 12 c3 + 10 c23), a21 = 0,

a31 = −(−10 c3 + 3 + 10 c23) c3 (450 a43 c23 − 75 a43 c3 + 255 a42 c3

(5 c3 − 3) (−2 + 5 c3)

+250 a42 c33 + 2− 45 a42 − 15 c3 + 35 c23 − 25 c33

(5 c3 − 3) (−2 + 5 c3)

−850 a43 c33 + 500 a43 c

43)

(5 c3 − 3) (−2 + 5 c3),

a32 = −(−10 c3 + 3 + 10 c23) c3 (−11 c3 + 2 + 19 c23 − 10 c33 + 102 c3 a42(5 c3 − 3) (−2 + 5 c3)


−18 a42 − 180 a42 c23 + 100 a42 c

33 − 30 a43 c3 + 180 a43 c

23 − 340 a43 c

33

(5 c3 − 3) (−2 + 5 c3)

+200 a43 c43)

(5 c3 − 3) (−2 + 5 c3),

a41 = −(100 a43 c23 − 25 c23 + 100 c23 − 120 a43 c3120 a43 c3 − 120 c3 a42

10 (3 − 12 c3 + 10 c23)

++30 c3 + 30 a42 + 30 a43 − 8)

10 (3 − 12 c3 + 10 c23), a42 = a42, a43 = a43,

b′1 =10 c23 − 8 c3 + 1

12 c3 (5 c3 − 3), b′2 =

25 (−12 c23 + 6 c3 − 1 + 8 c23)

12 (5 c3 − 3) (−10 c3 + 3 + 10 c23),

b′3 =25 (10 c24 − 12 c4 + 3

48 (−4 + 5 c4) (11 c4 − 4), b′4 = 0, c1 = 0, c2 =

5 c3 − 3

5 (2 c3 − 1),

c3 = c3, c4 = 1, b1 =10 c23 − 8 c3 + 1

24 c3 (5 c3 − 3),

b2 =5 (2 c3 − 1) (10 c23 − 9 c3 + 2)

24 (5 c3 − 3) (−10 c3 + 3 + 10 c23,

b3 = − −1 + c324 c3 (5 c3 − 3)

, b4 = 0, b′′1 =10 c23 − 8 c3 + 1

12 c3 (5 c3 − 3),

b′′2 =16 c43 − 32 c33 + 24 c23 − 8 c3 + 1)

(−2 + 5 c3) (5 c3 − 3) (−10 c3 + 3 + 10 c23),

b′′3 = − 1

12 c3 (−1 + c3) (−10 c3 + 3 + 10 c23),

b′′4 =3− 12 c3 + 10 c23

12 (−2 + 5 c3) (−1 + c3) (−4 + 5 c4).

Letting c3 =23 , the local truncation error in two free parameters given by

‖ t(6)g ‖2=1

21600

(

− 61200 a42 − 401400 a43 + 596000 a243 + 18746

+ 202500 a242 + 1875000 a42 a43)

1

2 . (60)

By using minimize command in Maple we obtaina42 = −0.0176230497844326, a43 = 0.0364465787986446 and the minimum localtruncation error is 0.005065254267. For the optimized value in fractional form


then we choose a42 = − 150 , and a43 = 1

25 . Finally, all the coefficients of four-stage fifth-order RKTG method denoted by RKTG5 can be written as follows(see Table 1):

5. Numerical Experiments

In this subsection, some of the problems involving y′′′ = f(x, y, y′) are testedupon. The numerical results are compared with the results obtained when thesame set of problems is reduced to a system of first-order equations and is solvedusing the existing RK of the same order.

• RKTG5: the four-stage fifth-order RKTG method derived in this paper.

• RK5B: the six-stage fifth-order RK method given by Butcher [13].

• RKF5: the six-stage fifth-order RK method given by Lambert [14].

• DOPRI5: the seven-stage fifth-order RK method derived by Dormand[11].

• RK4: the fourth-order classical RK method as given in Butcher [13].

Problem 1: (Homogeneous Linear Problem)

y′′′(x) = −25 y′(x),

y(0) = 0, y′(0) = 0, y′′(0) = 1,

The exact solution is given by y(x) = 125 − 1

25 cos(5x).

Problem 2: (Inhomogeneous Linear Problem)

y′′′(x) = y′(x) + cos2(x)− 1,

y(0) = 0, y′(0) = 0, y′′(0) = 1,

The exact solution is given by

y(x) = −1

5e−x +

4

5ex − 1

20sin(2x) +

1

2x− 1 .


Problem 3: (Homogeneous Non-linear Problem)

y′′′(x) =3 y′(x)

4( y(x))4,

y(0) = 1, y′(0) =1

2, y′′(0) = −1

4,

The exact solution is given by y(x) =√x+ 1 .

Problem 4: (Inhomogeneous Non-linear Problem)

y′′′(x) = y2(x) + cos2(x)− y′(x)− 1,

y(0) = 0, y′(0) = 1, y′′(0) = 0,

The exact solution is given by y(x) = sin(x) .

Problem 5: (Non-linear System)

y′′′1 (x) =1

2e4x y3(x) y

′2(x),

y′′′2 (x) =8

3e2x y1(x) y

′3(x),

y′′′3 (x) = 27 y2(x) y′1(x)

y1(0) = 1, y′1(0) = −1, y′′1 (0) = 1,

y2(0) = 1, y′2(0) = −2, y′′2 (0) = 4,

y3(0) = 1, y′3(0) = −3, y′′3 (0) = 9,

The exact solution is given by

y1(x) = e−x,

y2(x) = e−2x,

y3(x) = e−3x.

Problem 6: (Inhomogeneous Non-linear Problem)

y′′′(x) = 6 y′(x) y2(x),

y(0) = 1, y′(0) = −1, y′′(0) = 2,

The exact solution is given by y(x) = 11+x

.


RK4RK5FRK5B

DOPRI5RKTG5

log10(Function Evaluations)

log 1

0(M

AXERR)

2.42.221.81.61.41.210.8

−1

−2

−3

−4

−5

−6

−7

−8

Figure 1: Comparison for RKTG5, RK5B, RK5F, DOPRI5 and RK4Problem 1 with Xend=1

RK4RK5FRK5B

DOPRI5RKTG5


log 1

0(M

AXERR)

3.232.82.62.42.221.81.6

0

−1

−2

−3

−4

−5

−6

−7

−8

−9


6. An Application to a Problem in Thin Film Flow

Here, we will use the suggested method to a famous problem in engineering andphysics based on the thin film flow of a liquid. Many researchers in the litera-ture explain more on this problem. Momoniat and Mahomed [16], constructedsymmetry reduction and numerical solution of a third-order ODE from thinfilm flow. Tuck and Schwartz [17], discussed the movement of a thin film ofviscous fluid over a solid surface and taken into account Tension, gravity, aswell as viscosity. The problem was evaluated and solved using 3rd-order ODE


RK4RK5FRK5B

DOPRI5RKTG5


log 1

0(M

AXERR)

2.42.221.81.61.41.210.8

−2

−4

−6

−8

−10

−12


RK4RK5FRK5B

DOPRI5RKTG5


log 1

0(M

AXERR)

3.232.82.62.42.221.81.6

−2

−3

−4

−5

−6

−7

−8

−9

−10


as follows:

d3y

dx3= f(y) (61)

Many forms of the function were studied by [17]. for the drainage dry surfaceit has the form of f(y) can be stated as:

d3y

dx3= −1 +

1

y2. (62)


RK4RK5FRK5B

DOPRI5RKTG5


log 1

0(M

AXERR)

2.82.62.42.221.81.61.41.2

1

0

−1

−2

−3

−4

−5

−6

−7

−8


RK4RK5FRK5B

DOPRI5RKTG5


log 1

0(M

AXERR)

2.82.62.42.221.81.61.41.2

0

−1

−2

−3

−4

−5

−6

−7

−8

−9


When the surface is pre-wetted by a thin film with thickness ω > 0 (whereω > 0 is very small), the function f is given by

f(y) = −1 +1 + ω + ω2

y2− ω + ω2

y3. (63)

Problems concerning the flow of thin films of viscous fluid with a free surfacein which surface tension effects play a role typically lead to 3rd-order ODEsgoverning the shape of the free surface of the fluid, y = y(x). As indicated by[17], one such equation is

y′′′ = y−k, x ≥ x0 (64)


with initial conditions

y(x0) = δ, y′(x0) = σ, y′′(x0) = ξ, (65)

where δ, σ, and ξ are constants, is of specific significance since it portrays thedynamic balance amongst surface and gooey strengths in a thin fluid layer inthe disregard of gravity. For compare and contrast, we utilized Runge-Kuttastrategies which are 5th-order (RK5B and DOPRI5) strategies, individually.To utilize Runge-Kutta techniques we write (1) as a system of three 1st-orderequations. Following [18], we can write (64) as the following system:

dy1dx

= y2(x),dy2dx

= y3(x),dy3dx

= y−k1 (x), (66)

where

y1(0) = 1, y2(0) = 1, y3(x) = 1, (67)

we have taken x0 = 0 and δ = σ = ξ=1. Unfortunately, for general k, (64) can-not be solved analytically. However, we can use these reductions to determinean the efficient way to solve (1) numerically. Here, we are focusing on the casesk = 2 and k= 3.

The results are displayed in Tables 2 to 3 for the case k= 2 and Tables 4 and5 for the case k= 3.

yi

xi

y i

10.80.60.40.20

2.6

2.4

2.2

2

1.8

1.6

1.4

1.2

1

Figure 7: Plot of the solution yi for problem (64) for k = 2, h = 0.01


yi

xi

y i

10.80.60.40.20

2.6

2.4

2.2

2

1.8

1.6

1.4

1.2

1

Figure 8: Plot of the solution yi for problem (64) for k = 3, h = 0.01

RK4DOPRI5RKTG5

log10(h)

log 1

0(FunctionEvaluations)

-1-1.5-2-2.5-3-3.5-4

5

4

3

2

1

0

Figure 9: Plot of graph for function evaluations against step-size h fork = 3, h = 1/10i, i = 1...4.

7. Discussion and Conclusion

In this review, we have inferred the order conditions for a RK techniques whichcan be utilized to unravel 3rd-order ODEs specifically. A fourth-stage 5th-orderRKTG5 has been introduced and the comparison are made with existing RKmethods and we used in numerical comparison the criteria based on comput-ing the maximum error in the solution

(

max(| y(tn) − yn |))

which is equalto the maximum between absolute errors of the actual solutions and computedsolutions. The numerical outcomes are plotted in Figures 1, 2, ... , 6. Those Fig-ures show the proficiency bends where the common logarithm of the maximumglobal error throughout the integration versus computational cost measured by


the number of function evaluations. In Figures 7 and 8, we plot the numeri-cal arrangement, yi for k = 2 and k = 3, individually, with h = 0.01. Figure9 demonstrates that the new RKTG5 technique requires less capacity assess-ments than the RK4 and DOPRI5 strategies. This is on account of when issue(64) is unraveled utilizing RK4 and DOPRI5 technique, it should be decreasedto a system of 1st-order equations which is three times the dimension. Fromnumerical outcomes, we saw that the new RKTG5 strategy is more proficientcompared with existing RK strategies and its has demonstrated that the newtechnique is more precise and able when solving 3rd-order ODEs of the formy′′′ = f(x, y, y′) straightforwardly.


Tab

le1:

TheRKTG5Meth

od:

00

0

150

0150

0

23−

49

4860

301

4860

0−

127

727

0

1750

−150

125

0310

−235

935

0

148

542

3112

0124

25

84

956

0124

125

336

27

56

548


Table 2: Comparison of RK5B, DOPRI5 and RKTG5 methods whensolving Problem (64) with h = 0.1 and k = 2

x Exact Solution RK5B DOPRI5 RKTG5

0.0 1.000000000 1.0000000000 1.0000000000 1.00000000000.2 1.221211030 1.2212103654 1.2212100041 1.22121000390.4 1.488834893 1.4888507091 1.4888347796 1.48883477970.6 1.807361404 1.8074895468 1.8073613979 1.80736139880.8 2.179819234 2.1803393022 2.1798192349 2.17981923711.0 2.608275822 2.6097383114 2.6082748696 2.6082748735


x Exact Solution RK5B DOPRI5 RKTG5

0.0 1.000000000 1.0000000000 1.0000000000 1.00000000000.2 1.221211030 1.2212103651 1.2212100045 1.22121000450.4 1.488834893 1.4888507105 1.4888347799 1.48883477990.6 1.807361404 1.8074895516 1.8073613977 1.80736139770.8 2.179819234 2.1803393119 2.1798192339 2.17981923391.0 2.608275822 2.6097383271 2.6082748676 2.6082728676


x RK5B DOPRI5 RKTG5

0.0 1.0000000000 1.0000000000 1.00000000000.2 1.2211557749 1.2211551421 1.22115513940.4 1.4881300287 1.4881052848 1.48810528070.6 1.8044424216 1.8042625503 1.80426254590.8 2.1721917263 2.1715228023 2.17152279871.0 2.5927035854 2.5909582657 2.5909582638



x RK5B DOPRI5 RKTG5

0.0 1.0000000000 1.0000000000 1.00000000000.2 1.2211557725 1.2211551424 1.22115514240.4 1.4881300313 1.4881052842 1.48810528420.6 1.8044424292 1.8042625481 1.80426254810.8 2.1721917529 2.1715227981 2.17152279811.0 2.5927036287 2.5909582591 2.5909582591


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