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This paper presents the use of differential transformation method (DTM), an approximating technique for solving linear higher order boundary value problems. Using DTM, approximate solutions of order seven and eight boundary value problems were developed. Approximate results are given for some examples to illustrate the efficiency and accuracy of the method. The results from this method are compared with the exact solutions.

Higher order boundary value problems arise in the study of hydrodynamics and hydro magnetic stability, astronomy, fluid dynamics, astrophysics, engineering and applied physics. The boundary value problems of higher order have been investigated due to their mathematical importance and the potential for applications in diversified applied sciences [

Explicit weighting coefficients are formulated to implement the Generalized Differential Quadrature Rule (GDQR) for eighth-order differential equations. [

A great deal of interest has been focused on the applications of differential transformation method (DTM) to solve various scientific models [

A kth order differential transformation of a function y ( x ) = f ( x ) is defined about a point x = x 0 as:

Y ( K ) = [ d k y ( x ) d x k ] x = x 0 (2.1)

where k belongs to the set of non-negative integers, denoted as the K-domain.

The function y ( x ) may be expressed in terms of the differential transforms y ( K ) as:

y ( x ) = ∑ k = 0 ∞ [ ( x − x 0 ) k k ! ] Y ( k ) (2.2)

Upon combining 2.1 and 2.2, we obtain: y ( x ) = ∑ k = 0 ∞ 1 k ! ( x − x 0 ) k [ d k y ( x ) d x k ] x = x 0

Which is actually the Taylor’s series for y ( x ) at about = x 0 .

From the basic definition of the differential transformation, one can obtain certain laws of transformational operations, some of these, are listed in the following:

1) If z ( x ) = u ( x ) ± v ( x ) , then Z ( k ) = U ( k ) ± V (k)

2) If z ( x ) = α u ( x ) , then Z ( k ) = α U (k)

3) If z ( x ) = d u ( x ) d x then, Z ( k ) = ( k + 1 ) U ( k + 1 )

4) If z ( x ) = d 2 y ( x ) d x 2 then, Z ( k ) = ( k + 1 ) ( k + 2 ) U ( k + 2 )

5) If z ( x ) = d m y ( x ) d x m then, Z ( k ) = ( k + 1 ) ( k + 2 ) ⋯ ( k + m ) U ( k + m )

6) If z ( x ) = u ( x ) v ( x ) then Z ( k ) = ∑ l = 0 k V ( l ) U ( k − l )

7) If z ( x ) = x m then, Z ( k ) = ∂ ( k − n ) where, ∂ ( k − n ) = { 1 k = n 0 k ≠ n

8) If z ( x ) = e λ x then, Z ( k ) = λ k !

9) If z ( x ) = sin ( ω x + α ) then, Z ( k ) = ω k ! sin ( π k 2 + α )

10) If z ( x ) = cos ( ω x + α ) then, Z ( k ) = ω k ! cos ( π k 2 + α )

1) even order boundary value problems

Consider the special (2m) order BVP of the form

y ( 2 m ) ( x ) = f ( x , y ) , 0 < x < b (3.1)

With boundary conditions

y ( 2 m ) ( 0 ) = α 2 j , j = 0 , 1 , 2 , ⋯ , ( m − 1 ) (3.2)

y ( 2 m ) ( b ) = β 2 j , j = 0 , 1 , 2 , ⋯ , ( m − 1 ) (3.3)

2) odd order boundary value problems

Consider the special (2m + 1) order BVP of the form

y ( 2 m + 1 ) ( x ) = f ( x , y ) , 0 < x < b (3.4)

With boundary conditions

y ( 2 j + 1 ) ( 0 ) = ϒ 2 j + 1 , j = 0 , 1 , 2 , ⋯ , ( m ) (3.5)

y ( 2 j + 1 ) ( b ) = ϒ 2 j + 1 , j = 0 , 1 , 2 , ⋯ , ( m ) (3.6)

It is interesting to point out that y ( x ) and f ( x , y ) are assumed real and as many times differentiable as required for x ∈ [ 0 , b ]

Let the differential transform of the deflection function y ( x ) be defined from Equation (2.1) as:

Y ( K ) = 1 k ! [ d k y ( x ) d x k ] x = x 0 (4.1)

where x_{0} = 0. Also the deflection function may be expressed in terms of Y ( K ) from Equation (2.2) as:

y ( x ) = ∑ k = 0 ∞ [ ( x − x 0 ) k k ! ] Y ( k ) (4.2)

Now, using the transformation operations which has been formed in sec.2, one can obtain by taking the differential transform of Equations (3.1) and (3.4) respectively and some simplification, the following recurrence equations as m = 0 , 1 , 2 , ⋯

Y ( 2 m + k ) = ∑ k = 0 ∞ [ ( 2 m ) ! ( 2 m + k ) ! ] Y ( . , . ) (4.3)

Y ( 2 m + k + 1 ) = ∑ k = 0 ∞ [ ( 2 m + 1 ) ! ( 2 m + k + 1 ) ! ] Y ( . , . ) (4.4)

where Y ( . , . ) denotes the transformed function of linear and non linear function f ( x , y ) . It may be noted that Equation (4.2) is independent of the boundary conditions. The differential transforms of the boundary conditions at x = 0 are obtained from Equations (3.2) and (3.5) in the cases even order (odd order) boundary value problems respectively with the definition 4.1 as:

Y ( 2 j ) = 1 2 j α 2 j , j = 0 , 1 , 2 , ⋯ , ( 2 m − 1 ) (4.5)

Y ( 2 j + 1 ) = 1 2 j + 1 γ 2 j , j = 0 , 1 , 2 , ⋯ , 2 m (4.6)

Substituting from 4.5 and 4.6 into 4.3 and 4.4 and using 4.2, yields for j = 0 , 1 , 2 , ⋯ , ( m − 1 )

y ( x ) = ∑ k = 0 ∞ [ 1 ( 2 j ) ! α 2 j ] Y ( k ) x k

And for j = 0 , 1 , 2 , ⋯ , m

y ( x ) = ∑ k = 0 ∞ [ 1 ( 2 j + 1 ) ! γ 2 j + 1 ] Y ( k ) x k

Noting that y ( 2 r + 1 ) ( 0 ) = A r , r = 0 , 1 , 2 , ⋯ , ( m − 1 ) , and y ( 2 r ) ( 0 ) = B r , r = 0 , 1 , 2 , ⋯ , m , are constants that will be approximated at the end point x = b .

NUMERICAL EXAMPLES

Example 1:

y 8 ( x ) + x y ( x ) = − ( 48 + 15 x + x 3 ) e x , 0 < x < 1 (1)

with boundary conditions

y ( 0 ) = 0 , y ( 1 ) = 0 y 1 ( 0 ) = 1 , y 1 ( 1 ) = − e y 2 ( 0 ) = 0 , y 2 ( 1 ) = − 4 e y 3 ( 0 ) = − 3 , y 3 ( 1 ) = − 9 e (2)

whose analytical solution is y ( x ) = x ( 1 − x ) e x

transforming using DTM

Y ( k + 8 ) = k ! ( k + 8 ) ! [ − 48 k ! + 15 ∑ l = 0 k ∂ ( l − 1 ) ( k − l ) ! − ∑ l = 0 k ∂ ( l − 3 ) ( k − l ) ! − ∑ l = 0 k ∂ ( l − 1 ) U ( k − l ) ]

With boundary conditions

Y ( 0 ) = 0 , Y ( 1 ) = 1 , Y ( 2 ) = 0 , Y ( 3 ) = − 3 3 ! , Y ( 4 ) = A , Y ( 5 ) = B , Y ( 6 ) = C , Y ( 7 ) = D

At k = 0

Y ( 8 ) = 0 ! ( 8 ) ! [ − 48 0 ! + 15 ∑ l = 0 0 ∂ ( 0 − 1 ) ( 0 ) ! − ∑ l = 0 0 ∂ ( 0 − 3 ) ( 0 ) ! − ∑ l = 0 0 ∂ ( 0 − 1 ) U ( 0 ) ] = − 1 840

k = 1

Y ( 9 ) = 1 ! ( 9 ) ! [ − 48 1 ! + 15 ∑ l = 0 1 ∂ ( 1 − 1 ) ( 1 ) ! − ∑ l = 0 1 ∂ ( l − 3 ) ( 1 − l ) ! − ∑ l = 0 1 ∂ ( l − 1 ) Y ( 1 − l ) ] = − 1 5760

k = 2

Y ( 10 ) = 2 ! ( 10 ) ! [ − 48 2 ! + 15 ∑ l = 0 2 ∂ ( 1 − 1 ) ( 2 − l ) ! − ∑ l = 0 2 ∂ ( l − 3 ) ( 2 − l ) ! − ∑ l = 0 2 ∂ ( l − 1 ) Y ( 2 − l ) ] = − 7 10 !

k = 3

Y ( 11 ) = 3 ! ( 11 ) ! [ − 48 3 ! + 15 ∑ l = 0 3 ∂ ( 1 − 1 ) ( 3 − l ) ! − ∑ l = 0 3 ∂ ( l − 3 ) ( 3 − l ) ! − ∑ l = 0 3 ∂ ( l − 1 ) Y ( 3 − l ) ] = − 69 11 !

k = 4

Y ( 12 ) = 4 ! ( 12 ) ! [ − 48 4 ! + 15 ∑ l = 0 4 ∂ ( l − 1 ) ( 4 − l ) ! − ∑ l = 0 4 ∂ ( l − 3 ) ( 4 − l ) ! − ∑ l = 0 4 ∂ ( l − 1 ) Y ( 4 − l ) ] = − 1 5702400

k = 5

Y ( 13 ) = 5 ! ( 13 ) ! [ − 48 5 ! + 15 ∑ l = 0 5 ∂ ( l − 1 ) ( 5 − l ) ! − ∑ l = 0 5 ∂ ( l − 3 ) ( 5 − l ) ! − ∑ l = 0 5 ∂ ( l − 1 ) Y ( 5 − l ) ] = − 183 − A 13 !

k = 6

Y ( 14 ) = 6 ! ( 14 ) ! [ − 48 6 ! + 15 ∑ l = 0 6 ∂ ( l − 1 ) ( 6 − l ) ! − ∑ l = 0 6 ∂ ( l − 3 ) ( 6 − l ) ! − ∑ l = 0 6 ∂ ( l − 1 ) Y ( 6 − l ) ] = − 258 − B 14 !

k = 7

Y ( 15 ) = 7 ! ( 15 ) ! [ − 48 7 ! + 15 ∑ l = 0 7 ∂ ( l − 1 ) ( 7 − l ) ! − ∑ l = 0 7 ∂ ( l − 3 ) ( 7 − l ) ! − ∑ l = 0 7 ∂ ( l − 1 ) Y ( 7 − l ) ] = − 363 − C 15 !

k = 8

Y ( 16 ) = 8 ! ( 16 ) ! [ − 48 8 ! + 15 ∑ l = 0 8 ∂ ( l − 1 ) ( 8 − l ) ! − ∑ l = 0 8 ∂ ( l − 3 ) ( 8 − l ) ! − ∑ l = 0 8 ∂ ( l − 1 ) Y ( 8 − l ) ] = − 504 − D 16 !

y ( x ) = ∑ k = 0 n Y ( k ) x k

y ( x ) = x − 3 3 ! x 3 + A x 4 + B x 5 + C x 6 + D x 7 − 1 840 x 8 − 1 5760 x 9 − 79 10 ! x 10 − 69 11 ! x 11 − 1 5702400 x 12 − 183 + A 13 ! x 13 − 258 + B 14 ! x 14 − 363 + C 15 ! x 15 − 504 + D 16 ! x 16 ^{ }

Using the boundary conditions given in (2), the required equation is (

Y ( x ) = x − 3 3 ! x 3 − 0.180317402 x 4 − 0.492347647 x 5 + 0.272873875 x 6 − 0.094455951 x 7 − 1 840 x 8 − 1 5760 x 9 − 79 10 ! x 10 − 69 11 ! x 11 − 1 5702400 x 12 − 181.8196826 13 ! x 13 − 257.5076524 14 ! x 14 − 363.2728476 15 ! x 15 − 503.905544 16 ! x 16

Example 2:

Consider a 7^{th} order linear boundary value problem

X | Analytical solution (h = 0.1) | DTM (h = 0.1) | Error |
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0.0 | 0.0000 | 0.0000 | 0.0000 |

0.1 | 0.099465382 | 0.099477308 | 1.1926 × 10^{−5} |

0.2 | 0.195424441 | 0.19557019 | 1.45749 × 10^{−4} |

0.3 | 0.283470349 | 0.284021212 | 5.50863 × 10^{−4} |

0.4 | 0.358037927 | 0.359304348 | 1.266421 × 10^{−3} |

0.5 | 0.412180317 | 0.414365019 | 2.184702 × 10^{−3} |

0.6 | 0.437308512 | 0.440411098 | 0.028230781 |

0.7 | 0.422888068 | 0.426705169 | 3.817101 × 10^{−3} |

0.8 | 0.356086548 | 0.360307456 | 4.220908 × 10^{−3} |

0.9 | 0.22136428 | 0.225683179 | 4.318899 × 10^{−3} |

1.0 | 0 | 4.365206367 x 10^{−}^{3} | 4.365206367 × 10^{−3} |

U 7 ( x ) = x u ( x ) + e x ( x 2 − 2 x − 6 ) , 0 ≤ x ≤ 1 (1)

Subject to the boundary conditions

U ( 0 ) = 1 , U ( 1 ) = 0 U 1 ( 0 ) = 0 , U 1 ( 1 ) = − e U 2 ( 0 ) = − 1 , U 2 ( 1 ) = − 2 e U 3 ( 0 ) = − 2 (2)

Whose analytical solution is U ( x ) = ( 1 − x ) e x

Transformed formular is

U ( k + 7 ) = k ! ( k + 7 ) ! [ ∑ l = 0 k ∂ ( l − 1 ) U ( k − l ) + ∑ l = 0 k ∂ ( l − 2 ) ( k − l ) ! − 2 ∑ l = 0 k ∂ ( l − 1 ) ( k − l ) ! − 6 k ! ]

Transformed boundary conditions are

U ( 0 ) = 1 , U ( 1 ) = 0 , U ( 2 ) = − 1 2 ! , U ( 3 ) = − 2 3 ! , U ( 4 ) = P , U ( 5 ) = Q , U ( 6 ) = R

at k = 1

U ( 8 ) = 1 ! ( 8 ) ! [ ∑ l = 0 1 ∂ ( l − 1 ) U ( 1 − l ) + ∑ l = 0 1 ∂ ( l − 2 ) ( 1 − l ) ! − 2 ∑ l = 0 1 ∂ ( l − 1 ) ( 1 − l ) ! − 6 1 ! ] = − 1 5760

k = 2

U ( 9 ) = 2 ! ( 9 ) ! [ ∑ l = 0 2 ∂ ( l − 1 ) U ( 2 − l ) + ∑ l = 0 2 ∂ ( l − 2 ) ( 2 − l ) ! − 2 ∑ l = 0 2 ∂ ( l − 1 ) ( 2 − l ) ! − 6 2 ! ] = − 1 45360

k = 3

U ( 10 ) = 3 ! ( 10 ) ! [ ∑ l = 0 3 ∂ ( l − 1 ) U ( 3 − l ) + ∑ l = 0 3 ∂ ( l − 2 ) ( 3 − l ) ! − 2 ∑ l = 0 3 ∂ ( l − 1 ) ( 3 − l ) ! − 6 3 ! ] = − 1 403200

k = 4

U ( 11 ) = 4 ! ( 11 ) ! [ ∑ l = 0 4 ∂ ( l − 1 ) U ( 4 − l ) + ∑ l = 0 4 ∂ ( l − 2 ) ( 4 − l ) ! − 2 ∑ l = 0 4 ∂ ( l − 1 ) ( 4 − l ) ! − 6 4 ! ] = − 1 3991680

k = 5

U ( 12 ) = 5 ! ( 12 ) ! [ ∑ l = 0 5 ∂ ( l − 1 ) U ( 5 − l ) + ∑ l = 0 5 ∂ ( l − 2 ) ( 5 − l ) ! − 2 ∑ l = 0 5 ∂ ( l − 1 ) ( 5 − l ) ! − 6 5 ! ] = 1 + 30 P 119250400

k = 6

U ( 13 ) = 6 ! ( 13 ) ! [ ∑ l = 0 6 ∂ ( l − 1 ) U ( 6 − l ) + ∑ l = 0 6 ∂ ( l − 2 ) ( 6 − l ) ! − 2 ∑ l = 0 6 ∂ ( l − 1 ) ( 6 − l ) ! − 6 6 ! ] = 1 + 60 Q 518918400

k = 7

U ( 14 ) = 7 ! ( 14 ) ! [ ∑ l = 0 7 ∂ ( l − 1 ) U ( 7 − l ) + ∑ l = 0 7 ∂ ( l − 2 ) ( 7 − l ) ! − 2 ∑ l = 0 7 ∂ ( l − 1 ) ( 7 − l ) ! − 6 7 ! ] = 22 + 7 ! R 14 !

U ( x ) = ∑ k = 0 n U ( k ) x k = 1 − 1 2 x 2 − 1 3 x 3 + P x 4 + Q x 5 + R x 6 − 1 840 x 7 − 1 5760 x 8 − 1 45360 x 9 − 1 403200 x 10 − 1 3991680 x 11 + 1 + 30 P 119750680 x 12 + 1 + 60 Q 518918400 x 13 + 22 + 7 ! R 14 ! x 14 ^{ }

Using the conditions given in (2), the required equation is (

U ( x ) = 1 − 1 2 x 2 − 1 3 x 3 − 0.125004443 x 4 − 0.033324638 x 5 − 6.948728418 × 10 − 3 x 6 − 1 840 x 7 − 1 5760 x 8 − 1 45360 x 9 − 1 403200 x 10 − 1 3991680 x 11 − 2.75013329 119750400 x 12 − 0.99947828 518918400 x 13 − 13.02159123 14 ! x 14

X | Analytical solution (h = 0.1) | DTM (h = 0.1) | Error |
---|---|---|---|

0.0 | 1 | 1 | 0 |

0.1 | 0.994653826 | 0.994653825 | 1 × 10^{−}^{9} |

0.2 | 0.977122206 | 0.977122201 | 5 × 10^{−}^{9} |

0.3 | 0.944901165 | 0.944901147 | 1.8 × 10^{−}^{8} |

0.4 | 0.895094878 | 0.895094776 | 4.2 × 10^{−}^{8} |

0.5 | 0.824360635 | 0.824360563 | 7.2 × 10^{−}^{8} |

0.6 | 0.72884752 | 0.728847377 | 1.43 × 10^{−}^{7} |

0.7 | 0.604125812 | 0.604125704 | 1.08 × 10^{−7} |

0.8 | 0.445108185 | 0.445108094 | 9.1 × 10^{−}^{8} |

0.9 | 0.245960311 | 0.245960253 | 5.8 × 10^{−}^{8} |

1.0 | 0 | 5.621870471 × 10^{−}^{5} | 5.6 × 10^{−}^{5} |

In this paper, the differential transformation method is used to find the solution of higher order boundary value problems (order seven and eight). The results show that the convergence and accuracy of the method for numerically analysed eight order boundary value problem are in agreement with the analytical solutions. The method is easy to apply and can be applied easily to similar problems that engineering problems. Further work can be done on higher orders.

The authors declare no conflicts of interest regarding the publication of this paper.

Ogunrinde, R.B. and Ojo, O.M. (2018) Application of Differential Transformation Method to Boundary Value Problems of Order Seven and Eight. American Journal of Computational Mathematics, 8, 269-278. https://doi.org/10.4236/ajcm.2018.83022