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Integral-based solution for a class of second order boundary value problems
ELSiVIER
Applied
Mathematics
and Computation
98 (1999)
4348
Integral-based solution for a class of second order boundary value problems H.J. Al-Gahtani King Fahd
University
of Petroleum and Miner&, Civil Engineering Box 800. Dhahran 31261. Saudi Arabia
Department,
Abstract
An integral formulation for the solution of a classof secondorder boundary value problemswhich are describedby the equation d2y/dx2+ P(x,y, dy/&, d2y/dx2) = 0, x E (O,a), is presented.The resultingintegral equationsare then solved by expressing the dependentvariable y as a power serieswhich made the computation of various integrals possible.The proposedmethod is tested through someexamplesto show the applicability of the method to solve a wide range of secondorder differential equations including the nonlinear ones. 0 1999 Elsevier ScienceInc. All rights reserved. Keywords; equations
Boundary
element
method;
Integral
equations:
Nonlinear
ordinary
differential
1. Introduction Many problems in applied science and engineering are modelled by ordinary differential equations. Classical methods for obtaining an explicit formula for the solution are limited to certain types of equations. Particularly in nonlinear ordinary differential equations, obtaining such solutions becomes difficult and sometimes it is impossible and therefore one has to resort to semi-analytical or numerical methods. The objective of this paper is to present a boundary integral method [l] for solving a class of boundary value problems which are represented by second order ordinary differential equations of the following form:
0096-3003/99/$ ~- see front PII: s0096-3003(97)l0150-3
matter
0 1999 Elsevier
Science
Inc. All rights
reserved
H.J. Al-Gahtani
44
g++,Y>$$)
I Appl.
= 0,
Math.
Comput.
98 (1999)
43-48
x E (O,a),
(1)
where P is a polynomial function of x,y, dy/dx, d2y/dx2 and therefore the differential equation could be linear or nonlinear depending on the form of P. The proposed method consists of two steps: 1. Casting Eq. (1) in an integral form (Eqs. (7) and (8)). 2. Solving the resulting integral equations by expressing the dependent variable y as a power series, the coefficients of which can be obtained by equating the terms of similar powers. Finally, the proposed method is tested by three examples.
2. Integral equations
In order to cast Eq. (1) in an integral form, let us multiply both sides of Eq. (1) by a weighing function y* and integrate, by parts, twice over the range from x = 0 to x = a to get
Let us choose y* such that it satisfies the following equation:
where d(x - 0 = Dirac delta function. following equation y(C)
=
[gy*];-
An integration
Using Eq. (3) in Eq. (2) we get the
[yg],+pf&=o.
of Eq. (3) yields [2]
where sgn(x - <) = 1 for
x
> 5 and sgn(x - t) = -1 for x < r, and therefore
y* = -$sgn (x - <)1(x - 0.
(6)
Substituting for y* and dy*/dx from Eq. (5) and Eq. (6) into Eq. (4) yields the following equation:
H.J. Al-Gahtani
I Appl.
Math.
Comput.
98 (1999)
43-48
45
y(~)=ly(0)+~y(a)+Qqo)+~-ady Tz(4 2 2 2dx +;
V
oc(x - ()P dx - /“(x <
1
- ()P dx .
Applying Eq. (7) at 5 = 0 and [ = a yields the following written in a matrix form:
two equations
3. Solution of the integral equations
If the polynomial P is a function of x only, a closed form solution can be obtained easily by carrying out the integration in Eq. (8) to obtain the nonspecified boundary conditions, then using the result in Eq. (7) to get the solution [3]. If P, however, happens to be in addition a function of y and /or its derivatives then the following procedure is proposed: First, we represent y(t) by a power series, i.e. Y(5) = TUkCk.
(9)
k=O
For simplicity, let us assume that either y or dyldx (or their linear combination) is prescribed at 4 = 0 and 5 = a. Then, inserting Eq. (9) into the righthand side of Eq. (8) will transform the difficult integrals on the right-hand side of Eq. (8) into more readily solvable integrals and therefore the two equations can be solved for the remaining boundary values in terms of the coefficients ak. In order to obtain the coefficients ak, k B 0, use Eq. (9) and the results obtained from Eq. (8) in Eq. (7) and integrate the integrals term by term to obtain an equation of the following form:
where pk is a polynomial in ak the form of which depends on the form of P that we started with. Equating the coefficients of similar powers of r from both sides, i.e. ak = pkr leads to the complete determination of the coefficients ak, k 3 0, hence obtaining the required solution of Eq. (1). It should be noted that if P is nonlinear, pk will also be nonlinear and therefore the determination
H.J. Al-Gahtani
46
I Appl. Math.
Comput.
98 (1999)
43-48
of the coefficients ak will involve solving nonlinear algebraic equations. check the proposed method, three examples are given below.
To
Example 1. For the purpose of explaining the procedure of the method in detail, we will start with the following simple differential equation
d2y Q-Y=% so that P is simply equal to y. Let us consider the boundary conditions of y(0) = 0 and dy/dx(O) = 1. Then inserting y as expressed by Eq. (9) in Eq. (8) along with the above boundary conditions and solving Eq. (8) for the other two boundary conditions to obtain y(1) = 1 +$+$+$+$+$+S+...
(12)
Using Eqs. (12) and (13) along with the given boundary conditions and the power series expression of y in Eq. (7) and collecting coefficients of same power of 5, yield the following equation a0 +
al<
+ a2C2
+
a3C3
+
ad4
+ a&
+
..
(14) Equating the coefficients of similar powers of r, we obtain a2k
=
Substituting
0,
Eq. (15) into Eq. (4) we obtain the solution y(5) in a series form
y(5) =z l),eZk+l, C2k y
(16)
which is the Taylor expansion for the closed form solution ~(5)
= sinh (0
Example 2. Consider the following cients
(17) differential equation with variable coeffi-
dy (x- l)Qd2y-x-&+y-1’0 along with the boundary condition
of y(0) = 0 and dyldx(0) = 1.
H.J. Al-Gahtani
I Appl.
Math.
To apply the proposed method, Eq. (l), i.e. !&($-g)
-y+l
Following
Comput.
98 (1999)
43-48
41
let us first put Eq. (18) in the form of
=o.
the same procedure, we get y and dy/&
both evaluated at x = 1:
$(l)=ao+~+~+!$!+l!p+...
(21)
and the process of using Eq. (20) and Eq. (21) in Eq. (7) and equating coefficients of similar terms yields the solution
(22)
6 24 120’ which is the Taylor expansion for the closed form solution y=2x-e”+l.
Example 3. In order to demonstrate the applicability of the method to solve nonlinear ordinary differential equations, let us consider the following nonlinear differential equation
(24) Assuming the boundary conditions ing the same procedure, we get
of y(0) = 0, dy/dx( 1) = l/2 and apply-
ao+a~~+u*52+u3~3+a454+a5~5+‘~’ =
-5
+
%$
+
c1 +
;la2)
<3 _
Cal
-
2a:
-
3ulu3)
(4
(25)
6 _
(a2
-
3a2a3
-
2W4)
p
+
. .
5
Equating coefficients of similar powers of r yields: a0 = 0,
a] = -1,
47
=
;,
u3
=
0,
a4
=
I 5,
a5
=
-!
(26)
and hence the solution becomes
y(g=-~+c+14-e+... 2 4 5 whose closed form is
(27)
48
H.J. Al-Gahtani
y = +x3 - ln(x + I), which is the analytical solution although, the coefficients on functions of &, obtaining the algebraic equations. However,
I Appl. Math.
Comput.
98 (1999)
43-48
(28)
for the given differential equation [4]. Note that the right-hand side of Eq. (25) are nonlinear coefficients ak did not require solving nonlinear this will not be the case in general.
Acknowledgements
The author is indebted to King Fahd University of Petroleum for providing excellent research facilities.
and Minerals
References [I]
P.K. Banerjee, R. Butterheld, Boundary Element Methods in Engineering Science, McGrawHill, London, 1981. [2] D.E. Beskos, Boundary Element Methods in Structural Analysis, ASCE, New York, 1989. [3] R. Butterfield, New concepts illustrated by old problems, in: P.K. Banerjee, R. Butterfield (Ed.), Development in Boundary Element Methods, London, 1982. [4] F. Ayres, Theory and Problems of Differential Equations, Schaum’s Outline Series, McGrawHill, New York, 1974.
Applied
Mathematics
and Computation
98 (1999)
4348
Integral-based solution for a class of second order boundary value problems H.J. Al-Gahtani King Fahd
University
of Petroleum and Miner&, Civil Engineering Box 800. Dhahran 31261. Saudi Arabia
Department,
Abstract
An integral formulation for the solution of a classof secondorder boundary value problemswhich are describedby the equation d2y/dx2+ P(x,y, dy/&, d2y/dx2) = 0, x E (O,a), is presented.The resultingintegral equationsare then solved by expressing the dependentvariable y as a power serieswhich made the computation of various integrals possible.The proposedmethod is tested through someexamplesto show the applicability of the method to solve a wide range of secondorder differential equations including the nonlinear ones. 0 1999 Elsevier ScienceInc. All rights reserved. Keywords; equations
Boundary
element
method;
Integral
equations:
Nonlinear
ordinary
differential
1. Introduction Many problems in applied science and engineering are modelled by ordinary differential equations. Classical methods for obtaining an explicit formula for the solution are limited to certain types of equations. Particularly in nonlinear ordinary differential equations, obtaining such solutions becomes difficult and sometimes it is impossible and therefore one has to resort to semi-analytical or numerical methods. The objective of this paper is to present a boundary integral method [l] for solving a class of boundary value problems which are represented by second order ordinary differential equations of the following form:
0096-3003/99/$ ~- see front PII: s0096-3003(97)l0150-3
matter
0 1999 Elsevier
Science
Inc. All rights
reserved
H.J. Al-Gahtani
44
g++,Y>$$)
I Appl.
= 0,
Math.
Comput.
98 (1999)
43-48
x E (O,a),
(1)
where P is a polynomial function of x,y, dy/dx, d2y/dx2 and therefore the differential equation could be linear or nonlinear depending on the form of P. The proposed method consists of two steps: 1. Casting Eq. (1) in an integral form (Eqs. (7) and (8)). 2. Solving the resulting integral equations by expressing the dependent variable y as a power series, the coefficients of which can be obtained by equating the terms of similar powers. Finally, the proposed method is tested by three examples.
2. Integral equations
In order to cast Eq. (1) in an integral form, let us multiply both sides of Eq. (1) by a weighing function y* and integrate, by parts, twice over the range from x = 0 to x = a to get
Let us choose y* such that it satisfies the following equation:
where d(x - 0 = Dirac delta function. following equation y(C)
=
[gy*];-
An integration
Using Eq. (3) in Eq. (2) we get the
[yg],+pf&=o.
of Eq. (3) yields [2]
where sgn(x - <) = 1 for
x
> 5 and sgn(x - t) = -1 for x < r, and therefore
y* = -$sgn (x - <)1(x - 0.
(6)
Substituting for y* and dy*/dx from Eq. (5) and Eq. (6) into Eq. (4) yields the following equation:
H.J. Al-Gahtani
I Appl.
Math.
Comput.
98 (1999)
43-48
45
y(~)=ly(0)+~y(a)+Qqo)+~-ady Tz(4 2 2 2dx +;
V
oc(x - ()P dx - /“(x <
1
- ()P dx .
Applying Eq. (7) at 5 = 0 and [ = a yields the following written in a matrix form:
two equations
3. Solution of the integral equations
If the polynomial P is a function of x only, a closed form solution can be obtained easily by carrying out the integration in Eq. (8) to obtain the nonspecified boundary conditions, then using the result in Eq. (7) to get the solution [3]. If P, however, happens to be in addition a function of y and /or its derivatives then the following procedure is proposed: First, we represent y(t) by a power series, i.e. Y(5) = TUkCk.
(9)
k=O
For simplicity, let us assume that either y or dyldx (or their linear combination) is prescribed at 4 = 0 and 5 = a. Then, inserting Eq. (9) into the righthand side of Eq. (8) will transform the difficult integrals on the right-hand side of Eq. (8) into more readily solvable integrals and therefore the two equations can be solved for the remaining boundary values in terms of the coefficients ak. In order to obtain the coefficients ak, k B 0, use Eq. (9) and the results obtained from Eq. (8) in Eq. (7) and integrate the integrals term by term to obtain an equation of the following form:
where pk is a polynomial in ak the form of which depends on the form of P that we started with. Equating the coefficients of similar powers of r from both sides, i.e. ak = pkr leads to the complete determination of the coefficients ak, k 3 0, hence obtaining the required solution of Eq. (1). It should be noted that if P is nonlinear, pk will also be nonlinear and therefore the determination
H.J. Al-Gahtani
46
I Appl. Math.
Comput.
98 (1999)
43-48
of the coefficients ak will involve solving nonlinear algebraic equations. check the proposed method, three examples are given below.
To
Example 1. For the purpose of explaining the procedure of the method in detail, we will start with the following simple differential equation
d2y Q-Y=% so that P is simply equal to y. Let us consider the boundary conditions of y(0) = 0 and dy/dx(O) = 1. Then inserting y as expressed by Eq. (9) in Eq. (8) along with the above boundary conditions and solving Eq. (8) for the other two boundary conditions to obtain y(1) = 1 +$+$+$+$+$+S+...
(12)
Using Eqs. (12) and (13) along with the given boundary conditions and the power series expression of y in Eq. (7) and collecting coefficients of same power of 5, yield the following equation a0 +
al<
+ a2C2
+
a3C3
+
ad4
+ a&
+
..
(14) Equating the coefficients of similar powers of r, we obtain a2k
=
Substituting
0,
Eq. (15) into Eq. (4) we obtain the solution y(5) in a series form
y(5) =z l),eZk+l, C2k y
(16)
which is the Taylor expansion for the closed form solution ~(5)
= sinh (0
Example 2. Consider the following cients
(17) differential equation with variable coeffi-
dy (x- l)Qd2y-x-&+y-1’0 along with the boundary condition
of y(0) = 0 and dyldx(0) = 1.
H.J. Al-Gahtani
I Appl.
Math.
To apply the proposed method, Eq. (l), i.e. !&($-g)
-y+l
Following
Comput.
98 (1999)
43-48
41
let us first put Eq. (18) in the form of
=o.
the same procedure, we get y and dy/&
both evaluated at x = 1:
$(l)=ao+~+~+!$!+l!p+...
(21)
and the process of using Eq. (20) and Eq. (21) in Eq. (7) and equating coefficients of similar terms yields the solution
(22)
6 24 120’ which is the Taylor expansion for the closed form solution y=2x-e”+l.
Example 3. In order to demonstrate the applicability of the method to solve nonlinear ordinary differential equations, let us consider the following nonlinear differential equation
(24) Assuming the boundary conditions ing the same procedure, we get
of y(0) = 0, dy/dx( 1) = l/2 and apply-
ao+a~~+u*52+u3~3+a454+a5~5+‘~’ =
-5
+
%$
+
c1 +
;la2)
<3 _
Cal
-
2a:
-
3ulu3)
(4
(25)
6 _
(a2
-
3a2a3
-
2W4)
p
+
. .
5
Equating coefficients of similar powers of r yields: a0 = 0,
a] = -1,
47
=
;,
u3
=
0,
a4
=
I 5,
a5
=
-!
(26)
and hence the solution becomes
y(g=-~+c+14-e+... 2 4 5 whose closed form is
(27)
48
H.J. Al-Gahtani
y = +x3 - ln(x + I), which is the analytical solution although, the coefficients on functions of &, obtaining the algebraic equations. However,
I Appl. Math.
Comput.
98 (1999)
43-48
(28)
for the given differential equation [4]. Note that the right-hand side of Eq. (25) are nonlinear coefficients ak did not require solving nonlinear this will not be the case in general.
Acknowledgements
The author is indebted to King Fahd University of Petroleum for providing excellent research facilities.
and Minerals
References [I]
P.K. Banerjee, R. Butterheld, Boundary Element Methods in Engineering Science, McGrawHill, London, 1981. [2] D.E. Beskos, Boundary Element Methods in Structural Analysis, ASCE, New York, 1989. [3] R. Butterfield, New concepts illustrated by old problems, in: P.K. Banerjee, R. Butterfield (Ed.), Development in Boundary Element Methods, London, 1982. [4] F. Ayres, Theory and Problems of Differential Equations, Schaum’s Outline Series, McGrawHill, New York, 1974.