# An Approximate Solution for Boundary Value Problems in Structural

Authors: A. Barari; M. Omidvar; D. D. Ganji; Abbas Tahmasebi Poor
Journal: Mathematical Problems in Engineering
 (2008)

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## Abstract

Variational iteration method (VIM) is applied to solve linear and nonlinear boundary value problems with particular significance in structural engineering and fluid mechanics. These problems are used as mathematical models in viscoelastic and inelastic flows, deformation of beams, and plate deflection theory. Comparison is made between the exact solutions and the results of the variational iteration method (VIM). The results reveal that this method is very effective and simple, and that it yields the exact solutions. It was shown that this method can be used effectively for solving linear and nonlinear boundary value problems.


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## Body

## 1. Introduction

This paper discusses the analytical approximate
solution for fourth-order equations with nonlinear boundary conditions
involving third-order derivatives. The general form of the equation for a fixed
positive integern,n≥2,
is a differential equation of order 2n:
(1.1)y(2n)+f(x,y)=0 subject to the boundary conditions (1.2)y(2j)(a)=A2j,y(2j)(b)=B2j,j=0(1)n−1, where −∞<a≤x≤b<∞,A2j,B2j,j=0(1)n−1 are finite constants.It is assumed thaty is sufficiently
differentiable and that a unique solution of (1.1) exists. Problems of this kind
are commonly encountered in plate-deflection theory and in fluid mechanics for
modeling viscoelastic and inelastic flows [1–3]. Usmani [1, 2] discussed sixth
order methods for the linear differential equation y(4)+P(x)y=q(x) subject to the boundary conditions y(a)=A0, y″(A)=A2, y(b)=B0,y″(b)=B2.
The method described in [1] leads to five diagonal linear systems and involves
p′,p′′,q′,q′′ at a and b, while the method described
in [2] leads to nine diagonal linear systems.Ma and Silva [4] adopted iterative solutions for (1.1)
representing beams on elastic foundations. Referring to the classical beam
theory, they stated that if u=u(x) denotes the configuration of the deformed
beam, then the bending moment satisfies the relation M=-EIu///, where E is the Young modulus of elasticity and
I is the inertial moment. Considering the deformation caused by a load f=f(x), they deduced, from a free-body diagram, that f=-v/ and v=M/=-EIu///, where v denotes the shear force. For u representing an elastic beam of length L=1, which is clamped at its left side x=0, and resting on an elastic bearing at its right
side x =1, and adding a load f along its length to cause deformations
(Figure 1), Ma and Silva [4] arrived at
the following boundary value problem assuming an EI=1: (1.3)u(iv)(x)=f(x,u(x)),0<x<1, the boundary conditions were taken as (1.4)u(0)=u/(0)=0,(1.5)u//(1)=0,u///(1)=g(u(1)), where f∈C([0,1]×ℝ) and g∈C(ℝ) are real functions. The physical
interpretation of the boundary conditions is that u///(1) is the shear force at x=1, and the second condition in (1.5) means that the
vertical force is equal to g(u(1)), which denotes a relation, possibly nonlinear,
between the vertical force and the displacement u(1). Furthermore, since u//(1)=0 indicates that there is no bending moment at x=1, the beam is resting on the bearing g.Figure 1
Beam on elastic bearing.Solving (1.3) by means of iterative procedures, Ma and
Silva [4] obtained solutions and argued that the accuracy of results depends
highly upon the integration method used in the iterative process.With the rapid development of nonlinear science,
many different methods were proposed to solve differential equations, including
boundary value problems (BVPS). These two methods are the homotopy perturbation
method (HPM) [5–7] and the variational iteration method (VIM) [8–17]. In this
paper, it is aimed to apply the variational iteration method proposed
by He [14] to different forms of (1.1) subject to boundary conditions of physical
significance.


## 2. Basic Idea of He’s Variational Iteration Method

To clarify the
basic ideas of He's VIM, the following differential equation is considered:(2.1)L[u(t)]+N[u(t)]=g(t), where L is a linear
operator, N is a nonlinear
operator, and g(t) is an
inhomogeneous term. According to VIM, a correction functional could be written as
follows:(2.2)un+1(t)=un(t)+∫0tλ(τ)(Lun(τ)+Nu˜n(τ)−g(τ))dτ,where λ is a general Lagrange multiplier which can be
identified optimally via the variational theory. The subscript n indicates the nth approximation and u˜n is considered as a restricted variation, that
is, δu˜n=0.For fourth-order boundary
value problem with suitable boundary conditions, Lagrangian multiplier can be identified by
substituting the problem into (2.2), upon making it stationary leads to the following:
(2.3)d4dτ4λ=0,−λ‴+1|τ=x=0,λ″|τ=x=0.Solving the system of (2.3) yields(2.4)λ=16(τ−x)3 and the variational iteration formula is obtained in the form
(2.5)un+1(x)=un(x)+∫0x16(τ−x)3(un(4)(τ)+f(τ,un,un′,un′′,un′′′))dτ.


## 3. The Applications of VIM Method

In this section, the VIM is applied to different
forms of the fourth-order boundary value problem introduced in through (1.1).Example 3.1.
Consider the following linear
boundary value problem:(3.1)u(4)(x)=4ex+u(x),0<x<1,subject to the boundary conditions(3.2)u(0)=1,u′(0)=2,u(1)=2e,u′(1)=3e.
The exact
solution for this problem is (3.3)u(x)=(1+x)ex. According to (2.5),
the following iteration formulation is achieved:(3.4)un+1(x)=un(x)+∫0x16(τ−x)3(un(4)(τ)−un(τ)−4eτ)dτ. Now it is assumed
that an initial approximation has the form (3.5)u0(x)=ax3+bx2+cx+d, where a,b,c, and d are unknown constants to be further determined.
By the iteration formula (3.4), the
following first-order approximation may be written: (3.6)u1(x)=u0(x)+∫0x16(τ−x)3(u0(4)(τ)−u0(τ)−4eτ)dτ=ax3+bx2+cx+d+∫0x16(τ−x)3(−aτ3−bτ2−cτ−d−4eτ)dτ=1840ax7+1360bx6+1120cx5+124dx4+(−23+a)x3+(b−2)x2+(c−4)x+4ex+d−4. Incorporating
the boundary conditions (3.2), into u1(x),
the following coefficients can
be obtained:(3.7)a=−2289756301681+916440301681e,b=4575063301681−1516680301681e,c=2,d=1. Therefore, the
following first-order approximate solution is derived: (3.8)u1(x)=(−272593016810+1091301681e)x7+(152502136201720−4213301681e)x6+160x5+124x4+(−7472630905043+916440301681e)x3+(3971701301681−1516680301681e)x2−2x−3+4ex. Comparison of
the first-order approximate solution with exact solution is tabulated in
Table 1, showing a remarkable agreement.
Similarly, the following second-order approximation is obtained:(3.9)u2(x)=u1(x)+∫0x16(τ−x)3(u1(4)(τ)−u1(τ)−4eτ)dτ=16652800ax11+11814400bx10+1362880cx9+140320dx8+(1840a−11260)x7+(1360b−1180)x6+(−130+1120c)x5+(−16+124d)x4+(−43+a)x3+(b−4)x2+(c−8)x−8+8ex+d,a=−12706529114180681628862391+8553568161600012042109902241e,c=2,b=8416302814865227209620797−15745272661440012042109902241e,d=1. Therefore, the second-order
approximate solution may be written as
(3.10)u2(x)=(−5775695051920612456798703840+1285709512042109902241e)x11+(168326056297382449827194815360−8677950112042109902241e)x10+1181440x9+140320x8+(−73116379754331809346911580+10182819240012042109902241e)x7+(796188357327181795463486920−43736868504012042109902241e)x6−160x5−18x4+(−13615367597368681628862391+8553568161600012042109902241e)x3+(7507464331677227209620797−15745272661440012042109902241e)x2−6x−7+8ex. Again, the obtained
solution is of distinguishing accuracy, as indicated in
Table 2 and Figure 2.Table 1
Comparison of the
first-order approximate solution with exact solution.
xUEU1Error01.0000000001.0000000000.0000E + 0000.11.2156880101.2156815246.4860E−0060.21.4656833101.4656608902.2420E−0050.31.7548164501.7547739234.2527E−0050.42.0885545772.0884929796.1598E−0050.52.4730819062.4730072657.4641E−0050.62.9153900802.9153127347.7346E−0050.73.4233796023.4233125926.7010E−0050.84.0059736704.0059294044.4266E−0050.94.6732459114.6732298911.6020E−0051.02e2e0.0000E + 000Table 2
Comparison of the second-order
approximate solution with exact solution.
xUEU2Error01.0000000001.0000000000.0E + 0000.11.2156880101.2156880082.0E−0090.21.4656833101.4656833055.0E−0090.31.7548164501.7548164446.0E−0090.42.0885545772.0885545661.1E−0080.52.4730819062.4730819024.0E−0090.62.9153900802.9153900641.6E−0080.73.4233796023.4233796002.0E−0090.84.0059736704.0059736502.0E−0080.94.6732459114.6732459301.9E−0081.02e2e0.0E + 000Figure 2
Comparison between
different solutions.Example 3.2.
Consider
the following linear boundary value problem:(3.11)u(4)(x)=u(x)+u″(x)+ex(x−3),0<x<1, subject to the boundary conditions(3.12)u(0)=1,u′(0)=0,u(1)=0,u′(1)=−e. The exact
solution for this problem is(3.13)u(x)=(1−x)ex. According to (2.5), the iteration
formulation may be written as (3.14)un+1(x)=un(x)+∫0x16(τ−x)3(un(4)(τ)−un(τ)−uu′′(τ)−eτ(τ−3))dτ. Now it is assumed that an initial
approximation has the form(3.15)u0(x)=ax3+bx2+cx+d. Where a,b,c,
and d are unknown constants to be further determined.
By the iteration formula (3.14), the
following first-order approximation is developed:(3.16)u1(x)=u0(x)+∫0x16(τ−x)3(u0(4)(τ)−u0(τ)−u0′′(τ)−eτ(τ−3))dτ=ax3+bx2+cx+d+∫0x16(τ−x)3(−aτ3−bτ2−(6a+c)τ−2b−d−eτ(τ−3))dτ=1840ax7+1360bx6+(120a+1120c)x5+(112b+124d)x4+(23+a)x3+(b+52)x2+(ex+6+c)x−7ex+7+d. Incorporating the boundary
conditions (3.12), into u1(x),
it can be written as(3.17)a=7904470323149−2950080323149e,b=−12770295323149+4640400323149e,c=0,d=1. Therefore, the following first-order
approximate solution is obtained:(3.18)u1(x)=(1129213877788−3512323149e)x7+(−8513537755576+12890323149e)x6+(790447646298−147504323149e)x5+(−252174417755576+386700323149e)x4+(24359708969447−2950080323149e)x3+(−23924845646298+4640400323149e)x2+(6+ex)x+8−7ex. Comparison of
the first-order approximate solution with exact solution is tabulated in
Table 3,
again showing a clear agreement. Even higher accurate solutions could be
obtained without any difficulty.
Similarly, the following second-order
approximation can be written as(3.19)u2(x)=u1(x)+∫0x16(τ−x)3(u1(4)(τ)−u1(τ)−u1′′(τ)−eτ(τ−3))dτ=16652800ax11+11814400bx10+(1362880c+130240a)x9+(110080b+140320d)x8+(15040c+1420a+11260)x7+(1720d+1144+1180b)x6+(112+120a+1120c)x5+(12+124d+112b)x4+(3+a)x3+(b+212)x2+(24+3ex+c)x+27−27ex+d. Incorporating
the boundary conditions, (3.12), into u2(x),
yields (3.20)a=3818047893001104289712004667−1409850288000004289712004667e,c=0,b=−6294953010820654289712004667+2307900373632004289712004667e,d=1. The following
second-order approximate solution is then achieved in the following form:(3.21)u2(x)=(3470952630001259441782042260160−6357550012869136014001e)x11+(−41966353405471518883564084520320+38159728412869136014001e)x10+(3818047893001112972089102113008−1398661000012869136014001e)x9+(−2513691492323593172961188028173440+228958370404289712004667e)x8+(11497040799049975405037125880420−3356786400004289712004667e)x7+(−415373822050043514765440560040+12821668742404289712004667e)x6+(23337258558473351476544056004−70492514400004289712004667e)x5+(−1203224346103459102953088112008+192325031136004289712004667e)x4+(3946739253141114289712004667−1409850288000004289712004667e)x3+(−11689066500661238579424009334+2307900373632004289712004667e)x2+(3ex+24)x+28−27ex. The obtained solution is of evident accuracy, as shown in
Table 4 and Figure 3.Table 3
Comparison of the
first-order approximate solution with exact solution.
xUEU1Error01.00000000001.00000000000.0000000E + 0000.10.99465382620.99479315471.3932850E−0040.20.97712220640.97759490404.7269760E−0040.30.94490116560.94577762308.7645740E−0040.40.89509481880.89632972501.2349062E−0030.50.82436063550.82580874401.4481085E−0030.60.72884752000.73029192801.4444080E−0030.70.60412581210.60532408001.1982679E−0030.80.44510818560.44586254007.5435440E−0040.90.24596031110.24621930002.5898890E−0041.00.00000000000.00000000000.0000000E + 000Table 4
Comparison of the
second-order approximate solution with exact solution.
xUEU2Error01.00000000001.00000000000.00000E + 0000.10.99465382620.99465775803.93180E−0060.20.97712220640.97713577801.35716E−0050.30.94490116560.94492689002.57244E−0050.40.89509481880.89513211003.72912E−0050.50.82436063550.82440588004.52445E−0050.60.72884752000.72889453004.70100E−0050.70.60412581210.60416665004.08379E−0050.80.44510818560.44513528002.70944E−0050.90.24596031110.24597013009.81890E−0061.00.00000000000.00000000000.00000E + 000Figure 3
Comparison between
different solutions.Example 3.3.
Consider
the following nonlinear boundary value problem:(3.22)u(4)(x)=u2(x)+g(x),0<x<1, subject to the boundary conditions(3.23)u(0)=0,u′(0)=0,u(1)=1,u′(1)=1,where(3.24)g(x)=−x10+4x9−4x8−4x7+8x6−4x4+120x−48. The exact solution for this problem
is(3.25)u(x)=x5−2x4+2x2. According to (2.5), the iteration
formulation is written as follows: (3.26)un+1(x)=un(x)+∫0x16(τ−x)3(un(4)(τ)−un2(τ)−g(τ))dτ. Now it is assumed
that an initial approximation has the form(3.27)u0(x)=ax3+bx2+cx+d, where a,b,c,
and d are unknown constants to be further determined.
By the iteration
formula (3.26), the following first-order approximation is obtained:(3.28)u1(x)=u0(x)+∫0x16(τ−x)3(u0(4)(τ)−u02(τ)+τ10−4τ9+4τ8+4τ7−8τ6+4τ4−120τ+48)dτ=−124024x14+14290x13−12970x12−11980x11+(15040a2+1630)x10+11512abx9+(−1420+11680b2+1840ac)x8+(1420bc+1420ad)x7+(1180bd+1360c2)x6+(1+160cd)x5+(124d2−2)x4+ax3+bx2+cx+d. Incorporating the boundary
conditions (3.23), into u1(x),
results in the following values: (3.29)a=−0.006871650809;b=2.005929593;c=0,d=0. The following
first-order approximate solution is then achieved:(3.30)u1(x)=−4.162504162×10−5x14+2.331002331×10−4x13−3.367003367×10−4x12−5.050505050×10−4x11+1.587310956×10−3x10−9.116433669×10−6x9+1.4139007×10−5x8+x5−2x4−6.871650809×10−3x3+2.005929593x2. Comparison of
the first-order approximate solution with exact solution is tabulated in
Table 5,
which once again shows an excellent agreement.
Similarly, the
following second-order approximation may be written:(3.31)u2(x)=u1(x)+∫0x16(τ−x)3(u1(4)(τ)−u12(τ)+τ10−4τ9+4τ8+4τ7−8τ6+4τ4−120τ+48)dτ. Incorporating
the boundary conditions, (3.23), into u2(x),
yields (3.32)a=−8.269548014E−7;b=2.000000763;c=0,d=0. The following second-order
approximate solution is obtained:(3.33)u2(x)=−1.093855974×10−9x9+1.817×10−9x8−2x4−1.117934793×10−8x21+1.463705892×10−9x20+6.586694874×10−8x19+2.000000763x2−8.269548014×10−7x3−1.047931585×10−7x18−3.536760165×10−8x17+1.453571773×10−7x16−5.173598972×10−13x28−2.569735395×10−14x31+1.252296566×10−13x30−2.016131906×10−13x29+2.007605778×10−15x32+2.564345160×10−12x27+3.603899741×10−9x22+3.025×10−13x14−1.392179800×10−10x12+6.103539401×10−10x11+x5+9.879565106×10−12x24−2.268156651×10−12x26−5.281071651×10−12x25−3.917282540×10−10x23−1.335600908×10−13x15−6.059998643×10−10x10. The obtained solution is once again of remarkable accuracy, as shown in
Table 6 and Figure 4.Table 5
Comparison of the
first-order approximate solution with exact solution.
xUEU1Error00.00000000000.00000000000.0000000E + 0000.10.01981000000.01986242435.2424300E−0050.20.07712000000.07730221071.8221070E−0040.30.16623000000.16657813793.4813790E−0040.40.27904000000.27954909725.0909720E−0040.50.40625000000.40687472656.2472650E−0040.60.53856000000.53921782706.5782700E−0040.70.66787000000.66845113855.8113850E−0040.80.78848000000.78887270233.9270230E−0040.90.89829000000.89843569641.4569640E−0041.01.00000000001.00000000000.0000000E + 000Table 6
Comparison of the
second-order approximate solution with exact solution.
xUEU2Error00.00000000000.00000000000.000E + 0000.10.01981000000.01981000686.800E−0090.20.07712000000.07712002392.390E−0080.30.16623000000.16623004644.640E−0080.40.27904000000.27904006926.920E−0080.50.40625000000.40625008748.740E−0080.60.53856000000.53856009619.610E−0080.70.66787000000.66787009069.060E−0080.80.78848000000.78848006706.700E−0080.90.89829000000.89829002922.920E−0081.01.00000000001.00000000121.200E−009Figure 4
Comparison between different solutions.


## 4. Conclusion

This study
showed that the variational iteration method is remarkably effective for
solving boundary value problems. A fourth-order differential equation with
particular engineering applications was solved using the VIM in order to prove
its effectiveness. Different forms of the equation having boundary conditions
of physical significance were considered. Comparison between the approximate
and exact solutions showed that one iteration is enough to reach the exact
solution. Therefore the VIM is able to solve partial differential equations using
a minimum calculation process. This method is a very promoting method, which promises
to find wide applications in engineering problems.



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*Source: 394103-2008-07-02.xml*