# Numerical Solution of High Order Bernoulli Boundary Value Problems

Authors: F. Costabile; A. Napoli
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

For the numerical solution of high order boundary value problems with special boundary conditions a general procedure to determine collocation methods is derived and studied. Computation of the integrals which appear in the coefficients is generated by a recurrence formula and no integrals are involved in the calculation. Several numerical examples are presented to demonstrate the practical usefulness of the proposed method.


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## Body

## 1. Introduction

Higher order differential equations arise in a variety of different areas of science, engineering, and technology (see [1, 2]) since they model a wide spectrum of phenomena.Particularly, the solutions of fifth-order BVPs model viscoelastic flows [3] and the seventh-order BVPs model induction motors with two rotor circuits [4, 5]. Ordinary differential equations of sixth and eighth order arise in modeling instability when an infinite horizontal layer of fluid is heated from below and is subject to the action of rotation [6]. Moreover, high order boundary value problems arise in hydrodynamic, hydromagnetic stability [7], and other branches of applied sciences.In [8] the authors presented a class of collocation methods for the numerical solution of high order boundary value problems:
(1)y(n)(x)=f(x,y(x)),x∈I=[a,b],(2)B[x,y]=g,x∈∂I,
where n>1, y(x)=(y(x),y′(x),…,y(q)(x)), 0≤q<n, and B is a linear operator on the boundary ∂I, g∈Rn.The idea in [8] is the following: the differential problem (1)-(2) is written in the following equivalent integral form:
(3)y(x)=Pn-1[y,x]+∫ab‍Gn-1(x,t)f(t,y(t))dt,
where Pn-1[y,x] is the unique polynomial which satisfies the boundary conditions
(4)B[x,Pn-1]=g
and Gn-1(x,t) is a kernel (Green) function. Gn-1(x,t) is such that B[x,Gn-1]=0 and it is differentiable under the integral sign such that (3) satisfies (1).Thus, from (3) and (4) we obtain a collocation polynomial which approximates the solution of problem (1)-(2).In the present work the authors use this technique to derive collocation methods for the numerical solution of (1) with the particular boundary conditions
(5)y(a)=β0,y(k)(b)-y(k)(a)=βk+1,k=0,…,n-2
with βk, k=0,…,n-1, being real constants.Conditions (5) are called theBernoulli boundary conditions, since they are related to the Bernoulli interpolation problem [9]. They have physical and engineering interpretation [10], but to the authors’ knowledge, they have not been considered previously in the literature.In [9, 10] the BVP (1)–(5) is considered: in [10] a nonconstructive proof of the existence and uniqueness of solution is given, and in [9] Picard’s method is applied in connection with Newton’s method for the numerical solution of the problem.The present paper is organized as follows: in Section2 we summarize some theoretical results on the existence and uniqueness of solution for problem (1)–(5). Then, in Section 3, we present the method for the numerical solution of this type of problems, which produces smooth, global approximations in the form of polynomial functions. In Section 4 we give an a priori estimation of error and, in Section 5, we present some particular cases. In Section 6 we propose an algorithm to compute the numerical solution of (1)–(5) in the nodal points and then, in Section 7, we present some numerical examples of both linear and nonlinear BVPs which confirm the theoretical results.


## 2. Preliminaries

LetBk(x) be the Bernoulli polynomial of degree k [11] and let us set
(6)Sk(t)=Bk(t)-Bk(0).
Moreover, let
(7)h=b-a,Δfa(k)=f(k)(b)-f(k)(a).
The following theorems hold.Theorem 1 (see [12]).
Letf∈C(n+1)[a,b]. Then
(8)f(x)=f(a)+∑k=1n‍Sk(x-ah)hk-1k!Δfa(k-1)+Rn[f,x],
where Rn[f,x] is the remainder term
(9)Rn[f,x]=∫ab‍Gn(x,t)f(n+1)(t)dt
with Gn(x,t) being Peano’s kernel:
(10)Gn(x,t)=1n![(x-t)+n-∑k=1n‍Sk(x-ah)hk-1k×(nk-1)(b-t)n-k+1∑k=1n‍Sk(x-ah)hk-1k].Theorem 2 (see [12]).
Iff∈C(n-1)[a,b], then the polynomial
(11)Pn[f,x]=f(a)+∑k=1n‍Sk(x-ah)hk-1k!Δfa(k-1)
satisfies the Bernoulli interpolation problem
(12)Pn[f,a]=f(a),ΔPn(k)=Pn(k)[f,b]-Pn(k)[f,a]=f(k)(b)-f(k)(a)=Δfa(k),kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk=0,…,n-1.The proof of the existence and uniqueness of solution of (1)–(5) is based on (3) [8], under the hypothesis that the function f(x,y) satisfies the Lipschitz condition
(13)|f(x,y1(x))-f(x,y2(x))|≤∑k=0q‍Lk|y1(k)(x)-y2(k)(x)|
in a certain domain interval of [a,b]×Rq+1.


## 3. The Collocation Method

Lety(x) be the solution of (1)–(5). If xi, i=1,…,m, are m distinct points in [a,b] and y(x)∈C(n+m)[a,b], using Lagrange interpolation, we get
(14)y(n)(x)=∑i=1m‍li(x)y(n)(xi)+R¯m(y,x),
where
(15)R¯m(y,x)=(x-x1)⋯(x-xm)m!y(n+m)(ξx),ξx∈(a,b)
and li(t) are the fundamental Lagrange polynomials on the m points xi.Inserting (14) into (3), in view of (1), we obtain
(16)y(x)=Pn-1[y,x]+∑i=1m‍f(xi,y(xi))∫ab‍Gn-1(x,t)li(t)dt+∫ab‍Gn-1(x,t)R¯m(y,t)dt.
Hence the following identity holds:
(17)y(x)=Pn-1[y,x]+∑i=1m‍pn,i(x)f(xi,y(xi))+Tn,m(y,x),
where
(18)pn,i(x)=∫ab‍Gn-1(x,t)li(t)dt,i=1,…,m,(19)Tn,m(y,x)=∫ab‍Gn-1(x,t)R¯m(y,t)dt.This suggests defining the polynomials(20)yn,m(x)=Pn-1[y,x]+∑i=1m‍pn,i(x)f(xi,yn,m(xi)),
where yn,m(x)=(yn,m(x),yn,m′(x),…,yn,m(q)(x)), 0≤q≤n-1.Theorem 3.
The polynomial of degreen+m implicitly defined by (20) satisfies the relations
(21)yn,m(a)=y(a),(22)yn,m(k)(b)-yn,m(k)(a)=βk+1,k=0,…,n-2,(23)yn,m(n)(xi)=f(xi,yn,m(xi)),i=1,…,m;
that is, yn,m(x) is a collocation polynomial for (1)–(5) on the nodes xi, i=1,…,m.Proof.
From (18), pn,i(a)=pn,i(b)=0, i=0,…,m, and thus relations (21) follow from direct computation. To prove (22) we derive Gn-1(x,t)  k times, k=1,…,n-2, with respect to x, and using the well-known relation [11] Bs′(x)=sBs-1(x), s>0, we get
(24)∂k∂xkGn-1(x,t)={g1(x,t)=(x-t)n-k(n-k)!-∑j=kn‍Bj-k(x-ah)hj-k-1(b-t)n-j-1(j-k)!(n-j-1)!x≥tg2(x,t)=-∑j=kn‍Bj-k(x-ah)hj-k-1(b-t)n-j-1(j-k)!(n-j-1)!x<t.
From the property of Bernoulli polynomials Bs(1)=(-1)sBs(0), we have g1(a,t)=g2(b,t); thus
(25)pn,i(k)(a)=∫ab‍g1(a,t)li(t)dt=∫ab‍g2(b,t)li(t)dt=pn,i(k)(b).
Hence
(26)yn,m(k)(b)-yn,m(k)(a)=y(k)(b)-y(k)(a)+∑i=1m‍(pn,i(k)(b)-pn,i(k)(a))f(xi,yn,m(xi))=y(k)(b)-y(k)(a).
From this, (22) follows. Next, by deriving yn,m(x)  n times, we obtain
(27)yn,m(n)(x)=Pn-1(n)[y,x]+∑k=1n-1‍pn,k(n)(x)f(xk,yn,m(xk))=∑k=1n-1‍lk(x)f(xk,yn,m(xk))
and this implies (23).


## 4. The Error

In what follows for ally∈C(q)[a,b] we define the norm ∥y∥=max⁡0≤s≤q{max⁡a≤t≤b|y(s)(t)|} [14] and the constants
(28)L=∑k=0qLk,R=max⁡a≤x≤b|R¯m(y,x)|.
Further, we define
(29)Qm=max⁡0≤s≤q{max⁡a≤x≤b∑i=1m‍|pni(s)(x)|}Dn,s=max⁡a≤x≤b∑k=sn-1‍|Bk-s((x-a)/h)|(k-s)!(n-k)!,Δ=max⁡0≤s≤q{hn-s+1(n-s+1)!+hn-s-1Dn,s}.An a priori estimation of the global error is possible.Theorem 4.
With the previous notations, suppose thatLQm<1. Then
(30)∥y-yn,m∥≤RΔ1-LQm.Proof.
By deriving (17) and (20) we get
(31)y(s)(x)-yn,m(s)(x)=∑i=1n-1‍pni(s)(x)[f(xi,y(xi))-f(xi,yn,m(xi))]+∂s∂xs∫ab‍Gn-1(x,t)R¯n,m(y,t)dt.
Now, since
(32)∂s∂xs∫ab‍Gn-1(x,t)R¯n,m(y,t)dt=1(n-s)!∫ax‍(x-t)n-sR¯n,m(y,t)dt-∑k=sn-1‍Bk-s(x-ah)hk-s-1(k-s)!(n-k+1)!×∫ab‍(b-t)n-k+1R¯n,m(y,t)dt,
we have
(33)|∂s∂xs∫ab‍Gn-1(x,t)R¯n(y,t)dt|≤hn-s+1(n-s+1)!R+hn-s-1RDn,s.
Thus
(34)|y(s)(x)-yn,m(s)(x)|≤∑i=1n-1‍|pni(s)(x)|∑k=0q‍Lk|y(s)(xi)-yn(s)(xi)|+hn-s+1(n-s+1)!R+hn-s-1RDns≤L∥y-yn∥Qm+RΔ
and inequality (30) follows.


## 5. Particular Cases

Now we present explicitly the formulas for some values ofn.For the computation ofpn,i(x) we need ∫ax‍tkli(t)dt and ∫xb‍tkli(t)dt. Letting
(35)Fi1(x)=∫ax‍li(t)dt,Mi1(x)=∫xb‍li(t)dt,Fik(x)=∫axFi,k-1(t)dt,Mik(x)=∫xb‍Mi,k-1(t)dt,k≥2,
and integrating by parts k times, we obtain
(36)∫ax‍tkli(t)dt=∑j=0k‍(-1)jk!(k-j)!xk-jFi,j+1(x),∫xb‍tkli(t)dt=∑j=0k‍k!(k-j)!xk-jMi,j+1(x).


### 5.1. The Fifth-Order Case

Now we consider the case of the fifth-order BVP(37)y(v)(x)=f(x,y(x)),x∈[0,1],y(0)=β0,y(k)(1)-y(k)(0)=βk+1,k=0,…,3.
In this case Green’s function is
(38)G4(x,t)={14![t4(1-x)+t3(2x2-2x)+t2(-2x3+3x2-x)+t(x4-2x3+x2)]t≤x14![-t4x+t3(2x2+2x)-t2(2x3+3x2+x)+t(x4+2x3+x2)-x4]x≤t,4!p5,i(x)=(x4-2x3+x2)[Fi2(x)-Mi2(x)]-2x(2x2-3x+1)[Fi3(x)+Mi3(x)]+12x(x-1)[Fi4(x)-Mi4(x)]+24(1-x)Fi5(x)-24xMi5(x).
Hence
(39)y5,m(x)=P4[y,x]+∑i=1m‍p5,i(x)f(xi,y5,m(xi)).
By deriving (44) we get
(40)y5,m(s)(x)=P4(s)[y,x]+∑i=1m‍p5,i(s)(x)f(xi,y5,m(xi))s=1,…,5,
where p5,i(s)(x) can be easily computed using the same technique as for p5,i(x).


### 5.2.  The Seventh-Order Case

Consider(41)y(vii)(x)=f(x,y(x)),x∈[0,1],y(0)=β0,y(k)(1)-y(k)(0)=βk+1,k=0,…,5.In this case Green’s function is(42)G6(x,t)=16!{t6(1-x)+3t5(x2-x)+5t4(-x3+32x2-x2)+5t3(5x4-2x3+x2)+t2(-3x5+152x4-5x3+x2)+t(x6-3x5+52x4-x22)t≤x-t6x+3t5(x2+x)+5t4(-x3-32x2-x2)+5t3(5x4+2x3+x2)+t2(-3x5+152x4-5x3+x2)+t(x6-3x5+52x4-x22)-x6x≤t.
Hence
(43)6!p7,i(x)=(x6-3x5+52x4-x22)[Fi2(x)-Mi2(x)]-x(6x4-15x3+10x2-1)[Fi3(x)+Mi3(x)]+30(x-1)2[Fi4(x)-Mi4(x)]-60x(2x2-3x+1)[Fi5(x)+Mi5(x)]+360x(x-1)[Fi6(x)-Mi6(x)]+6!(1-x)Fi7(x)-6!xMi7(x),(44)y7,m(x)=P6[y,x]+∑i=1m‍p7,i(x)f(xi,y7,m(xi)).


### 5.3. Ordern=8,9,10

Forn=8, we have
(45)7!p8,i(x)=(x84-x7+76x6-712x4+x26)[Fi1(x)+Mi1(x)]+(x7-72x6+72x5-76x3+x6)[Fi2(x)-Mi2(x)]-7x(x5-3x4+52x3-x2)[Fi3(x)+Mi3(x)]-7x(6x4-15x3+10x2-1)[Fi4(x)-Mi4(x)]+210x2(x-1)2[Fi5(x)+Mi5(x)]-420x(2x2-3x+1)[Fi6(x)-Mi6(x)]+2520x(x-1)[Fi7(x)+Mi7(x)]+7!(1-x)Fi8(x)+7!xMi8(x).Forn=9, we get
(46)8!p9,i(x)=(x8-4x7+143x6-73x4+23x2)[Fi2(x)-Mi2(x)]+4x(-2x6+7x5-7x4+73x2-x3)[Fi3(x)+Mi3(x)]+28x2(2x4-6x3+5x4-1)[Fi4(x)-Mi4(x)]+56x(-5x4+15x3-10x2+1)[Fi5(x)+Mi5(x)]+1680x2(x-1)2[Fi6(x)-Mi6(x)]+3360x(-2x2+3x-1)[Fi7(x)+Mi7(x)]+20160x(x-1)[Fi8(x)-Mi8(x)]+8!(1-x)Fi9(x)-8!xMi9(x).Forn=10, we obtain
(47)9!p10,i(x)=(x105-x9+32x8-75x6+x4-310x3)×[Fi1(x)+Mi1(x)]-x(x8-92x7+6x6-215x4+2x2-310)×[Fi2(x)-Mi2(x)]+3x2(3x6-12x5+14x4-7x2+2)[Fi3(x)+Mi3(x)]+12x(-6x6+21x5-21x4+7x2-1)[Fi4(x)-Mi4(x)]+252x2(2x4-6x3+5x2-1)[Fi5(x)+Mi5(x)]+504x(-6x4+15x3-10x2+1)[Fi6(x)-Mi6(x)]+15120x2(x-1)2[Fi7(x)+Mi7(x)]+30240x(-2x2+3x-1)[Fi8(x)-Mi8(x)]+181440x(x-1)[Fi9(x)+Mi9(x)]+9!(1-x)Fi10(x)+9!xMi10(x).


## 5.1. The Fifth-Order Case

Now we consider the case of the fifth-order BVP(37)y(v)(x)=f(x,y(x)),x∈[0,1],y(0)=β0,y(k)(1)-y(k)(0)=βk+1,k=0,…,3.
In this case Green’s function is
(38)G4(x,t)={14![t4(1-x)+t3(2x2-2x)+t2(-2x3+3x2-x)+t(x4-2x3+x2)]t≤x14![-t4x+t3(2x2+2x)-t2(2x3+3x2+x)+t(x4+2x3+x2)-x4]x≤t,4!p5,i(x)=(x4-2x3+x2)[Fi2(x)-Mi2(x)]-2x(2x2-3x+1)[Fi3(x)+Mi3(x)]+12x(x-1)[Fi4(x)-Mi4(x)]+24(1-x)Fi5(x)-24xMi5(x).
Hence
(39)y5,m(x)=P4[y,x]+∑i=1m‍p5,i(x)f(xi,y5,m(xi)).
By deriving (44) we get
(40)y5,m(s)(x)=P4(s)[y,x]+∑i=1m‍p5,i(s)(x)f(xi,y5,m(xi))s=1,…,5,
where p5,i(s)(x) can be easily computed using the same technique as for p5,i(x).


## 5.2.  The Seventh-Order Case

Consider(41)y(vii)(x)=f(x,y(x)),x∈[0,1],y(0)=β0,y(k)(1)-y(k)(0)=βk+1,k=0,…,5.In this case Green’s function is(42)G6(x,t)=16!{t6(1-x)+3t5(x2-x)+5t4(-x3+32x2-x2)+5t3(5x4-2x3+x2)+t2(-3x5+152x4-5x3+x2)+t(x6-3x5+52x4-x22)t≤x-t6x+3t5(x2+x)+5t4(-x3-32x2-x2)+5t3(5x4+2x3+x2)+t2(-3x5+152x4-5x3+x2)+t(x6-3x5+52x4-x22)-x6x≤t.
Hence
(43)6!p7,i(x)=(x6-3x5+52x4-x22)[Fi2(x)-Mi2(x)]-x(6x4-15x3+10x2-1)[Fi3(x)+Mi3(x)]+30(x-1)2[Fi4(x)-Mi4(x)]-60x(2x2-3x+1)[Fi5(x)+Mi5(x)]+360x(x-1)[Fi6(x)-Mi6(x)]+6!(1-x)Fi7(x)-6!xMi7(x),(44)y7,m(x)=P6[y,x]+∑i=1m‍p7,i(x)f(xi,y7,m(xi)).


## 5.3. Ordern=8,9,10

Forn=8, we have
(45)7!p8,i(x)=(x84-x7+76x6-712x4+x26)[Fi1(x)+Mi1(x)]+(x7-72x6+72x5-76x3+x6)[Fi2(x)-Mi2(x)]-7x(x5-3x4+52x3-x2)[Fi3(x)+Mi3(x)]-7x(6x4-15x3+10x2-1)[Fi4(x)-Mi4(x)]+210x2(x-1)2[Fi5(x)+Mi5(x)]-420x(2x2-3x+1)[Fi6(x)-Mi6(x)]+2520x(x-1)[Fi7(x)+Mi7(x)]+7!(1-x)Fi8(x)+7!xMi8(x).Forn=9, we get
(46)8!p9,i(x)=(x8-4x7+143x6-73x4+23x2)[Fi2(x)-Mi2(x)]+4x(-2x6+7x5-7x4+73x2-x3)[Fi3(x)+Mi3(x)]+28x2(2x4-6x3+5x4-1)[Fi4(x)-Mi4(x)]+56x(-5x4+15x3-10x2+1)[Fi5(x)+Mi5(x)]+1680x2(x-1)2[Fi6(x)-Mi6(x)]+3360x(-2x2+3x-1)[Fi7(x)+Mi7(x)]+20160x(x-1)[Fi8(x)-Mi8(x)]+8!(1-x)Fi9(x)-8!xMi9(x).Forn=10, we obtain
(47)9!p10,i(x)=(x105-x9+32x8-75x6+x4-310x3)×[Fi1(x)+Mi1(x)]-x(x8-92x7+6x6-215x4+2x2-310)×[Fi2(x)-Mi2(x)]+3x2(3x6-12x5+14x4-7x2+2)[Fi3(x)+Mi3(x)]+12x(-6x6+21x5-21x4+7x2-1)[Fi4(x)-Mi4(x)]+252x2(2x4-6x3+5x2-1)[Fi5(x)+Mi5(x)]+504x(-6x4+15x3-10x2+1)[Fi6(x)-Mi6(x)]+15120x2(x-1)2[Fi7(x)+Mi7(x)]+30240x(-2x2+3x-1)[Fi8(x)-Mi8(x)]+181440x(x-1)[Fi9(x)+Mi9(x)]+9!(1-x)Fi10(x)+9!xMi10(x).


## 6. Algorithms

To calculate the approximate solution of problem (1)–(5) by (20) at x∈[a,b], we need the values yj(k)=yn,m(k)(xj), j=1,…,m, and k=0,…,q. These values can be calculated by solving the following system:
(48)yi(k)=Pn-1(k)[y,xi]+∑j=1m‍pnj(k)(xi)f(xj,yj)i=1,…,m,k=0,…,q
with yj=(yj,yj′,…,yj(q)), 0≤q≤n-1.To solve it, if we put(49)Y=(Y0,…,Yq)T,Yk=(y1(k),…,ym(k)),F(Y)=(Fm,…,Fm)T∈Rq+1,Fm=(f1,…,fm),fi=f(xi,yi),C=(B0,…,Bq)T,Bk=(Pm(k)[y,x1],…,Pn-1(k)[y,xm]),A=(A00⋯00⋱⋮⋮⋱00⋯0Aq),Ak=(p1,1(k)⋯p1,m(k)⋮⋮pm,1(k)⋯pm,m(k))
with pi,j(k)=pn,j(k)(xi), k=0,…,q, we write (48) as
(50)Y-AF(Y)=C
or, equivalently, Y=G(Y), where
(51)G(Z)=AF(Z)+C.For the existence and the uniqueness of the solution of (50) the following result holds.Proposition 5.
LetL be defined as in (28). If T=L∥A∥∞<1, the system (50) has a unique solution which can be calculated by an iterative method
(52)Y(ν+1)=G(Y(ν)),ν=1,2,…
with a fixed Y(0)∈Rs, s=m(q+1), and G defined as in (51). Moreover, if Y is the exact solution of the system,
(53)∥Y(ν)-Y∥∞≤Tν1-T∥Y(1)-Y(0)∥∞.Proof.
IfV=(V0,…,Vq)T, Vk=(v1(k),…,vm(k)) and W=(W0,…,Wq)T, Wk=(w1(k),…,wm(k)), then ∥G(V)-G(W)∥∞≤∥A∥∞L∥V-W∥∞; hence G is contractive. Thus the result follows from the well-known contraction mapping theorem.To calculate the elementsA0,…,Aq of the matrix A we need the values Fis(xj) and Mis(xj), s=1,…,n, i,j=1,…m, where Fis(x) and Mis(x) are defined in (35).Sinceli(t)=∏k=1,k≠im((t-xk)/(xi-xk))‍, it suffices to compute
(54)∫cxj=tk‍∫ctk-1‍⋯∫ct1‍rm,i(t)dtdt1⋯dtk-1,
where c=a or c=b, r0,0(t)=1, and
(55)rm,i(t)=(t-x1)⋯(t-xi-1)(t-xi+1)⋯(t-xm)i=1,2,…,m.
Let us define
(56)g0,1,c(i)(x)=x-c
and, for s=1,…,m-1,
(57)gs,j,c(i)(x)=∫cx=tj‍∫ctj-1‍⋯∫ct1‍(t-z1(i))(t-z2(i))⋯(t-zs(i))dtdt1⋯dtj-1,
where
(58)zj(i)={xjifj<ixj+1ifj≥ij=1,…,m-1.
We can easily compute
(59)g0,j,c(i)(x)=(x-c)jj!.
For the computation of (57) we use the recursive algorithm [15]
(60)gs,j,c(i)(x)=(x-zs(i))gs-1,j,c(i)(x)-jgs-1,j+1,c(i)(x).
Thus, if Wi=∏k=1,k≠im‍(xi-xk), we get
(61)Fik(xj)=gm-1,k,a(i)(xj)Wi,Mik(xj)=(-1)kgm-1,k,b(i)(xj)Wi.


## 7. Numerical Examples

Now we present some numerical results obtained by applying method (20) to find numerical approximations of the solutions of some test problems. As the true solutions are known, we considered the error function E(x)=|y(x)-yn,m(x)|. To solve the nonlinear system (48) we used the so-called modified Newton method [16] (the same Jacobian matrix is used for more than one iteration) and algorithm (60) for the computation of the entries of the matrix. Equidistant points are used as nodal points. Analogous results are obtained in the considered examples by using as nodes the zeros of Chebyshev polynomials of first and second kind.Example 1.
Consider the following(62)y(v)(x)=y(x)-(15+10x)ex,x∈[0,1],y(0)=0,y(1)=0,y′(1)-y′(0)=-(e+1),y′′(1)-y′′(0)=-4e,y′′′(1)-y′′′(0)=3-9e,
with solution y(x)=x(1-x)ex. Figure 1 shows the graph of the error function E(x) for two different values of m.Error function of problem (62) for m=4 (a) and for m=6 (b).
(a)
(b)Example 2.
Consider the following(63)y(v)(x)=-24e-5y+481+x5,x∈[0,1],y(0)=0,y(1)=log⁡2,y′(1)-y′(0)=-12,y′′(1)-y′′(0)=34,y′′′(1)-y′′′(0)=-74,
with solution y(x)=log⁡(x+1). The graph of E(x), for two different numbers of nodes, is plotted in Figure 2.Error function of problem (63) for m=4 (a) and for m=7 (b).
(a)
(b)Example 3.
Consider(64)y(vii)(x)=-y-ex(2x2+12x+35),x∈[0,1],y(0)=y(1)=0,y′(1)-y′(0)=-(e+1),y′′(1)-y′′(0)=-4e,y′′′(1)-y′′′(0)=3(1-3e),y(iv)(1)-y(iv)(0)=8(1-2e),y(v)(1)-y(v)(0)=5(3-5e),
with solution y(x)=x(1-x)ex. Figure 3 shows the graph of E(x).Error function of problem (64) for m=4 (a) and for m=8 (b).
(a)
(b)Note that the equation in (64) is the same as that in Example 2 of [1], but the boundary conditions are different.Example 4.
Consider(65)y(ix)(x)=y(x)-9ex,x∈[0,1],y(0)=1,y(1)=0,y′(1)-y′(0)=-e,y′′(1)-y′′(0)=1-2e,y′′′(1)-y′′′(0)=2-3e,y(iv)(1)-y(iv)(0)=3-4e,y(v)(1)-y(v)(0)=4-5e,y(vi)(1)-y(vi)(0)=5-6e,y(vii)(1)-y(vii)(0)=6-7e,
with solution y(x)=(1-x)ex. Figure 4 shows the graph of E(x).Error function of problem (65) for m=4 (a) and for m=8 (b).
(a)
(b)The equation in (65) is the same as that in [1, 2], but the boundary conditions are different.Example 5.
Consider(66)y(x)(x)=e-xy2(x),x∈[0,1],y(0)=1,y(k)(1)-y(k)(0)=e-1,k=0,…,8
with solution y(x)=ex. Figure 5 shows the graph of E(x).Error function of problem (66) for m=3 (a) and for m=6 (b).
(a)
(b)Note that the equation in (66) is the same as that in [2], but the boundary conditions are different. The conditions in [2] are the so-called Lidstone-type conditions. Problems of this type have been analyzed in [13] using a similar technique.


## 8. Conclusions

This paper presents a class of collocation methods fornth order differential equations with Bernoulli boundary conditions. For two positive integers n,m a polynomial of degree n+m approximating the exact solution is given explicitly. Numerical experiments support theoretical results. Further developments can be done, concerning particularly numerical estimates of the error.



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