A note on the Taylor s decomposition on four points for a third-order differential equation

A note on the Taylor’s decomposition on four points for

a third-order differential equation

Allaberen Ashyralyev

_{and Doghonay Arjmand}

_{Dep. of Math., Fatih University, Istanbul, Turkey}

2

_{Dep. of Elect. and Electr. Eng., Bogazici University, Istanbul, Turkey}

August 10, 2006

Abstract

Taylor’s decomposition on four points is presented. Three-step difference schemes generated by the Taylor’s decomposition on four points for the numerical solutions of an initial-value problem, a boundary value problem and a nonlocal boundary value problem for a third order differential equation are constructed. Numerical examples are given.

Key Words: Taylor’s decomposition on four points; Third-order differential equation; Three-step difference schemes; Approximation order

AMC 2000: 65J, 65M

1

Introduction

In the present paper we consider the boundary-value problem of the form (

d3_y(t)

dt3 + a(t)y(t) = f (t), 0 < t < T,

y(0) = y0, y0(0) = y00, y(T ) = yT,

(1.1) the initial-value problem of the form

(

d3_y(t)

dt3 + a(t)y(t) = f (t), 0 < t < T,

y(0) = y0, y0(0) = y00, y00(0) = y000,

(1.2) and the nonlocal boundary value problem of the form

(

d3_y(t)

dt3 + a(t)y(t) = f (t), 0 < t < T,

y(0) = y(T ), y0_{(0) = y}0_{(T ), y}00_{(0) = y}00_{(T ),} (1.3)

assuming a(t) and f (t) to be such that problems (1.1), (1.2) and (1.3) have a unique smooth solutions defined on [0, T ].

Modern computers allow the implementation of highly accurate difference schemes for differential equations. Hence, a task of current interest is the construction and investigation of highly accurate difference schemes for differential equations.

In the papers [1]-[2] the two-step difference schemes generated by an exact difference scheme or by the Taylor’s decomposition on two points for the numerical solutions of an initial-value problem and a boundary-value problem for a second-order differential equation are constructed. In the paper [3] the four-step difference schemes generated by the Taylor’s decomposition on five points for the numerical solutions of an initial-value problem and a boundary value problem for a fourth-order differential equation are presented.

The uniform grid space

[0, T ]τ= {tk = kτ, 0 ≤ k ≤ N, N τ = T }

is considered for the construction of the three-step difference schemes for the approximate solutions of problems (1.1), (1.2) and (1.3). It is well known that applying the approxima-tion of d3_dty(t3k) on four points tk+2, tk±1, tk ∈ [0, T ]τ we cannot construct better than the

following first order of accuracy difference scheme    τ−3_(u k+2− 3uk+1+ 3uk− uk−1) + a(tk)uk= f (tk), 1 ≤ k ≤ N − 2, u0= y(0), uN = y(T ), τ−1(u1− u0) = y0(0) (1.4) for the numerical solution of the boundary-value problem (1.1), the following first order of accuracy difference scheme

           τ−3_(u k+2− 3uk+1+ 3uk− uk−1) + a(tk)uk= f (tk), 1 ≤ k ≤ N − 2, u0= y(0), τ−1(u1− u0) = y0(0), τ−2_(u 0− 2u1+ u2) = y00(0) (1.5)

for the numerical solution of the initial-value problem (1.2) and the following first order of accuracy difference scheme

           τ−3_(u k+2− 3uk+1+ 3uk− uk−1) + a(tk)uk= f (tk), 1 ≤ k ≤ N − 2, u0= uN, − (uN + u0) + (uN −1+ u1) = 0, (uN − u0) − 2(uN −1− u1) + (uN −2− u2) = 0 (1.6)

for the numerical solution of the nonlocal boundary-value problem (1.3).

In the present paper Taylor’s decomposition on four points is presented. The use of this formula gives the three-step difference schemes of fourth-order of accuracy for the approximate solutions of problems (1.1)- (1.3). Numerical examples are given.

2

Taylor’s decomposition on four points

The utilization of Taylor’s decomposition on four points in the construction of the three-step difference schemes of the fourth order of accuracy for the approximate solutions of problems (1.1) (1.2) and (1.3) is based on the following theorem.

Theorem 2.1 Let the function ν(t) (0 ≤ t ≤ T ) have a sixth continuous derivative and tk+2, tk±1, tk∈ [0, T ]τ. Then the following relation holds:

τ−3(ν(tk+2) − 3ν(tk+1) + 3ν(tk) − ν(tk−1)) (2.1) −1 2ν 000_(t k) −1 2v 000_(t k+1) = o(τ4).

Proof. Using the Taylor’s formula, we get

τ−3(ν(tk+2) − 3ν(tk+1) + 3ν(tk) − ν(tk−1)) (2.2) = ν000_(t k) + ν(4)(tk)τ 2 + ν (5)_(t k)τ 2 4 + v (6)_(t k)τ 3 12+ o(τ 4_).

Now, we will obtain α, β, γ such that τ−3_(ν(t

k+2) − 3ν(tk+1) + 3ν(tk) − ν(tk−1))

−αν000_(t

k) − βv000(tk−1) − γv000(tk+1) − dv000(tk+2) = o(τ4).

The use of the Taylor’s formula gives

αν000(tk) + βν000(tk−1) + γν000(tk+1) + dν000(tk+2) = (α + β + γ + d)ν000(tk) + (−β + γ + 2d)ν(4)(tk)τ +(1 2β + 1 2γ + 2d)ν (5)_(t k)τ2+ (− 1 6β + 1 6γ + 8 6d)v (6)_(t k)τ3+ o(τ4).

Using the last formula and formula (2.2), we obtain τ−3_(ν(t k+2) − 3ν(tk+1) + 3ν(tk) − ν(tk−1)) −αν000_(t k) − βv000(tk−1) − γv000(tk+1) − dv000(tk+2) = (1 − α − β − γ − d)ν000_(t k) + (1 2 + β − γ − 2d)ν (4)_(t k)τ +(1 4 − 1 2β − 1 2γ − 2d)ν (5)_(t k)τ2+ ( 1 12+ 1 6β − 1 6γ − 8 6d)v (6)_(t k)τ3+ o(τ4).

Then equating to zero the coefficient of the lowest power of τ in the last formula, we obtain the system of equations _

      α + β + γ + d = 1, −β + γ + 2d = 1₂, 1 2β + 1 2γ + 2d = 1 4, −1₆β +1₆γ +8₆d = ₁₂1. From this it follows that α = 1

3

Fourth order of approximation difference scheme of

the boundary-value problem (1.1)

The construction of the three-step difference scheme of the fourth order of accuracy for the approximate solution of problem (1.1) is based on the Taylor’s decomposition on four points (2.1) and on the fourth order of approximation for v0_(0).

Theorem 3.1 Let the function ν(t) have a fifth continuous derivative. Then the following relation holds: v0(0) − τ−1(−3 2v(0) + 2v(τ ) − 1 2v(2τ )) (3.1) +τ2_{− 1 12v 000_{(0) −}1 4v 000_{(τ )} = o(τ}4_),

Proof. Using the Taylor’s formula, we obtain v0_{(0) − τ}−1₍₋3 2v(0) + 2v(τ ) − 1 2v(2τ )) (3.2) −1 3τ 2_v000_{(0) −}1 4τ 3_v(4)_{(0) = o(τ}4_).

Now, we will obtain α, β, γ such that −1

3τ

2_v000_{(0) −} 1

4τ

3_v(4)_{(0) = τ}2_[αν000_{(0) + βν}000_{(τ )] + o(τ}4_).

Using the last formula and the Taylor’s formula, we obtain −1 3τ 2_v000_{(0) −}1 4τ 3_v(4)_{(0) − τ}2_[αν000_{(0) + βν}000_{(τ )] + o(τ}4₎ = µ −1 3 − (α + β) ¶ τ2_v000_{(0) +} µ −1 4 − β ¶ τ3_v(4)_{(0) + o(τ}4_),

Then equating to zero the coefficient of the lowest power of τ in the last formula, we obtain

the system of equations _½

α + β = −1 3,

β = −1 4.

From that it follows α = −1

12 and β = −14. Then formula (3.1) follows from this and

formula (3.2). Theorem 3.1 is proved.

Applying the Taylor’s decomposition on four points (2.1), formula (3.1) and equation (1.1), we obtain

τ−3_(y(t

k+2) − 3y(tk+1) + 3y(tk) − y(tk−1)) +

1 2a(tk)y(tk) + 1 2a(tk+1)y(tk+1) =1 2f (tk) + 1 2f (tk+1) + o(τ 4_{), 1 ≤ k ≤ N − 2,} y0_{(0) − τ}−1₍₋3 2y(0) + 2y(τ ) − 1 2y(2τ ))

+τ2{ 1 12a(0)y(0) + 1 4a(τ )y(τ )} + τ 2_{− 1 12f (0) − 1 4f (τ )} = o(τ 4_),

Neglecting the last small terms in the last two formulas, we obtain the fourth order of accuracy difference scheme

       τ−3_(u k+2− 3uk+1+ 3uk− uk−1) +1₂a(tk)uk+1₂a(tk+1)uk+1 = 1 2f (tk) +12f (tk+1), tk= kτ, 1 ≤ k ≤ N − 2, u0= y(0), uN = y(T ), (−3

2−121τ3a(0))u0+ (2 −41τ3a(τ ))u1−12u2

= τ y0_{(0) + τ}3_{−1

12f (0) −14f (τ )}

(3.3)

for the approximate solution of problem (1.1). For numerical analysis we consider the boundary-value problem

(

d3_y(t)

dt3 + y(t) = −t6+ 3t5− 3t4− 119t3+ 180t2− 72t + 6,

0 ≤ t ≤ 1, y(0) = y0_{(0) = 0, y(1) = 0}

for the third order differential equations with the exact solution y(t) = t3_{(1 − t)}3_.

For approximate solutions of this boundary-value problem, we use the first order of accu-racy difference scheme (1.4) and the fourth order of accuaccu-racy difference scheme (3.3) with different values for τ , namely τ = 1

20,401,801. The errors of the numerical solutions defined

as

EN = max

0≤k≤N|y(tk) − uk|

are given in the following table.

Difference schemes E20 E40 E80

Difference scheme (1.4) 0.00160609 0.00073474 0.00035026 Difference scheme (3.3) 0.00000476 0.00000029 0.00000001

Thus, the fourth order of accuracy difference scheme (3.3) is more accurate comparing with the first order of accuracy difference scheme (1.4).

4

Fourth order of approximation difference scheme of

the initial-value problem (1.2)

The construction of the three-step difference scheme of the fourth order of accuracy for the approximate solution of problem (1.2) is based on the Taylor’s decomposition on four points (2.1), on the formula (3.1) and on the fourth order of approximation for v00_(0).

Theorem 4.1 Let the function ν(t) have a fifth continuous derivative. Then the following relations hold: v00_{(0) − τ}−2_{(v(0) − 2v(τ ) + v(2τ )) (4.1)} +τ {3 8v 000_{(0) +}2 3v 000_{(τ ) −} 1 24v 000_{(2τ )} = o(τ}4_),

Proof. Using the Taylor’s formula, we obtain v00_{(0) − τ}−2_{(v(0) − 2v(τ ) + v(2τ )) (4.2)} +{τ v000(0) + 7 12τ 2_v(4)_{(0) +}1 4τ 3_v(5)_{(0)} = o(τ}4_).

Now, we will obtain α, β, γ such that τ v000_{(0) +} 7 12τ 2_v(4)_{(0) +} 1 4τ 3_v(5)₍₀₎ = τ [αν000(0) + βv000(τ ) + γν000(2τ )] + o(τ4). Using the last formula and the Taylor’s formula, we obtain

τ v000(0) + 7 12τ 2_v(4)_{(0) +} 1 4τ 3_v(5)₍₀₎ −τ [αν000_{(0) + β}000_{(τ ) + γν}000_{(2τ )] + o(τ}4_{) = (1 − (α + β + γ)) τ v}000₍₀₎ + µ 7 12− (β + 2γ) ¶ τ2_v(4)_{(0) +} µ 1 4− µ 1 2β + 2γ ¶¶ τ3_v(5)_{(0) + o(τ}4_).

Then equating to zero the coefficient of the lowest power of τ in the last formula, we obtain

the system of equations _

  α + β + γ = 1, β + 2γ = 7 12, 1 2β + 2γ = 14.

Solving the last system, we get α = 3

8, β = 23and γ = −241. From this and formula (4.2)

follows formula (4.1). Theorem 4.1 is proved.

Applying the Taylor’s decomposition on four points (2.1), formula (4.1) and equation (1.2), we obtain

τ−3_(y(t

k+2) − 3y(tk+1) + 3y(tk) − y(tk−1)) +1

2a(tk)y(tk) + 1 2a(tk+1)y(tk+1) =1 2f (tk) + 1 2f (tk+1) + o(τ 4_{), 1 ≤ k ≤ N − 2,} y0_{(0) − τ}−1₍₋3 2y(0) + 2y(τ ) − 1 2y(2τ )) +τ2{ 1 12[a(0)y(0) − f (0)] + 1

4[a(τ )y(τ ) − f (τ )] = o(τ

4_),

y00_{(0) − τ}−2_{(y(0) − 2y(τ ) + y(2τ )) − τ}

½ 3 8[a(0)y(0) − f (0)] +2 3[a(τ )y(τ ) − f (τ )] − 1

24[a(2τ )y(2τ ) − f (2τ )]} = o(τ

4₎

Neglecting the last small terms in the last three formulas, we obtain the fourth order of accuracy difference scheme

               τ−3_(u k+2− 3uk+1+ 3uk− uk−1) +1₂a(tk)uk+1₂a(tk+1)uk+1 =1 2f (tk) +12f (tk+1), 1 ≤ k ≤ N − 2, u0= y(0), y0_{(0) − τ}−1₍₋₍3 2+ τ3 112a(0))u0 +(2 − τ3 1 4a(τ ))u1−12u2) − τ2{121f (0) +14f (τ )} = 0, y00_{(0) − τ}−2_{((1 + τ}3 3

8a(0))u0+ (−2 + τ3 23a(τ ))u1

+(1 − τ3 1

24a(2τ ))u2) − τ {−38f (0) −23f (τ ) +241f (2τ )} = 0

(4.3)

for the approximate solution of problem (1.2).

For numerical analysis we consider the initial-value problem (

d3_y(t)

dt3 + y(t) = −t6+ 3t5− 3t4− 119t3+ 180t2− 72t + 6,

0 ≤ t ≤ 1, y(0) = y0_{(0) = y}00_{(0) = 0}

for the fourth order differential equations. The exact solution of this problem is: y(t) = t3_{(1 − t)}3_.

For approximate solutions of this initial-value problem, we use the first order of accuracy difference scheme (1.5) and the fourth order of accuracy difference scheme (4.3) with dif-ferent values for τ , namely τ = 1

20,401,801. The errors of the numerical solutions are given

in the following table.

Difference schemes E20 E40 E80

Difference scheme (1.5) 0.06204958 0.03387875 0.01767174 Difference scheme (4.3) 0.00000483 0.00000029 0.00000001

Thus, the fourth order of accuracy difference scheme (4.3) is more accurate comparing with the first order of accuracy difference scheme (1.5).

5

Fourth order of approximation difference scheme of

the nonlocal boundary value problem (1.3)

The construction of the three-step difference scheme of the fourth order of accuracy for the approximate solution of problem (1.3) is based on the Taylor’s decomposition on four points (2.1), on the formulas (3.1), (4.1) and on the fourth order of approximation for v0_{(T ) and v}00_{(T ).}

Theorem 5.1 Let the function v(t) have a fifth continuous derivative. Then the following relations hold: v0_{(T ) − τ}−1₍3 2v(T ) − 2v(T − τ ) + 1 2v(T − 2τ )) (5.1) +τ2₍₋ 1 12v 000_{(T ) −}1 4v 000_{(T − τ )) = o(τ}4_),

v00_{(T ) − τ}−2_{(v(T ) − 2v(T − τ ) + v(T − 2τ )) (5.2)} +τ {−3 8v 000_{(T ) −}2 3v 000_{(T − τ ) +} 1 24v 000_{(T − 2τ )} = o(τ}4_),

Proof. First, we will prove (5.1). Using the Taylor’s formula, we obtain

v0(T ) − τ−1(3 2v(T ) − 2v(T − τ ) + 1 2v(T − 2τ )) − 1 3τ 2_v000_{(T ) +}1 4τ 3_v(4)_{(T ) = o(τ}4_).

Now we will obtain α, β such that −1

3τ

2_v000_{(T ) +}1

4τ

3_v(4)_{(T ) = τ}2_[αv000_{(T ) + βv}000_{(T − τ )] + o(τ}4_).

Using the last and Taylor’s formula we obtain −1 3τ 2_v000_{(T ) +}1 4τ 3_v(4)_{(T ) − τ}2_[αv000_{(T ) + βv}000_{(T − τ )] + o(τ}4₎ = (−1 3 − (α + β))τ 2_v000_{(T ) + (}1 4 + β)τ 3_v(4)_{(T ) + o(τ}4_).

Then equating the coefficient of the lowest power of τ in the last equation we obtain

the system of equations _½

α + β = −1 3,

β = −1 4.

Thus we find α = −1

12 and β = −14. Second, we will prove (5.2). Using the Taylor’s

formula, we obtain v00(T ) − τ−2(v(T ) − 2v(T − τ ) + v(T − 2τ )) −τ v000(T ) + 7 12τ 2_v(4)_{(T ) −}1 4τ 3_v(5)_{(T ) = o(τ}4_),

Now,we will obtain α, β and γ such that −τ v000_{(T ) +} 7

12τ

2_v(4)_{(T ) −}1

4τ

3_v(5)_{(T ) = τ {αv}000_{(T ) + βv}000_{(T − τ ) + γv}000_{(T − 2τ )} + o(τ}4_),

Using the last and Taylor’s formula we obtain −τ v000_{(T ) +} 7 12τ 2_v(4)_{(T ) −}1 4τ 3_v(5)_{(T ) − τ {αv}000_{(T ) + βv}000_{(T − τ ) + γv}000_{(T − 2τ )} + o(τ}4₎ = (−1−(α+β +γ))τ v000_{(T )+(}7 12−(−β −2γ))τ 2_v(4)_{(T )+(−}1 4−( 1 2β +2γ))v (5)_{(T )+o(τ}4_),

Then equating the coefficients of the lowest power of τ in the last equation we obtain the system of equations

   α + β + γ = −1, −β − 2γ = 7 12, 1 2β + 2γ = −14.

From that it follows α = −3

8, β = −23, and γ =241.Theorem 5.2 is proved.

Applying the Taylor’s decompositiın on four points,formulas (5.1), (5.2) and equation (3.1), we obtain

τ−3(y(tk+2) − 3y(tk+1) + 3y(tk) − y(tk−1)) +1

2a(tk)y(tk) + 1 2a(tk+1)y(tk+1) =1 2f (tk) + 1 2f (tk+1) + o(τ 4_{), 1 ≤ k ≤ N − 2,} (3 2y(T ) − 2y(T − τ ) + 1 2y(T − 2τ )) + τ 3₍ 1 12y 000_{(T ) +}1 4y 000_{(T − τ ))} −(−3 2y(0) + 2y(τ ) − 1 2y(2τ )) − τ 3₍1 12y 000_{(0) +}1 4y 000_{(τ )) = o(τ}4_),

(y(T ) − 2y(T − τ ) + y(T − 2τ )) + τ3_{3

8y 000_{(T ) +}2 3y 000_{(T − τ ) −} 1 24y 000_{(T − 2τ )}}

−(y(0) − 2y(τ ) + y(2τ )) − τ3_{−3

8y 000_{(0) −}2 3y 000_{(τ ) +} 1 24y 000_{(2τ )} = o(τ}4_),

Neglecting the last small terms in the last three formulas, we obtain the fourth order of accuracy difference scheme

                           τ−3_(u k+2− 3uk+1+ 3uk− uk−1) +1₂a(tk)uk+1₂a(tk+1)uk+1 =1 2f (tk) +12f (tk+1), 1 ≤ k ≤ N − 2, u0= uN, (−3

2−121τ3a(0))u0+ (2 −41τ3a(τ ))u1−12u2

−(3

2−121τ3a(T ))uN − (−2 −41τ3a(T − τ ))uN −1−12uN −2

= 1

12τ3(f (T ) − f (0)) +14τ3(f (T − τ ) − f (τ )),

(1 +3

8τ3a(0))u0+ (−2 +23τ3a(τ ))u1+ (1 −241τ3a(2τ ))u2

−(1 −3

8τ3a(T ))uN − (−2 −23τ3a(T − τ ))uN −1

−(1 + 1

24τ3a(T − 2τ ))uN −2= τ3{38(f (T ) + f (0))

+2

3(f (T − τ ) + f (τ )) − 241(f (T − 2τ ) + f (2τ ))}

(5.3)

for the approximate solution of problem (1.3) .

Now, for numerical analysis we consider the nonlocal boundary value problem (

d3_y(t)

dt3 + y(t) = −t6+ 3t5− 3t4− 119t3+ 180t2− 72t + 6,

0 ≤ t ≤ 1, y(0) = y(1), y0_{(0) = y}0_{(1), y}00_{(0) = y}00₍₁₎

for the third order differential equations. The exact solution of this problem is y(t) = t3_{(1 − t)}3_.

For approximate solutions of this nonlocal boundary-value problem, we use the first order of accuracy difference scheme (1.6) and the fourth order of accuracy difference scheme

(5.3) with different values for τ , namely τ = 1

20,401,801. The errors of the numerical solutions

are defined as EN = max 0≤k≤N|y(tk) − uk|. Difference schemes E20 E40 E80 Difference scheme (1.6) 0.16010542 0.11498498 0.06658580 Difference scheme (5.3) 0.00000312 0.00000019 0.00000001

Hence, the fourth order of accuracy difference scheme (5.3) is more accurate when compared with the first order of accuracy difference scheme (1.6).

It is known that various boundary value problems for the partial differential equations can be reduced to the boundary value problem

(

d3_y(t)

dt3 + A(t)y(t) = f (t), 0 < t < T,

y(0) = y0, y0(0) = y00, y(T ) = yT

(5.4) and the initial-value problem

(

d3_y(t)

dt3 + A(t)y(t) = f (t), 0 < t < T,

y(0) = y0, y0(0) = y00, y00(0) = y000

(5.5) and the nonlocal boundary value problem

(

d3_y(t)

dt3 + A(t)y(t) = f (t), 0 < t < T,

y(0) = y(T ), y0_{(0) = y}0_{(T ), y}00_{(0) = y}00_{(T )} (5.6)

for the differential equations in a Hilbert space H, with the self-adjoint positive definite operators A(t). The use of Taylor’s decomposition on four points permits to construct the difference schemes of the fourth order of accuracy for the approximate solutions of problems (5.4)-(5.6). Operator method of [4] permits to establish the stability of these difference schemes.

References

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[4] A. Ashyralyev, P.E. Sobolevskii, New Difference Schemes for Partial Differential Equa-tions. Birkhauser Verlag: Basel, Boston, Berlin. 2004.

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[7] D. Bazarov, H. Soltanov, Some Local and Nonlocal Boundary Value Problems for Equa-tions of Mixed and Mixed-Composite Types, Ylim: Ashgabat, 1995. (Russian). [8] A.G. Latrous, A. Memou, A three-point boundary value problem with an integral

condition for a third-order partial differential equation, Abstract and Applied Analysis, 2005 (1)(2005), 33-43.