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Galerkin ﬁnite element method for forced

Burgers’ equation

Sarita Thakar and Sunil S Kumbhar∗

Department of Mathematics, Shivaji University, Kolhapur (Maharashtra), India

email: sunilmath25@gmail.com, saritakolhapur@gmail.com

Journal of Mathematical Modeling

Vol. 7, No. 4, 2019, pp. 445-467 JMM

Abstract. In this paper second order explicit Galerkin ﬁnite element

method based on cubic B-splines is constructed to compute numerical solu-

tions of one dimensional nonlinear forced Burgers’ equation. Taylor series

expansion is used to obtain time discretization. Galerkin ﬁnite element

method is set up for the constructed time discretized form. Stability of the

corresponding linearized scheme is studied by using von Neumann analysis.

The accuracy, eﬃciency, applicability and reliability of the present method

is demonstrated by comparing numerical solutions of some test examples

obtained by the proposed method with the exact and numerical solutions

available in literature.

Keywords:Forced Burgers’ equation, cubic B-splines, Galerkin Finite Element

Method, Taylor series, von Neumann analysis.

AMS Subject Classiﬁcation: 65M60, 65N30.

1 Introduction

The Burgers’ equation is a one dimensional form of Navier-Stokes equation.

It was ﬁrstly introduced by Harry Batman and was taken later by J.M.

Burger as a model of turbulent ﬂuid motion. This equation arises in various

ﬁelds such as Fluid Dynamics, Nonlinear Acoustics, Gas Dynamics, Traﬃc

Flow, etc. In this paper, we consider the one dimensional nonlinear forced

Burgers’ equation,

ut+uux−νuxx =F(x, t), a ≤x≤b, t > 0,(1)

∗Corresponding author.

Received: 13 May 2019 / Revised: 26 July 2019 / Accepted: 24 October 2019.

DOI: 10.22124/jmm.2019.13259.1265

c

2019 University of Guilan http://jmm.guilan.ac.ir

446 S. Thakar, S.S. Kumbhar

with the initial condition

u(x, 0) = f(x), a ≤x≤b, (2)

and the boundary conditions,

u(a, t) = g1(t), u(b, t) = g2(t), t > 0,(3)

where ν > 0 is the coeﬃcient of kinematic viscosity, f(x), g1(t), g2(t) and

the forcing term F(x, t) are known functions. The control function F(x, t) is

assumed to be diﬀerentiable with respect to time. In the literature one can

see that, several numerical methods are available for Burgers’ equation with

F(x, t) = 0. The Galerkin ﬁnite element method based on cubic B-splines

is constructed by [2] to obtain numerical solutions of Eq. (1) with F(x, t) =

0. This method is implicit and unconditionally stable. Time-splitting of

homogeneous form of Eq. (1) is done by [6] to obtain system of partial

diﬀerential equations. Galerkin ﬁnite element methods based on quadratic

and cubic B-splines are constructed to obtain numerical solution of the

splitting system. Cubic B-spline and modiﬁed cubic B-spline collocation

methods are discussed by [7] and [12] respectively. Refs. [15] and [16] set

up least squares algorithms with cubic and quadratic B-splines. Ref. [5]

converted (1) with F(x, t) = 0 to a system of nonlinear ordinary diﬀerential

equations by method of discretization in time and space. Quadratic B-

spline Galerkin ﬁnite element method is employed on the resulting system.

Weighted avarage diﬀerential quadrature method is developed by [11]. The

Crank-Nicolson type ﬁnite diﬀerence method for Eq. (1) with F(x, t) =

0 is discussed by [14]. In constructing all the methods discussed above,

the homogeneous form of Eq. (1) is converted into system of ordinary

diﬀerential equations. The solution to this system of ordinary diﬀerential

equations is obtained by constructing ﬁrst or second order ﬁnite diﬀerence

schemes.

The numerical solutions of Eq. (1) based on multiquadratic quasi-

interpolation operator and radial basis function network schemes are ob-

tained by [9]. In these methods the solution or its space derivative is quasi

interpolated by using Hardy basis functions. Both the methods are con-

ditionally stable. Stability of both the methods depends upon the shape

parameters and the number of collocation points.

In the present paper we propose unconditionally stable cubic B-spline

Galerkin ﬁnite element method for Eqs. (1)-(3). The time discretization of

Eq. (1) is considered at the beginning. The recurrence relation is obtained

by using forward diﬀerence approximations. Galerkin ﬁnite element method

is then applied to construct a solution. The cubic B-splines has small

Galerkin FEM for forced Burgers’ equation 447

support and therefore many elements of the matrices in the ﬁnal assembled

system of Galerkin method are zero. In fact matrices in the ﬁnal assembled

system are septadiagonal and computations with these matrices requires

less computational cost. On the other hand Lagrange polynomials with the

Gauss-Legendre points are deﬁned on the whole domain.

The paper is organized as follows. In Section 2second order ﬁnite dif-

ference scheme is constructed and Galerkin ﬁnite element method is applied

to this second order ﬁnite diﬀerence scheme. Stability analysis of the corre-

sponding linearized method is discussed in Section 3. Numerical solution of

some test examples obtained by proposed method are reported in Section 4.

These solutions are compared with exact solutions and numerical solutions

available in the literature.

2 Method of solution

The domain [a, b] is partitioned uniformly as a=x0< x1< x2< . . . <

xN=binto Nnumber of ﬁnite elements with equal length h= (b−a)/N

and xj=x0+jh,j= 0,1,2, . . . , N . The time discretization of Eq. (1) is

obtained by using following forward second order Taylor series formula.

un

t=un+1 −un

∆t−∆t

2un

tt,(4)

where tn=t0+n∆t,un=u(x, tn), un

t=ut(x, tn) and un

tt =utt(x, tn).

The discretization of Eq. (1) with F(x, t) = 0 is obtained earlier by [3].

Diﬀerentiating Eq. (1) w.r.t. twe get,

utt =−∂x(uut) + ν∂2

xut+Ft(x, t),(5)

where ∂xand ∂2

xdenote the ﬁrst and second order partial derivatives with

respect to x. Substitution of utand utt from Eqs. (1) and (5) respectively

into Eq. (4) gives

−un∂xun+ν∂2

xun+F(x, tn)

=(un+1 −un)

∆t−∆t

2−un

t∂xun−un∂xun

t+ν∂2

xun

t+Ft(x, tn).

Using forward diﬀerence approximation to un

tin above equation and after

the simpliﬁcation we obtain

1−∆t

2−∂xun−un∂x+ν∂2

x(un+1 −un)

∆t

=−un∂xun+ν∂2

xun+F(x, tn) + ∆t

2Ft(x, tn).(6)

448 S. Thakar, S.S. Kumbhar

Eq. (6) gives the recurrence relation

1 + ∆t

2∂xun+un∂x−ν∂2

xun+1

=1 + ν∆t

2∂2

xun+ ∆tF (x, tn) + (∆t)2

2Ft(x, tn).(7)

The truncation error is given by (T.E.) = P DE −F DE [8]. From Eqs. (1)

and (6) we have

T.E. =nun

t+unun

x−νun

xx −F(x, tn)o−

n1−∆t

2−∂xun−un∂x+ν∂2

x(un+1 −un)

∆t+un∂xun

−ν∂2

xun−F(x, tn)−∆t

2Ft(x, tn)o

=−(∆t)2

12 (2un

ttt + 3un

xun

tt + 3unun

xtt −3νun

xxtt)− · · · .

Therefore Eq. (7) is a second order explicit scheme in the variable t.

Assume that the solution u(x, t) of the Burgers’ equation (1) is of the

form

u(x, t) =

N+1

X

m=−1

δm(t)φm(x),(8)

where δm(t), m=−1,0,1, . . . , N + 1 are the time dependent functions to

be determined and φm(x), m=−1,0,1, . . . , N + 1 are cubic B-splines given

by [10]

φm(x) = 1

h3

(x−xm−2)3,[xm−2, xm−1],

h3+ 3h2(x−xm−1)+3h(x−xm−1)2

−3(x−xm−1)3,[xm−1, xm],

h3+ 3h2(xm+1 −x)+3h(xm+1 −x)2

−3(xm+1 −x)3,[xm, xm+1],

(xm+2 −x)3,[xm+1, xm+2],

0, o.w.,

(9)

Using boundary conditions (3) we obtain

δ−1(t) = g1(t)−4δ0(t)−δ1(t),(10)

δN+1(t) = g2(t)−δN−1(t)−4δN(t).(11)

Galerkin FEM for forced Burgers’ equation 449

The solution given by Eq. (8) now becomes

u(x, t) = g1(t)φ−1(x) + g2(t)φN+1(x) +

N

X

i=0

δi(t)Bi(x),(12)

where

B0(x) = φ0(x)−4φ−1(x), B1(x) = φ1(x)−φ−1(x),

Bj(x) = φj(x),for j= 2,3, . . . , N −2,

BN−1(x) = φN−1(x)−φN+1(x), BN(x) = φN(x)−4φN+1 (x).

From Eq. (12) we have

ux(x, t) = g1(t)φ0

−1(x) + g2(t)φ0

N+1(x) +

N

X

i=0

δi(t)B0

i(x),(13)

uxx(x, t) = g1(t)φ00

−1(x) + g2(t)φ00

N+1(x) +

N

X

i=0

δi(t)B00

i(x).(14)

Deﬁne

hi(x, t) =

g1(t)φ−1(x)Bi(x)0, i = 0,1,2,

g2(t)φN+1(x)Bi(x)0, i =N−2, N −1, N ,

0, otherwise,

(15)

R1(x, tn, tn+1) = g1(tn)−g1(tn+1)φ−1(x)

+∆t

2ν(g1(tn) + g1(tn+1))φ00

−1(x)−2g1(tn)g1(tn+1)φ−1(x)φ0

−1(x),(16)

SN(x, tn, tn+1) = g2(tn)−g2(tn+1)φN+1(x)

+∆t

2ν(g2(tn) + g2(tn+1))φ00

N+1(x)−2g2(tn)g2(tn+1)φN+1(x)φ0

N+1(x).

(17)

Using Eqs. (12)-(16) in Eq. (7), on element [x0, x1] we have

2

X

i=0 hBi(x) + ∆t

2hi(x, tn) +

2

X

j=0

δj(tn)(BiBj)0−νB00

ii·δi(tn+1)

=

2

X

i=0 hBi(x)−∆t

2hi(x, tn+1)−νB00

i(x)i·δi(tn) + R1+ ∆tF (x, tn)

+(∆t)2

2Ft(x, tn).(18)

450 S. Thakar, S.S. Kumbhar

The solution u(x, t) and its derivatives on the typical element [xl, xl+1] for

l= 1,2, . . . , N −2 are

u(x, t) =

l+2

X

i=l−1

δi(t)Bi(x),(19)

ux(x, t) =

l+2

X

i=l−1

δi(t)B0

i(x),(20)

uxx(x, t) =

l+2

X

i=l−1

δi(t)B00

i(x).(21)

On [xl, xl+1] for l= 1,2, . . . , N −2, Eq. (7) becomes

l+2

X

i=l−1hBi(x) + ∆t

2

l+2

X

j=l−1

δj(tn)(BiBj)0−νB00

ii·δi(tn+1)

=

l+2

X

i=l−1hBi(x) + ν∆t

2B00

i(x)i·δi(tn)+∆tF (x, tn) + (∆t)2

2Ft(x, tn).(22)

On [xN−1, xN] Eq. (17) is used in Eq. (7) to obtain

N

X

i=N−2hBi(x) + ∆t

2hi(x, tn) +

N

X

j=N−2

δj(tn)(BiBj)0−νB00

ii·δi(tn+1)

=

N

X

i=N−2hBi(x)−∆t

2hi(x, tn+1)−νB00

i(x)i·δi(tn) + SN+ ∆tF (x, tn)

+(∆t)2

2Ft(x, tn).(23)

The element wise Galerkin weak formulation is obtained by the following

procedure. On multiplying Eq. (18) by the weight function Bk(x), k=

0,1,2 and integrating by parts on the interval [x0, x1] we get

hA1+∆t

2hn

1−B1+νC1i·δn+1

1

=hA1−∆t

2hn+1

1+νC1i·δn

1+Rn

1+Fn

1,(24)

where δn

1= (δn

0, δn

1, δn

2)T,

Rn

1=140(g1(n∆t)−g1((n+ 1)∆t))

h

49

40

1

Galerkin FEM for forced Burgers’ equation 451

+∆t

2(ν(g1(n∆t) + g1((n+ 1)∆t))

10h

51

54

3

+2g1(n∆t)g1((n+ 1)∆t))

168

97

70

1

),

Fn

1= ∆t

Rx1

x0F(x, n∆t)B0(x)dx

Rx1

x0F(x, n∆t)B1(x)dx

Rx1

x0F(x, n∆t)B2(x)dx

+(∆t)2

2

Rx1

x0Ft(x, n∆t)B0(x)dx

Rx1

x0Ft(x, n∆t)B1(x)dx

Rx1

x0Ft(x, n∆t)B2(x)dx

.

In order to compute Fn

1,F(x, t) and Ft(x, t) are evaluated at t=n∆tand

then Rx1

x0F(x, n∆t)Bidx,Rx1

x0Ft(x, n∆t)Bidx,i= 0,1,2 are computed. We

have also

A1=h

140

476 644 56

644 1088 128

56 128 20

,C1=1

10h

222 108 −24

108 192 24

−24 24 18

,

hn

1=−g1(n∆t)

840

1235 1586 89

758 1244 98

−1 26 5

,

and for i, j = 1,2,3, (i, j )th element of matrix B1, (B1)ij is computed by

the formula

(B1)ij =Zx1

x0

Bj−1B0B0

i−1dx, Zx1

x0

Bj−1B1B0

i−1dx, Zx1

x0

Bj−1B2B0

i−1dxδn

1.

Thus the elements of matrix B1are

(B1)11 =1

840(280,−4292,−878)δn

1,(B1)12 =−1

840(4292,13616,2264)δn

1,

(B1)13 =−1

840(878,2264,380)δn

1,(B1)21 =1

840(11944,13528,796)δn

1,

(B1)22 =1

840(13528,17920,1408)δn

1,(B1)23 =1

840(796,1408,160)δn

1,

(B1)31 =1

840(2596,4828,610)δn

1,(B1)32 =1

840(4828,10624,1600)δn

1,

(B1)33 =1

840(610,1600,280)δn

1.

Multiply Eq. (22) by the weight function Bk(x), k=l−1, l, l + 1, l + 2

and integrate by parts on the interval [xl, xl+1]. Because the overall con-

tribution of the terms Bi(x)Bj(x)Bk(x)|xl+1

xland B0

i(x)Bk(x)|xl+1

xlvanishes

in the assembled system, we exclude them from the ﬁnal expression. Thus

452 S. Thakar, S.S. Kumbhar

on the element [xl, xl+1] we have

hAl+1 +∆t

2Bl+1 +νCl+1i·δn+1

l+1

=hAl+1 −ν∆t

2Cl+1i·δn

l+1 +Fn

l+1,(25)

where for l= 1,2, . . . , N −2, δn

l+1 = (δn

l−1, δn

l, δn

l+1, δn

l+2)T,

Fn

l+1 = ∆t

Rxl+1

xlF(x, n∆t)Bl−1(x)dx

Rxl+1

xlF(x, n∆t)Bl(x)dx

Rxl+1

xlF(x, n∆t)Bl+1(x)dx

Rxl+1

xlF(x, n∆t)Bl+2(x)dx

+(∆t)2

2

Rxl+1

xlFt(x, n∆t)Bl−1(x)dx

Rxl+1

xlFt(x, n∆t)Bl(x)dx

Rxl+1

xlFt(x, n∆t)Bl+1(x)dx

Rxl+1

xlFt(x, n∆t)Bl+2(x)dx

,

Al+1 =h

140

20 129 60 1

129 1188 933 60

60 933 1188 129

1 60 129 20

,Cl+1 =1

10h

18 21 −36 −3

21 102 −87 −36

−36 −87 102 21

−3−36 21 18

and for i, j = 1,2,3,4; (i, j)th element of matrix Bl+1, (Bl+1)ij is com-

puted by the formula

(Bl+1)ij = Zxl+1

xl

Bj+l−2Bl−1B0

i+l−2dx, Zxl+1

xl

Bj+l−2BlB0

i+l−2dx,

Zxl+1

xl

Bj+l−2Bl+1B0

i+l−2dx, Zxl+1

xl

Bj+l−2Bl+2B0

i+l−2dx!δn

l+1.

Thus the elements of matrix Bl+1 are

(Bl+1)11 =−1

840 (280,1605,630,5)δn

l+1,(Bl+1 )12 =−1

840 (1605,10830,5349,108)δn

l+1,

(Bl+1)13 =−1

840 (630,5349,3468,129)δn

l+1,(Bl+1 )14 =−1

840 (5,108,129,10)δn

l+1,

(Bl+1)21 =−1

840 (150,1305,792,21)δn

l+1,(Bl+1 )22 =−1

840 (1305,17640,17541,1314)δn

l+1,

(Bl+1)23 =−1

840 (792,17541,25002,2781)δn

l+1,(Bl+1 )24 =−1

840 (21,1314,2781,420)δn

l+1,

(Bl+1)31 =1

840 (420,2781,1314,21)δn

l+1,(Bl+1 )32 =1

840 (2781,25002,17541,792)δn

l+1,

(Bl+1)33 =1

840 (1314,17541,17640,1305)δn

l+1,(Bl+1 )34 =1

840 (21,792,1305,150)δn

l+1,

(Bl+1)41 =1

840 (10,129,108,5)δn

l+1,(Bl+1 )42 =1

840 (129,3468,5349,630)δn

l+1,

(Bl+1)43 =1

840 (108,5349,10830,1605)δn

l+1,(Bl+1 )44 =1

840 (5,630,1605,280)δn

l+1.

On multiplying (23) by the weight function Bk(x), k=N−2, N −1, N

Galerkin FEM for forced Burgers’ equation 453

and integrating by parts on the interval [xN−1, xN] we get

hAN+∆t

2hn

N−BN+νCNi·δn+1

N

=hAN−∆t

2hn+1

N+νCNi·δn

N+Sn

N+Fn

N,(26)

where δn

N= (δn

N−2, δn

N−1, δn

N)T,

Sn

N=140(g2(n∆t)−g2((n+ 1)∆t))

h

1

40

49

+∆t

2(ν(g2(n∆t) + g2((n+ 1)∆t))

10h

3

54

51

−

2g2(n∆t)g2((n+ 1)∆t))

168

1

70

97

),

Fn

N= ∆t

RxN

xN−1F(x, n∆t)BN−2(x)dx

RxN

xN−1F(x, n∆t)B1(x)dx

RxN

xN−1F(x, n∆t)B2(x)dx

+(∆t)2

2

RxN

xN−1Ft(x, n∆t)BN−2(x)dx

RxN

xN−1Ft(x, n∆t)B1(x)dx

RxN

xN−1Ft(x, n∆t)B2(x)dx

,

AN=h

140

20 128 56

128 1088 644

56 644 476

,CN=1

10h

18 24 −24

24 192 108

−24 108 222

,

hn

N=g2(n∆t)

840

5 26 −1

98 1244 758

89 1586 1235

,

and for i, j = 1,2,3, (i, j)th element of matrix BN, (BN)ij is computed

by the formula

(BN)ij = ZxN

xN−1

Bj+N−3BN−2B0

i+N−3dx, ZxN

xN−1

Bj+N−3BN−1B0

i+N−3dx,

ZxN

xN−1

Bj+N−3BNB0

i+N−3dx!δn

N.

454 S. Thakar, S.S. Kumbhar

Thus the elements of matrix BNare

(BN)11 =−1

840(280,1600,610)δn

N,(BN)12 =−1

840(1600,10624,4828)δn

N,

(BN)13 =−1

840(610,4828,2596)δn

N,(BN)21 =−1

840(160,1408,796)δn

N,

(BN)22 =−1

840(1408,17920,13528)δn

N,(BN)23 =−1

840(796,13528,11944)δn

N,

(BN)31 =1

840(380,2264,878)δn

N,(BN)32 =1

840(2264,13616,4292)δn

N,

(BN)33 =1

840(878,4292,−280)δn

N.

Since φ−1(x) is zero on [xl, xl+1], l= 1,2,· · · , N −1, Rn

k,k= 2,3, . . . , N

are zero vectors. Similarly φN+1(x) is zero on [xl, xl+1], l= 0,1, . . . , N −2,

and therefore Sn

k,k= 1,2, . . . , N −1 are zero vectors. Also for k=

2,3, . . . , N −1, hn

kare zero matrices. Combining the contributions from

Eqs. (24), (25) and (26) in usual way we obtain the system of (N+ 1) ×

(N+ 1) algebraic equations

hA+∆t

2hn−B+νCi·δn+1

=hA−∆t

2hn+1 +νCi·δn+Rn+Sn+Fn,(27)

where δn= (δn

0, δn

1, . . . , δn

N)T. The initial solution δ0is obtained from

the initial condition (2). Since δ0has (N+ 1) components, the system of

(N+1) equations is obtained by evaluating (12) at distinct points x=yj∈

(a, b), j = 0,1, . . . , N and t= 0. Thus we have

u(yj,0) = f(yj) = g1(0)φ−1(yj) + g2(0)φN+1(yj) +

N

X

i=0

δ0

iBi(yj).(28)

The solution of this system of equations is the initial solution δ0. The

recurrence relation (27) generates δn+1 and the solution of Eqs. (1)-(3) at

t=tn+1 is given by (12) as

u(x, tn+1) = g1(tn+1)φ−1(x) + g2(tn+1)φN+1 (x) +

N

X

i=0

δn+1

iBi(x).(29)

3 Stability analysis

Since problem (1)-(3) is nonlinear, its discretization (7) and the algebraic

scheme (27) are nonlinear. To study the stability of scheme (27), the cor-

responding linearized scheme is considered for von Neumann analysis. The

Galerkin FEM for forced Burgers’ equation 455

linearized form of (27) is obtained by assuming that the solution uin the

nonlinear term uuxis locally constant and is equal to U. Thus the linear

system corresponding to scheme (27) is

hA+U∆t

2B∗+ν∆t

2Ci·δn+1 =hA−ν∆t

2Ci·δn+Rn+Sn+Fn,(30)

where B∗is obtained by combining contributions from Rxl+1

xlB0

i(x)Bj(x)dx

in usual way. The error equation corresponding to above equation is

hA+U∆t

2B∗+ν∆t

2Ci·n+1 =hA−ν∆t

2Ci·n,(31)

where, nis error in the solution at t=tn. Matrices A,B∗and Care

septadiagonal matrices and general row of these matrices are

A:h

140(1,120,1191,2416,1191,120,1)

B∗:1

20(1,56,245,0,−245,−56,−1)

C:−1

10h(3,72,45,−240,45,72,3)

The lth error equation in (31) is

α1n+1

l−3+α2n+1

l−2+α3n+1

l−1+α4n+1

l+α5n+1

l+1 +α6n+1

l+2 +α7n+1

l+3

=α8n

l−3+α9n

l−2+α10n

l−1+α11n

l+α10n

l+1 +α9n

l+2 +α8n

l+3 (32)

where n

jis the error in δjat t=n∆t, 0 ≤j≤N , 4≤l≤N−3,

α1=r1+r2−3r3, α2= 120r1+ 56r2−72r3,

α3= 1191r1+ 245r2−45r3, α4= 2416r1+ 240r3,

α5= 1191r1−245r2−45r3, α6= 120r1−56r2−72r3,

α7=r1−r2−3r3, α8=r1+ 3r3, α9= 120r1+ 72r3,

α10 = 1191r1+ 45r3, α11 = 2416r1−240r3,

r1=h

140, r2=U∆t

40 , r3=ν∆t

20h.

The Fourier growth factor is deﬁned as

n

l=ξneilkh (33)

456 S. Thakar, S.S. Kumbhar

where kis mode number and his the length of ﬁnite element. Using (33)

Eq. (32) gives

[(a−c)−ib]ξn+1 = [a+c]ξn,(34)

where

a= [2 cos(3kh) + 240 cos(2kh) + 2382 cos(kh) + 2416] r1,

b= [2 sin(3kh) + 112 sin(2kh) + 490 sin(kh)] r2,

c= [6 cos(3kh) + 144 cos(2kh) + 90 cos(kh)−240] r3.

From Eq. (34) the ampliﬁcation factor

ξ=a+c

(a−c)−ib.

Since r3>0, c≤0 and hence |ξ| ≤ 1 and therefore the linearized scheme

(30) is unconditionally stable.

4 Numerical experiments

In this section we illustrate seven test examples to support the method.

Mathematica 10.0 software is used to compute numerical solutions and

errors in it. The L2and L∞errors are deﬁned as,

L2=v

u

u

th

N

X

j=0

|Uexact

j−unumer

j|2, L∞= max

1≤j≤N|Uexact

j−unumer

j|

where, Uexact

jand unumer

jare exact and numerical solutions at x=xj

respectively.

Example 1. In this test example we consider Eq. (1) with the initial

condition u(x, 0) = sin(πx), boundary conditions u(0, t)=0, u(1, t)=0

and F(x, t) = 0. The exact solution of this problem is

u(x, t)=2πν P∞

n=1 ane−n2π2νt nsin(nπx)

a0+P∞

n=1 ane−n2π2νt ncos(nπx)

where the Fourier coeﬃcients an,n= 0,1,2,..., are given by

a0=Z1

0

e−(2πν)−1(1−cos(πx))dx,

an= 2 Z1

0

e−(2πν)−1(1−cos(πx)) cos(nπx)dx.

Galerkin FEM for forced Burgers’ equation 457

Table 1: Comparison of numerical and exact solution for ν= 0.01 of Ex-

ample 1.

x t [6] (CBGM) [7] Present Exact

∆t= 0.0001 ∆t= 0.0001 ∆t= 0.0001

h= 0.0125 h= 0.0125 h= 0.0125

0.25 0.4 0.34191 0.34192 0.34192 0.34191

0.6 0.26896 0.26897 0.26897 0.26896

0.8 0.22148 0.22148 0.22148 0.22148

1.0 0.18819 0.18819 0.18819 0.18819

3.0 0.07511 0.07511 0.07511 0.07511

0.50 0.4 0.66071 0.66071 0.66071 0.66071

0.6 0.52942 0.52942 0.52942 0.52942

0.8 0.43914 0.43914 0.43914 0.43914

1.0 0.37442 0.37442 0.37442 0.37442

3.0 0.15018 0.15018 0.15018 0.15018

0.75 0.4 0.91027 0.91027 0.91027 0.91026

0.6 0.76724 0.76725 0.76724 0.76724

0.8 0.64740 0.64740 0.64740 0.64740

1.0 0.55605 0.55605 0.55605 0.55605

3.0 0.22481 0.22483 0.22481 0.22481

The numerical solutions obtained by the proposed method, solutions

obtained in [6,7,12,15] and the exact solution for ν= 0.01 and 0.1 are

shown in Table 1and Table 2respectively. From Table 1and Table 2, it is

observed that the numerical solutions obtained by the proposed method are

compatible with numerical solutions available in the literature for ν= 0.01,

whereas for ν= 0.1 numerical solutions obtained by proposed method are

better than the solutions obtained in [6,12,15] even for large value of ∆t

and h.

Example 2. Consider Eq. (1) with initial condition u(x, 0) = 4x(1 −x)

and boundary conditions u(0, t) = 0; u(1, t) = 0 with F(x, t) = 0. The

numerical solution obtained by the proposed method, solution obtained

in [5,12,16] and exact solution for ν= 0.01 are listed in Table 3. It is

observed that the numerical solutions obtained by the proposed method

are better than the solutions obtained in [5,16] even for large values of ∆t

and h. The obtained numerical solutions are slightly less accurate than

solutions listed in [12] in the neighborhood of shock.

458 S. Thakar, S.S. Kumbhar

Table 2: Comparison of numerical and exact solution for ν= 0.1 of Example

1.

x t [12] [15] [6] (CBGM) Present Exact

∆t= 0.0025 ∆t= 0.0001 ∆t= 0.0001 ∆t= 0.0025

h= 0.025 h= 0.0125 h= 0.0125 h= 0.025

0.25 0.4 0.30892 0.30752 0.30890 0.30889 0.30889

0.6 0.24077 0.24042 0.24074 0.24074 0.24074

0.8 0.19572 0.19555 0.19568 0.19568 0.19568

1.0 0.16261 0.16234 0.16257 0.16257 0.16256

3.0 0.02718 - 0.02720 0.02720 0.02720

0.50 0.4 0.56970 0.55953 0.56964 0.56963 0.56963

0.6 0.44729 0.44797 0.44721 0.44721 0.44721

0.8 0.35930 0.35739 0.35924 0.35924 0.35924

1.0 0.29195 0.29134 0.29191 0.29192 0.29192

3.0 0.04016 - 0.04020 0.04020 0.04021

0.75 0.4 0.62520 0.64561 0.62541 0.62544 0.62544

0.6 0.48694 0.48267 0.48719 0.48722 0.48721

0.8 0.37365 0.37533 0.37390 0.37392 0.37392

1.0 0.28724 0.28585 0.28746 0.28747 0.28747

3.0 0.02974 - 0.02977 0.02977 0.02977

Example 3. In this example we consider Eq. (1) with the initial condition

u(x, 0) = 2νπ sin(π x)

α+cos(πx);α > 1, boundary conditions u(0, t) = 0; u(1, t)=0

and F(x, t) = 0. The exact solution of this problem is

u(x, t) = 2νπe−π2ν t sin(πx)

α+e−π2νt cos(πx).

The numerical solutions and L2and L∞errors for α= 2, h = 0.025, ν =

1.0,0.5,0.2,0.1 and ∆t= 0.0001 at t= 0.001 are shown in Table 4and

Table 5. From Table 4and Table 5, it is seen that for ν= 0.5,1.0 the

numerical solutions obtained by the present method are much better than

solutions obtained in [12] whereas for ν= 0.1,0.2 they are better than

solutions in [12].

Example 4. Consider Eq. (1) with the initial condition

u(x, 1) = x

1 + e1

4ν(x2−1

4),

and boundary conditions u(0, t) = u(1.2, t) = 0 and F(x, t) = 0. The exact

Galerkin FEM for forced Burgers’ equation 459

Table 3: Comparison of numerical and exact solution for ν= 0.01 of Ex-

ample 2.

x t [12] [5] [16] Present Exact

∆t= 0.001 ∆t= 0.0001 ∆t= 0.0001 ∆t= 0.001

h= 0.025 h= 0.025 h= 0.0125 h= 0.025

0.25 0.4 0.36225 0.36225 0.36218 0.36226 0.36226

0.6 0.28202 0.28199 0.28197 0.28204 0.28204

0.8 0.23044 0.23039 0.23040 0.23045 0.23045

1.0 0.19468 0.19463 0.19465 0.19469 0.19469

3.0 0.07613 0.07611 0.07613 0.07613 0.07613

0.50 0.4 0.68368 0.68371 0.68364 0.68368 0.68368

0.6 0.54832 0.54835 0.54829 0.54832 0.54832

0.8 0.45371 0.45374 0.45368 0.45371 0.45371

1.0 0.38567 0.38568 0.38564 0.38568 0.38568

3.0 0.15218 0.15216 0.15217 0.15218 0.15218

0.75 0.4 0.92052 0.92047 0.92047 0.92044 0.92050

0.6 0.78300 0.78302 0.78297 0.78288 0.78299

0.8 0.66272 0.66276 0.66270 0.66267 0.66272

1.0 0.56932 0.56936 0.56930 0.56931 0.56933

3.0 0.22782 0.22773 0.22773 0.22774 0.22774

solution of this example is

u(x, t) = (x

t)

1+(t

t0)1

2ex2

4νt

,

where t0=e1

8ν. The comparison of numerical solutions with exact so-

lutions for h= 0.005, ν = 0.005 and ∆t= 0.001 is given in the Table

6. The L2and L∞errors are computed for ν= 0.005, h = 0.005 and

∆t= 0.001 at diﬀerent time levels and their comparison with [11,12] is

shown in Table 7. In this example, Table 6shows that solutions produced

by the present method are better than solutions in [12] and are much close

to exact solutions even for small value of ν. It is also observed from Ta-

ble 7that, the L2and L∞errors by the proposed method are less than

the errors obtained in [11,12]. We have computed the numerical solutions

for ν= 0.01,0.001, N = 100,300 and ∆t= 0.001 at diﬀerent time levels.

These solutions have been depicted in Figure 1and Figure 2. It is noted

that steepness in the solution curves increases as value of νdecreases and

for suﬃciently small value of νsolutions becomes discontinuous.

460 S. Thakar, S.S. Kumbhar

Table 4: Comparison of numerical and exact solutions for α= 2, h = 0.025

and ∆t= 0.0001 at t= 0.001 of Example 3.

x ν = 1 ν= 0.5

[12] Present Exact [12] Present Exact

0.1 0.653547 0.653544 0.653544 0.327870 0.327870 0.327870

0.2 1.305540 1.305533 1.305534 0.655071 0.655069 0.655069

0.3 1.949376 1.949363 1.949364 0.978416 0.978412 0.978413

0.4 2.565949 2.565924 2.565925 1.288469 1.288464 1.288463

0.5 3.110778 3.110738 3.110739 1.563074 1.563063 1.563064

0.6 3.492910 3.492665 3.492866 1.756654 1.756642 1.756642

0.7 3.549585 3.549595 3.549595 1.787204 1.787206 1.787206

0.8 3.049957 3.050138 3.050134 1.537649 1.537696 1.537696

0.9 1.816379 1.816666 1.816660 0.916786 0.916863 0.916860

L∞×1042.85 0.056 - 0.744 0.030 -

L2×1041.07 0.021 - 0.279 0.011 -

Table 5: Comparison of numerical and exact solutions for α= 2, h = 0.025

and ∆t= 0.0001 at t= 0.001 of Example 3.

x ν = 0.2ν= 0.1

[12] Present Exact [12] Present Exact

0.1 0.131412 0.131412 0.131412 0.065750 0.065750 0.065750

0.2 0.262581 0.262581 0.262581 0.131383 0.131383 0.131383

0.3 0.392263 0.392262 0.392262 0.196281 0.196281 0.196281

0.4 0.516710 0.516709 0.516710 0.258576 0.258576 0.258576

0.5 0.627081 0.627079 0.627079 0.313850 0.313849 0.313849

0.6 0.705122 0.705120 0.705120 0.352972 0.352972 0.352972

0.7 0.717882 0.717882 0.717882 0.359443 0.359443 0.359443

0.8 0.618129 0.618137 0.618136 0.309579 0.309581 0.309580

0.9 0.368802 0.368815 0.368814 0.184751 0.184754 0.184754

L∞×1051.22 0.123 - 0.308 0.063 -

L2×1064.57 0.454 - 1.15 0.229 -

Example 5. Consider Burgers’ equation (1) with F(x, t) = kx

(2βt+1)2,k > 0,

β≥0 and the initial condition u(x, 0) = kx,k > β. The exact solution of

this problem for ν= 1 is obtained by Rao and Yadav [4] as follows

u(x, t) = A0x

(2βt + 1) , A0=β+pβ2+k.

Galerkin FEM for forced Burgers’ equation 461

Figure 1: Numerical solutions of Example 4for ν= 0.01,∆t= 0.001 and

N= 100.

Figure 2: Numerical solutions of Example 4for ν= 0.001,∆t= 0.001 and

N= 300.

The boundary conditions are given from the exact solution. The numerical

solutions of this example obtained for k= 5, β= 2 in x∈[−1,1] is

used to compute L2and L∞errors. The comparison of L2and L∞errors

obtained in [9] is given in Table 8. From Table 8it is seen that, the

numerical solutions obtained by the present method are compatible with

solutions in [9]. We have computed L2and L∞errors in the solution for

N= 20, k = 100 and β= 1 at t= 1 for diﬀerent values of ∆t. These errors

are listed in Table 9. It is observed that L2and L∞errors decreases with

decreasing value of ∆t. This shows that proposed scheme do not amplify

errors with increase in number of iterations and the scheme is numerically

462 S. Thakar, S.S. Kumbhar

Table 6: Comparison of numerical and exact solutions for ν= 0.005,∆t=

0.001 and h= 0.005 of Example 4.

x t [12] Present Exact

0.2 1.7 0.1176452 0.1176452 0.1176452

2.5 0.0799990 0.0799990 0.0799990

3.0 0.0666658 0.0666658 0.0666658

3.5 0.0571422 0.0571422 0.0571422

0.4 1.7 0.2351690 0.2351677 0.2351677

2.5 0.1599771 0.1599769 0.1599769

3.0 0.1333211 0.1333209 0.1333209

3.5 0.1142780 0.1142779 0.1142779

0.6 1.7 0.2958570 0.2959101 0.2959097

2.5 0.2381299 0.2381207 0.2381207

3.0 0.1994839 0.1994806 0.1994805

3.5 0.1712257 0.1712242 0.1712242

0.8 1.7 0.0006381 0.0006465 0.0006465

2.5 0.1021325 0.1020955 0.1020957

3.0 0.2088032 0.2088360 0.2088359

3.5 0.2145938 0.2145869 0.2145869

Table 7: Comparison of L2and L∞errors for ν= 0.005,∆t= 0.001 and

h= 0.005 of Example 4.

t[11]β= 0.5 [12] Present

∆t= 0.01, h = 0.001 ∆t= 0.001, h = 0.005 ∆t= 0.001, h = 0.005

L∞×104L2×104L∞×104L2×104L∞×104L2×104

1.7 13.47279 3.84209 0.994 0.252 0.006 0.0017

2.5 15.54700 4.91345 0.549 0.151 0.002 0.0008

3.0 15.52891 5.15077 0.414 0.118 0.023 0.0029

3.5 15.21961 5.25855 0.486 0.117 0.572 0.0754

stable.

Example 6. In this example, we consider the initial condition u(x, 0) = 0,

Galerkin FEM for forced Burgers’ equation 463

Table 8: Comparison of L2and L∞errors for k= 5, β = 2,∆t= 0.01 and

N= 10 of Example 5.

L∞L2

t= 5 t= 10 t= 5 t= 10

[9] IMQQI 2.816 ×10−91.876 ×10−10 2.020 ×10−91.345 ×10−10

Present 2.811 ×10−91.872 ×10−10 2.854 ×10−91.901 ×10−10

Table 9: Comparison of L2and L∞errors for k= 100, β = 1 and N= 20

at t= 1 of Example 5.

∆t0.01 0.005 0.001

L22.88 ×10−55.90 ×10−61.41 ×10−6

L∞2.85 ×10−55.89 ×10−61.39 ×10−6

Figure 3: Numerical solutions of Example 7for ν= 0.1,∆t= 0.01 and

N= 50.

boundary conditions u(0, t) = u(π, t) = 0 and F(x, t) = Asin(x); A > 0.

The exact solution of this example is obtained by [1]. The numerical solu-

tions obtained by the proposed method, exact solutions and solutions given

in [9] for ν= 1, A = 20 and ν= 0.1, A = 1 are listed in Table 10 and Table

11, respectively. The proposed method produces better solutions than so-

lutions obtained by DMQQI method and are compatible with the solutions

obtained by IMQQI method for ν= 1. For ν= 0.1 the solutions com-

puted by the proposed method are compatible with the solutions obtained

by DMQQI and IMQQI methods.

464 S. Thakar, S.S. Kumbhar

Table 10: Comparison of numerical results for ν= 1, A = 20,∆t= 0.001

at t= 3.0 of Example 6.

x[9] IMQQI [9] DMQQI Present Exact

N= 20 N= 30 N= 20 N= 30

0.5 2.1481 2.1474 2.1481 2.1481 2.1481

1.0 4.1562 4.1558 4.1563 4.1563 4.1562

1.5 5.8928 5.8924 5.8928 5.8928 5.8928

2.0 7.2404 7.2400 7.2403 7.2404 7.2404

2.5 8.0358 8.0298 8.0307 8.0303 8.0302

3.0 4.5143 4.4965 4.5155 4.5125 4.5140

Table 11: Comparison of numerical results for ν= 0.1, A = 1,∆t= 0.001

at t= 3.0 of Example 6.

x[9] IMQQI [9] DMQQI Present Exact

N= 30 N= 60 N= 30 N= 60

0.5 0.4851 0.4851 0.4853 0.4853 0.4824

1.0 0.9392 0.9391 0.9392 0.9392 0.9331

1.5 1.3300 1.3320 1.3320 1.3320 1.3221

2.0 1.6271 1.6371 1.6372 1.6372 1.6222

2.5 1.8122 1.8322 1.8315 1.8323 1.8102

3.0 1.6207 1.6476 1.6633 1.6551 1.6155

Example 7. In this test problem we consider Eq. (1) with F(x, t) = 0,

initial condition u(x, 0) = sin 2πx, 0≤x≤1 and boundary conditions

u(0, t) = u(1, t) = 0. We have computed the numerical solutions for ν=

0.1,0.01 and 0.001 at t= 0,1,2. These solutions are depicted in Figure

3, Figure 4, and Figure 5. Similar solutions are reported in [13]. It is

observed that when value of νdecreases, steepness of the curves increases

and eventually solution becomes discontinuous.

5 Conclusion

The cubic B-spline Galerkin ﬁnite element method is successfully imple-

mented to the one dimensional nonlinear forced Burgers’ equation. Cubic

Galerkin FEM for forced Burgers’ equation 465

Figure 4: Numerical solutions of Example 7for ν= 0.01,∆t= 0.01 and

N= 100.

Figure 5: Numerical solutions of Example 7for ν= 0.001,∆t= 0.0001 and

N= 300.

B-splines are redeﬁned to accommodate the boundary conditions. Burgers’

equation is discretized in time by using Taylors series expansion and then

Galerkin ﬁnite element method is constructed for discretized equation. The

Von Neumann stability analysis shows that the corresponding linearized

scheme is unconditionally stable. some numerical test examples are solved

to support the proposed method. It is seen that the method is eﬃcient and

reliable for solving the one dimensional forced Burgers’ equation.

Acknowledgements

The authors expresses sincere thanks to the anonymous reviewers for their

careful reading of the manuscript and many valuable comments and sug-

466 S. Thakar, S.S. Kumbhar

gestions that improves quality of the paper.

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