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In this paper second order explicit Galerkin finite element method based on cubic B-splines is constructed to compute numerical solutions 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.
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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+uuxνuxx =F(x, t), a xb, 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 xb, (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= (ba)/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
tt
2un
tt,(4)
where tn=t0+nt,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
unxun+ν∂2
xun+F(x, tn)
=(un+1 un)
tt
2un
txununxun
t+ν∂2
xun
t+Ft(x, tn).
Using forward difference approximation to un
tin above equation and after
the simplification we obtain
1t
2xununx+ν∂2
x(un+1 un)
t
=unxun+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
2xun+unxν∂2
xun+1
=1 + νt
22
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
xνun
xx F(x, tn)o
n1t
2xununx+ν∂2
x(un+1 un)
t+unxun
ν∂2
xunF(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
(xxm2)3,[xm2, xm1],
h3+ 3h2(xxm1)+3h(xxm1)2
3(xxm1)3,[xm1, 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
δ1(t) = g1(t)4δ0(t)δ1(t),(10)
δN+1(t) = g2(t)δN1(t)4δN(t).(11)
Galerkin FEM for forced Burgers’ equation 449
The solution given by Eq. (8) now becomes
u(x, t) = g1(t)φ1(x) + g2(t)φN+1(x) +
N
X
i=0
δi(t)Bi(x),(12)
where
B0(x) = φ0(x)4φ1(x), B1(x) = φ1(x)φ1(x),
Bj(x) = φj(x),for j= 2,3, . . . , N 2,
BN1(x) = φN1(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)φ1(x)Bi(x)0, i = 0,1,2,
g2(t)φN+1(x)Bi(x)0, i =N2, N 1, N ,
0, otherwise,
(15)
R1(x, tn, tn+1) = g1(tn)g1(tn+1)φ1(x)
+t
2ν(g1(tn) + g1(tn+1))φ00
1(x)2g1(tn)g1(tn+1)φ1(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)0νB00
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=l1
δi(t)Bi(x),(19)
ux(x, t) =
l+2
X
i=l1
δi(t)B0
i(x),(20)
uxx(x, t) =
l+2
X
i=l1
δi(t)B00
i(x).(21)
On [xl, xl+1] for l= 1,2, . . . , N 2, Eq. (7) becomes
l+2
X
i=l1hBi(x) + t
2
l+2
X
j=l1
δj(tn)(BiBj)0νB00
ii·δi(tn+1)
=
l+2
X
i=l1hBi(x) + νt
2B00
i(x)i·δi(tn)+∆tF (x, tn) + (∆t)2
2Ft(x, tn).(22)
On [xN1, xN] Eq. (17) is used in Eq. (7) to obtain
N
X
i=N2hBi(x) + t
2hi(x, tn) +
N
X
j=N2
δj(tn)(BiBj)0νB00
ii·δi(tn+1)
=
N
X
i=N2hBi(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
1B1+νC1i·δn+1
1
=hA1t
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(nt)g1((n+ 1)∆t))
h
49
40
1
Galerkin FEM for forced Burgers’ equation 451
+t
2(ν(g1(nt) + g1((n+ 1)∆t))
10h
51
54
3
+2g1(nt)g1((n+ 1)∆t))
168
97
70
1
),
Fn
1= ∆t
Rx1
x0F(x, nt)B0(x)dx
Rx1
x0F(x, nt)B1(x)dx
Rx1
x0F(x, nt)B2(x)dx
+(∆t)2
2
Rx1
x0Ft(x, nt)B0(x)dx
Rx1
x0Ft(x, nt)B1(x)dx
Rx1
x0Ft(x, nt)B2(x)dx
.
In order to compute Fn
1,F(x, t) and Ft(x, t) are evaluated at t=ntand
then Rx1
x0F(x, nt)Bidx,Rx1
x0Ft(x, nt)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(nt)
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
Bj1B0B0
i1dx, Zx1
x0
Bj1B1B0
i1dx, Zx1
x0
Bj1B2B0
i1dxδn
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=l1, 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 νt
2Cl+1i·δn
l+1 +Fn
l+1,(25)
where for l= 1,2, . . . , N 2, δn
l+1 = (δn
l1, δn
l, δn
l+1, δn
l+2)T,
Fn
l+1 = ∆t
Rxl+1
xlF(x, nt)Bl1(x)dx
Rxl+1
xlF(x, nt)Bl(x)dx
Rxl+1
xlF(x, nt)Bl+1(x)dx
Rxl+1
xlF(x, nt)Bl+2(x)dx
+(∆t)2
2
Rxl+1
xlFt(x, nt)Bl1(x)dx
Rxl+1
xlFt(x, nt)Bl(x)dx
Rxl+1
xlFt(x, nt)Bl+1(x)dx
Rxl+1
xlFt(x, nt)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
336 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+l2Bl1B0
i+l2dx, Zxl+1
xl
Bj+l2BlB0
i+l2dx,
Zxl+1
xl
Bj+l2Bl+1B0
i+l2dx, Zxl+1
xl
Bj+l2Bl+2B0
i+l2dx!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=N2, N 1, N
Galerkin FEM for forced Burgers’ equation 453
and integrating by parts on the interval [xN1, xN] we get
hAN+t
2hn
NBN+νCNi·δn+1
N
=hANt
2hn+1
N+νCNi·δn
N+Sn
N+Fn
N,(26)
where δn
N= (δn
N2, δn
N1, δn
N)T,
Sn
N=140(g2(nt)g2((n+ 1)∆t))
h
1
40
49
+t
2(ν(g2(nt) + g2((n+ 1)∆t))
10h
3
54
51
2g2(nt)g2((n+ 1)∆t))
168
1
70
97
),
Fn
N= ∆t
RxN
xN1F(x, nt)BN2(x)dx
RxN
xN1F(x, nt)B1(x)dx
RxN
xN1F(x, nt)B2(x)dx
+(∆t)2
2
RxN
xN1Ft(x, nt)BN2(x)dx
RxN
xN1Ft(x, nt)B1(x)dx
RxN
xN1Ft(x, nt)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(nt)
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
xN1
Bj+N3BN2B0
i+N3dx, ZxN
xN1
Bj+N3BN1B0
i+N3dx,
ZxN
xN1
Bj+N3BNB0
i+N3dx!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 φ1(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
2hnB+νCi·δn+1
=hAt
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)φ1(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)φ1(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+Ut
2B+νt
2Ci·δn+1 =hAνt
2Ci·δn+Rn+Sn+Fn,(30)
where Bis 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+Ut
2B+νt
2Ci·n+1 =hAνt
2Ci·n,(31)
where, nis error in the solution at t=tn. Matrices A,Band 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
l3+α2n+1
l2+α3n+1
l1+α4n+1
l+α5n+1
l+1 +α6n+1
l+2 +α7n+1
l+3
=α8n
l3+α9n
l2+α10n
l1+α11n
l+α10n
l+1 +α9n
l+2 +α8n
l+3 (32)
where n
jis the error in δjat t=nt, 0 jN , 4lN3,
α1=r1+r23r3, α2= 120r1+ 56r272r3,
α3= 1191r1+ 245r245r3, α4= 2416r1+ 240r3,
α5= 1191r1245r245r3, α6= 120r156r272r3,
α7=r1r23r3, α8=r1+ 3r3, α9= 120r1+ 72r3,
α10 = 1191r1+ 45r3, α11 = 2416r1240r3,
r1=h
140, r2=Ut
40 , r3=νt
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
[(ac)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
(ac)ib.
Since r3>0, c0 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 Lerrors are defined as,
L2=v
u
u
th
N
X
j=0
|Uexact
junumer
j|2, L= max
1jN|Uexact
junumer
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 anen2π2νt nsin(nπx)
a0+P
n=1 anen2π2νt ncos(nπx)
where the Fourier coefficients an,n= 0,1,2,..., are given by
a0=Z1
0
e(2πν)1(1cos(πx))dx,
an= 2 Z1
0
e(2πν)1(1cos(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) = 2νπeπ2ν t sin(πx)
α+eπ2νt cos(πx).
The numerical solutions and L2and Lerrors 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(x21
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 Lerrors 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 Lerrors 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,
β0 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 Lerrors. The comparison of L2and Lerrors
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 Lerrors 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 Lerrors 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 Lerrors 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 Lerrors for k= 5, β = 2,t= 0.01 and
N= 10 of Example 5.
LL2
t= 5 t= 10 t= 5 t= 10
[9] IMQQI 2.816 ×1091.876 ×1010 2.020 ×1091.345 ×1010
Present 2.811 ×1091.872 ×1010 2.854 ×1091.901 ×1010
Table 9: Comparison of L2and Lerrors for k= 100, β = 1 and N= 20
at t= 1 of Example 5.
t0.01 0.005 0.001
L22.88 ×1055.90 ×1061.41 ×106
L2.85 ×1055.89 ×1061.39 ×106
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, 0x1 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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In this article, we construct solutions of a nonhomogeneous Burgers equation subject to certain unbounded initial profiles. In an interesting study, Kloosterziel [1] represented the solution of an initial value problem (IVP) for the heat equation, with initial data in , as a series of the self-similar solutions of the heat equation. This approach quickly revealed the large time behavior for the solution of the IVP. Inspired by Kloosterziel [1]'s approach, we express the solution of the nonhomogeneous Burgers equation in terms of the self-similar solutions of a linear partial differential equation with variable coefficients. Finally, we also obtain the large time behavior of the solution of the nonhomogeneous Burgers equation.
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In the present paper numerical solutions of the one-dimensional Burgers' equation are obtained by a method based on collocation of cubic B-splines over finite elements. The accuracy of the proposed method is demonstrated by three test problems. The numerical results are found in good agreement with exact solutions. Time-space integration of the Burgers' equation yields a system of difference equation which is shown to be unconditionally stable.
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In this study, a least-squares quadratic B-spline finite element method for calculating the numerical solutions of the one-dimensional Burgers-like equations with appropriate boundary and initial conditions is presented. Three test problems have been studied to demonstrate the accuracy of the present method. Results obtained by the method have been compared with the exact solution of each problem and are found to be in good agreement with each other. A Fourier stability analysis of the method is also investigated.