Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 38, pp. 1–12.

ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu

DOI: 10.58997/ejde.2024.38

A SECOND ORDER CONVERGENT DIFFERENCE SCHEME

FOR THE INITIAL-BOUNDARY VALUE PROBLEM OF

ROSENAU-BURGERS EQUATION

SITONG DONG, XIN ZHANG, YUANFENG JIN

Abstract. We construct a two-level implicit nonlinear finite difference scheme
for the initial boundary value problem of Rosenau-Burgers equation based on

the method of order reduction. We discuss conservation, unique solvability,

and convergence for the difference scheme. The new scheme is shown to be
second-order convergent in time and space. Finally, numerical simulations

illustrate our theoretical analysis.

1. Introduction

In the study of the dynamics of dense discrete systems, the case of wave-wave
and wave-wall interactions cannot be described using the well-known KDV equation
which was suggested by Korteweg and de Vries in 1895 [1]. To overcome this
shortcoming, Rosenau [2, 3] proposed an equation in the form

ut + uxxxxt + ux + uux = 0. (1.1)

The theoretical results on the existence, uniqueness and regularity of the solutions
to (1.1) have been investigated by Park [4]. But it is difficult to find the analytical
solution for (1.1). Much attention has devoted to numerical solutions of (1.1) by
various numerical methods [5, 6, 7, 8].

On the other hand, for the further consideration of dissipation in space for dy-
namic systems, such as the phenomenon of bore propagation and water waves, the
viscous term −uxx needs to be included

ut + uxxxxt − uxx + ux + uux = 0. (1.2)

This equation is usually called the Rosenau–Burgers equation. Many scholars had
proposed difference schemes for the Rosenau–Burgers Equation. Hu et al [9] consid-
ered a nonlinear Crank-Nicolson difference scheme for the Rosenau-Burgers equa-
tion by the Newton iterative algorithm. Pan and Zhang [10] discussed a three-level
linear implicit difference scheme for the Rosenau-Burgers equation which is un-
conditionally stable. Ahmat and Abduwali [11] investigated two class of modified
local Crank- Nicolson schemes for the Rosenau-Burgers equation,which has simple

2020 Mathematics Subject Classification. 65M06, 65M12.
Key words and phrases. Rosenau-Burgers equation; nonlinearized difference scheme;

conservation law; convergence.
©2024. This work is licensed under a CC BY 4.0 license.

Submitted December 2, 2023. Published July 4, 2024.

1



2 S. DONG, X. ZHANG, Y. JIN EJDE-2024/38

structure and high accuracy. Omrani [12] explored the Galerkin–Crank–Nicolson
discrete method. Guo et al [13] considered a two-level implicit nonlinear discrete
scheme, which preserves the energy decline property of the Rosenau–Burgers equa-
tion. Luo et al [14] used a three-level linear implicit finite difference scheme with
energy dissipation which has second-order and fourth-order in time and space. For
the two-dimensional case, Rouatbi et al [15] presented a linearized Crank–Nicolson
difference scheme.

In this article, we consider the initial-boundary value problem of the Rosenau-
Burgers equation by the finite difference method.

ut + uxxxxt − uxx + ux + uux = 0, 0 < x < L, 0 < t ≤ T, (1.3)

u(x, 0) = φ(x), 0 ≤ x ≤ L, (1.4)

u(0, t) = 0, u(L, t) = 0, uxx(0, t) = 0, uxx(L, t) = 0, 0 ≤ t ≤ T, (1.5)

where φ(x) is a given function. The initial-boundary value problem of the Rose-
nau–Burgers equation contains at least one derivative boundary condition. There-
fore, a difference scheme must be established at the node adjacent to this boundary
point that is consistent with the difference schemes on the other nodes. To achieve
this, we employ the method of order reduction to get a difference scheme with con-
vergence order two in both space and time. Additionally, we rigorously prove the
convergence of this particular difference.

This article is structured as follows. In Section 2, we introduce some useful
notation and lemmas. In Section 3, we describe a conservative two-level implicit
nonlinear finite difference scheme for the Rosenau–Burgers equation. The scheme
has second-order accuracy in space and time. In Section 4, we analize the unique
solvability is analyzed. In Section 5, we prove the convergence and stability for the
difference schem. Finally, a numerical example illustrates our theoretical results.

2. Preliminaries

In this section, we introduce notation and lemmas that will be used throughout
this article. To partition the domain [0, L] × [0, T ], we use positive integers M
and N . Let h = L/M and τ = T/N . Denote xi = ih, 0 ≤ i ≤ M ; tk = kτ ,
0 ≤ k ≤ N ;Ωh = {xi : 0 ≤ i ≤ M}, Ωτ = {tk : 0 ≤ k ≤ N}, Ωhτ = Ωh × Ωτ . For
each grid function v = {vki : 0 ≤ i ≤M, 0 ≤ k ≤ N} defined on Ωhτ , we denote

δ+x v
k
i =

1

h
(vki+1 − vki ), δ

2
xv

k
i =

1

h
(δ+x v

k
i − δ+x v

k
i−1),∆xv

k
i =

1

2h
(vki+1 − vki−1),

v
k+ 1

2
i =

1

2
(vki + vk+1

i ), vki+ 1
2
=

1

2
(vki + vki+1), δtv

k+ 1
2

i =
1

τ
(vk+1

i − vki ),

Let Vh = {v : v = (v0, v1, . . . , vM−1, vM )} and V̊h = {v : v ∈ Vh, v0 = vM = 0}
be the spaces of grid functions on Ωh. For any grid functions u, v ∈ Vh, we define
the discrete inner product

(u, v) = h

M−1∑
i=1

uivi,

and the corresponding norms (seminorm)

∥u∥ =
√

(u, u), |u|1 =

√
(δ+x u, δ

+
x u), ∥u∥∞ = max

1≤i≤M
|ui|



EJDE-2024/38 A DIFFERENCE SCHEME FOR ROSENAU-BURGERS EQUATION 3

In addition, we define the function

ψ(u, v)i =
1

3
[ui∆xvi +∆x(uv)i], 1 ≤ i ≤M.

Lemma 2.1 ([16]). For each grid function u, v ∈ V̊h, we have

∥v∥∞ ≤
√
L

2
|v|1, |v|1 ≤ 2

h
∥v∥, ∥v∥ ≤ L√

6
|v|1,

(δ2xu, v) = −(δ+x u, δ
+
x v).

And for an arbitrary ε > 0, we have

∥v∥2∞ ≤ ε|v|21 +
1

4ε
∥v∥2.

Lemma 2.2 ([17]). For each grid functions u ∈ Vh and v ∈ V̊h we have

(ψ(u, v), v) = 0, (∆xu, u) = 0.

Lemma 2.3 (Gronwall inequality [17]). Let {F k}∞i=0 and {gk}∞i=0 be two non-
negative sequences and satisfy

F k+1 ≤ (1 + cτ)F k + τgk, k = 0, 1, 2, . . . ,

then

F k ≤ exp(ckτ)
(
F 0 + τ

k−1∑
l=0

gl
)
, k = 0, 1, 2, . . .

3. Nnonlinear conservative difference scheme

In this section, we use the method of reduction of order to establish a difference
scheme for the problem (1.3)-(1.5), and illustrate the truncation errors in detail,
then we analyze the conservation of the difference scheme.

3.1. Construction of difference scheme. Let v = uxx, then problem (1.3)–(1.5)
is equivalent to

ut + vxxt − v + ux + uux = 0, 0 < x < L, 0 < t ≤ T, (3.1)

v = uxx, 0 < x < L, 0 < t ≤ T, (3.2)

u(x, 0) = φ(x), 0 < x < L, (3.3)

u(0, t) = 0, u(L, t) = 0, uxx(0, t) = 0, uxx(L, t) = 0, 0 ≤ t ≤ T . (3.4)

According to (3.1) and (3.4),

v(0, t) = 0, v(L, t) = 0, 0 ≤ t ≤ T . (3.5)

We define the grid functions

U = {Uk
i : 0 ≤ i ≤M, 0 ≤ k ≤ N} and V = {V k

i : 0 ≤ i ≤M, 0 ≤ k ≤ N},
where Uk

i = u(xi, tk) and V
k
i = v(xi, tk).

Considering (3.1) at the point (xi, t
k+ 1

2 ) and (3.4) at the point (xi, tk),

ut(xi, tk+ 1
2
) + vxxt(xi, tk+ 1

2
)− v(xi, tk+ 1

2
) + ux(xi, tk+ 1

2
)

+ u(xi, tk+ 1
2
)ux(xi, tk+ 1

2
) = 0, 1 ≤ i ≤M − 1, 1 ≤ k ≤ N − 1,

(3.6)

v(xi, tk) = uxx(xi, tk), 1 ≤ i ≤M − 1, 1 ≤ k ≤ N. (3.7)



4 S. DONG, X. ZHANG, Y. JIN EJDE-2024/38

By using the Taylor expansion,

δtU
k+ 1

2
i + δtδx

2V
k+ 1

2
i − V

k+ 1
2

i +∆xU
k+ 1

2
i + ψ(Uk+ 1

2 , Uk+ 1
2 )i = R

k+ 1
2

i ,

1 ≤ i ≤M − 1, 0 ≤ k ≤ N − 1,
(3.8)

V k
i = δ2xU

k
i +Qk

i , 1 ≤ i ≤M − 1, 0 ≤ k ≤ N, (3.9)

there exist constants c1 and c2 such that

|Rk+ 1
2

i | ≤ c1(τ
2 + h2), 1 ≤ i ≤M − 1, 0 ≤ k ≤ N − 1,

|Qk
i | ≤ c2h

2, 1 ≤ i ≤M − 1, 0 ≤ k ≤ N,

|δtQ
k+ 1

2
i | ≤ c2(τ

2 + h2), 1 ≤ i ≤M − 1, 0 ≤ k ≤ N − 1.

(3.10)

Notice that the initial-boundary conditions (3.3)–(3.5) become

U0
i = φ(xi), 1 ≤ i ≤M − 1,

Uk
0 = 0, Uk

M = 0, 0 ≤ k ≤ N,

V k
0 = 0, V k

M = 0, 0 ≤ k ≤ N.

(3.11)

Omitting the small terms in (3.8) and (3.9) and combining with (3.11), we can
derive the difference scheme for (3.1)-(3.4),

δtu
k+ 1

2
i + δtδx

2v
k+ 1

2
i − v

k+ 1
2

i +∆xu
k+ 1

2
i + ψ(uk+

1
2 , uk+

1
2 )i = 0,

1 ≤ i ≤M − 1, 0 ≤ k ≤ N − 1,
(3.12)

vki = δ2xu
k
i , 1 ≤ i ≤M − 1, 0 ≤ k ≤ N, (3.13)

u0i = φ(xi), 1 ≤ i ≤M − 1, (3.14)

uk0 = 0, ukM = 0, 0 ≤ k ≤ N, (3.15)

vk0 = 0, vkM = 0, 0 ≤ k ≤ N. (3.16)

3.2. Conservation law.

Theorem 3.1. Let {uki , vki : 0 ≤ i ≤M, 0 ≤ k ≤ N} be the solutions of (3.12)–(3.16).
Denote

Ek = ∥uk∥2 + ∥vk∥2 + 2τ

k−1∑
l=0

|ul+ 1
2 |21, 0 ≤ k ≤ N.

Then we have

Ek = E0, 1 ≤ k ≤ N.

Proof. Take the inner product of (3.12) with uk+
1
2 ,

(δtu
k+ 1

2 , uk+
1
2 ) + (δtδ

2
xv

k+ 1
2 , uk+

1
2 )− (vk+

1
2 , uk+

1
2 ) + (∆xu

k+ 1
2 , uk+

1
2 )

+
1

3
(ψ(uk+

1
2 , uk+

1
2 ), uk+

1
2 ) = 0, 0 ≤ k ≤ N − 1.

From Lemma 2.2 it follows that

1

2τ
(∥uk+1∥2 − ∥uk∥2) + 1

2τ
(∥vk+1∥2 − ∥vk∥2) + |uk+ 1

2 |21 = 0.

This equation can be written as

∥uk+1∥2 + ∥vk+1∥2 + 2τ |uk+ 1
2 |21 = ∥uk∥2 + ∥vk∥2.



EJDE-2024/38 A DIFFERENCE SCHEME FOR ROSENAU-BURGERS EQUATION 5

Therefore, replacing the superscript k with l and summing for l from 0 to k − 1
yields

Ek = ∥uk∥2 + ∥vk∥2 + 2τ

k−1∑
l=0

|ul+ 1
2 |21 = Ek−1 = · · · = E0. □

4. Solvability

In this section, we analyze the unique solvability of the difference scheme estab-
lished in (3.12)-(3.16).

Theorem 4.1 (Browder theorem [16]). Let (H, (·, ·)) be a finite dimensional inner
product space, ∥ · ∥ be the associated norm, and Π : H → H be continuous operator,

∃α > 0, ∀z ∈ H, ∥z∥ = α, Re(Π(z), z) ≥ 0.

Then there exists a z∗ ∈ H satisfying ∥z∗∥ ≤ α such that Π(z∗) = 0.

Theorem 4.2. The difference scheme (3.12)-(3.16) has at least one solution.

Proof. Let
uk = (uk0 , u

k
1 , . . . , u

k
M ), vk = (vk0 , v

k
1 , . . . , v

k
M ).

It is easy to obtain u0 from (3.13) and (3.14). From (3.13)-(3.16), v0 can be found
by computing an associated system of linear equations. Suppose that {uk, vk} have

been determined, then we may regard {uk+ 1
2 , vk+

1
2 } as unknowns and

uk+1
i = 2u

k+ 1
2

i − uki , vk+1
i = 2v

k+ 1
2

i − vki , 0 ≤ i ≤M.

We denote

ωi = u
k+ 1

2
i , zi = v

k+ 1
2

i , 0 ≤ i ≤M,

Then
uk+1
i = 2wi − uki , vk+1

i = 2zi − vki , 0 ≤ i ≤M.

From (3.12), (3.13), (3.15) and (3.16), the system of equations can be considered
with respect to {wi}Mi=0 and {zi}Mi=0:

2

τ
(ωi − uki ) +

2

τ
(δ2xzi − δ2xv

k
i )− zi +∆xωi + ψ(ω, ω)i = 0,

1 ≤ i ≤M − 1,
(4.1)

zi = δ2xωi, 1 ≤ i ≤M − 1, (4.2)

ω0 = 0, ωM = 0, (4.3)

z0 = 0, zM = 0. (4.4)

We define Π(ω) : V̊h → V̊h by

Π(ω)i =

{
2
τ (ωi − uki ) +

2
τ (δ

2
xzi − δ2xv

k
i )− zi +∆xωi + ψ(ω, ω)i, 1 ≤ i ≤M − 1,

0, i = 0,M.

where (v0, v1, . . . , vM ) is determined by (3.13) and (3.15). Then Π(ω) is a contin-

uous function in V̊h. Taking the inner product of Π(ω) with ω, using Lemma 2.1
yields

(Π(ω), ω) =
2

τ
[∥ω∥2 − (uk, ω)] +

2

τ
[∥z∥2 − (δ2xu

k, δ2xω)] + |ω|21

≥ 2

τ
[(∥ω∥2 − ∥uk∥∥ω∥) + (∥z∥2 − ∥vk∥∥z∥)]



6 S. DONG, X. ZHANG, Y. JIN EJDE-2024/38

≥ 2

τ
[(∥ω∥2 − ∥uk∥∥ω∥) +

(
∥z∥2 − ∥vk∥ · 6

L2
∥ω∥

)
]

=
2

τ
(∥ω∥ − ∥uk∥ − 6

L2
∥vk∥)∥ω∥.

Thus, when ∥ω∥ = ∥uk∥+ 6
L2 ∥vk∥, (Π(ω), ω) ≥ 0. By Theorem 4.1, there exists

an ω∗ ∈ V̇h satisfying ∥ω∥ ≤ ∥uk∥+ 6
L2 ∥vk∥ such that (Π(ω∗)) = 0. Consequently,

the difference scheme (3.12), (3.13), (3.15) and (3.16) has at least one solution
uk+1 = 2ω∗ − uk.

Observing (4.2) and (4.4), when (ω∗
1 , ω

∗
2 , . . . , ω

∗
M−1) is known, (z

∗
1 , z

∗
2 , . . . , z

∗
M−1)

can be determined by (4.2) and (4.4) uniquely. Thus

vk+1
i = 2zi − vki , 1 ≤ i ≤M − 1 . □

Theorem 4.3. When τ < 4/c43, the solution of the difference scheme (3.12)–(3.16)
is unique.

Proof. Suppose that both {u(1), v(1) ∈ V̊h} and {u(2), v(2) ∈ V̊h} are two solutions
of (4.1)-(4.4). Then

2

τ
(u

(1)
i − uki ) +

2

τ
(δ2xv

(1)
i − δ2xv

k
i )− v

(1)
i +∆xu

(1)
i + ψ(u(1), u(1))i = 0,

1 ≤ i ≤M − 1,
(4.5)

v
(1)
i = δ2xu

(1)
i , 1 ≤ i ≤M − 1, (4.6)

u
(1)
0 = 0, u

(1)
M = 0, (4.7)

v
(1)
0 = 0, v

(1)
M = 0. (4.8)

and

2

τ
(u

(2)
i − uki ) +

2

τ
(δ2xv

(2)
i − δ2xv

k
i )− v

(2)
i +∆xu

(2)
i + ψ(u(2), u(2))i = 0,

1 ≤ i ≤M − 1,
(4.9)

v
(2)
i = δ2xu

(2)
i , 1 ≤ i ≤M − 1, (4.10)

u
(2)
0 = 0, u

(2)
M = 0, (4.11)

v
(2)
0 = 0, v

(2)
M = 0. (4.12)

Let

ui = u
(1)
i − u

(2)
i , vi = v

(1)
i − v

(2)
i , 0 ≤ i ≤M.

Subtracting (4.9)-(4.12) from (4.5)–(4.8) leads to

2

τ
ui +

2

τ
δ2xvi − vi +∆xui + [ψ(u(1), u(1))i − ψ(u(2), u(2))i] = 0,

1 ≤ i ≤M − 1,
(4.13)

vi = δ2xui, 1 ≤ i ≤M − 1, (4.14)

u0 = 0, uM = 0, (4.15)

v0 = 0, vM = 0. (4.16)

Taking the inner product of (4.13) with u yields

2

τ
∥u∥2 + 2

τ
(δ2xv, u)− (v, u) + (∆xu, u) = −(ψ(u(1), u(1))− ψ(u(2), u(2)), u) (4.17)



EJDE-2024/38 A DIFFERENCE SCHEME FOR ROSENAU-BURGERS EQUATION 7

In view of the definition of u and Lemma 2.2, we have

−(ψ(u(1), u(1))− ψ(u(2), u(2)), u) = −(ψ(u, u(1)), u)

= −h
3

M−1∑
i=1

[ui∆xu
(1)
i +∆x(uu

(1))i]ui

=
h

3

M−1∑
i=1

[u
(1)
i ∆x(u

2)i + (uu(1))i∆xui]

≤ 1

3

(
2∥u∥∞∥u(1)∥ |u|1 + ∥u∥∞∥u(1)∥ |u|1

)
≤ c3∥u∥∞|u|1 .

Then it follows from (4.17) that

2

τ
∥u∥2 + 2

τ
∥v∥2 + |u|21 ≤ c3∥u∥∞|u|1.

Using Lemma 2.1 gives

2

τ
∥u∥2 + |u|21 ≤ c3

(
ε|u|1 +

1

2ε
∥u∥

)
|u|1

≤ c3ε|u|21 + c3ε|u|21 +
1

4c3ε
(
c3
2ε

)2∥u∥2

= 2c3ε|u|21 +
c3

16ε3
∥u∥ .

Let ε = 1
2c3

. Then 2
τ ∥u∥

2 ≤ c3
4

2 ∥u∥2. When τ < 4
c34 , we obtain ∥u∥ = 0. □

5. Convergence

In this section, we analyze the convergence of the difference scheme (3.12)-(3.16).
Let

c0 = max
(x,t)∈[0,L]×[0,T ]

{|u(x, t)|, |ux(x, t)|}.

The convergent result is summarized as follows.

Theorem 5.1. Let {u(x, t), v(x, t)} be the solutions of (3.1)-(3.5) and {uki , vki :
0 ≤ i ≤ M, 0 ≤ k ≤ N} be the solutions of the difference scheme (3.12)-(3.16).
Also let

eki = Uk
i − uki , fki = V k

i − vki , 0 ≤ i ≤M, 0 ≤ k ≤ N,

and

c4 = 3c0
2 +

c0
2L2

2
+

3L

4
, c5 = e

3
2MT

√
6τM1,

M = (c4 + 1), M1 = c2
2 + c1

2.

If τ and h satisfy τ2 + h2 ≤ 1/c5 and 2c4τ ≤ 1/3, then the error estimate is

|ek|1 ≤ c5(τ
2 + h2), 0 ≤ k ≤ N.



8 S. DONG, X. ZHANG, Y. JIN EJDE-2024/38

Proof. Subtracting (3.12)-(3.16) from (3.8), (3.9) and (3.11), a system of error
equations is

δte
k+ 1

2
i + δtδ

2
xf

k+ 1
2

i − f
k+ 1

2
i +∆xe

k+ 1
2

i + [ψ(Uk+ 1
2 , Uk+ 1

2 )i − ψ(uk+
1
2 , uk+

1
2 )i]

= R
k+ 1

2
i , 1 ≤ i ≤M − 1, 0 ≤ k ≤ N − 1,

(5.1)

fki = δ2xe
k
i +Qk

i , 1 ≤ i ≤M − 1, 0 ≤ k ≤ N, (5.2)

e0i = 0, 1 ≤ i ≤M − 1, (5.3)

ek0 = 0, ekM = 0, 0 ≤ k ≤ N, (5.4)

fk0 = 0, fkM = 0, 0 ≤ k ≤ N. (5.5)

It follows from (5.2) that

f
k+ 1

2
i = δ2xe

k+ 1
2

i +Q
k+ 1

2
i , 1 ≤ i ≤M − 1, 0 ≤ k ≤ N − 1.

Taking the inner product of (5.1) with δte
k+ 1

2
i yields

∥δtek+
1
2 ∥2 + (δtδ

2
xf

k+ 1
2 , δte

k+ 1
2 )− (fk+

1
2 , δte

k+ 1
2 ) + (∆xe

k+ 1
2 , δte

k+ 1
2 )

+ (ψ(Uk+ 1
2 , Uk+ 1

2 )− ψ(uk+
1
2 , uk+

1
2 ), δte

k+ 1
2 ) = (Rk+ 1

2 , δte
k+ 1

2 ),

for 1 ≤ k ≤ N − 1.
From Lemma 2.1, it follows that

−(δtδ
2
xf

k+ 1
2 , δte

k+ 1
2 ) = −(δtf

k+ 1
2 , δtδ

2
xe

k+ 1
2 ) = −(δtf

k+ 1
2 , δt(f

k+ 1
2 −Qk+ 1

2 ))

= −(δtf
k+ 1

2 , δtf
k+ 1

2 ) + (δtf
k+ 1

2 , δtQ
k+ 1

2 )

≤ −∥δtfk+
1
2 ∥2 + ∥δtfk+

1
2 ∥ t∥δtQk+ 1

2 ∥,

(fk+
1
2 , δte

k+ 1
2 ) = (δ2xe

k+ 1
2 +Qk+ 1

2 , δte
k+ 1

2 )

≤ − 1

2τ
(|ek+1|21 − |ek|21) + ∥Qk+ 1

2 ∥ ∥δtek+
1
2 ∥,

and

−(∆xe
k+ 1

2 , δte
k+ 1

2 ) ≤ ∥∆xe
k+ 1

2 ∥ ∥δtek+
1
2 ∥,

From the definition of ψ(u, v)i and applying Lemma 2.2, we obtain

− (ψ(Uk+ 1
2 , Uk+ 1

2 )− ψ(uk+
1
2 , uk+

1
2 ), δte

k+ 1
2 )

= −(ψ(Uk+ 1
2 , ek+

1
2 ) + ψ(ek+

1
2 , Uk+ 1

2 )− ψ(ek+
1
2 , ek+

1
2 ), δte

k+ 1
2 ).

Note that

(ψ(ek+
1
2 , ek+

1
2 ), δte

k+ 1
2 ) = (ψ(ek+

1
2 , ek +

τ

2
δte

k+ 1
2 ), δte

k+ 1
2 )

= (ψ(ek+
1
2 , ek), δte

k+ 1
2 ) +

τ

2
(ψ(ek+

1
2 , δte

k+ 1
2 ), δte

k+ 1
2 )

= (ψ(ek+
1
2 , ek), δte

k+ 1
2 ).

Thus we obtain

− (ψ(Uk+ 1
2 , Uk+ 1

2 )− ψ(uk+
1
2 , uk+

1
2 ), δte

k+ 1
2 )

= −(ψ(Uk+ 1
2 , ek+

1
2 ) + ψ(ek+

1
2 , Uk+ 1

2 )− ψ(ek+
1
2 , ek), δte

k+ 1
2 )



EJDE-2024/38 A DIFFERENCE SCHEME FOR ROSENAU-BURGERS EQUATION 9

= −h
3

M−1∑
i=1

[U
k+ 1

2
i ∆xe

k+ 1
2

i + 2∆x(U
k+ 1

2 ek+
1
2 )i + e

k+ 1
2

i ∆xU
k+ 1

2
i ]δte

k+ 1
2

i

+
h

3

M−1∑
i=1

[e
k+ 1

2
i ∆xe

k
i +∆x(e

k+ 1
2 ek)i]δte

k+ 1
2

i

= −h
3

M−1∑
i=1

(3U
k+ 1

2
i ∆xe

k+ 1
2

i + e
k+ 1

2
i+1 δxU

k+ 1
2

i+ 1
2

+ e
k+ 1

2
i−1 δxU

k+ 1
2

i− 1
2

+ e
k+ 1

2
i ∆xU

k+ 1
2

i )δte
k+ 1

2
i

+
h

3

M−1∑
i=1

(2e
k+ 1

2
i ∆xe

k
i +

1

2
eki+1δxe

k+ 1
2

i+ 1
2

+
1

2
eki−1δxe

k+ 1
2

i− 1
2

)δte
k+ 1

2
i

≤ c0|ek+
1
2 |1 · ∥δtek+

1
2 ∥+ c0∥ek+

1
2 ∥ · ∥δtek+

1
2 ∥

+
1

3
(2∥ek+ 1

2 ∥∞ · |ek|1 + ∥ek∥∞ · |ek+ 1
2 |1)∥δtek+

1
2 ∥

≤ c0|ek+
1
2 |1 · ∥δtek+

1
2 ∥+ c0∥ek+

1
2 ∥ · ∥δtek+

1
2 ∥+

√
L

2
|ek+ 1

2 |1 · ∥δtek+
1
2 ∥

≤ ε1∥δtek+
1
2 ∥2 + c0

2

4ε1
|ek+ 1

2 |21 + ε2∥δtek+
1
2 ∥2 + c0

2

4ε2
∥ek+ 1

2 ∥21

+ ε3∥δtek+
1
2 ∥2 + 1

4ε3

L

4
|ek+ 1

2 |21

≤ (ε1 + ε2 + ε2)∥δtek+
1
2 ∥2 + (

c0
2

4ε1
+
c0

2

4ε2

L2

6
+

L

16ε3
)|ek+ 1

2 |21.

Let ε1 = ε2 = ε3 = 1
12 . Then

−(ψ(Uk+ 1
2 , Uk+ 1

2 )− ψ(uk+
1
2 , uk+

1
2 ), δte

k+ 1
2 ) ≤ 1

4
∥δtek+

1
2 ∥2 + c4|ek+

1
2 |21.

Substituting above equations into equation (5.1) gives

∥δtek+
1
2 ∥2

≤ −∥δtfk+
1
2 ∥2 + ∥δtfk+

1
2 ∥ ∥δtQk+ 1

2 ∥ − 1

2τ
(|ek+1|21 − |ek|21)

+ ∥Qk+ 1
2 ∥ · ∥δtek+

1
2 ∥+ ∥∆xe

k+ 1
2 ∥ · ∥δtek+

1
2 ∥+ 1

4
∥δtek+

1
2 ∥2 + c4|ek+

1
2 |21

≤ 1

4
∥δtQk+ 1

2 ∥ − 1

2τ
(|ek+1|21 − |ek|21) + ∥Qk+ 1

2 ∥2 + 1

4
∥δtek+

1
2 ∥2 + |ek+ 1

2 |21

+
1

4
∥δtek+

1
2 ∥2 + 1

4
∥δtek+

1
2 ∥2 + c4|k+

1
2 |21 + ∥Rk+ 1

2 ∥2 + 1

4
∥δtek+

1
2 ∥2.

(5.6)

From (3.10), simplifying and rearranging (5.6) leads to

|ek+1|12 ≤ |ek|12 + 2τ
c22
4
(τ2 + h2)2 + 2τc22(h

2)2 + 2τc1
2(τ2 + h2)2

+ τ(c4 + 1)|ek+1|21 + τ(c4 + 1)|ek|21.
Then

(1− τM)|ek+1|12 ≤ (1 + τM)|ek|12 + 2τM1(τ
2 + h2)2, 0 ≤ k ≤ N − 1.

If τM ≤ 1
3 , then

|ek+1|12 ≤ (1 + 3τM)|ek|12 + 6τM1(τ
2 + h2)2, 0 ≤ k ≤ N − 1.



10 S. DONG, X. ZHANG, Y. JIN EJDE-2024/38

Using Gronwall’s inequality we obtain

|ek|21 ≤ e3Mkτ6τM1(τ
2 + h2)2 ≤ e3MT 6τM1(τ

2 + h2)2, 0 ≤ k ≤ N. □

From Theorem 5.1 and Lemma 2.1, we find that

∥ek∥∞ ≤
√
L

2
|ek|1 ≤ c5

√
L

2
(τ2 + h2), 0 ≤ k ≤ N.

To demonstrate the results in the previous sections and confirm the accuracy of
our method, we consider the Rosenau-Burgers equation with the following initial-
boundary value problem.

ut + uxxxxt − uxx + ux + uux = 0, x ∈ [0, 1], t ∈ [0, 1], (5.7)

u(x, 0) = u0(x) = x4(1− x)4, x ∈ [0, 1], (5.8)

u(0, t) = u(1, t) = 0, uxx(0, t) = uxx(1, t) = 0. (5.9)

Since we do not know the exact solution of (5.7)-(5.9), to obtain the error esti-
mate, we consider the solution on mesh h = 1/256 as the reference solution, which
is fine enough as a referenced exact solution for obtaining the error estimation.

When the exact solution is known, we define the discrete error in the L∞-norm
as follows

E∞(h, τ) = max
1≤i≤M,0≤k≤N

|Uk
i − uki |,

where Uk
i and uki represent the exact solution and the numerical solution, respec-

tively. Furthermore, denote the spatial and temporal convergence orders, respec-
tively, as

Orderh∞ = log2
E∞(2h, τ)

E∞(h, τ)
, Orderτ∞ = log2

E∞(h, 2τ)

E∞(h, τ)
.

Figure 1 shows the three dimensional image of numerical solutions. From Figure
2, we can know that with the time going by, the waveform changes and the peak
value decreases. It shows that our scheme is stable.

Table 1 shows that the numerical results are conservative at different t where Ek

is defined in Theorem 3.1. Table 2 and Table 3 list the errors and corresponding
convergence orders. It shows that the maximal errors reduce with the decrease of
the spatial step h and time step τ . The convergence orders are both two in space
and time in L∞-norm, which are consistent with our theoretical results. It reveals
that the numerical method in this paper is accurate and efficient.

Table 1. Discrete energy Ek at h=1/256 and τ = 0.01.

t Ek

t=0.1 0.004792856601

t=0.3 0.004792856601

t=0.5 0.004792856601

t=0.7 0.004792856601

t=0.9 0.004792856601

Acknowledgments. This work was supported by the Natural Science Foundation
of Jilin Province (No. YDZJ202101ZYTS136).



EJDE-2024/38 A DIFFERENCE SCHEME FOR ROSENAU-BURGERS EQUATION 11

Figure 1. Three dimensional image of numerical solutions at
h=1/256 and τ=0.1.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

4
10

-3

t=0

t=0.5

t=1

Figure 2. Numerical solutions with h=0.01 and τ=0.1 at various times.

Table 2. Errors and temporal convergence orders when t = 1.0.

h τ E∞(h, τ) Orderτ∞
h=1/128 τ = 1/8 3.3233e− 09 –

h=1/128 τ = 1/16 8.2855e− 10 2.0040

h=1/128 τ = 1/32 2.0475e− 10 2.0168

h=1/128 τ = 1/64 4.9024e− 11 2.0623

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34 (1986), 42.



12 S. DONG, X. ZHANG, Y. JIN EJDE-2024/38

Table 3. Errors and spatial convergence orders when t = 1.0.

h τ E∞(h, τ) Orderh∞
h = 1/8 τ=0.1 4.7197e− 06 –

h = 1/16 τ=0.1 1.1560e− 06 2.0295

h = 1/32 τ=0.1 2.8428e− 07 2.0238

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Sitong Dong
Department of Mathematics, Yanbian University, Yanji 133002, China

Email address: 394018165@qq.com

Xin Zhang

Department of Mathematics Science, Harbin Engineering University, Harbin 150000,

China
Email address: zhangxin2022@hrbeu.edu.cn

Yuanfeng Jin
Department of Mathematics, Yanbian University, Yanji 133002, China

Email address: yfkim@ybu.edu.cn


	1. Introduction
	2. Preliminaries
	3. Nnonlinear conservative difference scheme
	3.1. Construction of difference scheme
	3.2. Conservation law

	4. Solvability
	5. Convergence
	Acknowledgments

	References