Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 30, pp. 1–30.

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

DOI: 10.58997/ejde.2025.30

LONG-TIME BEHAVIOR OF SOLUTIONS TO THE 2D

MAGNETIC BÉNARD PROBLEM IN POROUS MEDIA ON

UNBOUNDED DOMAINS

DANG THANH SON

Abstract. In this article, we study the long time behavior of solutions to the
2D magnetic Bénard problem in porous media, considering on an arbitrary

(bounded or unbounded) domain satisfying Poincaré inequality. We first prove

the existence of a weak solution and a global attractor for the problem. For
r = 1, 2, 3, we derive estimates for Hausdorff as well as fractal dimensions

of the global attractors. We then show an upper semicontinuity of global

attractors and final study the exponential stability of a stationary solution to
the problem.

1. Introduction

The study of fluid flow through porous media frequently employs Darcy’s Law to
model the momentum balance, which establishes a linear relationship between the
flow rate and the pressure drop within the medium. This relationship is described
as − k

µ∇p = u, where k is the permeability, µ is the dynamic viscosity, and p is

the pressure. However, in scenarios involving high velocities or non-Newtonian flu-
ids, deviations from this linearity occur, necessitating more comprehensive models
such as the Darcy-Forchheimer law. This law introduces a nonlinear correction to
account for increased pressure drops at higher velocities and leads to the so-called
Brinkman-Forchheimer-extended-Darcy equations (a generalisation actually) read

∂tu− ν∆u+ (u · ∇)u+ α|u|2qu+ β|u|2ru+∇p = f,

∇ · u = 0.

This system has been extensively studied, with investigations on its long-time be-
havior, the existence of global attractors, and various qualitative properties under
different boundary conditions and parameter settings (see, e.g., [5, 27, 33, 34]).

In the context of magnetohydrodynamic (MHD) flows with thermal effects, we
extend this framework to study the following problem on an arbitrary (bounded or

2020 Mathematics Subject Classification. 35B40, 35B41, 35Q35, 37L30, 76W05.
Key words and phrases. Magnetic Bénard problem; global attractor; porus medium

fractal and Hausdroff dimensions; upper semicontinuity; stationary solution.
©2025. This work is licensed under a CC BY 4.0 license.

Submitted December 2, 2024. Published March 20, 2025.

1



2 D. T. SON EJDE-2025/30

unbounded) domain Ω ⊂ R2 with boundary ∂Ω:

∂tu−R−1
e ∆u+ (u · ∇)u− S(B · ∇)B + αu+ β|u|r−1u

+∇
(
p+

S

2
|B|2

)
= θe2 + f, in Ω, t > 0,

∂tB +R−1
m ∇⊥(curlB) + (u · ∇)B − (B · ∇)u = Ψ, in Ω, t > 0,

∂tθ − κ∆θ + (u · ∇)θ = u2 + h, in Ω, t > 0,

∇ · u = ∇ ·B = 0, in Ω, t > 0,

u = 0, B · n = 0, curlB = 0, θ = 0 on ∂Ω, t > 0,

u(x, 0) = u0(x), B(x, 0) = B0(x), θ(x, 0) = θ0(x) in Ω.

(1.1)

In this system, the unknowns are the fluid particle velocity u = (u1, u2), the
magnetic field B = (B1, B2), the temperature θ (or the density in the modeling of
geophysical fluids), and the fluid pressure p = p(x, t). The pressure is the standard
pressure and can be obtained by applying the divergence-free condition for veloc-
ity, taking the divergence, and then inverting the Laplacian operator. The term
|B|2/2 represents the magnetic pressure, e2 = (0, 1) denoting the unit vector in
the direction of gravity. The unit outward normal on ∂Ω is represented by n, and
the parameters Re and Rm denote the Reynolds and magnetic Reynolds numbers,
respectively. The parameter S = M2/(ReRm) incorporates the Hartman number
M , while κ > 0 signifies the heat conductivity coefficient. Additionally, f,Ψ and h
represent external time-dependent forces, respectively, while the forcing term θe2
describes the buoyancy force’s action on fluid motion. And

curlu =
∂u2
∂x1

− ∂u1
∂x2

, for every vector function u,

∇⊥ϕ =
( ∂ϕ
∂x2

,− ∂ϕ

∂x1

)
, for every scalar function ϕ,

∇⊥(curlu) = ∇(∇ · u)−∆u.

In mathematics, the system (1.1) can be considered a modification (by an absorp-
tion term αu+ β|u|r−1u) of the classical Bénard problem combine with Maxwell’s
equations of electromagnetism. Herein, the positive constants α and β hold sig-
nificance as the Darcy and Forchheimer coefficients, respectively, representing the
permeability of porous medium and its porosity-related proportionality. The pa-
rameter r, constituting the absorption exponent, is in [1,∞). It models the convec-
tion of an incompressible flow, which occurs in a horizontal layer of conductive fluid
heated from below, with the presence of a magnetic field. Moreover, the model (1.1)
is recognized to be more accurate when the flow velocity is too large for the Darcy’s
law to be valid alone, and in addition, the porosity is not too small. The nonlin-
earity of the form |u|r−1u can be found in tidal dynamics as well as non-Newtonian
fluid flows (see [4] and the references therein). In instances where thermal effects
on the fluid can be negligible (i.e. θ ≡ 0) and α = β = 0, then (1.1) reduces to
the Magnetohydrodynamics (MHD) equations, which govern the dynamics of the
velocity and the magnetic field in electrically conducting fluids such as plasmas
(as expounded in [17]) and reflect the basic physics conservation laws. In the past
years, the existence and long-time behavior of solutions to the MHD have attracted
the attention of many mathematicians. There are many results on the existence
of solutions and existence of attractors for MHD, see e.g. [3, 13, 18, 37, 42] and



EJDE-2025/30 GLOBAL ATTRACTOR FOR 2D MAGNETIC BÉNARD PROBLEM 3

references therein. Furthermore, the regularity of solutions has been studied ex-
tensively in recent years (see e.g. [12, 21, 25, 26, 28]). We also refer the interested
reader to [39] for recent results on time optimal control problem associated with the
two-dimensional MHD equations with memory. In case θ ≡ 0, α = 0 and 4 ≤ r < 5,
β > 0, the authors in [31] have recently proved the existence of global attractor for
3D MHD with damping.

On the other hand, in case where the fluid remains unaffected by the magnetic
field (i.e. B ≡ 0) and α = β = 0, system (1.1) transforms into the Bénard problem.
Numerous studies have been dedicated to exploring the global well-posedness and
the existence of attractors for the Bénard problem, as evidenced by works such as
[2, 11, 20, 24, 32, 41] and related references.

Turning attention to the magnetic Bénard problem without velocity damping
(i.e. α = β = 0), significant attention has been focused on the 2D case. Bian et
al. established the global well-posedness of weak or strong solutions for the initial
boundary value problems under various boundary conditions and providing a com-
prehensive analysis of stability and instability within a fully nonlinear, dynamical
setting from a mathematical point of view as stated in the references [7, 8, 9, 43].
Furthermore, a comprehensive analysis of the long-time behavior of solutions to
an optimal control problem for the magnetic Bénard problem in a two-dimensional
bounded domain, with a focus on distributed control adjustments, is presented in
[38].

The study of long-time behavior of nonlinear dynamical system is an interesting
branch of applied mathematics and it is essential in understanding many natural
phenomena. For an extensive study on infinite dimensional dynamical systems in
mathematical physics, we refer to [14, 40]. And as is well known, a useful way for
studying the long-time behavior of solutions is to use the theory of attractors. The
classical global attractor for autonomous dynamical systems is an invariant compact
set which attracts all bounded sets and contains some important information about
the long-time behavior of the solutions.

The aim of this article is to continue the study of the long-time behavior of weak
solution to problem (1.1) in some domains not necessarily bounded. More precisely,
the domain Ω can be an arbitrary bounded or unbounded open set in R2 without
any regularity assumption on its boundary ∂Ω and with the assumption that the
Poincaré inequality holds on it, i.e., there exists λ1 > 0 such that∫

Ω

|ϕ(x)|2dx ≤ 1

λ1

∫
Ω

|∇ϕ(x)|2dx for all ϕ ∈ H1
0 (Ω). (1.2)

We also require that the domain Ω satisfy the cone condition so that Lemma 2.2 in
Section 2 is valid on Ω (see [1, Chapter 5] for details). We will discuss the existence
and long-time behavior of solutions in terms of the existence of a global attractor
and the stability of stationary solutions. Here, the existence and uniqueness of
solutions are studied by Galerkin approximation method. To prove the existence
of a global attractor, the usual approach is to obtain a bounded absorbing set in
a more regular space and then use the compactness of the Sobolev embeddings.
However, because the domain considered may be unbounded, the Sobolev embed-
ding is no longer compact, and therefore the usual method in bounded domains no
longer works. To overcome this difficulty, we exploit the energy equation method
introduced by Ball in [6]. Next, following the general lines of the approach in [36],
we show that the solution map S(t) : H → H is Fréchet differentiable with respect



4 D. T. SON EJDE-2025/30

to the initial data for the absorption exponent r = 1, 2, 3 and hence we obtain
estimates for the Hausdorff and Fractal dimensions of such attractors under an
additional condition on the domain. Then, we establish an upper semicontinuity
of global attractors for the problem (1.1). We take an expanding sequence of sim-
ply connected, bounded and smooth subdomains {Ωm}m ⊂ Ω. If Am and A are
the global attractors of (1.1) corresponding to Ωm and Ω, respectively, then we
show that for large enough m, the global attractor Am enters into any neighbor-
hood U(A) of A. Finally, the existence of a stationary solution is established by a
corollary of the Brouwer fixed point theorem.

This article is organized as follows. In Section 2, for the convenience of the read-
ers, we recall and prove some auxiliary results on the 2D magnetic Bénard problem
in porous media. In the next Section, we discuss the existence and uniqueness of
a weak solution. In Section 4, for (f,Ψ, h) ∈ V ′, we show that problem (1.1) pos-
sesses a global attractor by using the energy equation method. For the absorption
exponent r = 1, 2, 3, the estimates for Hausdorff as well as Fractal dimensions of
the global attractor for the problem (1.1) is obtained in Section 5. In Section 6,
we prove an upper semicontinuity of global attractors for the problem (1.1). The
existence and exponential stability of a stationary solution is shown in the last
Section.

2. Preliminaries

We first recall several function spaces necessary to write system (1.1). We denote

Lp(Ω) = Lp(Ω)2, H1(Ω) = H1(Ω)2, H1
0(Ω) = H1

0 (Ω)
2.

The spaces used in the theory of the magnetic Bénard problem are a combination
of spaces used for the Bénard problem and spaces used in the theory of Maxwell
equations. They are

V1 = {u ∈ C∞
0 (Ω)2 : ∇ · u = 0},

V1 = closure of V1 in the H1
0(Ω) norm = {u ∈ H1

0(Ω),∇ · u = 0},
H1 = closure of V1 in the L2(Ω) norm = {u ∈ L2(Ω),∇ · u = 0 and u · n|∂Ω = 0},
V2 = {B ∈ C∞(Ω̄)2 : ∇ ·B = 0 and B · n|∂Ω = 0},
V2 = closure of V2 in the H1(Ω) norm = {B ∈ H1(Ω),∇ ·B = 0 and B · n|∂Ω = 0},
H2 = closure of V2 in the L2(Ω) norm = H1,

V3 = H1
0 (Ω), H3 = L2(Ω),

V = V1 × V2 × V3, H = H1 ×H2 ×H3,

L̃p = closure of V1 in the Lp(Ω) norm = {u ∈ Lp(Ω),∇ · u = 0 and u · n|∂Ω = 0}.

For i = 1, 2, we define

(u, v)i :=

∫
Ω

2∑
j=1

ujvjdx, ∀u, v ∈ Hi,

(θ, η)3 :=

∫
Ω

θηdx, ∀θ, η ∈ H3,

and the associated norms | · |2j = (·, ·)j , j = 1, 3.



EJDE-2025/30 GLOBAL ATTRACTOR FOR 2D MAGNETIC BÉNARD PROBLEM 5

The inner product and norm in V1 are

((u, ũ))1 =

2∑
i=1

∫
Ω

∇ui · ∇ũi.dx, ∀u, ũ ∈ V1,

∥u∥1 = ((u, u))
1/2
1 , ∀u ∈ V1,

which is a norm in H1
0(Ω), thanks to Poincaré inequality. The inner product and

norm in V2 are

((B, B̃))2 =

∫
Ω

curlB · curl B̃dx, ∀B, B̃ ∈ V2,

∥B∥2 = ((B,B))
1/2
2 , ∀B ∈ V2.

Since the domain Ω is simply-connected, the above bilinear form is actually a scalar
product on V2, it defines a norm which is equivalent to that induced by H1(Ω) on
V2 (see [18]). The inner product and norm in V3 are

((θ, θ̃))3 =

∫
Ω

∇θ · ∇θ̃.dx, ∀θ, θ̃ ∈ V3,

∥θ∥3 = ((θ, θ))
1/2
3 , ∀θ ∈ V3.

Using the same notation again for simplicity, we define the inner product and norm
in V by

((z, z̃)) = ((u, ũ))1 + S.((B, B̃))2 + γ((θ, θ̃))3, ∀z = (u,B, θ), z̃ = (ũ, B̃, θ̃) ∈ V,

∥z∥ = ((z, z))1/2, ∀z ∈ V.

The inner products and norms in H1, H2 and H3 are the usual ones inherited from
L2(Ω) and L2(Ω), respectively. We define the inner product and norm in H by

(z, z̃) = (u, ũ) + S.(B, B̃) + γ.(θ, θ̃), ∀z = (u,B, θ), z̃ = (ũ, B̃, θ̃) ∈ H,

|z| = (z, z)1/2, ∀z ∈ H.

We define γ > 0 so that

γ ≥ 4Re

λ21κ
. (2.1)

This constant γ is chosen so that an operator to be defined later (related to the
linear part of the system of equations) is coercive under the norm defined. Moreover,
since S and γ are positive, the inner products and norms defined above for H and
V are equivalent to the usual ones defined on these product spaces.

It follows from (1.2) and the equivalence of norms in H1 and V2 that there exists
c0 > 0 such that

λ1|u|2 ≤ ∥u∥21, c0|B|2 ≤ ∥B∥22, λ1|θ|2 ≤ ∥θ∥23 (2.2)

for all u ∈ V1, B ∈ V2 and θ ∈ V3. Furthermore, by applying the Riesz representa-
tion theorem, we can identify the dual space H ′ with H and obtain the following
relation: V ⊂ H ≡ H ′ ⊂ V ′, where the injections are continuous and each space is
dense in the following ones. We also use ⟨·, ·⟩ to denote the induced duality between

the spaces V and its dual V ′ = V ′
1 × V ′

2 × V ′
3 as well as L̃p and its dual L̃p′

, where
1
p + 1

p′ = 1.



6 D. T. SON EJDE-2025/30

Let us define the bilinear forms ai : Vi×Vi → R, for i = 1, 2, 3 and a : V ×V → R
by

a1(u, ũ) = ((u, ũ))1 =

∫
Ω

2∑
i=1

∇ui · ∇ũi dx,

a2(B, B̃) = ((B, B̃))2 =

∫
Ω

curlB · curl B̃ dx,

a3(θ, θ̃) = ((θ, θ̃))3 =

∫
Ω

∇θ · ∇θ̃ dx,

a(z, z̃) = R−1
e a1(u, ũ) +R−1

m Sa2(B, B̃) + κγa3(θ, θ̃).

The bilinear form a is coercive since

min
(
R−1

e , R−1
m , κ

)
∥z∥2 ≤ a(z, z) ≤ max

(
R−1

e , R−1
m , κ

)
∥z∥2. (2.3)

We define σ : V × V → R by

σ(z, z̃) = −
∫
Ω

(θe2 · ũ+ γu2θ̃)dx.

By using the Hölder and Poincaré inequalities, we deduce that

|σ(z, z̃)| ≤ cσ∥z∥∥z̃∥. (2.4)

To obtain the coercivity of bilinear form σ, we assume that for all u ∈ V1 and
θ ∈ V3,

|(u2, θ)| ≤ ϵ∥u∥1∥θ∥3, (2.5)

where ϵ is a positive constant such that

ϵ ≤
(R−1

e κ

4γ

)1/2

≤ λ1R
−1
e κ

4
. (from (2.1)) (2.6)

Lemma 2.1. For all z ∈ V , we have the estimate

a(z, z) + σ(z, z) ≥ δ

2
∥z∥2

where δ := min(R−1
e , R−1

m , κ).

Proof. Using Young’s inequality in (2.5), we have

γ|(u2, θ)| ≤ γϵ∥u∥1∥θ∥3 ≤ γϵ2

κ
∥u∥21 +

γκ

4
∥θ∥23.

From (2.6), we obtain

γ|(u2, θ)| ≤
R−1

e ∥u∥21 + γκ∥θ∥23
4

. (2.7)

From the definition of γ in (2.1), by using (2.2) and the Hölder inequality, we obtain

|(θe2, u)| ≤ |θ||u| ≤ 1

λ1
∥u∥1∥θ∥3

≤ 2−1(Re)
−1/2∥u∥1(γκ)1/2∥θ∥3

≤ R−1
e ∥u∥21 + γκ∥θ∥23

4
.



EJDE-2025/30 GLOBAL ATTRACTOR FOR 2D MAGNETIC BÉNARD PROBLEM 7

From (2.7) and the inequality above, we deduce that

|σ(z, z)| = |(θe2, u)|+ γ|(u2, θ)| ≤
R−1

e ∥u∥21 + γκ∥θ∥23
2

.

Using the definition of a and the inequality above, we obtain

a(z, z) + σ(z, z) ≥ a(z, z)− |σ(z, z)| ≥ R−1
e

2
∥u∥21 +R−1

m S∥B∥2 + κ

2
γ∥θ∥23

≥ 1

2
min(R−1

e , R−1
m , κ)∥z∥2.

The proof is complete. □

We next define the trilinear forms

b(u, v, w) =

2∑
i,j=1

∫
Ω

ui
∂vj
∂xi

wj dx, b̄(u, θ, η) =

2∑
i=1

∫
Ω

ui
∂θ

∂xi
η dx,

whenever the integrals make sense, and B : V × V × V → R by

B(z1, z2, z3) = b(u1, u2, u3) + γb̄(u1, θ2, θ3)− Sb(B1, B2, u3)

+ Sb(u1, B2, B3)− Sb(B1, u2, B3), ∀zi = (ui, Bi, θi) ∈ V.

It is easy to check that if u, v, w ∈ Vi, i = 1, 2 and θ, η ∈ V3, then

b(u, v, w) = −b(u,w, v) and b̄(u, θ, η) = −b̄(u, η, θ). (2.8)

Hence

b(u, v, v) = 0 and b̄(u, θ, θ) = 0. (2.9)

From the relation of B(z1, z2, z2) and from (2.8), (2.9), we obtain

B(z1, z2, z2) = 0, ∀z1, z2 ∈ V,

B(z1, z2, z3) = −B(z1, z3, z2), ∀z1, z2, z3 ∈ V.
(2.10)

The following results is well-known.

Lemma 2.2 ([1, Theorem 5.8]). Let Ω ⊂ R2 satisfying the cone condition. Then
there exists the constant K depending on the dimensions of the cone C providing
the cone condition for Ω such that for all ϕ ∈ H1(Ω),

∥ϕ∥L4(Ω) ≤ K∥ϕ∥1/2L2(Ω)∥ϕ∥
1/2
H1(Ω).

By using similar arguments as in [3, Lemma 2.3] for MHD equations, we have
the following result.

Lemma 2.3. For any open set Ω ⊂ R2 satisfying the cone condition and z, z̃ ∈ V ,
we have

|B(z, z, z̃)| ≤ cb|z|∥z∥∥z̃∥.

Next, for r > 0 and for all u, v ∈ L̃r+1, we define Cr(u, v) = |u|r−1v and Cr(u) =
Cr(u, u). We have the following crucial properties of the nonlinearity Cr.

Lemma 2.4. For every r ≥ 1 and for all functions u, v ∈ L̃r+1

⟨Cr(u)− Cr(v), u− v⟩ ≥ 0. (2.11)



8 D. T. SON EJDE-2025/30

Proof. We have

⟨Cr(u)− Cr(v), u− v⟩
= ⟨|u|r−1u− |v|r−1v, u− v⟩
= ⟨|u|r−1, |u− v|2⟩+ ⟨|v|r−1, |u− v|2⟩+ ⟨v|u|r−1 − u|v|r−1, u− v⟩

= ∥|u|
r−1
2 (u− v)∥2

L̃2 + ∥|v|
r−1
2 (u− v)∥2

L̃2

+ ⟨u · v, |u|r−1 + |v|r−1⟩ − ⟨|u|2, |v|r−1⟩ − ⟨|v|2, |u|r−1⟩.

(2.12)

Moreover, we also have

⟨u · v, |u|r−1 + |v|r−1⟩ − ⟨|u|2, |v|r−1⟩ − ⟨|v|2, |u|r−1⟩

=
1

2
⟨|u|r−1 − |v|r−1, |u|2 − |v|2⟩ − 1

2
∥|u|

r−1
2 (u− v)∥2

L̃2 −
1

2
∥|v|

r−1
2 (u− v)∥2

L̃2

≥ −1

2
∥|u|

r−1
2 (u− v)∥2

L̃2 −
1

2
∥|v|

r−1
2 (u− v)∥2

L̃2 ,

since ⟨|u|r−1 − |v|r−1, |u|2 − |v|2⟩ ≥ 0 for all u, v ∈ L̃r−1. Applying the inequality
above to (2.12) we deduce

⟨Cr(u)− Cr(v), u− v⟩ ≥ 1

2
∥|u|

r−1
2 (u− v)∥2

L̃2 +
1

2
∥|v|

r−1
2 (u− v)∥2

L̃2 .

By using Hölder’s inequality, from u, v ∈ L̃r+1 we obtain

|u|
r−1
2 (u− v) and |v|

r−1
2 (u− v) are bounded in L̃2

and it yields (2.11). □

Furthermore, for all u ∈ L̃r+1, Cr(u) is Gateaux differentiable with Gateaux
derivative

C′
r(u)v =


v, for r = 1,

|u|v + u
|u| (u · v), if u ̸= 0, for r = 2,

0, if u = 0, for r = 2,

|u|r−1v + (r − 1)
(
u|u|r−3(u · v)

)
, for r ≥ 3,

(2.13)

for v ∈ L̃r+1. For u, v ∈ L̃r+1, it can be easily seen that

⟨C′
r(u)v, v⟩ =

∫
Ω

|u(x)|r−1|v(x)|2dx+(r−1)

∫
Ω

|u(x)|r−3|u(x) ·v(x)|2dx ≥ 0 (2.14)

for r ≥ 1. Note that for r = 2 also the same result holds, since in that case, the
second integral ∫

x∈Ω:u(x) ̸=0

1

|u(x)|
|u(x) · v(x)|2dx ≥ 0.



EJDE-2025/30 GLOBAL ATTRACTOR FOR 2D MAGNETIC BÉNARD PROBLEM 9

Moreover, for r ≥ 3, Cr(u) is twice Gateaux differentiable with second order
Gateaux derivative

C′′(u)(v ⊗ w)

=



(r − 1)|u|r−3[(u · w)v + (u · v)w + (w · v)u]
+(r − 1)(r − 3) u

|u|5−r (u · v)(u · w), for u ̸= 0, 3 ≤ r < 5

(r − 1)(r − 3) u
|u|5−r (u · v)(u · w), for u ̸= 0, 3 ≤ r < 5,

(r − 1)|u|r−3[(u · w)v + (u · v)w + (w · v)u]
+(r − 1)(r − 3)|u|r−5(u · v)(u · w)u, for r ≥ 5,

(2.15)

for all u, v, w ∈ L̃r+1.
We now define g : V → R by

g(z) = ⟨f, u⟩V ′
1 ,V1

+ S⟨Ψ, B⟩V ′
2 ,V2

+ γ⟨h, θ⟩V ′
3 ,V3

= ⟨Φ, z⟩V ′,V ,

where Φ = (f,Ψ, h). By using Schwarz’s inequality, we have

|g(z)| ≤ ∥Φ∥V ′∥z∥
with ∥Φ∥2V ′ = ∥f∥2V ′

1
+ S∥Ψ∥2V ′

2
+ γ∥h∥2V ′

3
. Finally, to prove the existence of a

stationary solution, we need the following lemma.

Lemma 2.5 ([10]). Let X be a finite dimensional Hilbert space with scalar product
[·, ·] and norm [·] and let P be a continuous mapping from X into itself such that

[P (ξ), ξ] > 0 for [ξ] = k > 0.

Then there exists ξ ∈ X, [ξ] < k, such that P (ξ) = 0.

3. Existence and uniqueness of a weak solution

Taking the inner product of the first equation of (1.1) with ũ ∈ V1, we obtain

(∂tu, ũ) +R−1
e ((u, ũ))1 + b(u, u, ũ)− Sb(B,B, ũ) + α(u, ũ) + β⟨Cr(u), ũ⟩ − (θe2, ũ)

= ⟨f, ũ⟩V ′
1 ,V1

.

We take the inner product of the second equation of (1.1) with SB̃ (B̃ ∈ V2) to
obtain

S(∂tB, B̃) + SR−1
m ((B, B̃))2 + Sb(u,B, B̃)− Sb(B, u, B̃) = S⟨Ψ, B̃⟩V ′

2 ,V2
.

And taking the inner product of the third equation of (1.1) with γθ̃ ∈ V3, we obtain

γ(∂tθ, θ̃) + γκ((θ, θ̃))3 + γb̄(u, θ, θ̃)− γ(u2, θ̃) = γ⟨h, θ̃⟩V ′
3 ,V3

.

This suggests the following weak formulation of problem (1.1).

Problem For z0 = (u0, B0, θ0) ∈ H given, find weak solution z = (u,B, θ) such
that

z ∈ L2(0, T ;V ) ∩ L∞(0, T ;H), u ∈ Lr+1(0, T ; L̃r+1), ∂tz ∈ L
r+1
r (0, T ;V ′),

∀T > 0,

(∂tz(t), z̃) + a(z(t), z̃) + σ(z(t), z̃) + B(z(t), z(t), z̃)

+ α(u, ũ) + β⟨Cr(u), ũ⟩ = ⟨Φ, z̃⟩, ∀z̃ = (ũ, B̃, θ̃) ∈ V, a.e. t ∈ (0, T ),

z(0) = z0 in H.

(3.1)



10 D. T. SON EJDE-2025/30

We have the result of the existence of a weak solution to Problem (3.1).

Theorem 3.1. For z0 = (u0, B0, θ0) ∈ H, T > 0 and (f,Ψ, h) ∈ L2(0, T ;V ′),
there exists a unique weak solution z = (u,B, θ) to the system (1.1) in the sense of
(3.1).

Proof. The existence of a weak solution to Problem (3.1) in (0, T ) is based on
Galerkin approximations, a priori estimates, and the compactness method. As it
is standard, we only provide some basic a priori estimates that we shall frequently
use later.

From the first equation in (3.1), by using Lemma 2.1 and (2.10) to obtain

1

2

d

dt
|z(t)|2 + δ

2
∥z(t)∥2 + α|u(t)|2 + β∥u(t)∥r+1

L̃r+1
≤ 1

δ
∥Φ∥2V ′ +

δ

4
∥z(t)∥2. (3.2)

Integrating over [0, t] ⊂ [0, T ] the inequality above, we obtain

|z(t)|2 + δ

2

∫ t

0

∥z(s)∥2ds+ 2α

∫ t

0

|u(s)|2ds+ 2β

∫ t

0

∥u(s)∥r+1

L̃r+1
ds

≤ |z(0)|2 + 2

δ

∫ t

0

∥Φ∥2V ′ds, ∀t ∈ [0, T ].

(3.3)

This implies the estimates of z in the space L∞(0, T ;H) ∩ L2(0, T ;V ) and u ∈
Lr+1(0, T ; L̃r+1).

Next we estimate the boundedness of ∂tz. From (2.3), (2.4) and Lemma 2.3, it
can be easily seen that ∂tz is bounded in L2(0, T ;V ′). Moreover,

∥|u|r−1u∥
r+1
r

L̃
r+1
r

= ∥u∥r+1

L̃r+1

yields that

|u|r−1u remains bounded in L
r+1
r (0, T ; L̃

r+1
r ).

Note that

• V1 ↪→ L̃r+1 for every r ≥ 1, then L̃
r+1
r ↪→ V ′

1 ,

• L2(0, T ) ↪→ L
r+1
r (0, T ) for every r > 0,

thus ∂tu is bounded in L
r+1
r (0, T ;V ′

1). Therefore ∂tz is bounded in L
r+1
r (0, T ;V ′).

To prove the uniqueness, assume that z1 = (u1, B1, θ1) and z2 = (u2, B2, θ2) are
two solutions of our problem, and set y = (w, ϕ, η) = (u1 − u2, B1 − B2, θ1 − θ2).
Then y = (w, ϕ, η) satisfies

∂tw −R−1
e ∆w + (u1 · ∇)u1 − (u2 · ∇)u2 + S

[
(B2 · ∇)B2 − (B1 · ∇)B1

]
+ αw + β (Cr(u1)− Cr(u2)) = ηe2,

∂tϕ+R−1
m ∇⊥(curlϕ) + (u1 · ∇)B1 − (u2 · ∇)B2 + (B2 · ∇)u2 − (B1 · ∇)u1 = 0,

∂tη − κ∆η + (u1 · ∇)θ1 − (u2 · ∇)θ2 = w2,

∇ · w = ∇ · ϕ = 0,

w(x, 0) = u10(x)− u20(x), ϕ(x, 0) = B10(x)−B20(x),

η(x, 0) = θ10(x)− θ20(x).

(3.4)
We multiply the first, second and third equations of (3.4) by w, Sϕ and γη, respec-
tively, then integrate over Ω and add the resulting equations, using the definition



EJDE-2025/30 GLOBAL ATTRACTOR FOR 2D MAGNETIC BÉNARD PROBLEM 11

of B and Lemma 2.1 to obtain

1

2

d

dt
|y(t)|2+ δ

2
∥y(t)∥2+α|w(t)|2+β⟨Cr(u1)−Cr(u2), w⟩ ≤ B(z2, z2, y)−B(z1, z1, y).

By Lemma 2.3, we obtain

|B(z2, z2, y)− B(z1, z1, y)| = | − B(y, z1, y)| ≤ cb|y|∥z1∥∥y∥ ≤ δ

2
∥y∥2 + c2b

2δ
|y|2∥z1∥2.

From this with (2.11) we deduce that

d

dt
|y(t)|2 ≤ c2b

2δ
|y|2∥z1∥2 or |y(t)|2 ≤ |y(0)|2 exp

(
c2b
2δ

∫ t

0

∥z1(s)∥2ds
)
.

The last estimate implies the uniqueness (if z10 = z20) and the continuous depen-
dence of solutions on the initial data. □

Remark 3.2. (i) It can be easily seen that z ∈ L∞(0, T ;H) implies z ∈ L∞(0, T ;V ′).

Thus, z and ∂tz ∈ L
r+1
r (0, T ;V ′) and then using [19, Theorem 2, Section 5.9.2], we

have u ∈ C([0, T ];V ′). The reflexivity of the space H and [15, Proposition 1.7.1]
gives u ∈ Cw([0, T ];H).

(ii) Note that ∂tz ∈ L2(0, T ;V ′), for r = 2 and ∂tz ∈ L4/3(0, T ;V ′), for r = 3.
Thus applying [19, Theorem 3, Section 5.9.2], we have z ∈ C([0, T ];H), for r = 2.
For r = 3, one can show that z ∈ C([0, T ];H) satisfying the energy equality (3.3)
(see [22, Theorem 4.1]).

4. Existence of a global attractor

In this section, we discuss the existence of a global attractor for Problem (3.1)
in two dimensional bounded domains. We assume that Φ ∈ V ′ is independent of
t in (3.1). Thanks to Theorem 3.1 we can define a continuous semigroup {S(t) =
(S1(t), S2(t), S3(t))}t≥0 in H by

S(t)(u0, B0, θ0) =
(
S1(t)u0, S2(t)B0, S3(t)θ0

)
= (u(t), B(t), θ(t)), t ≥ 0

where z(0) = (u(0), B(0), θ(0)) = (u0, B0, θ0) = z0 ∈ H. It is easy to see that the
map S(t) : H → H, for t ≥ 0, is Lipschitz continuous on bounded subsets of H.

Moreover, by setting zn(t) = S(t)(u0n , B0n , θ0n) and z(t) = S(t)(u0, B0, θ0), the
following lemma shows weak continuity of the semigroup {S(t)}t≥0, which is needed
to prove the asymptotic compactness of the semigroup by using the energy equation
method introduced by Ball [6].

Lemma 4.1. Let z0n = (u0n , B0n , θ0n) be a sequence in H converging weakly to an
element z0 ∈ H. Then

zn(t) → z(t) weakly in H, ∀t ≥ 0, (4.1)

zn(t) → z(t) weakly in L2(0, T ;V ), ∀T > 0, (4.2)

un(t) → u(t) weakly in Lr+1(0, T ; L̃r+1), ∀T > 0. (4.3)

The proof of the above lemma closely follows the argument presented in [36,
Lemma 2.1], with only slight adjustments. Therefore, we omit the proof here for
conciseness.

We now prove that the semigroup S(t) has a compact global attractor A. For
the general theory of global attractors, we refer the reader [29, 35, 40] for details.
The main result in this section is as follows.



12 D. T. SON EJDE-2025/30

Theorem 4.2. Under the conditions of Theorem 3.1, there exists a global attractor
A in H for the semigroup {S(t)}t≥0 associated with Problem (3.1).

To prove this theorem, we need to show that the semigroup {S(t)}t≥0 possesses
an absorbing set B bounded in H and is asymptotically compact in H, that is,

(1) given a bounded set B ⊂ H, there exists an entering time tB > 0 such that
S(t)B ⊂ B for all t ≥ tB ,

(2) for {zn}n is bounded and tn → ∞, then {S(tn)zn}n is precompact in H.

We first prove the existence of a bounded absorbing set for semigroup {S(t)}t≥0

generated by Problem (3.1) in H. For the sake of brevity, in the following lemma
we only give some formal calculations, the rigorous proof is done by using Galerkin
approximations.

Lemma 4.3. The semigroup {S(t)}t≥0 generated by Problem (3.1) has a bounded
absorbing set in H, that is, there exists a positive constant ρ and a time t0(|z0|, λ)
such that for the solution z(t) = S(t)z0,

|z(t)| ≤ ρ for all t ≥ t0(|z0|, δ, λ),
where δ is in Lemma 2.1 and λ = min(λ1, c0).

Proof. From (3.2), by using (2.2), we have

d

dt
|z(t)|2 + δλ

2
|z(t)|2 ≤ 2

δ
∥Φ∥2V ′ ,

and an application of Gronwall’s inequality yields

|z(t)|2 ≤ |z0|2e−
δλ
2 t +

2

δ2λ
∥Φ∥2V ′ .

Therefore, if we choose ρ2 = 4
δ2λ∥Φ∥

2
V ′ , then

|z(t)| ≤ 2

δ
√
λ
∥Φ∥V ′

for all t ≥ t0(|z0|, δ, λ), and so the proof is complete. □

We next show the asymptotic compactness of the semigroup {S(t)}t≥0 generated
by Problem (3.1). Let us define a symmetric bilinear form [·, ·] : V × V → R by

[z, z̃] = a(z, z̃) + σ(z, z̃)− δλ

4
(z, z̃), ∀z, z̃ ∈ V. (4.4)

From (2.3) and (2.4), we have

[z, z] +
δλ

4
|z|2 = a(z, z̃) + σ(z, z̃) ≤

(
max

(
R−1

e , R−1
m , κ

)
+ cσ

)
∥z∥2.

Thus,

[z]2 ≡ [z, z] ≤
(
max

(
R−1

e , R−1
m , κ

)
+ cσ

)
∥z∥2. (4.5)

By using the definition of |z| and (2.2) to obtain

δλ

4
|z|2 =

δλ

4
(|u|2 + S|B|2 + γ|θ|2)

≤ δ

4
(λ1|u|2 + Sc0|B|2 + γλ1|θ|2)

≤ δ

4
(∥u∥21 + S∥B∥22 + γ∥θ∥23) =

δ

4
∥z∥2.



EJDE-2025/30 GLOBAL ATTRACTOR FOR 2D MAGNETIC BÉNARD PROBLEM 13

Using this and Lemma 2.1 we obtain

[z]2 ≥ δ

2
∥z∥2 − δλ

4
|z|2 ≥ δ

4
∥z∥2. (4.6)

Putting together (4.5) and (4.6) we obtain

δ

4
∥z∥2 ≤ [z]2 ≤

(
max

(
R−1

e , R−1
m , κ

)
+ cσ

)
∥z∥2, ∀z ∈ V. (4.7)

Thus, [·, ·] defines an inner product in V with norm [·] = [·, ·]1/2 equivalent to ∥ · ∥.

Lemma 4.4. The semigroup {S(t)}t≥0 generated by Problem (3.1) is asymptotically
compact in H.

Proof. Let B be a bounded subset of H, and consider {zn}n ⊂ B and {tn}n, tn ≥
0, tn → ∞. Set

B = {z ∈ H : |z| ≤ ρ}, (4.8)

where ρ is the positive constant in Lemma 4.3, and then we find that there exists
a time tB > 0 such that

S(t)B ⊂ B, for all t ≥ tB .

Hence for tn large enough (tn ≥ tB), we have

S(tn)zn ∈ B. (4.9)

Thus, the sequence {S(tnk
)znk

}nk
is weakly precompact in H, and since B is closed

and convex, we have

{S(tnk
)znk

}nk
→ w weakly in H, (4.10)

for some subsequence {S(tnk
)znk

}nk
of {S(tn)zn}n and w ∈ B. Similarly, for each

T > 0, one can show that S(tn −T )zn ∈ B, for all tn ≥ T + tB . Terefore, we obtain
that S(tn−T )zn is precompact in H, and by using a diagonal argument and passing
to a further subsequence (if necessary), we can assume that

{S(tnk
− T )znk

}nk
→ wT weakly in H,

for all T ∈ N with wT = (uT , BT , θT ) ∈ B. Using the weak continuity of S(t)
established in Lemma 4.1 and (4.10), for all ξ ∈ H, we have

(w, ξ) = lim
k→∞

(
S(tnk

)znk
, ξ
)

= lim
k→∞

(
S(T )S(tnk

− T )znk
, ξ
)

=
(
S(T ) lim

k→∞
S(tnk

− T )znk
, ξ
)
=

(
S(T )wT , ξ

)
.

Therefore w = S(T )wT for all T ∈ N. Moreover, by using the weakly lower-
semicontinuity property, from (4.10) we obtain

|w| ≤ lim inf
k→∞

|S(tnk
)znk

|.

Our next aim is to show that

lim sup
k→∞

|S(tnk
)znk

| ≤ |w|. (4.11)

Then we have

lim
k→∞

|S(tnk
)znk

| = |w|,



14 D. T. SON EJDE-2025/30

and this, together with the weak convergence, will imply the strong convergence in
H of S(tnk

)znk
to w.

Now, setting z̃ = z in (3.1), using (2.10) and (4.4) to obtain

1

2

d

dt
|z(t)|2 + δλ

4
|z(t)|2 = ⟨Φ, z⟩ −

(
α|u(t)|2 + β∥u(t)∥r+1

L̃r+1
+ [z]2

)
,

then, using the variation of constant formula, we obtain

|z(T )|2 = e−
δλ
2 T |z0|2 + 2

∫ T

0

e−
δλ
2 (T−s)

[
⟨Φ, z(s)⟩

−
(
α|u(s)|2 + β∥u(s)∥r+1

L̃r+1
+ [z(s)]2

) ]
ds

(4.12)

which can be written as

|S(T )z0|2 = e−
δλ
2 T |z0|2 + 2

∫ T

0

e−
δλ
2 (T−s)

[
⟨Φ,S(s)z0⟩

−
(
α|S1(s)u0|2 + β∥S1(s)u0∥r+1

L̃r+1
+ [S(s)z0]2

) ]
ds,

(4.13)

for all z0 ∈ H and t ≥ 0. Thus, for T ∈ N and tn > T , we have by letting
z0 = S(tn − T )zn in (4.13), we obtain

|S(tn)zn|2

= |S(T )S(tn − T )zn|2

= e−
δλ
2 T |S(tn − T )zn|2 + 2

∫ T

0

e−
δλ
2 (T−s)

〈
Φ,S(s)S(tn − T )zn

〉
ds

− 2

∫ T

0

e−
δλ
2 (T−s)

[
α|S1(s)S1(tn − T )un|2 + β∥S1(s)S1(tn − T )un∥r+1

L̃r+1

]
ds

− 2

∫ T

0

e−
δλ
2 (T−s)[S(s)S(tn − T )zn]

2ds.

(4.14)
We now estimate the terms in (4.14). In the same way that we have obtained in
(4.9), for each T > 0,

S(tn − T )zn ∈ B, ∀tn ≥ T + tB .

From (4.8), we find that

lim sup
k→∞

e−
δλ
2 T |S(tn − T )zn|2 ≤ e−

δλ
2 T ρ2. (4.15)

Next, by using the weak continuity result in (4.2) and the convergence {S(tnk
−

T )znk
}k → wT weakly in H, we have

S(·)S(tnk
− T )znk

→ S(·)wT weakly in L2(0, T ;V ). (4.16)

Now, we consider∫ T

0

∥e− δλ
2 (T−s)Φ∥2V ′ds =

∫ T

0

e−δλ(T−s)∥Φ∥2V ′ds = ∥Φ∥2V ′

(1− e−δλT

δλ

)
<∞

and hence the mapping s 7→ e−
δλ
2 (T−s)Φ ∈ L2(0, T ;V ′). Thus, we find that

lim
k→∞

∫ T

0

e−
δλ
2 (T−s)

〈
Φ,S(s)S(tnk

− T )znk

〉
ds =

∫ T

0

e−
δλ
2 (T−s)

〈
Φ,S(s)wT ⟩ds.

(4.17)



EJDE-2025/30 GLOBAL ATTRACTOR FOR 2D MAGNETIC BÉNARD PROBLEM 15

From (4.7), the norm [·] is equivalent to ∥ · ∥. Also

0 < e−
δλ
2 T ≤ e−

δλ
2 (T−s) ≤ 1, ∀s ∈ [0, T ]

and therefore
( ∫ T

0
e−

δλ
2 (T−s)[·]2ds

)1/2
defines a norm on L2(0, T ;V ) equivalent to

the norm
( ∫ T

0
∥ · ∥2ds

)1/2
. Using (4.16) and the weakly lower-semicontinuity prop-

erty of the norm, we obtain∫ T

0

e−
δλ
2 (T−s)[S(s)wT ]

2ds ≤ lim inf
k→∞

∫ T

0

e−
δλ
2 (T−s)[S(s)S(tnk

− T )znk
]2ds.

Therefore,

lim sup
k→∞

[
− 2

∫ T

0

e−
δλ
2 (T−s)[S(s)S(tnk

− T )znk
]2ds

]
= −2 lim inf

k→∞

∫ T

0

e−
δλ
2 (T−s)[S(s)S(tnk

− T )znk
]2ds

≤ −2

∫ T

0

e−
δλ
2 (T−s)[S(s)wT ]

2ds.

(4.18)

We also have that(∫ T

0

e−
δλ
2 (T−s)| · |2ds

)1/2

and
(∫ T

0

e−
δλ
2 (T−s)∥ · ∥r+1

L̃r+1
ds
) 1

r+1

define norms on L2(0, T ;H) and Lr+1(0, T ; L̃r+1), which are equivalent to the norms( ∫ T

0
| · |2ds

)1/2
and

( ∫ T

0
∥ · ∥r+1

L̃r+1
ds
) 1

r+1 , respectively. Using (4.3) and the weakly

lower-semicontinuity property of the norm, we obtain

lim sup
k→∞

[
− 2

∫ T

0

e−
δλ
2 (T−s)

[
α|S1(s)S1(tnk

− T )unk
|2

+ β∥S1(s)S1(tnk
− T )unk

∥r+1

L̃r+1

]
ds
]

= −2 lim inf
k→∞

∫ T

0

e−
δλ
2 (T−s)

[
α|S1(s)S1(tnk

− T )unk
|2

+ β∥S1(s)S1(tnk
− T )unk

∥r+1

L̃r+1

]
ds

≤ −2

∫ T

0

e−
δλ
2 (T−s)

[
α|S1(s)uT |2 + β∥S1(s)uT ∥r+1

L̃r+1

]
ds.

(4.19)

Collecting (4.15) and (4.17)-(4.19) in (4.14) and then taking lim sup in (4.14), we
obtain

lim sup
k→∞

|S(tnk
)znk

|2

≤ e−
δλ
2 T ρ2 + 2

∫ T

0

e−
δλ
2 (T−s)⟨Φ,S(s)wT ⟩ds

− 2

∫ T

0

e−
δλ
2 (T−s)

(
α|S1(s)uT |2 + β∥S1(s)uT ∥r+1

L̃r+1
+ [S(s)wT ]

2
)
ds.

(4.20)



16 D. T. SON EJDE-2025/30

On the other hand, we obtain in (4.13) applied to w = S(T )wT that

|w|2 = |S(T )wT |2

= e−
δλ
2 T |wT |2 + 2

∫ T

0

e−
δλ
2 (T−s)⟨Φ,S(s)wT ⟩ds

− 2

∫ T

0

e−
δλ
2 (T−s)

(
α|S1(s)uT |2 + β∥S1(s)uT ∥r+1

L̃r+1
+ [S(s)wT ]

2
)
ds.

(4.21)

From (4.20) and (4.21) we find that

lim sup
k→∞

|S(tnk
)znk

|2 ≤ e−
δλ
2 T ρ2+ |w|2−e− δλ

2 T ≤ |w|2+e− δλ
2 T ρ2, ∀T ∈ N. (4.22)

Let us take T → ∞ in (4.22) to obtain

lim sup
k→∞

|S(tnk
)znk

|2 ≤ |w|2

and yields (4.11). This establishes that {S(tn)zn}n is precompact in H and hence
that {S(t)}t≥0 is asymptotically compact in H. □

5. Dimension of the attractor

In this section, we estimate the bounds for the Hausdorff as well as fractal
dimensions of the global attractor A obtained in previous section. Due to this
technical difficulty, we consider the cases r = 1, 3 only in this section.

Let z(t) be the unique weak solution of the Problem (3.1), we consider the
linearized system

(∂ts(t), s̃) + a(s(t), s̃) + σ(s(t), s̃) + B(s(t), z(t), s̃) + B(z(t), s(t), s̃)

+ α(v, ṽ) + β⟨C′
r(u)v, ṽ⟩V ′

1 ,V1
= 0, ∀s̃ = (ṽ, ψ̃, ζ̃) ∈ V, a.e. t ∈ (0, T ),

s(0) = s0 in H,

(5.1)

where C′
r(·) is defined in (2.13). As in the case of nonlinear Problem (3.1), one can

show that there exists a unique solution s = (v, ψ, ζ) ∈ L2(0, T ;V ) ∩ L∞(0, T ;H)

with v ∈ Lr+1(0, T ; L̃r+1) and ∂ts ∈ L
r+1
r (0, T ;V ′).

We define a map Λ(t; z0) : H → H by setting Λ(t; z0)s0 = s(t). In the next
Lemma, we show that the map Λ(t; z0) is bounded and the semigroup {S(t)}t≥0 is
uniformly differentiable on global attractor A, i.e.,

lim
ε→0

sup
0<|z̄0−z0|<ε, z̄0,z0∈A

|S(t)z̄0 − S(t)z0 − Λ(t; z0)(z̄0 − z0)|
|z̄0 − z0|

= 0. (5.2)

Theorem 5.1. Let z̄0, z0 ∈ H. Then for r = 1, 3, there exists a constantM(|z̄0|, |z0|)
such that

|S(t)z̄0 − S(t)z0 − Λ(t; z0)(z̄0 − z0)| ≤M |z̄0 − z0|, (5.3)

where the linear operator Λ(t; z0) for t > 0 is the solution operator of the prob-
lem (5.1). Or in other words, for every t > 0, the map S(t) : H → H is
Fréchet differentiable with respect to the initial data, and its Fréchet derivative
Dz0(S(t)z0)s0 = Λ(t; z0)s0. Moreover, (5.2) is satisfied.

Proof. We denote z̄(t) = S(t)z̄0, z(t) = S(t)z0 and let s(t) be the solution of (5.1)
with s0 = z̄0 − z0. From (3.3), we easily find that∫ t

0

∥z(s)∥2ds ≤ 2

δ

(
|z0|2 +

2

δ
t∥Φ∥2V ′

)
, ∀t ∈ [0, T ]. (5.4)



EJDE-2025/30 GLOBAL ATTRACTOR FOR 2D MAGNETIC BÉNARD PROBLEM 17

Writing ω(t) = z̄(t) − z(t), t ≥ 0 and using the same arguments as in proving the
uniqueness of solution in Theorem 3.1, we deduce

d

dt
|ω(t)|2 + δ

2
∥ω(t)∥2 ≤ 2δ−1c2b |ω(t)|2∥z(s)∥2

and

|ω(t)|2 ≤ |ω(0)|2 exp
(
2δ−1c2b

∫ t

0

∥z(s)∥2ds
)
.

Thus, using (5.4), we obtain

|ω(t)|2 ≤ |ω(0)|2 exp
(
4δ−2c2b

(
|z0|2 + 2δ−1t∥Φ∥2V ′

) )
, ∀t ∈ [0, T ]. (5.5)

Now, from (5.4) and (5.5), we have

δ

2

∫ t

0

∥ω(s)∥2ds

≤ |ω(0)|2 + 2δ−1c2b

∫ t

0

|ω(s)|2∥z(s)∥2ds

≤ |ω(0)|2 + 2δ−1c2b

∫ t

0

∥z(s)∥2|ω(0)|2 exp
[
4δ−2c2b

(
|z0|2 + 2δ−1s∥Φ∥2V ′

) ]
ds

≤ |ω(0)|2
[
1 + 4δ−2c2b

(
|z0|2 + 2δ−1t∥Φ∥2V ′

)
exp

[
4δ−2c2b

(
|z0|2 + 2δ−1t∥Φ∥2V ′

) ]
≤ |ω(0)|2 exp

[
8δ−2c2b

(
|z0|2 + 2δ−1t∥Φ∥2V ′

) ]
.

(5.6)
Let η(t) = (uη(t), Bη(t), θη(t)) be defined by

η(t) = z̄(t)− z(t)− s(t) = ω(t)− s(t), t ≥ 0.

Evidently, for all η̃ = (uη̃, Bη̃, θη̃) ∈ V , η(t) satisfies

(∂tη(t), η̃) + a(η(t), η̃) + σ(η(t), η̃) + α(uη(t), uη̃) + β⟨Cr(ū)− Cr(u)− C′
r(u)v, uη̃⟩

= −B(z̄(t), z̄(t), η̃) + B(z(t), z(t), η̃) + B(s(t), z(t), η̃) + B(z(t), s(t), η̃),
a.e. t ∈ (0, T ),

η(0) = 0 in H.

It is easy to see that

− B(z̄(t), z̄(t), η̃) + B(z(t), z(t), η̃) + B(s(t), z(t), η̃) + B(z(t), s(t), η̃)
= −B(z(t), η(t), η̃)− B(η(t), z(t), η̃)− B(ω(t), ω(t), η̃)

and consequently, by Lemma 2.1, for all t > 0,

1

2

d

dt
|η(t)|2 + δ

2
∥η(t)∥2 + α|uη(t)|2 ≤ B(η(t), η(t), z(t))− B(ω(t), ω(t), η(t))

− β⟨Cr(ū)− Cr(u)− C′
r(u)v, uη⟩.

(5.7)

By using the Lemma 2.3 and the Young inequality, we have

B(η(t), η(t), z(t))− B(ω(t), ω(t), η(t))
≤ cb(|η(t)|∥η(t)∥∥z(t)∥+ |ω(t)|∥ω(t)∥∥η(t)∥)

≤ δ

6
∥η(t)∥2 + 3δ−1c2b(|η(t)|2∥z(t)∥2 + |ω(t)|2∥ω(t)∥2).



18 D. T. SON EJDE-2025/30

To eatimate the term −β⟨Cr(ū)−Cr(u)−C′
r(u)v, uη⟩, we consider the cases r = 1, 3

separately. For r = 1, it can be easily seen that

−β⟨Cr(ū)− Cr(u)− C′
r(u)v, uη⟩ = −β|uη(t)|21.

It should be noted that〈 u1
|u1|

(u1 · v)−
u2
|u2|

(u2 · v), w
〉

=
〈 u1
|u1|

[(u1 − u2) · v], w
〉
+

〈( u1
|u1|

− u2
|u2|

)
(u2 · v), w

〉
=

〈 u1
|u1|

[(u1 − u2) · v], w
〉
+

〈u1(|u1| − |u2|) + (u1 − u2)|u1|
|u1||u2|

(u2 · v), w
〉

≤ 2⟨|u1 − u2||v|, |w|⟩

(5.8)

for all u1 ̸= 0, u2 ̸= 0 and v, w ∈ L̃3 (one can also obtain same estimates for u1 = 0
or u2 = 0). For r = 2, by using Taylor’s formula (see [16, Theorem 7.9.1]), Hölder’s,
Ladyzhenskaya’s, Young’s inequalities, (2.14) and (5.8), we have

− β⟨Cr(ū)− Cr(u)− C′
r(u)v, uη⟩

= −β
〈 ∫ 1

0

C′
r(ϵū+ (1− ϵ)u)(ū− u)dϵ− C′

r(u)v, uη
〉

= −β⟨C′
r(u)uη, uη⟩+ β

〈 ∫ 1

0

[C′
r(u)− C′

r(ϵū+ (1− ϵ)u)] (ū− u)dϵ, uη
〉

= −β⟨C′
r(u)uη, uη⟩+ β

∫ 1

0

〈[
|ū| − |ϵū+ (1− ϵ)u|

]
· (ū− u), uη

〉
dϵ

+ β

∫ 1

0

〈 ū
|ū|

[ū · (ū− u)]− ϵū+ (1− ϵ)u

|ϵū+ (1− ϵ)u|
[(ϵū+ (1− ϵ)u) · (ū− u)] , uη

〉
dϵ

≤ 3β

∫ 1

0

(1− ϵ)⟨|ū− u|2, |uη|⟩dϵ

≤ 3β

2
∥ū− u∥2

L̃4 |uη|1

≤ α|uη|21 +
9β2

16α
|ω|2∥ω∥2.

For r = 3, once again by using (2.14), Taylor’s formula (see [16, Theorem 7.9.1]),
Hölder’s, Ladyzhenskaya’s, Young’s, Poincaré’s inequalities and (2.15), we have

− β⟨Cr(ū)− Cr(u)− C′
r(u)v, uη⟩

= −β
〈
C′
r(u)(ū− u) +

1

2

∫ 1

0

C′′
r (ϵū+ (1− ϵ)u)dϵ(ū− u)⊗ (ū− u)− C′

r(u)v, uη
〉

= −β⟨C′
r(u)uη, uη⟩+ 3β

∫ 1

0

⟨[(ϵū+ (1− ϵ)u) · (ū− u)] (ū− u), uη⟩ dϵ

= 3β

∫ 1

0

∥ϵū+ (1− ϵ)u∥L̃4∥ū− u∥2
L̃4∥uη∥L̃4dϵ

≤ 3(23/4)β(∥ū∥L̃4 + ∥u∥L̃4)|ω|∥ω∥|uη|1/21 ∥uη∥1/21

≤ δ

12
∥η∥2 + 38/3

2
δ−1/3β4/3(∥ū∥L̃4 + ∥u∥L̃4)

4/3|ω|4/3∥ω∥4/3|η|2/3



EJDE-2025/30 GLOBAL ATTRACTOR FOR 2D MAGNETIC BÉNARD PROBLEM 19

≤ δ

12
∥η∥2 + β

2
(∥ū∥L̃4 + ∥u∥L̃4)

4|η|2 + δ−1/2β3/235/2|ω|2∥ω∥2.

Collecting the estimates above and applying to (5.7), we obtain

d

dt
|η(t)|2 + δ

2
∥η(t)∥2

≤


6δ−1c2b(|η(t)|2∥z(t)∥2 + |ω(t)|2∥ω(t)∥2), for r = 1,

6δ−1c2b |η(t)|2∥z(t)∥2 +
(
6δ−1c2b +

9β2

8α

)
|ω(t)|2∥ω(t)∥2, for r = 2,[

6δ−1c2b∥z(t)∥2 + β(∥ū(t)∥4
L̃4 + ∥u(t)∥4

L̃4)
]
|η(t)|2

+
(
6δ−1c2b + 2δ−1/2β3/235/2

)
|ω(t)|2∥ω(t)∥2, for r = 3.

(5.9)

For r = 3, we deduce from (3.3) that∫ t

0

(
∥ū(s)∥4

L̃4 + ∥u(s)∥4
L̃4

)
ds ≤ 1

β
(|z0|2 + 2δ−1t∥Φ∥2V ′), ∀t ≥ 0. (5.10)

By integrating (5.9) from 0 to t, and using that η(0) = 0, we obtain

|η(t)|2 ≤
(
6δ−1c2b + 2δ−1/2β3/235/2

) ∫ t

0

|ω(s)|2∥ω(s)∥2ds

+

∫ t

0

[
6δ−1c2b∥z(s)∥2 + β(∥ū(s)∥4

L̃4 + ∥u(s)∥4
L̃4)

]
|η(s)|2ds

for all t ≥ 0, and consequently, by the Gronwall inequality and (5.4)-(5.6), (5.10),

|η(t)|2 ≤
(
6δ−1c2b + 2δ−1/2β3/235/2

) ∫ t

0

|ω(s)|2∥ω(s)∥2ds

× exp
(∫ t

0

[
6δ−1c2b∥z(s)∥2 + β(∥ū(s)∥4

L̃4 + ∥u(s)∥4
L̃4)

]
ds
)

≤
(
12δ−2c2b + 4(δ−1β)3/235/2

)
|ω(0)|4

× exp
[ (

12δ−2c2b + 12δ−1c2b + 1
) (

|z0|2 + 2δ−1t∥Φ∥2V ′

) ]
.

Thus, by the definition of η, it is immediate that

|z̄(t)− z(t)− s(t)|
|z̄0 − z0|

= χ(t)|z̄0 − z0|

where

χ(t) =
(
12δ−2c2b + 4(δ−1β)3/235/2

)1/2
× exp

[(
6δ−2c2b + 6δ−1c2b +

1

2

)(
|z0|2 + 2δ−1t∥Φ∥2V ′

)]
,

and hence the differentiability of the semigroup S(t) with respect to the initial data
as well as (5.2) and (5.3) follows. The cases of r = 1, 2 can be proved in a similar
way. □

In the next Theorem, we show that the global attractor obtained in Theorem
4.2 has finite Hausdorff and fractal dimensions. To estimate the dimension of the
global attractor A, we need the assumption

R2 \ Ω contains a semicone (5.11)

to ensure that we can use the generalized Lieb-Thirring inequality in the general
case (see [23]).



20 D. T. SON EJDE-2025/30

Theorem 5.2. Assume the conditions of Theorem 3.1 and conditions (5.11) hold.
For r = 1, 3, then the global attractor A in Theorem 4.2 has finite Hausdorff and
fractal dimensions, which can be estimated as

dimHau(A) ≤ 1 +
M
K
, (5.12)

dimFra(A) ≤ 2
(
1 +

2M
K

)
, (5.13)

where

M :=
1

12

(
λ1R

−1
e + c0R

−1
m + κλ1

)
,

K := 6µR1/2
e max

(
Re;

2

7
Rm; 2κ−1

)[
4(R1/2

e +R1/2
m )2 + 4R3/2

m + κ−3/2
]
∥Φ∥2V ′ .

Proof. First, we can rewrite (5.1) as

(∂ts(t), s̃) = ⟨F ′(z)s, s̃⟩

= −
[
a(s(t), s̃) + σ(s(t), s̃) + B(s(t), z(t), s̃) + B(z(t), s(t), s̃)

+ α(v, ṽ) + β⟨C′
r(u)v, ṽ⟩

]
, ∀s̃ = (ṽ, ψ̃, ζ̃) ∈ V, a.e. t ∈ (0, T ),

s(0) = s0 in H.

(5.14)

Then we define the numbers qm for m ∈ N by

qm = lim sup
t→∞

sup
z0∈A

sup
ξi∈H, |ξi|≤1, i=1,...,m

1

t

∫ t

0

Tr(F ′(S(s)z0) ◦Qm(s))ds (5.15)

where Qm(s) = Qm(s; z0, ξ1, . . . , ξm) is the orthogonal projector of H onto the
space spanned by

{Λ(t; z0)ξ1, . . . ,Λ(t; z0)ξm}

and the trace (denoted by Tr) of F ′(S(s)z0)◦Qm(s) in (5.15) is defined at least a.e.
in t. From [40, section V.3.4] (see Proposition V.2.1 and Theorem V.3.3), we infer
that if qm < 0, for some m ∈ N, then the global attractor A has finite Hausdorff
and fractal dimensions estimated respectively as

dimHau(A) ≤ m, (5.16)

dimFra(A) ≤ m
(
1 + max

1≤i≤m

(qi)+
|qm|

)
. (5.17)

To estimate the numbers qm, let z0 ∈ A and ξ1, . . . , ξm ∈ H. Set z(t) = S(t)z0
and ηi = Λ(t, z0)ξi, t ≥ 0. Let

{
w̃i(t), C̃i(t), ϕ̃i(t)

}
i=1,...,m

, t ≥ 0 be a basis for

span{η1(t), . . . , ηm(t)} such that {w̃i(t)}i=1,...,m is orthonormal inH1, {C̃i(t)}i=1,...,m

is orthonormal in H2 and {ϕ̃i(t)}i=1,...,m is orthonormal in H3. Set

φi = (wi, Ci, ϕi) =
( w̃i√

3
,
C̃i√
3S
,
ϕ̃i√
3γ

)
.

An easy computation shows that {φi}i=1,...,m is orthonormal in H. Since ηi(t) ∈ V
for a.e. t ≥ 0, we can assume that φi(t) ∈ V for a.e. t ≥ 0 (by the Gram–Schmidt



EJDE-2025/30 GLOBAL ATTRACTOR FOR 2D MAGNETIC BÉNARD PROBLEM 21

orthogonalization process). Then, from Lemma 2.1, (2.10), (2.14) and (5.14), we
have

Tr(F ′(S(s)z0) ◦Qm(s))

=

m∑
i=1

⟨F ′(z(s))φi, φi⟩V ′,V

= −
m∑
i=1

[
a(φi, φi) + σ(φi, φi) + B(φi, z, φi) + α(wi, wi) + β⟨C′

r(u)wi, wi⟩
]

≤
m∑
i=1

[
− 1

2

(
1

Re
∥wi∥21 +

S

Rm
∥Ci∥22 + γκ∥ϕi∥23

)
+ |B(φi, z, φi)|

]
.

(5.18)

Now let

ρ(x) =

m∑
i=1

(
R−1/2

e |wi(x)|2 + SR−1/2
m |Ci(x)|2 + γκ1/2|ϕi(x)|2

)
,

then by calculating similarly as in [3, Theorem 5.1], we find that

∣∣ m∑
i=1

[
b(wi, u, wi)− Sb(Ci, u, Ci)

]∣∣ ≤ (R1/2
e +R1/2

m )∥u∥1∥ρ∥L2 ,

∣∣ m∑
i=1

S
[
b(wi, B, Ci)− b(Ci, B,wi)

]∣∣ ≤ S1/2(ReRm)1/4∥B∥2∥ρ∥L2 .

(5.19)

Applying Cauchy’s and Young’s inequalities, we obtain

γ|wi(x) · ∇θ(x)||ϕi(x)| ≤ γ|θ(x)||wi(x)||ϕi(x)|

≤ γ1/2R
1/4
e

2κ1/4
|∇θ(x)|

( 1

R
1/2
e

|wi(x)|2 + γκ1/2|ϕi(x)|2
)
.

Integrating this expression in x, summing up in i from 1 up to m, and using the
definition of ρ, we deduce that

∣∣ m∑
i=1

γb̄(wi, θ, ϕi)
∣∣

= γ
∣∣ m∑
i=1

∫
Ω

wi(x) · ∇θ(x)ϕi(x)dx
∣∣

≤ γ1/2R
1/4
e

2κ1/4

∫
Ω

|∇θ(x)|
m∑
i=1

( 1

R
1/2
e

|wi(x)|2 + γκ1/2|ϕi(x)|2
)
dx

=
γ1/2R

1/4
e

2κ1/4
∥θ∥3∥ρ∥L2 .

(5.20)



22 D. T. SON EJDE-2025/30

Hence, from (5.19) and (5.20), we obtain∣∣ m∑
i=1

B(φi, z, φi)
∣∣

=
∣∣ m∑
i=1

[
b(wi, u, wi)− Sb(Ci, B,wi) + Sb(wi, B,Ci)− Sb(Ci, u, Ci)

+ γb̄(wi, θ, ϕi)
]∣∣

≤ ∥ρ∥L2

[
(R1/2

e +R1/2
m )∥u∥1 + S1/2(ReRm)1/4∥B∥2 +

γ1/2R
1/4
e

2κ1/4
∥θ∥3

]
.

(5.21)

From the definition of ρ, w̃i, C̃i and ϕ̃i, we observe that

ρ(x) =
1

3

m∑
i=1

(
R−1/2

e |w̃i(x)|2 +R−1/2
m |C̃i(x)|2 + κ1/2|ϕ̃i(x)|2

)
.

Then the generalized Lieb-Thirring inequality (see [23]) can be applied to the or-

thonormal finite families {w̃i}i, {C̃i}i and {ϕ̃i}i (by condition (5.11). This guaran-
tees the existence of a constant µ independent of the number of functions m (but
depending on the shape of Ω) such that

∥ρ∥2L2

≤ 1

3

( 1

Re
∥

m∑
i=1

|w̃i(x)|2∥2L2 +
1

Rm
∥

m∑
i=1

|C̃i(x)|2∥2L2 + κ∥
m∑
i=1

|ϕ̃i(x)|2∥2L2

)
≤ µ

3

m∑
i=1

( 1

Re
∥w̃i∥21 +

1

Rm
∥C̃i∥2H1 + κ∥ϕ̃i∥23

)
≤ µ

m∑
i=1

( 1

Re
∥wi∥21 +

S

Rm
∥Ci∥22 + γκ∥ϕi∥23

)
.

(5.22)

Inserting (5.22) into (5.21) and using Young’s inequality, we obtain∣∣ m∑
i=1

B(φi, z, φi)
∣∣

≤ µ
[
(R1/2

e +R1/2
m )∥u∥1 + S1/2(ReRm)1/4∥B∥2 +

γ1/2R
1/4
e

2κ1/4
∥θ∥3

]2
+

1

4µ
∥ρ∥2L2

≤ 3µ
[
(R1/2

e +R1/2
m )2∥u∥21 + S(ReRm)1/2∥B∥22 +

γR
1/2
e

4κ1/2
∥θ∥23

]
+

1

4

m∑
i=1

( 1

Re
∥wi∥21 +

S

Rm
∥Ci∥22 + γκ∥ϕi∥23

)
.

Applying this inequality to (5.18), we obtain

Tr(F ′(S(s)z0) ◦Qm(s))

≤ 3µ
[
(R1/2

e +R1/2
m )2∥u∥21 + S(ReRm)1/2∥B∥22 +

γR
1/2
e

4κ1/2
∥θ∥23

]
− 1

4

m∑
i=1

( 1

Re
∥wi∥21 +

S

Rm
∥Ci∥22 + γκ∥ϕi∥23

)
.

(5.23)



EJDE-2025/30 GLOBAL ATTRACTOR FOR 2D MAGNETIC BÉNARD PROBLEM 23

Since {φi}i=1,...,m is orthonormal in H, we see that |wi|2 = S|Ci|2 = γ|ϕi|2 = 1/3.
Then, by (2.2), we deduce from (5.23) that

Tr(F ′(S(s)z0) ◦Qm(s))

≤ 3µ
[
(R1/2

e +R1/2
m )2∥u∥21 + S(ReRm)1/2∥B∥22 +

γR
1/2
e

4κ1/2
∥θ∥23

]
− 1

4

m∑
i=1

( λ1
Re

|wi|2 +
Sc0
Rm

|Ci|2 + γκλ1|ϕi|2
)

= 3µ
[
(R1/2

e +R1/2
m )2∥u∥21 + S(ReRm)1/2∥B∥22 +

γR
1/2
e

4κ1/2
∥θ∥23

]
− m

12

( λ1
Re

+
c0
Rm

+ κλ1

)
.

For all m ∈ N, we have

qm ≤ 3µ lim sup
t→∞

sup
z0∈A

1

t

∫ t

0

[
(R1/2

e +R1/2
m )2∥u(s)∥21 + S(ReRm)1/2∥B(s)∥22

+
γR

1/2
e

4κ1/2
∥θ(s)∥23

]
ds− m

12

( λ1
Re

+
c0
Rm

+ κλ1

)
.

(5.24)

From (1.1), we multiply the first, second and third equations by u, SB and γθ,
respectively. Then integrating over Ω, using (2.9), (2.5) and (2.6), we obtain the
energy estimates

1

2

d

dt
|u|2 +R−1

e ∥u∥21 − Sb(B,B, u) + α|u|2 + β∥u∥r+1

L̃r+1

≤ R−1
e

4
∥u∥21 +

Re

λ21
∥θ∥23 +

R−1
e

8
∥u∥21 + 2Re∥f∥2V ′

1

≤ 3R−1
e

8
∥u∥21 +

γκ

4
∥θ∥23 + 2Re∥f∥2V ′

1
,

1

2

d

dt
S|B|2 +R−1

m S∥B∥22 − Sb(B, u,B) ≤ 7R−1
m S

8
∥B∥22 +

2RmS

7
∥Ψ∥2V ′

2
,

1

2

d

dt
γ|θ|2 + κγ∥θ∥23 ≤ ϵ2γ

2κ
∥u∥21 +

κγ

2
∥θ∥23 +

κγ

8
∥θ∥23 +

2γ

κ
∥h∥2V ′

3

≤ 5κγ

8
∥θ∥23 +

R−1
e

2
∥u∥21 +

2γ

κ
∥h∥2V ′

3
.

Adding the inequalities above and using (2.8), we deduce that

d

dt
|z(t)|2 + R−1

e

4
∥u(t)∥21 +

R−1
m S

4
∥B(t)∥22 +

κγ

4
∥θ(t)∥23

≤ 2Re∥f∥2V ′
1
+

4RmS

7
∥Ψ∥2V ′

2
+

4γ

κ
∥h∥2V ′

3

≤ 2max

(
Re;

2

7
Rm;

2

κ

)
∥Φ∥2V ′ .

It follows that

lim sup
t→∞

sup
z0∈A

1

t

∫ t

0

∥u(s)∥21ds ≤ 8Re max
(
Re;

2

7
Rm;

2

κ

)
∥Φ∥2V ′ ,



24 D. T. SON EJDE-2025/30

lim sup
t→∞

sup
z0∈A

1

t

∫ t

0

S∥B(s)∥22ds ≤ 8Rm max
(
Re;

2

7
Rm;

2

κ

)
∥Φ∥2V ′ ,

lim sup
t→∞

sup
z0∈A

1

t

∫ t

0

γ∥θ(s)∥23ds ≤ 8κ−1 max
(
Re;

2

7
Rm;

2

κ

)
∥Φ∥2V ′ .

Therefore,

lim sup
t→∞

sup
z0∈A

1

t

∫ t

0

[
(R1/2

e +R1/2
m )2∥u(s)∥21 + S(ReRm)1/2∥B(s)∥22

+
γR

1/2
e

4κ1/2
∥θ(s)∥23

]
ds

≤ 2R1/2
e max

(
Re;

2

7
Rm;

2

κ

) [
4(R1/2

e +R1/2
m )2 + 4R3/2

m + κ−3/2
]
∥Φ∥2V ′ .

(5.25)

Applying (5.25) to (5.24), we obtain

qm ≤ −mK +M, ∀m ∈ N

where K and M are as in the statement of the theorem. If m′ ∈ N is defined by

m′ − 1 ≤ M
K

< m′

then qm′ < 0 and thus from (5.16), we find that

dimHau(A) ≤ m′ ≤ 1 +
M
K

and obtain (5.12). Furthermore, if m′′ ∈ N is defined by

m′′ − 1 ≤ 2M
K

< m′′

then using [40, Lemma VI.2.2], we have

qm′′ < 0 and
(qi)+
|qm′′ |

≤ 1, for all i = 1, . . . ,m′′.

From (5.17), we obtain

dimFra(A) ≤ 2m′′ ≤ 2
(
1 +

2M
K

)
and have (5.13), which completes the proof. □

6. Upper semicontinuity of global attractor

In this section, we verify the upper semicontinuity of global attractors for (1.1).
Let {Ωm}∞m=1 be an expanding sequence of simply connected bounded and smooth
subdomains of Ω (for example, one can take Ω = R×(−L,L)) such that ∪∞

m=1Ωm =
Ω. Throughout this section, we differentiate the H spaces defined in Ω and Ωm

as HΩ and HΩm
, respectively, and similar modifications are made for other spaces

also. Then, we consider

(∂tzm(t), z̃) + a(zm(t), z̃) + σ(zm(t), z̃) + B(zm(t), zm(t), z̃)

+ α(um, ũ) + β⟨Cr(um), ũ⟩ = ⟨Φm, z̃⟩, ∀z̃ = (ũ, B̃, θ̃) ∈ VΩm
, a.e. t ∈ (0, T ),

zm(0) = z0,m in HΩm
,

(6.1)



EJDE-2025/30 GLOBAL ATTRACTOR FOR 2D MAGNETIC BÉNARD PROBLEM 25

where

Φm(x) =

{
Φ(x), x ∈ Ωm,

0, x ∈ Ω \ Ωm,
and z0,m(x) =

{
z0(x), x ∈ Ωm,

0, x ∈ Ω \ Ωm.

We also assume that Φ ∈ HΩ. Let zm ∈ L∞(0, T ;HΩm
) ∩ L2(0, T ;VΩm

) be
the unique solution of the system (6.1) with ∂tzm ∈ L2(0, T ;V ′

Ωm
) and hence in

C([0, T ];HΩm). Then, by the same arguments in Section 4, the system (6.1) pos-
sesses an attractor Am ∈ HΩm

. To check whether the global attractor A and Am

of Problem (3.1) corresponding to Ω and Ωm, respectively, have the upper semi-
continuity when m→ ∞, we follow the results for the 2D Navier-Stokes equations
in [44].

We now prove the following important lemma.

Lemma 6.1. If z0,m ∈ Am, m = 1, 2, . . . then there exists z0 ∈ A such that up to
a subsequence,

z0,m → z0 strongly in H. (6.2)

Proof. First, we show that for given z0,m ∈ Am, m = 1, 2, . . . , there exists z0 ∈ A
such that up to a subsequence,

Sm(·)z0,m → S(·)z0 weakly in L2(−T, T ;VΩ), (6.3)

Sm(t)z0,m → S(t)z0 weakly in HΩ for each t ∈ R. (6.4)

Indeed, by using (4.8), we have that

|Sm(t)z0,m| ≤ ρ, for given z0,m ∈ Am and t ∈ R. (6.5)

By the same estimate as (3.3), we deduce that for each T > 0, the sequence
{Sm(t)z0,m}m∈N is uniformly bounded in L∞(−T, T,HΩ) ∩ L2(−T, T ;VΩ) and the
sequence {∂tSm(t)z0,m}m∈N is bounded in L2(−T, T ;V ′

Ω). From these uniform
bounds, we can extract a subsequence, which is denoted again by {z0,m}m∈N such
that

Sm(t)z0,m → z weak-star in L∞(−T, T ;HΩ),

Sm(t)z0,m → z weakly in L2(−T, T ;VΩ),
∂tSm(t)z0,m → ∂tz weakly in L2(−T, T ;V ′

Ω),

Sm(t)u0,m → u weakly in Lr+1(−T, T ; L̃r+1
Ω ).

(6.6)

From the first convergence in (6.6), we see that

Sm(t)z0,m → z weakly in L2(−T, T ;HΩ).

By the same arguments as in [36, Lemma 2.1], one can show that z(·) is a weak
solution to Problem (3.1) defined on R and z ∈ C(R;HΩ). Hence, we have z(t) =
S(t)z(0) and obtain (6.3). Let us now fix t∗ ∈ [−T, T ]. From (6.5), we have that
the sequence {Sm(t∗)z0,m}m∈N is bounded in HΩ. Therefore there exists z̃ ∈ HΩ,
a subsequence {Sm(t∗)z0,m}m∈N (still denoted by {Sm(t∗)z0,m}m∈N) such that

Sm(t∗)z0,m → z̃ weakly in HΩ.

Thus z(t) is a solution of Problem (3.1) with z(t∗) = z̃. Since t∗ ∈ [−T, T ] is
arbitrary, we obtain

Sm(t)z0,m → S(t)z(0) weakly in HΩ, for each t ∈ R. (6.7)



26 D. T. SON EJDE-2025/30

Next, we show that z0 ∈ A. By using the weakly lower-semicontinuity of norm,
(6.5) and (6.7), we obtain

|S(t)z(0)| = lim inf
m→∞

|Sm(t)z0,m| ≤ ρ, for all t ∈ R,

which implies that the solution S(t)z(0) defined on R and bounded. Hence,

S(t)z(0) ∈ A for all t ∈ R

and in particular z(0) = z0 ∈ A. Finally, by repeating the arguments in Lemma
4.4, we prove (6.2) as follows. From (6.4), we see that there exists z0 ∈ A such that
up to a subsequence,

z0,m → z0 weakly in HΩ,

and the weakly lower-semicontinuity property of the HΩ norm gives

lim inf
m→∞

|z0,m| ≥ |z0|. (6.8)

From (4.12), we also have that

|zm(t)|2 = e−
δλ
2 (t−τ)|zm(τ)|2

+ 2

∫ t

τ

e−
δλ
2 (t−s)

[
⟨Φ, z(s)⟩ − (α|u(s)|2 + β∥u(s)∥r+1

L̃r+1
+ [z(s)]2)

]
ds,

and the fact that z0,m = Sm(T )Sm(−T )z0,m for each T ∈ R, thus

|z0,m|2

= |Sm(T )Sm(−T )z0,m|2

= e−
δλ
2 (T−T0)|Sm(T0)Sm(−T )z0,m|2 + 2

∫ T

T0

e−
δλ
2 (T−s)⟨Φ,Sm(s)Sm(−T )z0,m⟩ds

− 2

∫ T

T0

e−
δλ
2 (T−s)

[
α|S1m(s)S1m(−T )u0,m|2 + β∥S1m(s)S1m(−T )u0,m∥r+1

L̃r+1

]
ds

− 2

∫ T

T0

e−
δλ
2 (T−s)[Sm(s)Sm(−T )z0,m]2ds.

Since z0,m ∈ Am, the solution Sm(t)z0,m of the system (6.1) is bounded on R
and Sm(t)z0,m ∈ Am for all t ∈ R. Hence, for each T ≥ 0, we deduce that
Sm(−T )z0,m ∈ Am. Thus, from (6.4), there exists zT ∈ A such that, up to a
subsequence,

Sm(t)Sm(−T )z0,m → zT weakly in HΩ, for all t ∈ R.

By a calculation similar to (4.12), we obtain

lim sup
m→∞

|z0,m|2 ≤ |z0|2+(ρ2−|S(T ∗)zT |2)e−
δλ
2 (T−T∗) ≤ |z0|2+ρ2e−

δλ
2 (T−T∗) (6.9)

for all T > T ∗. Since HΩ is a Hilbert space, passing T to ∞ in (6.9) and using
(6.8) to obtain (6.2). The proof is complete. □

We now state and prove the main result in this section as follows.

Theorem 6.2. Assume that Φ ∈ H. Let A and Am be the global attractors corre-
sponding to the systems (3.1) and (6.1), respectively. Then

lim
m→∞

distHΩ
(Am,A) = 0, (6.10)



EJDE-2025/30 GLOBAL ATTRACTOR FOR 2D MAGNETIC BÉNARD PROBLEM 27

where distHΩ
(Am,A) = supz∈Am

distHΩ
(z,A) is the Hausdorff semidistance of

space HΩ.

Proof. We use contradiction to prove this Theorem. Let us assume that (6.10) does
not hold. Then there exists a fixed ε0 > 0 and a sequence zm ∈ Am such that

distHΩ
(zm,A) ≥ ε0 > 0, m = 1, 2, . . . . (6.11)

While by Lemma 6.1, we find that there exists a subsequence {zmk
}k ⊂ {zm}m

such that

lim
k→∞

distHΩ(zmk
,A) = 0,

which is a contradiction to (6.11) and completes the proof. □

7. Existence and exponential stability of a stationary solution

In this section, we study the existence, uniqueness and exponential stability of
stationary solutions of problem (1.1) with the external force Φ ∈ L2(Ω)× L2(Ω)×
L2(Ω). We first consider the following steady state system associated with the
equation (1.1).

−R−1
e ∆u+ (u · ∇)u− S(B · ∇)B + αu+ β|u|r−1u+∇

(
p+

S

2
|B|2

)
= θe2 + f(x), in Ω,

R−1
m ∇⊥(curlB) + (u · ∇)B − (B · ∇)u = Ψ(x), in Ω,

−κ∆θ + (u · ∇)θ = u2 + h(x), in Ω,

∇ · u = ∇ ·B = 0, in Ω,

u = 0, B · n = 0, curlB = 0, θ = 0 on ∂Ω.

(7.1)

Taking the inner product of the first equation of (7.1) with ũ ∈ V1, we obtain

R−1
e ((u, ũ))1+b(u, u, ũ)−Sb(B,B, ũ)+α(u, ũ)+β⟨Cr(u), ũ⟩−(θe2, ũ) = ⟨f, ũ⟩V ′

1 ,V1
.

We take the inner product of the second equation of (7.1) with SB̃ (B̃ ∈ V2) to
obtain

SR−1
m ((B, B̃))2 + Sb(u,B, B̃)− Sb(B, u, B̃) = S⟨Ψ, B̃⟩V ′

2 ,V2
.

And taking the inner product of the third equation of (7.1) with γθ̃ ∈ V3, we obtain

γκ((θ, θ̃))3 + γb̄(u, θ, θ̃)− γ(u2, θ̃) = γ⟨h, θ̃⟩V ′
3 ,V3

.

Then we give the following weak formulation of problem (7.1).

Definition 7.1. A triple (u∗, B∗, θ∗) := z∗ ∈ V with u∗ ∈ L̃r+1 is called a weak
solution of the system (7.1) if

a(z∗, z̃) + σ(z∗, z̃) + B(z∗, z∗, z̃) + α(u∗, ũ) + β⟨Cr(u∗), ũ⟩ = (Φ, z̃),

for all z̃ = (ũ, B̃, θ̃) ∈ V .

Theorem 7.2. Suppose that |Φ|
δ2λ < 1

4cb
(where δ, cb and λ are in Lemmas 2.1,

2.3 and 4.3, respectively). Then, there exists a unique weak stationary solution of
system (1.1).



28 D. T. SON EJDE-2025/30

Proof. Let {wj = (uwj
, Bwj

, θwj
)}j be a Hilbert basis of V such that Vm = span{wj}j

is dense in V1 ∩ L̃r+1 × V2 × V3. For each integer m ≥ 1, we find the approximate
stationary solution in the form

zm(t) =

m∑
j=1

ξmj(t)wj

where
a(zm(t), z̃) + σ(zm(t), z̃) + B(zm(t), zm(t), z̃) + α(um(t), ũ)

+ β⟨Cr(um(t)), ũ⟩ = (Φ, z̃)
(7.2)

for all z̃ ∈ Vm. We apply Lemma 2.5 to prove the existence of zm as follows. Let
Rm : Vm → Vm be defined by

((Rmz, z̃)) = a(z, z̃) + σ(z, z̃) + B(z, z, z̃) + α(u, ũ) + β⟨Cr(u), ũ⟩ − (Φ, z̃)

for all z, z̃ ∈ Vm. For all z ∈ Vm, by using (2.2), Lemma 2.1 and (2.10), we have

((Rmz, z)) = a(z, z) + σ(z, z) + α|u|2 + β∥u∥r+1

L̃r+1
− (Φ, z)

≥ δ

2
∥z∥2 + α|u|2 + β∥u∥r+1

L̃r+1
− 1

λ1/2
∥z∥|Φ|

≥ δ

2
∥z∥2 − 1

λ1/2
∥z∥|Φ|.

Thus, if we take k = 2δ−1λ−1/2|Φ|, then ((Rmz, z)) ≥ 0 for all z ∈ Vm satisfying
∥z∥ = k. Thus, there exists a solution zm ∈ Vm with Rm(zm) = 0. From (7.2), we
replace z̃ with zm to obtain

∥zm∥ ≤ 2δ−1λ−1/2|Φ|, (7.3)

hence we can extract a subsequence of zm (still denoted by zm) such that zm → z∗

weakly in V . Moreover, applying the Aubin-Lions lemma (see [30]), we can conclude
that

|u|r−1u→ |u∗|r−1u∗ weakly in L̃
r+1
r .

Therefore, we have that z∗ is a weak stationary solution to problem (1.1). Now let
z∗1 and z∗2 be two stationary solutions to problem (1.1) and set y∗ = z∗1 − z∗2 , then
by using the same arguments as in proving the uniqueness of solution in Theorem
3.1 and (7.3), we deduce that

δ

2
∥y∗∥2 ≤ B(z∗2 , z∗2 , y∗)− B(z∗1 , z∗1 , y∗) ≤

cb
λ1/2

∥z∗1∥∥y∗∥2 ≤ 2(δλ)−1cb|Φ|∥y∗∥2.

Thus, we obtain (
δ − 4cb

δλ
|Φ|

)
∥y∗∥2 ≤ 0

and obtain the uniqueness of stationary solutions. □

Theorem 7.3. Assume that the assumptions of Theorem 7.2 hold. Then the unique
stationary solution z∗ of problem (1.1) is exponentially stable.

Proof. Notice that we can write the solution z(t) to problem (1.1) in the form
z(t) = z∗ + y(t), by repeating some arguments as in proving the uniqueness of
solution in Theorem 3.1 and (7.3), we deduce that

1

2

d

dt
|y(t)|2 + δ

2
∥y(t)∥2 ≤ cb|y(t)|∥z∗∥∥y(t)∥ ≤ 2cb

δλ
|Φ|∥y(t)∥2.



EJDE-2025/30 GLOBAL ATTRACTOR FOR 2D MAGNETIC BÉNARD PROBLEM 29

Then

d

dt
|y(t)|2 +

(
δ − 4cb

δλ
|Φ|

)
∥y(t)∥2 ≤ 0 or

d

dt
|y(t)|2 + ϑ|y(t)|2 ≤ 0

where ϑ := λ−1
(
δ − 4cb

δλ |Φ|
)
> 0 and this yields

|y(t)|2 = |z(t)− z∗| ≤ |z(0)− z∗|e−ϑt.

The proof is complete. □

Acknowledgements. The authors would like to thank the anonymous referees for
their helpful comments and suggestions.

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Dang Thanh Son

Foundation Sciences Faculty, Telecommunications University, 101 Mai Xuan Thuong,
Nha Trang, Khanh Hoa, Vietnam

Email address: dangthanhson@tcu.edu.vn, dangthanhson1810@gmail.com


	1. Introduction
	2. Preliminaries
	3. Existence and uniqueness of a weak solution
	4. Existence of a global attractor
	5. Dimension of the attractor
	6. Upper semicontinuity of global attractor
	7. Existence and exponential stability of a stationary solution
	Acknowledgements

	References