Mathematics in Applied Sciences and Engineering https://ojs.lib.uwo.ca/mase
Volume 4, Number 2, June 2023, pp.128-143 https://doi.org/10.5206/mase/15658

ON THE WEAK SOLUTION OF THE VON-KARMAN MODEL WITH

THERMOELASTIC PLATES

JAOUAD OUDAANI AND MUSTAPHA RAÏSSOULI

Abstract. In this article we aim to study the dynamic Von-Karman model coupled with thermoelastic

equations without rotational terms, subject to a thermal dissipation. We establish the existence as

well as the uniqueness of a weak solution related to the dynamic model. At the end, we apply the

finite difference method for approximating the solution of our problem.

1. Introduction

Nonlinear oscillation of an elastic model, for dynamic von-Karman model without rotational terms,

subject to a thermal dissipation [4] describes the phenomenon of small nonlinear vibration with a vertical

displacement to the elastic plates. The case of nonlinear thermoelastic plate interaction coupled with

thermal dissipation plays an interesting place in this subject and will be our fundamental target in the

present paper. The model with clamped boundary conditions, in the note account of rotational terms,

can be formulated as follows ([4]):

Find (u,φ,θ) ∈ L2
(

[0,T] ,H20 (ω)
)
×H20 (ω) ×H10 (ω) such that

(P0)




utt + ∆
2u + µ∆θ − [φ + F0,u] = p(x) in ω × [0,T] ,

kθt −η∆θ −µ∆ut = 0 in ω × [0,T] ,
u|t=0 = u0, (ut)|t=0 = u, θ|t=0 = θ0 in ω,

u = ∂νu = 0 on Γ × [0,T] ,
θ = 0 on Γ,

and

(Q)
{

∆2φ + [u,u] = 0 in ω × [0,T] ,
φ = 0, ∂νφ = 0 on Γ × [0,T] .

Here, u is the displacement, φ denotes the Airy stress function and θ is the thermal function, ω is the

surface plate, u0, u1, θ0 refer to the initial data and [., .] stands for the Monge-Ampère symbol defined

through ([1])

[φ,u] = ∂11φ∂22u + ∂11u∂22φ− 2∂12φ∂12u. (1.1)
The parameters µ,η > 0 are fixed real numbers and k > 0 measures the capacity of the heat/thermal.

The plate is subject to the internal force F0 and the external force p. In [4], the authors studied the

problem of the von-Karman model for the case 0 ≤ k ≤ 1.
Our fundamental target in this paper is to explore a condition that should be satisfied by the

external/internal loads and the initial data for ensuring the existence and the uniqueness of a weak

solution for to the von-Karman evolution, without rotational terms nor clamped boundary conditions,

subject to the thermal dissipation when k > 0 and 0 < µ ≤ 2η. The present approach turns out of to

Received by the editors 31 December 2022; accepted 25 April 2023; published online 9 June 2023.

2020 Mathematics Subject Classification. Primary 74B20,74F10; Secondary 74F05,74K05.

Key words and phrases. Von-Karman model, thermoelastic plate, rotational inertia, finite difference method.

128

https://ojs.lib.uwo.ca/mase
https://dx.doi.org/https://doi.org/10.5206/mase/15658


ON THE WEAK SOLUTION OF THE VON-KARMAN MODEL WITH THERMOELASTIC PLATES 129

construct an iterative process converging, in an appropriate sense, to the unique solution of the initial

problem.

This paper will be organized as follows. In Section 2 we present the mathematical structure of the

model that will be studied in the sequel together with some basic tools and results. In Section 3 we use

an iterative method that will be a good tool for establishing the existence and the uniqueness of a weak

solution of the dynamical plate problem without rotational terms, subject to a thermal dissipation.

Section 4 deals with a numerical simulation for approaching the solution of the initial problem.

2. Preliminaries and main results

Throughout this paper, we denote by ω a nonempty bounded domain in R2 with regular boundary
Γ = ∂ω. We suppose that the parameters k,µ,η in the problem (P0) are such that k > 0 and 0 < µ ≤ 2η.

Let p ≥ 1 be a real number and m ≥ 1 be an integer. The notation |.|p,ω refers to the standard norm
of Lp(ω) while ‖.‖m,ω stands for the classical norm of Hm(ω). For u ∈ H20 (ω), we put

‖u‖ =: |∆u|2,ω =
(∫

ω

(∆u)2
)1

2

,

which is obviously a norm in H20 (ω) ([4, 5]). Otherwise, we set

‖u‖20 =: ‖u‖
2

+ |ut|
2
2,ω . (2.1)

We state the following result which will be needed in the sequel, see [7].

Theorem 2.1. Let f ∈ L2(ω). Then the following problem

(R)




∆2v = f in ω,

v = 0 on Γ,

∂νv = 0 on Γ,

has one and only one solution v ∈ H20 (ω) ∩H4(ω) satisfying

‖v‖≤ c0 |f|2,ω ,

where c0 > 0 is a constant depending only on mes(ω).

Let us mention the following remark.

Remark 2.1. (i) Under the condition that f ∈ L2
(

[0,T] ,L2(ω)
)

, the solution of (R) belongs to the set

L2
(

[0,T] ,H20 (ω) ∩H4(ω)
)

. (ii) We mention again that the constant c0 > 0 in Theorem 2.1 depends

only on mes(ω) and does not depend on f.

The following result will be needed later, see [7, 3].

Theorem 2.2. Let g ∈ L2
(

[0,T] ,L2(ω)
)

, u0 ∈ L2(ω) and k,η,µ > 0. Then the following problem :

(D)




kut −η∆u = µg in ω × [0,T] ,
u|t=0 = u0 in ω,

u = 0 on Γ × [0,T] ,

has one and only one solution u ∈ C
(

[0,T] ; H2(ω) ∩H10 (ω)
)
∩C1

(
[0,T] ; L2(ω)

)
.

We have the following result as well.



130 J. OUDAANI AND M. RAÏSSOULI

Proposition 2.3. Let f ∈ H2(ω), k > 0 and 0 < µ ≤ 2η. Then the solution u of (D), when g = −∆f,
satisfies the following inequality

∀t ∈ [0,T], 0 ≤ k |u|22,ω + (2η −µ)
∫ t
0

|∇u|22,ω ≤ k |u0|
2
2,ω + µ

∫ t
0

|∇f|22,ω . (2.2)

Proof. Since u is the solution of the problem (D), with g = −∆f, then kut − η∆u = −µ∆f and so
k〈ut,u〉−η〈∆u,u〉 = −µ〈∆f,u〉, where 〈., .〉 refers to the standard inner product of L2(ω). This latter
equation is equivalent to

k

2

d

dt
|u|22,ω + η |∇u|

2
2,ω = µ〈∇f,∇u〉. (2.3)

By Hölder inequality in L2(ω) and the standard inequality ab ≤ 1
2
a2 + 1

2
b2, valid for any a,b > 0, we

can write

〈∇f,∇u〉≤
∣∣∣〈∇f,∇u〉∣∣∣ ≤ |∇f|2,ω |∇u|2,ω ≤ 12 |∇f|22,ω + 12 |∇u|22,ω .

Substituting this in (2.3) we get

k

2

d

dt
|u|22,ω + η |∇u|

2
2,ω ≤

µ

2
|∇f|22,ω +

µ

2
|∇u|22,ω .

Integrating side by side this latter inequality with respect to t > 0, and using the fact that (u)|t=0 = u0
in ω, we obtain

k

2
|u|22,ω + η

∫ t
0

|∇u|22,ω ≤
k

2
|u0|

2
2,ω +

µ

2

∫ t
0

|∇f|22,ω +
µ

2

∫ t
0

|∇u|22,ω .

This implies (2.2) and the proof is finished. �

The following result will be needed as well, see [4].

Theorem 2.4. Let f ∈ L2([0,T] ,L2(ω)) and (u0,u1) ∈ H20 (ω) ×L2(ω). Then the problem

(S1)




utt + ∆
2u = f in ω × [0,T] ,

u = ∂νu = 0 on Γ × [0,T] ,
u|t=0 = u0, (ut)|t=0 = u in ω,

has a unique solution such that (u,ut) ∈ C0([ 0,T ] ,H20 (ω) ×L2(ω)).

For the sake of simplicity, we set

F1(u,φ) =
[
φ + F0,u

]
. (2.4)

With this, the following result may be stated.

Proposition 2.5. Let u,v ∈ H20 (ω) be with small norms and F0 ∈ H4(ω) be such that ‖F0‖4,ω <
1
4

.

Let φ,ϕ ∈ H20 (ω) be the solutions of ∆2φ = − [u,u] and ∆2ϕ = − [v,v], respectively. Then there exists
0 < c1 < 1 such that ∣∣∣ [u,φ] − [v,ϕ] ∣∣∣

2,ω
≤ c1‖u−v‖

and ∣∣∣F1(u,φ) −F1(v,ϕ)∣∣∣
2,ω
≤ c1 ‖u−v‖ .

Proof. According to [4], we have∣∣∣ [u,φ] − [v,ϕ] ∣∣∣
2,ω
≤ c0

(
‖u‖2 + ‖v‖2

)
‖u−v‖ ,



ON THE WEAK SOLUTION OF THE VON-KARMAN MODEL WITH THERMOELASTIC PLATES 131

for some c0 > 0 depending only on mes(ω). If we assume that ‖u‖ ≤ c and ‖v‖ ≤ c, for some c > 0
enough small, then we get ∣∣∣ [u,φ] − [v,ϕ] ∣∣∣

2,ω
≤ 2c0c2 ‖u−v‖ .

Otherwise, by using (1.1) it is not hard to check that∣∣∣ [F0,u−v ] ∣∣∣
2,ω
≤ 4‖F0‖4,ω .

It follows that we have∣∣∣F1(u,φ) −F1(v,ϕ)∣∣∣
2,ω

≤
∣∣∣ [φ + F0,u] − [ϕ + F0,v ] ∣∣∣

2,ω
,

≤
∣∣∣ [φ,u] − [ϕ,v ] ∣∣∣

2,ω
+
∣∣∣ [F0,u−v ] ∣∣∣

2,ω
,

≤
(

2c0c
2 + 4‖F0‖4,ω

)
‖u−v‖ .

If ‖F0‖4,ω <
1
4

and

0 < c <

√
1 − 4‖F0‖4,ω

2c0
,

then

0 < 2c0c
2 < c1 =: 2c0c

2 + 4‖F0‖4,ω < 1.
In summary, the proposition is completely proved. �

Remark 2.2. According to Remark 2.1,(ii), the constant c1 in Proposition 2.5 depends only on mes(ω)

and ‖F0‖4,ω.

Now, we are in the position to state and establish the following main result.

Theorem 2.6. Let f ∈ L2
(
[0,T],L2(ω)

)
, θ0 ∈ H10 (ω) and (u0,u) ∈ H20 (ω) × L2(ω). The following

problem:

(S)




utt + ∆
2u + µ∆θ = f in ω × [0,T] ,

kθt −η∆θ = µ∆ut in ω × [0,T] ,
u = ∂νu = θ = 0 on Γ × [0,T] ,
(u)|t=0 = u0, (ut)|t=0 = u, (θ)|t=0 = θ0 in ω,

has one and only one solution (u,θ) ∈ L2([ 0,T ] ,H20 (ω)×H10 (ω)) satisfying that ut ∈ L2([ 0,T ] ,L2(ω))
and, for any t ∈ [0,T],

‖u‖20 + k |θ|
2
2,ω + 2η

∫ t
0

|∇θ|22,ω ≤ e
T
(
‖u0‖

2
+ |u|22,ω + k |θ0|

2
2,ω +

∫ T
0

|f|22,ω
)
. (2.5)

Further, the so-called energy equality holds true:

‖u‖20 + 2η
∫ t
0

|∇θ|22,ω + k |θ|
2
2,ω = ‖u0‖

2
+ |u|22,ω + k |θ0|

2
2,ω + 2

∫ t
0

∫
ω

fut. (2.6)

Proof. To prove our result, we will study the problem (S) by considering the nth-order approximate
solution and its associate variational problem. We divide the proof into fourth steps.

Step 1: Let { ek,e1k } be a basis in the space H
2
0 (ω) × H10 (ω). The n-order Galerkin approximate

solution to the problem (S1), with clamped boundary conditions on the interval [ 0,T ], is a function
(un(t),θn(t)) of the form, [1, 6],

un(t) =

n∑
k=1

hk(t)ek and θ
n(t) =

n∑
k=1

lk(t)e
1
k, n = 1, 2, 3, ...,



132 J. OUDAANI AND M. RAÏSSOULI

where (hk, lk) ∈ W2,∞(0,T; R) ×W1,∞(0,T; R). Let (un,φn,θn) be a solution of (P0) and (Q) corre-
sponding to the initial data (un0,θn0) and un1 such that the two following requirements are satisfied:

(un0,θn0) converges to (u0,θ0) in L
2([ 0,T ] ,H20 (ω) ×H

1
0 (ω)) (2.7)

(un1) converges to u in L
2([ 0,T ] ,L2(ω)). (2.8)

Now, let us consider the iterative problem (Sn) associated to the problem (S) given by:

(Sn)




untt + ∆
2un + µ∆θn = f in ω × [0,T] ,

kθnt −η∆θn = µ∆unt in ω × [0,T] ,
un = ∂νu

n = θn = 0 on Γ × [0,T] ,
(un)|t=0 = un0, (u

n
t )|t=0 = un1, (θ

n)|t=0 = θn0 in ω,

We now multiply the first equation of (Sn) by unt and the second equation by θn and we then integrate
both them over ω, with the help of some standard integral rules, we get



∫
ω

unttu
n
t +

∫
ω

∆un∆unt + µ

∫
ω

∆θnunt =

∫
ω

funt

k

∫
ω

θnt θ
n + η

∫
ω

(∇θn)2 = µ
∫
ω

∆unt θ
n.

Since (unt ,θ
n) ∈ H10 (ω) ×H10 (ω) and

∫
ω

∆θnunt =

∫
ω

θn∆unt , the two last equations imply that




1

2

d

dt

(
|unt |

2
2,ω + ‖u

n‖2
)

+ µ

∫
ω

θn∆unt =

∫
ω

funt ,

k

2

d

dt
|θn|22,ω + η |∇θ

n|22,ω = µ
∫
ω

θn∆unt .

From these two latter equalities we deduce that we have

1

2

d

dt

(
|unt |

2
2,ω + ‖u

n‖2
)

+
k

2

d

dt
|θn|22,ω + η |∇θ

n|22,ω =
∫
ω

funt .

Integrating this latter equality over [0, t], and using (2.1) with the fact that

un|t=0 = un0, (u
n
t )|t=0 = un1, θ

n
|t=0 = θn0,

we get

1

2

(
‖un‖20 + k |θ

n|22,ω
)

+ η

∫ t
0

|∇θn|22,ω =
1

2

(
|un1|

2
2,ω + ‖un0‖

2
+ k |θn0|

2
2,ω

)
+

∫ t
0

∫
ω

funt . (2.9)

Let s ∈ [0,T]. By the Hölder inequality in L2(ω), with (2.1), we have∫ t
0

∫
ω

funt ≤
∫ t
0

|f|2,ω |u
n
t |2,ω

≤
1

2

∫ t
0

|f|22,ω +
1

2

∫ t
0

|unt |
2
2,ω

≤
1

2

∫ T
0

|f|22,ω +
1

2

∫ t
0

(
‖un‖20 + k |θ

n|22,ω + 2η
∫ s
0

|∇θn|22,ω
)
.

(2.10)



ON THE WEAK SOLUTION OF THE VON-KARMAN MODEL WITH THERMOELASTIC PLATES 133

Combining (2.9) and (2.10), we have shown that

‖un‖20 + k |θ
n|22,ω + 2η

∫ t
0

|∇θn|22,ω ≤ |un1|
2
2,ω + ‖un0‖

2
+ k |θn0|

2
2,ω

+

∫ T
0

|f|22,ω +
∫ t
0

(
‖un‖20 + k |θ

n|22,ω + 2η
∫ s
0

|∇θn|22,ω
)
. (2.11)

Step 2: For 0 ≤ s ≤ t, we put

I(s) = ‖un‖20 + k |θ
n|22,ω + 2η

∫ s
0

|∇θn|22,ω .

The inequality (2.11), yields

I(s) −
∫ s
0

I(z)dz = ‖un‖20 + k |θ
n|22,ω + 2η

∫ s
0

|∇θn|22,ω −
∫ s
0

(
‖un‖20 + k |θ

n|22,ω + 2η
∫ z
0

|∇θn|22,ω
)
dz

≤ |un1|
2
2,ω + ‖un0‖

2
+ k |θn0|

2
2,ω +

∫ T
0

|f|22,ω .

It follows that

d

ds

(
e−s

∫ s
0

I(z)dz
)

= e−s
(
I(s) −

∫ s
0

I(z)dz
)

≤ e−s
(
|un1|

2
2,ω + ‖un0‖

2
+ k |θn0|

2
2,ω +

∫ T
0

|f|22,ω
)
.

(2.12)

Now, if we remark that

|un1|
2
2,ω + ‖un0‖

2
+ k |θn0|

2
2,ω +

∫ T
0

|f|22,ω = I(0) +
∫ T
0

|f|22,ω

does not depend on s, and we integrate (2.12) over [0, t], then we get∫ t
0

d

ds

(
e−s

∫ s
0

I(z)dz

)
ds ≤

(∫ t
0

e−sds
)(
|un1|

2
2,ω + ‖un0‖

2
+ k |θn0|

2
2,ω +

∫ T
0

|f|22,ω

)
,

from which we deduce

e−t
∫ t
0

I(z)dz ≤ (1 −e−t)
(
|un1|

2
2,ω + ‖un0‖

2
+ k |θn0|

2
2,ω +

∫ T
0

|f|22,ω
)
.

It follows that∫ t
0

I(z)dz ≤ (1−e
−t)

e−t

(
|un1|

2
2,ω + ‖un0‖

2
+ k |θn0|

2
2,ω +

∫ T
0

|f|22,ω
)

= (et − 1)
(
|un1|

2
2,ω + ‖un0‖

2
+ k |θn0|

2
2,ω +

∫ T
0

|f|22,ω
)

≤ (eT − 1)
(
|un1|

2
2,ω + ‖un0‖

2
+ k |θn0|

2
2,ω +

∫ T
0

|f|22,ω
)
.

This, with (2.11), yields

‖un‖20 + k |θ
n|22,ω + 2η

∫ t
0

|∇θn|22,ω ≤
(
|un1|

2
2,ω + ‖un0‖

2
+ k |θn0|

2
2,ω +

∫ T
0

|f|22,ω
)

+ (eT − 1)
(
|un1|

2
2,ω + ‖un0‖

2
+ k |θn0|

2
2,ω +

∫ T
0

|f|22,ω
)
,

and therefore

‖un‖20 + k |θ
n|22,ω + 2η

∫ t
0

|∇θn|22,ω ≤ e
T
(
|un1|

2
2,ω + ‖un0‖

2
+ k |θn0|

2
2,ω +

∫ T
0

|f|22,ω
)
. (2.13)



134 J. OUDAANI AND M. RAÏSSOULI

According to (2.7) and (2.8), the sequences (un0,θn0) and (un1) are, respectively, bounded in the

spaces L2
(

[0,T] ,H20 (ω) × H10 (ω) × L2(ω)
)

and L2
(

[0,T] ,L2(ω) × L2(ω)
)
. This, with (2.13), imply

that the sequences (un,θn) and (unt ) are also bounded, respectively, in L
2
(

[0,T] ,H20 (ω) × H10 (ω) ×
L2(ω)

)
and L2

(
[0,T] ,L2(ω) × L2(ω)

)
. These latter Banach spaces are reflexive and therefore there

exists a subsequence (unl,θnl ) such that (unl,θnl ) ⇀ (u,θ) weakly in L2
(

[0,T] ,H20 (ω) × L2(ω)
)

and(
(unl )t,∇θnl

)
⇀
(
(u)t,∇θ

)
weakly in L2

(
[0,T] ,L2(ω) ×L2(ω)

)
.

Step 3: In this step, we will establish that (u,θ), previously defined, is a weak solution of the problem

(S), by following the same way as in [8].
Let ϕj ∈ C1(0,T) be such that ϕj(T) = 0 for any 1 ≤ j ≤ j0, and we set

ψ =

j0∑
j=1

ψj ⊗ej, ϕ =
j0∑
j=1

ϕj ⊗e1j.

According to (Sn), with further elementary manipulations and operations, we may infer that

−
∫ T
0

∫
ω
u
nl
t ψt + µ

∫ T
0

∫
ω
∇θnl∇ψ +

∫ T
0

∫
ω

∆unl ∆ψ =

∫ T
0

∫
ω
fψ −

∫
ω
unl1ψ(0) (2.14)

and ∫ T
0

(
−k

∫
ω
θnlϕt + η

∫
ω
∇θnl∇ϕ−µ

∫
ω
∇unl∇ϕt

)
= −k

∫
ω
θnl0ϕ(0) −µ

∫
ω
∇unl1∇ϕ(0). (2.15)

Letting nl → +∞ in (2.14) and (2.15) we deduce that the two following equalities

−
∫ T
0

∫
ω
utψt + µ

∫ T
0

∫
ω
∇θ∇ψ +

∫ T
0

∫
ω

∆u∆ψ =

∫ T
0

∫
ω
fψ −

∫
ω
uψ(0)

and ∫ T
0

(
−k

∫
ω
θϕt + η

∫
ω
∇θ∇ϕ−µ

∫
ω
∇u∇ϕt

)
= −k

∫
ω
θ0ϕ(0) −µ

∫
ω
∇u∇ϕ(0),

hold true for all ψ ∈ L2([0,T],H20 (ω)), with ψt ∈ L2([0,T],H1(ω)) and ϕ ∈ L2([0,T],H10 (ω)), with
ϕt ∈ L2([0,T],L2(ω)), such that ψ(T) = ϕ(T) = 0. This means that (u,θ) is a weak solution of the
problem (S).

Further, by analogous way as for proving (2.13), we may show that for all t ∈ [0,T], we have the
following inequality

‖u‖20 + k |θ|
2
2,ω + 2η

∫ t
0

|∇θ|22,ω ≤ e
T
(
|u|22,ω + ‖u0‖

2
+ k |θ0|

2
2,ω +

∫ T
0

|f|22,ω
)
. (2.16)

Step 4: We now show the uniqueness. Let (u1,θ1) and (u2,θ2) are two solutions of (S). Then
(u1 −u2,θ1 −θ2) is a solution of the following problem


(u1 −u2)tt + ∆2(u1 −u2) + µ∆(θ1 −θ2) = 0 in ω × [0,T] ,
k(θ1 −θ2)t −η∆(θ1 −θ2) = µ∆(u1 −u2)t in ω × [0,T] ,
θ1 −θ2 = u1 −u2 = ∂ν(u1 −u2) = 0 on Γ × [0,T] ,
(u1 −u2)|t=0 = 0, ((u1 −u2)t)|t=0 = 0, (θ1 −θ2)|t=0 = 0 in ω.

According to (2.16) we have

‖u1 −u2‖0 + k |θ1 −θ2|
2
2,ω + 2η

∫ t
0

|∇(θ1 −θ2)|
2
2,ω

≤ eT
(
|u1 −u2|

2
2,ω + ‖(u1)0 − (u2)0‖

2
+ k |(θ1)0 − (θ2)0|

2
2,ω

)
.

Then we deduce u1 = u2 and θ1 = θ2.

Finally, by (2.9) we have for all n ≥ 0

1

2

(
‖un‖0 + k |θ

n|22,ω
)

+ η

∫ t
0

|∇θn|22,ω =
1

2

(
|un1|

2
2,ω + ‖un0‖

2
+ k |θn0|

2
2,ω

)
+

∫ t
0

∫
ω

funt ,



ON THE WEAK SOLUTION OF THE VON-KARMAN MODEL WITH THERMOELASTIC PLATES 135

from which by letting n → +∞ we get (2.6), so completing the proof. �

3. Iterative approach and convergence result

To establish the existence and the uniqueness of a weak solution for (P0), without rotational terms
i.e. α = 0, we will use the following iterative approach.

Let n ≥ 2 and let 0 6= u1 ∈ H20 (ω) be given. We define φn−1 ∈ H20 (ω) as the unique solution of
∆2φn−1 = − [un−1,un−1] and (un,θn) as the solution of the following problem :

(Pn)




(un)tt + ∆
2un = F(un−1,φn−1,θn) in ω × [0,T] ,

k(θn)t −η∆θn = µ∆(un)t in ω × [0,T] ,
un = ∂νun = θn = 0 on Γ × [0,T] ,
(un)|t=0 = u0, ((un)t)|t=0 = u, (θn)|t=0 = θ0 in ω,

where we set F(u,φ,θ) = F1(u,φ) −µ∆θ + p and, F1 is defined by (2.4).
The main result of this section is recited in the following.

Theorem 3.1. Let p ∈ L2(ω), (u0,u) ∈ H20 (ω) × L2(ω) and θ0 ∈ H10 (ω). We suppose that all the
following norms

‖F0‖4,ω , |p|2,ω , ‖u0‖
2
, |u|22,ω and ‖θ0‖

2
1,ω

are small enough, and 0 < µ ≤ 2η. Then the problem (P0), without rotational forces, has one and only
one weak solution (u,φ,θ) in L2

(
[0,T] ,H20 (ω) ×H20 (ω) ×H10 (ω)

)
such that ut ∈ L2

(
[0,T] ,L2(ω)

)
.

Proof. We divide it into four steps.

Step 1: Let us consider the problem (Pn) with 0 6= u1 does not depend on t. For the sake of simplicity
we use the notation

‖(u,θ)‖∗ = ‖u‖
2
0 + k |θ|

2
2,ω + 2η

∫ t
0

|∇θ|22,ω ,

where, ‖.‖0 is defined by (2.1). Let c0 > 0 be the constant defined by Proposition 2.5. For ‖F0‖4,ω <
1
4

we choose c =: c(‖F0‖4,ω ,c0,T) > 0 such that

0 < 4c0c < 1, and 0 < c <

√
1 − 4‖F0‖4,ω

2c0
.

We also choose u1 (independent on t) such that 0 < ‖u1‖2,ω < c < 1.
By using an induction method, we will prove that the two following inequalities

‖u‖20 =: ‖un‖
2

+ |(un)t|
2
2,ω ≤‖u1‖

2
2,ω and ‖φn‖2,ω ≤‖u1‖2,ω (3.1)

are satisfied for all n ≥ 1 and t ∈ [0,T].
Since u1 does not depend on t, then we have

‖u1‖
2
0 =: ‖u1‖

2
+ |(u1)t|

2
2,ω = ‖u1‖

2
2,ω .

Now, let φ1 be the solution of ∆
2φ1 = − [ u1,u1 ]. Theorem 2.1 tells us that there exists c0 > 0 such

that

‖φ1‖2,ω ≤ c0 |[ u1,u1 ]|1,ω ,
and by using the same way as in the proof of Proposition 2.5, with ‖u1‖2,ω < c and 0 < 4c0c < 1, we
may deduce that

‖φ1‖2,ω ≤ 4c0 ‖u1‖
2
2,ω ≤ 4c0c‖u1‖2,ω ≤‖u1‖2,ω .

Hence, the inequalities (3.1) are satisfied for n = 1.

Assume that for k = 2, ...,n and t ∈ [0,T], we have

‖uk‖
2
0 ≤‖u1‖

2
2,ω and ‖φk‖2,ω ≤‖u1‖2,ω .



136 J. OUDAANI AND M. RAÏSSOULI

According to Theorem 2.1 and Proposition 2.5, with Remark 2.1 and Remark 2.2, we have

‖φn‖2,ω ≤ c0 |[ un,un ]|1,ω ≤ 4c0 ‖un‖
2 ≤ 4c0c‖un‖≤ c1 ‖un‖ .

Since un+1 is a solution of (Pn+1), Proposition 2.6, Proposition 2.5 and Theorem 2.1 imply that, there
exist c1 > 0, with

0 < c1 =: 2c0c
2 + 4‖F0‖4,ω < 1, (3.2)

such that

‖(un+1,θn+1)‖∗ ≤ e
T
(
‖u0‖

2
+ k |θ0|

2
2,ω + |u|

2
2,ω +

∫ T
0

|F1(un,φn) + p|22,ω
)

≤ eT
(
‖u0‖

2
+ k |θ0|

2
2,ω + |u|

2
2,ω + 2

∫ T
0

(
|F1(un,φn)|22,ω + |p|

2
2,ω

))
≤ eT

(
‖u0‖

2
+ k |θ0|

2
2,ω + |u|

2
2,ω + 2c

2
1

∫ T
0

‖un‖
2

+ 2T |p|22,ω
)

≤ eT
(
‖u0‖

2
+ |u|22,ω + k |θ0|

2
2,ω + 2Tc

2
1 ‖u1‖

4
2,ω + 2T |p|

2
2,ω

)
.

This, with the fact that 0 < c1 < 1, ‖u1‖ < 1 and c21 ‖u1‖
4
2,ω ≤ c1 ‖u1‖

2
2,ω, implies that

‖(un+1,θn+1)‖∗ ≤ e
T
(
‖u0‖

2
+ |u|22,ω + k |θ0|

2
2,ω + 2Tc1 ‖u1‖

2
2,ω + 2T |p|

2
2,ω

)
.

If we choose c > 0 and ‖F0‖4,ω small enough then c1 defined by (3.2) is also small enough and so
0 < c2 =: 2Te

Tc1 < 1. We can then write

‖(un+1,θn+1)‖∗ ≤ e
T
(
‖u0‖

2
+ |u|22,ω + k |θ0|

2
2,ω + 2T |p|

2
2,ω

)
+ c2 ‖u1‖

2
2,ω . (3.3)

In another part we can write

‖u0‖
2

+ |u|22,ω + k |θ0|
2
2,ω + 2T |p|

2
2,ω ≤

(1 − c2)
eT

‖u1‖
2
2,ω , (3.4)

since the left quantity of this inequality was assumed to be small enough. Otherwise, it is not hard to

check that

‖un+1‖
2
0 =: ‖un+1‖

2
+ |(un+1)t|

2
2,ω ≤‖(un+1,θn+1)‖∗ (3.5)

and

‖φn‖2,ω ≤ c1 ‖un‖2,ω ≤‖u1‖2,ω .
According to (3.3), (3.5) and (3.5) we deduce that we have

‖un+1‖
2
0 ≤ e

T
(
‖u0‖

2
+ |u|22,ω + k |θ0|

2
2,ω + 2T |p|

2
2,ω

)
+ c2 ‖u1‖

2
2,ω

≤ eT
(1 − c2)
eT

‖u1‖
2
2,ω + c2 ‖u1‖

2
2,ω = ‖u1‖

2
2,ω .

Furthermore, we have

‖φn+1‖2,ω ≤ c0 |[ un+1,un+1 ]|1,ω ,
which with, ‖u1‖2,ω < c and 0 < 4c0c < 1, immediately yields

‖φn+1‖2,ω ≤ 4c0 ‖un+1‖
2 ≤ 4c0 ‖u1‖

2
2,ω ≤ 4c0c‖u1‖2,ω ≤‖u1‖2,ω .

In summary, we have shown that for all n ≥ 1 and any t ∈ [0,T] one has

‖un‖
2
0 ≤‖u1‖

2
2,ω and ‖φn‖2,ω ≤‖u1‖2,ω .

Moreover we have

k |θn|
2
2,ω + 2η

∫ t
0

|∇θn|
2
2,ω ≤‖(un,θn)‖∗ ≤‖u1‖

2
2,ω .



ON THE WEAK SOLUTION OF THE VON-KARMAN MODEL WITH THERMOELASTIC PLATES 137

Step 2: For n ≥ 2, let (un,θn) be a solution of (Pn). Let 2 ≤ m ≤ n. It is not hard to see that
θnm =: θn −θm and unm =: un −um satisfy the following:


unmtt + ∆

2unm + µ∆θnm = F1(un−1,φn−1) −F1(um−1,φm−1) in ω × [0,T] ,
kθnmt −η∆θnm = µ∆(unm)t in ω × [0,T] ,
unm = θnm = ∂νu

nm = 0 on Γ × [0,T] ,
(unm)|t=0 = ((u

nm)t)|t=0 = ((θ
nm)t)|t=0 = 0 in ω.

According to Proposition 2.5 and Theorem 2.1 we deduce that

‖φn−1 −φm−1‖2,ω ≤ 4c0c‖un−1 −um−1‖ .

Using Proposition 2.6 and Proposition 2.5 again we have

‖(un −um,θn −θm)‖∗ ≤ e
T

∫ T
0

|F1(un−1,φn−1) −F1(um−1,φm−1)|
2
2,ω

≤ c1eT
∫ t
0

‖un−1 −um−1‖
2
.

This, with 0 < c3 = e
Tc1 <

1
T

, yields

‖(un −um,θn −θm)‖∗ ≤ c3
∫ t
0

‖(un−1 −um−1,θn−1 −θm−1)‖∗

≤ (c3)m−2
∫ t
0

...

∫ t
0

‖(un−m+2 −u1,θn−m−2 −θ1‖∗

≤ (c3)m−2
∫ t
0

...

∫ t
0

n−m+1∑
k=0

(c3)
k

∫ t
0

...

∫ t
0

‖(u2 −u1θ2 −θ1)‖∗

≤ (c3)m−2
∫ t
0

...

∫ t
0

n−m+1∑
k=0

(c3)
k

∫ t
0

...

∫ t
0

(
‖(u2,θ2)‖∗ + ‖(u1,θ1)‖∗

)
≤ (c3T)m−2

n−m+1∑
k=0

(c3T)
k
(
4‖u1‖

2
2,ω

)
.

It follows that we have∫ T
0

‖(un −um,θn −θm)‖∗ ≤ T(c3T)
m−2

n−m+1∑
k=0

(c3T)
k(4‖u1‖

2
2,ω),

and so we infer that

‖φn −φm‖2,ω ≤ 4c0c‖un −um‖ .
The sequence (un,φn−1)n≥2 is a Cauchy sequence in the Banach space L

2
(

[0,T] ,H20 (ω) ×H20 (ω)
)
.

It follows that (un,φn−1) converges to (u,φ) in L
2
(

[0,T] ,H20 (ω)×H20 (ω)
)

and (un)t converges to (u)t
in L2

(
[0,T] ,L2(ω)

)
.

Step 3: Now, let us rewrite that θnm =: θn −θm is a solution of the following problem


kθnmt −η∆θnm = µ∆(unm)t in ω × [0,T] ,
θnm = 0, on Γ × [0,T] ,
((θnm)t)|t=0 = 0 in ω.

Using Theorem 2.2, Proposition 2.3 and inequality (2.2), we have

k |θn−1 −θm−1|22,ω + (2η −µ)
∫ t
0
|∇(θn−1 −θm−1)|22,ω ≤ µ

∫ t
0

(|∇(un−1 −um−1)t|2,ω)
2.



138 J. OUDAANI AND M. RAÏSSOULI

We deduce that (θn) is a Cauchy sequence in the Banach space L
2([0,T] ,H10 (ω)) and so (θn) converges

to θ in L2([0,T] ,H10 (ω)). Otherwise, by Proposition 2.5 we may deduce that F1(un−1,φn−1) converges

to F1(u,φ) in (L
2(ω))2.

Thanks to Theorem 2.4, we have (un, (un)t) ∈ C0([0,T] ,H20 (ω) × L2(ω)) with (un)|t=0 = u0 and
((un)t)|t=0 = u1, and so (u)|t=0 = u0, ((u)t)|t=0 = u.

For showing that (u,θ) is a weak solution of the problem (P0), we follow the same way as in [8].
Let

{
ej,e

1
j

}
be a basis in the space H20 (ω) × H10 (ω) and let ϕj ∈ C1(0,T) , 1 ≤ j ≤ j0, be such that

ϕj(T) = 0. We set

ψ =

j0∑
j=1

ϕj ⊗ej, ϕ =
j0∑
j=1

ϕj ⊗e1j.

As in the proof of Theorem 2.6 (Step 3), we have

−
∫ T
0

∫
ω

(un)tψt +µ

∫ T
0

∫
ω

∇θn∇ψ+
∫ T
0

∫
ω

∆un∆ψ =

∫ T
0

∫
ω

(F1(un−1,φn−1)+p)ψ−
∫
ω

u1ψ(0) (3.6)

and ∫ T
0

(
−k

∫
ω

θnϕt + η

∫
ω

∇θn∇ϕ−µ
∫
ω

∇un∇ϕt
)

= −k
∫
ω

θ0ϕ(0) −µ
∫
ω

∇u1∇ϕ(0). (3.7)

Letting n →∞ in (3.6) and (3.7) we deduce that, for all ψ ∈ L2([0,T],H20 (ω)), ψt ∈ L2([0,T],L2(ω)),
ϕ ∈ L2([0,T],H10 (ω)) and ϕt ∈ L2([0,T],L2(ω)) with ψ(T) = ϕ(T) = 0, we have

−
∫ T
0

∫
ω

utψt + µ

∫ T
0

∫
ω

∇θ∇ψ +
∫ T
0

∫
ω

∆u∆ψ =

∫ T
0

∫
ω

(F1(u,φ) + p)ψ −
∫
ω

u1ψ(0),

and ∫ T
0

(
−k

∫
ω

θϕt + η

∫
ω

∇θ∇ϕ−µ
∫
ω

∇u∇ϕt
)

= −k
∫
ω

θ0ϕ(0) −µ
∫
ω

∇u1∇ϕ(0).

Hence, (u,θ) is a weak solution of the problem (S1) with f = F1(u,φ) + p.

Summarizing, we have proved that (u,φ,θ) is a weak solution of the thermoelastic von-Karman

evolution.

Step 4: We now prove the uniqueness. Assume that there exist two weak solutions (u1,φ1,θ1) and

(u2,φ2,θ2) in L2([0,T] ,H10 (ω) × H20 (ω) × H10 (ω)) such that, for some c > 0 small enough, we have∥∥u1∥∥ ≤ c and ∥∥u2∥∥ ≤ c. Then u12 =: u1 −u2 and θ12 =: θ1 −θ2 satisfy the following problem
(P3)




u12tt + ∆
2u12 = F(u1,φ1,θ1) −F(u2,φ2,θ2) in ω × [0,T] ,

kθ12t −η∆θ12 = µ∆u12t in ω × [0,T] ,
u12 = ∂νu

12 = θ12 = 0 on Γ × [0,T] ,
u12(x1,x2, 0) = 0, (u

12)t(x1,x2, 0) = 0 in ω.

(θ12)t(x1,x2, 0) = 0 in ω.

It follows that (u1 −u2,θ1 −θ2) is a solution of the problem (P3). Proposition 2.5, Proposition 2.6 and
Theorem 2.1 ensure that there exists c1 > 0 such that

∥∥(u1 −u2,θ1 −θ2)∥∥∗ ≤ eT
∫ T
0

∣∣F1(u1,φ1) −F1(u2,φ2)∣∣22,ω
≤ c1eT

∫ T
0

∥∥u1 −u2∥∥2 ≤ c1eT ∫ T
0

∥∥(u1 −u2,θ1 −θ2)∥∥∗ .



ON THE WEAK SOLUTION OF THE VON-KARMAN MODEL WITH THERMOELASTIC PLATES 139

Since c1 is small enough so 0 < c3 = e
Tc1 <

1
T

and therefore∫ T
0

∥∥(u1 −u2,θ1 −θ2)∥∥∗ ≤ Tc3
∫ T
0

∥∥(u1 −u2,θ1 −θ2)∥∥∗ ,
which, with 0 < Tc3 < 1, immediately yields u

1 = u2, θ1 = θ2 and then φ1 = φ2.

In conclusion, the dynamic von-Karman equations coupled with thermal dissipation, without rota-

tional inertia, has one and only one weak solution (u,φ,θ) in L2
(

[0,T] ,H20 (ω)×H20 (ω)×H10 (ω)
)

. The

proof of the theorem is finished. �

We end this section by stating the following result.

Proposition 3.2. Let (u,φ,θ) ∈ L2
(

[0,T] ,H20 (ω) ×H20 (ω) ×H10 (ω)
)

be the unique solution of (P0).
Let φ0 ∈ H20 (ω) be the unique solution of ∆2φ0 = − [u0,u0]. Then the following equality

Ẽ
(
u(t),ut(t),φ

)
+ k |θ|22,ω + 2η

∫ t
0

|∇θt|
2
2,ω = Ẽ1(u0,u,φ0) + k |θ0|

2
2,ω ,

holds true for any t ∈ [0,T], where we set

Ẽ
(
u(t),ut(t),φ

)
=: |ut|

2
2,ω + ‖u‖

2
+

1

2

∫
ω

(
|∆φ|2 − 2 [u,F0] u− 4pu

)
and

Ẽ1(u0,u,φ0) =: |u|
2
2,ω + ‖u0‖

2
+

1

2

∫
ω

(
|∆φ0|

2 − 2 [u0,F0] u0 − 4pu0
)
.

Proof. By virtue of Theorem 2.6, for any t ∈ [0,T] we have the following equality

‖u‖20 + 2η
∫ t
0

|∇θ|22,ω + k |θ|
2
2,ω = ‖u0‖

2
+ |u|22,ω + k |θ0|

2
2,ω

+ 2

∫ t
0

∫
ω

F1(u,φ)ut + 2

∫ t
0

∫
ω

p(x1,x2)ut.

(3.8)

First, let us observe that we have∫ t
0

∫
ω

p(x1,x2)ut =

∫
ω

p(x1,x2)u(t) −
∫
ω

p(x1,x2)u0.

Otherwise, with ∆2φ = [u,u], we have∫ t
0

∫
ω

F1(u,φ)ut =

∫ t
0

∫
ω

[u,φ + F0] ut =

∫ t
0

∫
ω

[u,φ] ut +

∫ t
0

∫
ω

[u,F0] ut,

=
1

2

∫ t
0

∫
ω

d

dt
([u,u] φ) +

1

2

∫ t
0

∫
ω

d

dt
([u,F0] u),

= −
1

4

∫
ω

|∆φ|2 +
1

4

∫
ω

|∆φ0|
2

+
1

2

∫
ω

[u,u] F0 −
1

2

∫
ω

[u0,u0] F0.

Substituting these into (3.8), we get

‖u‖20 + 2η
∫ t
0

|∇θ|22,ω + k |θ|
2
2,ω +

1

2

∫
ω

|∆φ|2 −
∫
ω

[u,u] F0 − 2
∫
ω

p(x1,x2)u

= ‖u0‖
2

+ |u|22,ω + k |θ0|
2
2,ω +

1

2

∫
ω

|∆φ0|
2 −

∫
ω

[u0,u0] F0 − 2
∫ t
0

∫
ω

p(x1,x2)u0.

The proof of the proposition is finished. �



140 J. OUDAANI AND M. RAÏSSOULI

4. Numerical application

In this section we will investigate a numerical resolution of our initial problem in the aim to illustrate

the previous study.

4.1. Preliminaries. We take ω =]0, 1[×]0, 1[⊂ R2 and let T > 0. For solving numerically the problem
(P0), we use the finite difference method by considering a uniform mesh of width h. For this, let us de-
note by ωh the set of all mesh points inside the domain ω with internal points: (xi,yj) = (ih,jh), i,j =

1, ...N−1, h = 1/(N + 1), ∆t = 1/T. Otherwise, we denote by ωh the set of boundary mesh points and
by uh the finite-difference that approximates u. In [2], the author discussed a numerical study about

the convergence and stability for the conservative finite difference schemes related to the dynamic von

Karman plate equations.

In the aim to approximate numerically the unique weak solution of our problem, we use the discrete

model of von-Karman evolution presented in [2, 9]:

(∗)




δ2t u
n
ij + µ(δ

2
x + δ

2
y)θ

n
ij + ∆

2
hu

n
ij = [ u

n
ij v

n
ij + Fij ] + pij in ωh,

kδtθ
n
ij −η(δ

2
x + δ

2
y)θ

n
ij −µδt(δ

2
x + δ

2
y)u

n
ij = 0 in ωh,

∆2hv
n
ij = − [ u

n
ij u

n
ij ] in ωh,

u0ij = (ϕ0)ij, δtu
0
ij = (ϕ1)ij, θ

0
ij = (θ0)ij in ωh,

unij = v
n
ij = θ

n
ij = 0 on ωh,

∂νu
n
ij = ∂νv

n
ij = 0 on ωh,

with the following discrete differential operators:

δ2t u
n
ij =

un+1ij − 2u
n
ij + u

n−1
ij

(∆t)2
,

δtu
n
ij =

un+1ij −u
n
ij

∆t
,

∆2hu
n
ij = h

−4 [ uij−2 + uij+2 + ui−2j + ui+2j − 8(uij−1 + uij+1 + ui−1j + ui+1j)
+ 2(ui−1j−1 + ui−1j+1 + ui+1j−1 + ui+1j+1) − 20uij ] ,

δ2xu
n
ij =

uni+1j − 2u
n
ij + u

n
i−1j

h2
,

δ2yu
n
ij =

unij+1 − 2u
n
ij + u

n
ij−1

h2
,

δ2xyu
n
ij =

uni+1j+1 −u
n
i+1j−1 −u

n
i−1j+1 + u

n
i−1j−1

(2h)2
,

[ unij, v
n
ij ] = δ

2
xu

n
ijδ

2
yv
n
ij − 2δ

2
xyu

n
ijδ

2
xyv

n
ij + δ

2
yu

n
ijδ

2
xv
n
ij.

Summarizing the above, we have in fact transformed the above problem to the numerical resolution

into 2 steps, as itemized below:

Step 1: We first utilize the numerical procedure of 13-point formula of finite difference discussed in [6].

This method is used for illustrating the weak solution of the next problem:


∆2v = f1 in ω,

v = g1 on Γ,

∂νv = g2 on Γ.

Step 2: Afterwards, we adopt the discrete model of the von-Karman evolution (∗) for approaching the
solution of the thermoelastic model coupled with the dynamic von-Karman evolution.



ON THE WEAK SOLUTION OF THE VON-KARMAN MODEL WITH THERMOELASTIC PLATES 141

4.2. Non-coupled approach. In [6], the author discussed a numerical analysis of finite-difference

method about the numerical resolution of the Biharmonic equation. Such method, which is known as

the non-coupled method of 13-point, may be summarized by the following result:

Proposition 4.1. The 13-point approximation of the Biharmonic equation for approaching the unique

solution v of the problem (P) is defined by:

Lhvij = h
−4
{
vij−2 + vij+2 + vi−2j + vi+2j − 8

(
vij−1 + vij+1 + vi−1j + vi+1j

)
+ 2
(
vi−1j−1 + vi−1j+1 + vi+1j−1 + vi+1j+1

)
− 20vij

}
= f1

(
xi,yj

)
,

for i,j = 1, 2, ...,N − 1, where we set vij = v(xi,yj).

When the mesh point (xi,yj) is adjacent to the boundary ωh, then the undefined values of vh are

conventionally calculated by the following approximation of ∂νv:

vi−2,j =
1

2
vi+1,j −vij +

3

2
vi−1,j −h(∂xv)i−1,j,

vi,j−2 =
1

2
vi,j+1 −vij +

3

2
vi,j−1 −h(∂yv)i,j−1,

vi+2,j =
1

2
vi+1,j −vij +

3

2
vi−1,j −h(∂xv)i+1,j,

vi,j+2 =
1

2
vi,j+1 −vij +

3

2
vi,j−1 −h(∂yv)i,j+1.

The following example illustrates the previous theoretical study.

Example 1. Let us consider the following body forces:

p(x,y) = 10−2(x− 1)2(y − 1)2(e−x
2−y2 ),

u1 = 10
−3(y3(x− 4)2)(e−x

2−y2 ),

θ0 = 10x
2(x−y − 1)(e−(x−1)

2−(y−1)2 ),

u0 = 10
−3x2(y − 3)3(e−x

2−y2 ),

F0 = 10
−3x(e−x

2−y2 ) sin2(πx).

Displacement of plate, T = 0.1s Displacement of plate, T = 50s



142 J. OUDAANI AND M. RAÏSSOULI

Thermal value, T = 0.1s Thermal value, T = 50s

Contour displacement valueT = 50s Contour thermal valueT = 50s

Acknowledgements: The authors would like to thank the anonymous referee for his/her valuable

comments which improved the final version of this manuscript.

References

[1] S. S. Antman and T. Von-Karman, A panorama of Hungarian mathematics in the twentieth centuray I, Bolyai Soc.

Math. Studies 14(2006), 373-382.

[2] S. Bilbao, A family of conservative finite difference schemes for the dynamical von Karman plate equations, Numer.

Meth. Partial. Diff. Eqns. 24(1) (2007), 193-218. https://www.doi.org/10.1002/num.20260.

[3] H. Brezis, Analyse Fonctionnelle, Théorie et Application., Masson, Paris (1983).

[4] I. Chueshov and I. Lasiecka, Von Karman Evolution, Well-posedness and Long Time Dynamics, New York, Springer,

2010.

[5] P. G. Ciarlet and R. Rabier, Les Equations de von Karman., Lecture Notes in Mathematics., Vol. 826, New York,

Springer, 1980.

[6] M. M. Gubta and R. P. Manohar, Direct solution of Biharmonic equation using non coupled approach, J. Computa-

tional Physics 33/2 (1979) 236-248. http://dx.doi.org/10.1016/0021-9991(79)90018-4.

[7] J. L. Lions., E. Magenes, Problèmes aux Limites non Homogènes et Applications., Vol.1, Gauthier-Villars, Dunod,

Paris (1968).

[8] J. L. Lions, Quelques Methodes de Résolution des Problèmes aux Limites non Linéaires., Dunod, Paris (2002).

[9] D. C. Pereira, C. A. Raposo, A. J. Avila, Numerical solution and exponential decay to von Karman system with fric-

tional damping, Int. J. Math. Information Sci. 8(4) (2014), 1575-1582. http://dx.doi.org/10.12785/amis/080411.

https://www.doi.org/10.1002/num.20260
http://dx.doi.org/10.1016/0021-9991(79)90018-4
http://dx.doi.org/10.12785/amis/080411


ON THE WEAK SOLUTION OF THE VON-KARMAN MODEL WITH THERMOELASTIC PLATES 143

J. Oudaani, Ibn Zohr University, Poly-Disciplinary Faculty, Department of Mathematics and Managment,

Code Postal 638, Ouarzazate, Morocco.

Email address: oudaani1970@gmail.com

M. Räıssouli, corresponding author, Department of Mathematics, Science Faculty, Moulay Ismail Univer-

sity, Morocco.

Email address: raissouli.mustapha@gmail.com


	1. Introduction
	2. Preliminaries and main results
	3. Iterative approach and convergence result
	4. Numerical application
	4.1. Preliminaries
	4.2. Non-coupled approach

	References