International Journal of Analysis and Applications

Volume 19, Number 2 (2021), 193-204

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-19-2021-193

GLOBAL EXISTENCE AND UNIQUENESS OF THE WEAK SOLUTION IN

THIXOTROPIC MODEL

AMIRA RAHAI, AMAR GUESMIA∗

Laboratory of Applied Mathematics and History and Didactics of Mathematics (LAMAHIS), Department of

Mathematics, University 20 august 1955 Skikda, Algeria

∗Corresponding author: guesmiaamar19@gmail.com

Abstract. In this paper, we study global existence, uniqueness and boundedness of the weak solution for

the system (P ) which is formulated by two subsystems (P1) and (P2), the first describes the thixotropic

problem and the second describes the diffusion degradation of c, using Galerkin’s method, Lax-Milgran’s

and maximum principle. Moreover we show that the unique solution is positive.

1. Introduction

The phenomenon of thixotropy has recently attracted a great deal of attention. The term was first

applied [3] to an ”isothermal reversible sol-gel transformation”. As the gel state is often merely one of high

viscosity, the definition has been made more general, and the term is then applied [5] to any ” isothermal

reversible decrease of viscosity with increase of rate of shear”.

Colloidal solutions provide the more common examples of thixotropy and may be divided into three important

classes :

• Solutions in Newtonian liquids of lyophilic substances whose molecules are of great length, e.g.,

gelatine, starch and many synthetic polymers.

• Suspensions of solid particles such as pigments in oils, or clays in water.

Received November 10th, 2020; accepted December 7th, 2020; published February 1st, 2021.

2010 Mathematics Subject Classification. Primary 90C57, 90C59 Secondary 90C49.

Key words and phrases. thixotropic; global solution; boundedness; positive solution.

©2021 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

193

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-19-2021-193


Int. J. Anal. Appl. 19 (2) (2021) 194

• Concentrated emulsions ( [11], [12]) of oil droplets in water; foams of gas bubbles in water (with, of

course, stabilising agents).

Thixotropic fluids are used widely in civil engineering, food, cosmetic as well as pharmaceutical industries,

and impact every aspect of our lives. As emulsions, suspensions, or polymeric gels, they are very differ-

ent from each other compositionally, but most of them have one thing in common, i.e., the existence of

microstructures. The microstructures are changeable and may comprise a network of flocculated colloidal

particles, tangles of polymers, or a spatial arrangement of suspended particles or drops [1].

Thixotropic fluids have a lot of special characters, such as aging, rejuvenation, and viscosity bifurcation [14]

and by rate dependent properties associated to their structural level. The behavior of these substances un-

der rheological tests have been analyzed in many scientific works ( [2], [9], [10], [13], [15]), which was firstly

proposed by Moore [8] in 1959. All of these scientific works were presented a qualitative explanation of the

break down and build-up processes of the structure.

In this paper, we are interested in the study of the global existence and uniqueness of weak positive solution

for the elliptic-parabolic model’s.

Our model is defined as follows:

(P)




(P1)




ut + ∆u−λdiv
[
u

∇(c−u0)√
β+|∇(c−u0)|2

]
+ u = u0 (t,x) ∈ R+ × Ω

u = 0 ∂Ω

u (0,x) = u0 x ∈ Ω

(P2)


 −∆c + τc = 0 x ∈ Ωc = g ∂Ω

Where u (t,x) is a function denotes the speed of fluid in the position x ∈ Ω ⊂ R2 or R3, Ω is a bounded

convex domain with smooth boundary ∂Ω ∈ H
3
2 (∂Ω), λ > 0 is the viscosity of the fluid, β > 0 is a parameter

constant, c denotes the concentration of chemical signal that stimulates the fluid. The parameter τ is a time

constant and it is expressed on the one hand the movement of fluid and secondly the diffusion degradation

of c.

To simplify the solution of the system (P), a decomposition of (P) into two subsystem (P1) and (P2) are

adopted. Galerkin’s method is very important to help us to demonstrate the existence and uniqueness of a

weak solution for system (P1) . To prove the existence and uniqueness of a weak solution for system (P2),

we use Lax-Milgram’s theorem and maximum principle. However this theorem can not be applied directly

because it is nonhomogenous system. For this reason an adoptation of Trace theorem it used to simplify the

system(P2) . Therefore we have the existence and uniqueness of a weak solution for system (P). Moreover

we show that the solution is positive.



Int. J. Anal. Appl. 19 (2) (2021) 195

The following initial-boundary conditions on u0 and g assumptions are used to prove the proposed solution

of (P)

• H1 : g ∈ L
1
2 (∂Ω) .

• H2 : g ∈ L
3
2 (∂Ω) .

• H3 : u0 ∈ L2 (Ω) .

• H4 : u0 ≥ 0 and g ≥ 0.

If the hypothesis H1 is satisfies and using the theorem of trace, one can find a lifting of this trace which

we denote R (g) ∈ H10 (Ω) . Thus by definition it verefies γ0 (R (g)) = g. Now we looking for c having the

form c = c̃ + R (g) reduves the problem (P2) to c̃ .

(
P̃2

)
 −∆c̃ + τc̃− ∆R (g) + τR (g) = 0 x ∈ Ωc̃ = 0 on ∂Ω

Definition 1.1. We say (u, c̃) ∈ L2
(
0,T,H10 (Ω)

)
×H10 (Ω) with ut ∈ L2

(
0,T,H−1(Ω)

)
is a weak solution

of the problem (P) if and only if

(1.1) 〈ut,v〉 + B (u,v,t) = (u0,v)

(1.2) a (c̃,q,t) = l (q)

where




B (u,v,t) = −
∫

Ω
∇u∇vdx +

∫
Ω

(δτc + 1) uvdx− δ
∫

Ω
u∇u0∇vdx

a (c̃,q,t) =
∫

Ω
(∇c̃∇q + τc̃q) dx

l (q) = −
∫

Ω
(∇R (g)∇q + τR (g) q) dx

for all (v,q) ∈
(
H10 (Ω)

)2
, 0 ≤ t ≤ T,

(1.3) u (0,x) = u0 ∈ L2 (Ω)

and

(1.4) δ =
λ√

β + |∇(c−u0)|
2
.

Remark 1.1. Note that u ∈ C
(
[0,T] ,L2 (Ω)

)
as u ∈ L2

(
0,T,H10 (Ω)

)
and ut ∈ L2

(
0,T,H−1(Ω)

)
. Then

equation 1.3 makes sense.



Int. J. Anal. Appl. 19 (2) (2021) 196

2. Existence of weak solution of the problem (P)

In this section, we are interested in the study of the existence and uniqueness of weak solution of the

problem (P1), which its variational formulat is given by equation 1.1 using Galerkin’s method and use the

theorem of Lax-Milgram to study the existence and uniqueness of weak solution of the problem (P2), which

its variational formulat is given by equation 1.2 . So we have the existence and uniqueness of weak solution

of the problem (P) .

2.1. Existence of weak solution of the problem (P2).

Theorem 2.1. If the hypothesis H1 holds. Then the problem (P2) has only one solution c ∈ H1 (Ω) for any

q ∈ H1 (Ω) .

By applying the theorem of Lax-Milgram, the solution c̃ of the problem 1.2 exists and it is unique. So

(P2) has unique solution.

Remark 2.1. Elliptic regularity theorem remains valid provided that the boundary condition g is in the space

L
3
2 (∂Ω) which is the image by the operator trace space H2 (Ω) .

Remark 2.2. [17] If c ∈ H2 (Ω) and ( c is a solution of problem (P2)) this implies that c ∈ W 1,q (Ω) (

H2 (Ω) ↪→ W 1,q (Ω) for 1 ≤ q ≤ 2 ).

Using the Maximum Principle one can show that the solution of the problem (P2) is positive as follows.

Multiplying the first equation of (P2) by q ∈ H10 (Ω) , we obtain other variational formulat for problem (P2)

(
P̃3

)∫
Ω

(∇c∇q + τcq) dx = 0.

Proposition 2.1. [16] If g ∈ L
3
2 (∂Ω) and c ∈ H1 (Ω) ∩ C

(
Ω
)
then the problem

(
P̃3

)
have a positive

solution c.

Proof. As ∂Ω is smooth enough and g ∈ L
3
2 (∂Ω) then c ∈ H2 (Ω) . And as Ω ⊂ R2 or R3, by embedding of

Sobolev spaces ( H2
(
Ω
)
↪→ C

(
Ω
)

) this implies that c ∈ C
(
Ω
)
. If c = g ≥ 0 on ∂Ω, then c− = min (c, 0) ∈

H10 (Ω) . So, we have

∫
Ω

cc−dx =

∫
Ω

(
c−
)2
dx

∫
Ω

∇c∇c−dx =
∫

Ω

(
∇c−

)2
dx,

Since the support of functions c− and c+ = max (c, 0) is set A (x) = {x/u (x) = 0} .

This implies that ∇u = 0 on A (x) . As c = c+ + c−, thus we have



Int. J. Anal. Appl. 19 (2) (2021) 197

0 =

∫ ((
∇c−

)2
+ τ

(
c−
)2)

dx ≥ min (1,τ)
∥∥c−∥∥2

H10 (Ω)

Finally, we find c− = 0. �

2.2. Existence of weak solution of the problem (P1).

Lemma 2.1. i) For all v ∈ H10 (Ω) then B (., ., t) is continuous in H10 (Ω) ×H10 (Ω) , there exists a constant

positive M such that

(2.1) |B (u,v,t)| ≤ M ‖u‖H1(Ω) ‖v‖H1(Ω)

ii) For any u ∈ H10 (Ω) and H2 is hold. Then exists a constant positive α such that

(2.2) B (u,u,t) ≥ α‖u‖2H10 (Ω) .

Proof. i) We use the Cauchy-Shwartz inequality and C ∈ H1 (Ω) ↪→ Lq (Ω) for any q ∈
[
1, 2n

n−2

[
with n = 2

or n = 3, we obtain i) as follows

|B (u,v,t)| ≤ ‖∇u‖L2(Ω) ‖∇v‖L2(Ω) +
[
|δτ|‖c‖L2(Ω) + 1

]
‖u‖L2(Ω) ‖v‖L2(Ω)

+ |δ|‖u‖L2(Ω) ‖u0‖L2(Ω) ‖∇v‖L2(Ω)
≤ M ‖u‖H1(Ω) ‖v‖H1(Ω) .

ii) Making use of −∆c + τc = 0 the expression of B (u,u,t) becomes

B (u,u,t) = −
∫

(∇u)2 dx +
∫

(δτc + 1) u2dx− δ
∫
u∇u∇u0dx

= −
∫

(∇u)2 dx +
∫

(δτc + 1) u2dx− δ
2

∫
(∇u)2∇u0dx

=
∫ (
−1 − δ

2
∇u0

)
(∇u)2 dx +

∫
(δτc + 1) u2dx

≥‖∇u‖2L2(Ω) .

Finally, by Poincarre inequality yields,

B (u,u,t) ≥ α‖u‖2H10 (Ω)

�

2.2.1. Galerkin approximations. To demonstrate the existence of weak solution of the problem (P1) via the

method of Galerkin, we assume wk = wk (x) are smooth function verifying

(2.3) {wk}
∞
k=1 is an arthogonal basis of H

1
0 (Ω)

and

(2.4) {wk}
∞
k=1 is an arthonormal basis of L

2 (Ω) .



Int. J. Anal. Appl. 19 (2) (2021) 198

Consider a positive integer m. We will look for a function um : [0,T] → H10 (Ω) of the form

(2.5) um : =

m∑
k=1

dkm (t) wk

which satisfies

(2.6) dkm (0) = (u0,wk)

and

(2.7)
〈
u
′

m,wk

〉
+ B (um,wk, t) = (u0,wk) , 0 ≤ t ≤ T and k = 1, ...,m

where u
′

= ut and here (., .) denotes the scalar product in L
2 (Ω) .

Theorem 2.2. (construction of the approximate solution ) For each integer m, there exists a unique function

um of the form equation 2.5 satisfying equation 2.6 and equation 2.7 .

Proof. Assuming um has the structure equation 2.5. Substituting equation 2.5 into equation 2.7 and using

equation 2.4 we obtained

(2.8) d′km (t) +
∑m
l=1 d

l
mB (wl,wk, t) = d

k
m (0) , 0 ≤ t ≤ T and k = 1, ...,m

According to standard existence theory for ordinary differential equations, there exists a unique absolutely

continuous functions dm (t) =
(
d1m,d

2
m, ...,d

m
m,
)

satisfying equation 2.6 and equation 2.8. So um of the form

equation 2.5 satisfies equation 2.6 and equation 2.7 for all t ∈ [0,T] . �

2.2.2. Energy estimates. We propose now to send m to infinity and show a subsequence of our solutions um

of the approximation problems equation 2.6 and equation 2.7 converges to a weak solution of (P1). For this

we will need some uniform estimates.

Theorem 2.3. ( Energy estimates ) [17]. There exists a constant C, depending only on Ω, T and c, such

that

(2.9) max0≤t≤T ‖um‖L2(Ω) + ‖um‖L2(0,T,H10 (Ω)) + ‖u
′
m‖L2(0,T,H−1(Ω)) ≤ C‖u0‖L2(Ω) for m = 1, 2, ...

Proof. (1) Multiplying equation 2.7 by dkm (t), summing for k = 1, ...,m, and then recalling equation 2.5

we find

(2.10) (u′m,um) + B (um,um, t) = (u0,um)



Int. J. Anal. Appl. 19 (2) (2021) 199

for all 0 ≤ t ≤ T . From Lemme 2.1, there exists constant α > 0 such that

(2.11) α‖um‖
2
H10 (Ω)

≤ B (um,um, t)

for all 0 ≤ t ≤ T, m = 1, ... Furthermore |(u0,um)| ≤ 12 ‖u0‖
2
L2(Ω) +

1
2
‖um‖

2
L2(Ω) , and (u

′
m,um) =

d
dt

(
‖um‖

2
L2(Ω)

)
for a.e. 0 ≤ t ≤ T . Consequently equation 2.10 yields the inequality

(2.12)
d

dt

(
‖um‖

2
L2(Ω)

)
+ 2α‖um‖

2
H10 (Ω)

≤ C1 ‖um‖
2
L2(Ω) + C2 ‖u0‖

2
L2(Ω)

for all 0 ≤ t ≤ T and appropriate constants C1 and C2.

(2) Now write

(2.13) ϕ (t) : = ‖um‖
2
L2(Ω)

and

(2.14) ζ (t) : = ‖u0‖
2
L2(Ω) .

Then equation 2.12 implies

(2.15) ϕ′ (t) ≤ C1ϕ (t) + C2ζ (t)

for a.e. 0 ≤ t ≤ T. Thus the differential form of Gronwall’s inequality yields the estimate

(2.16) ϕ (t) ≤ eC1t
(
ϕ (0) + C2

∫ T
0

ζ (s) ds

)
(0 ≤ t ≤ T) .

Since ϕ (0) = ‖um (0)‖
2
L2(Ω) ≤ ‖u0‖

2
L2(Ω) by equation 2.6, we obtain from equations 2.13 - 2.16

the estimate

(2.17) max
0≤t≤T

‖um‖L2(Ω) ≤ C‖u0‖L2(Ω) .

(3) Integrate inequality equation 2.12 from 0 to T and we employ the inequality equation 2.17 to find

‖um‖
2
L2(0,T,H10 (Ω))

=

∫ T
0

‖um‖
2
H10 (Ω)

dt ≤ C‖u0‖
2
L2(Ω) .



Int. J. Anal. Appl. 19 (2) (2021) 200

(4) Fix any v ∈ H10 (Ω), with ‖v‖
2
H10 (Ω)

≤ 1, and write v = v1 + v2, where v1 ∈ span (wk)
k=m
k=1 , and(

v2,wk
)

= 0 (k = 1, ...,m) . We use equation 2.7, we deduce for all 0 ≤ t ≤ T that

(
u′m,v

1
)

+ B
(
um,v

1, t
)

=
(
u0,v

1
)

then equation 2.5 implies

〈u′m,v〉 = (u
′
m,v) =

(
u′m,v

1
)

=
(
u0,v

1
)
−B

(
um,v

1, t
)
,

consequently

|〈u′m,v〉| ≤ C
(
‖u0‖

2
L2(Ω) + ‖um‖H10 (Ω)

)
.

Simce
∥∥v1∥∥2

H10 (Ω)
≤‖v‖2H10 (Ω) ≤ 1. Thus

‖u′m‖H−1(Ω) ≤ C
(
‖u0‖

2
L2(Ω) + ‖um‖H10 (Ω)

)
,

and therefore

‖u′m‖
2
L2(0,T,H−1(Ω)) =

∫ T
0

‖u′m‖
2
H−1(Ω) dt ≤ C

∫ T
0

(
‖u0‖

2
L2(Ω) + ‖um‖H10 (Ω)

)
dt ≤ C‖u0‖L2(Ω) .

�

2.2.3. Existence and uniqueness. Next we pass to limit as m → ∞, to build a weak solution of our initial

boundary-value problem (P1) .

Theorem 2.4. (Existence of weak solution). Under hypothesis H2 and H3, there exists a weak solution of

(P1) .

Proof. (1) According to the energy estimates equation 2.9, we see that the sequence {um}
∞
m=1 is bounded

in L2
(
0,T,H10 (Ω)

)
and {u′m}

∞
m=1is bounded in L

2
(
0,T,H−1(Ω)

)
. Consequently there exists a

subsequence which is also noted by {um}
∞
m=1 and a function u ∈ L

2
(
0,T,H10 (Ω)

)
, with u′ ∈

L2
(
0,T,H−1(Ω)

)
, such that

(2.18)
um ⇀ u weakly in L

2
(
0,T,H10 (Ω)

)
u′m ⇀ u

′ weakly in L2
(
0,T,H−1(Ω)

)
.



Int. J. Anal. Appl. 19 (2) (2021) 201

(2) Next fix an integer N and choose a function v ∈ C1
(
0,T,H10 (Ω)

)
having the form

(2.19) v (t) =

N∑
k=1

dk (t) wk

where
{
dk
}N
k=1

are given smooth functions. We choose m ≥ N, multiply equation 2.7 by dk (t) ,

sum for k = 1, ...,N, and then integrate with respect to t to find

(2.20)

∫ T
0

〈u′m,v〉 + B (um,v,t) dt =
∫ T

0

(u0,v) dt.

we recall equation 2.18 to find upon passing to weak limits that

(2.21)
∫T

0
〈u′,v〉 + B (u,v,t) dt =

∫T
0

(u0,v) dt ∀v ∈ L2
(
0,T,H10 (Ω)

)
.

As functions of the from equation 2.19 are dense in L2
(
0,T,H10 (Ω)

)
. Hence in particular

(2.22) 〈u′,v〉 + B (u,v,t) = (u0,v) ∀v ∈ H10 (Ω) and∀t ∈ [0,T] ,

and from remark 1.1 we have u ∈ C
(
0,T,L2(Ω)

)
.

(3) In order to prove u (0) = u0, we first note from equation 2.21 that

(2.23)

∫ T
0

−〈u,v′〉 + B (u,v,t) dt =
∫ T

0

(u0,v) dt + (u (0) ,v (0))

for each v ∈ C1
(
0,T,H10 (Ω)

)
with v (t) = 0. Similary, from equation 2.20 we deduce

(2.24)

∫ T
0

−〈um,v′〉 + B (um,v,t) dt =
∫ T

0

(u0,v) dt + (um (0) ,v (0)) .

we use again equation 2.18, we obtain

(2.25)

∫ T
0

−〈u,v′〉 + B (u,v,t) dt =
∫ T

0

(u0,v) dt + (u0,v (0)) ,

since um (0) → u0 in L2(Ω). Comparing equation 2.23 and equation 2.25, we conclude u (0) = u0.

�

Theorem 2.5. (Uniqueness of a weak solutions) A weak solution of (P1) is unique.



Int. J. Anal. Appl. 19 (2) (2021) 202

Proof. We suppose there exists two weak solutions u1 and u2. We put

U = u2 −u1

then U is also a solution of (P1) with U0 = (u2 −u1) (0) ≡ 0. Setting v = U in identity equation 2.19 we

have

d

dt

(
1

2
‖U‖2L2(U) + B (U,U,t)

)
= 0.

From Lemma 2.1 we have B (U,U,t) ≥ α‖U‖2H10 (U) ≥ 0, so
d
dt

( 1
2
‖U‖2L2(U)) ≤ 0, then integrate with

respect to t to find

‖U‖2L2(U) ≤‖U0‖
2
L2(U) = 0,

thus U ≡ 0. �

2.3. Global solution of problem (P). Our main results in this paper are stated as follows.

Theorem 2.6. i) if c > c0 > 0 and B (u,u,t) ≥
∫

Ω
u0udx. then the solution (u,c) of problem (P) is global.

ii) if c > c0 > 0 and B (u,u,t) ≥
∫

Ω
u0udx. then the solution (u,c) of problem (P) is global. Furthermore

there exists σ > 0 such that ‖u‖L2(Ω) ≤ e
σt‖u0‖L2(Ω) .

Proof. we put

(2.26) Z (t) =
1

2

∫
Ω

u2dx

we derivate the equation 2.26 and we use first equations of (P1) and (P2) to find

i) we have

dZ

dt
=

∫
Ω

u0udx−B (u,u,t) ≤ 0

therefore

Z (t) ≤ Z (0) .

ii) we have

dZ
dt

=
∫

Ω
u0udx−B (u,u,t) =

∫
Ω
u0udx +

∫
Ω
∇u2dx−

∫
Ω

(δτc + 1) u2dx + δ
∫

Ω
u∇u∇u0dx

=
∫

Ω
u0udx +

∫
Ω

(
1 + δ∇u0

2

)
∇u2dx−

∫
Ω

(δτc + 1) u2dx

≤ |δτc0 + 1|‖u‖
2
L2(Ω) = σZ (t) .

This implies that



Int. J. Anal. Appl. 19 (2) (2021) 203

Z (t) ≤ Z (0) eσt.

�

Proposition 2.2. [16] Let u0 ∈ L2 (Ω) and u ∈ C
(
[0,T] ; L2 (Ω) ∩L2

(
[0,T] ; H10 (Ω)

))
is the unique weak

positive solution of (P1). If u0 ≥ 0 in Ω, then u ≥ 0 in ]0,T[ × Ω.

Proof. If u0 ≥ 0 on ∂Ω. Therefore u− = min (u, 0) ∈ L2
(
[0,T] ; H10 (Ω)

)
. We obtain for all 0 ≤ t ≤ T

1

2

d

dt

∫
Ω

(
u−
)2
dx +

∫
Ω

B
(
u−,u−, t

)
dx =

∫
Ω

u0u
−dx

Using the Lemma 2.1 and integrating with respect to t from 0 to T, we get

1

2

d

dt

∫
Ω

(
u−
)2
dx + α

∫ T
0

‖u (s)‖2H10 (Ω) ds ≤
1

2

d

dt

∫
Ω

(
u− (0)

)2
dx = 0.

Since u− (0) = (u0)
−

= 0. So u− = 0. �

Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication

of this paper.

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	1. Introduction
	2. Existence of weak solution of the problem ( P) 
	2.1. Existence of weak solution of the problem ( P2) 
	2.2. Existence of weak solution of the problem ( P1) 
	2.3. Global solution of problem ( P) 

	References