Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 16, pp. 1–16.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

BOUNDEDNESS AND GLOBAL SOLVABILITY FOR A

CHEMOTAXIS-HAPTOTAXIS MODEL WITH

p-LAPLACIAN DIFFUSION

CHANGCHUN LIU, PINGPING LI

Abstract. We consider a chemotaxis-haptotaxis system with p-Laplacian dif-

fusion in three dimensional bounded domains. It is asserted that for any p > 2,

under the appropriate assumptions, the chemotaxis-haptotaxis system admits
a global bounded weak solution if for initial data satisfies certain conditions.

1. Introduction

In this article, we study the problem

ut = div(|∇u|p−2∇u)− χ∇ ·
( u

(1 + u)α
∇v
)
− ξ∇ ·

( u

(1 + u)β
∇w
)

+ µu(1− u− w), x ∈ Ω, t > 0,

vt −∆v + v = f(u)g(w), x ∈ Ω, t > 0,

wt = −vw, x ∈ Ω, t > 0,

(|∇u|p−2∇u− χ u

(1 + u)α
∇v − ξ u

(1 + u)β
∇w) · ~n|∂Ω =

∂v

∂~n

∣∣
∂Ω

= 0,

u(x, 0) = u0(x), v(x, 0) = v0(x), w(x, 0) = w0(x), x ∈ Ω,

(1.1)

where Ω is a bounded domain in R3 with the smooth boundary, p > 2, χ, ξ ≥ 0,
µ > 0, α > 0, β ≥ 0, f and g satisfy

|f(s)| ≤ K1|s|q1 +K2, |g(s)| ≤ K3|s|q2 +K4, 0 < q1 ≤ 1, 0 < q2 <∞, (1.2)

with some positive constants Ki. u represent the cancer cell density, v is the matrix
degrading enzyme concentration and w stands for the extracellular matrix density.
A chemotaxis model was first introduced by Keller and Segel [4] in 1970, later,
many modified chemotaxis models have been widely studied by many researchers.
Recently, Chen and Tao [1] considered the chemotaxis-haptotaxis system

ut = ∆u− χ∇ · (u∇v)− ξ∇ · (v∇w) + µu(1− u− w), x ∈ Ω, t > 0,

vt = ∆v − v + f(u)g(w), x ∈ Ω, t > 0,

wt = −vw + ηw(1− u− w), x ∈ Ω, t > 0.

2010 Mathematics Subject Classification. 92C17, 35K65, 35K92.

Key words and phrases. Chemotaxis-haptotaxis; p-Laplacian diffusion; global solution.
c©2020 Texas State University.

Submitted February 12, 2019. Published February 10, 2020.

1



2 C. LIU, P. LI EJDE-2020/16

They showed that for any given suitably regular initial data the problem posses a
unique global-in-time classical solution which is uniformly bounded (see also Tao
and Winkler [11]).

Xu, Zhang and Jin [13] studied the problem

ut = ∆um − χ∇ ·
( u

(1 + u)α
· ∇v

)
− ξ∇ ·

( u

(1 + u)β
· ∇w

)
+ µu(1− u− w), x ∈ Ω, t > 0,

vt −∆v + v = u, x ∈ Ω, t > 0,

wt = −vw, x ∈ Ω, t > 0.

It is shown that under zero-flux boundary conditions, for any m > 0, the above
problem admits a global bounded weak solution. They also discussed the large time
behavior of solutions for the fast diffusion case, and showed that if 0 < m ≤ 1, for
appropriately large µ, for any initial datum, the solution (u, v, w) goes to a steady
(1, 1, 0) as t→∞.

Zheng [15] investigated the Keller-Segel-Stokes system with nonlinear diffusion

nt + u · ∇n = ∆nm −∇ · (n∇c),
ct + u · ∇c = ∆c− c+ n,

ut +∇P = ∆u+ n∇φ,
∇u = 0.

He proved that if m > 4/3, then for any sufficiently regular nonnegative initial data
there exists at least one global bounded solution for system, which in view of the
known results for the fluid-free system is an optimal restriction on m.

Tao and Li [7] studied the chemotaxis-Navier-Stokes system

nt + u · ∇n = ∇ · (|∇n|p−2∇n)−∇ · (nχ(c)∇c) ,
ct + u · ∇c = ∆c− nf(c),

ut + (u · ∇)u = ∆u+∇P + n∇φ,
∇u = 0.

They show that if p > 2, under appropriate assumptions on f and χ, for all suf-
ficiently smooth initial data (n0, c0, u0), the system has at least one global weak
solution. The relevant equations have also been studied in [8, 14].

Liu and Li [6] studied the problem

ut = ∇ · (D(u,∇u)∇u)− χ∇ · (u · ∇v)− ξ∇ · (u · ∇w)

+ µu(1− u− w), x ∈ Ω, t > 0,

vt −∆v + v = u, x ∈ Ω, t > 0,

wt = −vw, x ∈ Ω, t > 0.

They proved that the problem admits a global bounded weak solution.
Li [5] considered an attraction-repulsion chemotaxis system with p-Laplacian

diffusion

ut = ∇ · (|∇u|p−2∇u)− χ∇ · (u · ∇v)− ξ∇ · (u · ∇w), x ∈ Ω, t > 0,

0 = ∆v + αu− βv, x ∈ Ω, t > 0,

0 = ∆w + γu− δw, x ∈ Ω, t > 0.



EJDE-2020/16 BOUNDEDNESS AND GLOBAL SOLVABILITY 3

Now we state our assumptions:

u0,∆
√
w0 ∈ L∞(Ω), |∇u0|p−2∇u0 ∈ L2(Ω), v0, w0 ∈W 2,∞(Ω),

u0, v0, w0 ≥ 0, |∇u0|p−2∇u0 · ~n =
∂v0

∂~n

∣∣
∂Ω

=
∂w0

∂~n

∣∣
∂Ω

= 0,

∂Ω ∈ C2,γ with 0 < γ < 1.

(1.3)

In section 2, the boundary conditions become equivalent to

|∇u|p−2∇u · ~n|∂Ω =
∂v

∂~n

∣∣
∂Ω

=
∂w

∂~n

∣∣
∂Ω

= 0. (1.4)

This article is organized as follows: in Section 2, we prove some lemmas on the
regularized problem of the system (1.1). In Section 3, the main result on the
existence of a weak solution.

2. Regularized problem

We consider a regularized problem for solving system (1.1),

uεt = div((|∇uε|2 + ε)
p−2
2 ∇uε)− χ∇ ·

( uε
(1 + uε)α

∇vε
)

− ξ∇ ·
( uε

(1 + uε)β
∇wε

)
+ µuε(1− uε − wε), x ∈ Ω, t > 0,

vεt −∆vε + vε = f(uε)g(wε), x ∈ Ω, t > 0,

wεt = −vεwε, x ∈ Ω, t > 0,(
(|∇uε|2 + ε)

p−2
2 ∇uε − χ

uε
(1 + uε)α

∇vε − ξ
uε

(1 + uε)β
∇wε

)
· ~n
∣∣
∂Ω

=
∂vε
∂~n

∣∣∣
∂Ω

= 0,

uε(x, 0) = uε0(x), vε(x, 0) = vε0(x), wε(x, 0) = wε0(x), x ∈ Ω,

(2.1)

where

uε0, vε0, wε0 ∈ C2+γ(Ω), uε0, vε0, wε0 ≥ 0,

∂wε0
∂~n

∣∣∣
∂Ω

=
∂uε0
∂~n

∣∣∣
∂Ω

=
∂vε0
∂~n

∣∣∣
∂Ω

= 0,

uε0 → u0, vε0 → v0, wε0 → w0, uniformly,

‖∇iuε0‖L∞ ≤ 2‖∇iu0‖L∞ , ‖|∇uε0|p−2∇uε0‖L2 ≤ 2‖|∇u0|p−2∇u0‖L2 ,

‖∇ivε0‖L∞ ≤ 2‖∇iv0‖L∞ , ‖∇iwε0‖L∞ ≤ 2‖∇iw0‖L∞ , (i = 0, 1, 2),

‖∇
√
wε0‖L∞ ≤ 2‖∇

√
w0‖L∞ .

(2.2)

From ODE theory,

wε = wε0 exp
(
−
∫ t

0

vε(x, s) ds
)
. (2.3)

Through the direct calculation,

∇wε = e−
∫ t
0
vε(x,s) ds

[
∇wε0 − wε0

∫ t

0

∇vε(x, s) ds
]
, (2.4)



4 C. LIU, P. LI EJDE-2020/16

and

∆wε = ∆wε0e
−

∫ t
0
vε(x,s) ds − 2e−

∫ t
0
vε(x,s) ds∇wε0

∫ t

0

∇vε(x, s) ds

+ wε0e
−

∫ t
0
vε(x,s) ds

∣∣∣ ∫ t

0

∇vε(x, s) ds
∣∣∣2

− wε0e−
∫ t
0
vε(x,s) ds

∫ t

0

∆vε(x, s) ds.

(2.5)

Tao and Winkler [10] stated that

−∆wε(x, t) ≤ ‖wε0‖L∞vε(x, t) + ‖∆wε0‖L∞ + 4‖∇
√
wε0‖2L∞ +

1

e
‖wε0‖L∞ . (2.6)

Thanks to ∂wε0
∂~n

∣∣
∂Ω

= 0 and ∂vε
∂~n

∣∣
∂Ω

= 0 and using (2.4), we posses ∂wε
∂~n

∣∣
∂Ω

= 0.

Considering the zero-flux boundary conditions of the system (2.1), the boundary
conditions are equivalent to

∂uε
∂~n

∣∣∣
∂Ω

=
∂vε
∂~n

∣∣∣
∂Ω

=
∂wε
∂~n

∣∣∣
∂Ω

= 0. (2.7)

Based on a fixed point argument similar the one in [9] or [7], the local classical
solution existence result of problem (2.1) can be proved.

Lemma 2.1. Let p > 2. Under assumption (2.2), then there exists Tmax ∈ (0,+∞]
such that (2.1) admits a unique classical solution (uε, vε, wε) ∈ C2+α,1+α

2 (Ω ×
(0, Tmax)) and for all (x, t) ∈ Ω× (0, Tmax),

uε > 0, vε > 0, wε ≥ 0. (2.8)

Moreover, either Tmax =∞ or

lim sup
t→Tmax

(‖uε(·, t)‖L∞ + ‖vε‖W 1,∞) =∞ .

Take

τ = min{1, Tmax

2
} ≤ 1. (2.9)

Note that if Tmax ≥ 2 then if τ = 1, and if Tmax ≤ 2 then τ ≤ 1.

Lemma 2.2. Let p > 2 and (uε, vε, wε) be the classical solution of problem (2.1)
in (0, Tmax). Under assumptions (2.2), one obtains

sup
t∈(0,Tmax)

‖wε(·, t)‖L∞ ≤ ‖wε0‖L∞ ≤ 2‖w0‖L∞ , (2.10)

sup
t∈(0,Tmax)

‖uε(·, t)‖L1 +
µ

2
sup

t∈(τ,Tmax)

∫ t

t−τ

∫
Ω

u2
ε dx ds ≤ C, (2.11)

sup
t∈(0,Tmax)

‖vε‖2W 1,2 + sup
t∈(τ,Tmax)

∫ t

t−τ
(‖vε‖2W 2,2 + ‖vεt‖2L2) ds ≤ C, (2.12)

where C is independent of ε.

Proof. According to (2.3),

wε = wε0e
−

∫ t
0
v(x,s) ds ≤ |wε0| ≤ ‖wε0‖L∞ ≤ 2‖w0‖L∞ ,



EJDE-2020/16 BOUNDEDNESS AND GLOBAL SOLVABILITY 5

which implies that (2.10) holds. Integrating the first equation in (2.1) over Ω and
combining (2.7) with (2.8), we obtain

d

dt

∫
Ω

uεdx = µ

∫
Ω

uε(1− uε − wε)dx ≤ µ
∫

Ω

uεdx− µ
∫

Ω

u2
εdx.

By the above inequality and the Young inequality, we have

d

dt

∫
Ω

uεdx+
µ

2

∫
Ω

u2
εdx ≤

µ

2
|Ω|. (2.13)

Utilizing [2, Lemma 2.4] and thanks to (2.13), one obtains

sup
t∈(0,Tmax)

‖uε(·, t)‖L1 +
µ

2
sup

t∈(τ,Tmax)

∫ t

t−τ

∫
Ω

u2
ε dx ds ≤ C.

Multiplying the second equation in (2.1) by vε, integrating over Ω, and using the
Young inequality, we obtain

1

2

d

dt

∫
Ω

v2
εdx+

∫
Ω

|∇vε|2dx+

∫
Ω

|vε|2dx =

∫
Ω

f(uε)g(wε)vεdx. (2.14)

By (1.2) and (2.10), we obtain

d

dt

∫
Ω

v2
εdx+

∫
Ω

|∇vε|2dx+

∫
Ω

v2
εdx ≤ C((‖wε‖q2∞ + 1)

∫
Ω

|f(uε)vε|dx. (2.15)

Using the Hölder inequality, we have

d

dt

∫
Ω

v2
εdx+

∫
Ω

|∇vε|2dx+

∫
Ω

v2
εdx

≤ C
(∫

Ω

|f(uε)|6/5dx
)5/6(∫

Ω

|vε|6dx
)1/6

.

(2.16)

From (1.2), the Sobolev imbedding theorem, and the Young inequality, we have

d

dt

∫
Ω

v2
εdx+

1

2

∫
Ω

|∇vε|2dx+
1

2

∫
Ω

v2
εdx ≤ C

∫
Ω

u2
εdx+ C. (2.17)

It follows from (2.11), (2.15) and [2, Lemma 2.4] that

sup
t∈(0,Tmax)

‖vε‖2L2 + sup
t∈(τ,Tmax)

∫ t

t−τ
‖vε‖2W 1,2ds ≤ C. (2.18)

Multiplying the second equation in (2.1) by vεt, integrating it over Ω, applying the
Young inequality, and combining with (1.2), we obtain∫

Ω

v2
εtdx+

1

2

d

dt

∫
Ω

(v2
ε + |∇vε|2)dx =

∫
Ω

f(uε)g(wε)vεtdx

≤ C
∫

Ω

u2
εdx+

1

2

∫
Ω

v2
εtdx+ C.

Therefore,

sup
t∈(τ,Tmax)

∫ t

t−τ
‖vεt‖2L2ds ≤ C. (2.19)

Similarly, multiplying the second equation in (2.1) by ∆vε, integrating it over Ω,
and using the Young inequality, we obtain

d

dt

∫
Ω

|∇vε|2dx+

∫
Ω

|∆vε|2dx+

∫
Ω

|∇vε|2dx ≤ C
∫

Ω

u2
εdx+ C. (2.20)



6 C. LIU, P. LI EJDE-2020/16

By means of (2.11), (2.20) and [2, Lemma 2.4],

sup
t∈(0,Tmax)

‖∇vε‖2L2 + sup
t∈(τ,Tmax)

∫ t

t−τ
‖∇vε‖2W 1,2ds ≤ C. (2.21)

Thanks to (2.18), (2.19) and (2.21), we obtain

sup
t∈(0,Tmax)

‖vε‖2W 1,2 + sup
t∈(τ,Tmax)

∫ t

t−τ
(‖vε‖2W 2,2 + ‖vεt‖2L2) ds ≤ C.

The proof is complete. �

Lemma 2.3. Let p > 2 and (uε, vε, wε) be the classical solution of problem (2.1)
in (0, Tmax). Under assumptions (2.2), we have

sup
t∈(τ,Tmax)

∫ t

t−τ

∫
Ω

|∇u
p−1
p

ε |p dx ds ≤ C, (2.22)

where C is independent of ε.

Proof. Multiplying the first equation in (2.1) by 1 + lnuε, integrating over Ω, and
using (2.7), we have

d

dt

∫
Ω

uε lnuεdx

=

∫
Ω

(1 + lnuε)div((|∇uε|2 + ε)
p−2
2 ∇uε)dx

− χ
∫

Ω

(1 + lnuε)∇
( uε

(1 + uε)α
∇vε

)
dx

− ξ
∫

Ω

(1 + lnuε)∇
( uε

(1 + uε)β
∇wε

)
dx

+ µ

∫
Ω

(1 + lnuε)uε(1− uε − wε)dx

= −
∫

Ω

|∇uε|2

uε
· (|∇uε|2 + ε)

p−2
2 dx+ χ

∫
Ω

(1 + uε)
−α∇uε∇vεdx

+ ξ

∫
Ω

(1 + uε)
−β∇uε∇wεdx+ µ

∫
Ω

(1 + lnuε)uε(1− uε − wε)dx

= −
∫

Ω

|∇uε|2

uε
· (|∇uε|2 + ε)

p−2
2 dx+ χ

∫
Ω

(
∇
∫ uε

0

(1 + s)−αds
)
∇vεdx

+ ξ

∫
Ω

(
∇
∫ uε

0

(1 + s)−βds
)
∇wεdx+ µ

∫
Ω

(1 + lnuε)uε(1− uε − wε)dx

= −
∫

Ω

|∇uε|2

uε
· (|∇uε|2 + ε)

p−2
2 dx− χ

∫
Ω

(∫ uε

0

(1 + s)−αds
)

∆vεdx

− ξ
∫

Ω

(∫ uε

0

(1 + s)−βds
)

∆wεdx+ µ

∫
Ω

(1 + lnuε)uε(1− uε − wε)dx.

(2.23)

By (2.8) and p > 2,( p

p− 1

)p ∫
Ω

∣∣∇u p−1
p

ε

∣∣pdx =

∫
Ω

|∇uε|p

uε
dx

≤
∫

Ω

|∇uε|2

uε
(|∇uε|2 + ε)

p−2
2 dx.

(2.24)



EJDE-2020/16 BOUNDEDNESS AND GLOBAL SOLVABILITY 7

It is not difficult to show that

µ

∫
Ω

(1 + lnuε)uε(1− uε − wε)dx ≤ C. (2.25)

Note that

w1 ≡ 2‖∆w0‖L∞ + 16‖∇
√
w0‖2L∞ +

2

e
‖w0‖L∞ .

Applying (2.6), (2.23), (2.24) and (2.25), considering the Poincaré inequality and
the inequalities

∫ uε
0

(1 + s)−αds ≤ uε and
∫ uε

0
(1 + s)−βds ≤ uε, one obtains

d

dt

∫
Ω

uε lnuεdx+
( p

p− 1

)p ∫
Ω

∣∣∇u p−1
p

ε

∣∣pdx
≤ −χ

∫
Ω

(∫ uε

0

(1 + s)−αds
)

∆vεdx− ξ
∫

Ω

(∫ uε

0

(1 + s)−βds
)

∆wεdx

+ µ

∫
Ω

(1 + lnuε)uε(1− uε − wε)dx

≤ χ
∫

Ω

uε|∆vε|dx+ ξ

∫
Ω

uε

(
‖wε0‖L∞vε + ‖∆wε0‖L∞

+ 4‖∇
√
wε0‖2L∞ +

1

e
‖wε0‖L∞

)
dx+ C

≤ 1

2
χ

∫
Ω

u2
εdx+

1

2
χ

∫
Ω

|∆vε|2dx+ ξ‖w0‖L∞
∫

Ω

u2
εdx+ ξ‖w0‖L∞

∫
Ω

v2
εdx

+ ξw1

∫
Ω

uεdx+ C

≤
(1

2
χ+ ξ‖w0‖L∞

)∫
Ω

u2
εdx+

1

2
χ

∫
Ω

|∆vε|2dx+ ξ‖w0‖L∞
∫

Ω

v2
εdx+ C

≤
(1

2
χ+ ξ‖w0‖L∞

)∫
Ω

u2
εdx+ max

{1

2
χ, ξ‖w0‖L∞

}
×
∫

Ω

(v2
ε + |∆vε|2 + |∇vε|2)dx+ C.

(2.26)

Utilizing (2.11), (2.12), (2.26) and [2, Lemma 2.4], we have

sup
t∈(0,Tmax)

∫
Ω

uε lnuεdx+
( p

p− 1

)p
sup

t∈(τ,Tmax)

∫ t

t−τ

∫
Ω

∣∣∇u p−1
p

ε

∣∣p dx ds ≤ C. (2.27)

If uε ≥ 1, it follows from (2.27) that

sup
t∈(τ,Tmax)

∫ t

t−τ

∫
Ω

∣∣∇u p−1
p

ε

∣∣p dx ds ≤ C( p

p− 1

)−p
. (2.28)

If 0 < uε < 1, using |xq−1 log x| ≤ (e(q − 1))−1, inequality (2.27) becomes

sup
t∈(τ,Tmax)

∫ t

t−τ

∫
Ω

∣∣∇u p−1
p

ε

∣∣p dx ds ≤ (C +
|Ω|
e

)( p

p− 1

)−p
. (2.29)

The proof is complete. �

Lemma 2.4. Let p > 2 and (uε, vε, wε) be the classical solution of problem (2.1)

in (0, Tmax). Under assumptions (2.2), if χ
µ <

d+1
6d ( 3

M )
1
d+2 , for any 0 < d ≤ 4, we



8 C. LIU, P. LI EJDE-2020/16

have

sup
t∈(0,Tmax)

∫
Ω

ud+1
ε dx

+
(µ

2
(d+ 1)− 6d+1dd+2

(d+ 1)d+1
· χ

d+2

µd+1
M
)

sup
t∈(τ,Tmax)

∫ t

t−τ

∫
Ω

ud+2
ε dx ds ≤ C,

(2.30)

where M satisfies∫ t

t−τ

∫
Ω

|∆vε|d+2 dx ds ≤M
(

sup
t∈(τ,Tmax)

∫ t

t−τ

∫
Ω

ud+2
ε dx ds+ ‖vε0‖d+2

W 2,d+2

)
.

Proof. Multiplying the first equation in (2.1) by udε for any d > 0 and integrating
over Ω, we have

1

d+ 1

d

dt

∫
Ω

ud+1
ε dx

=

∫
Ω

udε div((|∇uε|2 + ε)
p−2
2 ∇uε)dx− χ

∫
Ω

udε∇ ·
( uε

(1 + uε)α
∇vε

)
dx

− ξ
∫

Ω

udε∇ ·
( uε

(1 + uε)β
∇wε

)
dx+ µ

∫
Ω

ud+1
ε (1− uε − wε)dx

= −d
∫

Ω

(|∇uε|2 + ε)
p−2
2 ud−1

ε |∇uε|2dx− χ
∫

Ω

udε∇ ·
( uε

(1 + uε)α
∇vε

)
dx

− ξ
∫

Ω

udε∇ ·
( uε

(1 + uε)β
∇wε

)
dx+ µ

∫
Ω

ud+1
ε (1− uε − wε)dx.

(2.31)

By the Sobolev embedding and (2.12),

sup
t∈(0,Tmax)

‖vε‖Ld+2 ≤M1 sup
t∈(0,Tmax)

‖vε‖H1 ≤ C, (2.32)

where M1 is the Sobolev embedding constant. Using (2.6), (2.31), (2.32), the
inequalities

∫ uε
0
sd(1 + s)−αds ≤

∫ uε
0
sdds ≤ 1

d+1u
d+1
ε and

∫ uε
0
sd(1 + s)−βds ≤∫ uε

0
sdds ≤ 1

d+1u
d+1
ε , and the Young inequality, we have

1

d+ 1

d

dt

∫
Ω

ud+1
ε dx+ d

∫
Ω

ud−1
ε |∇uε|pdx+ µ

∫
Ω

ud+2
ε dx

≤ −χ
∫

Ω

udε∇ ·
( uε

(1 + uε)α
∇vε

)
dx

− ξ
∫

Ω

udε∇ ·
( uε

(1 + uε)β
∇wε

)
dx+ µ

∫
Ω

ud+1
ε dx

= −χ
∫

Ω

(
d

∫ uε

0

sd(1 + s)−αds
)

∆vεdx

− ξ
∫

Ω

(
d

∫ uε

0

sd(1 + s)−βds
)

∆wεdx+ µ

∫
Ω

ud+1
ε dx

≤ χd

d+ 1

∫
Ω

ud+1
ε |∆vε|dx+

ξd

d+ 1

∫
Ω

ud+1
ε

(
‖wε0‖L∞vε + ‖∆wε0‖L∞

+ 4‖∇
√
wε0‖2L∞ +

1

e
‖wε0‖L∞

)
dx+ µ

∫
Ω

ud+1
ε dx

≤ χd

d+ 1

∫
Ω

ud+1
ε |∆vε|dx+

2ξd‖w0‖L∞
d+ 1

∫
Ω

ud+1
ε vεdx



EJDE-2020/16 BOUNDEDNESS AND GLOBAL SOLVABILITY 9

+
(
µ+

ξdw1

d+ 1

)∫
Ω

ud+1
ε dx

≤ χd

d+ 1

( 6χd

µ(d+ 1)

)d+1
∫

Ω

|∆vε|d+2dx+
µ

2

∫
Ω

ud+2
ε dx+ C.

The above inequality implies

d

dt

∫
Ω

ud+1
ε dx+ d(d+ 1)

∫
Ω

ud−1
ε |∇uε|pdx+

µ

2
(d+ 1)

∫
Ω

ud+2
ε dx

≤ 6d+1dd+2

(d+ 1)d+1

χd+2

µd+1

∫
Ω

|∆vε|d+2dx+ C.

(2.33)

According to [3, Lemma 2.4], there exists a constant M such that∫ t

t−τ

∫
Ω

|∆vε|d+2 dx ds ≤M
(

sup
t∈(τ,Tmax)

∫ t

t−τ

∫
Ω

ud+2
ε dx ds+ ‖vε0‖d+2

W 2,d+2

)
. (2.34)

On the basis of [2, Lemma 2.4], (2.33) and (2.34),

sup
t∈(0,Tmax)

∫
Ω

ud+1
ε dx+ d(d+ 1) sup

t∈(τ,Tmax)

∫ t

t−τ

∫
Ω

ud−1
ε |∇uε|p dx ds

+
(µ

2
(d+ 1)− 6d+1dd+2

(d+ 1)d+1
· χ

d+2

µd+1
M
)

sup
t∈(τ,Tmax)

∫ t

t−τ

∫
Ω

ud+2
ε dx ds ≤ C.

(2.35)

The proof is complete. �

Lemma 2.5. Let p > 2 and (uε, vε, wε) be the classical solution of problem (2.1)
in (0, Tmax). Under assumption (2.2), if χ

µ <
2
9 ( 3
M )2/5, we have

sup
t∈(0,Tmax)

(‖vε‖W 1,∞ + ‖wεt‖L∞) ≤ C, (2.36)

where C is independent of ε.

Proof. According to Duhamel’s principle,

vεt = e−tet∆vε0 +

∫ t

0

e−(t−s)e(t−s)∆f(uε)g(wε)ds,

where {et∆}t≥0 is the Neumann heat semigroup (see for [12]) in Ω. Taking d = 4

in Lemma 2.4, noticing (1.2), and applying the fact
∫∞

0
e−ss−3/10ds convergences

to a constant, we have for any t ∈ (0, Tmax),

‖vε(·, t)‖L∞

≤ e−t‖vε0‖L∞ +

∫ t

0

e−(t−s)(t− s)− 3
2 ·

1
5 ‖f(uε)g(wε)‖L5ds

≤ e−t‖vε0‖L∞ + sup
s∈(0,Tmax)

‖f(uε(·, s))g(wε(·, s))‖L5

∫ ∞
0

e−ss−3/10ds

≤ 2‖v0‖L∞ + C
(∫

Ω

u5
ε + C|Ω|

)1/5
∫ ∞

0

e−ss−3/10ds ≤ C.

(2.37)

Similarly we have

‖∇vε(·, t)‖L∞ ≤ C. (2.38)



10 C. LIU, P. LI EJDE-2020/16

Considering the third equation of the problem and utilizing (2.1), (2.10) and (2.37),
one has

‖wεt‖L∞ = ‖ − vεwε‖L∞ ≤ C. (2.39)

From (2.37), (2.38) and (2.39), we have

‖vε‖W 1,∞ + ‖wεt‖L∞ ≤ C.

The proof is complete. �

Lemma 2.6. Let p > 2 and (uε, vε, wε) be the classical solution of problem (2.1)
in (0, Tmax). Under assumptions (2.2), if χ

µ <
2
9 ( 3
M )2/5, then

sup
t∈(0,Tmax)

‖uε‖L∞ ≤ C, (2.40)

where C is independent of ε.

Proof. we define

mk = 2mk−1 + 2− p, k ∈ N = {1, 2, 3, . . . }, (2.41)

with

m0 > p− 2. (2.42)

Utilizing (2.41) and (2.42), one posses a nonnegative strictly increasing sequence
{mk}k∈N such that

mk →∞ as k →∞, (2.43)

C12k ≤ mk ≤ C22k, for all k ∈ N, (2.44)

where C1 and C2 are constants independent of k. Let

θk = 2
mk + p− 2

mk + p′ − 2
> 2, (2.45)

where p′ = p
p−1 . It is obvious that

θ′k =
θk

θk − 1
∈ (1, 2) and

1

θ′k
>

1

2
>
p′

4
. (2.46)

Note that

Mk = sup
t∈(0,∞)

∫
Ω

ûε
mk(x, t)dx, k ∈ N, (2.47)

where ûε(x, t) = max{uε(x, t), 1} for x ∈ Ω and t ∈ [0, Tmax). Multiplying the first
equation in (2.1) by mku

mk−1
ε and integrating over Ω, we obtain

d

dt

∫
Ω

umkε dx = −mk(mk − 1)

∫
Ω

umk−2
ε (|∇uε|2 + ε)

p−2
2 |∇uε|2dx

+ χmk(mk − 1)

∫
Ω

umk−1
ε (1 + uε)

−α∇uε∇vεdx

+ ξmk(mk − 1)

∫
Ω

∇
(∫ uε

0

smk−1(1 + s)−βds
)
· ∇wεdx

+

∫
Ω

µmku
mk
ε (1− uε − wε)dx.

(2.48)



EJDE-2020/16 BOUNDEDNESS AND GLOBAL SOLVABILITY 11

It is not difficult to check that mk − 1 ≥ 1 if k ≥ 1 according to (2.41), (2.42) and
(2.43). Using (2.6), (2.7), (2.8) and (2.48), noticing that

∫ uε
0
smk−1(1 + s)−βds ≤∫ uε

0
smk−1ds = 1

mk
umkε , and combining the Young inequality, we have

d

dt

∫
Ω

umkε dx+mk(mk − 1)

∫
Ω

umk−2
ε |∇uε|pdx

≤ d

dt

∫
Ω

umkε dx+mk(mk − 1)

∫
Ω

umk−2
ε (|∇uε|2 + ε)

p−2
2 |∇uε|2dx

≤ χmk(mk − 1)

∫
Ω

umk−1
ε |∇uε| · |∇vε|dx

− ξmk(mk − 1)

∫
Ω

(∫ uε

0

smk−1(1 + s)−βds
)

∆wεdx+

∫
Ω

µmku
mk
ε dx

≤ mk(mk − 1)

2

∫
Ω

umk−2
ε |∇uε|pdx+ 2

1
p−1χ

p
p−1mk(mk − 1)

×
∫

Ω

(
u

2−mk
p +mk−1

ε |∇vε|
)p′

dx

+
(
ξ (2‖w0‖L∞C + w1) + µ

)
mk(mk − 1)

∫
Ω

umkε dx.

The above inequality indicates that

d

dt

∫
Ω

umkε dx+
mk(mk − 1)

2

( p

mk + p− 2

)p ∫
Ω

|∇u
mk−2

p +1
ε |pdx

=
d

dt

∫
Ω

umkε dx+
mk(mk − 1)

2

∫
Ω

umk−2
ε |∇uε|pdx

≤ 2
1
p−1χ

p
p−1mk(mk − 1)

∫
Ω

(
u

2−mk
p +mk−1

ε |∇vε|
)p′

dx

+ (ξ (2‖w0‖L∞C + w1) + µ)mk(mk − 1)

∫
Ω

umkε dx.

(2.49)

The remaining part of the proof can be done in the same way as that in the proof
of [6, Lemma 2.6], we omit the details. �

Lemma 2.7. Let p > 2 and (uε, vε, wε) be the classical solution of problem (2.1)
in (0, Tmax). Under assumptions (2.2), if χ

µ <
2
9 ( 3
M )2/5, s for all T > 0, we have

∫ T

0

‖uεt(·, t)‖p
′

(W 1,p(Ω))∗ds ≤ CT + CT p
′+1, (2.50)

where C is independent of ε.

Proof. For t ∈ (0,∞) and ϕ ∈ C∞(Ω), multiplying the first equation in (2.1) by ϕ,
integrating over Ω, applying (2.8), (2.36), (2.40), considering the Hölder inequality



12 C. LIU, P. LI EJDE-2020/16

and using the inequalities |∇ϕ| ≤ C and |ϕ| ≤ C, we have∣∣∣ ∫
Ω

uεtϕdx
∣∣∣

=
∣∣∣− ∫

Ω

(|∇uε|2 + ε)
p−2
2 ∇uε∇ϕdx+ χ

∫
Ω

uε
(1 + uε)α

∇vε∇ϕdx

+ ξ

∫
Ω

uε
(1 + uε)β

∇wε∇ϕdx+ µ

∫
Ω

uε(1− uε − wε)ϕdx
∣∣∣

≤
∫

Ω

|(|∇uε|2 + ε)
p−2
2 ∇uε| · |∇ϕ|dx+ χ

∫
Ω

|uε∇vε| · |∇ϕ|dx

+ ξ

∫
Ω

|uε∇wε| |∇ϕ|dx+ µ

∫
Ω

uε|ϕ|dx+ µ

∫
Ω

u2
ε|ϕ|dx

+ µ

∫
Ω

uεwε|ϕ|dx

≤ C‖(|∇uε|2 + ε)
p−2
2 ∇uε‖Lp′ + C‖∇wε‖Lp′ + C.

(2.51)

Using (2.4) and (2.36), we have∫ T

0

∫
Ω

|∇wε|p
′
dx ds ≤ CT + CT p

′+1. (2.52)

It follows from (2.22) and (2.40) that∫ T

0

∫
Ω

|∇uε|p dx ds =

∫ T

0

∫
Ω

| p

p− 1
u

1
p
ε ∇u

p−1
p

ε |p dx ds ≤ CT. (2.53)

Hence, by (2.51), (2.52), (2.53) and that

|(|∇uε|2 + ε)
p−2
2 ∇uε|p

′
≤ |(|∇uε|2 + ε)

p−1
2 |p

′

≤ |(|∇uε|2 + 1)
p−1
2 |p

′

= (|∇uε|2 + 1)
p
2 ≤ 2

p
2 (|∇uε|p + 1),

we have ∫ T

0

‖uεt(·, t)‖p
′

(W 1,p(Ω))∗ds ≤ CT + CT p
′+1.

The proof is complete. �

3. Existence of a global bounded weak solution

By Lemmas 2.1, 2.5 and 2.6, we have Tmax = ∞, τ = 1. We know that prob-
lem (2.1) admits a unique classical global solution (uε, vε, wε) which belongs to

C2+γ,1+ γ
2 (Ω× (0,∞)) with uε ≥ 0 ,vε ≥ 0 and wε ≥ 0.

Lemma 3.1. Let p > 2, under assumptions (2.2) and χ
µ < 2

9 ( 3
M )2/5, we assert

there exists a nonnegative triplete (u, v, w) such that for any T > 0,

uε(·, t)→ u(·, t), a.e. in QT , (3.1)

uε
?
⇀ u, in L∞(QT ), (3.2)

uε → u, in Lploc(QT ), (3.3)

∇uε ⇀ ∇u, in Lploc(QT ), (3.4)

(|∇uε|2 + ε)
p−2
2 ∇uε ⇀ |∇u|p−2∇u, in Lp

′

loc(QT ), (3.5)



EJDE-2020/16 BOUNDEDNESS AND GLOBAL SOLVABILITY 13

vε → v, uniformly, (3.6)

vε ⇀ v, in W 2,1
r (QT ) for any r > 1, (3.7)

wε → w, uniformly, (3.8)

∇wε → ∇w, uniformly, (3.9)

∆wε
?
⇀ ∆w, in L∞(QT ), (3.10)

as ε→ 0.

Proof. Utilizing (2.53), Lemma 2.6, Lemma 2.7 and Aubin-Lions lemma, there
exists a nonnegative function u such that, as ε→ 0,

uε → u, in Lploc(Ω× [0, T )) and a.e. in QT . (3.11)

By the Fubin-Tonelli theorem and (3.11), we have (3.1) and (3.3). Then (3.11) and
Lemma 2.6 yield (3.2). Equations (2.53) and (3.11) imply (3.4). Combining (3.2)
and (3.11), applying Lemma 2.6, as in [7], one obtains that

|∇uε|p−2∇uε ⇀ |∇u|p−2∇u, in Lp
′

loc(Ω× [0, T )),

which implies (3.5), and

(|∇uε|2 + ε)
p−2
2 ∇uε ⇀ |∇u|p−2∇u, in Lp

′

loc(Ω× [0, T )).

So (3.5) holds. It follows from Lemma 2.6 and (1.2) that for any r > 0,

sup
t∈(1,∞)

∫ t

t−1

‖f(uε)g(wε)‖rLrds ≤ C. (3.12)

Then, by [3, Lemma 2.4] and (3.12), we have

sup
t∈(1,∞)

∫ t

t−1

(‖vε‖rW 2,r + ‖vεt‖rLr ) ds ≤ C

which means

‖vε‖W 2,1
r (Q1(t))

≤ C, (3.13)

where Q1(t) = Ω × (t − 1, t). Then there exists a nonnegative v such that (3.7)

holds according to (3.13). Because of (3.13) and W 2,1
r (QT ) ↪→ C2− 5

r ,1−
5
2r (QT ), for

any r > 5/2, one obtains vε → v uniformly. This shows (3.6). Equations (2.3) and

(3.6) yield (3.8). Using W 1,1
r (QT ) ↪→ C1− 5

r ,1−
5
2r (QT ) for any r > 5 and (3.13), we

obtain

∇vε → ∇v, uniformly. (3.14)

Thus, by (2.4), (3.6) and (3.14), (3.9) holds. It follows from the second equation in
(2.1), Lemma 2.5, and Lemma 2.6 that∫ t

0

∆vε(x, s) ds =

∫ t

0

(vεt + vε − f(uε)g(wε)) ds

= vε(x, t)− vε0 +

∫ t

0

(vε − f(uε)g(wε)) ds.

≤ C + Ct.

(3.15)

Equations (2.5) and (3.15), and Lemma 2.5 guarantee that

|∆wε| ≤ Ct+ Ct2 + C. (3.16)



14 C. LIU, P. LI EJDE-2020/16

Using (2.4) and Lemma 2.5, one obtains

|∇wε| ≤ C + Ct. (3.17)

Thanks to (3.16) and (3.17), we find that for any T > 0,

sup
t∈(0,T )

∫
Ω

‖∇wε‖W 1,∞(QT )dx ≤ C + CT + CT 2. (3.18)

Hence, (3.8), (3.9) and (3.18) indicate that (3.10) holds. �

Let

H1 = {u ∈ L∞((0,∞);L1
loc(Ω× [0,∞));ut ∈ Lp

′
([0,∞); (W 1,p(Ω))∗)};

H2 = {v ∈ L∞((0,∞);W 1,∞(Ω)); vt,∆v ∈ Lrloc([0,∞);Lr(Ω)) for any r > 1};
H3 = {w ∈ L∞(Ω× (0,∞));wt ∈ L∞(Ω× (0,∞));w ∈ L∞loc([0,∞);W 2,∞(Ω))}.

Definition 3.2. A nonnegative triplet (u, v, w) is a weak solution of system (1.1),
if (u, v, w) ∈ H1 ×H2 ×H3 and satisfies∫∫

QT

uϕt dx ds+

∫
Ω

u(x, 0)ϕ(x, 0)dx

=

∫∫
QT

|∇u|p−2∇u · ∇ϕdx ds+ χ

∫∫
QT

u

(1 + u)α
∇v · ∇ϕdx ds

+ ξ

∫∫
QT

u

(1 + u)β
∇w · ∇ϕdx ds+ µ

∫∫
QT

u(1− u− w)ϕdx ds ,∫∫
QT

vϕt dx ds+

∫
Ω

v(x, 0)ϕ(x, 0)dx

=

∫∫
QT

(∇v∇ϕ+ (v − f(u)g(w))ϕ) dx ds,−
∫∫

QT

wϕt dx ds

−
∫

Ω

w(x, 0)ϕ(x, 0)dx+

∫∫
QT

vwϕdx ds = 0,

for any given T > 0, for any ϕ ∈ C∞(QT ) with ∂ϕ
∂~n

∣∣∣
∂Ω

= 0 and ϕ(x, T ) = 0.

Theorem 3.3. Let p > 2. Under assumptions (1.2) for a sufficiently small χ/µ,
system (1.1) admits a nonnegative weak solution (u, v, w) with u ∈ H1, v ∈ H2,
w ∈ H3.

Proof. For any ϕ ∈ C∞(QT ) with ∂ϕ
∂~n

∣∣∣
∂Ω

= 0 and ϕ(x, T ) = 0, we obtain∫∫
QT

uεϕt dx ds+

∫
Ω

uε(x, 0)ϕ(x, 0)dx

=

∫∫
QT

(|∇uε|2 + ε)
p−2
2 ∇uε · ∇ϕdx ds+ χ

∫∫
QT

uε
(1 + uε)α

∇vε · ∇ϕdx ds

+ ξ

∫∫
QT

uε
(1 + uε)β

∇wε · ∇ϕdx ds+ µ

∫∫
QT

uε(1− uε − wε)ϕdx ds ,



EJDE-2020/16 BOUNDEDNESS AND GLOBAL SOLVABILITY 15∫∫
QT

vεϕt dx ds+

∫
Ω

vε(x, 0)ϕ(x, 0)dx

=

∫∫
QT

(∇vε∇ϕ+ (vε − f(uε)g(wε))ϕ) dx ds,−
∫∫

QT

wεϕt dx ds

−
∫

Ω

wε(x, 0)ϕ(x, 0)dx+

∫∫
QT

vεwεϕdx ds = 0,

according to the fact that (uε, vε, wε) is the classical solution of the problem (2.1).
Applying Lemma 3.1, one has∫∫

QT

uϕt dx ds+

∫
Ω

u(x, 0)ϕ(x, 0)dx

=

∫∫
QT

|∇u|p−2∇u · ∇ϕdx ds+ χ

∫∫
QT

u

(1 + u)α
∇v · ∇ϕdx ds

+ ξ

∫∫
QT

u

(1 + u)β
∇w · ∇ϕdx ds+ µ

∫∫
QT

u(1− u− w)ϕdx ds,∫∫
QT

vϕt dx ds+

∫
Ω

v(x, 0)ϕ(x, 0)dx

=

∫∫
QT

(∇v∇ϕ+ (v − f(u)g(w))ϕ) dx ds,−
∫∫

QT

wϕt dx ds

−
∫

Ω

w(x, 0)ϕ(x, 0)dx+

∫∫
QT

vwϕdx ds = 0,

as ε→ 0. Then (u, v, w) in Lemma 3.1 is the desired global weak solution of (1.1).
From the previous parts it follows that u ∈ H1, v ∈ H2, and w ∈ H3. �

Acknowledgements. This work was supported by the Jilin Scientific and Tech-
nological Development Program (number 20170101143JC).

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Changchun Liu

School of Mathematics, Jilin University, Changchun 130012, China

Email address: liucc@jlu.edu.cn

Pingping Li

School of Mathematics, Jilin University, Changchun 130012, China
Email address: 1522201343@qq.com


	1. Introduction
	2. Regularized problem
	3. Existence of a global bounded weak solution
	Acknowledgements

	References