Electronic Journal of Differential Equations, Vol. 2008(2008), No. 141, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
EXISTENCE OF GLOBAL SOLUTIONS FOR A SYSTEM OF
REACTION-DIFFUSION EQUATIONS HAVING A TRIANGULAR
MATRIX
EL HACHEMI DADDIOUAISSA
Abstract. We consider the system of reaction-diffusion equations
ut − a∆u = β − f (u, v) − αu,
vt − c∆u − d∆v = g(u, v) − σv.
Our aim is to establish the existence of global classical solutions using the
method used by Melkemi, Mokrane, and Youkana [19].
1. Introduction
In this manuscript, we consider a reaction-diffusion system that arises in the
study of physical, chemistry, and various biological processes including population
dynamics [3, 6, 7, 9, 14, 23, 24]. The system of equations is
∂u
− a∆u = β − f (u, v) − αu (x, t) ∈ Ω × R+
∂t
∂v
− c∆u − d∆v = g(u, v) − σv (x, t) ∈ Ω × R+ ,
∂t
with the boundary conditions
∂u
∂v
=
= 0 on ∂Ω × R+ .
∂η
∂η
and the initial data
0 ≤ u(0, x) = u0 (x);
0 ≤ v(0, x) = v0 (x)
in Ω.
(1.1)
(1.2)
(1.3)
(1.4)
n
where Ω is a smooth open bounded domain in R , with boundary ∂Ω of class
C 1 and η is the outer normal to ∂Ω. The constants of diffusion a, c, d are such
that a > 0, d > 0, c > 0 and a > d, c2 < 4ad which is the parabolic condition,
and α, σ are positive constants, β ≥ 0, and f, g are nonnegative functions of class
C(R+ × R+ ), such that
(H1) For all τ ≥ 0, f (0, τ ) = 0;
(H2) For all ξ ≥ 0 and all τ ≥ 0, 0 ≤ f (ξ, τ ) ≤ ϕ(ξ)(µ + τ )r ;
(H3) For all ξ ≥ 0 and all τ ≥ 0, g(ξ, τ ) ≤ ψ(τ )f (ξ, τ ) + φ(τ ),
2000 Mathematics Subject Classification. 35K57, 35K45.
Key words and phrases. Reaction-diffusion systems; Lyapunov functional; global solution.
c
2008
Texas State University - San Marcos.
Submitted August 19, 2007. Published October 16, 2008.
1
2
E. H. DADDIOUAISSA
EJDE-2008/141
where r, µ are positive constants such that r ≥ 1, µ ≥ 1, ϕ, ψ and φ are nonnegative
functions of class C(R+ ), such that
lim
τ →+∞
ψ(τ )
φ(τ )
= lim
= 0.
τ →+∞ τ
τ
c
.
φ(0) > β
a−d
In addition we suppose that
c
c
c
c
g(ξ,
ξ) +
f (ξ,
ξ) ≥
[(σ − α)ξ + β],
a−d
a−d
a−d
a−d
(1.5)
(1.6)
∀ξ ≥ 0.
(1.7)
Melkemi et al [19] established the existence of global solutions, (eventually uniformly bounded in time) using a novel approach that involved the use of a Lyapunov function for system (1.1)–(1.4) when c = 0. Here, we apply the same method
to the study of system (1.1)–(1.4) when c > 0; that is, for a model that involves a
triangular matrix.
2. Existence of local and positive solutions
First we convert system (1.1)–(1.4) into an abstract first order system in X =
C(Ω) × C(Ω) of the form
U 0 (t) = AU (t) + F (U (t)),
t > 0,
U (0) = U0 ∈ X,
(2.1)
(2.2)
where
AU (t) = (a∆u, c∆u + d∆v),
F (U (t)) = (β − f (u, v) − αu, g(u, v) − σv).
Since F is locally Lipschitz in U and X, for every initial data U0 ∈ X, system
(2.1)–(2.2) admits a unique strong local solution on ]0, T ∗ [, where T ∗ is the eventual
blowing-up time, (see Kirane [14], Friedman [8], Henry [12], Pazy [20]).
c
Multiplying (1.1) by a−d
, and subtracting the resulting equation from (1.2) leads
to the system
∂u
− a∆u = Λ(u, z),
∂t
∂z
− d∆z = Υ(u, z),
∂t
(x, t) ∈ Ω × R+
(2.3)
(x, t) ∈ Ω × R+ .
(2.4)
where
Λ(u, z) = β − f (u, v) − αu
c
c
Υ(u, z) = g(u, v) +
f (u, v) +
(αu − β) − σv,
a−d
a−d
c
z=v−
u,
a−d
with the boundary conditions
∂u
∂z
=
= 0,
∂η
∂η
on ∂Ω × R+
(2.5)
EJDE-2008/141
EXISTENCE OF GLOBAL SOLUTIONS
3
and initial data
u(0, x) = u0 (x)
in Ω,
(2.6)
c
z(0, x) = z0 (x) = v0 (x) −
u0 (x) in Ω.
(2.7)
a−d
If we assume (1.7) and (H1) then a simple application of a comparison theorem to
system (2.3)-(2.4) implies (see [14]) that for positive initial data u0 ≥ 0 and z0 ≥ o
we have that
c
u(t, x) ∀(x, t) ∈ Ω×]0, T ∗ [.
u(t, x) ≥ 0, v(t, x) ≥
a−d
3. Existence of global solutions
Before we establish the existence of a global solution, we introduce some notation.
Here, we let
Z
1
|u(x)|p dx and kuk∞ = max |u(x)|.
kukpp =
x∈Ω
|Ω| Ω
denote the usual norms in spaces Lp (Ω), L∞ (Ω) and C(Ω). Applying the comparison principle we get that
β
) = K.
(3.1)
α
To establish the uniform boundedness of v, it is sufficient to show the uniform
boundedness of z. This task is carried out using a result found in Henry [12, pp.
35-62], from which it sufficient to derive a uniform estimate for kΥ(u, z)kp ; that is,
finding a constant C such that
u(t, x) ≤ max(ku0 k∞ ,
kΥ(u, z)kp ≤ C.
(3.2)
Here C is a nonnegative constant independent of t, for some p > n/2. The key
is to establish the uniform boundedness of kvkp on ]0, T ∗ [, taking into account
assumptions (H2) and (H3). When p ≥ 2, we put
Γ(p) =
l=
pΓ + 1
,
p−1
2βρ
,
Γ(p)σ
ω=[
Γ = (a − d)2 [1 +
1
],
4ad
S2
p
+ 2 + µ](p − 1)
2
4adR
R
(3.3)
where ρ > 0,
(a − d)ρ
ρ
, R=
.
Γ(p)l
Γ(p)(l + ρK)
Using these definitions and notation we can state the following key proposition
S=
Proposition 3.1. Assume that p ≥ 2 and let
Z
Z
1
GN (t) = N
udx + (v + ω)p exp(−
ln(Γ(p)[l + ρ(K − u)]))dx,
Γ(p)
Ω
Ω
(3.4)
where (u, v) is the solution of (1.1)–(1.4) on ]0, T ∗ [. Then under the assumptions
(H3) and (1.5) there exist two positive constants N and s such that
dGN
≤ −(p − 1)σGN + s.
dt
The proof of the above proposition requires some lemmas.
(3.5)
4
E. H. DADDIOUAISSA
EJDE-2008/141
Lemma 3.2. If (u, v) be a solution of (1.1) − (1.4) then
Z
Z
d
f (u, v)dx ≤ β|Ω| −
u(t, x)dx.
dt Ω
Ω
(3.6)
Proof. We integrate both sides of (1.1),
d
u(t, x)
dt
satisfied by u, which is positive and then we find (3.6).
f (u, v) = β − αu −
Lemma 3.3. Assume that p ≥ 2, then under the assumptions (H1)–(H3), and
(1.5) there exists N1 , such that
[p(g(ξ, τ ) − φ(τ ))(τ + ω)p−1 − θf (ξ, τ )(τ + ω)p ] ≤ N1 f (ξ, τ )
(3.7)
for all 0 ≤ ξ ≤ K and τ ≥ 0, θ > 0.
Proof. from the assumption (H3) and (1.5), we conclude that there exists τ0 > 0,
such that for all 0 ≤ ξ ≤ K, τ ≥ τ0 , one finds
ψ(τ )
− θ](τ + ω)p f (ξ, τ ) ≤ 0
τ +ω
now if τ is in the compact interval [0, τ0 ], then the continuous function
[p
χ(ξ, τ ) = pψ(τ )(τ + ω)p−1 − θ(τ + ω)p
is bounded.
Lemma 3.4. For all τ ≥ 0 and ω ≥ 1, we have
βρ
(τ + ω)p − σp(τ + ω)p−1 τ + pφ(τ )(τ + ω)p−1 ≤ −(p − 1)σ(τ + ω)p + M1 (3.8)
Γ(p)l
where M1 is positive constant.
Proof. Let us put
βρ
Π=
(τ + ω)p − σp(τ + ω)p−1 τ
Γ(p)l
βρ
βρ
ω
βρ
≤[2
− σp]τ (τ + ω)p−1 + [2
−
](τ + ω)p
Γ(p)l
Γ(p)l τ + ω Γ(p)l
since l =
2βρ
Γ(p)σ ,
then
Π + pφ(τ )(τ + ω)p−1 ≤ −(p − 1)σ(τ + ω)p + [
p(σω + φ(τ )) 1
− ]σ(τ + ω)p
σ(τ + ω)
2
using (1.5) and Lemma 3.3, we can establish Proposition 3.1.
Proof of Proposition 3.1. Let
h(u) = −
1
ln(Γ(p)[l + ρ(K − u)])
Γ(p)
so that
Z
GN (t) = N
udx + L(t)
Ω
where
Z
L(t) =
Ω
eh(u) (v + ω)p dx.
EJDE-2008/141
EXISTENCE OF GLOBAL SOLUTIONS
5
Differentiating L with respect to t and using Greens formula, we obtain that
L0 (t) = I + J
where
Z
[a(h00 (u) + h02 (u))(v + ω)p + pch0 (u)(v + ω)p−1 ]eh(u) ∇u2 dx
I =−
Ω
Z
[p(a + d)h0 (u)(v + ω)p−1 + p(p − 1)c(v + ω)p−2 ]eh(u) ∇u∇vdx
−
ZΩ
−
p(p − 1)d(v + ω)p−2 eh(u) ∇v 2 dx
Ω
and
Z
J=
βh0 (u)(v + ω)p eh(u) dx
Z
+ [pg(u, v)(v + ω)p−1 − h0 (u)f (u, v)(v + ω)p ]eh(u) dx
ZΩ
Z
−
αh0 (u)u(v + ω)p eh(u) dx −
σp(v + ω)p−1 veh(u) dx.
Ω
Ω
Ω
We see that I involves a quadratic form with respect to ∇u, ∇v.
D =[a(h00 (u) + h02 (u))(v + ω)p + pch0 (u)(v + ω)p−1 ]∇u2
+ [p(a + d)h0 (u)(v + ω)p−1 + p(p − 1)c(v + ω)p−2 ]∇u∇v
+ p(p − 1)d(v + ω)p−2 ∇v 2
which is nonnegative if
δ =[p(a + d)h0 (u)(v + ω)p−1 + p(p − 1)(v + ω)p−2 ]2
− 4p(p − 1)d(v + ω)p−2 [a(h00 (u) + h02 (u))(v + ω)p
+ pch0 (u)(v + ω)p−1 ] ≤ 0.
Indeed,
δ =p2 (a + d)2 h02 (u)(v + ω)2p−2 + 2(a + d)cp2 (p − 1)h0 (u))(v + ω)2p−3
+ c2 p2 (p − 1)2 (v + ω)2p−4 − 4p(p − 1)ad(h00 (u) + h02 (u))(v + ω)2p−2
− 4cdp2 (p − 1)h0 (u)(v + ω)2p−3
and from (3.3) we have that v + ω ≥ 1. It follows that
δ ≤[p2 (a + d)2 h02 (u) − 4p(p − 1)ad(h00 (u) + h02 (u))](v + ω)2p−2
+ [2(a + d)cp2 (p − 1)h0 (u) + c2 p2 (p − 1)2
− 4cdp2 (p − 1)h0 (u)](v + ω)2p−3 .
Let us try
T = p2 (a + d)2 h02 (u) − 4p(p − 1)ad(h00 (u) + h02 (u))
the choice of h(u) implies
ρ
ρ
≤ h0 (u) ≤
,
Γ(p)[l + ρK]
Γ(p)l
(3.9)
6
E. H. DADDIOUAISSA
EJDE-2008/141
and, consequently,
T =−
ρ2 4ad(a − d)2 p2
≤ −4adp2 R2 ≤ 0.
[Γ(p)(l + ρ(K − u))]2
In addition
δ ≤[2(a + d)cp2 (p − 1)h0 (u) + c2 p2 (p − 1)2 − 4cdp2 (p − 1)h0 (u)](v + ω)2p−3
+ T (v + ω)(v + ω)2p−3
≤[(p − 1)c2 + 2(a − d)c
4adR2 ω 2
ρ
−
]p (p − 1)(v + ω)2p−3 + T v(v + ω)2p−3 .
Γ(p)l
(p − 1)
If we replace w by its value (3.3) and use the parabolic condition c2 − 4ad < 0, we
deduce that
δ ≤[p(c2 − 4ad) − (c − S)2 − 4adR2 µ]p2 (p − 1)(v + ω)2p−3
+ T v(v + ω)2p−3 ≤ 0,
that is, I ≤ 0.
We can control the second term by observing that
Z
J ≤ [βh0 (u)(v + ω)p − σp(v + ω)p−1 v + pφ(v)(v + ω)p−1 ]eh(u) dx
Ω
Z
+ [p(g(u, v) − φ(v))(v + ω)p−1 − h0 (u)f (u, v)(v + ω)p ]eh(u) dx.
Ω
Using (3.9),
Z
βρ
J≤ [
(v + ω)p − σp(v + ω)p−1 v + pφ(v)(v + ω)p−1 ]eh(u) dx
Γ(p)l
Ω
Z
ρ
+ [p(g(u, v) − φ(v))(v + ω)p−1 −
f (u, v)(v + ω)p ]eh(u) dx.
Γ(p)[l + ρk]
Ω
Applying Lemmas 3.3 and 3.4, one finds that
Z
Z
Z
p h(u)
h(u)
J ≤ −(p − 1)σ (v + ω) e
dx + M1
e
dx + N1
f (u, v)eh(u) dx.
Ω
Ω
In addition we see that
h(u) ≤ −
Ω
2βρ
1
ln
,
Γ(p)
σ
and, consequently,
1
J ≤ −(p − 1)σL + M1 |Ω|e− Γ(p) ln
2βρ
σ
1
+ N1 e− Γ(p) ln
2βρ
σ
Z
f (u, v)dx.
Ω
Letting
1
2βρ
1
2βρ
M = M1 |Ω|e− Γ(p) ln σ , N = N1 e− Γ(p) ln σ
and using Lemma 3.2 we conclude that
Z
d
J ≤ − (p − 1)σL + M + N [β|Ω| −
u(t, x)dx]
dt Ω
Z
Z
d
≤ − (p − 1)σGN + (p − 1)σN
udx + M + N β|Ω| − N
u(t, x)dx
dt Ω
Ω
Z
d
≤ − (p − 1)σGN + |Ω|N [(p − 1)Kσ + β] + M − N
u(t, x)dx.
dt Ω
EJDE-2008/141
EXISTENCE OF GLOBAL SOLUTIONS
7
It follows that
dGN
≤ −(p − 1)σGN + s
dt
where s = |Ω|N [(p − 1)Kσ + β] + M .
We can now establish the main result of this manuscript.
Theorem 3.5. Under the assumptions (H1)-(H3) and (1.5), the solutions of (1.1)–
(1.4) are global and uniformly bounded in [0, +∞[.
Proof. Multiplying (3.5) by e(p−1)σt and integrating, implies the existence of a
positive constant C > 0 independent of t such that
GN (t) ≤ C.
Since
1
eh(u) ≥ e− Γ(p) ln(Γ(p)[l+ρk)]) ,
for all p ≥ 2, we have
Z
1
1
(v + ω)p dx ≤ e Γ(p) ln(Γ(p)[l+ρk)) GN (t) ≤ Ce Γ(p) ln Γ(p)[l+ρk] = C(p).
Ω
Consequently,
Z
(v + µ)p dx ≤ C(p),
Ω
Z
v p dx ≤ C(p).
Ω
Now we chose p > n/2 and we search for a bound for kΥ(u, v)kp . We put
A1 = max ψ(τ ),
0≤τ ≤τ0
A2 = max ϕ(ξ),
0≤ξ≤K
A3 = max φ(τ )
0≤τ ≤τ0
where τ0 = max(τ1 , τ2 ) such that
τ ≥ τ1
⇒
ψ(τ ) ≤ τ,
τ ≥ τ2
⇒
φ(τ ) ≤ τ.
Using (H1)–(H3) implies
g(u, v) ≤ ψ(v)f (u, v) + φ(v) ≤ ψ(v)ϕ(u)(µ + v)r + φ(v) ≤ A2 ψ(v)(µ + v)r + φ(v).
Since 0 ≤ u ≤ K, we have
Z
Z
g(u, v)p dx ≤ [A2 ψ(v)(µ + v)r + φ(v)]p dx.
Ω
Ω
Now we use the inequality
(x + y)q ≤ 2q−1 (xq + y q )
all x, y ≥ 0 and q ≥ 1, to obtain the following sequence of estimates
Z
g(u, v)p dx
Ω
Z
≤
2p−1 [Ap2 ψ(v)p (µ + v)rp + φ(v)p ]dx
Ω
Z
Z
i
h Z
v p dx
≤ 2p−1 Ap2
Ap1 (µ + τ0 )rp dx +
v p (µ + v)rp dx + |Ω|Ap3 +
v≥τ0
v≤τ0
v≥τ0
Z
Z
i
h
p
p
p
p−1
r
(r+1)p
v p dx
≤2
A2 A1 (µ + τ0 )
|Ω| + A2
(µ + v)
dx + |Ω|A3 +
v≥τ0
≤ 2p−1 [(A2 A1 (µ + τ0 )r )p |Ω| + Ap2 C((r + 1)p) + |Ω|Ap3 + C(p)] = Egp
v≥τ0
8
E. H. DADDIOUAISSA
and
Z
Ω
EJDE-2008/141
f (u, v)p dx ≤ Ap2 C(rp) = Efp .
We conclude that
c
[kf (u, v)kp + αkukp + β|Ω|] + σkvkp
a−d
p
[Ef + (αK + β)|Ω|] + σ p C(p).
kΥ(u, v)kp ≤ kg(u, v)kp +
c
a−d
We conclude that the unique solution of (1.1)–(1.4) is globally and uniformly
bounded in [0, +∞[×Ω.
≤ Eg +
Acknowledgments. The author wants to thank Professors A. Youkana and M.
Mechab for their kind advice, and the anonymous referee for his/her comments that
improved the final version of this article.
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El Hachemi Daddiouaissa
Department of Mathematics University Kasdi Merbah, UKM Ouargla 30000, Algeria
E-mail address: dmhbsdj@gmail.com