S. B A D R A O U I (Guelma)
BEHAVIOUR OF GLOBAL SOLUTIONS FOR A SYSTEM OF REACTION-DIFFUSION EQUATIONS FROM COMBUSTION THEORY
Abstract. We are concerned with the boundedness and large time be- haviour of the solution for a system of reaction-diffusion equations mod- elling complex consecutive reactions on a bounded domain under homoge- neous Neumann boundary conditions. Using the techniques of E. Conway, D. Hoff and J. Smoller [3] we also show that the bounded solution converges to a constant function as t → ∞. Finally, we investigate the rate of this convergence.
1. Introduction. In this paper we investigate the asymptotic behaviour of global solutions for the following reaction-diffusion system:
(1.1) ∂Y
1∂t = d
0∆Y
1− d
1Y
1Y
2f
1(T ), x ∈ Ω, t > 0, (1.2) ∂Y
2∂t = d
2∆Y
2+ d
3Y
1Y
2f
1(T )
− d
4Y
2f
2(T ) − d
5Y
2− d
6Y
22, x ∈ Ω, t > 0,
(1.3) ∂T
∂t = d
7∆T + d
8Y
1Y
2f
1(T )
+ d
9Y
2f
2(T ) + d
10Y
2+ d
11Y
22, x ∈ Ω, t > 0, (1.4) ∂Y
1∂ν = ∂Y
2∂ν = ∂T
∂ν = 0, x ∈ ∂Ω, t > 0, (1.5) (Y
1, Y
2, T )(x, 0) = (Y
10, Y
20, T
0)(x), x ∈ Ω,
1991 Mathematics Subject Classification: 35K57, 35B40, 35B45.
Key words and phrases: reaction-diffusion equations, boundedness, global existence, large time behaviour.
[133]
where Ω is a bounded domain in R
nwith boundary ∂Ω, such that ∂Ω is a C
mhypersurface separating Ω from R
n/Ω (m ≥ 1), d
j(j = 0, 1, . . . , 11) are positive constants, f
i(i = 1, 2) are given by the Arrhenius law
f
i(T ) = B
iexp(−E
i/T ),
where B
i, E
iare constants, and E
idenotes the activation energy.
This system of reaction-diffusion equations arises as a model of chain branching and chain breaking kinetics of reactions with complex chemistry.
Here Y
1is the concentration of fuel, Y
2is the concentration of radicals, and T is the dimensionless temperature. Y
1, Y
2and T depend on x and t where (x, t) ∈ Ω × R
+.
Under suitable conditions (see (CD) in Section 3), it is expected that (1.1)–(1.5) has a unique global solution (Y
1, Y
2, T ) and this solution tends to an equilibrium state uniformly in x as t → ∞.
We will show that (Y
1(t), Y
2(t), T (t)) approaches an equilibrium state (0, 0, T
∞) in C
µ(Ω)
3as t → ∞ for every µ ∈ [0, 2), where T
∞is a con- stant, and we will consider the rate of this convergence, by means of in- tegral equations, fractional powers of operators, Poincar´e’s inequality and some imbedding theorems.
2. Preliminary results. We state some results needed in the sequel.
Lemma 2.1. Let (E, k · k
E) and (F, k · k
F) be two real Banach spaces with continuous inclusion E ⊂ F . Let A be a linear operator generating a strongly continuous semigroup G(t) in E such that:
(i) G(t)E ⊂ F for all t > 0,
(ii) there exists θ ∈ [0, 1) such that kG(t)ϕk
F≤ ct
−θkϕk
Efor all t > 0.
Moreover , let p > 1/(1 − θ), f ∈ L
ploc(R
+, E) and sup
t≥0kf k
Lp(t,t+1;E)< ∞. Let u be a mild solution on R
+of du
dt = Au(t) + f (t).
If u ∈ L
∞(0, ∞; E), then u(t) ∈ F for all t > 0 and u ∈ C
B(δ, ∞; F ) for all δ > 0, where C
B(δ, ∞; F ) is the space of all continuous functions u : (δ, ∞) → F such that sup{ku(t)k
F: t ≥ δ} < ∞.
For the proof, see [4].
Lemma 2.2. Let G(t) be the semigroup generated by the operator d∆
in L
p(Ω). Then for all 1 ≤ p < q ≤ ∞ and all ϕ ∈ L
p(Ω) we have G(t)ϕ ∈ L
q(Ω) and
kG(t)ϕk
q≤ c(p, q)t
−(n/2)(1/p−1/q)kϕk
p.
For the proof, see [2].
3. Global existence and positivity. Throughout this paper, the following assumptions are in force:
(CD) (i) d
j(j = 0, 1, . . . , 11) are positive constants,
(ii) Y
10, Y
20and T
0are nonnegative measurable functions such that 0 ≤ Y
10(x), Y
20(x), T
0(x) ≤ M
0for almost every x ∈ Ω, for some positive constant M
0.
Theorem 3.1. Assume (CD). Then there exists a unique nonnegative global solution (Y
1, Y
2, T ) for (1.1)–(1.5) which is smooth in Ω × (0, ∞).
P r o o f. For each 1 < p < ∞ and j ∈ {1, 2, 3} define the linear operator A
j,pon L
p(Ω) by
(3.1) D(A
j,p) = {u ∈ W
2,p(Ω) : (∂u/∂ν)|∂Ω = 0}, A
1,pu = d
0∆u, A
2,pu = d
2∆u, A
3,pu = d
7∆u,
where W
2,p(Ω) is the usual Sobolev space. It is well known that A
j,pgen- erates a compact, analytic contraction semigroup G
j,p(t), t ≥ 0, of bounded linear operators on L
p(Ω) (see, e.g., Amann [2]).
For the local existence we write (1.1)–(1.3) as a system of integral equa- tions via the variation of constants formula. For simplicity we set
F
1(Y
1, Y
2, T )(t)(·) = − d
1Y
1(t)Y
2(t)f
1(T (t))(·),
F
2(Y
1, Y
2, T )(t)(·) = (d
3Y
1(t)Y
2(t)f
1(T (t)) − d
4Y
2(t)f
2(T (t))
− d
5Y
2(t) − d
6Y
22(t))(·),
F
3(Y
1, Y
2, T )(t)(·) = (d
8Y
1(t)Y
2(t)f
1(T (t)) + d
9Y
2(t)f
2(T (t)) + d
10Y
2(y) + d
11Y
22(t))(·),
for x ∈ Ω, t > 0; we then have Y
1(t) = G
1,p(t)Y
10+
t
\
0
G
1,p(t − τ )F
1(Y
1(τ ), Y
2(τ ), T (τ )) dτ, (3.2)
Y
2(t) = G
2,p(t)Y
20+
t
\
0
G
2,p(t − τ )F
2(Y
1(τ ), Y
2(τ ), T (τ )) dτ, (3.3)
T (t) = G
3,p(t)T
0+
t
\
0
G
3,p(t − τ )F
3(Y
1(τ ), Y
2(τ ), T (τ )) dτ.
(3.4)
For each α > 0 define the operator B
j,p= I − A
j,p. Then the fractional
powers B
−αj,p= (I −A
j,p)
−αexist and are injective, bounded linear operators
on L
p(Ω) (see Pazy [8]). Let B
αj,p= (B
j,p−α)
−1and X
j,pα= D(B
j,pα), the
domain of B
j,pα. Then X
j,pαis a Banach space with the graph norm kuk
α=
kB
αj,pwk
p, and for α > β ≥ 0, X
j,pαis a dense subspace of X
pβwith the
inclusion X
j,pα⊂ X
j,pβcompact (we use the convention X
p0= L
p(Ω)). Also if 0 ≤ α < 1 we have
(3.5) X
j,pα⊂ C
µ(Ω) for every 0 ≤ µ < mα − n/p.
Note that this inclusion is valid even if p = 1 (see Henry [5], p. 39).
In addition, G
j,pand B
j,pαhave the properties summarised in the follow- ing lemma.
Lemma 3.2. The operators G
pand B
pαsatisfy (i) G
j,p(t) : L
p(Ω) → X
j,pαfor all t > 0,
(ii) G
j,p(t)B
αj,pu = B
j,pαG
j,p(t)u for every u ∈ X
j,pα,
(iii) kG
j,p(t)uk
α≤ C
1(α)t
−αe
−tkuk
pfor every t > 0 and u ∈ L
p(Ω), (iv) k(G
j,p(t) − I)uk
p≤ C
2(α)t
αkuk
αfor 0 < α ≤ 1 and u ∈ X
j,pα. The proof can be found in Pazy [8].
Select 0 < α < 1 and p > 1 so that (3.5) holds, and use the tech- niques of Pazy [8] to show that there exists a unique noncontinuable solution (Y
1, Y
2, T ) to (3.2)–(3.4) for Y
10∈ X
1,pα, Y
20∈ X
2,pαand T
0∈ X
3,pα. The solution satisfies
Y
1∈ C([0, δ]; X
1,pα) ∩ C
1((0, δ); L
p(Ω)), Y
2∈ C([0, δ]; X
2,pα) ∩ C
1((0, δ); L
p(Ω)), T ∈ C([0, δ]; X
3,pα) ∩ C
1((0, δ); L
p(Ω)),
for some δ > 0; and we have kY
1(t)k
∞+ kY
2(t)k
∞+ kT (t)k
∞→ ∞ as t → t
maxif t
max< ∞.
Suppose now that (Y
10, Y
20, T
0) ∈ L
∞(Ω)
3and let {Y
10k}
∞k=1be a se- quence in X
1,pα, {Y
20k}
∞k=1a sequence in X
2,pαand {T
0k}
∞k=1a sequence in X
3,pαsuch that Y
10k, Y
20k, T
0k≥ 0 and kY
10k− Y
10k
p→ 0, kY
20k− Y
20k
p→ 0 and kT
0k− T
0k
p→ 0 as t → ∞. Using the equation (3.2) and the properties of A
1,pstated in Lemma 3.2, it follows for α ≤ β < 1 that
kY
1kk
β≤ C
βt
−βkY
10kk
p+
t
\
0
C
β(t − τ )
−βkF
1(Y
1k(τ ), Y
1k(τ ), Y
1k(τ ))k
pdτ for all t ∈ [0, t
kmax), where t
kmaxis the maximal time of existence for the system (1.1)–(1.5) with initial conditions 0 ≤ (Y
10k, Y
20k, T
0k) ∈ X
1,pα× X
2,pα× X
3,pα. From these estimates one can deduce the existence of a C
βsuch that
max{kY
1k(t)k
β, kY
2k(t)k
β, kT
k(t)k
β} ≤ C
βt
−βfor all t ∈ [0, δ], k ≥ 1; thus {(Y
1k(t), Y
2k(t), T
k(t))}
∞k=1is contained in
a bounded subset of X
1,pβ× X
2,pβ× X
3,pβfor each t ∈ (0, δ]. So by the
compact imbedding of X
j,pβin X
j,pα(j = 1, 2, 3) for α < β < 1, we see that
for each t ∈ (0, δ] the sequences {Y
1k(t)}
∞k=1, {Y
2k(t)}
∞k=1and {T
k(t)}
∞k=1contain convergent subsequences {Y
1k,i(t)}
∞i=1, {Y
2k,i(t)}
∞i=1and {T
k,i(t)}
∞i=1in X
1,pα, X
2,pαand X
3,pαrespectively.
Now define Y
1(t) = lim
i→∞
Y
1k,i(t), Y
2(t) = lim
i→∞
Y
2k,i(t), T (t) = lim
i→∞
T
k,i(t) for each t ∈ [0, δ]. Then (Y
1(t), Y
2(t), T (t)) satisfies (3.2)–(3.4) for each t ∈ [0, δ]. Replacing [0, t
max) with [δ, t
max) and (Y
10, Y
20, T
0) by (Y
1(δ), Y
2(δ), T (δ)) and using the results already established when (Y
10, Y
20, T
0) ∈ X
1,pα× X
2,pα× X
3,pα, we find that there is a unique, classical noncontinuable solution (Y
1(t), Y
2(t), T (t)) on Ω × [0, t
max), for every (Y
10, Y
20, T
0) ∈ (L
∞(Ω))
3.
Since F
1(0, Y
2, T ) ≥ 0, F
2(Y
1, 0, T ) ≥ 0 and F
3(Y
1, Y
2, 0) ≥ 0 it follows that Y
1(t), Y
2(t) and T (t) have nonnegative values on Ω (see [10]), and by the maximum principle we have
(3.6) kY
1(t)k
∞≤ kY
10k
∞for all t ∈ [0, t
max).
Multiplying (1.2) by Y
2p−1and integrating the result over Ω × (0, t) we obtain
1 n
d dt
\
Ω
Y
2pdx ≤ c
\
Ω
Y
2pdx, where c = d
3kY
10k
∞kf
1(T (t))k
∞, hence
\
Ω
Y
2pdx ≤ |Ω| kY
20k
∞e
nptfor all t < t
max. We can then deduce
(3.7) kY
2(t)k
∞≤ e
ctkY
20k
∞for all t < t
max.
From the expression of F
3(Y
1, Y
2, T ) and (3.7) we can find two positive numbers c
1and c
2such that
(3.8) kF
3(Y
1(T ), Y
2(T ), T (t))k
∞≤ e
ct(c
1+ c
2e
ct) for all t < t
max, where c
1= B
1d
8kY
10k
∞+ d
9B
2+ d
10and c
2= d
11kY
20k
∞.
From (3.4) and (3.8) we obtain kT (t)k
∞≤ kT
0k
∞+
t
\
0
e
cτ(c
1+ c
2e
cτ) dτ, from which we have
(3.9) kT (t)k
∞≤ kT
0k
∞+ c
1c (e
ct− 1) + c
22c (e
2ct− 1) for all t < t
max.
Inequalities (3.6), (3.7) and (3.9) contradict the fact that t
max< ∞, hence
t
max= ∞.
4. Boundedness of the solution. In fact, the solution obtained in Theorem 3.1 is uniformly bounded over Ω × (0, ∞).
Theorem 4.1. Assume (CD). Then there exists a positive number M such that
(4.1) (4.2)
0 ≤ Y
1(x, t) ≤ kY
10k
∞for x ∈ Ω, t ≥ 0, 0 ≤ Y
2(x, t), T (x, t) ≤ M for x ∈ Ω, t ≥ 0.
P r o o f. The function Y
1is uniformly bounded by kY
10k
∞by the maxi- mum principle.
Let B(x, t) = d
3Y
1(x, t)f
1(T (x, t))− d
4f
2(T (x, t))− d
5− d
6Y
2(x, t). Then we can write
∂Y
2∂t = d
2∆Y
2+ B(x, t)Y
2with B(x, t) ≤ a (for example a = d
3kY
10kB
1) and B(x, t) is locally Lipschitz in (x, t). Moreover, Y
2∈ L
∞(R
+, L
1(Ω)). In fact, integrating (1.1) over Ω × (0, t) we obtain
(4.3)
\
Ω
Y
1(x, t) dx =
\
Ω
Y
10(x) dx − d
1 t\
0
\
Ω
Y
1(x, τ )Y
2(x, τ )f
1(T (x, τ )) dx dτ, which implies
(4.4)
t
\
0
\
Ω
(Y
1Y
2f
1(T ))(x, τ ) dx dτ ≤ |Ω|
d
1kY
10k
∞for all t ≥ 0, where |Ω| is the Lebesgue measure of Ω. Similarly, we get (4.5)
\
Ω
Y
2(x, t) dx
≤
\
Ω
Y
20(x) dx + d
3 t\
0
\
Ω
(Y
1Y
2f
1(T ))(x, τ ) dx dτ for all t ≥ 0.
From (4.4) and (4.5) we obtain (4.6) kY
2(t)k
1≤ |Ω|
kY
20k
∞+ d
3d
1kY
10k
∞for all t ≥ 0.
An application of the result of Alikakos ([1], §3) shows that Y
2(t) is uniformly bounded over Ω × (0, ∞):
(4.7) kY
2(t)k
∞≤ K for all t ≥ 0,
for some K > 0.
Now, integrating (1.3) over Ω × (0, t) we obtain
\
Ω
T (x, t) dx =
\
Ω
T
0(x) dx + d
8 t\
0
\
Ω
(Y
1Y
2f
1(T ))(x, τ ) dx dτ (4.8)
+ d
9 t\
0
\
Ω
(Y
2f
2(T ))(x, τ ) dx dτ
+ d
10 t\
0
\
Ω
Y
2(x, τ ) dx dτ + d
11 t\
0
\
Ω
Y
22(x, τ ) dx dτ.
Integrating (1.2) over Ω × (0, t) we obtain (4.9)
\
Ω
Y
2(x, t) dx + d
4t
\
0
\
Ω
(Y
2f
2)(x, τ ) dx dτ + d
5t
\
0
\
Ω
Y
2(x, τ ) dx dτ
+ d
6 t\
0
\
Ω
Y
22(x, τ ) dx dτ = d
3 t\
0
\
Ω
(Y
1Y
2)(x, τ ) dx dτ +
\
Ω
Y
20(x) dx, from which we deduce that
(4.10)
∞
\
0
\
Ω
(Y
2f
2)(x, τ ) dx dτ < ∞ and
∞
\
0
\
Ω
Y
22(x, τ ) dx dτ < ∞.
From (4.4)–(4.7) and (4.10) in (4.8) we obtain (4.11)
\
Ω
T (x, t) dx ≤ C for all t ≥ 0, i.e., T ∈ L
∞(R
+, L
1(Ω)).
To prove that T ∈ L
∞(R
+, L
∞(Ω)) we distinguish two cases. We define S
p(t) ≡ G
3,p(t).
Case 1: n = 1, i.e., Ω = (a, b) ⊂ R. In this case we take E := L
1(Ω) and F = C(Ω). Then Lemma 2.2 shows that
(4.12) kS
1(t)ϕk
∞≤ ct
−1/2kϕk
1for all ϕ ∈ L
1(Ω).
Take α = 3/4; from Lemma 2.2 and (3.5) we have S
1(t)L
1(Ω) ⊂ C(Ω).
Applying Lemma 2.1, we conclude that T ∈ C
B(δ, ∞; C(Ω)) for all δ > 0, hence from the result concerning the local existence we obtain
kT (t)k
∞≤ C for all t ≥ 0.
Case 2: n ≥ 2. Let q
1= 1, q
r= n/(n − r) and E = L
qr(Ω), F =
L
qr+1(Ω) for r ∈ {1, . . . , n−1}. We have T ∈ C
B(R
+, L
q1(Ω)), S
q1(t)L
q1(Ω) ⊂
L
q2(Ω) and kS
q1(t)ϕk
q2≤ ct
−1/2kϕk
q1. Application of Lemma 2.1 gives
T ∈ C
B(R
+, L
q2(Ω)). Next we take E = L
q2(Ω) and F = L
q3(Ω) to ob-
tain T ∈ C
B(R
+, L
q3(Ω)). Continuing this process we finally have T ∈
C
B(R
+, L
n(Ω)). In the last iteration we take E = L
n(Ω) and F = C(Ω).
As S
n(t)L
n(Ω) ⊂ X
3,nαand kS
n(t)ϕk
∞≤ ct
−1/2kϕk
nfor all ϕ ∈ L
n(Ω) and T ∈ C
B(R
+, L
n(Ω)), from Lemma 2.1 we conclude that T ∈ C
B(R
+; C(Ω)).
5. Asymptotic behaviour. First, let us establish a preparatory lemma. Consider the problem
(P) ∂u/∂t + Au = ϕ(t),
u(0) = u
0,
where −A generates an analytic semigroup G(t) in a Banach space (X, k · k) with Re σ(A) > a > 0. We have the following lemma.
Lemma 5.1. Let X be a Banach space. If ϕ ∈ L
∞(R
+, X) and the problem (P) has a bounded global solution u ∈ L
∞(R
+, X) then for all 0 <
α < 1 we have
(A) sup
t≥δkA
αu(t)k ≤ C(α, δ) for any δ > 0, and
(B) the function t 7→ A
αu(t) is H¨ older continuous from [δ, ∞) to X for any δ > 0.
P r o o f. The solution u of (P) satisfies the integral equation u(t) = G(t)u
0+
t
\
0
G(t − τ )ϕ(τ ) dτ, t > 0.
Applying A
αto both sides yields kA
αu(t)k ≤ kA
αG(t)u
0k +
t
\
0
kA
αG(t − τ )ϕ(τ )k dτ.
From this and Lemma 3.2, we obtain kA
αu(t)k
p≤ C
1(α)t
−αe
−atku
0k +
t
\
0
C
1(α)(t − τ )
−αe
−a(t−τ )kϕ(τ )k dτ
≤ C
1(α)ku
0k + C
1(α)M Γ (1 − α)a
α−1.
Here Γ is the gamma function of Euler. Hence kA
αu(t)k is uniformly bounded on [δ, ∞) for any δ > 0.
To prove (B), we have
kA
αu(t + h) − A
αu(t)k ≤ k(G(h) − I)A
αG(t)u
0k +
t+h
\
t
kA
αG(t + h − τ )ϕ(τ )k dτ
+
t
\
0
k(G(h) − I)A
αG(t − τ )ϕ(τ )k dτ.
Set
I
1= k(G(h) − I)A
αG(t)u
0k, I
2=
t+h
\
t
kA
αG(t + h − τ )ϕ(τ )k dτ,
I
3=
t
\
0
k(G(h) − I)A
αG(t − τ )ϕ(τ )k dτ.
From the inequalities of Lemma 3.2, there exist two constants C
1(α), C
2(α) such that
I
1≤ C
1(α + β)C
2(α)t
−1e
−atku
0kh
β, I
2≤ M C
1(α)h
1−α,
I
3≤ M C
1(α + β)C
2(β)Γ (1 − α − β)a
α+β−1h
β,
where M = sup
t≥0kϕ(t)k
pfor every 0 < β < 1. Taking β < 1 − α, we then have for all t ≥ δ,
kA
αu(t + h) − A
αu(t)k ≤ C(α, ku
0k) max{h
β, h
1−α}.
Remark . As a consequence of this lemma, the function t 7→ A
αu(t) is uniformly continuous.
The following proposition is also useful in the sequel.
Proposition 5.2. For any δ > 0, the family {Y
1(t) : t ≥ δ} is relatively compact in C(Ω).
P r o o f. We have ∂Y
1/∂t = d
0∆Y
1+ F
1(Y
1, Y
2, T ) where F
1(Y
1, Y
2, T ) =
−d
1Y
1Y
2f
1(T ). There is a positive constant N such that kF
1(Y
1, Y
2, T )k
∞≤ N for all t ≥ 0. Let 0 < ε < 1 and t > ε. Then we can write Y
1(t) = G
1,∞(ε)Y
1(t−ε)+[Y
1(t)−G
1,∞(ε)Y
1(t−ε)], where G
1,∞(t) is the semigroup generated by d
0∆ with homogeneous Neumann boundary conditions in the Banach space C(Ω). We set
Y
1ε(t) = G
1,∞(ε)Y
1(t − ε) and Y
1ε(t) = Y
1(t) − G
1,∞(ε)Y
1(t − ε).
Then {Y
1ε(t) : t ≥ δ} is relatively compact in C(Ω) since {Y
1(t − ε) : t ≥ δ}
is bounded and G
1,∞(δ) is a compact operator. Also, kY
1ε(t)k
∞=
t
\
t−ε
G
1,∞(t − s)F
1(Y
1, Y
2, T )(s) ds
∞
≤ εN,
therefore {Y
1(t) : t ≥ 1} is totally bounded, hence {Y
1(t) : t ≥ 1} is relatively
compact in C(Ω). As {Y
1(t) : δ ≤ t ≤ 1} is compact in C(Ω), it follows
that {Y
1(t) : t ≥ δ} is relatively compact in C(Ω). The same holds true for
{Y
2(t) : t ≥ δ} and {T (t) : t ≥ δ}.
Theorem 5.3. Under the assumptions (CD) we have
(5.1) lim
t→∞
kY
1(t)k
∞= 0, lim
t→∞
kY
2(t)k
∞= 0 and there exists a positive constant T
∞such that
(5.2) lim
t→∞
kT (t) − T
∞k
∞= 0.
P r o o f. From (1.1) we have
(5.3) d
dt
\
Ω
Y
1(x, t) dx = −d
1\
Ω
(Y
1(t)Y
2(t)f
1(T (t)))(x) dx ≤ 0, hence the function t 7→
T
Ω
Y
1(x, t) dx is nonincreasing. Let Y
1be a constant such that
(5.4) lim
t→∞
\
Ω
Y
1(x, t) dx = Y
1. From (1.2) we have
(5.5) d dt
\
Ω
Y
2(x, t) dx =
\
Ω
(d
3Y
1Y
2f
1(T )−d
4Y
2f
2(T )−d
5Y
2−d
6Y
22)(x, t) dx.
From (5.3) and (5.5) we deduce (5.6) d
dt
\
Ω
1 d
1Y
1+ 1 d
3Y
2(x, t) dx
= −
\
Ω
d
4d
3Y
2f
2(T ) + d
5d
3Y
2+ d
6d
3Y
22(x, t) dx ≤ 0, from which we infer that there is a constant K such that
(5.7) 1
d
1\
Ω
Y
1(x, t) dx + 1 d
3\
Ω
Y
2(x, t) dx → K as t → ∞.
Combining (5.1) and (5.7) we conclude that there is a positive constant Y
2such that
(5.8) lim
t→∞
\
Ω
Y
2(x, t) dx = Y
2.
Integrating (5.6) over (0, ∞) we conclude that there is a constant C such that
(5.9)
∞
\
0
\
Ω
Y
2(x, τ ) dx dτ ≤ C.
Combining (5.8) and (5.9) we find that Y
2= 0, whence
(5.10) lim
t→∞
\
Ω
Y
2(x, t) dx = 0.
As Y
2(x, t) ≥ 0, the invariance principle of La Salle [5] and (5.10) imply lim
t→∞kY
2(t)k
∞= 0.
Multiplying (1.1) by Y
1and integrating over Ω and using Poincar´e’s inequality we obtain
d dt
\
Ω
Y
12(x, t) dx ≤ −c
\
Ω
Y
12(x, t) dx for some positive constant c > 0, from which we deduce (5.11) kY
1(t)k
22≤ e
−ctkY
10k
22.
Also, as a consequence of the maximum principle we have (5.12) kY
1(t)k
∞≤ kY
1(s)k
∞for t ≥ s > 0.
According to Proposition 5.2, {Y
1(t) : t ≥ δ} is relatively compact in C(Ω) for all δ > 0; so from this, (5.11) and (5.12) we have
(5.13) lim
t→∞
kY
1(t)k
∞= 0.
Multiplying (1.2) by Y
2and integrating over Ω × (0, t) we have (5.14) kY
2(t)k
22+ 2d
2t
\
0
k∇Y
2(τ )k
22dτ + 2d
4 t\
0
\
Ω
Y
22f
2(T ) dx dτ
+ 2d
5 t\
0
kY
2(τ )k
22dτ + 2d
6 t\
0
\
Ω
Y
23dx dτ
= kY
20k
22+ 2d
3 t\
0
\
Ω
Y
1Y
22f
1(T ) dx dτ.
Similarly for (1.3), (5.15) kT (t)k
22+ 2d
7t
\
0
k∇T (τ )k
22dτ
= kT
0k
22+ 2d
8 t\
0
\
Ω
Y
1Y
2T f
1(T ) dx dτ
+ 2d
9 t\
0
\
Ω
Y
2T f
2(T ) dx dτ
+ 2d
10 t\
0
\
Ω
Y
2T dx dτ + 2d
11 t\
0
\
Ω
Y
22T dx dτ.
By (4.4) and as Y
1, Y
2and T are uniformly bounded, it follows from (5.14)
and (5.15) that ∇Y
2, ∇T ∈ L
2(R, L
2(Ω)), i.e.
(5.16)
∞
\
0
k∇Y
1(τ )k
22dτ < ∞,
∞
\
0
k∇Y
2(τ )k
22dτ < ∞,
∞
\
0
k∇T (τ )k
22dτ < ∞.
For the equation (1.1) for example, we define the operator B
pas follows:
D(B
p) = {u ∈ W
2,p(Ω) : (∂u/∂ν)|∂Ω = 0}, B
pu = (−d
0∆ + a)u, with a fixed positive real number a > 0. It is well known that −B
pgenerates an analytic semigroup and Re σ(B
p) > a > 0. Also, if we set ϕ(t) = aY
1(t)+
F
1(Y
1, Y
2, T )(t), then ϕ ∈ L
∞(R
+, L
p(Ω)). Application of Lemma 5.1 then implies
(5.17) sup
t≥δ
kB
αpY
1(t)k
p≤ C(p, α, δ) for any δ > 0, and
(5.18) t 7→ B
pαY
1(t) is uniformly continuous from [δ, ∞) to L
p(Ω) for any δ > 0.
The same holds for Y
2and T .
By (5.18) we find that t 7→ k∇Y
1(t)k
2, t 7→ k∇Y
2(t)k
2and t 7→ k∇T (t)k
2are uniformly continuous on [δ, ∞) by choosing p = 2 and suitable α ∈ (0, 1) and m. From this and (5.16), Lemma 5.1 gives
(5.19) lim
t→∞
k∇Y
1(t)k
2= 0, lim
t→∞
k∇Y
2(t)k
2= 0, lim
t→∞
k∇T (t)k
2= 0.
The interested reader can see [7] for details.
Since {T (t) : t ≥ δ} is compact in C(Ω) it follows that there is a sequence {t
k} such that
tk
lim
→∞T (t
k) = T
∞in C(Ω),
where T
∞is a constant. Owing to the Poincar´e inequality (see [11]) we have λ
T (t) − |Ω |
−1
\
Ω
T (x, t) dx
2
2
≤ k∇T (t)k
22.
Here λ is the smallest positive eigenvalue of −∆ with homogeneous Neumann boundary conditions on ∂Ω. Since the limit T
∞is uniquely determined we have
t→∞
lim T (t) = T
∞in C(Ω).
6. Rates of convergence. In this section we study the rates of con- vergence obtained in Theorem 5.3.
Theorem 6.1. Assume (CD). Then for given µ ∈ [0, 2), there exist
K
1(µ), K
2(µ), K(µ) > 0 and ̺, σ, ω > 0 such that
kY
1(t)k
Cµ(Ω)≤ K
1(µ)e
−̺(t−t∗), kY
2(t)k
Cµ(Ω)≤ K
2(µ)e
−σ(t−t∗), kT (t)k
Cµ(Ω)≤ K(µ)e
−ω(t−t∗),
for some t
∗> 0, as t → ∞, where 0 < σ < d
5, ̺ = min{σ, d
0λ}, ω = min{σ, d
7λ} and λ is the smallest positive eigenvalue of −∆ with homoge- neous Neumann boundary condition on ∂Ω.
Let us recall the following two lemmas.
Lemma 6.2. For 1 < p < ∞ and d > 0, let L
pbe the operator defined by D(L
p) = {u ∈ W
2,p(Ω) : (∂u/∂ν)|∂Ω = 0}, L
pu = −d∆u, and let the operators Q
0, Q
+: L
p(Ω) → L
p(Ω) be defined by
Q
0u = 1
|Ω|
\
Ω
u(x) dx, Q
+u = u − Q
0u.
Define the operator L
p+as L
p+≡ L
p|Q
+L
p(Ω), the restriction of L
pto Q
+L
p(Ω). Then there exists a constant C
3(α) > 0 such that for u ∈ L
p(Ω) and t > 0,
kL
αp+e
−tLp+Q
+uk
p≤ C
3(α)q(t)
−αe
−dλtkQ
+uk
p,
where q(t) = min{t, 1} and λ is the smallest positive eigenvalue of −∆ with homogeneous Neumann boundary conditions on ∂Ω.
Lemma 6.2 is proved by Rothe [9].
Lemma 6.3. For α ∈ [0, 1) and β > 0, there exists a constant C(α, β) > 0 such that
t
\
0
q(ξ)
−αe
βξdξ ≤ C(α, β)e
βt. For the proof, see [6].
Proof of Theorem 6.1. For 1 < p < ∞ we take the operators D(A
p) = D(B
p) = D(R
p) = {u ∈ W
2,p(Ω) : (∂u/∂ν)|∂Ω = 0}, A
pu = −d
0∆u, B
p= −(d
2∆ − d
5)u, R
p= −d
7∆u.
By Theorem 5.3 we already know that Y
1(t) → 0 and Y
2(t) → 0 in C(Ω) as t → ∞, hence for any ε > 0 there exists a constant t
∗> 0 such that (6.1) d
3Y
1f
1(T ) < ε for all t ≥ t
∗.
We take 0 < ε < d
5.
(I) The decay rate of kY
2(t)k
p. From (1.2) and (6.1) we get
(6.2) ∂Y
2∂t ≤ d
2∆Y
2− (d
5− ε)Y
2, t > t
∗.
Multiplying both sides by Y
2p−1for p ∈ [1, ∞), integrating over Ω and using Green’s formula, we obtain
(6.3) d
dt kY
2(t)k
pp≤ −p(d
5− ε)kY
2(t)k
ppfor t > t
∗, which leads to
kY
2(t)k
p≤ kY
2(t
∗)k
pe
−(d5−ε)(t−t∗)for t > t
∗. The H¨older inequality then yields
(6.4) kY
2(t)k
p≤ M |Ω|
1/pe
−(d5−ε)(t−t∗)for t > t
∗, where M is the positive number appearing in (4.2).
(II) The decay rate of kY
2(t)k
Cµ(Ω). To investigate the decay rate of kY
2(t)k
Cµ(Ω), we treat the following integral equation which is equivalent to (1.2) with (1.4) for t > t
∗:
∂Y
2∂t = d
2∆Y
2− d
5Y
2+ F (Y
1, Y
2, T ),
where F (Y
1, Y
2, T ) = d
3Y
1Y
2f
1(T ) − d
4Y
2f
2(T ) − d
6Y
22. Let G
p(t) be the semigroup generated by −B
p. Then
(6.5) Y
2(t) = G
p(t − t
∗)Y
2(t
∗) +
t
\
t∗
G
p(t − τ )F (τ ) dτ, t > t
∗. From Lemma 3.2(iii), we obtain
(6.6) kB
pαY
2(t)k
p≤ C
1(α)q(t − t
∗)
−αe
−d5(t−t∗)kY
2(t
∗)k
p+ d
3M B
1J
1(t), where
J
1(t) =
t
\
t∗
kG
p(t − τ )k
Lp(Ω)→Lp(Ω)kY
2(τ )k
pdτ.
It is sufficient to estimate J
1(t). By (6.4) and Lemmas 3.2 and 6.3 we have
J
1(t) ≤ C
1(α)M |Ω|
1/pt−t\∗
0
q(t − t
∗− τ )
−αe
−(d5−ε)τdτ (6.7)
≤ C
1(α)M |Ω|
1/pe
−(d5−ε)(t−t∗)for t ≥ t
∗.
Consequently, the imbedding D(B
αp) ⊂ C
µ(Ω) ensures the existence of a
constant K
2(µ) > 0 such that for every 0 < σ < d
5there is t
∗> 0 such that
(6.8) kY
2(t)k
Cµ(Ω)≤ K
2(µ)e
−σ(t−t∗)for t > t
∗.
(III) The decay rate of kY
1(t)k
Cµ(Ω). First we write Y
1(t) = Q
0Y
1(t) + Q
+Y
1(t), where
Q
0Y
1(t) = 1
|Ω|
\
Ω
Y
1(x, t) dx, Q
+Y
1(t) = Y
1(t) − Q
0Y
1(t).
We see from (4.3) and the fact that Y
1→ 0 as t → ∞ in C(Ω) that Q
0Y
1(t) ≡ 1
|Ω|
\
Ω
Y
1(x, t) dx (6.9)
= 1
|Ω|
\
Ω
Y
10(x, t) dx − d
1|Ω|
t
\
0
\
Ω
(Y
1Y
2f
1)(x, τ ) dx dτ
= d
1|Ω|
∞
\
t
\
Ω
(Y
1Y
2f
1)(x, τ ) dx dτ
≤ d
1|Ω| B
1M
0∞
\
t
\
Ω
Y
2(x, τ ) dx dτ
≤ d
1B
1M
0∞
\
t
e
−σ(τ −t∗)dτ
≤ 1
̺ d
1B
1M
0e
−̺(t−t∗)for t > t
∗.
Next, we study the decay of Q
+Y
1(t). We consider the integral equation associated with (1.1) and apply A
αp+Q
+to get
A
αp+Q
+Y
1(t) = G
1,p+(t − t
∗)Q
+Y
1(t
∗) − d
1 t\
t∗
G
1,p+(t − τ )(Q
+Y
1Y
2f
1)(τ ) dτ for t > t
∗. By Lemma 6.2, we get
kA
αp+Q
+Y
1(t)k
p≤ C
3(α)q(t − t
∗)
−αe
−d0λ(t−t∗)kQ
+Y
1(t
∗)k
p(6.10)
+ M B
1d
1C
3(α)kQ
+kJ
2(t) for t > t
∗, where J
2(t) =
Tt
t∗
q(t−τ )
−αe
−d0λ(t−τ )kY
2(τ )k
pdτ for t > t
∗and kQ
+k is the norm of the linear operator Q
+: L
p(Ω) → L
p(Ω). Here Q
+Y
1(t
∗) ∈ D(A
αp+) because t > t
∗and by the smoothness of Y
1(t
∗). By Lemma 6.3,
J
2(t) =
t
\
t∗
q(t − τ )
−αe
−d0λ(t−τ )kY
2(τ )k
pdτ (6.11)
≤ M |Ω|
1/pt−t∗
\
0