Electronic Journal of Differential Equations, Vol. 2010(2010), No. 75, pp. 1–15.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
REACTION-DIFFUSION SYSTEM OF EQUATIONS IN
NON-STATIONARY MEDIUM AND ARBITRARY NON-SMOOTH
DOMAINS
SIKIRU ADIGUN SANNI
Abstract. A system of non-linear partial differential equations describing
one-step irreversible reaction, reactant to product, in a non-stationary medium
and non-smooth domain is considered. After obtaining the necessary a priori
estimates, the existence of a unique local strong solution to the system is
proved using a fixed point theorem.
1. Introduction
We consider the semilinear parabolic system of partial differential equations
∇.v̄ = 0
in ΩT
(1.1)
∂v̄
1
− ν∆v̄ = −∇.(v̄ ⊗ v̄) − ∇p in ΩT
∂t
ρ
∂u
− k∆u = −∇.(v̄u) + Qwf (u) in ΩT
∂t
∂w
− d∆w = −∇.(v̄w) − wf (u) in ΩT
∂t
v̄ = 0̄, u = w = 0 on ∂Ω × [0, T )
v̄(x, 0) = v̄0 (x),
u(x, 0) = u0 (x),
w(x, 0) = w0 (x)
(1.2)
(1.3)
(1.4)
(1.5)
(1.6)
where 0̄ is the zero vector in R3 , ⊗ is the matrix multiplication defined by the tensor
v̄ ⊗ v̄ := vi vj (i, j = 1, 2, 3) and ΩT := Ω × [0, T ). Notice then that ∇.(v̄ ⊗ v̄) =
∂vj
∂
∂
∂xi (v̄i v̄j ) = ∂xj (v̄i v̄j ) = vi ∂xi = v̄.∇v̄ (using (1.1)).
In applications, the system models a single-step irreversible reaction, reactant
→ product in non-stationary incompressible medium. v̄(x, t) is the velocity of the
medium; ν and ρ are the kinematic viscosity and the density of the medium respectively. u(x, t) is the temperature in the reaction vessel, w(x, t) is the mass fraction
of the reactant, 1−w(x, t) is the mass fraction of the product, k the positive thermal
conductivity and d the reactant diffusivity. Qwf (u) and −wf (u) are the reaction
kinetics, determined by a positive, uniformly bounded and differentiable function
2000 Mathematics Subject Classification. 35B40, 35K57, 80A25.
Key words and phrases. Irreversible reaction; reactant diffusivity; thermal conductivity;
a priori estimates; Banach’s fixed point theorem.
c
2010
Texas State University - San Marcos.
Submitted November 27, 2009. Published May 21, 2010.
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S. A. SANNI
EJDE-2010/75
f (u). Furthermore, f 0 (u) is assumed to be Lipschitz continuous. It is assumed that
Ω is an open and bounded arbitrary non-smooth domain in R3 . Theoretically, the
reactant decomposes at a rate which is proportional to w(x, t)f (u), where f (u) is
the approximate number of molecules that have sufficient energy for the reaction
to begin. In this paper, we shall assume that
0 ≤ f (u) ≤ B
0
0
|f (u)| ≤ B ,
0
(1.7)
0
|f (u) − f (ũ)| ≤ L|u − ũ|
(1.8)
For further information on chemical kinetics and combustion, the reader is referred to Buckmaster[3], Buckmaster and Ludford [4], and Frank-Kamenetskii [9].
Several combustion models assumed some smoothness on the boundary vis-a-vis
stationary media. Authors of these models include Avrin [1, 2], Daddiouaissa [5],
De Oliviera et al [6], Fitzgibbon and Martin [8], Henry [10], Konach [11], Sanni
[13], Sattinger [14], and some literature cited in them.
In this paper, we establish the existence of a unique local-in-time strong solution to the system (1.1)-(1.6), in arbitrary non-smooth domains. Clearly, the
inclusion of the Navier-Stokes equations in the system implies that the medium is
non-stationary.
Using Leray projector [15], the problem (1.1)-(1.6) can be reduced to that of
finding only (v̄, u, w) by a variational formulation. We are thus motivated to define:
Definition 1.1. We call a solution (v̄, u, w) of the system (1.1)-(1.6) a strong
solution, provided (v̄, u, w) ∈ X 3 , where X is defined by
X := L∞ [0, T ; H01 (Ω)] ∩ H 1 [0, T ; H01 (Ω)] ∩ W 1,∞ [0, T ; L2 (Ω)]
(1.9)
2. A priori estimates
We will need the following Sobolev embedding theorem, stated and proved in [7,
pp. 265-266].
Theorem 2.1. Assume that Ω ⊂ Rn is open and bounded. Suppose U ∈ W01,p (Ω)
for some 1 ≤ p < n. Then we have the estimate
kU kLq (Ω) ≤ Ck∇U kLp (Ω)
(2.1)
for each q ∈ [1, p∗], the constant C depending only on p, q, n and Ω, where p∗ :=
is the Sobolev conjugate.
np
n−p
Notice that the hypothesis of Theorem 2.1 requires no smoothness assumption
on the boundary Ω.
We now set out to obtain a priori estimates required to prove the existence of a
unique local strong solution to the system (1.1)-(1.6). We first state and prove the
following Lemmas.
Lemma 2.2. Let u ∈ H 1 (Ω) and v, w, p ∈ H01 (Ω). Then
Z
(2.2)
uwp dx ≤ kuk2L2 (Ω) + C(Ω)−1 kwk2H 1 (Ω) kpk2H 1 (Ω)
0
0
Ω
Z
uwp dx ≤ kuk2L2 (Ω) + kpk2H 1 (Ω) + C(Ω)−3 kwk4H 1 (Ω) kpk2L2 (Ω)
(2.3)
0
0
Z Ω
uwp dx ≤ kuk2L2 (Ω) + C(Ω)−1 kwk2H 1 (Ω) T −1/2 kpk2L2 (Ω) + T 1/2 kpk2H 1 (Ω)
Ω
0
0
(2.4)
EJDE-2010/75
REACTION-DIFFUSION SYSTEM
3
Z
u (pw − p̃w̃) dx
Ω
h
i (2.5)
≤ kuk2L2 (Ω) + C(Ω)−1 kpk2H 1 (Ω) kw − w̃k2H 1 (Ω) + kp − p̃k2H 1 (Ω) kw̃k2H 1 (Ω)
0
0
0
0
Z
u (pw − p̃w̃) dx
Ω
h
≤ kuk2L2 (Ω) + C(Ω)−1 kpk2H 1 (Ω) T −1/2 kw − w̃k2L2 (Ω)
0
i
+ T 1/2 kw − w̃k2H 1 (Ω) + kp − p̃k2H 1 (Ω) T −1/2 kw̃k2L2 (Ω) + T 1/2 kw̃k2H 1 (Ω)
0
0
0
(2.6)
Z
uvwp dx ≤ k∇uk2L2 (Ω) + C(Ω)−1 kvk4H 1 (Ω) kuk2L2 (Ω)
0
Ω
(2.7)
+ C(Ω) T −1/2 kpk2L2 (Ω) + T 1/2 kpk2H 1 (Ω) kwk2H 1 (Ω)
0
0
Z
u(vwp − v̄˜w̃p̃)dx
Ω
≤ C(Ω)−1 kvk4H 1 (Ω) + kw̃k4H 1 (Ω) kuk2L2 (Ω)
0
0
h
+ k∇uk2L2 (Ω) + C(Ω) kwk2H 1 (Ω) T −1/2 kp − p̃k2L2 (Ω) + T 1/2 kp − p̃k2H 1 (Ω)
0
0
i
−1/2
2
1/2
2
2
2
+ T
kp̃kL2 (Ω) + T kp̃kH 1 (Ω) kw − w̃kH 1 (Ω) + kv − v̄˜kH 1 (Ω)
0
0
0
(2.8)
Proof. 1. Proof of (2.2). By Hölder’s inequality,
Z
uwp dx ≤ kukL2 (Ω) kwkL4 (Ω) kpkL4 (Ω)
Ω
(2.9)
≤ C(Ω)kukL2 (Ω) kwkH01 (Ω) kpkH01 (Ω)
by Sobolev embedding theorem. Then (2.2) follows easily from (2.9) by Cauchy’s
inequality with .
2. Proof of (2.3) and (2.4).
Z
Z
1
p2 w2 dx (by Cauchy’s inequality with )
uwp dx ≤ kuk2L2 (Ω) +
4
Ω
Ω
1
2
≤ kukL2 (Ω) + kpkL2 (Ω) kpkL6 (Ω) kwk2L6 (Ω) (by Hölder’s inequality)
4
≤ kuk2L2 (Ω) + C(Ω)(4)−1 kpkL2 (Ω) kpkH01 (Ω) kwk2H 1 (Ω) ,
0
(2.10)
by Sobolev embedding theorem. Then (2.3) and (2.4) follow by applying Cauchy’s
inequality with 2 and T 1/2 , respectively, to the appropriate factors of the second
term on the right side of (2.10).
3. Proof of (2.5) and (2.6).
Z
Z
Z
u(pw − p̃w̃)dx =
up(w − w̃)dx +
uw̃(p − p̃)dx.
(2.11)
Ω
Ω
Ω
Then (2.5) and (2.6) follows by applying (2.2) and (2.4) to (2.11) respectively.
4
S. A. SANNI
EJDE-2010/75
4. Proof of (2.7). By Young’s inequality and then by Hölder’s inequality,
Z
Z
Z
1
1
uvwp dx ≤
u2 v 2 dx +
w2 p2 dx
2
2
Ω
Ω
Ω
1
(2.12)
≤ kukL2 (Ω) kukL6 (Ω) kvk2L6 (Ω) + kwk2L6 (Ω) kpkL2 (Ω) kpkL6 (Ω)
2
≤ kukL2 (Ω) kukH01 (Ω) kvk2H 1 (Ω) + kwk2H 1 (Ω) kpkL2 (Ω) kpkH01 (Ω) ,
0
0
by By Sobolev embedding theorem. (2.7) follows by applying Cauchy’s inequalities
with and T 1/2 to the first and second terms on the right side of (2.12) respectively.
5. Proof of (2.8).
Z
u(vwp − v̄˜w̃p̃)dx
Ω
Z
Z
Z
(2.13)
=
uvw(p − p̃)dx +
uv p̃(w − w̃)dx +
up̃w̃(v − v̄˜)dx .
Ω
Ω
Ω
Then (2.8) follows by applying (2.7) to the each term on the right side of (2.13).
This concludes the proof of Lemma 2.2
Lemma 2.3. Let (1.1)-(1.4) hold. Suppose v̄0 , u0 , w0 ∈ H01 (Ω) ∩ H 2 (Ω), then
k∂t v̄0 k2L2 (Ω) + k∂t u0 k2L2 (Ω) + k∂t w0 k2L2 (Ω)
≤ C(k∇v̄0 k1H (Ω)2 + k∇u0 k1H (Ω)2 + k∇w0 k1H (Ω)2 )(1 + kv̄0 k2H 1 (Ω) ),
(2.14)
0
where C = C(ν, k, d, B, ρ, Ω, Q)
Proof. Taking (1.1) and (1.3) on t = 0 and multiplying the corresponding equation
to (1.3) by ∂t u0 , we estimate
Z
|∂t u0 |2 dx
Ω
Z
Z
Z
=−
∂t u0 .v̄0 .∇u0 dx + k
∂t u0 ∆u0 dx + Q
∂t u0 w0 f (u0 )dx
Ω
Ω
Ω
Z
Z
Z
1
QB |w0 |2 dx + k
|∆u0 |2 dx
≤ 2
|∂t u0 |2 dx +
(2.15)
4
0
Ω
Ω
(Integrating by parts, using Cauchy’s inequality with and (1.7))
Z
C(Q, B, k, Ω)
≤ 2
|∂t v̄0 |2 dx +
Ω
2
1
× kv̄0 kH 1 (Ω) k∇u0 kH (Ω)2 + kw0 k2H 1 (Ω) + k∇2 u0 k2L2 (Ω) ,
0
0
by Hölder and Poincare’s inequalities and using that k∆v̄0 kL2 (Ω) ≤ k∇2 v̄0 kL2 (Ω) .
Choosing > 0 sufficiently small and simplifying, we deduce
i
h
(2.16)
k∂t u0 k2L2 (Ω) ≤ C(Q, B, k, Ω) (1 + kv̄0 k2H 1 (Ω) )k∇u0 k1H (Ω)2 + kw0 k2H 1 (Ω)
0
0
Evaluating (1.1), (1.2) and (1.4) at t = 0, we obtain analogous estimates to (2.16),
viz:
k∂t w0 k2L2 (Ω)
k∂t v̄0 k2L2 (Ω) ≤ C(ν, Ω)(1 + kv̄0 k2H 1 (Ω) )k∇v̄0 k1H (Ω)2
0
i
h
2
≤ C(d, B, Ω) (1 + kv̄0 kH 1 (Ω) )k∇w0 k1H (Ω)2 + kw0 k2H 1 (Ω)
0
Combining (2.16), (2.17) and (2.18), we deduce (2.14).
0
(2.17)
(2.18)
EJDE-2010/75
REACTION-DIFFUSION SYSTEM
5
Theorem 2.4. Let v̄0 , u0 , w0 ∈ H01 (Ω) ∩ H 2 (Ω). Suppose (v̄, u, w) is a strong
solution of the system (1.1)-(1.6). Then we have the estimate
sup k∂t v̄k2L2 (Ω) + kv̄k2H 1 (Ω) + k∂t uk2L2 (Ω) + kuk2H 1 (Ω) + k∂t wk2L2 (Ω) + kwk2H 1 (Ω)
0
[0,T ]
0
0
+ k∇ (∂t v̄) k2L2 [0,T ;L2 (Ω)] + k∇(∂t u)k2L2 [0,T ;L2 (Ω)] + k∇ (∂t w) k2L2 [0,T ;L2 (Ω)]
CT [(1 + kv̄0 k2H 1 (Ω) )(G(v̄0 , u0 , w0 ) + 1)]3
0
≤
{1 − 2CT [(1 + kv̄0 k2H 1 (Ω) )(G(v̄0 , u0 , w0 ) + 1)]2 }3/2
0
=: Σ = constant,
(2.19)
for
T < {2C[(1 + kv̄0 k2H 1 (Ω) )(k∇v̄0 k1H (Ω)2 + k∇u0 k1H (Ω)2 + k∇w0 k1H (Ω)2 ) + 1]2 }−1 ,
0
(2.20)
where
G(v̄0 , u0 , w0 ) = k∇v̄0 k1H (Ω)2 + k∇u0 k1H (Ω)2 + k∇w0 k1H (Ω)2
and C = C(ν, k, d, Q, B, B 0 , Ω).
We will use (1.3) and the corresponding conditions in (1.5) and (1.6) to obtain
estimates for u; and thereafter, for brevity, state analogous estimates for v̄ and w.
Proof. 1. Multiplying (1.3) by ∂t u, integrating the ensuing equation by parts over
Ω and using (1.1) and (1.5), we deduce
Z
k d
2
|∂t u| dx +
|∇u| dx
2 dt
Ω
Ω
Z
Z
=
∇(∂t u).(v̄u)dx + Q
∂t uwf (u)dx
Z
2
Ω
Ω
(2.21)
1
≤ k∇(∂t u)k2L2 (Ω) + C(Ω) kv̄k1H (Ω)2 kuk1H (Ω)2 + Q2 B 2 kwk2H 1 (Ω)
0
+ k∂t uk2L2 (Ω) ,
using (2.2)) of Lemma 2.2. Simplifying, (2.21) yields
d k
kuk2H 1 (Ω)
0
dt 2
1
≤ k∇(∂t u)k2L2 (Ω) + C(Ω) kv̄k1H (Ω)2 kuk1H (Ω)2 + Q2 B 2 kwk2H 1 (Ω)
0
(2.22)
2. Differentiating (1.3) with respect to t yields
∂
(∂t u) − k∆(∂t u) = −∂t v̄.∇u + v̄.∇(∂t u) + Q∂t w.f (u) + Qw∂t uf 0 (u)
∂t
(2.23)
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S. A. SANNI
EJDE-2010/75
Multiply by ∂t u and integrating by parts over Ω and use (1.1), (1.5) to deduce:
1 d
(k∂t uk2L2 (Ω) ) + kk∇(∂t u)k2L2 (Ω)
2 dtZ
Z
=
∇(∂t u).∂t v̄.udx +
∇(∂t u).v̄.∂t udx
Ω
Z
Z
0
+Q
∂t u∂t uwf (u)dx + Q
∂t u∂t wf (u)dx
Ω
Ω
(2.24)
Ω
≤ k∇(∂t u)k2L2 (Ω) + k∇(∂t v̄)k2L2 (Ω) + C(Ω)−3 k∂t v̄k2L2 (Ω) kuk4H 1 (Ω)
0
k∇(∂t u)k2L2 (Ω)
+ QB
+
0
+
k∇(∂t u)k2L2 (Ω)
[k∂t uk2L2 (Ω)
QB(k∂t uk2L2 (Ω)
+
+
−3
+ C(Ω)
k∇(∂t u)k2L2 (Ω)
k∂t uk2L2 (Ω) kvk4H 1 (Ω)
0
+ C(Ω)−3 k∂t uk2L2 (Ω) kwk4H 1 (Ω) ]
0
k∂t wk2L2 (Ω) ),
using (2.3) of Lemma 2.2 and Young’s inequality.
3. Combining (2.22) and (2.24) we deduce
d
(k∂t uk2 2
+ kkuk2H 1 (Ω) ) + 2kk∇(∂t u)k2L2 (Ω)
0
dt h L (Ω)
≤ C k∇(∂t u)k2L2 (Ω) + k∇(∂t v̄)k2L2 (Ω) + k∂t uk2L2 (Ω)
+ 1 + −1 + −3 1 + k∂t v̄k2L2 (Ω) kv̄k2H 1 (Ω) + k∂t uk2L2 (Ω)
0
3 i
2
2
2
+ kukH 1 (Ω) + k∂t wkL2 (Ω) + kwkH 1 (Ω)
0
(2.25)
0
0
where C = C(Q, B, B , Ω).
4. Following steps 1-3 in respect of (1.1), (1.2), (1.4) and the corresponding
conditions in (1.5), we obtain analogous estimates to (2.25):
d
(k∂t v̄k2L2 (Ω) + νkv̄k2H 1 (Ω) ) + 2νk∇(∂t v̄)k2L2 (Ω)
0
dt
h
≤ C(Ω) k∇(∂t v̄)k2L2 (Ω) + (−1 + −3 ) k∂t v̄k2L2 (Ω) + kv̄k2H 1 (Ω)
i
0
(2.26)
,
d
(k∂t wk2L2 (Ω) + dkwk2H 1 (Ω) ) + 2dk∇(∂t w)k2L2 (Ω)
0
dt h
≤ C k∇(∂t w)k2L2 (Ω) + k∇(∂t v̄)k2L2 (Ω) + k∂t uk2L2 (Ω)
+ (1 + −1 + −3 ) 1 + k∂t v̄k2L2 (Ω) + kv̄k2H 1 (Ω) + k∂t uk2L2 (Ω)
0
3 i
2
2
2
+ kukH 1 (Ω) + k∂t wkL2 (Ω) + kwkH 1 (Ω)
,
0
(2.27)
0
0
where C = C(B, B , Ω).
5. Combining (2.25)-(2.27), choosing > 0 sufficiently small and simplifying, we
deduce
d
k∂t v̄k2L2 (Ω) + kv̄k2H 1 (Ω) + k∂t uk2L2 (Ω) + kuk2H 1 (Ω) + k∂t wk2L2 (Ω)
0
0
dt
+ kwk2H 1 (Ω) + k∇(∂t v̄)k2L2 (Ω) + k∇(∂t u)k2L2 (Ω) + k∇(∂t w)k2L2 (Ω)
0
≤ C(ν, k, d, Q, B, B 0 , Ω) 1 + k∂t v̄k2L2 (Ω) + kv̄k2H 1 (Ω) + k∂t uk2L2 (Ω)
0
EJDE-2010/75
REACTION-DIFFUSION SYSTEM
+ kuk2H 1 (Ω) + k∂t wk2L2 (Ω) + kwk2H 1 (Ω)
0
7
3
0
Solving the above equation, maximizing the left and right sides of the ensuing
inequalities and using Lemma 2.3 concludes the proof of the Theorem 2.1.
3. Existence of a Solution
We prove the existence of a unique local strong solution to the system (1.1)-(1.6),
in a subset K of the space X 3 equipped with the norm
k(η, ξ, ζ)kX 3
h
≤ kηk2L∞ [0,T ;H 1 (Ω)] + k∂t ηk2L∞ [0,T ;L2 (Ω)] + kξk2L∞ [0,T ;H 1 (Ω)]
0
0
+ k∂t ξk2L∞ [0,T ;L2 (Ω)] + kζk2L∞ [0,T ;H 1 (Ω)] + k∂t ζk2L∞ [0,T ;L2 (Ω)]
0
+ k∇(∂t η)k2L2 [0,T ;L2 (Ω)] + k∇(∂t ξ)k2L2 [0,T ;L2 (Ω)] + k∇(∂t ζ)k2L2 [0,T ;L2 (Ω)]
i 12
,
(3.1)
where X is defined by (1.9).
Theorem 3.1. Let v̄0 , u0 and w0 ∈ H01 (Ω) ∩ H 2 (Ω). Then there exists a unique
local strong solution to the system (1.1)-(1.6).
Proof. 1. The fixed point arguments for the system (1.1)-(1.6) are
∇.Q̄ = 0 in ΩT
(3.2)
∂ Q̄
1
− ν∆Q̄ = −∇.(v̄ ⊗ v̄) − ∇Y in ΩT
∂t
ρ
∂R
− k∆R = −∇.(v̄u) + Qwf (u) in ΩT
∂t
∂S
− d∆S = −∇.(v̄w) − wf (u) in ΩT
∂t
Q̄ = 0̄, R = S = 0 on ∂Ω × [0, T )
Q̄(x, 0) = v¯0 (x),
R(x, 0) = u0 (x),
S(x, 0) = w0 (x),
(3.3)
(3.4)
(3.5)
(3.6)
(3.7)
where Y is the pressure distribution corresponding to the solution (Q̄, R, S).
2. We next define a mapping
τ : X3 → X3
(3.8)
by setting τ [(v̄, u, w)] = (Q̄, R, S), whenever (Q̄, R, S) is derived from (v̄, u, w) via
(3.2)-(3.7). We will prove that for sufficiently small T > 0, τ is a contraction
mapping. Choose (v̄, u, w), (v̄˜, ũ, w̃) ∈ X 3 and define
τ [(v̄, u, w)] = (Q, R, S),
˜ , R̃, S̃).
τ [(v̄˜, ũ, w̃)] = (Q̄
˜ , R̃, S̃) of the system (3.2)-(3.7), we have
Thus, for two solutions (Q, R, S), and (Q̄
˜) = 0
∇.(Q̄ − Q̄
in ΩT
∂
˜ ) − ν∆(Q̄ − Q̄
˜ ) = −∇.(v̄ ⊗ v̄ − v̄˜ ⊗ v̄˜) − 1 ∇(Y − Ỹ ) in Ω
(Q̄ − Q̄
T
∂t
ρ
∂
(R − R̃) − k∆(R − R̃) = −∇.(v̄u − v̄˜ũ) + Q(wf (u) − w̃f (ũ)) in ΩT
∂t
(3.9)
(3.10)
(3.11)
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S. A. SANNI
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∂
(S − S̃) − d∆(S − S̃) = −∇.(v̄w − v̄˜w̃) − (wf (u) − w̃f (ũ)) in ΩT
∂t
˜ = 0̄, R − R̃ = S − S̃ = 0 on ∂Ω × [0, T )
Q̄ − Q̄
˜ )(x, 0) = 0̄, (R − R̃)(x, 0) = 0, (S − S̃)(x, 0) = 0
(Q̄ − Q̄
(3.12)
(3.13)
(3.14)
3. Multiplying (3.11) by ∂t (R − R̃), integrating the ensuing equation by parts
over Ω, using (3.13) and applying (2.5) of Lemma 2.2, we deduce
k d k∇(∂t (R − R̃))k2L2 (Ω)
k∂t (R − R̃)k2L2 (Ω) +
2 dt
Z
Z
˜
=
∇ ∂t (R − R̃) .(v̄u − v̄ ũ)dx + Q
∂t (R − R̃)(wf (u) − w̃f (ũ))dx
Ω
Ω
≤ k∇(∂t (R − R̃))k2L2 (Ω) + k∂t (R − R̃)k2L2 (Ω)
+ C(Ω)−1 kv̄k2H 1 (Ω) ku − ũk2H 1 (Ω) + kũk2H 1 (Ω) kv − v̄˜k2H 1 (Ω)
0
0
0
0
0
−1
2
2
+ C(Q, B, B , Ω)
kwkH 1 (Ω) ku − ũkH 1 (Ω) + kw − w̃k2H 1 (Ω) ,
0
0
(3.15)
0
where we have used some bounds in (1.7) and (1.8).
4. Further, we differentiate (3.11) with respect t to get
∂
(∂t (R − R̃)) − k∆(∂t (R − R̃))
∂t
= −∇. (∂t v̄u − ∂t v̄˜ũ + v̄∂t u − v̄˜∂t ũ) Q ∂t wf (u)
− ∂t w̃f (ũ) + w∂t uf 0 (u) − w̃∂t ũf 0 (ũ)
(3.16)
Multiplying (3.16) by ∂t (R − R̃), integrating by parts over Ω, and applying Young’s
inequality with , (2.6) and (2.8) as appropriate, we deduce
1 d
(k∂t (R − R̃)k2L2 (Ω) ) + kk∇(∂t (R − R̃))k2L2 (Ω)
2 dtZ
(∂t v̄u − ∂t v̄˜ũ + v̄∂t u − v̄˜∂t ũ).∇(∂t (R − R̃))dx
Z
+Q
∂t (R − R̃).(∂t wf (u) − ∂t w̃f (ũ) + w∂t uf 0 (u) − w̃∂t ũf 0 (ũ))dx
Ω
≤ 3k∇(∂t (R − R̃))k2L2 (Ω) + 2 + C(Ω)−1 kwk4H 1 (Ω) k∂t (R − R̃)k2L2 (Ω)
0
nh
0
−1
−1/2
2
+ C(Ω, B, B , Q, L)
T
k∂t v̄˜kL2 (Ω) + k∂t ũk2L2 (Ω) + k∂t wk2L2 (Ω)
+ k∂t uk2L2 (Ω) + T 1/2 k∇(∂t v̄˜)k2L2 (Ω) + k∇(∂t ũ)k2L2 (Ω) + k∇(∂t w)k2L2 (Ω)
i
+ k∇(∂t u)k2L2 (Ω)
ku − ũk2H 1 (Ω) + kv̄ − v̄˜k2H 1 (Ω) + kw − w̃k2H 1 (Ω)
0
0
0
h
−1/2
2
2
2
+ T
k∂t (v̄ − v̄˜)kL2 (Ω) + k∂t (u − ũ)kL2 (Ω) + k∂t (w − w̃)kL2 (Ω)
+ T 1/2 k∇(∂t (v̄ − v̄˜))k2L2 (Ω) + k∇(∂t (u − ũ))k2L2 (Ω)
i
o
+ k∇(∂t (w − w̃))k2L2 (Ω)
1 + kuk2H 1 (Ω) + kv̄k2H 1 (Ω) + kw̃k2H 1 (Ω) .
=
Ω
0
0
0
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REACTION-DIFFUSION SYSTEM
9
5. Combining the above inequality with (3.15), Choosing > 0 sufficiently small,
and simplifying, we deduce
d
k∂t (R − R̃)k2L2 (Ω) + kR − R̃k2H 1 (Ω) + k∇(∂t (R − R̃))k2L2 (Ω)
0
dt n
h
≤ C (1 + kwk2H 1 (Ω) )2 k∂t (R − R̃)k2L2 (Ω) + kR − R̃k2H 1 (Ω) + 1 + kv̄k2H 1 (Ω)
0
0
0
2
2
−1/2
2
2
2
+ kũkH 1 (Ω) + kwkH 1 (Ω) + T
k∂t v̄˜kL2 (Ω) + k∂t ũkL2 (Ω) + k∂t wkL2 (Ω)
0
0
+ k∂t uk2L2 (Ω) + T 1/2 k∇(∂t v̄˜)k2L2 (Ω) + k∇(∂t ũ)k2L2 (Ω) + k∇(∂t w))k2L2 (Ω)
i
ku − ũk2H 1 (Ω) + kv̄ − v̄˜k2H 1 (Ω) + kw − w̃k2H 1 (Ω)
+ k∇(∂t u)k2L2 (Ω)
0
0
0
h
−1/2
2
2
2
˜
k∂t (v̄ − v̄ )kL2 (Ω) + k∂t (u − ũ)kL2 (Ω) + k∂t (w − w̃)kL2 (Ω)
+ T
+ T 1/2 k∇(∂t (v̄ − v̄˜))k2L2 (Ω) + k∇(∂t (u − ũ))k2L2 (Ω)
i
o
+ k∇(∂t (w − w̃))k2L2 (Ω)
1 + kuk2H 1 (Ω) + kv̄k2H 1 (Ω) + kw̃k2H 1 (Ω) ,
0
0
0
(3.17)
where C = C(k, Ω, B, B 0 , Q, L).
˜ and S − S̃, which for
6. There exist analogous estimates to (3.17) for Q̄ − Q̄
brevity, we do not render here. If we combine these estimates with (3.17), we
deduce, after an application of the differential form of the Gronwall’s inequality,
the estimates:
˜ , R̃, S̃)k 3
k(Q̄, R, S) − (Q̄
X
= kτ [(v̄, u, w)] − τ [(v̄˜, ũ, w̃)]kX 3
1/2
h
i
≤ C T + T 1/2
exp 2−1 T (1 + kwk2H 1 (Ω) )
0
1/2
× 1 + k(v̄, u, w)k2X 3 + k(v̄˜, ũ, w̃)k2X 3
k(v̄, u, w) − (v̄˜, ũ, w̃)kX 3
(3.18)
where C = C(Q, B, B 0 , Ω, k, d, ν, L).
7. Notice that the bound in (3.18) is not uniform. Thus we need to prove the
existence of a unique solution in a subset of X 3 . Define a convex set
√
K := {(v̄, u, w)|(v̄, u, w) − (v̄0 , u0 , w0 ) ∈ X03 and k(v̄, u, w)kX 3 ≤ 2 Σ},
(3.19)
where X03 is the set where the initial and the boundary values are zero; and Σ =
constant is the bound in (2.19). We will show that, if T > 0 is sufficiently small,
then
τ [K] ⊆ K,
kτ [(v̄, u, w)] − τ [(v̄˜, ũ, w̃)]kX 3 ≤ γk(v̄, u, w) − (v̄˜, ũ, w̃)kX 3
(3.20)
for all (v̄, u, w), (v̄˜, ũ, w̃) ∈ K and some γ < 1. Using (2.19) and (3.7), we have
kτ [(v̄0 , u0 , w0 )]kX 3 = k(Q̄(x, 0), R(x, 0), S(x, 0))kX 3 = k(v̄0 , u0 , w0 )kX 3 ≤
√
Σ
(3.21)
10
S. A. SANNI
EJDE-2010/75
Therefore, for (v̄, u, w) ∈ K, using (3.18) and (3.21),
kτ [(v̄, u, w)]kX 3
≤ kτ [(v̄0 , u0 , w0 )]kX 3 + kτ [(v̄, u, w)] − τ [(v̄0 , u0 , w0 )]kX 3
1/2
h
i
√
≤ Σ + C T + T 1/2
exp 2−1 T (1 + kwk2H 1 (Ω) )
0
1/2
(3.22)
k(v̄, u, w) − (v̄0 , u0 , w0 )kX 3
× 1 + k(v̄, u, w)k2X 3 + k(v̄0 , u0 , w0 )k2X 3
1/2
√
√
≤ Σ + C T + T 1/2
exp 2−1 T (1 + 4Σ) (1 + 5Σ)1/2 (4 Σ)
√
≤ 2 Σ,
for T > 0 sufficiently small such that
1/2
4C T + T 1/2
exp[2−1 T (1 + 4Σ)](1 + 5Σ)1/2 ≤ 1
(3.23)
Thus τ [(v̄, u, w)] ∈ K, and hence τ (K) ⊆ K for T > 0 sufficiently small. Furthermore, if T is chosen sufficiently small such that
1/2
C T + T 1/2
exp[2−1 T (1 + 4Σ)](1 + 5Σ)1/2 = γ < 1,
(3.24)
then, (3.18) implies
kτ [(v̄, u, w)] − τ [(v̄˜, ũ, w̃)]kX 3 < γk(v̄, u, w) − (v̄˜, ũ, w̃)kX 3
(3.25)
for all (v̄, u, w), (v̄˜, ũ, w̃) ∈ K. Thus, the mapping τ is a strict contraction for
sufficiently small T > 0.
8. Given (v̄k , uk , wk ) (k = 0, 1, 2, . . . ), inductively define
(Q, R, S) := (v̄k+1 , uk+1 , wk+1 ) ∈ K
to be the unique weak solution of the linear initial boundary value problem
∇.v̄k+1 = 0 in ΩT
∂v̄k+1
1
− ν∆v̄k+1 = −∇.(v̄k ⊗ v̄k ) − ∇pk+1 in ΩT
∂t
ρ
∂uk+1
− k∆uk+1 = −∇.(v̄k uk ) + Qwk f (uk ) in ΩT
∂t
∂wk+1
− d∆wk+1 = −∇.(v̄k wk ) − wk f (uk ) in ΩT
∂t
v̄k+1 = 0̄, uk+1 = wk+1 = 0 on ∂Ω × [0, T )
v̄k+1 (x, 0) = v̄0 (x),
uk+1 (x, 0) = u0 (x),
wk+1 (x, 0) = w0 (x),
(3.26)
(3.27)
(3.28)
(3.29)
(3.30)
(3.31)
where Y := pk+1 is the pressure distribution corresponding to (v̄k+1 , uk+1 , wk+1 ).
By the definition of the mapping τ , we have (for k = 0, 1, 2, . . . ), using (3.26)(3.31) that
(v̄k+1 , uk+1 , wk+1 ) = τ [(v̄k , uk , wk )].
(3.32)
Consider the series
(v̄1 , u1 , w1 ) +
X
r≥2
[(v̄r , ur , wr ) − (v̄r−1 , ur−1 , wr−1 )]
(3.33)
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11
The partial sum of the first k + 1 terms of the series (3.33) is
(v̄1 , u1 , w1 ) +
k+1
X
[(v̄r , ur , wr ) − (v̄r−1 , ur−1 , wr−1 )] = (v̄k+1 , uk+1 , wk+1 )
(3.34)
r=2
Now, using (3.25), we have
k(v̄2 , u2 , w2 ) − (v̄1 , u1 , w1 )kX 3 = kτ [(v̄1 , u1 , w1 )] − τ [(v̄0 , u0 , w0 )]kX 3
< γk(v̄1 , u1 , w1 ) − (v̄0 , u0 , w0 )kX 3
k(v̄3 , u3 , w3 ) − (v̄2 , u2 , w2 )kX 3 = kτ [(v̄2 , u2 , w2 )] − τ [(v̄1 , u1 , w1 )]kX 3
< γ 2 k(v̄1 , u1 , w1 ) − (v̄0 , u0 , w0 )kX 3
(3.35)
(3.36)
By induction,
k(v̄k+1 , uk+1 , wk+1 ) − (v̄k , uk , wk )kX 3 < γ k k(v̄1 , u1 , w1 ) − (v̄0 , u0 , w0 )kX 3
√
(3.37)
< 4γ k Σ,
since (v̄1 , u1 , w1 ), (v̄0 , u0 , w0 ) are in K, defined by (3.19). Hence the series (3.33)
is absolutely convergent, since using (3.37), the series
X
√
4γ k Σ,
(3.38)
k=0
which converges, dominates
X
k(v̄1 , u1 , w1 )kX 3 +
k(v̄r , ur , wr ) − (v̄r−1 , ur−1 , wr−1 )kX 3 .
(3.39)
r≥2
This implies that the series (3.33) is convergent. Define
lim (v̄k+1 , uk+1 , wk+1 ) := (v̄, u, w).
k→∞
Thus (v̄k+1 , uk+1 , wk+1 ) → (v̄, u, w) uniformly in K. Thus
lim (v̄k+1 , uk+1 , wk+1 ) = (v̄, u, w) = lim τ [(v̄k , uk , wk )] = τ [(v̄, u, w)]
k→∞
k→∞
(3.40)
By (3.40), (v̄, u, w) ∈ K is the unique fixed point of τ .
9. As in [15], define
V := The closure of {ζ̄ ∈ Cc∞ (Ω) : ∇.ζ̄ = 0} in H01 (Ω).
(3.41)
In view of the previous steps of this section, we are motivated to give the following
definition.
Definition 3.2. The weak formulation of (1.1)-(1.6) is: For given (v̄0 , u0 , w0 ) ∈
[H01 (Ω) ∩ H 2 (Ω)]3 , find (v̄, u, w) ∈ K satisfying
Z
Z
Z
∂t v̄.ζ̄dx + ν
∇v̄ : ∇ζ̄dx = −
∇.(v̄ ⊗ v̄).ζ̄dx
(3.42)
Ω
Ω
Ω
Z
Z
Z
Z
∂t uξdx + k
∇u.∇ξdx = −
∇.(v̄u)ξdx + Q
wf (u)ξdx
(3.43)
Ω
ZΩ
ZΩ
Z
ZΩ
∂t wξdx + d
∇w.∇ξdx = −
∇.(v̄w)ξdx −
wf (u)ξdx
(3.44)
Ω
Ω
v̄(x, 0) = v̄0 (x),
Ω
u(x, 0) = u0 (x),
for each ζ ∈ V and each ξ ∈ H01 (Ω).
Ω
w(x, 0) = w0 (x),
(3.45)
12
S. A. SANNI
EJDE-2010/75
10. Before verifying that (v̄, u, w) is weak solution of (1.1)-(1.6), we first prove
the following Lemma.
Lemma 3.3. If (v̄k , uk , wk ) ∈ K, ξ ∈ H01 (Ω) and ζ̄ ∈ V , then
Z
Z
∇.(v̄k ⊗ v̄k ).ζ̄dx →
∇.(v̄ ⊗ v̄).ζ̄dx
Ω
Z
ZΩ
∇.(v̄k uk )ξdx →
∇.(v̄u)ξdx
ZΩ
ZΩ
∇.(v̄k wk )ξdx →
∇.(v̄w)ξdx
Ω
(3.46)
(3.47)
(3.48)
Ω
in L2 (Ω)
Z
wk f (uk )ξdx →
wf (u)ξdx
f (uk ) → f (u)
Z
Ω
(3.49)
(3.50)
Ω
Proof. (i). Proof of (3.46). Integrating by parts, we have
Z
Z
|
∇.(v̄k ⊗ v̄k ).ζ̄dx| = |
v̄k ⊗ v̄k : ∇ζ̄dx| ≤ kζkH01 (Ω) kuk k2H 1 (Ω) ,
Ω
0
Ω
(3.51)
by using (2.9) of Lemma 2.2. Equation (3.46) follows by taking limits on both sides
of (3.51). Further, the proofs of (3.47) and (3.48) follow by similar calculations.
(ii). Proof of (3.49). We have
Z
Z Z uk
Z
2
0
0
B |uk | + f (0)2 dx
|f (uk )|2 dx =
f
(r)dr
+
f
(0)
dx
≤
Ω
Ω
Ω
0
Z
(3.52)
0
2
0
≤ C1 (B , f (0)) (|uk | + 2|uk | + 1|)dx ≤ C2 (B , f (0), Ω)(kuk kL2 (Ω) + 1)2
Ω
R
where we have used the first inequality in (1.8) and the estimate Ω |uk |dx ≤
1
|Ω| 2 kuk kL2 (Ω) ). Then (3.49) follows by taking limits on both sides of (3.52).
(iii). Proof of (3.50). We estimate
Z
|
wk f (uk )ξdx| ≤ kwk kH01 (Ω) kf (uk )kL2 (Ω) kξkH01 (Ω) ,
(3.53)
Ω
using (2.9) of Lemma 2.2. Hence, (3.50) follows by taking limits in (3.53).
11. We now verify that (v̄, u, w) ∈ K is a weak solution of (1.1). Fix ζ ∈ V and
ξ ∈ H01 (Ω). Using (3.26)-(3.31), we have
Z
Z
Z
∂t v̄k+1 .ζ̄dx + ν
∇v̄k+1 : ∇ζ̄dx = −
∇.(v̄k ⊗ v̄k ).ζ̄dx
(3.54)
Ω
Ω
Ω
Z
Z
∂t uk+1 ξdx + k
∇uk+1 .∇ξdx
Ω
Ω
Z
Z
(3.55)
=−
∇.(v̄k uk )ξdx + Q
wk f (uk )ξdx
Ω
Ω
Z
Z
∂t wk+1 ξdx + d
∇wk+1 .∇ξdx
Ω
Ω
Z
Z
(3.56)
=−
∇.(v̄k wk )ξdx −
wk f (uk )ξdx
Ω
v̄k+1 (x, 0) = v̄0 (x),
uk+1 (x, 0) = u0 (x),
Ω
wk+1 (x, 0) = w0 (x) .
(3.57)
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REACTION-DIFFUSION SYSTEM
13
Letting k → ∞ in (3.54)-(3.57) and using Lemma 2.3 to handle the nonlinear terms
yield (3.42)-(3.45) as desired.
12. We next demonstrate how to obtain the pressure Y = pk+1 . First, we obtain
the boundary condition on pressure by taking (1.2) on the boundary and using
(1.5) to deduce
1
∇p = ν∆v̄ on ∂Ω
(3.58)
ρ
Following the steps in [12], we express (3.58) in terms of the standard normal
derivatives as
∂ 2 v̄
1 ∂p
(3.59)
= ν n̂. 2
ρ ∂n
∂n
where n̂(x) is the inward normal at x on ∂Ω.
Taking the divergence of (3.27) yields the equation satisfied by Y = pk+1 as
∆pk+1 = ρ∇.[∇.(v̄k ⊗ v̄k )],
(3.60)
which is a form of Poisson’s equation. Further, in sympathy with the boundary
condition (3.59), we impose the the boundary condition on the pressure pk+1 as
1 ∂pk+1
∂ 2 v̄
= ν n̂. 2
ρ ∂n
∂n
Hence, the formal solution of (3.60) subject to the condition (3.61) is
Z
pk+1 (x, t) = −ρ
G(x, y)∇.[∇.(v̄k (y, t) ⊗ v̄k (y, t))]dy
Ω
Z
∂ 2 v̄(y, t)
dS(y)
+ ρν
G(x, y)n̂.
∂n2
∂Ω
(3.61)
(3.62)
where G(x, y) is the Green’s function satisfying the Laplace’s equation in the form
∆G(x, y) = δ(x − y)
(3.63)
with the condition
∂G(x, y)
= 0 (x on ∂Ω).
∂n
where δ is the Dirac delta function.
1
, x, y ∈ Ω, we define
For n = 3, G(x, y) = |x−y|
Z
1
pk+1 (x, t) := −ρ
∇.[∇.(v̄k (y, t) ⊗ v̄k (y, t))]dy
Ω |x − y| + Z
1
∂ 2 v̄(y, t)
+ ρν
n̂.
dS(y),
|x − y| + ∂n2
Z ∂Ω
1
p (x, t) := −ρ
∇.[∇.(v̄(y, t) ⊗ v̄(y, t))]dy
Ω |x − y| + Z
1
∂ 2 v̄(y, t)
+ ρν
n̂.
dS(y)
|x − y| + ∂n2
Z∂Ω
1
p(x, t) := −ρ
∇.[∇.(v̄(y, t) ⊗ v̄(y, t))]dy
|x
−
y|
Ω
Z
∂ 2 v̄(y, t)
1
+ ρν
n̂.
dS(y)
∂n2
∂Ω |x − y|
(3.64)
(3.65)
(3.66)
(3.67)
14
S. A. SANNI
EJDE-2010/75
where > 0. Notice that
lim pk+1 (x, t) = pk+1 (x, t),
→ 0
lim p (x, t) = p(x, t)
→ 0
Hence, integrating twice by parts, using (1.1) and (1.5), we have
|pk+1 (x, t) − p (x, t)|
Z
1
= |ρ
∇.{∇.[v̄k (y, t) ⊗ v̄k (y, t) − v̄(y, t) ⊗ v̄(y, t)]}dy|
|x
−
y| + ZΩ
1
= |ρ
∇ ∇[
] : [v̄k (y, t) ⊗ v̄k (y, t) − v̄(y, t) ⊗ v̄(y, t)]dy|
|x − y| + Ω
(3.68)
which tends to 0 as k → ∞. Therefore
lim pk+1 (x, t) = p (x, t),
k→∞
(3.69)
From whence sending to 0, we obtain
lim pk+1 (x, t) = p(x, t),
k→ ∞
(3.70)
where, p(x, t) given by (3.67), is the pressure corresponding to the solution (v̄, u, w).
Indeed, (3.67) is the formal solution for the pressure p(x, t) satisfying
∆p = ρ∇.[∇.(v̄ ⊗ v̄)],
in terms of G(x, y), as obtained in [12].
(3.71)
4. Regularity
The Analysis so far carried out requires no smoothness assumption on the boundary. However, for smooth solution up to the boundary, one requires the boundary
∂Ω to be C ∞ . The lengthy proofs of the associated regularity theorems are currently being established by an analysis of certain difference quotients in another
paper.
Acknowledgments. The author would like to thank the anonymous referee whose
thoughtful comments improved the original version of this manuscript.
References
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[2] Avrin, J. D.; Qualitative theory of the Cauchy problem for a one-step reaction model on
bounded domains, SIAM J. Math. Anal., 22 (1991), 379-391.
[3] Buckmaster J.; The Mathematics of Combustion, SIAM, Philadelphia (1985).
[4] Buckmaster, J.; Ludford, G.; Theory of Laminar Flames, Cambridge University Press, Cambridge (1982).
[5] Daddiouaissa, E.; Existence of global solutions for a system of equations having a triangular
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Sikiru Adigun Sanni
Department of Mathematics, Statistics and Computer Science, University of Uyo, Uyo,
Akwa Ibom State, Nigeria
E-mail address: sikirusanni@yahoo.com