ISSN 0101-8205 www.scielo.br/cam
On a nonstationary nonlinear coupled system
∗GANG LI1, HUI WANG1 and JIANG ZHU2
1Department of Mathematics, Nanjing University of Information Science & Technology, 210044 Nanjing, China
2Laboratório Nacional de Computação Científica, MCT, Avenida Getúlio Vargas, 333, 25651-075 Petrópolis, RJ, Brazil
E-mails: [email protected] / [email protected] / [email protected]
Abstract. In this paper, a strongly nonlinear coupled elliptic-parabolic system modelling a class of engineering problems with heat effect is studied. Existence of a weak solution is first
established by Schauder fixed point theorem, where the coupled functionsσ (s),k(s)are assumed
to be bounded. The uniqueness of the solution is obtained by applying Meyers’ theorem and
assuming thatσ (s),k(s)are Lipschitz continuous. The regularity of the solution is then analyzed
in dimensiond≤2 under the assumptions onσ (s),k(s)∈C2(R)and the boundedness of their derivatives of second order. Finally, the blow-up phenomena of the system are studied.
Mathematical subject classification: 35J60, 35K55.
Key words:elliptic-parabolic system, existence, uniqueness, regularity, blow-up.
1 Introduction
In many engineering problems, cf. [1, 2, 3], and the references therein, we en-counter an incompressible quasi Newtonian flows with viscous heating which
#CAM-225/10. Received: 18/VI/10. Accepted: 18/IX/10.
can be modelled as:
(i) −2∇ ∙(σ (θ )D(u))+ ∇p = f
(ii) ∇ ∙u=0
(iii) θt − ∇ ∙(k(θ )∇θ )+u∙ ∇θ +cθ =σ (θ )|D(u)|2
(1)
whereuis the velocity, pthe pressure,θ the temperature. The viscosityσ is a function of θ,
D(u)= 1
2 ∇u+ ∇u
T
is the strain rate tensor, and|D(u)|2is the second invariant of D(u).
Problems of this type have received especial attention recently, cf. [2, 4, 5]. Very similar problems can be found in modelling turbulent flows, cf. [6, 7], thermistor problems, cf. [9, 10, 12, 15, 13, 14, 16, 19, 23, 22, 24, 26, 29, 31, 32], semiconductor devices, cf. [30, 17, 28], electromagnetic “induction heating” problems, cf. [11], and so on. The main difficulties in analysis of the system (1) come from the strongly coupled nonlinearity and the incompressibility (from numerical point of view). In this paper, we focus the first difficulty and consider its simplified scalar model:
(i) −∇ ∙(σ (θ )∇u)= f in×(0,T),
(ii) θt− ∇ ∙(k(θ )∇θ )+b∙ ∇θ+cθ=σ (θ )|∇u|2 in×(0,T),
(iii) u =0, θ =0 onŴ×(0,T),
(iv) θ (x,0)=θ0(x) in.
(2)
whereu,θ : ×(0,T) → Rare unknowns, is a bounded open subset of
Rd, d = 1, 2 or 3, Ŵits regular boundary, T is some positive given number. b,c are given vector and scalar functions. We will study the problem (1) in next work.
The model problem (2) can be also thought, from mathematical point of view, as a generalization of thermistor problem whereuis potential,θtemperature and
Following the works by Antontsev-Chipot [10] and Elliott-Larsson [16] for thermistor problem, we give in this paper a complete analysis such as exis-tence, uniqueness, regularity and blow-up of the problem (2). While in [24] the authors assumed f ∈ L2()to get the existence of the solution, we establish similar results with a weaker assumption on it (see Theorem 1 below), which also extends results in [10, 16], where simply f = 0 is involved in the par-tial differenpar-tial equation. By applying Meyers’ estimate from [21], regularity assumptions on the solution such that
∇ui,∇φ2∈ L2q/(q−n)(0,T;Lq()), i =1,2 and
∇φ1∈ L4q/(q−n)(0,T;Lq()), q >max(n,2)
etc. in [10] are not needed to reach the uniqueness of the solution. And a non-trivial extension of the blow-up analysis in [10] to the case of diffusion-convection-reaction is presented following the idea from [8].
The paper is constructed as follows. In section 2 we formulate the variational form of the problem. And the following two sections are devoted to analyze the existence and uniqueness of the weak solution. Then we study the regularity of the solution. Finally, we discuss the blow-up.
2 Variational formulation
We will use standard notation for the spaces and corresponding norms. Let
Wm,s() denotes the standard sobolev space, with its norm k ∙ k
Wm,s(), for m ≥ 0 and 1≤ s ≤ ∞. We write Hm() = Wm,s()whens = 2, with the normk ∙ kWm,2(), andLs()=W0,s()whenm=0, with the normk ∙ kLs(). W0m,s() is the closure of the space C0∞() for the normk ∙ kWm,s(). When
considering space-time functions v(x,t), (x,t) ∈ ×(0,T), we define the spaceLr(0,T;X)(1≤r <∞)(whereX is a Sobolev space on) as:
Lr(0,T;X):=
v :(0,T)→ X Z T
0
kv(t)krXdt <∞
.
We assume that θ0 ∈ L2(), and let V = H01(), V
′ = H−1() the dual space ofV. Then, for a given f ∈ L∞(0,T;V′), the variational formulation of the problem (2) can be defined as:
Find u ∈L∞(0,T;V), θ ∈L2(0,T;V)∩C([0,T];L2()), θt ∈L2(0,T;V′)such that
(i) a(θ;u, v)=(f, v), ∀v∈V
(ii) (θt, η)+b(θ;θ, η)=(σ (θ )|∇u|2, η), ∀η∈V ∩L∞()
(3)
where
a(θ;u, v)=(σ (θ )∇u,∇v), (4)
b(ξ;θ, η)=(k(ξ )∇θ,∇η)+(b∙ ∇θ, η)+(cθ, η), (5)
and(∙,∙)denotes the inner product of[L2()]dor the duality between[Ls()]d and[Ls′()]d,s′is the dual number ofs.
3 Existence of a weak solution We assume thatσ,k ∈C(R)satisfying
0< σ1≤σ (θ )≤σ2<∞; 0< k1≤k(θ )≤k2<∞ (6)
whereσi,ki are positive constants. C denotes a generic constant depending on ,d andσi,ki.
It is easy to see that, for any givenθ, andv, w∈V
a(θ;v, v)≥σ1k∇vk2L2(), (7)
|a(θ;v, w)| ≤σ2k∇vkL2()k∇wkL2() (8)
wherek∇ ∙ kL2 is equivalent to the normk ∙ kV by Poincaré inequality (cf. p. 11 in [25]).
Lemma 1. For any givenθ, the solution to(3.i)u satisfies that
k∇ukL∞(0,T;L2())≤CkfkL∞(0,T;V′). (9)
Moreover, if f ∈ L2(0,T;Lq())where q >d/2, then
kukL2(0,T;L∞())≤CkfkL2(0,T;Lq()). (10)
Proof. Letv = u in (3.i), and notice that (7), we can get (9). (10) is a
con-sequence of Theorem 8.16 in [20].
By (2.i), the right-hand side of (2.ii) can be written as
σ (θ )|∇u|2= ∇ ∙(σ (θ )u∇u)+ f u. (11)
Then, for anyη∈ V, we have
(σ (θ )|∇u|2, η) = −(∇ ∙(σ (θ )u∇u)+ f u, η)
= (σ (θ )u∇u,∇η)+(f u, η). (12)
By Lemma 1, if f ∈ L∞(0,T;V′)∩ L2(0,T;Lq()), then (12) defines an
element ofL2(0,T;V′). Thus, we can rewrite (3) equivalently as:
Find u ∈ L∞(0,T;V), θ ∈ L2(0,T;V)∩C([0,T];L2()), θt ∈ L2(0,T;V′)such that
(i) a(θ;u, v)=(f, v), ∀v ∈V
(ii) (θt, η)+b(θ;θ, η)=(σ (θ )|∇u|2, η), ∀η∈V
(13)
Problem (13) is easier to study since its trial and test function spaces are same. Since
(b∙ ∇θ, θ )= −(∇ ∙(bθ ), θ )= −(b∙ ∇θ, θ )−(∇ ∙bθ, θ )
or
(b∙ ∇θ, θ )= −1
2(∇ ∙bθ, θ ). (14)
So, if
c−1
whereλmindenotes the smallest eigenvalue of−1in, p>0 and if
b∈ [L3()]d, c∈L3/2(), (16)
we have, for anyξ, θ, η∈V b(ξ;η, η)≥k1k∇ηk2L2()+
c−1 2∇ ∙b
η, η
≥αk∇ηk2L2(), (17) |b(ξ;θ, η)| ≤ k2k∇θkL2()k∇ηkL2()+ kbkL3()k∇θkL2()kηkL6()
+kckL3/2()kθkL6()kηkL6()
≤ βk∇θkL2()k∇ηkL2()
(18)
whereα=min(k1,p), β =β(k2,b,c)is a constant. Then we can prove the following:
Theorem 1. (Existence) If f ∈ L∞(0,T;V′)∩L2(0,T;Lq()) (q > d/2), and b,c functions satisfying (15), then there exists a weak solution {u, θ} to problem(13)such that
k∇ukL∞(0,T;L2())≤CkfkL∞(0,T;V′), (19) kukL2(0,T;L∞())≤CkfkL2(0,T;Lq()), (20)
kθk2
C(0,T;L2())+ kθk 2
L2(0,T;V) ≤ Ckfk2
L∞(0,T;V′)kfk2L2(0,T;Lq())+ kθ0k2L2()
.
(21)
Proof. Choose ξ ∈ L2(0,T;L2()), since (7), (8) hold, we denote by
uξ ∈V the solution of
a(ξ;uξ, v)=(f, v), ∀v ∈V (22)
in view of the Lax-Milgram Theorem.
According the Theorem 2.1 in [10], there exists a uniqueθξ ∈ L2(0,T;V)∩ C(0,T;L2())withθξt ∈L2(0,T;V′)andθξ0=θ0the solution of
(θξt, η)+(k(ξ )∇θξ,∇η)+(b∙ ∇θξ, η)+(cθξ, η)
= −(σ (ξ )uξ∇uξ,∇η)+(f uξ, η), ∀η∈ V
together withk=k(ξ ). Let us consider the map
ξ →θξ = J(ξ ). (24)
This map carriesL2(0,T;L2())into itself. Moreover, since (15) holds, by (6) and (17), if choosingη=θξ in (23) and integrating int, we have
1 2kθξk
2
L2()+α Z t
0
k∇θξk2L2()dt ≤ 1
2kθξk 2
L2()+ Z t
0
(k(ξ )∇θξ,∇θξ)+(b∙ ∇θξ, θξ)+(cθξ, θξ) dt ≤ Z t 0
(σ (ξ )uξ∇uξ,∇θξ)dt + Z t 0
(f uξ, θξ)dt +
1 2kθ0k
2
L2().
(25)
By Hölder, Young’s inequalities, the first two terms of the right-hand side of the last equation follows,
Z t
0
(σ (ξ )uξ∇uξ,∇θξ)dt
≤ σ2 Z t
0
kuξkL∞()k∇uξkL2()∙ k∇θξkL2()dt
≤ α 4kθξk
2
L2(0,T;V)+C1kfk 2
L∞(0,T;V′)kfk2L2(0,T;Lq()),
(26) Z t 0
(f uξ, θξ)dt ≤ Z t 0
kfkV′ ∙ kuξkL∞()∙ kθξkVdt
≤ α 4kθξk
2
L2(0,T;V)+C2kfk 2
L∞(0,T;V′)kfk2L2(0,T;Lq()).
(27)
Hence, (25) follows kθξk2L2()+
Z t
0
k∇θξk2L2()dt
≤ C3 min 1 2, α 2 −1 n
kfk2L∞(0,T;V′)kfk2L2(0,T;Lq())+ kθ0k
2
L2() o
.
(28)
And again, choosingv ∈ L2(0,T;V),kvkL2(0,T;V) =1 in equation (23), one easily deduce
Therefore, provided we takeRlarge enough,ξ →θξ maps the ballBRof center
0 and radiusRinL2(0,T;L2())into itself. Moreover, since
W =
θ ∈ L2(0,T;V)|θt ∈ L2(0,T;V′)
is compactly imbedded inL2(0,T;L2()), and this mapping will be carried into a relatively compact set by (28), (29). We are then going to show that this map is continuous, it will be done by the Schauder fixed point theorem. We consider a sequenceξn ∈L2(0,T;L2())such that
ξn→ξ in BR (30)
definesuξnas in (22) andθξn= J(ξn). We will show that
θξn = J(ξn)→ J(ξ )=θξ in BR. (31)
For that, subtracting the equation satisfied by θξ from the one satisfied by
θξnwithη=θξn−θξ, we get d
dt(θξn−θξ), θξn−θξ
+ k(ξn)∇θξn−k(ξ )∇θξ,∇(θξn−θξ)
+ b∙ ∇(θξn−θξ), θξn−θξ
+(c(θξn−θξ), θξn−θξ)
= σ (ξn)|∇uξn|
2−σ (ξ )|∇u
ξ|2, θξn−θξ
.
(32)
By (6), (14), (15) and (17), if integrating int, we have, 1
2kθξn−θξk 2
L2()+α Z t
0
k∇(θξn−θξ)k2L2()dt ≤ 1
2kθξn−θξk 2
L2()+ Z t
0
(k(ξn)∇(θξn−θξ),∇(θξn−θξ))dt
−1 2
Z t
0
(∇ ∙b(θξn−θξ), θξn−θξ)dt
+ Z t
0
(c(θξn−θξ), θξn−θξ)dt
= Z t
0
((k(ξ )−k(ξn))∇θξ,∇(θξn−θξ))dt
+ Z t
0
((σ (ξn)uξn∇uξn−σ (ξ )uξ∇uξ),∇(θξn−θξ))dt
+ Z t
0
(f uξn− f uξ, θξn−θξ)dt
= I1+I2+I3.
Set
I4=
α
6 Z t
0
k∇(θξn−θξ)k2L2()dt, then by Hölder inequality and Poincaré inequality, we get
|I1| =
Z t
0
((k(ξ )−k(ξn))∇θξ,∇(θξn−θξ))dt
≤ I4+ 1
α
Z t
0
k(k(ξn)−k(ξ ))∇θξk2L2()dt, |I2| =
Z t
0
((σ (ξn)uξn∇uξn−σ (ξ )uξ∇uξ),∇(θξn−θξ))dt
≤ I4+ 1
α
Z t
0
kσ (ξn)uξn∇uξn−σ (ξ )uξ∇uξk2L2()dt, |I3| =
Z t
0
(f uξn− f uξ, θξn−θξ)dt ≤ Z t 0
kfkV′∙ kuξ
n−uξkL∞()∙ kθξn−θξkVdt
≤ I4+
kfk2L∞(0,T;V′)
α
Z t
0
kuξn−uξk2L∞()dt.
(34)
Thus, taking into account the Definition of I4, we have k(θξn−θξ)(t)k2L2()+
Z t
0
k∇(θξn−θξ)k2L2()dt
≤ 1 α min 1 2, α 2
−1 Z T
0
k(k(ξn)−k(ξ ))∇θξk2L2()dt
+ Z T
0
kσ (ξn)uξn∇uξn−σ (ξ )uξ∇uξk2L2()dt
+kfk 2
L∞(0,T;V′)
α min 1 2, α 2
−1 Z T
0
kuξn−uξk2L∞()dt
.
(35)
Since θξn is in a relatively compact set of BR, it is enough to show that θξ
is the only limit point forθξn. Letθξ′ be such limit point, i.e.
θξ′ = lim
nm→∞
θξn m in BR
provided that we have extracted another sequence of nm that still denoted by
nmwe can assume
Then, since|∇θξ|2 ∈ L1(QT), and by (6), we know almost everywhere by the
Lebesgue theorem, Z T
0
k(k(ξn m)−k(ξ ))∇θξk2L2()dt = Z T
0 Z
|k(ξn m)−k(ξ )|2|∇θξ|2d xdt→0.
Next, forn=nm the second integral in the right-hand side of (35) reads
Z T
0
kσ (ξn)uξn∇uξn−σ (ξ )uξ∇uξk2L2()dt ≤
Z T
0
kσ (ξn)uξn∇uξn−σ (ξn)uξn∇uξk2L2()dt +
Z T
0
kσ (ξn)uξn∇uξ−σ (ξn)uξ∇uξk2L2()dt +
Z T
0
kσ (ξn)uξ∇uξ−σ (ξ )uξ∇uξk2L2()dt = I +I I +I I I.
(37)
It is clear that
I ≤ C
Z T
0
kfk2Lq()k∇(uξn−uξ)k2L2()dt,
I I ≤ C
Z T
0
k(uξn−uξ)∇uξk2L2()dt,
I I I ≤ C
Z T
0
kfk2Lq()k(σ (ξn)−σ (ξ ))∇uξk2L2()dt.
By (36), (6) and Lemma 1, together with the Lebesgue theorem we can obtain
I I I →0. Next,uξnsatisfies
−∇ ∙(σ (ξn)∇uξn)= f; uξn =0 onŴ,
hence
σ (ξn)∇uξn,∇(uξn−uξ)
= σ (ξ )∇uξ,∇(uξn−uξ)
or
σ (ξn)∇(uξn−uξ),∇(uξn−uξ)
= (σ (ξ )−σ (ξn))∇uξ,∇(uξn−uξ)
,
which implies, for everyt,
Thus,
I ≤C
Z T
0
kfk2Lq()k(σ (ξn)−σ (ξ ))∇uξk2L2()dt→0 as above forI I I. By the Poincaré inequality, this implies
Z T
0
kuξn−uξk2L2()dt →0, and up to an extracted subsequence we can assume
uξn−uξ →0 a.e. on×(0,T); (39)
then the Lebesgue convergence theorem gives I I →0, which also implies the third integral in the right-hand side of (35) approaches to zero almost everywhere in×(0,T). Henceθξn→θξ =θξ′ inL2(0,T;L2()).
4 Uniqueness of the solution
Definition 1. We denote byRs for1<s <∞the class of regular subsets G in
Rd for which the Laplacian operator maps W1,s
0 (G)onto W
−1,s(G).
Remark 1. A boundedC1domain, for example, is of classR
s for everys ∈ (2,∞), see Theorem 4.6 in [34].
From now on, we assume that is of class Rr∗ for some r∗ > 2. For 1≤s ≤r∗, we define Ms ≥1 by
inf
u∈W01,s\{0} sup
v∈W01,s′\{0}
(∇u,∇v)
k∇ukLs()k∇vk
Ls′()
= 1
Ms
. (40)
It is easy to see thatM2=1 andMs′ =Ms. Lemma 2. ([32])If r ∈(2,r∗]is such that
Mr
σ2−σ1
σ1+σ2
<1, (41)
then, for anyθ, we have
inf
u∈W01,r\{0} sup
v∈W01,r′\{0}
a(θ;u, v)
k∇ukLr()k∇vk
Lr′ ()
where
γ = σ1+σ2
2Mr
1−Mr
σ2−σ1
σ1+σ2
>0. (43)
Similarly to [24, 32], we have
Lemma 3. Let f ∈ L∞(0,T;W−1,r()), where r is defined in Lemma2. Then,
for any givenθ, the solution to(13.i)u∈ L∞(0,T;W01,r())and satisfies that
k∇ukL∞(0,T;Lr())≤
1
γkfkL∞(0,T;W−1,r()) (44)
whereγ is defined by(43).
Lemma 4. Under the assumptions of Lemma3, ifb,c are bounded continuous functions,σ, k ∈ L∞(R), then solution to(13.ii)θ ∈Wr withθ (0)=θ0, where
Wr =
θ ∈ Lr(0,T;W1,r()):θt ∈ Lr(0,T;W−1,r()) .
Proof. Since u ∈ L∞(0,T;W1,r()), it follows thatσ (θ )|∇u|2 ∈ L∞(0,T;
Lr/2()) ֒→L∞(0,T;W−1,r()) ֒→Lr(0,T;W−1,r()). Following the idea
of Theorem 1 and Remark 5 in [21], and the similar proof in Chapter 4 of [18]
we can complete the proof.
To study the uniqueness of problem (13), we need to assume that: σ,k are Lipschitz continuous, i.e. there is a Lipschitz constant L, for any ξ1, ξ2 ∈ R such that,
|k(ξ1)−k(ξ2)|, |σ (ξ1)−σ (ξ2)| ≤L|ξ1−ξ2|. (45) Let(ui, θi),i =1,2, two weak solutions to problem (13), and setθˉ =θ1−
θ2,uˉ =u1−u2, noticing (4), we have∀v∈V,
a(θ1; ˉu, v)=a(θ1;u1, v)−a(θ1;u2, v)=a(θ2;u2, v)−a(θ1;u2, v). Therefore, lettingv= ˉu, by (7) and (45), we have
σ1k∇ ˉuk2L2() ≤ Lk∇ ˉukL2()k∇u2kLr()k ˉθkL2r/(r−2)()
or
On the other hand, subtracting the equation satisfied byθ1from the one satis-fied byθ2, we get, for anyη∈V,
(θtˉ, η)+b(θ1; ˉθ , η)
= b(θ2;θ2, η)−b(θ1;θ2, η)
+(σ (θ1)|∇u1|2−σ (θ2)|∇u2|2, η) =
k(θ2)−k(θ1)
∇θ2,∇η
+
σ (θ1)−σ (θ2)
|∇u1|2 +σ (θ2)∇ ˉu∙ ∇(u1+u2), η
.
Letη= ˉθ, and noticing (17), we have 1
2
d dtk ˉθ (t)k
2
L2()+αk∇ ˉθk 2
L2()
≤ ((k(θ1)−k(θ2))∇θ2,∇ ˉθ )+((σ (θ1)−σ (θ2))|∇u1|2,θ )ˉ +(σ (θ2)∇ ˉu∙ ∇(u1+u2),θ ).ˉ
(47)
By Hölder and Young inequalities, we easily can deduce that
(k(θ1)−k(θ2))∇θ2,∇ ˉθ )
≤ Lk∇θ2kLr()k∇ ˉθkL2()k ˉθkL2r/(r−2)() ≤ εk∇ ˉθk2L2()+Cεk∇θ2k
2
Lr()k ˉθk2L2r/(r−2)(),
((σ (θ1)−σ (θ2))|∇u1|
2,θ )ˉ
≤ Lk∇u1k
2
Lr()k ˉθk2
L2r/(r−2)(),
(σ (θ2)∇ ˉu∙ ∇(u1+u2),θ )ˉ
≤ σ2k∇ ˉukL2()k∇(u1+u2)kLr()k ˉθkL2r/(r−2)().
Combining the above estimates and (46), choosingε= α4, (47) follows 1
2
d dtk ˉθk
2
L2()+ 3α
4 k∇ ˉθk 2
L2()
≤ Ck∇θ2k2
Lr()+ k∇u1k2Lr()+ k∇u2k2Lr()
k ˉθk2
L2r/(r−2)().
(48)
By the Galiardo-Nirenberg interpolation inequality, (48) becomes
1 2
d dtk ˉθk
2
L2()+
3α
4 k∇ ˉθk
2
L2()
≤ Ck∇θ2k2Lr()+ k∇u1k2Lr()+ k∇u2k2Lr()
k ˉθk2(1− d r)
L2() k∇ ˉθk 2d
r
L2()
≤ εk∇ ˉθk2L2()+Cε
k∇θ2k2Lrr/(()r−d)+ k∇u1k2Lrr/(()r−d)+ k∇u2k2Lrr/(()r−d)
where we apply Young’s inequalityab ≤ εar/d +C
εbr/(r−d). Again choosing
ε = α4, and by the estimates in Lemma 3, 4, and by Sobolev inequality that
Lr(0,T)⊂L2r/(r−d)(0,T), Gronwall lemma implies that
k ˉθ (t)k2L2()+ Z t
0
k∇ ˉθ (τ )k2L2()dτ ≤Ck ˉθ (0)k 2
L2().
Thus, by (46), uniqueness of the solution follows. Therefore, with the above result, we can state:
Theorem 2. (Uniqueness)Under the conditions of Lemmas3and4with Lips-chitz assumption on k, σ as (45), there exists at most one weak solution to problem(2).
Remark 2.Here the uniqueness result is obtained without additional regularity assumptions on the solution as those required in [10].
5 Regularity of the solution
In this section we study the regularity of the solution to the problem (2) only on the dimension of spaced ≤ 2( in [33] we considered a simplified case of
k(θ )≡1). For our regularity estimates, we need to assume that
σ (s),k(s)∈C2(R), |σ′(s)| + |σ′′(s)| + |k′(s)| + |k′′(s)| ≤L′ (49) for alls ∈R, whereL′is some positive constant.
Then we have:
Theorem 3. (Regularity)Let T > 0 and assume thatθ0 ∈ H2()∩H01(),
b,c are bounded continuous functions. f ∈C2(0,T;Lr())where r is defined
in Lemma2. Then problem (2)has a unique solution u ∈ L∞(0,T;H2()),
θ ∈ C1(0,T;L2()) ∩C(0,T;H2()). Moreover, there is a constant C, depending on T, θ0, f, and on σ,k through the constantsσi,ki in(6), such
that for every t∈ [0,T], we have
Proof. We consider the initial value problem ( hereV =H2()∩H01())
θ (t)∈V,
(θt, η)+(k(θ )∇θ,∇η)+(b∙ ∇θ, η)+(cθ, η)
=(σ (θ )|∇u|2, η), ∀η∈V,t>0
(θ (0), η)=(θ0, η), ∀η∈V
(50)
whereu(t)is determined by the linear elliptic problem
u(t)∈ H01, (σ (θ )∇u,∇v)=(f, v) ∀v ∈ H01,t>0 (51) we are now to analyze the regularity of the solution.
Step 1. We first show some estimates ofu. By (10), we know
kukL∞() ≤CkfkLq() (52)
for someq > d/2. Next, by Lemma 3, there existes a 2 < r < r∗ such that k∇u(t)kLr()≤CkfkW−1,r(). This implies that for everyt,
ku(t)kW1,r()≤CkfkLr() ≤C (53)
We note that we should consider mostly the derivatives ofσ (θ )to obtain fur-ther estimates ofu. First, equation (51) implies that−σ (θ )1u−∇σ (θ )∙∇u= f, so that by Hölder inequality,
k1ukLr() =
f + ∇σ (θ )∙ ∇u
σ (θ )
Lr() ≤CkfkLr()+Ck∇θkL
r1()k∇ukLr2()
for anyr1,r2satisfying 1
r1 + 1
r2 = 1
r. We thus obtain
kukW2,r() ≤Ck1ukLr() ≤C(kfkLr()+ k∇θkLr1()k∇ukLr2()).
Hence, by Gagliardo-Nirenberg interpolation inequality and estimate (53), we will get
kukW2,r() ≤ C
kfkLr()+ kθkH2()kukδ
W1,r()kuk
1−δ
W2,r()
≤ C
kfkLr()+ kθkH2()kuk1−δ
W2,r()
≤
kfkLr()+ kθk1/δ
H2()
whereδ =1−d/r+d/r2=1−d/r1. In the last step we also used Young’s inequality
ab≤ε1−1/δa1/δ+εb1/(1−δ), ε >0, 0< δ <1, a,b>0
For the above estimates ofk∇θkLr1()to hold, it is required thatr1<∞, which in its turn is equivalent toδ <1. Thus, we have proved the preliminary estimate
ku(t)kHr()≤C
1+ kθ (t)k1H/δ2()
, ∀ δ ∈ [1−d/p,1) (54)
whereCis independent ofδ. Next, arguing as [16] treats, we get ku(t)kW1,∞()+ ku(t)k2
W1,4() ≤C
1+ kθ (t)kρH2()
(55) whereρ <1 ifd ≤2.
Step 2. We now estimateskθkH2()andkθtkL2(). First, without lost of gener-ality, we assume thatkθkW1,4()≤C (see Lemma 4). We note that it suffices to estimatekθtkL2(). In fact, equation (50) implies thatθt −k(θ )1θ +b∙ ∇θ +
cθ = P(σ (θ )|∇u|2+k′(θ )|∇θ|2), where P denotes the orthogonal projection ontoV. Hence,
kθkH2() ≤ Ck1θkL2()≤C
kθtkL2()+ kb∙ ∇θkL2()+ kcθkL2()
+ kσ (θ )|∇u|2kL2()+ kk′(θ )|∇θ|2kL2()
≤ CkθtkL2()+ kuk2
W1,4()+ kθk 2
W1,4()+ k∇θkL2()+ kθkL2()
.
By the interpolation inequality, we know the two terms in the last inequality can be estimaed bykθkH2(), and in view of the estimate (55), so we can get
kθ (t)kH2() ≤C 1+ kθt(t)kL2(). (56) For the further estimates ofθt, we differentiate equations (50) and (51) with respect tot. Beginning with (51), we have
(σ (θ )∇ut,∇v)= −(σ (θ )t∇u,∇v)+(ft, v), ∀v ∈ H01 (57) Becauseσ (θ )is continuous inC(R), and in view of (55), we have
kutkH1() ≤ C kσ′(θ )θt∇ukL2()+ kftkL2()
≤ C 1+ kθtkL2()kukW1,∞()
≤ C
1+ kθtk2L2()
.
Next, by (12), if differentiating equation (50), we have
(θtt, η)+((k(θ )∇θ )t,∇η)+(b∙ ∇θt, η)+(cθt, η)
= −((σ (θ )u∇u)t,∇η)+((f u)t, η) (59)
for each η ∈ V, where (k(θ )∇θ )t = k′(θ )θt∇θ +k(θ )∇θt. Similarly, if condition (15) is satisfied, withη=θt in above equation, we would get
1 2
d dtkθtk
2
L2()+αk∇θtk 2
L2()
≤ 1 2
d dtkθtk
2
L2()+(k(θ )∇θt,∇θt)− 1
2(∇ ∙bθt, θt)+(cθt, θt) = (σ (θ )u∇u)t,∇θt
+
(f u)t, θt
+
k′(θ )θt∇θ,∇θt
≤ Ck(σ (θ )u∇u)tk2
L2()+ k(f u)tk2L2()
+L′kθtk2L4()k∇θk 2
L4()+εk∇θtk 2
L2()
(60)
where
k(σ (θ )u∇u)tkL2()
≤ kσ′(θ )θtu∇ukL2()+ kσ (θ )ut∇ukL2()+ kσ (θ )u∇utkL2()
≤ CkθtkL2()kukL∞()kukW1,∞()+ kutkLr˜()kukW1,r()
+kukL∞()kutkH1()
1/r +1/r˜ = 1/2. By Sobolev’s inequality and known bounds foru andut in
(52), (53), (55), (58), we getk(σ (θ )u∇u)tkL2()≤C 1+ kθtk2
L2()
,thus (60) follows
d dtkθtk
2
L2()+ k∇θtk 2
L2() ≤C
1+ kθtk4L2()
(61) Sinceθ (0)= P(θ0), thus
kθ (0)kH2()≤Ck1θ (0)kL2() ≤Ckθ0kH2()≤C, then by (55),
kθt(0)kL2() ≤ kk(θ (0))1θ (0)kL2()+ kb∙ ∇θ (0)kL2()+ kcθ (0)kL2() +kk′(θ (0))|∇θ (0)|2kL2()+ kP(σ (θ (0))|∇u(0)|2)kL2() ≤ Ck1θ (0)kL2()+Cku(0)k2
Therefore, integrating (61) int, we get kθt(t)k2L2()+
Z t
0
k∇θt(τ )k2L2()dτ ≤C+C Z t
0
kθtk4L2()dτ By Gronwall’s lemma, we obtain
kθt(t)k2L2()+ Z t
0
k∇θt(τ )k2L2()dτ ≤exp Z t
0
kθt(τ )k2L2()dτ
(62) We are now to establish the estimates of the right hand side of the above inequality. Takingη=θt in (50),
kθtk2L2()+ 1 2
d
dt(k(θ )∇θ,∇θ )+(b∙ ∇θ, θt)+(cθ, θt)
= (σ (θ )|∇u|2, θt)+1 2(k
′(θ )|∇θ|2, θt) ≤ Ckuk2
W1,4()kθtkL2()+
L′ 2k∇θk
2
L4()kθtkL2()
(63)
In view of (56) and Sobolev inbedding inequality, if integrating int, we arrive at Z t
0
kθtk2L2()dτ + k∇θk 2
L2() ≤Ck∇P(θ0)k2L2()+Ct ≤C (64) sinceθ (0)= P(θ0), where P is bounded with respect to the normk∇ ∙ kL2(), which implies (62) is bounded byC.
Substituting this result into (54),(55), (56), (58), we may conclude that ku(t)kHr()+ ku(t)kW1,∞()+ kut(t)kH1()
+ kθ (t)kH2()+ kθt(t)kL2() ≤C
(65)
Step 3. We next to estimate tkθt(t)kH2() and tkθtt(t)kL2(). We note that
θtt − ∇ ∙(k(θ )∇θ )t+b∙ ∇θt +cθt = P(σ (θ )|∇u|2)t, where
∇ ∙(k(θ )∇θ )t =k′′(θ )θt|∇θ|2+2k′(θ )∇θt∇θ +k′(θ )θt1θ +k(θ )1θt
and
k(σ (θ )|∇u|2)tk
L2()
≤ kσ′(θ )θt|∇u|2k
L2()+2kσ (θ )∇u∇utkL2() ≤ CkθtkL2()kuk2
W1,∞()+ kukW1,∞()kutkW1,2()
≤ C
so it is easy to get the following estimates by (65), kθt(t)kH2() ≤C 1+ kθtt(t)kL2()
(67) In order to obtain estimate oftkθtt(t)kL2(), we differentiate (50) with respect tot and letη =θtt, similar disposal like before, and in view of (66) and (67), we get
kθttk2L2()+
d dtk∇θtk
2
L2()≤C Hence, if multiplying bytand integrating int, it follows
Z t
0
τkθttk2L2()dτ+tk∇θtk2L2() ≤Ct+ Z t
0
k∇θtk2L2()dτ ≤C (68) by virtue of (62) and (64). We then differentiate (59) with respect totand let
η=θtt to have 1 2
d dtkθttk
2
2+((k(θ )∇θ )tt, θtt)+(b∙ ∇θtt, θtt)+(cθtt, θtt) = ((σ (θ )u∇u)tt, θtt)+((f u)tt, θtt)
where
(k(θ )∇θ )tt =k′′(θ )θt2∇θ+k′(θ )θtt∇θ+2k′(θ )θt∇θt +k(θ )∇θtt (f u)tt = fttu+2ftut+ f utt
we could use similar method as above to treat them separately, then, if condition (15) is satisfied, we obtain
d dtkθttk
2
L2()+αk∇θttk2L2() ≤C
k(σ (θ )u∇u)ttk2L2()+ kuttk2L2()
(69) Differentiating (57) with respect tot, we have
(σ (θ )∇utt,∇v)+(σ (θ )tt∇u+2σ (θ )t∇ut,∇v)=(ftt, v) ∀v∈ H01 withv=utt, it follows
kuttkH1() ≤ C
(kθttkL2()+ kθtkL2()kθtkL∞())kukW1,∞()
+kθtkL∞()kutkW1,2()+ kfttk2
L2()
≤ C 1+ kθttkL2()
For the term
(σ (θ )u∇u)tt =σ (θ )ttu∇u+σ (θ )utt∇u
+σ (θ )u∇utt +2σ (θ )tut∇u+2σ (θ )tu∇ut+2σ (θ )ut∇ut
we estimate similarly to [16] to getk(σ (θ )u∇u)ttkL2() ≤ C(1+ kθttkL2()), together with (70) show (69) could be estimated by
d dtkθttk
2
L2()+ k∇θttk2L2() ≤C
1+ kθttk2L2()
If we multiply byt2and integrate, and in view of (68), it follows
t2kθtt(t)k2L2()+ Z t
0
τ2k∇θttk2L2()dτ ≤C
1+ Z t
0
τkθttk2L2()dτ
≤C
which completes the proof.
6 Blow-up result
In this section we are interested to investigate under what condition the solution exists globally or finite time blow-up? We consider the problem as follows:
(i) −∇ ∙(σ (θ )∇u)= f in×(0,T)
(ii) θt − ∇ ∙(k(θ )∇θ )+b∙ ∇θ+cθ=σ (θ )|∇u|2 in×(0,T)
(iii) u=u0, x∈Ŵ, t >0
(iv) ∂θ/∂ν=0, x∈Ŵ, t >0
(v) θ (x,0)=θ0, x∈
(71)
where∂/∂νis the outward normal derivative of∂.
equations thatθ˜and corresponding solutionu˜satisfy,
(i) −∇ ∙(σ (θ )˜ ∇ ˜u)= f int× {t>0}
(ii) θt˜ − ∇ ∙(k(θ )˜ ∇ ˜θ )+cθ˜=σ (θ )˜ |∇ ˜u|2 int× {t>0}
(iii) u˜= ˜u0=u0(x−tb,t), x∈Ŵ,t>0
(iv) ∂θ /∂ν˜ =0, x∈Ŵ,t>0
(v) θ (˜ x,0)=θ0, x∈t
(72)
where we still writex,tif not causing any confusion andt =−tb. We note that this transformation does not change the shape of the boundaryŴofand initial valueθ0. In this case we see that ifθ˜ blow up so doesθ andvise versa, thus the convection term has no effect on whether solution is blow-up in finite time. So we turn our attention to problem (72). We assume that
θ0≥0, x ∈ (73)
0<k(s), σ (s) <∞,∀s ≥0; σ (s)differentiable, σ′(s)≥0,∀s ≥0 (74)
and Z ∞
0
ds
σ (s) <∞
From [10], ifdγ (x)is the superficial measure onŴ, then
λ→
Z
Ŵ
|u−λ|2dγ (x)
achieves its minimum value for
λ= ˉu = 1
|Ŵ| Z
Ŵ
udγ (x)
thus, if we setut =
1 |t|
Z
t
ud x, we have for some constantC,
Z
Ŵ
|u− ˉu|dγ (x) ≤ Z
Ŵ
|u−ut|dγ (x)
≤ C
Z
t
so we get Z
Ŵ
|u− ˉu|2dγ (x)≤C
Z
t
|∇u|2d x ∀u ∈ H1(t) (75) whereCdenotes the best constant. Then we have the following result:
Theorem 4.Assume that c is nonpositive function and
Z
Z ∞
θ0
ds
σ (s)d x <
1
C
Z ∞ 0
Z
Ŵ
| ˜u0− ˉ˜u0|2dγ (x)dt (76)
where
˜
u0=u0(x−tb,t), uˉ˜0= 1 |Ŵ|
Z
Ŵ
˜
u0dγ (x)
then problem(71)cannot have a smooth global solution.
Proof. The proof is similar to that of Theorem 5.1 in [10]. Remark 3. In one dimension case, whenk ≡ 1 andbis a function of x with 1
2∇ ∙b+c ≥ 0, we could show thatθ (x,t)blow up globally. In other words, ift∗denotes the blowup time when
θ (x,t)→ ∞, a.e.x ∈ whent→t∗
Indeed, consider=(0,1), then integrate the equation
(σ (θ )u′)′= f
it follows
u′ = Z 1
0
f(x,t)d x+C(t)
σ (θ )
thus the equation satisfied byθ reads
θt−θx x +b∙θx +cθ = Z 1
0
f(x,t)d x+C(t)
2
Differentiating inx and letη=θx, we see that
ηt−ηx x+b∙ηx +(∇ ∙b+c)η= −
σ′(θ )
Z 1 0
f(x,t)d x +C(t)
2
σ (θ )2 η
with initial boundary
η(x,t)=0, x =0,1 η(x,0)=(θ0)x Assuming that(θ0)x ∈ L∞(0,1), and noting that
σ′(θ )
Z 1 0
f(x,t)d x+C(t)
2
σ (θ )2 ≥0,
1
2∇ ∙b+c≥0 then it follows from the parabolic maximum principle that
|θx|∞≤ |(θ0)x|∞ i.e.
θ (x,t)= Z x
x0
θx(x,t)d x+θ (x0,t) shows that ifθ (x0,t)blows up, thenθblows up for anyx.
Remark 4. We note that the result of Theorem 4 is independent of f, then, without loss of generality, we specialize the problem (72) in one dimension with f =0 to show the sharpness of (76). Still consider = (0,1),θ (˜ 0) =
θ0=Const and look for a solutionθ˜ = ˜θ (t)depending ontonly. Set ˜
u0(−tb,t)=a0(t), u˜0(1−tb,t)=a1(t) then the equation satisfied byuleads to
˜
u(x,t)=a0(t)+x a1(t)−a0(t)
thus, the equation satisfied byθ becomes ˜
i.e.
Z t
0 ˜
θt+cθ˜ σ (θ )˜ dt =
Z t
0
a1(t)−a0(t) 2
dt
or
Z θ˜ θ0
ds
σ (s)+
Z t
0
cθ˜ σ (θ )˜ ds=
Z t
0
a1(s)−a0(s) 2
ds
In this case, the failure of (76) reads Z ∞
θ0
ds
σ (s) ≥
Z ∞ 0
a1(s)−a0(s) 2
ds
which implies that (72) has a global solution which is bounded when Z ∞
θ0
ds
σ (s) ≥
Z ∞ 0
a1(s)−a0(s) 2
ds
and is unbounded otherwise.
Acknowledgements. The authors would like to thank Professor Hongwei Wu from Southeast University, Mr. Zhifen Xu and Miss Danni Hu from Nanjing University of Information Science & Technology for their helpful suggestions and discussions.
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