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 functions σ(s), k(s) are assumed to be bounded. The uniqueness of the solution is obtained by applying Meyers' theorem and assuming that σ(s), k(s) are Lipschitz continuous. The regularity of the solution is then analyzed in dimension d < 2 under the assumptions on σ(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.
elliptic-parabolic system; existence; uniqueness; regularity; blow-up
On a nonstationary nonlinear coupled system* * This work was supported partially by the National Natural Science Foundation of China (40537034) and the Brazilian National Council for Scientific and Technological Development (CNPq).
Gang LiI; Hui WangI; Jiang ZhuII
IDepartment of Mathematics, Nanjing University of Information Science & Technology, 210044 Nanjing, China. E-mails: ligang@nuist.edu.cn / huispecial@gmail.com
IILaboratório Nacional de Computação Científica, MCT, Avenida Getúlio Vargas, 333, 25651-075 Petrópolis, RJ, Brazil. E-mail: jiang@lncc.br
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 functions σ(s), k(s) are assumed to be bounded. The uniqueness of the solution is obtained by applying Meyers' theorem and assuming that σ(s), k(s) are Lipschitz continuous. The regularity of the solution is then analyzed in dimension d < 2 under the assumptions on σ(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 encounter an incompressible quasi Newtonian flows with viscous heating which can be modelled as:
where u is the velocity, p the pressure, θ the temperature. The viscosity σ is a function of θ,
is the strain rate tensor, and |D(u)|2 is 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:
where u, θ: Ω × (0, T) → are unknowns, Ω is a bounded open subset of d, d = 1, 2 or 3, Γ its 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 where u is potential, θ temperature and f, b, c ≡ 0. It is necessary and important to understand well such a fundamental model problem (2) in simulating quasi Newtonian flows with viscous heating, turbulent flows, thermistor problems, semiconductor devices, electromagnetic "induction heating" problems, etc.
Following the works by Antontsev-Chipot [10] and Elliott-Larsson [16] for thermistor problem, we give in this paper a complete analysis such as existence, uniqueness, regularity and blow-up of the problem (2). While in [24] the authors assumed f ∈ L2( Ω) 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 partial differential equation. By applying Meyers' estimate from [21], regularityassumptions on the solution such that
∇ui, ∇ ϕ2∈ L2q/(q - n)(0,T; Lq( Ω)), i = 1,2
and
∇ ϕ1∈ L4q/(q - n)(0,T; Lq( Ω)), 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( Ω) denotes the standard sobolev space, with its norm , for m > 0 and 1 < s < ∞. We write Hm( Ω) = Wm,s( Ω) when s = 2, with the norm , and Ls( Ω) = W0,s( Ω) when m = 0, with the norm . ( Ω) is the closure of the space ( Ω) for the norm . When considering space-time functions v(x, t), (x, t) ∈ Ω × (0, T), we define the space Lr(0, T; X)(1 < r < ∞) (where X is a Sobolev space on Ω) as:
In a similar way we can define L ∞(0, T; X) and C(0, T; X). Vector variables are, in general, denoted with bold face.
We assume that θ0∈ L2( Ω), and let V = ( Ω), V' = H-1( Ω) the dual space of V. Then, for a given f ∈ L ∞(0, T; V'), the variational formulation of the problem (2) can be defined as:
where
and (·, ·) denotes the inner product of [L2( Ω)]d or the duality between [Ls( Ω)]d and [Ls'( Ω)]d, s' is the dual number of s.
3 Existence of a weak solution
We assume that σ, k ∈ C() satisfying
where σi, ki are positive constants. C denotes a generic constant depending on Ω, d and σi, ki.
It is easy to see that, for any given θ, and v, w ∈ V
where is equivalent to the norm ||·||V by Poincaré inequality (cf. p. 11 in [25]).
To symmetrize the trial and test function spaces of (3.ii), we note that in [24, 32, 33] Meyer's estimate was applied and some regularity assumption in 3D case was needed. Here we are doing in a different way which is based on the Maximum principle and that regularity assumption is then unnecessary.
Lemma 1. For any given θ, the solution to (3.i) u satisfies that
Moreover, if f ∈ L2(0, T;Lq( Ω)) where q > d/2, then
Proof. Let v = u in (3.i), and notice that (7), we can get (9). (10) is a consequence of Theorem 8.16 in [20].
□
By (2.i), the right-hand side of (2.ii) can be written as
Then, for any η ∈ V, we have
By Lemma 1, if f ∈ L ∞(0, T; V') ∩ L2(0, T; Lq( Ω)), then (12) defines an element of L2(0, T; V'). Thus, we can rewrite (3) equivalently as:
Problem (13) is easier to study since its trial and test function spaces are same.
Since
(b · ∇ θ, θ) = - ( ∇(b θ), θ) = -(b · ∇ θ, θ) - ( ∇ · b θ, θ)
or
So, if
where λmin denotes the smallest eigenvalue of - Δ in Ω, p > 0 and if
we have, for any ξ, θ, η ∈ V
where α = 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( Ω)) (q > d/2), and b, c functions satisfying (15), then there exists a weak solution {u, θ} to problem (13) such that
Proof. Choose ξ ∈ L2(0, T; L2( Ω)), since (7), (8) hold, we denote by u ξ ∈ V the solution of
in view of the Lax-Milgram Theorem.
According the Theorem 2.1 in [10], there exists a unique θ ξ ∈ L2(0, T; V) ∩C(0, T; L2( Ω)) with ∈ L2(0, T; V') and = θ0 the solution of
together with k = k( ξ). Let us consider the map
This map carries L2(0, T; L2( Ω)) into itself. Moreover, since (15) holds, by (6) and (17), if choosing η = θ ξ in (23) and integrating in t, we have
By Hölder, Young's inequalities, the first two terms of the right-hand side of the last equation follows,
Hence, (25) follows
And again, choosing v ∈ L2(0, T; V), = 1 in equation (23), one easily deduce
Therefore, provided we take large enough, ξ → θ ξ maps the ball BR of center 0 and radius in L2(0, T; L2( Ω)) into itself. Moreover, since
W = { θ ∈ L2(0, T; V)| θt ∈ L2(0, T; V')}
is compactly imbedded in L2(0, T; L2( Ω)), 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 sequence ξn ∈ L2(0, T; L2( Ω)) such that
defines as in (22) and = J( ξn). We will show that
For that, subtracting the equation satisfied by θ ξ from the one satisfied by with η = - θ ξ , we get
By (6), (14), (15) and (17), if integrating in t, we have,
Set
then by Hölder inequality and Poincaré inequality, we get
Thus, taking into account the Definition of I4, we have
Since is in a relatively compact set of BR, it is enough to show that θ ξ is the only limit point for . Let be such limit point, i.e.
= in BR
provided that we have extracted another sequence of nm that still denoted by nm we can assume
Then, since | ∇ θ ξ|2∈ L1(QT), and by (6), we know almost everywhere by the Lebesgue theorem,
Next, for n = nm the second integral in the right-hand side of (35) reads
It is clear that
By (36), (6) and Lemma 1, together with the Lebesgue theorem we can obtain III → 0. Next, satisfies
- ∇( σ( ξn) ∇ ) = f; = 0 on Γ,
hence
( σ( ξn) ∇ , ∇( - u ξ)) = ( σ( ξ) ∇u ξ , ∇( - u ξ))
or
( σ( ξn) ∇( - u ξ), ∇( - u ξ)) = (( σ( ξ) - σ( ξn)) ∇u ξ, ∇( - u ξ)),
which implies, for every t,
Thus,
as above for III. By the Poincaré inequality, this implies
and up to an extracted subsequence we can assume
then the Lebesgue convergence theorem gives II → 0, which also implies the third integral in the right-hand side of (35) approaches to zero almost everywhere in Ω × (0, T). Hence → θ ξ = in L2(0, T; L2( Ω)).
□
4 Uniqueness of the solution
Definition 1. We denote bys for 1 < s < ∞ the class of regular subsets G in d for which the Laplacian operator maps (G) onto W-1,s(G).
Remark 1. A bounded C1 domain, for example, is of class s for every s ∈ (2, ∞), see Theorem 4.6 in [34].
From now on, we assume that Ω is of class
r* for some r* > 2. For 1 < s < r*, we define Ms > 1 by
It is easy to see that M2 = 1 and Ms' = Ms.
Lemma 2. ([32]) If r ∈ (2, r*] is such that
then, for any θ, we have
where
Similarly to [24, 32], we have
Lemma 3. Let f ∈ L ∞(0, T; W-1,r( Ω)), where r is defined in Lemma 2. Then, for any given θ, the solution to (13.i) u ∈ L ∞(0, T; ( Ω)) and satisfies that
where γ is defined by (43).
Lemma 4. Under the assumptions of Lemma 3, if b, c are bounded continuous functions, σ, k ∈ L ∞(), then solution to (13.ii) θ ∈ Wr with θ(0) = θ0, where
Wr = { θ ∈ Lr(0, T; W1,r( Ω)) : θt ∈ Lr(0, T; W-1,r( Ω))}.
Proof. Since u ∈ L ∞(0, T; W1,r( Ω)), it follows that σ( θ)| ∇u|2∈ L ∞(0, T; Lr/2( Ω)) L ∞(0, T; W-1,r( Ω)) Lr(0,T; W-1,r( Ω)). 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∈ such that,
Let (ui, θi), i = 1, 2, two weak solutions to problem (13), and set = θ1- θ2, = u1 - u2, noticing (4), we have ∀ v ∈ V,
a( θ1; , v) = a( θ1; u1, v) - a( θ1; u2, v) = a( θ2; u2, v) - a( θ1; u2, v).
Therefore, letting v = , by (7) and (45), we have
or
On the other hand, subtracting the equation satisfied by θ1 from the one satisfied by θ2, we get, for any η ∈ V,
Let η = , and noticing (17), we have
By Hölder and Young inequalities, we easily can deduce that
Combining the above estimates and (46), choosing ε = , (47) follows
By the Galiardo-Nirenberg interpolation inequality, (48) becomes
where we apply Young's inequality ab < εar/d + C εbr/(r - d) . Again choosing ε = , 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
Thus, by (46), uniqueness of the solution follows. Therefore, with the above result, we can state:
Theorem 2. (Uniqueness) Under the conditions of Lemmas 3 and 4 with Lipschitz 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 space d < 2( in [33] we considered a simplified case of k( θ) ≡ 1). For our regularity estimates, we need to assume that
for all s ∈ , where L' is some positive constant.
Then we have:
Theorem 3. (Regularity) Let T > 0 and assume that θ0∈ H2( Ω) ∩ ( Ω), b,c are bounded continuous functions. f ∈ C2(0, T; Lr( Ω)) where r is defined in Lemma 2. Then problem (2) has a unique solution u ∈ L ∞(0, T; H2( Ω)), θ ∈ C1(0, T; L2( Ω)) ∩ C(0, T; H2( Ω)). Moreover, there is a constant C, depending on T, θ0, f, Ω and on σ, k through the constants σi, ki in (6), such that for every t ∈ [0, T], we have
Proof. We consider the initial value problem ( here V = H2( Ω) ∩( Ω))
where u(t) is determined by the linear elliptic problem
we are now to analyze the regularity of the solution.
Step 1. We first show some estimates of u. By (10), we know
for some q > d/2. Next, by Lemma 3, there existes a 2 < r < r* such that . This implies that for every t,
We note that we should consider mostly the derivatives of σ( θ) to obtain further estimates of u. First, equation (51) implies that - σ( θ) Δu - ∇ σ( θ)· ∇u = f, so that by Hölder inequality,
for any r1, r2 satisfying . We thus obtain
Hence, by Gagliardo-Nirenberg interpolation inequality and estimate (53), we will get
where δ = 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 of to hold, it is required that r1 < ∞, which in its turn is equivalent to δ < 1. Thus, we have proved the preliminary estimate
where C is independent of δ. Next, arguing as [16] treats, we get
where ρ < 1 if d < 2.
Step 2. We now estimates and . First, without lost of generality, we assume that < C (see Lemma 4). We note that it suffices to estimate . In fact, equation (50) implies that θt - k( θ) Δ θ + b · ∇ θ + c θ = P( σ( θ)| ∇u|2 + k'( θ)| ∇ θ|2), where P denotes the orthogonal projection onto V. Hence,
By the interpolation inequality, we know the two terms in the last inequality can be estimaed by , and in view of the estimate (55), so we can get
For the further estimates of θt, we differentiate equations (50) and (51) with respect to t. Beginning with (51), we have
Because σ( θ) is continuous in C(), and in view of (55), we have
Next, by (12), if differentiating equation (50), we have
for each η ∈ V, where (k( θ) ∇ θ)t = k'( θ) θt ∇ θ + k( θ) ∇ θt. Similarly, ifcondition (15) is satisfied, with η = θt in above equation, we would get
where
1/r + 1/ = 1/2. By Sobolev's inequality and known bounds for u and ut in (52), (53), (55), (58), we get , thus (60) follows
Since θ(0) = P( θ0), thus
then by (55),
Therefore, integrating (61) in t, we get
By Gronwall's lemma, we obtain
We are now to establish the estimates of the right hand side of the above inequality. Taking η = θt in (50),
In view of (56) and Sobolev inbedding inequality, if integrating in t, we arrive at
since θ(0) = P( θ0), where P is bounded with respect to the norm , which implies (62) is bounded by C.
Substituting this result into (54), (55), (56), (58), we may conclude that
Step 3. We next to estimate and . We note that θtt - ∇·(k( θ) ∇ θ)t + b· ∇ θt + c θt = P( σ( θ)| ∇u|2)t , where
∇(k( θ) ∇ θ)t = k"( θ) θt| ∇ θ|2 + 2k'( θ) ∇ θt∇ θ + k'( θ) θtΔ θ + k( θ) Δ θt
and
so it is easy to get the following estimates by (65),
In order to obtain estimate of , we differentiate (50) with respect to t and let η = θtt, similar disposal like before, and in view of (66) and (67), we get
Hence, if multiplying by t and integrating in t, it follows
by virtue of (62) and (64). We then differentiate (59) with respect to t and let η = θtt to have
where
we could use similar method as above to treat them separately, then, if condition (15) is satisfied, we obtain
Differentiating (57) with respect to t, we have
( σ( θ) ∇utt, ∇v) + ( σ( θ)tt∇u + 2 σ( θ)t∇ut, ∇v) = (ftt, v) ∀v ∈
with v = utt, it follows
For the term
( σ( θ)u ∇u)tt = σ( θ)ttu ∇u + σ( θ)utt ∇u
+ σ( θ)ttu ∇utt + 2 σ( θ)tut∇u + 2 σ( θ)tu ∇ut + 2 σ( θ)ut∇ut
we estimate similarly to [16] to get , together with (70) show (69) could be estimated by
If we multiply by t2 and integrate, and in view of (68), it follows
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:
where ∂/ ∂ ν is the outward normal derivative of ∂ Ω.
We note that our difficulty to treat problem (71) compared to generalconsidered problem lies in the convection term b · ∇ θ and reaction term cθ. From the physical point of view in [8], any solution θ to (71) can be writtenas θ(x, t) = (x - tb,t), by a variable transformation, we can obtain theequations that and corresponding solution satisfy,
where we still write x, t if not causing any confusion and Ωt = Ω - tb. We note that this transformation does not change the shape of the boundary Γ of Ω and initial value θ0. In this case we see that if blow up so does θ and vise 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
and
From [10], if d γ(x) is the superficial measure on Γ, then
achieves its minimum value for
thus, if we set
udx, we have for some constant C,
so we get
where C denotes the best constant. Then we have the following result:
Theorem 4. Assume that c is nonpositive function and
where
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, when k ≡ 1 and b is a function of x with ∇ b + c > 0, we could show that θ(x, t) blow up globally. In other words, if t* denotes the blowup time when
θ(x, t) → ∞, a.e. x ∈ Ω when t → t*
Indeed, consider Ω = (0,1), then integrate the equation
( σ( θ)u')' = f
it follows
thus the equation satisfied by θ reads
Differentiating in x and let η = θx, we see that
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
then it follows from the parabolic maximum principle that
| θx| ∞ < |( θ0)x| ∞
i.e.
shows that if θ(x0, t) blows up, then θ blows up for any x.
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 Ω = (0,1), (0) = θ0 = Const and look for a solution = (t) depending on t only. Set
0(-tb, t) = a0(t), 0(1 - tb, t) = a1(t)then the equation satisfied by u leads to
(x, t) = a0(t) + x(a1(t) - a0(t))
thus, the equation satisfied by θ becomes
t - c = σ()a1(t) - a0(t))2i.e.
or
In this case, the failure of (76) reads
which implies that (72) has a global solution which is bounded when
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.
Received: 18/VI/10.
Accepted: 18/IX/10.
#CAM-225/10.
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Publication Dates
-
Publication in this collection
06 Jan 2012 -
Date of issue
2011
History
-
Accepted
18 Sept 2010 -
Received
18 June 2010