On a Degenerate Evolution System Associated with the Bean Critical-State for Type II Superconductors

and Applied Analysis 3 Example 3. f p (x, s) = (2/p)a(x)s p/2 where a ∈ C2(Ω) with 0 < c 1 ≤ a(x) ≤ c 2 < ∞ satisfying (H.1). In particular, if a(x) ≡ 1, system (16)–(19) becomes (1)–(4). The following definition of solution of system (16)–(19) is based on Brezis [9, Definition 3.1]. Definition 4. One calls a vector field H = H(x, t), a solution of system (16)–(19), if the following hold: (i) H ∈ C([0, T]; L2(Ω)) and is absolutely continuous with respect to t in all compact subset of (0, T), and for a.e. t ∈ (0, T), H t ∈ L 2 (Ω,R3) (therefore for a.e. t ∈ (0, T),H is differentiable with respect to t); (ii) H(x, t) ∈ Hp 0 (curl, div0, Ω) for a.e. t ∈ (0, T); (iii) curl[f p (x, |curlH|) curlH] ∈ L2(Ω,R3) for a.e. t ∈ (0, T); (iv) (16) holds inL(Ω,R) for a.e. t ∈ (0, T) andH(x, 0) = H 0 (x) in Ω. (One notes that, for a.e. t ∈ (0, T),H is differentiable with respect to t.) Here we give Green’s formula which will be used frequently later (cf. [6]). Lemma 5. For A ∈ H(curl, Ω) and B ∈ W1,p 󸀠 (Ω,R3) where p 󸀠 is the conjugate exponent for p, that is to say, p󸀠 = p/(p−1), then one has ∫ Ω curlA ⋅ B dx − ∫ Ω A ⋅ curlB dx = ⟨^ × A,B⟩ ∂Ω , (20) where a ⋅ b denotes the usual inner product for vectors a and b in R3 and ⟨⋅, ⋅⟩ ∂Ω denotes the duality betweenW(∂Ω) andW 󸀠 ,p 󸀠 (∂Ω). 3. Existence, Uniqueness, and Regularity of Solution In this section, we will consider the existence and uniqueness for system (16)–(19) and also regularity undermore restrictive hypotheses on F. Theorem 6. Let p ≥ 2. Assume that Ω is a bounded domain inR without holes and withC boundary. Under hypotheses (H.1) and (H.2), system (16)–(19) has a unique solution in the sense of Definition 4. Moreover, one hasH t ∈ L 2 (Q T ) andH ∈ L ∞ (0, T;H p 0 (curl, div0, Ω)). The proof will be achieved by applying theorems due to Brezis [9, Theorems 3.4 and 3.6]. In order to do so, define a Hilbert space X = L2(Ω,R3) and a functional I p [V] for V ∈ X by


Introduction
The Bean critical-state model describes the hysteretic magnetization of type II superconductors under a varying external magnetic field (cf.Prigozhin [1] and de Gennes [2]).For the description of the classical Bean critical-state model, Yin et al. [3] proposed the following degenerate evolution system: where Ω is a bounded domain with  1,1 boundary Ω and has no holes in R 3 ,   = Ω × (0, ] ( > 0), and ^denotes the outward normal unit vector field to Ω, and  ≥ 2. In the Bean model, the electric field E and the current density J are characterized as follows.There exists a critical current   such that |J| ≤   in Ω and Thus if |J| reaches   , then |E| takes the value in [0, ∞).By scaling, we may assume that   = 1.The relation between |E| and |J| is followed from the Ampere law: as  → ∞.Here H is the magnetic field and  = |curl H| −2 .Thus model (1) provides a good approximation for the Bean model.For large , the resistivity  is small in a region That is to say,   becomes the superconductivity region as  → ∞.For more details, see Bean [3,4] and references therein.Though the authors in [3] considered system (1)-( 4), there are many mistakes and mistypes; for example, in Definition 2.1 (page 786), the differentiability of H with respect to the time variable  is not assumed, and they mistake the notion of the subdifferential (pages 788, 791-793) and so forth.
In this paper, we will extend the results of [3] to more general resistivity term of the form  = (, |curl H|) for some function .Since the resistivity  = (, |curl H|) may be of the form not only (, |curl H|) = |curl H| −2 but also (, |curl H|) = ()|curl H| −2 or more general type, we are convinced that the extension is meaningful.

Preliminaries and the Setting of the Problem
In this section, we introduce some spaces of vector fields which are used in this paper and set the problem.
By the Sobolev embedding theorem, we can get the following.

Lemma 2. Assume that Ω is bounded domain in R 3 without holes and with 𝐶
where  = 1 − 3/.
(One notes that, for a.e. ∈ (0, ), H is differentiable with respect to .) Here we give Green's formula which will be used frequently later (cf.[6]).

Existence, Uniqueness, and Regularity of Solution
In this section, we will consider the existence and uniqueness for system (16)-( 19) and also regularity under more restrictive hypotheses on F. The proof will be achieved by applying theorems due to Brezis [9,Theorems 3.4 and 3.6].In order to do so, define a Hilbert space  =  2 (Ω, R 3 ) and a functional   [V] for V ∈  by It is clear that the effective domain (  ) =   0 (curl, div0, Ω) ̸ = 0.
Lemma 7.Under hypothesis (H.1),   is a proper lower semicontinuous and convex functional.
Proof.First we show that   is lower semicontinuous; that is, if Thus we assume that it is trivial, so we may assume that  < ∞.Choosing a subsequence, if necessary, we may assume that   [V  ] < ∞ and as  → ∞.Thus we have V  ∈   0 (curl, div0, Ω) and it follows from (H.1) that Since Ω has no holes, it follows from Lemma 1 that {V  } is bounded in  1, (Ω, R 3 ).Passing to a subsequence, we may assume that V  → Ṽ weakly in  1, (Ω, R 3 ) and from the compactness of embedding from  1, (Ω, R 3 ) to   (Ω, R 3 ), V  → Ṽ strongly in   (Ω, R 3 ).Since  ≥ 2, we have Ṽ = V.Since    (, ) ≥ 0 for all  > 0 and    (, ) ≥ 0 for all  ≥ 0, taking the Taylor theorem into consideration, we have Since and curl V  → curl V weakly in   (Ω, R 3 ), it follows from the Hölder inequality that the last term of (25) tends to zero as  → ∞.Thus Next we show that   is convex; that is, for any V and W in  2 (Ω, R 3 ) and any 0 <  < 1, When V or W does not belong to is convex and for each  ∈ Ω, the function  →   (, ) is an increasing function and convex with respect to , for any ( Therefore it is easily shown that   is convex.Now we consider the subdifferential of the functional   (cf.Struwe [10, page 58]).The domain of the subdifferential   is defined by Then the multivalued subdifferential   at H ∈ (  ) is defined by Here we note that we have Next we will show that   [H] for H ∈ (  ) is singlevalued.In order to do so, define an operator   by Since ^× (V − H) = 0 on Ω, it follows from Lemma 5 and (33) that This implies that   ⊂   ; that is, (  ) ⊂ (  ) and
We claim that   has a minimizer in   0 (curl, div0, Ω).In fact, for any  > 0, there exists a constant (, ) > 0 such that Throughout this paper, |Ω| denotes the volume of Ω. Therefore if we choose  > 0 small enough, we can see that   is coercive.Since  =   0 (curl, div0, Ω) is a closed convex subset of  1, (Ω, R 3 ), if we show that   :  → R is lower semicontinuous and convex, it follows that   is weakly lower semicontinuous (cf.Takahashi [12, Lemma 1.3.9]).Since the convexity follows as before, we will show that   is lower semicontinuous.Let Passing to a subsequence, we may assume that curl V  → curl V a.e. in Ω.Since   is a continuous function,   (, |curl V  | 2 ) →   (, |curl V| 2 ) a.e. in Ω.Since   ≥ 0, it follows from the Fatou lemma that From this fact and the fact that Passing to a subsequence, we may assume that V  → V 0 weakly in  1, (Ω, R 3 ) and strongly in   (Ω, R 3 ) by compactness of embedding from  1, (Ω, R 3 ) to   (Ω, R 3 ) as above.Since divV  = 0 in Ω, we have divV 0 = 0 in Ω, and since ^× V  = 0 on Ω, we have ^× V 0 = 0 on Ω.Thus we have V 0 ∈   0 (curl, div0, Ω), and Thus a minimizer exists.So let H ∈   0 (curl, div0, Ω) be a minimizer of   .Then H satisfies the Euler-Lagrange equation weakly: and so H ∈ (  ) and is arbitrary, we have where ( +   ) denotes the range of  +   .We show that   =   .For any V ∈ (  ) and If F = F(, ) and H 0 = H 0 () satisfy (H.2), it follows from [9, Theorems 3.6 and 3.4] that the system has a unique solution H in the sense of Definition 4. Taking Lemma 8 into consideration, we can see that H is a unique solution of system ( 16)-(19).
Proof of Theorem 6.Let H be a solution of ( 16)-(19).Taking the inner product of ( 16) and H and then integrating over The first term of left hand side of (45) is equal to Since ^× H(, ) = 0 on Ω × [0, ], using Lemma 5 and (H.1), the second term of left hand side of (45) satisfies Applying the Schwarz inequality, the right hand side of (45) satisfies that, for any  > 0, there exists a constant (), Thus we have for any  ∈ [0, ].Taking the supremum on [0, ] and then choosing  > 0 small enough, there exists a constant  depending on , but independent of  such that sup Next, taking the inner product of ( 16) and H  and integrating over   , we have, using Lemma 5, If we note that it is shown that It follows from (H.1) that Thus we have Taking the supremum of the left hand side, there exists a constant  independent of  such that where the constant  is independent of .It follows from this inequality that we can see that H  ∈  2 (  ) and H ∈  ∞ (0, ;   0 (curl, div0, Ω)).This completes the proof of Theorem 6.
For more regularity of solution, we assume the following.
where the constant  depends on  and .
Next we show the second estimate (59).Taking the inner product of (60) and (   (, |V| 2 )V)  and then integrating over   , we have The first term of the left hand side of ( 68) is estimated as follows: Here we note that    ≥ 0. The second term of the left hand side of (68) is equal to For the estimate of the right hand side of (68), using the integration by parts and taking (H.1) into consideration, we have The first term of the right hand side of this inequality is already estimated in (58).This completes the proof of Theorem 9.
By Lemma 2, we have the following.
Proof.By Theorem 6, H ∈  ∞ (0, ; where the constant  is independent of .By the mean value theorem for integral, there exists  * ∈   such that Therefore we have

Limit Solution as 𝑝 → ∞
In this section, we consider the asymptotic behavior of the solution H = H(, ) (depending on ) as  → ∞.We assume the following.
then it is trivial, so we may assume that  < ∞, and passing to a subsequence, we may assume that Since Ω has no holes, it follows from [7] that Therefore ). Passing to a subsequence, we may assume that V → Ṽ weakly in  1,2 (Ω, R 3 ) and strongly in  2 (Ω, R 3 ).Thus we see that Ṽ = V.Since it is clear that divV = 0 in Ω and ^×V = 0 on Ω, it suffices to prove |curl V| ≤ 1 a.e. in Ω, in order to show that V ∈ ( ∞ ).
For any  > 0, define a set Since curl V  → curl V weakly in  2 (  , R 3 ), we have Thus |  | = 0. Since  > 0 is arbitrary, we see that |curl V()| ≤ 1 a.e. in Ω.From this we can also see that ).Now we have the following theorem.
In fact, from (91), we have where  1 is independent of .For any  > 0, put Thus we see that |  | = 0. Since  > 0 is arbitrary, we see that (2) holds.
(3) Clearly the first equation of ( 83) is equivalent to for any V ∈  2 0 (curl, div0, Ω) with |curl V| ≤ 1.Since H (  ) is a unique solution of Here we note that by hypothesis Letting   → ∞ in (103) and using the fact that    [H (  ) ] ≥ 0, we see that Thus we can see that H (∞) satisfies the first equation of ( 83). ( 4) For the initial condition, it follows from as  → +0.Hence H (∞) is a solution of (83).Since the solution of ( 83) is unique, we can replace   in Step 2 by any .
Now we show the Hölder continuity of the limit solution.