Abstract

We study a degenerate evolution system containing the -curl system in a bounded domain with initial and boundary conditions for the magnetic field under the influence of a system force . This is concerned with an approximation of Bean’s critical-state model for type II superconductors. We will show the existence, uniqueness, and regularity of solutions. Moreover we will get the properties of the limit solution as .

1. 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 is a bounded domain with boundary and has no holes in , , and denotes the outward normal unit vector field to , and . In the Bean model, the electric field and the current density are characterized as follows. There exists a critical current such that in and Thus if reaches , then takes the value in . By scaling, we may assume that . The relation between and is followed from the Ampere law:as . Here is the magnetic field and . Thus model (1) provides a good approximation for the Bean model. For large , the resistivity is small in a regionThat 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 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 for some function . Since the resistivity may be of the form not only but also or more general type, we are convinced that the extension is meaningful.

This paper is organized as follows. In Section 2, we introduce some spaces of vector fields and describe the setting of the problem. In Section 3, we consider the existence, uniqueness, and regularity of the solution for the problem. Finally, in Section 4, we examine the properties of the limit solution as . The result shows that the resistivity vanishes in the region .

2. 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.

Let be a bounded domain with boundary. Throughout this paper, we assume that has no holes. That is to say, the second Betti number of is equal to zero. This means thatBut we allow to be not simply connected. For these notations, see Dautray and Lions [5] or Amrouche and Seloula [6]. Let and define some spaces of vector fields defined in with values in . One hasWe note that, for , the tangential trace is well defined in (cf. [6]). And moreover, define

The following lemma follows from [6], Aramaki [7], and Pan [8].

Lemma 1. If has no holes, thenand the norm in given byis equivalent to the norm in . In particular, is a Banach space with the norm .

By the Sobolev embedding theorem, we can get the following.

Lemma 2. Assume that is bounded domain in without holes and with boundary. If , then where .

Throughout this paper we denote the norm of vector field in or by or , respectively.

Let . We assume that the function satisfies the following.(H.1) with satisfies that there exist independent of , and such that where and . Then it is clear that satisfiesand is convex with respect to variable.

We consider the initial and boundary value problem: where and are given vector fields. We assume the following.(H.2) satisfies in and on for a.e. , and satisfies that and .

We note that hypothesis (H.2) contains the compatibility conditions: in and on .

Example 3. where with satisfying (H.1). In particular, if , system (16)–(19) becomes (1)–(4).

The following definition of solution of system (16)–(19) is based on Brezis [9, Definition3.1].

Definition 4. One calls a vector field , a solution of system (16)–(19), if the following hold:(i) and is absolutely continuous with respect to in all compact subset of , and for a.e. , (therefore for a.e. , is differentiable with respect to );(ii) for a.e. ;(iii) for a.e. ;(iv)(16) holds in for a.e. and in .(One notes that, for a.e. , is differentiable with respect to .)

Here we give Green’s formula which will be used frequently later (cf. [6]).

Lemma 5. For and where is the conjugate exponent for , that is to say, , then one has where denotes the usual inner product for vectors and in and denotes the duality between and .

3. 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 .

Theorem 6. Let . Assume that is a bounded domain in without holes and with boundary. Under hypotheses (H.1) and (H.2), system (16)–(19) has a unique solution in the sense of Definition 4. Moreover, one has and .

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 and a functional for by It is clear that the effective domain .

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 in as , then . This is equivalent to show that if in as and , then . In fact, it is clear from the fact that if in , there exists a subsequence of such thatThus we assume that in and . We show that . If , it is trivial, so we may assume that . Choosing a subsequence, if necessary, we may assume that and as . Thus we have and it follows from (H.1) thatSince has no holes, it follows from Lemma 1 that is bounded in . Passing to a subsequence, we may assume that weakly in and from the compactness of embedding from to , strongly in . Since , we have . Since for all and for all , taking the Taylor theorem into consideration, we have Sinceand weakly in , 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 and in and any , When or does not belong to , , or , the inequality is trivial. Thus let and be in . Since a function is convex and for each , 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 where is the usual inner product of and in . If , then , so we can rewrite Then the multivalued subdifferential at is defined by Here we note that we have Next we will show that for is single-valued. In order to do so, define an operator by Since on , it follows from Lemma 5 and (33) thatThis implies that ; that is, and for .

Lemma 8. Under the hypothesis (H.1), is single-valued and .

Proof. We follow the arguments in Evans [11, page 571]. For any given , define a functionalon . We note that and is a Banach space with respect to the norm which is equivalent to norm according to the fact that has no holes. We claim that has a minimizer in . In fact, for any , there exists a constant such thatThroughout this paper, denotes the volume of . Therefore if we choose small enough, we can see that is coercive. Since is a closed convex subset of , if we show that 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 in and as . Then in . Passing to a subsequence, we may assume that a.e. in . Since is a continuous function, a.e. in . Since , it follows from the Fatou lemma that From this fact and the fact that in since , we have , so is weakly lower semicontinuous. Let be a minimizing sequence of . ThenSince is coercive, is bounded in . Passing to a subsequence, we may assume that weakly in and strongly in by compactness of embedding from to as above. Since in , we have in , and since on , we have on . Thus we have , andThus a minimizer exists. So let be a minimizer of . Then satisfies the Euler-Lagrange equation weakly:Since , we haveand so and . Since is arbitrary, we havewhere denotes the range of .
We show that . For any and , there exists such that . Since , we see that . On the other hand, . According to [11, Chapter  9, Section  6, Theorem  1(iv)], the equation has a unique solution in . Thus we have and . Hence is single-valued and . This completes the proof.

If and satisfy (H.2), it follows from [9, Theorems  3.6 and  3.4] that the systemhas a unique solution in the sense of Definition 4. Taking Lemma 8 into consideration, we can see that is a unique solution of system (16)–(19).

Proof of Theorem 6. Let be a solution of (16)–(19). Taking the inner product of (16) and and then integrating over , we have The first term of left hand side of (45) is equal toSince on , 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 , there exists a constant , Thus we have for any . Taking the supremum on and then choosing small enough, there exists a constant depending on , but independent of such that Next, taking the inner product of (16) and and integrating over , we have, using Lemma 5, If we note thatit 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 and . This completes the proof of Theorem 6.

For more regularity of solution, we assume the following.(H.3), , , and has a compact support in for every .

Then we have the following.

Theorem 9. Under (H.1)–(H.3), the solution of (16)–(19) satisfies and Moreover, the following estimates hold: where the constant depends on and .

Proof. For the brevity of notation, we write . Taking curl of (16), we haveSince on , we have on . Since near for every , it follows from (16) that Moreover we have Taking the inner product of (60) and and then integrating over , it follows from (61) and (62) that Therefore using the Hölder inequality, for any , there exists such that Since for , the first term of the right hand side of (64) is estimated byThus we have Taking the supremum of the left hand side and then choosing small enough, we have where the constant depends on , , and . Since for , we can get the first estimate (58).
Next we show the second estimate (59). Taking the inner product of (60) and and then integrating over , we have The first term of the left hand side of (68) is estimated as follows: Here we note that . 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 haveHere the constant depends on and . By the first inequality (58), there exists a constant depending on and such thatAdding the above inequalities, we get second inequality (59) without the term which we now estimate the term (73). From (16), we get 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.

Theorem 10. Under (H.1)–(H.3), let be a solution of (16)–(19). Then the following hold.(i) if .(ii) for any if .(iii) if where and .

Proof. By Theorem 6, . Using the Sobolev embedding theorem, (i) and (ii) are clear. Similarly if , is Hölder continuous with respect to with the Hölder exponent . We will show that is Hölder continuous with respect to with the Hölder exponent . Let , small and , and define a ball in centered at with radius . If , we consider in the following. From Schwarz’s inequality, we have where the constant is independent of . By the mean value theorem for integral, there exists such thatTherefore we haveThus for any ,

4. Limit Solution as

In this section, we consider the asymptotic behavior of the solution (depending on ) as . We assume the following.(H.4)One has .

Define a subspace of . For , define a functional It is clear that is proper convex functional. We will show that is lower semicontinuous. Let , in , and . Then it suffices to show that . If , then it is trivial, so we may assume that , and passing to a subsequence, we may assume that . Thus , and . Since has no holes, it follows from [7] that Therefore is bounded in . Passing to a subsequence, we may assume that weakly in and strongly in . Thus we see that . Since it is clear that in and on , it suffices to prove a.e. in , in order to show that . For any , define a setSince weakly in , we haveThus . Since is arbitrary, we see that a.e. in . From this we can also see that is a closed convex subset in .

Now we have the following theorem.

Theorem 11. Under (H.1), (H.2), and (H.4), has a unique limit as such that that is to say,for a.e. for all and

Proof.  
Step  1 (the uniqueness of the solution of (83)). We follow Yin [13]. Let and be two solutions of (83). Then we have Adding these inequalities, we haveThereforeand so we havefor a.e. . Hence in .
Step  2 (existence of the limit solution). (1) Since (H.4) implies thatfor any , we notice that the right hand sides of (50) and (56) are independent of . Therefore we have where and are independent of . Therefore for any subsequence , there exists a subsequence such that (2) We claim that .
In fact, from (91), we have where is independent of . For any , putThen it follows from the Hölder inequality and (98) that Thus we see that . Since is arbitrary, we see that (2) holds.
(3) Clearly the first equation of (83) is equivalent tofor any with . Since is a unique solution offor any with , Here we note that by hypothesisas . Since strongly in and weakly in , we see that, for a.e. , Letting in (103) and using the fact that , we see thatThus we can see that satisfies the first equation of (83).
(4) For the initial condition, it follows fromas . Hence is a solution of (83). Since the solution of (83) is unique, we can replace in Step by any .

Now we show the Hölder continuity of the limit solution.

Theorem 12. The above limit solution of (83) is Hölder continuous in .

Proof. We put for brevity. Since in for a.e. , we havein the sense of distribution. Since for a.e. , the tangent trace is well defined, and since on , weakly in for a.e. , we have on for a.e. . Moreover, by Step  2 in the proof of Theorem 11, for any , for a.e. . Thus satisfies that, for a.e. , Since for any , it follows from [6, Proposition  5.1 and Remark  5.1(ii)] that andwhere is independent of . By the Sobolev embedding theorem, for a.e. , for any . The Hölder continuity with respect to follows from the similar argument of the proof of Theorem 10.

Finally we assume that the function satisfies the following.(H.5) for .

Remark 13. When and , it is easily shown that satisfies (H.4).

Theorem 14. Under conditions (H.1), (H.2), and (H.5), there exists a nonnegative measurable function such that in the distribution sense, and

Proof. Put . Taking the inner product of (16) and , we haveSince by hypothesis (H.1), is bounded in . Therefore there exists a subsequence such that weakly in as . Since weakly in from (95), letting in (113) with , we have On the other hand, from (113) with , we have Hence, using (93) with , as . From (114) with , we have Since strongly in , we see thatas . Since weakly in , using the Hölder inequality and (H.5), we have Here the last inequality is based on the fact that . This implies that , so and are linearly dependent. Thus there exists a measurable function such that where if . Since in and weakly in and weakly in , we havein the distribution sense. Sincethe function is nonnegative. Moreover, if , thenSo we have . Hence . This completes the proof.

Remark 15. For the limit solution , the resistivity vanishes in the region . That is to say, the current is in the superconductivity state there.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.