Boundary Value Problems for the Magnetic Laplacian in Semiclassical Analysis
Zhongwei Shen

TL;DR
This paper extends boundary value problem analysis for the magnetic Laplacian in semiclassical analysis, establishing uniform estimates in Lipschitz domains, generalizing classical results to magnetic operators.
Contribution
It provides the first uniform nontangential maximal function estimates for the magnetic Laplacian in Lipschitz domains in the semiclassical regime.
Findings
Established uniform boundary estimates for magnetic Laplacian
Extended classical Laplacian results to magnetic operators
Results are valid even for smooth domains
Abstract
This paper is concerned with the magnetic Laplacian in semiclassical analysis, where is a semiclassical parameter. We study the Neumann and Dirichlet problems for the equation in a bounded Lipschitz domain . Under the assumption that the magnetic field is of finite type on , we establish the nontangential maximal function estimates for , which are uniform for . This extends a well-known result due to D. Jerison and C. Kenig for the Laplacian in Lipschitz domains to the magnetic Laplacian in the semiclassical setting. Our results are new even for smooth domains.
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Taxonomy
TopicsNonlinear Partial Differential Equations · Numerical methods in inverse problems · Advanced Harmonic Analysis Research
Boundary Value Problems for the Magnetic Laplacian
in Semiclassical Analysis
Zhongwei Shen
Abstract
This paper is concerned with the magnetic Laplacian in semiclassical analysis, where is a semiclassical parameter. We study the Neumann and Dirichlet problems for the equation in a bounded Lipschitz domain . Under the assumption that the magnetic field is of finite type on , we establish the nontangential maximal function estimates for , which are uniform for . This extends a well-known result due to D. Jerison and C. Kenig for the Laplacian in Lipschitz domains to the magnetic Laplacian in the semiclassical setting. Our results are new even for smooth domains.
Keywords. Schrödinger Operator; Magnetic Field; Semiclassical Analysis; Boundary Value Problem; Lipschitz Domain.
MR (2020) Subject Classification: 35P25.
1 Introduction
Let and . Consider the Schrödinger operator with a magnetic potential in semiclassical analysis,
[TABLE]
also called the magnetic Laplacian, where is a semiclassical parameter. In this paper, we are interested in boundary regularities of that are uniform in . More precisely, consider the Neumann problem in a bounded Lipschitz domain ,
[TABLE]
where denotes the outward unit normal to , and the Dirichlet problem,
[TABLE]
Let denote the magnetic field. In the main results of this paper, we shall assume that is of finite type on ; i.e., there exist an integer and such that
[TABLE]
for any . Recall that the nontangential maximal function is defined by
[TABLE]
for , where is a large (fixed) constant depending on . The following two theorems are the main results of this paper.
Theorem 1.1**.**
Let be a bounded Lipschitz domain in , . Let . Suppose is of finite type on . Then for any , a solution to the Neumann problem (1.2) satisfies the nontangential maximal function estimate,
[TABLE]
for , where . Moreover, we have
[TABLE]
The constants in (1.6)-(1.7) and depend only on , in (1.4), and .
Let , where and
[TABLE]
is a (tangential) differential operator on .
Theorem 1.2**.**
Assume that and satisfy the same conditions as in Theorem 1.1. Then for any , a solution to the Dirichlet problem (1.3) satisfies the estimate,
[TABLE]
for , where . Moreover, we have
[TABLE]
The constants in (1.8)-(1.9) and depend only on , in (1.4), and .
A few remarks are in order.
Remark 1.3**.**
In the case in and , the Neumann and Dirichlet problems (1.2) and (1.3) reduce to
[TABLE]
and
[TABLE]
respectively. In this case, it was proved by D. Jerison and C. Kenig [4, 5] that the weak solutions in to (1.10) and (1.11) satisfy the nontangential maximal function estimates,
[TABLE]
respectively, assuming that is a bounded Lipschitz domain. Our main results extend this classical work to the magnetic Laplacian in the semiclassical setting, where the estimates are uniform in . To the best of the author’s knowledge, Theorems 1.1 and 1.2 are new even for smooth domains.
Remark 1.4**.**
The auxiliary function in Theorems 1.1 and 1.2 is defined by
[TABLE]
where denotes the ball centered at with radius . This function, which was introduced by the present author in [9, 10], plays the role of a critical scaling in the study of Schrödinger operators with electrical and magnetic potentials. See earlier work in [2, 12, 14] as well as references for more recent work in [7, 8]. Note that if is a (matrix-valued) polynomial of degree , then
[TABLE]
It follows that
[TABLE]
where depend only on and . Under the finite-type condition on in Theorem 1.1, we have
[TABLE]
for . See Remark 10.2.
Remark 1.5**.**
Consider the Dirichlet-to-Neumann map associated with the operator ,
[TABLE]
where is the solution of the Dirichlet problem (1.3). It follows from Theorems 1.1 and 1.2 that
[TABLE]
for . Note that represents the tangential component of on . In fact, our proof yields the weighted estimates,
[TABLE]
for any (the bounding constants depend on ). See Theorem 7.9.
Remark 1.6**.**
Let be a solution of the Dirichlet problem (1.3) with . Then
[TABLE]
This follows readily from the maximum principle for subharmonic functions. No addition condition beyond is not needed. See Theorem 2.5.
Remark 1.7**.**
Suppose is of finite type on ; i.e., the inequality (1.4) holds for any . By compactness it follows that (1.4) holds in a neighborhood of . In this case, the estimates (1.6) and (1.8) hold if one modifies the definition of the nontangential maximal function by considering only points near ,
[TABLE]
This follows directly from the proofs of Theorems 1.1 and 1.2.
We now describe the main ingredients in the proofs of Theorems 1.1 and 1.2. First, we rewrite the equation as , where . By a localization argument, we reduce the problem to the equation in a bounded Lipschitz domain , assuming that for some and the magnetic field satisfies the following conditions. There exists such that
[TABLE]
for any , and
[TABLE]
for any . Note that when is replaced by , the condition (1.15) remains invariant, while (1.16) is satisfied for sufficiently large. Moreover, under the conditions (1.15)-(1.16), it is proved in [8] that
[TABLE]
for any .
Next, let be a weak solution of in . By an approximation argument, we may assume that and thus are smooth. To relate the norm of the normal component of on to the norm of its tangential components, we use two Rellich-type identities, adapted to the operator . See Lemmas 7.1 and 7.2. As a result, the proof for (1.13) with is reduced to the estimates,
[TABLE]
To establish (1.18), we study the Green function and the Neumann function for the operator in . Suppose (the case is handled by the method of descending). Using (1.17), we are able to show that under the assumptions (1.15)-(1.16),
[TABLE]
for any and . See Sections 5 and 6. The estimates in (1.18) follow from (1.19) by using the Rellich identities mentioned above and a duality argument.
Finally, with the Rellich estimates (1.13) at our disposal, we establish the nontangential maximal function estimates (1.6) and (1.8) with the help of (1.14). A key observation here is that although the commutator may not be zero in , the function nevertheless satisfies the interior estimate,
[TABLE]
if . See Section 4. This allows us to use the techniques developed for (1.18) to control the additional error terms.
2 Preliminaries
Throughout the section we assume that is a bounded Lipschitz domain and .
Lemma 2.1**.**
Let and . Then for ,
[TABLE]
and for ,
[TABLE]
Proof.
By differentiating , we obtain
[TABLE]
for . Hence, . Since , this yields (2.1).
Next, by differentiating (2.3), we see that
[TABLE]
which, together with (2.1) and , leads to (2.2). ∎
Lemma 2.2**.**
Let . Then
[TABLE]
where for , for , and depends only on and . Furthermore, if , then
[TABLE]
and
[TABLE]
Proof.
Let , where . It follows from (2.1) that . Since , by Sobolev inequality,
[TABLE]
By letting , we obtain (2.5) for . A density argument gives (2.5) for . The proofs for (2.6) and (2.7) are similar. ∎
Theorem 2.3**.**
Let be a solution of in . Then
[TABLE]
where and . The constant depends only on , and ,
Proof.
We give the proof for . The case is similar. Let , where . By (2.2), we have in . By differentiating in , as in the proof of the mean value property for harmonic functions, it follows that if ,
[TABLE]
Hence, for ,
[TABLE]
By letting , we obtain
[TABLE]
for any . This yields (2.8) by using Young’s inequality. ∎
Lemma 2.4**.**
Let be a solution of the Dirichlet problem,
[TABLE]
where . Then in , where is harmonic in and on .
Proof.
Let and , where . It follows from (2.2) that in . Thus, is subharmonic in . Note that
[TABLE]
By the maximum principle for subharmonic functions, this implies that
[TABLE]
Hence, in . By letting , we obtain in . ∎
Theorem 2.5**.**
Let be a bounded Lipschitz domain and . Let be a solution of the Dirichlet problem (2.9). Then
[TABLE]
for , where depends on . The constant in (2.10) depends only on , , the Lipschitz character of . If is , the estimate (2.10) holds for .
Proof.
By Lemma 2.4, we see that on , where is a harmonic function in such that on . As a consequence, the estimate (2.10) follows from the well-known results for harmonic functions in Lipschitz and domains [6]. ∎
3 The function
Let be a bounded Lipschitz domain in , . It follows that there exist and such that for any ,
[TABLE]
in a new coordinate system, obtained from the standard one through translation and rotation, where is a Lipschitz function with and . A constant is said to depend on the Lipschitz character of if depends on and the number of balls , centered on , which are needed to cover . By translation, without the loss of generality, we assume that for some , where depends on the Lipschitz character of .
We will impose the following conditions on in this and next few sections: and there exists such that
[TABLE]
for any ball . It follows from (3.2) that
[TABLE]
for any ball ; i.e., is a weight in . In particular, satisfies the doubling condition,
[TABLE]
for any , where depends only on and in (3.3). As a consequence, we also have
[TABLE]
if , where denotes the cube center at with side length .
For , let be defined by (1.12). Under the assumption that for any ,
[TABLE]
it follows by definition that for any .
Lemma 3.1**.**
Suppose satisfies (3.2) for any ball . Also assume that (3.6) holds for any . Then,
[TABLE]
for any . Moreover, for any ,
[TABLE]
[TABLE]
The constants depend only on in (3.2).
Proof.
Let . Using (3.6), it is not hard to see that
[TABLE]
This implies that . Also, by (3.2) and (3.10),
[TABLE]
As a result, we have proved (3.7).
The inequalities (3.8)-(3.9) were proved by the present author in [10] for any , under the assumption that is a weight in for some . With the conditions in the lemma, the same argument gives (3.8)-(3.9) for any . Note that the estimates (3.7)-(3.8) are scaling invariant; the constants do not depend on . ∎
It follows readily from (3.8)-(3.9) that
[TABLE]
for any .
Theorem 3.2**.**
Suppose satisfies (3.2) for any ball . Also assume that (3.6) holds for any . Then
[TABLE]
for any , where depends only on in (3.2) and the Lipschitz character of .
Proof.
See [8, Theorem 3.8]. We point out that the proof in [8] only uses the fact in . Outside of , there is no need to assume that is a curl of some vector field. This observation allows us to extend to a neighborhood of without extending in the proof of our main results. ∎
The next lemma gives a Caccioppoli inequality for the operator . The conditions on are not needed.
Lemma 3.3**.**
Let and , where either or . Suppose that and in , where . If , we also assume that either or on , where denotes the outward unit normal to . Then for ,
[TABLE]
where depends only on .
Proof.
The proof is similar to the case of . Using
[TABLE]
where , we obtain
[TABLE]
To finish the proof, we choose such that in and . ∎
Remark 3.4**.**
Let and . Suppose that and in . It follows from the proof of Lemma 3.3 that
[TABLE]
where depends only on .
Theorem 3.5**.**
Suppose satisfies the same conditions as in Theorem 3.2. Let and , where either or . Suppose and in . If , we also assume that either on or on . Then
[TABLE]
for any , where depends on , the Lipschitz character of , and in (3.2).
Proof.
We may assume for otherwise the estimate is trivial. Let . It follows from (3.12) that
[TABLE]
where we have used (3.14) for the last inequality. Let . Choose so that in and . By (3.17) and (3.9), this gives
[TABLE]
which leads to
[TABLE]
The estimate (3.16) now follows by iterating (3.18). ∎
4 Interior estimates
Suppose in . It follows by Theorem 2.3 that
[TABLE]
where depends only on . In this section we establish pointwise estimates for and under the conditions (3.2) and (3.6) on the magnetic field .
Theorem 4.1**.**
Assume satisfies the condition (3.2) for any ball . Also assume that (3.6) holds for any . Suppose in . Then for any ,
[TABLE]
[TABLE]
where depends on and in (3.2).
Proof.
It follows from (4.1) and (3.16) that
[TABLE]
for any . To see (4.3), note that by (3.9),
[TABLE]
for any . This, together with (4.2), gives (4.3). ∎
Theorem 4.2**.**
Under the same assumptions as in Theorem 4.1, we have
[TABLE]
where depends on in (3.2).
Proof.
Note that
[TABLE]
where the repeated index is summed from to . It follows that , where
[TABLE]
In view of (2.8), this implies that if ,
[TABLE]
where and . Using and , we obtain
[TABLE]
if . It then follows by an iteration argument that
[TABLE]
for some , where we have used (3.13) and (3.8) for the last step. The desired estimate now follows readily from (4.4) by choosing . ∎
Theorem 4.3**.**
Under the same assumptions as in Theorem 4.1, we have
[TABLE]
where depends on in (3.2).
Proof.
We consider two cases.
**Case 1. ** Assume that
[TABLE]
It follows from (4.7) that if ,
[TABLE]
where , , and the fact is used for the last step. By iterating the estimate (4.10), we obtain
[TABLE]
Note that under the assumption (4.9), we have and
[TABLE]
for any . This, together with (4.11), yields (4.8).
Case 2. Assume that
[TABLE]
Then . It follows from (4.6) that
[TABLE]
for any , where we have used (4.4) for the last inequality. By (4.5), we obtain
[TABLE]
which completes the proof. ∎
We introduce a modified nontangential maximal function. For , define
[TABLE]
for .
Corollary 4.4**.**
Assume satisfies the same conditions as in Theorem 4.1. Suppose and in . Let . Then
[TABLE]
for any . Consequently,
[TABLE]
for any , where depends on the Lipschitz character of and in (3.2).
Proof.
This follows readily from (4.3) and (4.8). ∎
5 The Green function
In this section we establish decay estimates for the Green function in a bounded Lipschitz domain .
Lemma 5.1**.**
Let and . Let be a solution of
[TABLE]
Then
[TABLE]
where depends on the Lipschitz character of .
Proof.
Let and . It follows from (2.2) and (5.1) that . Thus, is subharmonic in and on . Hence,
[TABLE]
where depends only on . By letting , we obtain (5.2). ∎
Consider the Dirichlet problem,
[TABLE]
Lemma 5.2**.**
For , the Dirichlet problem (5.3) has a unique weak solution in . Moreover, if , the solution satisfies
[TABLE]
where for and for . The constant in (5.4) depends at most on , and .
Proof.
Consider the bilinear form
[TABLE]
for . Since is bounded in , we have
[TABLE]
where , depends on , and we have used (2.5) for the second inequality. It follows that
[TABLE]
Thus, by the Lax-Milgram Theorem, (5.3) has a unique weak solution in for any . To see (5.4), we note that
[TABLE]
This, together with (2.5), gives (5.4). ∎
Remark 5.3**.**
Suppose . Since
[TABLE]
under the assumption and is Lipschitz, the weak solution of (5.3) is Hölder continuous in .
Remark 5.4**.**
Let . Choose such that on and . By considering and applying Lemma 5.2, one obtains a unique week solution in for the Dirichlet problem (1.3).
Theorem 5.5**.**
There is a continuous function with the following properties.
For any , , and ,
[TABLE]
where depends only on and the Lipschitz character of , and depends only on and . 2. 2.
For any , ,
[TABLE] 3. 3.
Fix . Then , on , and
[TABLE] 4. 4.
For , the weak solution of (5.3) is given by
[TABLE]
Proof.
The proof is similar to that in the case of second-order elliptic operators in divergence form [1, 3]. Fix and . Let denote the weak solution of (5.3) with , given by Lemma 5.2. Thus,
[TABLE]
for any . It follows from (5.4) that
[TABLE]
where for and for .
Next, we fix and . Let . For , let be the weak solution of (5.3). Since in and on , it follows by (5.2), (2.8) and (5.4) that
[TABLE]
Also, note that
[TABLE]
for any . In view of (5.12) and (5.15), we have
[TABLE]
This, together with (5.14), gives
[TABLE]
provided . By duality, we obtain
[TABLE]
Since and
[TABLE]
it follows from Lemma 5.1 and (5.17) that if ,
[TABLE]
Recall that for and for . Thus, we have proved that if ,
[TABLE]
for any .
Using the estimates (5.20) and (5.13), one can show that is bounded in for if , and in for any if . One can also show that is bounded in for . It follows that is bounded in for and . This implies that there exists a subsequence , where , and such that
[TABLE]
where . By Lemma 3.3, is bounded in for any fixed . It follows that and (5.10) holds.
Since , using (5.20) and the standard elliptic regularity theory, one may deduce that is equicontinuous in for any fixed . Thus, we may assume uniformly in for any fixed . As a result, (5.8) follows from (5.20). To show (5.9), we use (5.16) with and to obtain
[TABLE]
By letting , we arrive at (5.9).
Finally, let be the weak solution of (5.3) with . Since , is Hölder continuous in . By taking limits in (5.16), we obtain
[TABLE]
which, together with (5.9), gives (5.11). ∎
Theorem 5.6**.**
Let be a bounded Lipschitz domain. Suppose satisfies the condition (3.2) for any ball . Also assume that (3.6) holds for any . Let be a solution of (5.1), where and . Then for any ,
[TABLE]
where depends on , the Lipschitz character of and in (3.2).
Proof.
This follows readily from Lemma 5.1 and Theorem 3.5. ∎
Theorem 5.7**.**
Assume and satisfy the same conditions as in Theorem 5.6. Let be the Green function for the operator in , given by Theorem 5.5. Let . Then, for any and ,
[TABLE]
for , where depends on , , the Lipschitz character of and in (3.2). If , we have
[TABLE]
for any , where depends on , , , the Lipschitz character of and in (3.2).
Proof.
We gives the proof for . The proof for is the same. Fix and . Let and . We consider two cases.
Case 1. Suppose dist. Then . It follows from (4.2) that
[TABLE]
where we have used (5.8) for the last inequality. This gives (5.23)
Case 2. Suppose dist. Choose such that . By (5.22) we have
[TABLE]
where we have used (5.8) for the last inequality. Finally, note that by (3.8),
[TABLE]
It follows that
[TABLE]
Since is arbitrary, this, together with (5.25), gives (5.23). ∎
Theorem 5.8**.**
Let . Assume and satisfy the same conditions as in Theorem 5.6. For , let be a weak solution of (5.3). Then
[TABLE]
for any and , where depends on , , the Lipschitz character of , and in (3.2).
Proof.
By Theorem 5.5, we may write
[TABLE]
where
[TABLE]
To show (5.26), it suffices to show that for any ,
[TABLE]
and for any ,
[TABLE]
To see (5.27), we use the estimate
[TABLE]
for any and the fact that (5.23) holds for any . This gives
[TABLE]
for any , from which the inequality (5.27) follows readily. The proof for (5.28) is similar. ∎
6 The Neumann function
In this section we establish decay estimates for the Neumann function in a bounded Lipschitz domain .
Lemma 6.1**.**
Let and . Let be a solution of
[TABLE]
where denotes the outward unit normal to . Then
[TABLE]
where depends only on .
Proof.
Let and . Since in , by (2.2), we have in . Using on and (2.3), we see that on . By the even reflection, one can show that is a nonnegative sub-solution of a second-order elliptic equation in divergence form with bounded measurable coefficients in . More precisely, without the loss of generality, assume that
[TABLE]
where is a Lipschitz continuous function with and . Let be the map
[TABLE]
from to . Note that . For , define
[TABLE]
Then and
[TABLE]
for any nonnegative , where
[TABLE]
Note that is symmetric. The Lipschitz condition ensures that is bounded measurable and uniformly elliptic. As a result, by the De Giorgi-Nash-Moser estimates, we obtain
[TABLE]
where depends only on the Lipschitz character of . By letting , this leads to
[TABLE]
from which and the interior estimate (4.1), the estimate (6.2) follows by a simple covering argument. ∎
In the remaining of this section we will assume that there exists such that
[TABLE]
for any . It follows from (6.3) and (2.6) that
[TABLE]
for any , where for and for . The constant in (6.4) depends on , and in (6.3).
Consider the Neumann problem,
[TABLE]
Let and , we say is a weak solution of (6.5) if
[TABLE]
for any .
Lemma 6.2**.**
For and , the Neumann problem (6.5) possesses a unique weak solution in . Moreover, the solution satisfies
[TABLE]
where for and for . The constant in (6.7) depends at most on , and in (6.3).
Proof.
Consider the bilinear form in (5.5) for . Since is bounded in , we have
[TABLE]
where depends on and we have used the assumption (6.3) for the second inequality. It follows that
[TABLE]
for any , where . By the Lax-Milgram Theorem and (2.7), this implies that (6.5) has a unique weak solution in for any and . To see (6.7), we use (6.6) with , (6.4) and (2.7). ∎
Theorem 6.3**.**
There is a continuous function with the following properties.
For any , , and ,
[TABLE]
where depends on , and in (6.3), and depends also on . 2. 2.
For any , ,
[TABLE] 3. 3.
Fix . Then and
[TABLE] 4. 4.
For and , the weak solution of (6.5) is given by
[TABLE]
Proof.
With Lemmas 6.1 and 6.2 at our disposal, the proof is similar to that of Theorem 5.5. Indeed, fix and . Let denote the weak solution of (6.5) with and , given by Lemma 6.2. Thus,
[TABLE]
for any . Using (6.7) and (6.2), one can show that if ,
[TABLE]
for any . It follows that is bounded in for . This implies that there exists a sequence , where , and such that weakly in , where . The proof for (6.8)-(6.10) is the same as in the case of the Green function. To see (6.11), let be a weak solution of (6.5) with and . By letting in (6.6), we obtain
[TABLE]
This, together with (6.12), yields
[TABLE]
Under the assumptions that and , using the standard regularity theory for in a Lipschitz domain, one may show that is Hölder continuous in . By letting in (6.13), we see that
[TABLE]
In view of (6.9), this gives the representation formula (6.11). ∎
Theorem 6.4**.**
Let be a bounded Lipschitz domain. Suppose that satisfies the condition (3.2) for any ball . Also assume that (3.6) holds for any . Let be the Neumann function for the operator in , given by Theorem 6.3. Let . Then, for any and ,
[TABLE]
for , where depends on , , , and in (3.2). If , we have
[TABLE]
for any , where depends on , , , the Lipschitz character of and in (3.2).
Proof.
The proof is similar to that of Theorem 5.7. We only point out that under the condition 3.6, we have for any . In view of Theorem 3.2, this leads to
[TABLE]
which gives the condition (6.3) for any . ∎
Theorem 6.5**.**
Let . Suppose and satisfy the same conditions as in Theorem 6.4. For , let be a weak solution of (6.5) with . Then
[TABLE]
for any and , where depends on , , , the Lipschitz character of and in (3.2).
Proof.
The proof is similar to that of Theorem 5.8. ∎
7 Rellich estimates
Let be a bounded Lipschitz domain in and . Let , where
[TABLE]
for . It follows from integration by parts that if ,
[TABLE]
Lemma 7.1**.**
Let . Suppose and in . Then
[TABLE]
where the repeated indices are summed from to .
Proof.
Using (7.1), we obtain
[TABLE]
where . This gives
[TABLE]
where we have used the fact . Also note that by (7.1),
[TABLE]
From (7.4), we deduce that
[TABLE]
The equation (7.2) now follows readily from (7.3) and (7.5). ∎
Let , where
[TABLE]
for .
Lemma 7.2**.**
Let . Suppose and in . Then
[TABLE]
where the repeated indices are summed from to .
Proof.
Write the equation (7.2) as . Then (7.6) follows from the fact as well as the observation that
[TABLE]
∎
For the rest of this section, we shall further assume that is a smooth domain for some , where depends on the Lipschitz character of . We also assume that satisfies (3.2) for any and (3.6) for any . We emphases that although is assumed to be smooth, the dependence of constants on is through and its Lipschitz character.
Lemma 7.3**.**
Suppose that and in . Then for any ,
[TABLE]
[TABLE]
where depends on , and in (3.2).
Proof.
Fix . Recall that by (3.6). Let , where depends on the Lipschitz character of . Choose such that on , on , in and in . It follows from (7.2) that
[TABLE]
where we have used the fact for . We now multiply the inequality above by and integrate the resulting inequality in over . Using the fact that if , this leads to
[TABLE]
which yields (7.7) by using the Cauchy inequality. A similar argument, using (7.6), gives (7.8). ∎
Lemma 7.4**.**
Under the same assumptions as in Lemma 7.3, we have
[TABLE]
for any , where depends on .
Proof.
Fix and let . It follows from Remark 3.4 that
[TABLE]
By integrating the inequality above in over , we see that
[TABLE]
is bounded by the right-hand side of (7.9), where
[TABLE]
To handle the region , we use the fact that if ,
[TABLE]
It follows that if , then
[TABLE]
where . By integrating the inequality above in over , we see that
[TABLE]
is also bounded by the right-hand side of (7.9). ∎
Theorem 7.5**.**
Suppose . Let . Suppose in . Assume that either on or on . Then
[TABLE]
for any , where depends on .
Proof.
It follows from (7.9) that the left-hand side of (7.10) is bounded by
[TABLE]
This yields (7.10) by using Theorems 5.8 and 6.5 with . ∎
Lemma 7.6**.**
Let . Then
[TABLE]
where depends on and .
Proof.
Let and . Then
[TABLE]
for any . By applying the inequality above to and letting , we obtain
[TABLE]
This leads to
[TABLE]
By integrating the inequality above in over , we obtain (7.11). ∎
Lemma 7.7**.**
Under the same assumptions as in Lemma 7.3, we have
[TABLE]
[TABLE]
for any , where depends on , in (3.2) and the Lipschitz character of .
Proof.
To see (7.12), we use (7.7) and (7.11) to obtain
[TABLE]
where we have used (7.9) for the second inequality. The inequality (7.12) now follows readily by applying the Cauchy inequality to the last integral.
The proof for (7.13) is similar. By (7.8) and (7.9), we obtain
[TABLE]
which yields (7.13) by applying the Cauchy inequality to the last integral. ∎
Lemma 7.8**.**
Let . Suppose and in . Then for any ,
[TABLE]
where depends on and .
Proof.
We use a duality argument. For , let be the weak solution of in with the Neumann condition on . Note that
[TABLE]
It follows by the Cauchy inequality that
[TABLE]
We will show that
[TABLE]
which, together with (7.16), yields (7.15) by duality.
Since is smooth and , we have . By using Lemma 7.7, we obtain
[TABLE]
This, together with (6.16) with , gives (7.17).
The proof of (7.14) is similar. For , let be the weak solution of in with the Dirichlet condition on . Then
[TABLE]
Hence, by the Cauchy inequality,
[TABLE]
We claim that
[TABLE]
which, together with (7.18), leads to (7.14) by duality.
Finally, to prove (7.19), we apply the estimate (7.13). Since is smooth and , we have . Observe that since on , we have on . As a result,
[TABLE]
This, together with the estimate (5.26) with , gives (7.19) ∎
We are now ready to prove the main results of this section.
Theorem 7.9**.**
Let . Suppose and in . Then for any ,
[TABLE]
[TABLE]
where depends on and .
Proof.
We give the proof of (7.20). A similar argument yields (7.21).
First, by (7.12) and (7.15), we obtain
[TABLE]
Next, by (7.9) and (7.15), the third and fourth terms in the left-hand side of (7.20) are bounded by
[TABLE]
In view of (7.22), the integrals above are bounded by the right-hand side of (7.20). ∎
8 Nontangential maximal function estimates
Throughout this section, we assume that and is a smooth domain for some , where depends on the Lipschitz character of . We also assume that satisfies (3.2) for any and (3.6) for any . As before, the dependence of constants on is through and its Lipschitz character.
Lemma 8.1**.**
Let and . Let be defined by (4.13). Then
[TABLE]
where depends on and in (3.2).
Proof.
Fix . By a change of the coordinate system, we may assume that and
[TABLE]
where is a Lipchitz function with and . For with , define
[TABLE]
It is not hard to see that
[TABLE]
where denotes the Hardy-Littlewood maximal operator on , defined by
[TABLE]
for . As a result, by the boundedness of ,
[TABLE]
where .
Next, note that for with ,
[TABLE]
This, together with (8.2), leads to
[TABLE]
By a covering argument, we obtain
[TABLE]
for any , where we have used the Cauchy inequality for the last step. We now apply (8.3) to the function and then let . Using , we deduce that
[TABLE]
Finally, we apply (8.4) to , where is a function in with the properties that for and [11]. It follows that
[TABLE]
for any . This gives (8.1). ∎
Theorem 8.2**.**
Let be a solution of the Neumann problem,
[TABLE]
Let . Then
[TABLE]
where depends on in (3.2) and .
Proof.
In view of Corollary 4.4, it suffices to prove that
[TABLE]
where is defined by (4.13).
Let for . Then , where
[TABLE]
Let , where in and on . By Theorem 2.5,
[TABLE]
where we have used the Rellich estimate (7.20) for the last inequality.
Next, note that in and on . It follows by Theorem 7.5 that
[TABLE]
where we have used (8.9) for the last inequality. Since and , we obtain
[TABLE]
where we have used the Rellich estimate (7.20) with for the last step. In view of Lemma 8.1, this gives
[TABLE]
Finally, to prove (8.8), note that
[TABLE]
where we have used (8.10) and (8.11) for the second inequality. Moreover, if , then
[TABLE]
where we have used (8.1) for the first inequality and (7.20) with for the second. ∎
Theorem 8.3**.**
Let be a solution of the Dirichlet problem,
[TABLE]
Let . Then
[TABLE]
where depends on in (3.2) and .
Proof.
The proof is similar to that of Theorem 8.2, using (7.21) in the place of (7.20). ∎
9 An approximation argument
In this section we use an approximation argument to extend the results in the last section to Lipschitz domains in , . We assume that is a Lipschitz domain in , where and depends on the Lipschitz character of . As in the last section, we also assume that satisfies (3.2) for any and (3.6) for any .
Lemma 9.1**.**
Let be a bounded Lipschitz domain in . There exists a sequence of smooth domains with uniform Lipschitz characters such that with the following properties:
- •
There exists a sequence of homeomorphisms such that uniformly on as . Moreover,
[TABLE]
for any .
- •
The unit outward normal to , converges to for a.e. .
- •
There are positive functions on such that uniformly in , a.e. as , and
[TABLE]
for any measurable .
- •
There exists such that
[TABLE]
Proof.
See [13]. ∎
Lemma 9.2**.**
Let , where is a bounded Lipschitz domain in . Suppose that is a weak solution of the Neumann problem (8.6) for some , or a weak solution of the Dirichlet problem (8.12) for some . Then
[TABLE]
as , where is given by Lemma 9.1.
Proof.
We give the proof for the Neumann problem (8.6). The proof for (8.12) is similar.
Since and , we may rewrite (8.6) as
[TABLE]
for some and . Let , where and denotes the fundamental solution for in . Note that . It follows that
[TABLE]
as . Also observe that is harmonic in and . As a result, and have nontangential limits a.e. on and [4, 5]. By the dominated convergence theorem,
[TABLE]
as , which, together with (9.2), gives (9.1). ∎
Theorem 9.3**.**
Let be a solution of the Neumann problem (8.6) for some . Let . Then
[TABLE]
where depends on in (3.3), and the Lipschitz character of .
Proof.
Let be a weak solution of (8.6) with . Let be a sequence of smooth domains, given by Lemma 9.1. Since and , we have . This allows us to apply Theorem 8.2 and (7.20) to obtain
[TABLE]
where we have used to denote the nontangential maximal function of for the domain . Thanks to Lemma 9.1, the constant in (9.4) depends only on , the Lipschitz character of , and in (3.2). In particular, the estimate (9.4) is uniform with respect to . Using the observation for , we see that
[TABLE]
for . We now let in the inequality above. Note that by Lemma 9.2,
[TABLE]
Thus,
[TABLE]
By letting in the inequality above, we obtain (9.3). ∎
Theorem 9.4**.**
Let be a solution of the Dirichlet problem (8.12) for some . Let . Then
[TABLE]
where depends on in (3.3), and the Lipschitz character of .
Proof.
The proof is similar to that of Theorem 9.3, using Theorem 8.3 in the place of Theorem 8.2. ∎
10 Proof of Theorems 1.1 and 1.2
In this section, with Theorems 9.3 and 9.4 at our disposal, we use a localization argument to complete the proofs of our main results.
Let be a bounded Lipschitz domain in , . Assume and is of finite type on . It follows that there exist an integer and such that
[TABLE]
for any . By extension we may assume and (there is no need to extend ). By compactness we may further assume that (10.1) holds for any with dist.
Let .
Lemma 10.1**.**
Suppose (10.1) holds for any with . Then
[TABLE]
for any and any with . Moreover,
[TABLE]
for any and . The constants depend on , in (10.1) and .
Proof.
Let . Let denote the (matrix-valued) Taylor polynomial of at . Then, if ,
[TABLE]
for any , where depends on and . It follows that
[TABLE]
for any with . Also note that
[TABLE]
where we have used (10.1) for the last step. As a result, we see that
[TABLE]
if and is sufficiently small. This gives (10.3). Together with (10.4), it also yields (10.2) under the additional condition . Note that the inequality (10.2) with and implies the doubling condition,
[TABLE]
for any . The general case for (10.2) follows by a covering argument. ∎
Remark 10.2**.**
Suppose (10.1) holds for any with dist. Also assume that
[TABLE]
It follows by definition that for any . By letting in (10.2), we see that
[TABLE]
for any and . As a result, we obtain
[TABLE]
for any . By definition it is not hard to see that .
Proof of Theorem 1.1.
We first consider the case . Rewrite (1.2) as
[TABLE]
where . By Lemma 10.1, the magnetic field satisfies the condition
[TABLE]
for any . It follows that satisfies the same condition. Moreover, by the proof of Lemma 10.1, there exists , depending only on , in (10.1) and such that
[TABLE]
for any and . This implies that
[TABLE]
for any and , where depends only on , in (10.1) and .
Next, we fix and consider a family of Lipschitz domains , where and is sufficiently small. Let . Note that , satisfies (3.2) for any . Moreover, by (10.9),
[TABLE]
if and is sufficiently large. This allows us to apply Theorem 9.3 to obtain
[TABLE]
where and denotes the nontangential maximal function of with respect to the domain . We point out that the constant in (10.10) does not depend on , as the family of Lipschitz domains possesses uniform Lipschitz characters. It follows that
[TABLE]
for , where
[TABLE]
for . By integrating the inequality above in , we obtain
[TABLE]
By a covering argument, this gives
[TABLE]
where .
To bound for away from , observe that if ,
[TABLE]
where we have used the interior estimate (4.14) for the first inequality and (3.12) for the second. It follows that
[TABLE]
This, together with (10.11), leads to
[TABLE]
To bound away from , we let and apply Theorem 9.3 to the domain , where . This yields
[TABLE]
By integrating the inequality above in , we obtain
[TABLE]
Thus, by a covering argument,
[TABLE]
In view of (10.12), we have proved that
[TABLE]
where we have used the energy estimate for the last step.
Finally, note that
[TABLE]
where . Let . The desired estimates (1.6)-(1.7) for now follow from (10.13). Note that by Remark 10.2, if ,
[TABLE]
where depend only on , in (10.1) and .
The case can be handled by the method of descending. Indeed, let be a weak solution of the Neumann problem (1.2) with boundary data in a two-dimensional Lipschitz domain . Define and . Then in , where is a bounded Lipschitz domain in . Note that on , and that on . It follows from the case that
[TABLE]
Observe that if , then . The other estimates in Theorem 1.1 may be obtained in the same manner. We omit the details. ∎
Proof of Theorem 1.2.
The proof for the case is similar to that of Theorem 1.2, using Theorem 9.4. The case can also be treated by the method of descending. Let be a weak solution of the Dirichlet problem (1.3) with boundary data in a two-dimensional Lipschitz domain . Let , and be defined as in the proof of Theorem 1.1. Then is a weak solution of in with Dirichlet data , where on , and for . It is not hard to check that
[TABLE]
and
[TABLE]
where we have used (3.12). Moreover,
[TABLE]
where we have used the fact for any , if . As a result, we have proved that
[TABLE]
By choosing small so that , we obtain (1.8). The estimate (1.9) follows in a similar manner. ∎
Conflict of interest. The author declares that there is no conflict of interest.
**Data availability. ** Data sharing not applicable to this article as no datasets were generated or analyzed during the current study.
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