384 G. SWEERS side f C() there is a constant k>0suchthat k d (x ) u (x) for all x : () For smooth regularity theory shows that there is another consta

Transcription

1 Rend. Istit. Mat. Univ. Trieste Suppl. Vol. XXVIII, 383{49 (997) Hopf's Lemma and Two Dimensional Domains with Corners Guido Sweers () Summary. - Let IR beabounded domain that is smooth except for a nite number of corners. The aim of this paper is to obtain conditions such that the solution of the Poisson problem u = f 0 in (f 6= 0), with zero Dirichlet boundary condition, behaves similarly at the boundary as does the rst eigenfunction. That is: there exist c c > 0 such that c (x) u (x) c (x) : As a consequence we can improve some results for smooth domains to domains with corners, such as: the maximum principle for an equation with a potential which is unbounded near the boundary the anti-maximum of Clement and Peletier a Green function estimate of Zhao.. Introduction If is smooth enough and f 0 suciently regular one nds pointwise estimates for the solution u of ( u = f in () u =0 by the maximum principle and Hopf's boundary point lemma. Indeed they imply the following. For a nonnegative nonzero right hand () Indirizzo dell'autore: Department of Pure Mathematics, Delft University of Technology, PObox 503 { 600 GA Delft (The Netherlands). sweers@twi.tudelft.nl

2 384 G. SWEERS side f C() there is a constant k>0suchthat k d (x ) u (x) for all x : () For smooth regularity theory shows that there is another constant k 0 (0 ) such that u (x) k 0 d (x ) for all x (3) where the distance function d ( ) is dened by d (x ) = inf fjx yj y = g : (4) If there can be no doubt we write d (x) =d (x ). The standard boundary point lemma of E. Hopf for the Laplacian is as follows. See Theorem.7. in []. Hopf's Boundary Point Lemma. Let u satisfy u 0 in D and u M in D, u (P )=M for some Assume that P lies on the boundary of a ball B D. If u is continuous on D [ P and if the outward directional exists at P, then u (P ) > If is not smooth the estimates () and (3) are no longer true in general. Roughly speaking the estimate in () goes wrong for domains with corners pointing outside and (3) goes wrong for domains with inside pointing corners. Instead of comparing with d (x) in () and (3), one could try to compare with (x), the rst eigenfunction of ( = in =0 (5) For smooth there is not much dierence since c d (x) (x) c 0 d (x) for some c c 0 > 0. For the Laplacian in two dimensional domains the result goes back to C. Neumann in 888. For general elliptic operators (in general dimensions) the result was published in 95 independently by E. Hopf and by Oleinik. See page 56 of [] for references and more bibliographical details.

3 HOPF'S LEMMA AND TWO DIMENSIONAL 385 For nonsmooth the question is the following: do k k 0 (0 ) exist such that and k (x) u (x) for all x (6) u (x) k 0 (x) for all x? (7) We will show that for domains with nitely many corners that satisfy a uniform interior cone condition with large angle the answer is yes. If such a domain has a corner with a small angle the answer is a conditional yes. The condition will be sharp. Related estimates for plane domains with corners are obtained by Oddson in [9]. His estimates are not sucient for the results we areinterested in. In several other papers related estimates can be found. One should mention Kondrat'ev [3], [4] and also [7], [8]. However, collecting the literature on elliptic boundary value problems on angular domains is la mer a boire. The domains that we consider will have a nite number of corners. It would be interesting to see if similar results hold for Lipschitz domains. On such more general domains the conditions will depend on the Lipschitz coecient. In order to simplify notations we use: Definition. Set u v on A if and only if there exists k (0 ) such that u (x) k v (x) for all x A: Set u v on A if such a constant does not exist. Definition. Set u ' v on A if and only if there exists k (0 ) such that k v (x) u (x) k v (x) for all x A: We will use results of Kadlec [] and Grisvard [9]. For the pointwise results we are interested in, we will compare the solutions with appropriate sub- and supersolutions. In the last sections we will show how the estimates can be used to extend the results mentioned in the abstract. In the appendix one nds some of the results for conformal mappings that we use.

4 386 G. SWEERS. Some standard preliminaries Let us recall some regularity results for solutions u of () on a bounded domain R n. The function u will denote a solution of () in appropriate sense. R) If satises an exterior cone condition then (see Theorem 8.30 of [8]) f L q () with q> n implies u C 0(). R) isc (see Theorem 9.5 of [8]) then f L q (), q> implies u W q 0 () \ W q (). R3) If is convex (see [], [9]), then f L () implies u W 0 () \ W (). The following imbedding results hold: I) On general (Theorem 7.0 of [8]) u W q 0 () q>nimplies u C(). I) If satises a uniform interior cone condition, then uw q (), q> n implies u C() \ L (). I3) isc 0, then (Corollary 7. of [8]) u W q (), q> n implies u C(), (and q>nimplies u C ()). Notice that conditions on the outside of the domain, an exterior cone condition or convexity (which might be called exterior plane condition), imply the regularity of the solution. Conditions on the inside allow one to transfer integrability properties of rst or second order derivatives to continuity of the function itself. 3. Is the rst eigenfunction the lowest positive superharmonic function? On smooth domains the results that we mentioned in the introduction imply that any positive superharmonic function, that is continuous, lies above a positive constant times the rst eigenfunction. A proof of this result uses the smoothness of the boundary. However, it is not clear if this regularity is necessary. Let us state this open question as a conjecture.

5 HOPF'S LEMMA AND TWO DIMENSIONAL 387 Conjecture 3. Let R n beabounded domain and suppose that there is a rst eigenfunction C 0 \ C () (we x max = ). Then for every u W p loc (), p> n, with u 0 u 0 in, either u 0 or there exists c>0 such that u c in. This estimate from below by the rst eigenfunction could take care of some of the results in this paper. But since there does not seem to be a proof in the general case we will (only) show it under some conditions on the boundary. By the way, the estimate from above will not be true in general. The eigenvalue problem in general domains has recently been studied by Ba~nuelos in [3]. 4. Elementary domains 4. Notations Since results related with the Riemann Mapping Theorem are more easily stated using C instead of R we will use boldface for the complex alternative: for x R set x = x + ix C for A R set A = fx + ix C x Ag for h : R! R set h (x) =h (x)+ih (x) : We will start by considering the following domains in R D ( ) = x R jxj < jarg xj < : The angle of the domain D ( ) atzero(we will always measure from inside) equals. Related are the circular boundary: = the conical boundary: S = the related growth rate: = : Notice ( ) = [ S. We willalsouse n x D ( ) jxj = n x D ( ) jxj < D ( ) = x R x D ( ) : o o

6 388 G. SWEERS 4. Cones that are convex near 0 We start with domains = D ( ) when <and will be interested in the behavior near (0 0). Note that the corner at (0 0) is convex and hence there is no regularity problem (see Grisvard [9], [0]). D( ) for some (0 ). We will compare the solutions of 8 < : v =0 v =0 v =cos ' in D ( ) on S on (8) ( = in D ( ) =0 ( ) with the rst eigenvalue (take > 0), and for > ( u = r in D ( ) u =0 ( ) : (9) (0) Note that () has a solution u C 0 () if f L q () with q> if ( R ) satises an exterior cone condition at every boundary point. Lemma 4. Let D ( ) v be as above. Then for all (0 ) v ' on D ( ) : Proof. We can write both functions explicitly. We have v (r ') = r cos ' and by using a Bessel function of order, namely (r ') = J r cos ' with corresponding eigenvalue

7 = where HOPF'S LEMMA AND TWO DIMENSIONAL 389 ( is unique up to multiplication by a constant) J (s) = X m=0 () m s +m m! ( + m +) and the rst positive zeroofj (). Since r J r r on 0 the estimate in i) holds. Remember that J (s) solves s J 00 + sj 0 + s J =0. Theorem 5. Suppose > and x (0 ). Let D ( ) and u be as above. Then i. for +> we have u ' on D ( ), ii. for +< we have u ' r + on D ( ), iii. for += we have u ' ( ln r) on D ( ). Remark : From ii) and iii) it follows that for + one has u on D ( ) : Remark : The relation between the opening angle (0 )and the critical growth ( ) is = or reversed =. + For the semilinear problem u = f (u) with f (0) < 0 the angle is critical for the existence of a positive solution, see [4]. alpha / Pi psi Pi

8 390 G. SWEERS Proof. In all of the parts we will use the maximum principle as follows. If u v C () \ C() satisfy ( u v in D ( ) then Note that u v D ( ) u v in D ( ) : u on D ( ) \ D ( ) : We will compare u with several w i. All functions w i satisfy w i =0 on S : i) above. Weset = min +. Then it follows that + > and hence that r + r r compare with for r [0 ]. Now we w = k r cos ' r cos (') cos = () where we take k large enough such that w 0 on D ( ). Then, using the fact that cos = < 0, we nd w = r cos = r on D ( ) and w r cos ' on D ( ). Since ( r w on D ( ) 0=u w D ( ) the previous lemma and the maximum principle imply u w on D ( ) : i) below. We compare with w = r r + cos ': ()

9 HOPF'S LEMMA AND TWO DIMENSIONAL 39 It follows that 8 < : w = ( +) r cos ' r on D ( ) 0=w = u ( ) and hence by the previous lemma and the maximum principle we nd w u on D ( ) : ii) above. Set + w 3 = r cos + (( +)') cos + One nds w 3 = r ( +) cos ( r w 3 on D ( ) from which it follows that ii) below. Taking we obtain 8 < : w 4 = and hence u w 3 D ( ) u w 3 r + on D ( ) : : (3) and hence w 4 = r + cos ' (4) ( +) r cos ' r on D( ) 0 w 4 u D( ) r + w 4 u on D ( ) : iii) For + = we compare with w 5 = r (k ln r) cos ' + ' sin ' (5)

10 39 G. SWEERS with k large enough such that w 5 > 0onD ( ). Since w 5 = r we can use this function for both sides of the estimate. From ( w5 r w 5 in D ( ) it follows that Since w 5 u w 5 D ( ) w 5 u w 5 in D ( ) : w 5 r ln r cos ' w 5 in D ( ) the previous lemma implies the last estimate of the theorem. Corollary 6. Fix (0 ) and let u be a solution of 8 < : u = f u =0 u = g in D ( ) on S on with 0 g C ( ) and 0 f C (D ( )), such that for some and M one has If #>, then 0 6= f (x x ) M jx j # for x D ( ) : u ' on D ( ) : Remark 3: Note that if we do not assume a xed sign for f or g we nd that there is k>0 such that ju (x)j M (x) for all x D ( ) : Remark 4: The conditions on f imply that f L (D ( )). For 0 one has f L (D ( )). For < < 0 one nds in fact that f L p (D ( )) for all p< and hence f L (D ( )) : Z D( ) jfj p dx M p Z Z cx x =0 x =cx x p dx dx = cm p p + :

11 HOPF'S LEMMA AND TWO DIMENSIONAL 393 Hence there is a solution u W 0 (D ( )). Regularity results, see [8], show that u W ( 0 )forany 0 D ( ) with 0 not containing the corner points. In fact, since D ( ) is convex it follows from [] that u W (D ( )) see Theorem 3... of [9]. 4.3 Cones that are concave near 0 In this section we recall some results for domains with an entrant corner. We will use D ( ) that have a corner at (0 0) with angle ( ). D ( ) for ( ). From results of Grisvard [], see also Lemma and Theorem in [9], one has for f L p (D ( )), with p and q = N, p + q =, that there exists a solution u W0 (D ( )) and it satises where u =u f + nx p m= u f W p () c m f R m (') =sin m n p = q c m f r m m (') ' the entier of q : Taking p =+" in the above, with 0 <"<, one nds u =u f + c f r cos '

12 394 G. SWEERS with u f W +" (D ( )). Since W +" (D ( )) C D ( ) holds in a two dimensional domain with Lipschitz boundary we nd that ju f (x)j d r cos ' on D ( ). Hence ju (x)j r cos ' on D ( ). Note that the solution of (8) is given by v = r cos '. Next we will show that the solutions of (8), (9) and (0) with >, have the same behavior near 0. Lemma 7. Let v and be the solutions of respectively (8) and (9) on D ( ) with concave corner. Then v ' on D ( ) : Proof. See Lemma 4. Corollary 8. Let D ( ) and beasabove. Let f L p (D ( )) \ C (D ( )) with p> such that 0 6= f 0, and let 0 g C ( ). Then the solution u of 8 < : u = f in D ( ) u = 0 on S u = g on satises u ' on D ( ) : (6) Remark 5: Note that f (r) =r, with <<0, is in L p (D ( )) with p. If u is the solution of (0) then u satises u on D ( ). Remark 6: Again, if we skip the sign condition for f and g, we nd that for some k>0 ju (x)j k (x) for all x D ( ) : Proof. By the results of Grisvard we nd u r cos ' on D ( ) :

13 HOPF'S LEMMA AND TWO DIMENSIONAL 395 By the explicit formula for we obtain r cos ' on D ( ) which shows the estimate from above. The estimate from below follows by ( v =0u in D ( ) v u D ( ) the maximum principle and the previous lemma. 5. The basic result in general domains with corners The domains that we will consider satisfy the following assumptions. Condition 9. The domain is an open bounded subset of R such that: i. is the inside of a (closed) Jordan curve, say = e i' ' [0 ] ii. is Dini smooth except in nitely many points e it j k+m j= iii. at every point y (j) smooth corner = e it j the has a Dini iv. the corresponding angles, j = lim "#0 \ 0 (e i(t j") ) 0 (e i(t j +") ) which are measured from inside, lie in (0 ). We will assume that j (0 ) for j k, and j ( ) for k + j k + m.

14 396 G. SWEERS Remark : is called a Jordan curve if = (@B (0)) with (0)! continuous and one-one. A function is called Dini smooth if the derivative exists and is Dini continuous. A (part of the) boundary is called Dini smooth if there exists a Dini smooth parameterization with 0 6=0. The curve has a Dini smooth corner at e it j if j " = e it t (t j " t j ] + i " = e it t [t j t j + ") are Dini smooth arcs for some small ">0. Remark : The condition implies that is simply connected. This is not necessary. Most results in this paper have an obvious extension to bounded domains which boundary consists of nitely many non intersecting Jordan curves, all of which satisfy the items ii), iii) and vi) in Condition 9. Theorem 0. Suppose satises Condition 9 and that f L p ()\ C () with p> and 0 6= f 0. Let u W 0 () \ W p loc () be the solution of u = f in Then u =0 u on : (7) If moreover, f is such that for every i f ::: kg there exist # i > and " i M i > 0 with i x f (x) M i y (i) # i on B "i y (i) \ (8) then u ' on : (9) y (i) the estimates in (7) and (9) follow S k+m Proof. On n B i= by the strong S maximum principle. Hence it remains to show(7)and k+m (9) on \ B i= y (i). By Corollary A.5 there is an appropriate holomorphic mapping h i from to C, suchthat h 0 is bounded away from 0 and, and for some c>0 c D ( i ) h B " y (i) \ cd( i ) :

15 HOPF'S LEMMA AND TWO DIMENSIONAL 397 By Corollaries 6 and 8 we nd that the solution u h of ( uh (x) =f h inv (x) for x h () u h =0 satises, respectively without and with condition (8) " uh c D( i ) on u h ' c c D ( i ) : (0) D( i ) Since (0) also holds when u h is replaced by h(),we nd respectively that " uh h() u h ' h() on c D ( i ) : Lemma A. shows for some " 0 > 0 that " u on B " 0 y (i) \ : u ' 6. A semilinear equation Consider the equation u = u p with p ( 0). u =0 in () Theorem. Let be abounded domain in R that satises Condition 9. Let 0 be the angle of the smallest corner If u W q () \ W q 0 (), with q >, is a solution of (), then p> 0 q. Proof. We nd u L q () and hence, if denotes the angle of the boundary at 0, we nd that there is c>0 such that near 0 we have c jxj d (x) u (x) c jxj d (x) : Then u p L q ( \ B " (0)) for small ", if and only if pq +>.

16 398 G. SWEERS 7. A generalized maximum principle The classical maximum principle for the Dirichlet problem not only holds with a positive potential but also with a potential V with V >,where is the rst eigenvalue. A version of the classical maximum principle. Let be a bounded domain in R n with a C boundary. Suppose that V L with V >, and that u C \ C () satises u + Vu 0 in u =0 () then u 0. Z If is a C 0 -domain one is able, by using the Hardy inequality ju (x)j Z dx c H jru (x)j dx for all u W 0 () (3) d (x) (see [5] or [7]) to generalize the result above to potentials V that are unbounded near the Although the result seems to be standard we have not been able to locate a reference. Theorem. Let be abounded domain in R n with a C 0 -boundary. Let c H denote the best constant in (3). Suppose V C () is such that c H < c V d (). If u C \ C () satises () and u W 0 () then u 0: Proof. Set V = (jv jv ) and u = (juju). Set = fx u (x) < 0g and suppose that is nonempty. For x wehave uv u V + u V + u + =0which implies for all C 0 ( ) with 0 that Z (ru r V u) dx 0: Added in proof: [30]

17 HOPF'S LEMMA AND TWO DIMENSIONAL 399 Since u W 0 () we nd with (3) that 0 Z (ru ru V uu ) dx + c jru j c H Z dx 0: Hence u =0. Theorem 3. Let be abounded domain in R that satises Condition 9. Suppose that V C () is such that there are " K > 0 for which the following holds. i) d (x) V (x) for x ii) K d (x) " V (x) for x : If < c H then a function u C \ C () \ W 0 () that satises () is positive and moreover either u 0 or there is c>0 with u (x) c (x) for x : where is the rst eigenfunction of : For smooth domains and V bounded, optimal results in comparing u and are found in [7]. Proof. By Theorem it follows that for small the function u satises u 0. Then we have u + V + u = V u 0 in. By the standard strong maximum principle one nds u 0oru>0for every domain 0 with 0 and hence in. It remains to show the boundary behavior. With similar arguments as we used in the proof of Theorem 0, which are the results stated in the appendix, it is sucient toshow the boundary behavior in a neighborhood of 0 for = D ( ) withtheappropriate. First we consider the case where =. We may assume that " and we x = ". i) =. We use the function w 6 (x x )=x ( + x x ) x : (4) Then one has on the set where w 6 > 0 holds that w 6 + V + w 6 w 6 + Kx " w 6

18 400 G. SWEERS x ( +)+8x +Kx " which is negative forx (0 ) with := min r ( +) K +8 Set S = x D () x <x <. Then w 6 0 on the part where x = x. Fix in number = max! p p. Since u>0 in D () the w6 ( x ) u ( x ) jx j : (5) is well dened and positive. Now we are able to compare u and w 6 t on S. For large t>0wehave u>w 6 t in S and for every t>0we nd u > w 6 t So either u w 6 0 in S or there is a smallest t > 0such that for some x S u w 6 t in S u (x ) = w 6 (x ) t : Suppose that the second possibility holds. Since u > 0 on S and hence w 6 (x ) t = u (x ) > 0 there exists S, with S S smooth, such that w 6 t 0 on S. Then we nd on S, and hence on S that (+V + )(u (w 6 t )) V u (+V + ) w 6 + V + t 0: It follows by the strong maximum principle that u>w 6 t on S and hence a contradiction. That is u w 6 0inS: By another application of the sweeping principle of McNabb ([6]), now a shift of u(w 6 t) of at most p in the x -direction and repeating the argument above, we nd that u (x + s x ) w 6 (x x )ins when jsj p. Hence we have u (x x ) w 6 (x 0) for 0 <x < jx j < p :

19 HOPF'S LEMMA AND TWO DIMENSIONAL 40 The estimate in the theorem follows since w 6 (x 0) cx for some c>0 and all x 0 ii) 6=. With similar arguments as before we use Then r w 7 + V + w 7 w 7 (r ')=r ( + r r) cos ' : (6) r r cos ' Kr +" ( + r r)cos ' which is negative on D ( ) for 0 < r < r 0 := r replace S by ~ S = r0 D ( ) and with ~ =max w7 (r 0 ') u (r 0 ') j'j : r +Kr cos ' K. We The result of part i) shows that the quotient remains bounded when '! from inside. Finishing the argument as before we nd that u ~w 7, implying the estimate of the theorem. 8. The anti-maximum principle Clement and Peletier showed in [6] a result that reads for the Laplacian with zero Dirichlet boundary conditions as follows. Let be abounded domain in R n with a C boundary. Suppose f L p (), p>n, such that 0 6= f 0, andsuppose u satises the equation u u = f in (7) u =0 Then there exists >0, depending on f, such that if << +,

20 40 G. SWEERS i. u (x) < 0 for all (x) > 0 for all where n is the outward normal. Birindelli recently ([4]) extended the anti-maximum principle to general domains but only for right hand sides f which have its support outside of the non smooth boundary. We allow less general domains and more general f. Our estimate will be optimal. Theorem 4. Let be abounded domain in R that satises Condition 9. Suppose f L p ()\C (), withp> and such that 0 6= f 0. We assume that for all i f ::: kg there exists # i > and i >0 with jf (x)j M x y (i) # i for all x B y (i) \, (8) Suppose u satises the equation in (7). Then there exists > 0, depending on f, such that for << +, there exists c c > 0 with c (x) u (x) c (x) for all x : Remark : Without loss of generality we may suppose that p<+ i i for all i fk + ::: k+ mg. (9) The result is also optimal in the following sense. Proposition 5. Let =D ( ) for some (0 ) and take # with # 0. For f = r # we nd that for all ( ) the solution u of (7) changes sign. Remark : Note that f = r # L p () for some p>. For < one may take f =.

21 HOPF'S LEMMA AND TWO DIMENSIONAL 403 The proof basically follows [6]. First we suppose that not only satises Condition 9 but even is C smooth except for C smooth corners. It means that the boundary consists of nitely many curves i = i ([0 ]) with i C [0 ] and 0 i 6= 0. For such adomain one can solve () for f L p (), with <p+" and " small, in the following way. Lemma 6. Let be as above. Then there is ">0 and there exist f i g k+m i=k+ such that i. i C ii. support ( i ) B " y (i) \ iii. (x) i (x) (x) for x B " y (i) \ iv. for all f L +" the solution u W 0 () of () satises u =u + with u W +" () and c i R. k+m X i=k+ c i i Proof. By the remark following Corollary A.5 one nds holomorphic mappings h i that maps a neighborhood of y (i) in onto cd ( i )and moreover h i C. This implies that x 7! u (x) W p () is equivalent withx 7! u h inv i kuk W p () ' u h inv i (x) W p (h i ()) and even W p (h()) for u W p () : Since we assume (9) we nd by [9] for all i fk + ::: k+ mg a neighborhood of 0 (= h i y (i) ) with u h inv i (x) =~u i (x)+c i i (x) where ~u i W p i (cd ( i )) and i (x) = r cos ' i i (x). The function i is chosen such that it localizes i,thatis i C R + i i

22 404 G. SWEERS and for some 0 < < <cone has i on B (0) and i 0 on R nb (0). For u we nd that u =u + k+m X i=k+ c i i h i on with u W p (). Direct calculus shows i C (h ()). The estimate in 3) follows from Corollary 8. We will replace the function e that is used in [6] by Remember that one has e (x) = (x)+d (x) : (x) d (x) (x) (x) d (x) (x) for x near a `convex' corner, for x near a `concave' corner. The following Banach space (even a Banach lattice) will be used: with norm C e = u C 0 juj e kuk e =sup u (x) x e (x) Since e () d () wendthatc \C 0 is continuously imbedded in C e. Since e () () we nd i C e. Denote Y = L p () (p as above) with its standard norm and : X = u W p () u =0on@ [ i ] k+m i=k+ with norm k+m X kuk X = kuk + W p () i=k+ jc i j P k+m where u = u + c i=k+ i i. Since the set f i g is independent and i = W p () this norm is well dened. Theorem of [9] shows that kuk c W p () kuk + kuk L p () L p () : (30)

23 HOPF'S LEMMA AND TWO DIMENSIONAL 405 Since W p () is compactly imbedded in L p () we nd from Theorem 6. of [3] that the operator : u W p () u =0on@! L p () is a semi-fredholm operator that is, it has a closed range and a nite dimensional null space. Since i L p () we also nd that A =:X! Y is a semi-fredholm operator. Theorem of [9] implies that A L(X Y )hasaninverse (and hence A is a Fredholm operator of index 0). We will denote this inverse by T. Being the inverse of a bounded linear operator on Banach spaces (see Theorem 4. of [3]) it is bounded. Hence we nd ktfk X 'kfk Y for f L p () : By Theorem 7.6 of [8] it follows that the imbedding W p ()! C is compact. Hence the imbedding X! C [[ i ] k+m i=k+ is compact. Since i C e and since C \ C 0 is continuously imbedded in C e we nd that the operator T e := T : C e! C e is well dened and compact. We summarize. Lemma 7. The following imbedding results hold. i. X! C \ C 0 [ i ] k+m i=k+ is compact. ii. C \ C 0 [[ i ] k+m i=k+! C e is continuous. iii. C e! Y is continuous. Since T L(Y X) the operator T e L(C e C e ) is compact. Up to now we used that the boundary of consists of piecewise C -curves. We may replace C by Dini smoothness using a transformation h as in Corollary A.6. The function h is a dieomorphism from to a piecewise smooth domain h () which has the same corners as and with 0 <cjrhj c for some c>0. Instead of (7) one considers ( uh jrhj u h = jrhj f h in h () u h =0 : (3)

24 406 G. SWEERS By the strong maximum principle one nds that T e is positive and irreducible. Since the operator T e is compact, positive and irreducible we may use the Krein-Rutman Theorem with the De Pagter Theorem ([0], see also [5]) and nd an analogy of Lemma in [6]. The last statement of the next lemma follows from Theorem 0. Indeed, u T e u = g is solved by u = I T e g = and u T e g. X k T e g k=0 Lemma 8. We have: i. The spectral radius r (T e ) is positive (and r (T e )= ). ii. r (T e ) is a simple eigenvalue of T e with eigenvector, and is the only eigenvector with xed sign. iii. r (T e ) is a simple eigenvalue of Te with eigenvector,dened by (u) = Z (x) u (x) dx: iv. For every g C e, with 0 g 6= 0and >r(t e ), there exists exactly one solution u of u T e u = g and it satises u. Remark 3: Note that in contrary to Lemma of [6] we do not nd T e f e for f C e with 0 f 6= 0. This implies that we do not obtain strong positivity oft e in the sense of []. The operator T e would be strongly positive if T e (P e nf0g) P e where P e is the positive coneinc e. Similarly as in [6] one has the decomposition as in their Lemmata and 3.

25 HOPF'S LEMMA AND TWO DIMENSIONAL 407 Lemma 9. The space Y satises Y =[[ ] R (A I). For every f Y there exists f R (A I) such that with = (Tf) = ( ). f = + f (3) Let u X. Since X Y there is a unique decomposition u = + u (33) with u X \ R (A I). If u is a solution of (7) with f as in (3) we nd = (34) Au u = f : (35) Note that A I : X \ R (A I) X! R (A I) Y is an isomorphism for j j small. Hence there are constants M f, not depending on, suchthat ku k X <M f for all [ + ] : (36) We shall need an additional result for the inverse of this restriction of A I. Lemma 0. Suppose that f R (A I) satises (8). Then there ism 0 f such that for j j <the solution u of (35) satises ju (x)j M 0 f (x) for x : (37) Proof. First note that for all f Y which satisfy the conditions of Theorem 4, there exists M such that for all [0 ], with >0, the solution u of (7) satises ju (x)j M (x) : (38) Indeed, denoting by u the solution for, we nd ( ) u u 0 = u 0 ( ) u 0

26 408 G. SWEERS and hence by the maximum principle we have, uniformly for [0 ], that there are c M such that From (36) it follows that u u 0 + c M in : (39) ju (x)j cm e (x) in : (40) Since the rst eigenvalue large for " small, we maysolve ( (+) u = f in on = B " y (i) \ can be chosen u = u for < + and use (38) and the bound (8) for f near y (i) on to nd ju jm c in B " y (i) \ : (4) Together (40) and (4) show the estimate. Proof of Theorem 4. Let u X be decomposed as in (33) with and u as in (34-35). If f satises (8) then f satises (8). For ( + ) wendby Lemma 0 that u + M 0 f The result follows for 0 < small. Proof of Proposition 5. Fix ( ). The function u solves u = u + r #. First we show that there is c>0andr 0 > 0 such that u c for r<r 0 : (4) Since u C 0 D ( ) there is r 0 > 0 such that u + r # > r# for r<r 0. By the standard Hopf's boundary point Lemma there is c>0suchthatu + c > 0onD ( ) \fr = r 0 g. Hence we nd (u + c )=u + r # + c 0 in r 0 D ( ) u + c 0 (r 0 D ( ))

27 HOPF'S LEMMA AND TWO DIMENSIONAL 409 which implies that u + c > 0onr 0 D ( ). Now letv solve v = u in D ( ) v =0 ( ) : Due to (4) it follows that v c 0 in D ( ). By the maximum principle and Theorem 5 we nd that v c and that there is c > 0suchthatu v c r #+. Hence, since # + < 0 and u c r #+ c we nd that u is positive near0. 9. Green function estimate and 3G-Theorem Zhao in [9] obtained a two sided estimate for the Green function for on a -dimensional domain. His result is the following. There exist C>0 such that for all x y d (x) d (y) C G (x y) ln + C G (x y) : (43) jx yj See also [6]. This result is not true for Lipschitz domains. Zhao's proof needs Dini smooth boundary ([5]). Theorem. Let be abounded domain in R that satises Con- dition 9. Let " > 0 be such that min y (i) y (j) =" and let w be dened by 8 min d(x) d(y) ' (x) ' (y) w (x y) = for (x y) B " y (i) \ >< max d(x) d(y) ' (x) ' (y) >: for (x y) B " y (i) \ elsewhere. with i k, with k + i k + m,

28 40 G. SWEERS Then there exist C>0 such that for all x y C G (x y) ln +w (x y) (x) (y) jx yj C G (x y) : (44) Remark : The properties of w are such that one can dene a function ~w C with w ~w w on : Remark : Theorem 7.4 in [8] gives an asymptotic expansion of the Green function near a conical boundary point. Proof. By Koebe's distortion Theorem, see Corollary.4 of [], one nds the following. Let h map conformally to D = fz C jzj < g. Then one has Since 4 it follows that C G (x y) ln jh (x)j d (x) h 0 (x) jh (x)j G (x y) =G B (0) (h (x) h(y)) + d (x) d (y) h 0 (x) h 0 (y)! C G jh (x) h (y)j (x y) Stretching a corner, say in 0, with angle (0 ) one uses h :! C dened by h (z) =z with =. One nds for some c i > 0 that c (x) d (x) c jxj h 0 (x) c jxj c (x) d (x) : To nd estimates for jh (x) h (y)j we distinguish three cases. i. Both x and y are near a convex corner y (j) ( j k). That is = >. Then by Lemma A.3 c 3 jx yj ' (x) d (x) + ' (y) d (y) jh (x) h (y)j c 3 jx yj : (45)

29 HOPF'S LEMMA AND TWO DIMENSIONAL 4 and d (x)! when x! y (j) (corner). ' (x) nontangentially Hence one nds (x) (y) c 6 jx yj min d(x) d(y) ' (x) ' (y) d (x) d (y) h 0 (x) h 0 (y) c 6 (x) (y) jx yj jh (x) h (y)j min d(x) d(y) ' (x) ' (y) ii. Both x and y are near a concave corner y (i) (k+ j k+m). That is = <. Then by Lemma A.3 c 3 jx yj and Hence d (x) ' (x) + d (y) ' (y) jh (x) h (y)j c 3 jx yj : d (x)! 0 when x! y (i) (corner). ' (x) nontangentially c (x) (y) 4 max d(x) d(y) jx yj ' (x) ' (y) c (x) (y) 3 jx yj d(x) ' (x) + d(y) ' (y) d (x) d (y) h 0 (x) h 0 (y) c 4 (x) (y) jx yj jh (x) h (y)j max d(x) d(y) ' (x) ' : (y) iii. x is near y (i) and y is near y (j) 6= y (i). Then c (x) (y) 7 jx yj d (x) d (y) h 0 (x) h 0 (y) jh (x) h (y)j (46) c 7 (x) (y) jx yj :

30 4 G. SWEERS A. Auxiliary results related with the Riemann Mapping Remember that one may use a holomorphic mapping to transform one domain to another domain with possibly little change in the estimates we are interested in. We will start by stating such result. Next we will recall the relation between smoothness of the boundary and smoothness of related conformal mappings. A excellent reference for results in the last direction is the book ([]) by Ch. Pommerenke. A. Conformal transformation with bounded derivative Amappingz 7! h (z) that is conformal on and continuous on, changes problem () in ( (u h inv (x) )= h 0 h inv (x) f h inv (x) for x h () u h inv (x) =0 : (47) If h 0 is bounded away from 0 and we can compare solutions of () and ( uh (x) = f h inv (x) for x h () u h =0 : (48) in a uniform way. Similarly we may compare the eigenvalue problems. Let us denote by A, A the rst eigenfunction respectively eigenvalue on A. Since h() (h (x)) we have: Lemma A.. Let h :! C 0 <c inf = h() h 0 (x) h() (h (x)) for x x be conformal and satisfying h 0 (x) sup h 0 (x) c < : (49) x Let u respectively u h be the solutions of () and (48) for some f L p () with p> and f 0. Then c u h (x) u h inv (x) c u h (x) for x h () (50)

31 HOPF'S LEMMA AND TWO DIMENSIONAL 43 and h() (x) ' h inv (x) for x h () (5) c h() c A. Holomorpic mappings on domains with corners We will use conformal mappings h from D = fz C jzj < g onto. The smoothness of such a conformal mapping h is directly related with the smoothness We refer to a theorem of Kellogg- Warschawski, Theorem 3.6 on page 49 of []. Moreover, is Dini-smooth, then h 0 has a continuous extension to D which is nowhere equal to 0 on D. See Theorem 3.5 in []. In that case the inverse of h has the same regularity as h. Remember that Holder continuity implies Dini continuity. For the domains that we are interested in we use a Theorem by Lindelof and an extension by Warschawski. Both results are also found in [], see Theorem 3.9. The domain satises the assumptions of Condition 9. For h :! C without uniformly bounded derivative the following consequence of Koebe's distortion Theorem holds. Lemma A.. Suppose satises Condition 9. Let h :! C conformal with h (@) Then it follows that be d 4 h 0 (x) d (h 4d h 0 (x) : (5) Proof. Corollary.4 of [] states that for conformal f : D! C 4 jzj f 0 (z) d (f (D )) jzj f 0 (z) : (53) By Condition 9. and the Riemann Mapping Theorem there exists a conformal mapping f : D! C with f (D ) = and f (@D ) The claim follows by using(53)for f and h f. In order to handle the cones we need an estimate for h : D! C dened by h (z) =z with (0 ).

32 44 G. SWEERS Lemma A.3. Fix > 0. For (0 ) there exist c>0 such that for all x y D ( ( )) : c jx yj jxj + jyj jx y jc jx yj : For ( ) there exist c>0 such that for all x y D c jx yj () jxj + jyj jx y jc jx yj : : Proof. The result for > is a direct consequence of the result for (0 ). Hence we assume (0 ). For z D () the function z 7! z is well dened by re i' = r e i' with j'j <. Assume without loss of generality that jxj jyj. We set w = x y and w = x (y ). Notice that w is not well dened in general and if well dened it not necessarily equals w. However, in a small neighborhood of it behaves properly. Indeed, for w K with K = w C jarg wj jwj 3 we nd arg xy < and hence w = x (y ). Since w 7! w is conformal on a neighborhood of K there is c > 0 such thatforwk : c jw j jw j c jw j : (54) If w D nk then both jw j and jw j are bounded away from 0 and bounded from above. Hence (54) is satised for some (other) c. Using again 0 <<we nish by jxj + jyj jx y j jyjjw j c jyj jw j and jxj + jyj jx y jjyjjw j c jyjjw j : Lemma A.4. Let f map D conformally onto and assume that. satises Condition 9. Set min i6=j y (i) y (j) =". Dene i = i

33 HOPF'S LEMMA AND TWO DIMENSIONAL 45 Then 8 x y (i) i + y y (i) i jx yj for (x y) B " y (i) \ with i k >< f inv (x) f inv (y) ' x y (i) i + y y (i) i jx yj and f inv 0 (x) ' 8 >< >: >: jx yj x y (i) i for (x y) B " y (i) \ with k + i k + m elsewhere, for x B " y (i) \ with i k + m elsewhere. Remark : Power series type expansions at a corner are established by Wigley in [8]. Proof. We start with the rst estimate. For (x y) = S k+m i= B " y (i) \ one nds either x B y " (i) and y = B " y (i) for some i (or vice Sk+m versa), or x y n B i= y " (i). In the rst case the estimates hold since f inv (x) f inv (y) and jx yj are bounded away from 0. In the second case the estimates follow since u 7! f 0 (u) is uniformly bounded away from 0 and on Sk+m f n inv B i= y " (i) : The last result follows from an adaptation of Theorem 3.5 in []. It remains to show the estimate when both x and y belong to B " y (i) \. Suppose that B " y (i) \ y (i) +D ( ) after a possible

34 46 G. SWEERS rotation. Then we may dene g :! R, with g holomorphic on by g (x) = x y (i) i. By a theorem of Warschawski (Theorem 3.9 in []) one nds that jg f (u) g f (v)j 'ju vj for u v f inv B " y (i) \ : Hence and by using Lemma A.3 we nd if i > that for x y B " y (i) \ f inv (x) f inv (y) 'jg (x) g (y)j ' ' x (i) i y y y (i) i ' ' x y (i) i + y y (i) i jx yj : Similarly the result for i (0 ) can be shown. The second statement of Theorem 3.9 in [] shows that (g f) 0 (u) ' and since i > 0 hence for u h inv B " y (i) \ x y (i) i = g 0 (x) ' f 0 f inv (x) = inv 0 f (x) : i Corollary A.5. Let satisfy Condition 9. For every i f :::, k + mg there is a continuous mapping h i :! R such that i. h i :! C is conformal ii. h i (@) =@h i () iii. h i y (i) =0 iv. h i () \ B (0) = D ( i ) v. h 0 i can be extended continuously vi. 0 < h 0 i < on.

35 HOPF'S LEMMA AND TWO DIMENSIONAL 47 Remark : Assuming isc n with 0 << implies that h i C n on. See Theorem 3.6 of []. Proof. Since + i " and i " are Dini smooth arcs 0 is uniformly bounded near the corner. Hence there exists a domain that + i "= [, i "= 6=and@ n y (i) is Dini smooth. By the Riemann Mapping Theorem there exist a holomorphic mapping f from D onto with f(( 0)) = y (i) and f(( n@. Set g (z) = z for z with jarg zj z + g (z) = z i for z with jarg zj i : For suciently large constant c>0 the function h i dened by h i = c g inv g inv f inv satises the assumptions above. In a similar way one shows: Corollary A.6. Let satisfy Condition 9. Then there is a continuous mapping h :! R such that i. h :! C is conformal ii. h (@) iii. h 0 can be extended continuously iv. 0 < h 0 < on n h y (i) i k + m C at h y (i), with i k + m, has a corner with the same angle at y (i). References [] Amann H., Fixed point equations and nonlinear eigenvalue problems in ordered Banach spaces, SIAM Review, 8 (976), 60{709.

36 48 G. SWEERS [] Ancona A., On strong barriers and an inequality of Hardy for domains in R n, London Math. Soc., 34 (986), 74{90. [3] Ba~nuelos R., Sharp estimates for Dirichlet eigenfunctions in simply connected domains, J. Dierential Equations, 5 (996), 8{98. [4] Birindelli I., Hopf's lemma and anti-maximum principle in general domains, J. Dierential Equations, 9 (995), 450{47. [5] Chung K. L. and Zhao Z., From Brownian motion to Schrodinger's equation, Springer-Verlag, Berlin, 995. [6] Clement Ph. and Peletier L. A., An anti-maximum principle for second order elliptic operators, J. Dierential Equations, 34 (979), 8{9. [7] Davies E. B., Heat Kernels and Spectral Theory, Cambridge University Press, 989. [8] Gilbarg D. and Trudinger N. S., Elliptic partial dierential equations of second order, nd edition, Springer, Berlin Heidelberg New York, 983. [9] Grisvard P., Elliptic problems in nonsmooth domains, Pitman, Boston, 985. [0] Grisvard P., Singularities in boundary value problems, Masson / Springer-Verlag, Paris/Berlin, 99. [] Grisvard P., Singular behaviour of elliptic problems in non Hilbertian Sobolev spaces, J. Math. Pures Appl., 74 (995), 3{33. [] Kadlec J., On the regularity of the solution of the Poisson problem for a domain which boundary locally resembles the boundary of a convex domain, Czechoslovak Math. J., 4 (964), 386{393 (in Russian). [3] Kondrat'ev V. A., Boundary problems for elliptic equations in domains with conical or angular points, Trans. Moscow Math. Soc., 6 (967), 7{33. (translated from Trudy Moskov. Mat. Obshch., 6 (967), 09{9.) [4] Kondrat'ev V. A., On solutions of weakly nonlinear elliptic equations in the neighborhood of a conic point at the boundary, Dierential Equations, 9 (993), 46{5. (translated from Dierentsial'nye Uravneniya, 9 (993), 98{305.) [5] Kufner A., Weighted Sobolev Spaces, Wiley, Chichester, 985. [6] McNabb A., Strong comparison theorems for elliptic equations of second order, J. Appl. Math. Mech., 0 (96), 43{440. [7] Miller K., Barriers on cones for uniformly elliptic operators, Ann. Mat. Pura Appl., 76 (967), 93{05. [8] Nazarov S. A. and Plamenevsky B. A., Elliptic problems in domains with piecewise smooth domains, Walter de Gruyter, Berlin, 994. [9] Oddson J. K., On the boundary point principle for elliptic equations

37 HOPF'S LEMMA AND TWO DIMENSIONAL 49 in the plane, Bull. Amer. Math. Soc. (N.S.), 74 (968), 666{670. [0] de Pagter B., Irreducible compact operators, Math. Z., 9 (986), 49{53. [] Pommerenke Ch., Boundary behaviour of conformal maps, Springer-Verlag, Berlin etc., 99. [] Protter M. H. and Weinberger H. F., Maximum principles in dierential equations, Prentice Hall, Englewood Clis N.J., 967. [3] Schechter M., Principles of functional analysis, Academic Press, 97. [4] Sweers G., Semilinear elliptic problems on domains with corners, Comm. Partial Dierential Equations, 4 (989), 9{47. [5] Sweers G., Strong positivity in C( ) for elliptic systems, Math. Z., 09 (99), 5{7. [6] Sweers G., Positivity for a strongly coupled elliptic system by Green function estimates, J. Geom. Anal., 4 (994), -4. [7] Walter W., A theorem on elliptic dierential inequalities with an application to gradient bounds, Math. Z., 00 (989), 93{99. [8] Wigley N. M., Development of a mapping function at a corner, Pacic J. Math., 5 (965), 435{46. [9] Zhao Z., Green functions and conditional Gauge Theorem for a - dimensional domain, in: \Seminar on Stochastic Processes 987", Birkhauser, Boston-Basel-Berlin 988, 83{94. Reference added in proof: [30] Peetre J. and Rus I.A., Sur la positive de la fonction de Green, Math. Scand., (967), 80{89.