ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE

Transcription

1 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE S BERTONE, A CELLINA, AND E M MARCHINI Abstract In this paper we consider Hopf s Lemma the Strong Maximum Principle for supersolutions to N g i u x i u xi x i = 0 under suitable hypotheses that allow g i to assume value zero at zero 1 Introduction Let Ω R N be a connected, open bounded set, on Ω, consider the operator N 1 F u = g i u x i u xix i, where g i : [0, + [0, + are continuous functions Ω is called regular if, for every z Ω, there exists a tangent plane, continuously depending on z, Ω satisfies the interior ball condition at z if there exists an open ball B Ω with z B When F is elliptic, two classical results hold Hopf s Lemma: Let Ω be regular, let u be such that F u 0 on Ω Suppose that there exists z Ω such that uz < ux, for all x in Ω If, in addition, Ω satisfies the interior ball condition at z, we have u z < 0, ν where ν is the outer unit normal to B at z The Strong Maximum Principle: Let u be such that F u 0 on Ω, then if u it attains minimum in Ω, it is a constant In 197 Hopf proved the Strong Maximum Principle in the case of second order elliptic partial differential equations, by applying a comparison technique, see [11] For the class of quasilinear elliptic problems, many contributions have been given, 1991 Mathematics Subject Classification 35B05 Key words phrases Strong maximum principle The present version of the paper owes much to the precise kind remarks of two anonymous referees 1

2 S BERTONE, A CELLINA, AND E M MARCHINI to extend the validity of the previous results, as in [1, 5, 6, 8, 9, 10, 1, 13, 14, 15, 16, 18] In the case in equation 1 we have g i 1, for every i, then F u = u, we find the classical problem of the Laplacian, see [7, 9] On the other h, when there exists i {1,, N} such that g i 0 on an interval I = [0, T ] R, the Strong Maximum Principle does not hold Indeed, in this case, it is always possible to define a function u assuming minimum in Ω such that N g iu x i u xix i = 0 For instance, let g N t = 0, for every t [0, ] The function ux 1,, x N = { x N 1 4 if 1 x N 1 0 otherwise satisfies 1 in R N We are interested in the case when 0 g i t 1 it does not exist i such that g i 0 on an interval Since g i could assume value zero, the equation 1 is degenerate elliptic Our interest in the problem comes, partially, from the calculus of variations We can consider the equation N g i u x i u xix i = 0 as the Euler-Lagrange equation associated to the functional N 1 Ju = L uxdx = f i u x i u x i dx, Ω where L u is strictly convex in {u x1,, u xn : u x i t, for every i = 1,, N} Indeed, fix i Let f i be a solution to the differential equation Ω g i t = f i t + 5tf it + t f i t, for t [0, t ] Since we have that L u x i u x i = f i u x i + 5u x i f iu x i + u 4 x i f i u x i = g i u x i, div u L u = N L u x i u xix i = N g i u x i u xix i Since J is convex, solutions to the Euler-Lagrange equation are minima among those functions satisfying the same boundary conditions From our assumptions, it follows that v 0 is a solution The Strong Maximum Principle can be seen as a strong uniqueness result among solutions to a variational problem: two solutions u v are given, not crossing each other: if the two solutions meet at one point, they are identically equal For a functional J possessing rotational symmetry, simple necessary sufficient condition for the validity of the Strong Maximum Principle are presented in [4] The purpose of the present paper is to investigate the same problem in the class of functional that are convex, but non possessing rotational symmetry,

3 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 3 The results of [4] for the validity of the Strong Maximum Principle, for equations possibly non elliptic but arising from a functional having rotational symmetry, show that this validity depends only on the behaviour of the functions g i near zero In this paper, we prove, in section 3, a sufficient condition for the validity of the Hopf s Lemma of the Strong Maximum Principle; a remarkable feature of this condition is that it concerns only the behaviour of the function g i t that goes fastest to zero, as t goes to zero Hopf s lemma the Strong Maximum Principle are essentially the same result as long as we can build subsolutions whose level lines can have arbitrarily large curvature This need not be always possible for problems not possessing rotational symmetry This difficulty will be evident in sections 4 5 In these sections, a more restricted class of equations is considered, namely when all the functions g i, for i = 1,, N 1, are 1 only g N is allowed to go to zero In this simpler class of equations we are able to show that the condition g N t 3/ lim t 0 + tg N t > 0 is at once necessary for the validity of Hopf s Lemma sufficient for the validity of the Strong Maximum Principle Preliminary results We impose the following local assumptions Assumptions L: There exists t > 0 such that: i on [0, t ], for every i = 1,, N 1, 0 g N t g i t 1; ii g N is continuous on [0, t ]; positive differentiable on 0, t ]; iii on 0, t ], the function t g N t + g N tt is non decreasing Notice that, in case ii above is violated, the Strong Maximum Principle does not hold; that condition iii above includes the case of the Laplacian, g i t 1;, finally, that under these assumptions, g i could assume value zero at most for t = 0 Moreover, the strict convexity of L u in {u x1,, u xn : u x i t, for every i = 1,, N} follows by the fact that g i is positive in 0, t ] Since we will need general comparison theorems that depend on the global properties of the solutions, ie on their belonging to a Sobolev space, we will need also a growth assumption on g i assumption G to insure these properties of the solutions Assumption G: Each function f i as defined in, is bounded away from zero f i u x i u x i strictly convex is

4 4 S BERTONE, A CELLINA, AND E M MARCHINI Any function g i satisfying assumptions L on [0, t ] can be modified so as to satisfy assumption G on [0, + In fact, taking f i such that f i t = g i t, for t > t, it is enough to modify g i to t, + by setting g i t = g i t, for t > t Definition 1 Let Ω be open, let u W 1, Ω The map u is a weak solution to the equation F u = 0 if, for every η C0 Ω, L ux, ηx dx = 0 Ω u is a weak subsolution F u 0 if, for every η C0 Ω, η 0, L ux, ηx dx 0 Ω u is a weak supersolution F u 0 if, for every η C0 Ω, η 0, L ux, ηx dx 0 Ω We say that a function w W 1, Ω is such that w Ω 0 if w + W 1, 0 Ω The growth assumption G assures that, if u W 1, Ω, then L ux L Ω The strict convexity of L implies the following comparison lemma Lemma 1 Let Ω be an open bounded set, let v W 1, Ω be a subsolution let u W 1, Ω be a supersolution to the equation F u = 0 If v Ω u Ω, then v u ae in Ω The following technical lemmas will be used later Lemma Let n =,, N set h n a = g N t1 a n 1 1 a + g N taa For every 0 < t t t defined in assumptions L, h n a h n 1/n, for every a in [0, 1] Proof Since, on 0, t ], the function t g N t + g N tt is non decreasing, we have that h n a = g N t1 a g N n 1 t1 a t1 a n 1 n 1 + g Nta + g N tata 0 if only if a 1/n, so that h n a h n 1/n, for every a [0, 1] Lemma 3 For every 0 < t t t defined in assumptions L, we have that N xi g N t xi Proof We prove the claim by induction on N t g N N

5 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 5 Let N = Set a = sin θ 1 Applying Lemma we obtain that x1 x x1 x g N t + g N t = Nt1 a1 a + g N taa g N t Suppose that the claim is true for N 1, ie N 1 xi xi g N t Let us prove it for N Set g N t N 1 y 1 = cos θ N cos θ cos θ 1 y = cos θ N cos θ sin θ 1 y N 1 = sin θ N set a = sin θ N 1 Applying Lemma we obtain that N xi xi g N t = N 1 g N t yi 1 a 1 a + g N taa t 1 a + g N taa g N, N yi g N t1 a N 1 the claim is proved 3 A sufficient condition for the validity of Hopf s Lemma of the Strong Maximum Principle Consider the improper Riemann integral ξ as an extended valued function G, 0 g N ζ /N dζ = lim ζ ξ 0 Gξ = ξ 0 ξ ξ g N ζ /N dζ, ζ g N ζ /N dζ ζ where we mean that Gξ + whenever the integral diverges 3 Given a continuous function u, a direction ν a point z, we denote + u z = lim sup ν h 0 We wish to prove the following lemma uz + hν uz h

6 6 S BERTONE, A CELLINA, AND E M MARCHINI Lemma 4 Hopf s Lemma Let Ω R N be a connected, open bounded set Let u W 1, Ω C Ω be a weak solution to N g i u x i u xix i 0 In addition to the assumptions L G on g i, assume that Gξ + Suppose that there exists z Ω such that uz < ux, for all x in Ω that Ω satisfies the interior ball condition at z Then + u z < 0, ν where ν is the outer unit normal to B at z Note that the interior ball condition automatically holds if U is C As an example of an equation satisfying the assumptions of the theorem above, consider the Laplace equation u = 0 The functions g i 1 satisfy the assumptions L G, Gξ = Another example is obtained setting ξ 0 1 dζ = + ζ g N t = 1 lnt for 0 t 1/e The assumptions L G are satisfied; moreover, for 0 ξ /N 1/e, Gξ = ξ 0 dζ ζ lnζ /N = + Remark 1 In this paper we will consider the operator in polar coordinates Set F v = N g i vx i v xix i x 1 = cos θ N 1 cos θ cos θ 1 x = cos θ N 1 cos θ sin θ 1 x N = sin θ N 1 so that v xi = v x i [ xi v xix i = v + v 1 ] xi

7 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 7 When v is a radial function, F reduces to N [ F v = g i v xi xi v + v N v g i v xi xi + v 1 N g i v ] xi = xi 1 xi In general, we do not expect that the equation F v = 0 admits radial solutions However we will use the expression of F valid for radial functions in order to reach our results Proof of Lemma 4 a We assume that uz = 0 that B = BO, r Let { } ɛ = min ux : x BO, r/ > 0 ω = BO, r \ BO, r/ b We seek a radial function v W 1, ω Cω satisfying v is a weak solution to F v 0 in ω v > 0 in ω 4 v = 0 in BO, r v ɛ in BO, r/ v z < 0 Consider the Cauchy problem ζ = N 1 ζ g N ζ /N 5 ζr/ = ɛ r There exists a unique local solution ζ of 5, such that ζ ɛ r g N ζ /N dζ = ζ r/ N 1 ds = N 1 ln s r We claim that ζ is defined in [r/, + Indeed, suppose that ζ is defined in [r/, τ, with τ < + Since ζ > 0, ζ is an increasing function, so that τ < + if only if lim τ ζ = 0 But ζ g N ζ /N = lim dζ = lim N 1 ln, τ ɛ ζ τ r r a contradiction Hence, the solution ζ of 5 is defined in [r/, + Setting v = ζ, since, for every r/, r, we have that the function solves the problem ɛ r < v < 0, v = v g N v N r v sds + v N 1 = 0,

8 8 S BERTONE, A CELLINA, AND E M MARCHINI in particular, for every r/, r, v > 0 v η < 0; moreover vr = 0 vr/ ɛ Since v 0 t v 0, for every r/, r, by the hypotheses on g i by Lemma 3, we have that N xi F v = v g i v xi + v N g i v xi xi 1 N v v N g N v g N v v g N v N xi xi xi xi + v N 1 + v N 1 + v N 1 xi = The function v solves 4, indeed, v is in C ω it is such that F v 0 v > 0 in ω, vr = 0, vr/ ɛ v z < 0 c Since u, v W 1, ω Cω, v is a weak subsolution u is a weak solution to F u = 0, v ω u ω, applying Lemma 1, we obtain that v u in ω We have vz = uz v z = v z η ν If + u z > η, ν where + u ν has been defined in 3, then there exists h < 0 such that vz + hν > uz + hν, in contradiction with the fact that v u in ω From Hopf s Lemma we derive: Theorem 1 Strong Maximum Principle Let Ω R N be a connected, open bounded set Let u W 1, Ω C Ω be a weak supersolution to N g i u x i u xix i = 0 In addition to the assumptions L G on g i, assume that Gξ + Then, if u attains its minimum in Ω, it is a constant Proof a Assume min Ω u = 0 set C = {x Ω : ux = 0} By contradiction, suppose that the open set Ω \ C b Since Ω is a connected set, there exist s C R > 0 such that Bs, R Ω Bs, R Ω \ C Let p Bs, R Ω \ C Consider the line ps Moving p along this line, we can assume that Bp, dp, C Ω \ C that there exists a point z C such that dp, C = dp, z Set r = dp, C Wlog suppose that

9 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 9 p = O c The set Ω \ C satisfies the interior ball condition at z, hence Hopf s Lemma implies + u z < 0 ν But this is a contradiction: since u attains minimum at z Ω, we have that Duz = 0 4 A necessary condition for the validity of Hopf s Lemma In this the following section we consider the operator 6 F u = N 1 u xix i + gu x N u xn x N, We wish to provide a necessary condition for the validity of Hopf s Lemma in a class of degenerate elliptic equations Consider the case ξ gζ /N Gξ = dζ < + ζ 0 Theorem Let g, satisfying assumptions L G, be such that, on 0, t ], t defined in assumptions L, g t > 0, gt + g tt 1, gt 3/ lim t 0 + tg = 0 t Then there exist: an open regular region Ω R N ; a radial function u C Ω, such that F u 0 in Ω, a point z Ω such that uz = 0, uz < ux for every x Ω u ν z = 0 where ν is the outer unit normal to Ω at z 7 If, in addition, we assume that gt tg t is bounded in 0, t ] then Ω satisfies the interior ball condition at z Remark When lim t 0 + g tt exists, it follows that exists, that gt 3/ lim t 0 + tg t gt + g tt 1, on 0, t ]

10 10 S BERTONE, A CELLINA, AND E M MARCHINI Indeed, we have that lim gt + t 0 g tt = 0 + From ξ gt ξ gζ /N dt = dζ < +, 0 t 0 ζ it follows that lim t 0 + gt = 0 If lim t 0 + g tt > 0, there exists K > 0 such that, when 0 < t t, g tt K, so that gtt = gt K, a contradiction t 0 gs + g ss ds Kt, The map 1 gt = lnt k, with k >, for 0 t 1/e, satisfies the assumption gt 3/ lim t 0 + tg = 0 t The following lemma is instrumental to the proofs of the main results Lemma 5 Let g satisfies assumptions L G Suppose that for every 0 < t t, g t 0 gt + g tt 1 Set k 1 a = 1 a + agta k a = a 1 agta For every 0 < t t t defined in assumptions L, k 1 k are non increasing in [0, 1] Proof Since, for every 0 < t t, we have that, for every 0 a 1, g t 0 gt + g tt 1, k 1 a = 1 + gta + g tata 0 k a = 1 + gta 1 ag tat = 1 + gta + g tata g tat 0 Proof of Theorem a Let v be a radial function Setting a = sin θ N 1, 6 reduces to F v = v 1 a + agv a + v N + a + 1 agv a Let a = 1, we seek a solution to 8 v gv + N 1 v = 0

11 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 11 a radius R1 > 0 such that v R1 + 1 = 0 v < 0, for every [, R1 + 1 Consider the Cauchy problem ζ = N 1 ζ gζ 9 ζ = 1 We are interested in a negative solution ζ Define R1 to be the unique positive real solution to R1 + 1 G 1 N 1 ln = 0, ie R1 = e G 1 N 1 1 The solution ζ of 9, satisfies Gζ G 1 = ζ ζ gt dt = t N 1 ds = N 1 ln s Then, for every, R1 + 1, Gζ > 0 ζ < 0, while ζr1 + 1 = 0 Setting v = ζ v = R1+1 v sds, we obtain that v solves 8, for every, R1 + 1, v < v R1 + 1 = 0 v > vr1 + 1 = 0 b Set, for 1, R1], u = v + 1 Since, for the function v, we have at + 1 we obtain v gv + N 1 v = 0, 10 u gu + N 1 u + 1 = 0 This equality yields, for 1, R1, u gu + N 1 u = N 1u < 0 c Let 1/ < a < 1 We wish to find Ra R1 such that u is a solution to F u = u 1 a + agu a + u N + a + 1 agu a 0, for 1, Ra Since F u = u + 1gu [ N 1 1 a + agu a setting + 1gu N + a + 1 agu a ], k = N 1 1 a + agu a +1gu N + a + 1 agu a,

12 1 S BERTONE, A CELLINA, AND E M MARCHINI we obtain that F u 0 if only if k 0 We have k = N 1 [ 1 a + agu a ] gu [ N + a + 1 agu a ] + N 1a d d gu a + 1 d d gu [ N + a + 1 agu a ] + 1gu 1 a d d gu a = N 1 [ 1 a + agu a ] gu [ N + a + 1 agu a ] d d gu [ N + a + 1 agu a ] gu 1 a d d gu a+ [ N 1a d d gu a d d gu [ N + a + 1 agu a ] gu 1 a d ] d gu a Assume N Applying Lemma 5, we obtain from 11 we have N 1 [ 1 a + agu a ] gu [ N + a + 1 agu a ] N 1gu gu N 1 = 0; d d gu d d gu a 0, d d gu [ N + a + 1 agu a ] gu 1 a d d gu a 0; since N + a N 1a, from 11, we deduce [ N 1a d d gu a d d gu N + a + 1 agu a gu 1 a d ] d gu a [ N 1a d d gu a d ] d gu N + a d N 1a d gu a d d gu 0 The previous inequalities imply that k 0 It follows that F u 0, for every 1, Ra, if only if kra 0

13 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 13 We have that kra = RaN 1 1 a + agu Ra a Ra + 1gu Ra N + a + 1 agu Ra a N 1 [ Ra1 a gu Ra a ] cn 1 [ R11 a gu Ra a ] [ N 1 R11 a gu Ra ] We define Ra to be a solution to 1 R11 a gu Ra = 0 d In order to solve 1 for the unknown Ra, recalling that 1 a = cos θ N 1 = c, let gu r hr = R1 The function h is decreasing, differentiable with inverse differentiable We have that c = hr1 c, so that R1 c = h 1 c, R1 c is increasing in c lim R1 c 0 c = R1 Let 0 < c < 1/ be such that R1 c 1, so that, for c c, we have R1 c R1 c 1 We have obtained that, for every 1 a 1 c, there exists Ra such that 1 holds It follows that kra [ R11 a gu Ra ] so that the function u solves F u 0 for every 1, Ra = 0, e Set Ω = {x 1,, x N R N : 1, R1 c c < c } Ω R N is a connected, open bounded set u W 1, Ω C Ω is a weak solution to F u 0 The point z = R, 0,, 0 Ω is such that uz < ux, for all x Ω We wish to show that Ω is regular in a neighborhood of z = R, 0,, 0 Since d dc R1 c exists, in 0, c, to prove our claim it is sufficient to show that Recalling 10, we have that 13 Since lim c 0 d dc R1 c = 0 d dc R1 c = h 1 c 1 = h R1 c = R1 N 1 R1 c + 1 gu R1 c 3/ u R1 c g u R1 c lim u R1 c = 0 c 0

14 14 S BERTONE, A CELLINA, AND E M MARCHINI it follows that Since d dθ cos θ N 1 θn 1= π regular gt 3/ lim t 0 + tg = 0, t lim c 0 d dc R1 c = 0 = 1, this shows that d dθ R1 cos θ N 1 = 0, Ω is f To prove the validity of the interior ball condition at z = R, 0,, 0, it is enough to show that the second derivative of R1 c is bounded at c = 0, ie that 1 d c dc R1 c is bounded Set from 13 we obtain d R1 dc R1 c = N 1 from 10 tc = u R1 c, R1 c + 1 gtc 3/ tcg tc, dtc dc = u R1 c u R1 c d dc R1 c = From gt0 = 0, we obtain that R1 gtc1/ g tc d dc gtc1/ = g tc dtc = R1 gtc 1/ dc gtc 1/ = R1c gtc 3/ ctcg tc = gtc R1 tcg tc From condition 7 we obtain 1 d c dc R1 c M

15 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 15 O z R Ω R1 c 1 Figure 1 Ω in the case N = 5 A sufficient condition for the validity of the Strong Maximum Principle Consider the case ξ gζ /N Gξ = dζ < + 0 ζ We wish to prove the following theorem Theorem 3 The Strong Maximum Principle Let Ω R N be a connected, open bounded set Let u W 1, Ω C Ω be a weak supersolution to N 1 u xix i + gu x N u xnx N = 0, where g satisfies assumptions L G, on 0, t ] t defined in assumptions L, g t 0 gt + g tt 1 Moreover, suppose that there exists K > 0 such that, for every 0 < ξ /N t, we have 14 gξ /N K e Gξ 1

16 16 S BERTONE, A CELLINA, AND E M MARCHINI Then, if u attains its minimum in Ω, it is a constant Remark 3 When the function g satisfies the condition for every 0 < t t, then it satisfies for every 0 < ξ /N t g tt K gt 3/, gt + g tt 1 gξ /N K e Gξ 1, Indeed, since Gξ < +, we have that lim t 0 gt = 0 Hence, we can suppose that gt 1/K + 1, for 0 < t t, so that gt + g tt gt + K gt 3/ 1 Moreover, since g0 = 0, G0 = 0 gξ /N gξ /N K K e Gξ 1, ξ we obtain that gξ /N K e Gξ 1 Remark 4 Among the functions g such that gt 3/ lim t 0 + tg t exists, there exists K > 0 such that, for every 0 < ξ /N t, we have gξ /N K e Gξ 1 if only if gt 3/ lim t 0 + tg > 0 t An example of a map satisfying the assumptions of the theorem above, is given by 1 gt = lnt, for 0 t 1/e 4 For 0 ξ /N 1/e 4, we have Set Gξ = ξ 0 1 ζlnζ /N dζ = 1 lnξ /N, 1 gξ /N = lnξ /N 1 e lnξ /N 1 = e Gξ 1 Rλ, λ N = {x 1,, x N : x i λ, for i = 1,, N 1, x N λ N } Differently from the proof of Lemma 4 Theorem 1, we will build a subsolution that is not radially symmetric This construction is provided by next theorem

17 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 17 Theorem 4 Under the same assumptions on g as on Theorem 3, for every r > 0 every ɛ, there exist: l, l N ; an open convex region A Rl, l N ; a function v W 1, ω C 1 ω Cω, where ω = BA, r \ A, such that i 0 l Kr, 0 l N r/4; ii v is a weak solution to F v 0 in ω v > 0 in ω 15 v = 0 in BA, r v ɛ in A To give a sketch of the proof we consider the construction in R Using polar coordinates, θ setting a = sin θ, the operator F, when computed on a radial function w, reduces to F w = w 1 a + agw a + w a + 1 agw a We will use this expression of F, valid only for radial functions, to build a non-radial solution of F u 0 We set a = 1 w a solution to F w = 0 on [R1, R1 + r, where R1 is such that wr1 ɛ wr1 + r = 0 The radial function w would solve F w 0, for 0 a 1, but the domain we would obtain is too large for our purposes Fix a 1 < 1 find the smallest Ra 1 > 0 such that, setting w a1 = w Ra 1 + R1, the function w a1 is a solution to F w a1 0, for every a < a 1 We also set D 0 = {x 1, x : R1 < < R1 + r, a 1 sin θ 1} on D 0 we define v 0 x 1, x = wx 1, x This function v 0 is a restriction of the function v we seek Fix a < a 1 set D 1 = {x 1, x : Ra 1 < 1 < Ra 1 + r, a sin θ a }, where 1 = dx 1, x, O 1 O 1 = R1 Ra 1 1 a, a, see figure, on D 1, set v 1 x 1, x = w a1 1 x 1, x The formula { v0 x vx 1, x = 1, x on D 0 v 1 x 1, x on D 1 defines a function v that it is a weak solution to F v 0 on intd 0 D 1 Iterating this procedure, we construct a function v on a domain ω, as in figure, solving 15 Condition 14 implies that the domain ω is sufficiently small Proof of Theorem 4 Fix r; we can assume that ɛ is such that 0 < ɛ /r t that ɛ g r + 1 ɛ ɛ g r ln g r 1 4K

18 18 S BERTONE, A CELLINA, AND E M MARCHINI Fix the origin O 0 = 0,, 0, set polar coordinates as x 1 = cos θ N 1 cos θ cos θ 1 x = cos θ N 1 cos θ sin θ 1 x N = sin θ N 1 1 When w is a radial function, setting a = sin θ 1, F reduces to F w = w 1 a + agw a + w N + a + 1 agw a For a = 1, we seek a solution to 16 N 1 w + w gw = 0 such that w R1 = ɛ/r w < 0, for every [R1, R1 + r Consider the Cauchy problem ζ = N 1 ζ gζ 17 ζr1 = ɛ r We are interested in a negative solution ζ Define R1 to be the unique positive real solution to R1 + r G ɛ/r N 1 ln = 0, R1 ie r R1 = e G ɛ/r N 1 1 Consider the unique solution ζ of 17, such that ζr1 = ɛ/r, ie, such that ζ gt Gζ G ɛ/r = dt = N 1 ds = N 1 ln t s R1 ζr1 Then, for every [R1, R1+r, Gζ > 0 ζ < 0, while ζr1+r = 0 Setting w = ζ w = R1 R1+r w sds, we obtain that w solves 16, for every R1, R1 + r, ɛ/r = w R1 < w < w R1 + r = 0 0 = wr1 + r < w < wr1 ɛ Applying Lemma 5, we infer that the function w defined in 1 is actually a solution to F w = w 1 a + w N + a + gw a w a + w 1 a 0, for every 0 a 1 every R1, R1 + r

19 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 19 3 Let ā < 1 We wish to find the smallest Rā > 0 such that, setting wā = w Rā + R1, the function wā is a solution to 18 F wā = wā1 ā + wā N + ā + gwā ā wāā + wā 1 ā for every Rā, Rā + r Since, for the function w, we have at Rā + R1 we obtain This equality yields N 1 w wā + gw w = 0, N 1 Rā + R1 + gwā wā = 0 N + ā wā + āgwā āwā = wā N + ā Rā + R1 ān 1 gwā ā Rā + R1 gwā 1 ā wā + wā gwā ā = 1 āwā Rā + R1 Rā + R1gwā ā N 1 gwā Since F wā = N + ā wā + āgwā āwā + 1 ā wā + wā gwā, we obtain that F wā 0 if only if Rā + R1 N + ā + 1 āgwā ā 0, Set Since k = N 1 1 ā + āgw ā gwā ā 0 Rā + R1 N + ā + 1 āgwā ā N 1 1 ā + āgw ā gwā ā d d gwā d d gwā ā 0,

20 0 S BERTONE, A CELLINA, AND E M MARCHINI applying Lemma 5, we have that k = N + ā + 1 āgwā ā + Rā + R11 ā d d gwā ā N 1 1 gwā gwā d N 1 gwā d gwā āā N 1 gwā N 1 gwā d d gwā 1 ā + āgwā ā d 1 d gwā ā + āgw ā ā N 1 gwā d d gwā 1 ā + āgwā ā āgwā N 1 gwā 1 āgwā d d gwā 0 d d gwā āā Since the function k is non increasing, it follows that F wā 0, for every Rā, Rā + r, if only if R1 N + ā + 1 āgwārā ā N 1Rā 1 ā + āgw ā gwā Rā Rā ā = ɛ R1 N + ā + 1 āg r ā N 1Rā ɛ g 1 ā + āg ɛ r ā 0, r ie if only if Hence, we define ɛ Rā g r ɛ Rā = g r R1 N + ā + g N 1 ɛ r ā 1 ā + g ɛ r ā ā R1 N + ā + g N 1 ɛ r ā 1 ā + g ɛ r ā ā 1 ā 1 ā 4 The function wā defined in point 3 is a solution to F wā = wā1 a + wā N + a + gwā a wāa + wā 1 a 0,

21 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 1 for every a < ā Rā, Rā + r, Indeed, applying Lemma 5 we obtain that, for every Rā + R1 N + a + 1 agwā a N 1 1 a + agw ā gwā a Rā + R1 N + ā + 1 āgwā ā N 1 1 ā + āgw ā gwā ā 0 5 Assume we have a partition α of [0, π/], α = {0 = α n < < α 1 < α 0 = π/} This partition defines two partitions of [0, 1], given by c i = cos α i s i = sin α i Consider the sums S 1 α = n S α = n 1 R1 c i 1 R1 c i c i = R1c 1 + R1 c i c i+1 c i n n 1 Rs i 1s i 1 s i = R11 s 1 + Rs i s i s i+1, where, in the previous equalities, we have taken into account that R1 c n = R0 = 0 Our purpose is to provide a partition α corresponding estimates for S 1 α S α that are independent of ɛ The sums n 1 R1 c i c i+1 c i are Riemann sums for the integrals 1 c 1 R1 c dc Consider the first integral From we obtain that 1 c 1 R1 c dc = N 1 c + g n 1 Rs i s i s i+1 s1 0 ɛ r 1 c c c + g ɛ r 1 c 1 c ɛ g r R1 1 N 1 c 1 ɛ 1 dc R1g r c c 1 Rs ds N 1 c N 1 c + g ɛ r 1 c c + g ɛ r 1 c 1 c c dc

22 S BERTONE, A CELLINA, AND E M MARCHINI Set ɛ Sx 1 c = R1c + R1g db r b = R1 c + g ɛ c R1g r g + 1 ɛ c 1 r Evaluating the last term at the minimum point c = S x ɛ ɛ g = R1g r c r ɛ r 1 c 1 = g ɛ r, we obtain g ɛ r 1 r g ɛ r e Gɛ/r 1 R1g We fix c 1 = g ɛ r, so that α1 = arccos g ɛ r Consider the second integral From we obtain that Set s1 0 S ys = Since N + s + g Rs ds = s1 0 ɛ r s 1 s 1 s + g ɛ r s ɛ g r ɛ g r ɛ R11 s + g Rs ds ɛ r N + s + g 1 s R1 s1 N + s + g N 1 0 R1 s1 N + s + g N 1 0 r ɛ R11 s + g r = ɛ r s 1 s ɛ 1 s + g ɛ r s s R1 s N + b + g N 1 0 ɛ R1 N 1 [ g r ɛ ɛ 1 1 g g r r s 1 s ds ɛ r 1 s 1 s ds ɛ r 1 b 1 b db = 1 s + N 1 ln r, ] 1 + s 1 s

23 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 3 evaluating the last term at the point s 1 = sin α 1 = 1 g ɛ r, we obtain ɛ ɛ 1 g = R1 1 1 g + S x ɛ g r R1 N 1 r ɛ g r 1 1 g r ɛ r 1 ln 1 1 g ɛ r g ɛ r ɛ ɛ R1 g r g r + 1 ɛ ɛ g r ln g r To define the other points of the required partition α, consider the integrals Set 1 c 1 R1 c dc ɛ σ = R1g r s1 0 Rs ds By the basic theorem of Riemann integration, taking a partition α with mesh size small enough, the value of the Riemann sums differs from n 1 R1 c i c i+1 c i 1 R1 c dc c 1 by less than σ In particular we obtain n 1 Rs i s i s i+1 s1 0 Rs ds n 1 S 1 α = R1c 1 + R1 c i c i+1 c i 1 R1c 1 + c 1 R1 c dc + σ r n 1 S α = R11 s 1 + Rs i s i s i+1 s1 R11 s 1 + Rs ds + σ 0 r g ɛ r e Gɛ/r N 1 1 ɛ g r + 1 g g ɛ r e Gɛ/r N 1 1 Kr ɛ ɛ ln g r 4 r r

24 4 S BERTONE, A CELLINA, AND E M MARCHINI 6 With respect to the coordinates fixed at the beginning of the proof, consider x i 0 Set D 0 = {x 1,, x N : R1 < < R1 + r, a 1 sin θ N 1 a 0 = 1} on D 0 define the function v 0 x 1,, x N = wx 1,, x N By point 1, the function v 0 is of class C intd 0 satisfies, pointwise, the inequality F v 0 0 Having defined v 0, define v 1 as follows Set: for N =, for N >, O 1 = O 1 x 1, O 1 x = R1 Ra 1 1 a 1, a 1, O 1 θ 1,, θ N = O 1 x 1,, O 1 x N = R1 Ra 1 1 a 1 cos θ N cos θ 1, 1 a 1 cos θ N sin θ 1,, a 1, for N, 1 x 1,, x N = x 1 O 1 x x N O 1 x N sin θ 1 N 1x 1,, x N = Recalling the definition of w a1 in 3, consider, on D 1, set x N O 1 x N 1 x 1,, x N D 1 = {x 1,, x N : Ra 1 < 1 x 1,, x N < Ra 1 + r, a sin θ 1 N 1x 1,, x N a 1 } v 1 x 1,, x N = w a1 1 x 1,, x N The function v 1 is of class C intd 1 We claim that v 1 still satisfies F v 1 0 Remark that the set of the points O 1 is equal to O 1 = {x 1,, x N : = R1 Ra 1, sin θ N 1 = a 1 } that for every point p D 1, the corresponding point O 1 p is the projection of p on O 1, while 1 p = dp, O 1 Then we obtain 1 θ i = 0 for every i = 1,, N, O 1 x i x i 0 for every i = 1,, N 1, O 1 x N x N = 0

25 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 5 v 1 = w a 1 1 x1 1 Ox 1 1,, x N Ox 1 N = v 1 xix i = w a1 1 w a1 1 cos θ 1 N 1 cos θ 1, cos θ 1 N 1 sin θ 1,, sin θ 1 N 1, xi O 1 x i 1 + wa1 1 xi Ox 1 1 i 1 1 O1 x i x i = Then F v 1 = w a1 1 cos θ N 1 sin θ i 1 + w a1 1 1 cos θ N 1 sin θ i 1 O1 x i 1 x i N 1 v 1 xix i + v 1 xn x N gv 1 x N = w a1 1 cos θ 1 N 1 + sin θ 1 N 1 g w a1 1 sin θ 1 N 1 + w a1 1 N + sin θn cos θn 1 1 g w a1 1 sin θn w a1 N 1 1 Ox 1 i 0 1 x i since w a1 1 verifies equation 18 The sets D 0 D 1 intersect on sin θ N 1 x 1,, x N = sin θ 1 N 1 x 1,, x N = a1 For a point x 1,, x N in this intersection we have Hence, on D 0 D 1 if only if 1 x 1,, x N = x 1,, x N R1 Ra 1 the functions v 0 v 1 coincide: Ra1 1x1,, xn Ra1 + r R1 x 1,, x N R1 + r, v 1 x 1,, x N = w a1 1 x 1,, x N = w 1 x 1,, x N + R1 Ra 1 = The formula wx 1,, x N = v 0 x 1,, x N { v0 x vx 1,, x N = 1,, x N on D 0 v 1 x 1,, x N on D 1 defines a function v in C 0 intd 0 D 1 We claim that it is also in C 1 intd 0 D 1

26 6 S BERTONE, A CELLINA, AND E M MARCHINI In fact, we have v 0 x 1,, x N = w x 1,, x N x 1,, x N, x 1,, x N v 1 x 1,, x N = wa1 1x 1,, x N x1 Ox 1 1 x 1,, x N 1,, x N Ox 1 N, on D 0 D 1, 1 x1 Ox 1 1 x 1,, x N 1,, x N Ox 1 1 N = x 1,, x N x 1,, x N On intd 0 intd 1, the function v is of class C satisfies, pointwise, the inequality F v 0 We claim that v is also in W 1, intd 0 D 1 that it is a weak solution to F v 0 on intd 0 D 1 In fact, for every η C0 intd 0 D 1, applying the divergence theorem separately to intd 0 to intd 1, we obtain [div v L vxηx + L vx, ηx ] dx = intd 0 D 1 intd 0 intd 1 intd 0 D 0 {sin θ N 1= a 1} D 1 {sin θ 1 N 1 = a 1} [div v L vxηx + L vx, ηx ] dx = ηx L vx, n 0 x dl + ηx L vx, n 1 x dl = intd 1 ηx L vx, n 0 x dl+ ηx L vx, n 1 x dl, where n 0, n 1 are the outer unit normal respectively to intd 0, intd 1 The last term equals zero, since v C 1 intd 0 D 1 n 0 = n 1 on D 0 {sin θ N 1 = a1 } = D 1 {sin θ 1 N 1 = a 1 } Hence, when η 0, we have that intd 0 D 1 L vx, ηx dx 0, as we wanted to show Assuming defined O n θ 1,, θ N a function v C 1 intd 0 D n, consider O n 1 θ 1,, θ N = Ox n 1 1,, Ox n 1 N = O n θ 1,, θ N + Ra n Ra n 1 1 an 1 cos θ N cos θ 1, 1 a n 1 cos θ N sin θ 1,, a n 1

27 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 7 Set n 1 x 1,, x N = x 1 O n 1 x x N O n 1 x N, sin θ n 1 N 1 x x N Ox n 1 1,, x N = N n 1 x 1,, x N D n 1 = {x 1,, x N : Ra n < n 1 x 1,, x N < Ra n 1 + r, 0 = a n sin θ n 1 N 1 x 1,, x N a n 1 } define on D n 1 the function v n 1 x 1,, x N = w an 1 n 1 x 1,, x N Set D = intd 0 D n 1, the same considerations as before imply that the function v 0 x 1,, x N on D 0 vx 1,, x N = v n 1 x 1,, x N on D n 1 is such that v W 1, D C 1 D CD it is a weak solution to F v 0 on D This completes the construction of v as a weak solution to F v 0 on D 0 D n 1 Set O = O x 1,, O x N = 0,, 0, R1 l N We have that D 0 D n 1 {x 1,, x N : 0 x i l + r for i = 1,, N 1, O x N x N O x N + l N + r } Define the full domain ω the solution by symmetry with respect to the point O Figure shows this construction in dimension N = for n 1 = Hence the solution will be in W 1, ω C 1 ω Cω a weak solution of F v 0 on ω 7 The previous construction yields a region A centered in O, a corresponding region ω a function v that solves 15 The change of coordinates ˆx 1 = x 1,, ˆx N 1 = x N 1, ˆx N = x N ON, centers A at the origin proves the theorem In order to prove Theorem 3, we need this further lemma Lemma 6 Consider the sets A RO, l, l N, where A, O, l, l N have been defined in Theorem 3 Then, for every p R, dp, A < l N Proof Set q = O + l,, l, l N We prove that dq, A < l N Set p i = O + 0,, 0, l, 0,, 0 let Π N 1 the hyperplane passing through p 1,, p N Since A is convex p i A, we obtain that See Figure 3 dq, A < dq, Π N 1 < l N

28 8 S BERTONE, A CELLINA, AND E M MARCHINI ω A O D 0 r D 1 O D α O 1 α 1 O Figure The sets ω A in the case N = n = 3 Rl, l N p 1 l q A l N O p Figure 3 A RO, l, l N in the case N =

29 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 9 Proof of Theorem 3 a Suppose that u attains its minimum in Ω, assume min Ω u = 0 set C = {x Ω : ux = 0} By contradiction, suppose that the open set Ω \ C b Since Ω is a connected set, there exist s C R > 0 such that Bs, R Ω Bs, R Ω \ C Let p Bs, R Ω \ C Consider the line ps Moving p along this line, we can assume that Bp, dp, C Ω \ C, that there exists a point z C such that dp, C = dp, z Set c Fix r: dp, C 0 < r < 3N 1K { ɛr = min uz : z B p, dp, C 4 r }, we have that ɛr > 0, we set ɛ = min{ɛr, r ξ} d For r ɛ as defined in c, consider: l, l N, A v as defined in Theorem 4 Without loss of generality, since the set A is symmetric with respect to both coordinate axis, we can suppose that pz belongs to the first quadrant, ie that, for every i = 1,, N, z i p i, where z = z 1,, z N p = p 1,, p N Define the point q on the segment pz such that dq, p = dp, z r q = q l,, l, l N, Rq, l, l N = q +Rl, l N, A = q +A v x+q = vx We first claim that Rq, l, l N B p, dp, C r 4 Let t Rq, l, l N, then t can be written as q 1 α 1 l,, q N 1 α N 1 l, q N dp,c α N l N, with 0 α i 1, for i = 1,, N Since r <, we have that 3N 1K N N N dt, p = q i α i l i p i = dq, p + 4α i li 4α i l i q i p i dp, C r See Figure 4 + N 4li dp, C r + 16N 1K r + dp, r 4 < C r 4 Set Since A Rq, l, l N, we have obtained that A B p, dp, C r, 4 so that u ɛ in A By Lemma 6, we have that dq, A < l N r 4, dz, A dz, q + dq, A < 3 4 r,

30 30 S BERTONE, A CELLINA, AND E M MARCHINI B p, dp, C B p, dp, C r B 4 p, dp, C r R l l q z p Figure 4 The set R = Rq, l, l in the case N = so that z ω = BA, r \ A e The function v satisfies v is a weak solution to F v 0 in ω v > 0 in ω v = 0 in BA, r v ɛ in A Since u, v W 1, ω Cω, v is a weak subsolution u is a weak solution to F u = 0, v ω u ω, applying Lemma 1, we obtain that u v in ω But uz = 0 < v z, a contradiction References [1] H Berestycki, L Nirenberg, SRS Varadhan, The principal eigenvalue maximum principle for second-order elliptic operators in general domains, Communications on Pure Applied Mathematics , 1, 47 9 [] H Brezis, Analyse fonctionnelle, théorie et applications, Masson, Paris, 1983 [3] H Brezis, A Ponce, Remarks on the strong maximum principle, Differential Integral Equations , 1, 1 1 [4] A Cellina On the Strong Maximum Principle, Proc Amer Math Soc ,, [5] L Damascelli, F Pacella, M Ramaswamy, A strong maximum principle for a class of non-positone singular elliptic problems, Nonlinear Differential Equations Appl ,,

31 ON HOPF S LEMMA AND THE STRONG MAXIMUM PRINCIPLE 31 [6] JI Diaz, Nonlinear partial differential equations free boundaries, Pitman Research Notes in Mathematics, Vol 106, 1985 [7] L C Evans, Partial Differential Equations, Graduate Studies in Mathematics, Volume 19, American Mathematical Society, Providence, Rhode Isl [8] P Felmer, A Quaas, On the strong maximum principle for quasilinear elliptic equations systems, Adv Differential Equations 7 00, 1, 5 46 [9] D Gilbarg, N S Trudinger, Elliptic Partial Differential Equations of Second Order, Springer-Verlag, Berlin Heidelberg New York Tokyo, 1983 [10] S Granlund, Strong maximum principle for a quasi-linear equation with applications, Ann Acad Sci Fenn Ser A VI Phys , 5 pp [11] E Hopf, Elementare Bemerkungen über die L?sungen partieller Differentialgleichungen zweiter Ordnung vom elliptischen Typus, Sitzungsberichte Preussiche Akademie Wissenschaften, Berlin, 197, [1] P Pucci, J Serrin, The strong maximum principle revisited, J Differential Equations , 1, 1 66 [13] P Pucci, J Serrin, A note on the strong maximum principle for elliptic differential inequalities, J Math Pures Appl , 1, [14] P Pucci, J Serrin, H Zou, A strong maximum principle a compact support principle for singular elliptic inequalities, J Math Pures Appl , 8, [15] J Serrin, On the strong maximum principle for quasilinear second order differential inequalities, J Functional Anal 1970, [16] J Serrin, Commentary on Hopf strong maximum principle, CS Morawetz, JB Serrin, YG Sinai Eds, Selected Works of Eberhard Hopf with commentaries, Amer Math Soc, Providence, RI, 00 [17] K Taira, A strong maximum principle for degenerate elliptic operators, Comm Partial Differential Equations , 11, [18] J L Vázquez, A strong maximum principle for some quasilinear elliptic equations, Appl Math Optim , 3, Dipartimento di Matematica e Applicazioni, Universitá degli Studi di Milano-Bicocca, Via R Cozzi 53, 015 Milano address: simonebertone@unimibit Dipartimento di Matematica e Applicazioni, Universitá degli Studi di Milano-Bicocca, Via R Cozzi 53, 015 Milano address: arrigocellina@unimibit Dipartimento di Matematica e Applicazioni, Universitá degli Studi di Milano-Bicocca, Via R Cozzi 53, 015 Milano address: elsamarchini@unimibit