Large solutions of semilinear equations with Hardy potential
aa r X i v : . [ m a t h . A P ] J un LARGE SOLUTIONS OF SEMILINEAR EQUATIONSWITH HARDY POTENTIAL
MOSHE MARCUS Introduction
Assume that Ω is a bounded Lipschitz domain in R N . Denote(1.1) L µ = ∆ + µ/δ , δ ( x ) = dist ( x, ∂ Ω) , µ ∈ R . (NE) − L µ u + f ( u ) = 0 , (NE0) − ∆ u + f ( u ) = 0 . Assumptions on f :(F1) f ∈ C [0 , , f (0) = 0 , f and f ′ positive on (0 , ∞ ) h ( t ) := f ( t ) /t → ∞ sa t → ∞ . (F2) ∃ a > u of (NE0),(1.2) h ( u ( x )) ≤ a δ ( x ) − ∀ x ∈ Ω , where h ( t ) := f ( t ) /t, ∀ t > f satisfies (F1) and is convex on R + then h is non-decreasing.The Keller – Osserman condition on f is a key condition in the studyof equations such as (NE0). The condition is:(KO) ψ ( t ) := Z ∞ t (2 F ( s ) − / < ∞ ∀ t > F ( s ) = R s f . Since (F2) implies that the set of positive solutionsof (NE0) is locally uniformly bounded, conditions (F1) and (F2) implythe Keller – Osserman condition (se [12] and [2]).The converse does nothold in general, but it is still valid under an additional condition on f [14, Lemma 5.1]: Date : June 5, 2020.
Lemma 1.1.
Assume that f satisfies (F1) and the Keller - Ossermancondition (KO). Let V R denote the large solution of (NE0) in B R (0) .Given a number c > , suppose that there exists a constant C > such that (1.3) ψ (2 s ) ≤ C h ( s ) − / whenever h ( s ) ≥ c . Then (1.4) h ( V R (0)) ≤ C R − where C = N (2 + C ) . The existence of c > f .Consequently C depends only on f , N and the choice of c .Condition (1.3) holds for a large family of functions f including f ( t ) = t p , p > f ( t ) = e t − f also depends onthe space variable were obtained by Ancona in an appendix to [1].Naturally in that case, in general, the constant C depends also on thedomain.Further, by [18, Theorem 6.1], (F2) implies that there exist positiveconstants a , a such that, for every positive solution u of (NE) in Ω,(F2’) h ( u ( x ) /a ) ≤ a δ ( x ) − ∀ x ∈ Ω . The constants a , a depend on the constant in (1.4) and the constantin the strong Harnack inequality associated with L µ .If Ω is a bounded C domain, f is convex and satisfies conditions(F1) and (KO) then there exists a unique large solution of (NE0), say U Ω f , and it satisfies,(1.5) U Ω f ( x ) φ ( δ ( x )) → x → ∂ Ω , where φ is the inverse of the function ψ defined in (KO). This resultwas first proved in [2] under a stronger assumptions on f . Under theconditions stated above, it was established in [3] (see Theorem 3 andinequality (34) there).The function φ is the unique solution of the problem,(1.6) φ ′′ = F ( φ ) in R + , lim t → φ ( t ) = ∞ . In fact, under the same assumptions on f , uniqueness of the largesolution (but not (1.5)) holds in much more general domains. In partic-ular, if Ω is a bounded Lipschitz domain, f is convex and satisfies (F1)and (KO) then there exists a unique large solution of (NE0) [21, Thm.1.4]. ARGE SOLUTIONS 3
In this note we discuss the existence and uniqueness of large solutionsof equation (NE) when µ ≥ f and on thedomain. We establish the following results.Denote by C H (Ω) the Hardy constant in Ω relative to the potential δ − i.e. c H (Ω) = inf ϕ ∈ C ∞ c (Ω) R Ω |∇ ϕ | R Ω δ − ϕ . Theorem 1.2.
Let Ω be a bounded Lipschitz domain. Assume that f satisfies (F1) and (F2) and that it is convex on R + . Then for every µ ≥ , there exists a large solution U of (NE) such that U > U Ω f . Theorem 1.3.
Let Ω be a bounded Lipschitz domain. Assume that f satisfies (F1) and (F2) and that it is convex on R + .Under these assumptions, if ≤ µ < / then (NE) has a uniquelarge solution. Theorem 1.4.
Let Ω be a bounded C domain. Assume that f is con-vex and satisfies conditions(F1), (F2). In addition assume that:For every a > there exist c > and t > such that (1.7) ah ( t ) ≤ h ( ct ) , t > t For every a ∈ (0 , there exist c ∈ (0 , and t > such that (1.8) h ( ct ) ≤ ah ( t ) , t > t . There exists
A > such that (1.9) h ( φ ) ≤ Aδ − . Then, for every µ > , equation (NE) has a unique large solution ¯ U µ . Moreover there exist positive constants c , c such that (1.10) c φ ≤ ¯ U µ ≤ c ˜ φ where ˜ φ := h − ( δ − ) . Remark 1.5.
The conditions of Theorem 1.4 are satisfied in particularby the functions f ( u ) = u p , p > f ( u ) = e u − f is a power or, more generally,(1.11) b ≤ f ( t ) /t p ≤ b ∀ t > t for some positive constants, b , b , t then,( ∃ C > φ ≤ Cφ.
Equations of the form (NE) have been studied intensively in thelast decade. Among the earliest were the works of Bandle, Moroz
MOSHE MARCUS and Reichel [4] and [5]. The first of these papers dealt with powernonlinearities. The second studied the case f ( u ) = e u and provedexistence and uniqueness of the large solution, in smooth domains,when 0 < µ < c H (Ω) (= the Hardy constant in Ω). Recall that, ingeneral c H (Ω) ≤ / C domains. Underthe same conditions, equation (NE) was also studied by Du and Wei [8]who proved that, in this case, (NE) has a unique large solution for every µ > µ < / f ( u ) = u p , p > − (∆ u + µ/ | x | ) u + u p = 0 , p > ∈ ∂ Ω) have been studied by Guerch andVeron [11], F.C. Cirstea [7], Du and Wei [9] a.o.2.
Proof of Theorem 1.2
Under the assumptions of the theorem, equation (NE0) possesses aunique large solution, say U Ω f , (Marcus and Veron [21, Thm. 1.4]).Since µ > U Ω f is a subsolution of (NE).Let { D n } be a smooth exhaustion of Ω and denote by u n the solutionof the boundary value problem − L µ u + f ( u ) = 0 in D n u = U Ω f on ∂D n . Then { u n } is a monotone increasing sequence. For every set E ⋐ Ωlet n E be a number such that E ⋐ D n for all n ≥ n E . Condition(F2’)implies that { u n : n ≥ n E } is uniformly bounded in E . Thereforethe sequence converges to a solution U of (NE). Since U > U Ω f it followsthat U is a large solution of (NE). (cid:3) Remark 2.1.
The above argument shows tat if u is any positive so-lution of (NE0) there exists a solution ˜ u of (NE) such that ˜ u > u .Moreover, assuming (F1) and (KO), if Ω is a bounded domain suchthat (NE0) has a large solution then (NE) has a large solution. ARGE SOLUTIONS 5 Proof of Theorem 1.3
We start with some notation. Denote, T ( r, ρ ) = { ξ = ( ξ , ξ ′ ) ∈ R × R N − : | ξ | < ρ, | ξ ′ | < r } . By assumption Ω is a bounded Lipschitz domain. Consequently thereexist positive numbers r , κ such that, for every y ∈ ∂ Ω, there exist: (i)a set of Euclidean coordinates ξ = ξ y centered at y with the positive ξ axis pointing in the direction of the inner normal n y and (ii) a Lipschitzfunction F y on R N − with Lipschitz constant ≤ κ such that F y (0) = 0,and(3.1) Q y ( r , ρ ) := Ω ∩ T y ( r , ρ )= { ξ = ( ξ , ξ ′ ) : F y ( ξ ′ ) < ξ < ρ , | ξ ′ | < r } , where ξ = ξ y , T y ( r , ρ ) = y + T ( r , ρ ) and ρ = 10 κr . Without lossof generality, we assume that κ > ξ y is called a standard set of coordinates at y and T y ( r, ρ ) with 0 < r ≤ r and ρ = cκr , 2 < c ≤
10 is called astandard cylinder at y .Suppose there exist two large solution u , u . We may assume that u ≤ u . Otherwise we replace u by the solution lying between thesubsolution max( u , u ) and the supersolution u + u . Here we use thefact that, as f is convex, f (0) = 0 and f is increasing, f ( a ) + f ( b ) ≤ f ( a + b ) ∀ a, b > . In what follows y is kept fixed and we drop the superscript in Q y .Further, for 0 < a ≤ aQ = Q ( ar , aρ ) , Γ ,a = ∂ ( aQ ) ∩ ∂ Ω , Γ ,a = ∂ ( aQ ) ∩ Ω . Part 1.
Construction of a subsolution w of (NE) in ω := Q suchthat: (i) w ∈ C (¯ ω ∩ Ω),(ii) w = 0 on ∂ω ∩ Ω,(iii) wu → ξ → Γ ,a uniformly for | ξ ′ | < ar , ∀ a ∈ (0 , ).Let ω n := { ξ ∈ ω : F ( ξ ′ ) + ρ n < ξ } . Let v n be the solution of the boundary value problem, MOSHE MARCUS − L µ v n = 0 in ω n ,v n = u − j n on ∂ω n , where j n is a non-negative, continuous function on ∂ω n such that j n +1 ≤ j n ≤ u and j n = ( u on Γ ′ n = ∂ω n ∩ { ξ = F ( ξ ′ ) + ρ n } , ∂ω n ∩ { ξ > F ( ξ ′ ) + ρ n } . In particular v n = 0 on Γ ′ n and v n ≤ u in ω n .Let G µ,n be the Green function of L µ in Q n := { ξ ∈ Q : F ( ξ ′ ) + ρ n < ξ } . and let ξ = ( ρ , L µ in ω n , for every a ∈ (0 , /
2) there exists a constant C a such that, v n ( ξ ) ≤ C a G µ,n ( ξ, ξ ) in ω n,a := { ξ ∈ aQ : F ( ξ ′ ) + ρ n < ξ } . Recall that, if Ω is Lipschitz and ρ is sufficiently small then, by [15],the local Hardy constant in the strip Ω ρ := { x ∈ Ω : δ ( x ) < ρ } is1 /
4, although the Hardy constant in Ω may be lower. Therefore, for µ < / L µ has a Green function G µ,Q in Q . Since G µ,n is dominatedby G µ,Q and G µ,Q ( ξ, ξ ) is bounded in ω it follows that there exists aconstant C ′ ( a ) such that,(1*) sup ω n,a v n ≤ C ′ a ∀ a ∈ (0 , / . As { v n } increases, the sequence converges to a solution v of L µ v = 0in ω and the convergence is uniform in aQ for every a ∈ (0 , / { v n } is bounded by u which is continuous in { ξ ∈ ¯ ω : ξ > ǫ } ∀ ǫ ∈ (0 , ρ/ . Therefore, for every ǫ as above, the sequence converges uniformly inthis set. Thus v is continuous in { ξ ∈ ¯ ω, ξ > ǫ } and(2*) v = u on ∂ω ∩ Ω . Put w = u − v . Then w is a subsolution of (NE) in ω :(3*) − L µ w + f ( w ) < − L µ u + f ( u ) = 0 . By (1*) v = u − w ≤ C ′ ( a ) in aQ. ARGE SOLUTIONS 7
By assumption, u → ∞ as ξ → ∂ω ∩ ∂ Ω. Therefore(4*) 1 − wu → ξ → Γ ,a , ∀ a ∈ (0 , / . In conclusion w has the properties stated in Part 1. Part 2.
Completion of proof.
Given β ∈ (0 , ρ /
10) denote w β ( ξ ) = w ( ξ + β, ξ ′ ) in ω β := { ξ ∈ ω : ξ < ρ − β } . By (3*), w β is a subsolution, − (∆ + µδ ) w β + f ( w β ) < ω β , because in ω β : δ ( ξ ) = dist ( ξ, ∂ Ω) < dist (( ξ + β, ξ ′ ) , ∂ Ω). In addition w β = 0 on ∂ω β ∩ Ω and w β is bounded on ∂ω β ∩ ∂ Ω while u → ∞ as ξ → ∂ω β ∩ ∂ Ω. Thus w β < u on ∂ω β . By comparison principle w β < u in ω β . Letting β → w ≤ u in ω . Therefore, by(4*), lim sup ξ → Γ ,a u /u ≤ , ∀ a ∈ (0 , / . But, by assumption, u ≤ u . Hence(3.2) u /u → ξ → Γ ,a ∀ a ∈ (0 , / . This holds for every point y ∈ ∂ Ω, with rate of convergence independentof y . Therefore, for every ǫ > s > u ≤ u ≤ (1 + ǫ ) u in Ω s . Our assumptions on f imply that (1 + ǫ ) u is a supersolution of (NE).Therefore, by comparison principle, u ≤ u ≤ (1 + ǫ ) u in Ω. Letting ǫ → u = u . (cid:3) Proof of Theorem 1.4
Throughout this section we assume conditions (F1) and (F2) andthe convexity of f . These conditions imply that h is non-decreasing.As mentioned before (F2) implies (KO). We also assume that Ω is abounded domain of class C and that µ >
0. Other assumptions willbe mentioned as needed.Denote:(4.1) Ω ρ := { x ∈ Ω : δ ( x ) < ρ } , MOSHE MARCUS (4.2) ˜ φ ( δ ) := h − ( δ − ) , δ > . By (1.9) and (1.7), there exists
C > h ( φ ) ≤ Ah ( ˜ φ ) ≤ h ( C ˜ φ ) hence φ ≤ C ˜ φ. The proof of the theorem is based on several lemmas
Lemma 4.1.
Assume (1.7) . (i) There exists a maximal solution U max of (NE). It satisfies (4.4) h ( U max /a ) ≤ a δ − = a h ( ˜ φ ) U max is a large solution and (4.5) 1 C φ < U Ω f < U max . (ii) Every positive solution u of (NE) satisfies (4.6) u ≤ ¯ A ˜ φ where ¯ A is a constant depending only on the numbers a , a in (4.4) .Proof. In view of (F2) - which implies (F2’) - the family of positivesolutions of (NE) is uniformly bounded in every compact subset of thedomain. Therefore there exists a maximal solution U max and, by (F2’)and (4.2), it satisfies (4.4). U Ω f is a subsolution of (NE). The smallest solution between U Ω f and U max is a large solution of (NE). Therefore, by (1.5), we obtain (4.5).Let u be an arbitrary positive solution of (NE). By (4.4) and (1.7)there exists a constant A ′ such that(4.7) h ( u/a ) ≤ a δ − = a h ( ˜ φ ) ≤ h ( A ′ ˜ φ ) . Since h is monotone this implies (4.6) with ¯ A = A ′ a . (cid:3) Next we prove uniqueness of the large solution for every µ ≥ underthe conditions stated in Theorem 1.4 . The proof is based on severallemmas. Lemma 4.2. (i) Let D be a C subdomain of Ω such that Γ := ∂D ∩∩ ∂ Ω is the closure of a non-empty, relatively open subset of ∂ Ω , say O . Let U be a positive supersolution U of (NE) in D . (Here δ ( x ) =dist ( x, Γ) .) If U → ∞ as x → E , for every E ⋐ O then, (4.8) µ ≤ lim sup x → E h ( U ) δ , for every compact E ⊆ O. In particular, every large solution satisfies the above inequality for x → ∂ Ω . ARGE SOLUTIONS 9 (ii) Using the notation of part (i): if U is a positive supersolution of(NE) in D then (4.9) µ − ≤ lim sup x → E h ( U ) δ , for every compact E ⊆ O. Proof. (i) By negation, suppose that there exists ǫ positive such that µ > h ( U ) δ + ǫ in D ∩ Ω ρ for some ρ >
0. Then, in this set,0 ≤ − ∆ U − µδ − U + h ( U ) U < − ∆ U − ǫδ − U. Thus U is ∆-superharmonic in D ∩ Ω ρ and consequently has a (classical)measure boundary trace on O . This contradicts the assumption.(ii) If (4.9) is false, there exists ǫ > ρ > h ( u ) δ < µ − − ǫ in D ∩ Ω ρ . Hence,0 ≤ − ∆ u − µδ − u + h ( u ) u < − ∆ u − ( 14 + ǫ ) δ − u in D ∩ Ω ρ . It is known [15, Thm. 5] that if γ > there is no local positivesupersolution of the equation 0 = − ∆ u − γδ − u . This contradicts theprevious inequality (cid:3) Lemma 4.3.
Suppose that Ω is radially symmetric: a ball, the exteriorof a ball or an annulus. Then there exists ¯ b ∈ (0 , such that everyradially symmetric large solution U of (NE) satisfies (4.10) ¯ bU Ω f ≤ U in Ω . Proof.
First consider the case when Ω is a ball, say B R (0), or the ex-terior of a ball B ′ R (0) = R N \ B R (0).By Lemma 4.2 there exists a constant c > δ → h ( U ) /h ( ˜ φ ) > c for every radially symmetric large solution U = U ( δ ( x )) where δ ( x ) = R −| x | in B R and δ ( x ) = | x |− R in B ′ R . By (1.8), there exists b ∈ (0 , δ >
0, depending on c , such that ch ( ˜ φ ) ≥ h ( b ˜ φ ) , < δ < δ . Therefore, by (4.11), there exists a sequence { δ n } converging to zerosuch that(4.12) h ( U ( δ n )) ≥ ch ( ˜ φ ( δ n )) ≥ h ( b ˜ φ ( δ n )) . Hence, by (1.5) and (4.3), there exists ¯ b > n ,(4.13) U ( δ n ) ≥ b ˜ φ ( δ n ) ≥ b C φ ( δ n ) ≥ ¯ b U Ω f ( δ n ) . As U Ω f is a subsolution we obtain (4.10).Now, let Ω be an annulus, R < | x | < R . Then, by Lemma 4.2,there exist sequences { δ n } where δ n = δ ( x n ) = | x n | − R → { δ ′ n } where δ ′ n = δ ( x ′ n ) = R − | x ′ n | → c > U ( δ n ) ≥ ¯ bU Ω f ( δ n ) , U ( δ ′ n ) ≥ ¯ bU Ω f ( δ ′ n ) . Hence, by the comparison principle, U ( x ) ≥ ¯ bU Ω f ( x ) , when δ n ≤ | x | ≤ δ ′ n , for all sufficiently large n . This imolies (4.10). (cid:3) Lemma 4.4.
Suppose that Ω is radially symmetric and let F rad denotethe family of large, radially symmetric solutions. Then F rad containsa maximal element U rad and a minimal element u rad .Proof. The existence of a r.s. large solution follows from the fact that U Ω f is a r.s. large subsolution. The existence of a maximal elementfollows from the fact that, by (F2’), sup F rad finite everywhere in Ωand is itself a solution (see e.g. [MVbook,p.79]). Obviously sup F rad is r.s.. The existence of a minimal element follows from Lemma 4.3.Indeed if U , U ∈ F rad then min( U , U ) ∈ F rad is a supersolutionlarger than the subsolution ¯ bU Ω f . Therefore there exists U ∈ F rad such that U < min( U , U ). By a well known result, this fact impliesthat inf F rad is a solution (see [MVbook]) which obviously belongs to F rad . (cid:3) Lemma 4.5.
Let Ω be a ball or the exterior of a ball. Then F rad is asigleton: the unique radially symmetric large solution.Proof. By negation suppose that U rad = u rad . By (4.6) U rad ≤ ¯ A ˜ φ. By (4.13), there exists b > δ n ↓ b ˜ φ ( δ n ) ≤ u rad ( δ n ) . Hence U rad ( δ n ) ≤ M u rad ( δ n ) where M = ¯ Ab . ARGE SOLUTIONS 11
The function
M u rad is a supersolution of (NE). Hence, by the compar-ison principle,(4.14) U rad ≤ M u rad in Ω . Adapting a trick introduced in [19] we show that this leads to acontradiction.Put w = u rad − M ( U rad − u rad ) . By (4.14), ˜ w := M + 12 M u rad < w < u rad
The convexity of f and the assumption f (0) = 0 imply that ˜ w is asubsolution of (NE). On the other hand, using these facts it is easilyverified that w is a supersolution of (NE). It follows that there existsa radially symmetric large solution strictly smaller than u rad , whichbrings us to a contradiction. (cid:3) Completion of proof of the theorem.
Choose
R > P ∈ ∂ Ω there exists a ball B R ( P ′ ) such that ¯ B R ( P ′ ) ∩ ¯Ω = P . Let u R be the unique radiallysymmetric large solution of (NE) in B ′ R (0) = R N \ B R (0). (Of course,here δ ( x ) = dist ( x, ∂B ′ R (0)).) Denote, u PR ( x ) := u R ( x − P ′ ) ∀ x ∈ B ′ R ( P ′ ) := R N \ B R ( P ′ )and u PR,ǫ ( x ) := u R ( x − P ′ − ǫ n P ). Since µ > u PR,ǫ is a subsolution of (NE) in Ω and it is dominatedby U on ∂ Ω. Therefore, u PR,ǫ ( x ) ≤ U ( x ) in Ω. Letting ǫ →
0, we obtaim(4.15) u PR ≤ U in Ω . The function U ∗ := sup P ∈ ∂ Ω u PR is a subsolution of (NE) in Ω and U ∗ ( x ) → ∞ as Ω ∋ x → ∂ Ω . By (4.15), U ∗ ≤ U for every large solution U of (NE) in Ω. Thereforethe smallest solution of (NE) dominating U ∗ in Ω is the minimal largesolution, denoted by u min . Let { δ n } and b be as in (4.13) with respect to R N \ B R (0). Then u min ( x ) ≥ U ∗ ( x ) ≥ b ˜ φ ( δ n ) ∀ x ∈ Ω : δ ( x ) = δ n . This inequality and (4.6) imply that(4.16) U max ≤ M u min , M := ¯ Ab . Consequently, by the same argument as in the proof of Lemma 4.5, weconclude that U max = u min . (cid:3) References [1] A. Ancona and M. Marcus,
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Department of Mathematics, Technion, Haifa 32000, ISRAEL
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