Semilinear Poisson problems in Sobolev-Besov spaces on Lipschitz domains
Abstract
Extending recent work for the linear Poisson problem for the Laplacian in the framework of Sobolev-Besov spaces on Lipschitz domains by Jerison and Kenig [16], Fabes, Mendez and Mitrea [9], and Mitrea and Taylor [30], here we take up the task of developing a similar sharp theory for semilinear problems of the type [Delta]u-N(x, u) = F(x), equipped with Dirichlet and Neumann boundary conditions.
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Publ. Mat. 46 (2002), 353–403 SEMILINEAR POISSON PROBLEMS IN SOBOLEV-BESOV SPACES ON LIPSCHITZ DOMAINS Martin Dindoˇ s∗and Marius Mitrea† Abstract Extending recent work for the linear Poisson problem for the Laplacian in the framework of Sobolev-Besov spaces on Lipschitz domains by Jerison and Kenig [16], Fabes, Mendez and Mitrea [9], and Mitrea and Taylor [30], here we take up the task of developing a similar sharp theory for semilinear problems of the type ∆u−N(x, u)=F(x), equipped with Dirichlet and Neumann boundary conditions. 1. Introduction As evidenced by the large body of works (cf., e.g., the monographs [23], [13], [10], [11], [4], [12], [36], [1], [14] and the references therein) nonlinear elliptic boundary value problems have become lately the center of considerable attention, especially due to their pivotal role in such diverse disciplines as spectral and scattering theory, differential geometry, mathematical physics, etc. Meanwhile, and particularly more so in the last decade, this field of research has substantially profited from basic progress in many of the related areas in which some of its methods and techniques are rooted. In [28], [29], [30], [31], [27], the authors have initiated a program aimed at extending the Euclidean, constant coefficient theory from [39], [6], [16], [9] to the case of Lipschitz domains in Riemannian manifolds. One notable achievement in [30] is developing a sharp linear theory for the Poisson problem with Dirichlet and Neumann boundary conditions in Lipschitz domains for the Laplace-Beltrami operator, thus extending 2000 Mathematics Subject Classification. Primary: 35J65, 35B65; Secondary: 42B20, 46E35. Key words. Nonlinear equations, Lipschitz domains, elliptic PDE’s, boundary value problems. ∗This research was partially conducted by the author for the Clay Mathematics Institute. †Author partly supported by NSF.
354 M. Dindoˇ s, M. Mitrea the main results in [16], [9]. The aim of the present paper is to continue this line of work and produce a semilinear version of the aforementioned results, which is in the nature of best possible. A task similar in spirit but in a different functional analytic setting has been accomplished in [7], [8]. In order to be more specific, we momentarily digress for the purpose of introducing some notation and making some definitions. Let M be a smooth, connected, compact, boundaryless manifold, of real dimension dim M=n(unless explicitly mentioned otherwise, we shall assume that n≥3), equipped with a Riemannian metric tensor g= j,k gjk dxj⊗dxkwhose coefficients satisfy gjk ∈C1+γ,some γ>0.(1.1) The Laplace-Beltrami operator on Mis then given in local coordinates by ∆u:= div(grad u)=g−1/2∂j(gjkg1/2∂ku),(1.2) where we use the summation convention, take (gjk) to be the matrix inverse to (gjk), and set g:= det(gjk). Recall that Ω ⊂Mis called aLipschitz domain provided ∂Ω can be described in appropriate local coordinates by means of graphs of Lipschitz functions. Also, the Sobolev scale Lp s(M), 1 <p<∞,s≥0, is obtained by lifting Lp s(Rn):= {(I−∆)s/2f;f∈Lp(Rn)}to M. We denote by Lp s(Ω) the restriction of elements in Lp s(M) to the Lipschitz domain Ω. As is customary, we set Lp s,0(Ω) for the subspace consisting of restrictions to Ω of elements from Lp s(M) with support contained in ¯ Ω. For s>0 and 1 <p,q<∞ with 1/p +1/q = 1, we set Lp −s(Ω) := (Lq s,0(Ω))∗. As is well-known, if Bp,q s(∂Ω), 1 ≤p, q ≤∞,0<|s|<1, stands for the usual class of Besov spaces on ∂Ω, then the trace operator Tr is well-defined from Lp s(Ω) onto Bp,p s−1/p(∂Ω) for each 1 <p<∞and 1/p<s<1+1/p. For a more detailed exposition, the interested reader is referred to [32], [37], [3], [2], and especially [16] for the context of Lipschitz domains. Returning to the mainstream discussion, here we shall be concerned with the semilinear elliptic PDE ∆u−N(x, u)=F(x)inΩ⊂M(1.3) equipped with either Dirichlet or Neumann boundary conditions. Let us point out that when the nonlinearity N(x, u) is of class C1in uthen (1.3) can be rephrased as ∆u−a(x, u)u=fin Ω,(1.4)
Semilinear Poisson Problems 355 where a(x, u):=1 0 ∂N ∂u (x, tu)dt, and f(x):=F(x)+N(x, 0).(1.5) Under various growth and smoothness assumptions on Ω, N,Fand u, the PDE (1.3) has received a lot of attention in the literature lately. Extending some earlier work in [38], T. Runst and W. Sickel have considered in [32, Chapter 6] the case of (1.3) when Ω is smooth, the nonlinearities are of power type, and u,Fbelong to Sobolev-Besov-TriebelLizorkin spaces. In [15], V. Isakov and A. I. Nachman have treated (1.3) equipped with a Dirichlet boundary condition in the case of a two dimensional Euclidean Lipschitz domain Ω and when u∈L2 1(Ω) ∩C0(¯ Ω). In [20], J. Johnsen and T. Runst have addressed (1.3) in the framework of Sobolev-Besov-Triebel-Lizorkin spaces in the case of smooth domains and nonlinearities of composition type (i.e., when N(x, u)=N(u)). Related work can also be found in [18], [13], [5], [22]. In the case of Dirichlet boundary conditions, our main results (cf. Theorem 3.1 and Theorem 3.2) deal with the following situation: Ω is an arbitrary Lipschitz domain, u∈Lp s+1/p(Ω), F∈Lp s+1/p−2(Ω), 0 <s<1, 1<p<∞, and either N(x, u) has sublinear growth in uor a(x, u) (from (1.5)) has an admissible polynomial growth (including a limiting case of exponential behavior). Similar (albeit technically somewhat less refined) results hold in the case of Neumann boundary conditions, even in the presence of (sublinear, power type) nonlinearities in the boundary conditions; cf. Theorem 4.1 for a precise statement. We are interested in the maximal range of indices s,p, for which (1.3) is solvable in the context of Lipschitz domains on the Sobolev-Besov scales. In this respect, we would like to stress that our results are sharp in the sense that they reduce to an optimal linear theory in the absence of nonlinearities. That is, we solve ∆u−a(x, u)u=f∈Lp s+1/p−2(Ω), Tr u=g∈Bp,p s(∂Ω),u∈Lp s+1/p(Ω) (1.6) under admissible growth conditions on a(x, u) (of polynomial and exponential nature) and for the same range of indices s,pas in [30]. The emphasis in the approach we develop is on understanding how the solution of the linear Poisson problem depends on lower order perturbations of the Laplace-Beltrami operator. The main achievement in this regard is to prove that the nonlinear mapping V→ (∆ −V)−1has
356 M. Dindoˇ s, M. Mitrea sublinear growth in V. That is, in a suitable functional analytic context, (1.7) (∆ −V)−1≤C(1 + V)θ, where θ∈[0,1) and Cis independent of V≥0. See (2.6)–(2.7) for an exact formulation. In turn, this key estimate allows for a convenient implementation of the Schauder fixed-point theorem in the framework of Lebesgue spaces. The proof of (1.7) is a delicate argument which relies on the maximum principle, estimates for fractional powers of the Dirichlet Laplacian and interpolation. In the process, we also show that if V∈Ln/2is nonnegative, then the Schr¨odinger operator ∆−V:Lp s+1/p,0(Ω) −→ Lp s+1/p−2(Ω)(1.8) has the same optimal invertibility range as the unperturbed Laplacian ∆. The integrability exponent n/2 is natural, given the sharp unique continuation results from [17]. The layout of the paper is as follows. In Section 2 we refine the treatment of the linear Dirichlet Poisson problem from [30] by allowing less regular lower order terms and by deriving more precise estimates. This is then used in Section 3 to prove, among other things, the solvability of (1.6). Neumann boundary conditions are considered in Section 4. The case of of nonlinearities N(x, u) with sublinear growth in uis treated in Section 5. This section also contains a more detailed analysis of several relevant examples. Finally, in Section 6, sufficient conditions on the data are produced so that a solution of (1.6) obeying nontangential maximal function estimates can be found. Here we also analyze the case when boundary nonlinearities are allowed. Acknowledgements. It is a pleasure to thank Michael Taylor for bringing us together and for his constant support and interest in our work. The second named author would also like to thank Winfried Sickel and Jon Johnsen for some helpful conversations. 2. The linear theory revisited We shall retain as much as possible the notation introduced in Section 1. The aim of this section is to prove a refined version of the well-posedness result for the Dirichlet Poisson problem for ∆ on SobolevBesov spaces from [30].
Semilinear Poisson Problems 357 Theorem 2.1. Let Ω⊂Mbe an arbitrary Lipschitz domain and for a function Vsatisfying V≥0and V∈Lr(M)for some r≥n/2,(2.1) set L:= ∆ −V. Then there exists ε=ε(Ω) >0with the following significance. Let 0<s<1and 1<p<∞satisfy at least one of the following conditions: 2 1+ε<p< 2 1−εand 0<s<1; 1≤p< 2 1+εand 2 p−1−ε<s<1; 2 1−ε<p≤∞ and 0<s<2 p+ε, (2.2) and, in addition, 1 r−2 n<n−1 np −s n<1−1 r.(2.3) Then the Dirichlet Poisson problem (DP) Lu =f∈Lp s+1 p−2(Ω), Tr u=g∈Bp,p s(∂Ω), u∈Lp s+1 p (Ω), (2.4) has a unique solution, which also satisfies uLp s+1 p (Ω) ≤CfLp s+1 p−2(Ω) +gBp,p s(∂Ω) (2.5) for some C=C(Ω,V,p,s)>0. Introduce τ:= np/(n−1−sp)if sp < n−1and τ:= ∞if sp > n−1. Then, if 1/p ≤sand sp =n−1, there exists a constant C=C(Ω,p,s)> 0, independent of V, such that uLτ(Ω) ≤C(fLp s+1 p−2(Ω) +gBp,p s(∂Ω)).(2.6)
358 M. Dindoˇ s, M. Mitrea If sp =n−1then for some C>0independent of Vand any τ∈[2,∞), uLτ(Ω) ≤Cτ1−1/p(fLp s+1 p−2(Ω) +gBp,p s(∂Ω)).(2.7) On the other hand, if s<1/p, then the estimate uLτ(Ω) ≤C(1 + VLr(Ω))θ(fLp s+1 p−2(Ω) +gBp,p s(∂Ω)),(2.8) holds with θ:= ε+r 2(r−1) (1 p−s),if r≥1+(1−sp)/(2p−2), ε+1 2−n/r ((n−1)/p−s−n+2),if r<1+(1−sp)/(2p−2), (2.9) where ε>0is arbitrarily small and C=C(Ω,s,p,ε)>0is independent of V. In dimension n=2, results similar in spirit hold provided (2.2) is replaced by 1 2−ε<1 p<1 2+εand 0<s<1; 1 2+ε≤1 p<1and 1 p−1 2−ε<s<1; 0<1 p≤1 2−εand 0<s<1 p+1 2+ε, (2.10) where ε=ε(∂Ω) ∈(0,1 2]. More specifically, the estimates (2.5)–(2.7) hold unchanged, whereas (2.8) holds if also r>4/3. Finally, when ∂Ω∈C1, instead of (2.2) when n≥3,or(2.10) when n=2, we may simply allow s∈(0,1),p∈(1,∞). Parenthetically, we note that when r>nthen (2.3) is satisfied for any s∈(0,1), p∈(1,∞). Another simple yet useful remark is that, if ε>0 is small enough, then n≥3,r>n 2,s,pas in (2.2) with ε>0 small =⇒(2.3) holds.(2.11) Furthermore, in the two dimensional context, (2.12) r>4 3,s,pas in (2.10) with ε>0 small =⇒(2.3) holds automatically.
Semilinear Poisson Problems 359 1 p sε 1−ε 1 2 1+ε 2 1−ε 2 1 1 ✦✦✦✦✦✦ ✦ ✦✦✦✦✦✦ ✦ 0 Figure 1. The hexagon describing the well posedness region (2.2). Before presenting the proof of the Theorem 2.1, we need to discuss several auxiliary results. Lemma 2.2. Let Ωbe a Lipschitz domain (of dimension ≥2) and fix 1<p<∞,0<s<1,n/2≤r≤∞, so that (2.3) holds. Then the inclusion Lr(Ω) ·Lp s+1/p(Ω) !→Lp s+1/p−2(Ω)(2.13) is compact. Proof: Consider first the case when sp < n −1 and r>n/2. In this scenario, we shall show that there exists ε=ε(n, p, s, r)>0 such that Lr(Ω) ·Lp s+1/p(Ω) !→Lp s+ε+1/p−2(Ω).(2.14) In concert with Rellich’s selection lemma, this proves that, under the current assumptions, multiplication by an element from Lr(Ω) is a compact operator from Lp s+1/p(Ω) into Lp s+1/p−2(Ω). To see (2.14), if 1/p0:= 1/p −(s+1/p)/n it follows that 1 <p 0<∞ and Lp s+1/p(Ω) !→Lp0(Ω). Going further, if 1/p1:= 1/r +1/p0then 1<p 1<∞, thanks to (2.3), and we have Lr(Ω) ·Lp0(Ω) !→Lp1(Ω). Now, since Lp s+ε+1/p−2(Ω)=(Lp 2−s−ε−1/p,0(Ω))∗with 1/p +1/p=1,it
360 M. Dindoˇ s, M. Mitrea suffices to prove that Lp 2−s−ε−1/p(Ω) !→Lp2(Ω), where 1/p2+1/p1=1. This, however, is guaranteed by the estimate 1/p2>1/p−(2−s−1/p)/n. Some simple algebra now shows that this is always the case provided r>n/2. When sp ≥n−1, r>n/2, one proceeds analogously, choosing p0∞. Finally, when r=n/2, the well-definiteness of the inclusion (2.13) is proved similarly. Then, based on this, (2.14) and an approximation argument, it follows that (2.13) is compact as well. Lemma 2.3. Let Ωbe a Lipschitz domain and assume that s,psatisfy at least one of the conditions in (2.2). Also, fix a nonnegative function V∈Lr(M), with r≥n/2, and suppose that (2.3) holds. Then u∈Lp s+1/p(Ω),(∆ −V)u≤0,Tr u≥0=⇒u≥0.(2.15) In particular, if Vj∈Lr(M),r≥n/2, are such that 0≤V1≤V2, then (2.16) uj∈Lp s+1/p(Ω),Tr u1≥Tr u2≥0, (∆ −V1)u1≤(∆ −V2)u2≤0=⇒u1≥u2. Analogous results are valid for n=2, provided that s,psatisfy at least one of the conditions in (2.10). Finally, when ∂Ω∈C1, any s∈(0,1), p∈(1,∞)will do (in all dimensions) as long as (2.3) is satisfied. Proof: For an arbitrary, fixed u∈Lp s+1/p(Ω), let us set f:= (∆ −V)u∈ Lp s+1/p−2(Ω) and g:= Tr u∈Bp,p s(∂Ω). Thus, by assumption, f≤0 and g≥0. For starters, we make the claim that there exist fj∈C∞ comp(Ω), gj∈ Lip(∂Ω), such that fj≤0, gj≥0, and fj→fin Lp s+1/p−2(Ω), gj→gin Bp,p s(∂Ω). Indeed, the approximating sequence {gj}jcan be produced via a standard localization and mollifying procedure. There remains to prove that fbelongs to the closure of the convex set C:= {ψ∈ C∞ comp(Ω); ψ≤0}in Lp s+1/p−2(Ω). Assuming the opposite (and seeking a contradiction), the Hahn-Banach theorem ensures the existence of some Φ∈Lp s+1/p−2(Ω)∗=Lp 2−s−1/p,0(Ω), 1/p+1/p= 1, such that Φ,φ≤ 0<Φ,ffor each φ∈C. (Strictly speaking, what the Hahn-Banach theorem originally gives is the previous double inequality with ‘zero’ replaced by some real number λ. However, given that Cis in fact a cone, it is easy to see that we can always assume that λ= 0.) Now, the first inequality implies Φ ≥0 which, in turn, contradicts the second, given that f≤0. This finishes the proof of the claim made at the beginning of the paragraph.
Semilinear Poisson Problems 361 Next, granted the current assumptions, we will show in the course of the proof of Theorem 2.1 that u(is uniquely determined by, and) depends continuously on f,g. Consequently, thanks to a limiting argument (whose implementation is routine, given the claim above), it suffices to prove (2.15) in the case when f∈L∞(Ω) and g∈C0(∂Ω), which we will assume hereafter. It is not difficult to see that these extra assumptions entail u∈C0(¯ Ω) ∩C1 loc(Ω). With an eye on (2.15) and seeking a contradiction, assume next that there exists x0∈Ω such that u(x0)<0. If we let Obe the connected component of the set {x∈Ω; u(x)<0}which contains x0and recall that u|∂Ω>0, it follows that O⊂⊂Ω. Let ψ∈C∞(R) be a Lipschitz, non-decreasing, odd function such that ψ(t)=0for|t|≤1 and ψ(t)=tfor |t|≥2. Set ui(x):=i−1ψ(iu(x)), x∈¯ O,i=1,2,... . Then, so we claim, (2.17) ui∈C1 comp(O),u i≤0, and u−uiL∞(O),∇u−∇uiL2(O)−→ 0. Indeed, that uiis compactly supported in Oand ∇u−∇uiL2(O)→0 are consequences of u|∂O= 0 and definitions. As for ∇u−∇uiL2(O)→ 0, note that ∇ui(x)=ψ(iu(x))∇u(x)→∇u(x)asi→∞, for each x∈O. Since ψis Lipschitz and u∈C1(¯ O), Lebesgue’s dominated convergence theorem then yields the desired conclusion. Going further, O |∇u|2+V|u|2= lim O ∇u, ∇ui+Vuu i =−lim O ui(∆ −V)u≤0, (2.18) which contradicts the fact that uis not identically zero in O. This finishes the proof of the implication (2.15). Finally, to justify (2.16), it suffices to apply (2.15) to the difference u:= u1−u2(with the choice V:= V1). Lemma 2.4. For each Lipschitz domain Ω⊂Mand 1<p<∞there exists a constant C=C(M,Ω,p)>0such that for any u∈Lp n/p(Ω) and τ∈[2,∞) uLτ(Ω) ≤Cτ1−1/puLp n/p(Ω).(2.19) Furthermore, if n≥3and n−1<p<∞, then fLτ 2/τ,0(Ω) ≤Cτ1−1/pfLp n/p,0(Ω),for all τ∈[p, ∞),(2.20) where C=C(M,Ω,n,p)>0is independent of τ.
368 M. Dindoˇ s, M. Mitrea hence u(x∗)>1 2f(x∗)>0. Observe that the function uis a supersolution for the Laplace-Beltrami operator on Ω (since ∆u=Vf ≥0inΩ)so that, when considered on O, it must attain its maximum at a boundary point. Thus, there exists z∈∂O⊂Ω at which u(z) = maxx∈O u(z)≥ u(x∗)>1 2f(x∗). On the other hand, the definition of Oensures that f≡0on∂O. Consequently, ˙ T−1 Vf(z)=f(z)−u(z)<−1 2f(x∗)<0, so that |˙ T−1 Vf(z)|>1 2fL∞(Ω). This contradicts (2.51), and concludes the proof of (2.50). Turning now to the task of proving (2.49), recall that for each r/(r−1) ≤p≤∞, there exists a constant κ(V)>0, independent of p, such that ˙ TVfLp(Ω) ≤κ(V)fLp(Ω) and ˙ TVfL∞(Ω) ≤2fL∞(Ω),(2.53) uniformly for f∈C0(Ω). A standard interpolation inequality (cf., e.g., [13, p. 41]) then yields, for f∈C0(Ω) and p∈[r/(r−1),∞], ˙ TVfLp(Ω) ≤˙ TVfθ Lr/(r−1)(Ω)˙ TVf1−θ L∞(Ω) ≤κ(V)θ21−θfθ Lr/(r−1)(Ω)f1−θ L∞(Ω), (2.54) where θ:= r/(p(r−1)) ∈(0,1). In particular, θ!0asp∞. Now, given any f∈L∞(Ω), we can find a sequence of functions (fj)j in C0(Ω) such that fjL∞(Ω) ≤fL∞(Ω) and fj→fin Lq(Ω) for any q<∞. Then, by virtue of (2.54), we get TVfL∞(Ω) = lim p→∞ TVfLp(Ω) = lim p→∞ lim j→∞ ˙ TVfjLp(Ω) ≤lim p→∞ lim j→∞ κ(V)θ21−θfjθ Lr/(r−1)(Ω)fj1−θ L∞(Ω) ≤2fL∞(Ω). (2.55) This justifies (2.49) and shows that TVL(L∞(Ω)) ≤2, independently of V, thus proving (2.41). With (2.41) in hand and relying on Lemma 2.5, we see that (2.40) holds for each 2 ≤p<∞, uniformly in V. There remains the case 1 <p<2 which we treat next. Our strategy is based on duality and requires analyzing the action of the operator KV,p :=(−∆)1/p DTV(−∆)−1/p D=−(−∆)1/p D(∆ −V)−1 D(−∆)1−1/p D (2.56) on Lpspaces. Concretely, from Proposition 2.6 (with µ=2/p) it suffices to show that KV,pL(Lp(Ω)) is bounded uniformly in Vor, equivalently,
Semilinear Poisson Problems 369 that K∗ V,pL(Lp(Ω)) ≤C(p) uniformly in V, where 1/p+1/p= 1. Since K∗ V,p =−(−∆)1−1/p D(∆ −V)−1 D(−∆)1/p D =(−∆)1/p DTV(−∆)−1/p D=KV,p (2.57) by relying once again on Proposition 2.6, we see that matters are further reduced to proving that TVL(Lp 2/p,0(Ω)) ≤Cuniformly in V. However, having p∈(1,2) entails p∈(2,∞) and this is precisely the case already addressed. The proof of the lemma is therefore finished. In our next lemma we consider the action of the operator TVon Lpspaces. Lemma 2.8. Assume that n≥3. Then, if V∈Lr(Ω),r>n/2,is nonnegative, there holds TVL(Lp(Ω)) ≤C(1 + VLr(Ω))for each p>r/(r−1),(2.58) where C=C(Ω,r,p)>0is independent of V. The estimate (2.58) also holds when n=2provided r>4/3. Proof: Assume first that n≥3. To prove (2.58), we rely on the fact that TVf=f+(∆−V)−1 D(Vf) for f∈C∞ comp(Ω), and analyze the actions of the multiplicative operator MV, defined by MVf:= V·f, and (∆−V)−1 D separately. In this scenario, the crux of the matter is establishing the estimate (∆ −V)−1 DL(Lq(Ω),L qn/(n−2q)(Ω)) ≤C, q ∈(1,n/2),(2.59) for some C=C(q)>0 independent of V. Indeed, (2.59) in concert with the elementary estimate MVL(Lp(Ω),Lpr/(p+r)(Ω)) ≤VLr(Ω), readily yields (2.58) provided that one can choose q∈(1,n/2) so that 1/r+1/p ≤ 1/q ≤1/p +2/n. As this latter condition is easily checked, (2.58) will follow as soon as we justify (2.59). In this regard, a simple yet useful observation (whose proof amounts to an algebra exercise) is that (2.60) ∀q∈(1,n/2), ∃(s, 1/p) as in (2.2) such that Lq(Ω) !→Lp s+1/p−2(Ω). Consequently, one can always employ a factorization of the type (2.61) (∆ −V)−1 D:Lq(Ω) ι −→ Lp s+1/p−2(Ω) (∆−V)−1 D −→ Lp s+1/p,0(Ω) ι −→ Lqn/(n−2q)(Ω)
370 M. Dindoˇ s, M. Mitrea in order to justify that (∆ −V)−1 D∈L(Lq(Ω),L qn/(n−2q)(Ω)) for each q∈(1,n/2). Turning now to the actual task of proving (2.59), fix an arbitrary f∈Lq(Ω) which we write as f=f+−f−, where 0 ≤f+,f−≤|f|. Positivity of (−∆)−1 Dgives 0≤f±≤V(−∆)−1 Df±+f±.(2.62) Applying (V−∆)−1 Dto the above inequalities and invoking Lemma 2.3 yields 0≤(V−∆)−1 Df±≤(−∆)−1 Df±.(2.63) Together with the decomposition f=f+−f−and (2.61) with V=0, this allows us to write (V−∆)−1 DfLqn/(n−2q)(Ω) ≤(−∆)−1 Df+Lqn/(n−2q)(Ω) +(−∆)−1 Df−Lqn/(n−2q)(Ω) ≤2CfLq(Ω), (2.64) as desired. This establishes (2.59) and finishes the proof of (2.58) when n≥3. When n= 2 one can proceed in a similar fashion, the most notable difference being that (2.59) now becomes (∆ −V)−1 DL(Lq(Ω),L ∞(Ω)) ≤ C, for q>1. Armed with Lemma 2.7 and Lemma 2.8 we now turn to the task of establishing the following important estimates. Lemma 2.9. Assume that V∈Lr(Ω),r>n/2, is an arbitrary nonnegative function, and that 0<s<1,1<p<∞. (i) If s/(n−1) <1/p ≤s, then TVL(Lp s+1/p,0(Ω),Lnp/(n−1−sp)(Ω)) ≤C,(2.65) where C=C(Ω,s,p)>0is a finite constant independent of V. (ii) If sp>n−1then, for some constant C=C(Ω,r,p,s)>0independent of the function V, TVL(Lp s+1/p,0(Ω),L∞(Ω)) ≤C.(2.66) (iii) If sp =n−1, there exists C=C(Ω,r,p)>0independent of V such that TVL(Lp s+1/p,0(Ω),Lτ(Ω)) ≤Cτ1−1/p for any τ∈[2,∞).(2.67)
Semilinear Poisson Problems 371 (iv) If s<1 p<s n−1+n(r−1) r(n−1) ,n≥3, then (2.68) TVL(Lp s+1/p,0(Ω),Lnp/(n−1−sp)(Ω)) ≤ C(1+VLr(Ω) )ε+r 2(r−1) (1 p−s),if r≤1+(1−sp)/(2p−2), C(1+VLr(Ω) )ε+1 2−n/r ((n−1)/p−s−n+2),if r<1+(1−sp)/(2p−2), for each ε>0, where C=C(Ω,s,p,ε)>0is independent of V. When n=2, the same holds provided r>4/3. Proof: Consider first the situation when 0 <s/(n−1) <1/p ≤s<1 (which automatically entails n≥3). The case sp =1,1<p<∞has been dealt with in Lemma 2.7. The larger range we intend to addressed here is then a simple consequence of this and the factorization TV:Lp s+1/p,0(Ω)!→Lp∗ 2/p∗,0(Ω) TV −→ Lp∗ 2/p∗,0(Ω)!→Lnp/(n−1−sp)(Ω),(2.69) where 1/p∗:= (n−1)(1/p−s/(n−1))/(n−2), and both inclusions above are classical embedding results. As for (ii), i.e. the case sp > n−1, n≥2, we use (2.41) and the factorization TV:Lp s+1/p,0(Ω) !→L∞(Ω) TV −→ L∞(Ω).(2.70) Next we address the case sp =n−1 (which forces p>n−1). When n= 2, the conclusion we seek is an immediate consequence of Lemma 2.7 and Lemma 2.4. Similar ingredients can be used to handle the case n≥3 as well. The idea is to use the factorization TV:Lp n/p,0(Ω) ι1 !→Lτ 2/τ,0(Ω) TV −→ Lτ 2/τ,0(Ω) ι2 !→Lτ(Ω),(2.71) in concert with ι1≤Cτ1−1/p, (2.40), and the fact that ι2≤C, uniformly in τ∈[p, ∞). This finishes the analysis of the point (iii) in our lemma. At this stage, it remains to deal with the situation described in the point (iv) of the lemma, and our intention is to eventually interpolate between (2.40) and (2.58). To this end, recall first that the two classes, Lp s+1/p,0(Ω) and Lnp/(n−1−sp)(Ω), with 1 <p<∞and −1/p ≤s, are complex interpolation scales, in the sense that for each 1 <p j<∞,
372 M. Dindoˇ s, M. Mitrea sj≥−1/pj,j=0,1, Lp0 s0+1/p0,0(Ω),L p1 s1+1/p1,0(Ω)θ=Lp∗ s∗+1/p∗,0(Ω), Lnp0/(n−1−s0p0)(Ω),L np1/(n−1−s1p1)(Ω)θ=Lnp∗/(n−1−s∗p∗)(Ω), if 1/p∗:= (1 −θ)/p0+θ/p1,s ∗:= (1 −θ)s0+θs1,0<θ<1. (2.72) This observation and convexity arguments readily entail the following principle: if 1 <p i<∞,−1/pi≤si,τi:= npi/(n−1−sipi)>1, and TVL(Lpi si+1/pi(Ω),Lτi(Ω)) ≤Mi,i=1,2,3, then TVL(Lp s+1/p,0(Ω),Lnp/(n−1−sp)(Ω)) ≤Mλ1 1Mλ2 2Mλ3 3,(2.73) provided the point (s, 1/p) has baricentrical coordinates (λ1,λ 2,λ 3), λi≥0, relative to the triangle with vertices (si,1/pi), i=1,2,3. We shall use the above remark twice, first for the triangle with vertices at (ε, ε), (1 −ε, 1−ε), (1/(r+ε)−1,1−1/(r+ε)) with ε>0 sufficiently small. In this scenario, assuming that n≥3, M1and M2are controlled by Cεand M3≤Cε(1+VLr(Ω)). Also, λ3is a O(ε) variation of r(1/p −s)/(2(r−1)). This yields the first line in (2.68). The condition r>1+(1−sp)/(2p−2) guarantees that the point (s, 1/p) lies inside the triangle under discussion. The second application of the aforementioned remark is similar in spirit and requires a preparatory step. Specifically, as a result of (2.58) and the factorization TV:Lp s+1/p,0(Ω) !→Lnp/(n−1−sp)(Ω) TV −→ Lnp/(n−1−sp)(Ω)(2.74) it follows that (2.75) TVL(Lp s+1/p,0(Ω),Lnp/(n−1−sp)(Ω)) ≤C(1 + VLr(Ω)) if 0 <1 p−s n−1<n(r−1) r(n−1). Note that the intersection between 1 p−s n−1=n(r−1) r(n−1) with 1/p = 1 is the point with coordinates (n/r−1,1). Inspired by this observation, we write (2.73) for the triangle with vertices at (1−ε, 1−ε), (1/(r+ε)−1,1−1/(r+ ε)), (n/(r+ε)−ε(n−1)−1,1−ε) for some sufficiently small ε>0. This time, assuming that n≥3, we get M1≤Cε,M2,M 3≤Cε(1+VLr(Ω)), whereas λ2+λ3is a O(ε) variation of [(n−1)/p −s−n+2]/(2 −n/r). Availing ourselves of (2.58), (2.40), the second line of (2.68) follows. The
Semilinear Poisson Problems 373 role of r≤1+(1−sp)/(2p−2) and 1 p−s n−1<n(r−1) r(n−1) is to ensure that the point (s, 1/p) lies inside the triangle we are currently considering. The subcase of (iv) corresponding to n= 2 is handled similarly, and requires that r>4/3 (cf. Lemma 2.8). Finally, we have all the necessary ingredients in order to tackle the Proof of Theorem 2.1: Part II: Here we present the final arguments in the proof of (2.6), (2.8), (2.7), and also treat the case r=n/2. In what follows, if his an arbitrary real-valued function, we set h±:= max{±h, 0}=1 2(|h|±h),(2.76) so that h±≥0 and h=h+−h−. Also, |h±(x)−h±(y)|≤|h(x)−h(y)| so that (·)±:Bp,p s(∂Ω) −→ Bp,p s(∂Ω) are bounded, 1<p<∞,0<s<1.(2.77) For the time being, we continue to assume r>n/2. If we now denote by v(±) V∈Lp s+1/p(Ω) the solutions of Lv(±) V=(∆−V)v(±) V= 0 in Ω,Tr v(±) V=g±,(2.78) then, clearly, vV:= v(+) V−v(−) Vsolves LvV=(∆−V)vV= 0 in Ω,Tr vV=g.(2.79) Set also v(±) 0for the solutions of (2.78) with V= 0. According to Lemma 2.3 we have 0≤v(+) V≤v(+) 0and 0 ≤v(−) V≤v(−) 0.(2.80) Assuming sp < n −1, it follows that vVLnp/(n−1−sp)(Ω) ≤v(+) VLnp/(n−1−sp)(Ω) +v(−) VLnp/(n−1−sp)(Ω) ≤v(+) 0Lnp/(n−1−sp)(Ω) +v(−) 0Lnp/(n−1−sp)(Ω) ≤Cv(+) 0Lp s+1/p(Ω) +Cv(−) 0Lp s+1/p(Ω) ≤2CgBp,p s(∂Ω), (2.81) where the last constant Cin (2.81) is that appearing in the estimate (2.5) for V= 0. The case when sp > n−1 is similar, while the case sp =n−1 is proved with the help of Lemma 2.4. Similarly, by wVwe denote the solution to LwV=(∆−V)wV=fin Ω,Tr wV=0.(2.82)
374 M. Dindoˇ s, M. Mitrea Clearly, w0=∆ −1 Dfand wV=(∆−V)−1 Df, hence wVLτ(Ω) =(∆ −V)−1 D∆w0Lτ(Ω) ≤TVL(Lp s+1/p(Ω),Lτ(Ω))w0Lp s+1/p(Ω) ≤CTVL(Lp s+1/p(Ω),Lτ(Ω))fLp s+1/p−2(Ω). (2.83) The constant Cappearing in the last line is again the constant in the estimate (2.5) for V=0. Now,uV=vV+wVsolves the desired equation (2.4) and the estimates (2.6)–(2.7) follow from (2.81), (2.83) and Lemma 2.9. In the two dimensional situation, the well posedness range (2.10) is handled analogously, granted the work in [25], [26]. Here we only want to remark that, in order to justify (2.30) for u∈Lp s+1/p,0(Ω) and V∈ Lr(Ω), r>(1−|1/(2p)−s/2|)−1, observe first that for some ε>0 small, Vu∈L1+ε(Ω) !→Lq 2/q−2+ε(Ω), if q>1. In turn, this further entails u= pr1[Dir−1(Vu)] ∈Lq 2/q+ε(Ω) !→L2 1(Ω), thanks to the aforementioned references. Since, under the current assumptions, V|u|2∈L1(Ω) is also readily verified, this takes care of (2.30) when n=2. Finally, we are left with the analyzing the case r=n/2, a task which we take up next. For starters, we claim that it suffices to deal with the situation when the datum fis actually selected from Lq(Ω), where qis given by 1 q:= 2 n+n−1−sp np .(2.84) Indeed, given an arbitrary f∈Lp s+1/p−2(Ω), set w:=∆−1 Df∈Lp s+1/p,0(Ω). Then u=u0+wsolves (2.1), provided u0is a solution of (∆ −V)u0=Vw in Ω, Tr u0=g∈Bp,p s(∂Ω), u0∈Lp s+1 p (Ω). (2.85) Now, w∈Lnp/(n−1−sp)(Ω) by standard embedding results and, further, Vw ∈Lq(Ω) by H¨older’s inequality (note that (2.3) guarantees that q>1). This justifies the claim made at the beginning of the paragraph. Assuming next that f∈Lq(Ω), it follows that f±Lq(Ω) ≤fLq(Ω). Consider an approximating sequence Vj→Vin Ln/2(Ω), so that Vj≥0,
Semilinear Poisson Problems 375 Vj∈L∞(Ω), for each j, and let u± jbe the unique solution of (∆ −Vj)u± j=−f∓∈Lq(Ω), Tr u± j=g±∈Bp,p s(∂Ω), u± j∈Lp s+1 p (Ω). (2.86) As before, we get that u± jLnp/(n−1−sp)(Ω) ≤u± 0Lnp/(n−1−sp)(Ω), where u± 0solves (2.86) with V0:= 0. From this and the decomposition uj= u+ j−u− j, we conclude that there exists some constant C>0 independent of Vsuch that ujLnp/(n−1−sp)(Ω) ≤C(fLq(Ω) +gBp,p s(Ω)).(2.87) We now make the claim that (uj)j∈Nis a Cauchy sequence in Lnp/(n−1−sp)(Ω). To see this, for j, k ∈N, we note that w:= uj−uk solves the problem (∆ −Vj)w=(Vj−Vk)uk∈Lq(Ω),w∈Lp s+1 p,0(Ω).(2.88) Hence, by virtue of (2.87), we have uj−ukLnp/(n−1−sp)(Ω) ≤C(Vj−Vk)ukLq(Ω) ≤CVj−VkLn/2(Ω)ukLnp/(n−1−sp)(Ω) ≤CVj−VkLn/2(Ω)(fLq(Ω) +gBp,p s(Ω)). (2.89) From this, our claim follows. In order to continue, let ube the limit of the sequence (uj)j∈Nin Lnp/(n−1−sp)(Ω). We now intend to show that usolves (2.1). Denote by u=Tg(f) the solution operator of the Poisson problem (2.1) corresponding to V:= 0. In particular, uj=Tg(f+Vjuj). Since Vjuj→Vu in Lq(Ω) !→Lp s+1/p−2(Ω), we may conclude that uj=Tg(f+Vjuj)→ Tg(f+Vu)inLp s+1/p(Ω). From this we see that u=Tg(f+Vu), i.e., u solves (2.1). To finish the proof of the theorem we need to show that the solution we have just constructed is unique. By linearity, this comes down to proving that the operator ∆−V:Lp s+1 p,0(Ω)−→ Lp s+1 p−2(Ω)=Lp 2−s−1 p (Ω)∗,1 p+1 p=1,(2.90) is one-to-one. From what we have proved so far, this operator is onto so, given the invariance of the conditions (2.2)–(2.3) to the transformation (s, 1/p)→ (1 −s, 1−1/p), the desired result follows by duality.
376 M. Dindoˇ s, M. Mitrea Parenthetically, let us note that we could have reached the same conclusion using the ontoness of (2.90) plus the fact that, much as in (2.28), the operator (2.90) is Fredholm with index zero. 3. Nonlinearities with polynomial growth In this section we study the semilinear version of (2.4) in the case when the nonlinearity is allowed to have superlinear growth. In order to state our first result, recall that a function a(x, u) is called Carath´eodory if it is measurable in xand continuous in u. Theorem 3.1. Assume that Ω⊂Mis a connected Lipschitz domain in M, where dim M=n≥4. Also, suppose that a:Ω×R→Ris a Carath´eodory function such that 0≤a(x, u)≤k1(x)+k2(x)|u|m,(3.1) where 0≤kj∈Lqj(Ω) with qj≥1,j=1,2. For 1<p<∞,0<s<1, consider the following semilinear Poisson problem with Dirichlet boundary condition: ∆u−a(x, u)u=f∈Lp s+1/p−2(Ω), Tr u=g∈Bp,p s(∂Ω),u∈Lp s+1/p(Ω). (3.2) Then there exists ε=ε(Ω) >0such that the problem (3.2) has at least one solution provided the following is true: The pair (p, s)satisfies at least one of the conditions in (2.2),sp < n −1, and 0≤1 q1 ≤2 nand 1 q1 <1−n−1−sp np ,(3.3) 0≤m< np n−1−sp min 2 n,1−n−1−sp np −1 q2.(3.4) For sp ≥n−1the number mcould be taken arbitrarily large as long as q1,q 2>(2 n+n−1 np −s n)−1. Moreover, (3.5) ∀M>0∃K>0 so that fLp s+1/p−2(Ω) +gBp,p s(∂Ω) ≤M=⇒uLp s+1/p(Ω) ≤K. If sp ≥1the solution usatisfies the estimate uLnp/(n−1−sp)(Ω) ≤C(fLp s+1/p−2(Ω) +gBp,p s(∂Ω))(3.6) for some C=C(∂Ω,s,p,a)>0.
Semilinear Poisson Problems 377 If, in addition, the nonlinearity b(x, u):=a(x, u)usatisfies 0≤∂ ∂ub(x, u)≤˜ k1(x)+˜ k2(x)|u|m,(3.7) with 0≤˜ kj∈Lqj(Ω),j=1,2,qj’s and mas before, then the solution u of the boundary problem (3.2) is also unique. Analogous results are valid in dimensions n=2and n=3.If dim M=n=3the results stated above remain true without change, provided sp ≥1/2and p(1 + s)≥2. Otherwise 0≤m< 3p 2−sp 1 σ−1 q2,0≤1 q1 <1−2−sp 3p,(3.8) will suffice, where σ:= 3 2+1−2sp 2(2−sp),if 2sp < 1and p(3 + s)≥5, 3 21+2−p(1+s) 2−sp ,if p(1 + s)<2and p(3 + s)<5. (3.9) If n=2the previous results are valid provided (2.10) is used in place of (2.2), and q1>2,0≤m< p(q2−2) 2q2(1 −sp)for sp < 1, q1,q 2>1− 1 2p−s 2−1 ,m≥0,for sp ≥1. (3.10) By way of contrast, 1<p<∞,0<s<1, will do in all dimensions when ∂Ω∈C1. A few comments are in order here. (i) As far as the range of validity (described by means of (2.2), (2.10)) is concerned, our theorem is in the nature of best possible. This is because the aforementioned range is optimal for the associated linear problems. (ii) In some instances min (3.4) can be allowed to attain the value np n−1−sp 2 n−1 q2and the statement of the theorem remains true. In order to avoid further complications we decided not to include this case in the theorem given above. Interested reader can analyze this boundary case by techniques outlined in the proof that follows. (iii) Similar results are valid at the level of 2nd-order, formally selfadjoint, non-positive, strongly elliptic systems, at least when n=2 or n=3.
384 M. Dindoˇ s, M. Mitrea (a) sp =n−1and for each A>0there exists ˜ k∈Lt(Ω),t>n/2, such that 0≤∂ ∂ub(x, u)≤˜ k(x) exp(A|u|γ),with 0≤γ=γ(p)=p/(p−1),(3.44) (b) sp > n −1and (3.45) 0 ≤∂ ∂ub(x, u), ∀M>0 sup u∈[−M,M] ∂ ∂ub(x, u)∈Lt(Ω) for some t>σ, then the solution uof the boundary problem (3.38) is also unique. Finally, analogous results are valid in dimension n=2, provided (2.10) is used in place of (2.2). Moreover, 1<p<∞,0<s<1, sp ≥n−1, will do in all dimensions when ∂Ω∈C1. Proof: Consider first the somewhat simpler case sp > n −1. It follows from Theorem 2.1 that any solution to the equation (2.4) satisfies uL∞(Ω) ≤C(fLp s+1 p−2(Ω) +gBp,p s(∂Ω)),(3.46) with C>0 independent of V≥0. Introduce N:= C(fLp s+1 p−2(Ω) + gBp,p s(∂Ω)), then define ψN(u):= u, for |u|≤N, N, for u>N, −N, for u<−N, (3.47) and, finally, consider the Poisson problem: ∆u−a(x, ψN(u))u=f∈Lp s+1/p−2(Ω), Tr u=g∈Bp,p s(∂Ω),u∈Lp s+1/p(Ω). (3.48) The problem (3.48) satisfies all the assumptions in Theorem 3.1 and, hence, has at least one solution u. Moreover, matters can be arranged so that the solution also satisfies the estimate (3.46). This forces a(x, u(x)) = a(x, ψN(u(x))), thus usolves (3.38) as well. The proof of the uniqueness part parallels its counterpart in Theorem 3.1.
Semilinear Poisson Problems 385 Next, we turn our attention to the more interesting case sp =n−1. To begin with, define the vector space Xpby setting Xp={h:Ω−→ R; sup τ≥2 τ−1+1/phLτ(Ω) <∞},(3.49) and equip it with the norm hXp:= sup τ≥2 τ−1+1/phLτ(Ω).(3.50) It is not very difficult to check that Xpis indeed a Banach space. An essential property enjoyed by functions in Xpis that there exists κ= κ(M,Ω) >0 such that h∈Xp,α:= κ hp/(p−1) Xp =⇒exp(α|h|p/(p−1))∈L1(Ω).(3.51) The proof of this fact can be carried out exactly as the proof of Trudinger’s inequality; cf. [36, Chapter 13] for more details. Taking another look at the estimate (2.7), we learn that the solution uto the equation (2.4) belongs to Xpand there exists C>0 independent of Vsuch that uXp≤C(fLp s+1 p−2(Ω) +gBp,p s(∂Ω)).(3.52) To proceed from here, consider first the case when f∈L∞(Ω). Denote by u(±) Vthe solution to the boundary problem: (∆ −V)u(±) V=f∓in Ω,Tr u(±) V=g±.(3.53) In this situation, relying on Lemma 2.3 we infer that 0 ≤u(±) V≤u(±) 0, hence there exists h=u(+) 0+u(−) 0∈X psuch that for any V≥0 the solution uVof (2.4) satisfies |uV(x)|≤h(x),for any x∈Ω.(3.54) Let us now define O:= {u∈L1(Ω); |u|≤h},(3.55) so that, clearly, O⊂X p. Another simple yet useful observation is that for any τ∈[2,∞) the set Ois closed and convex in Lτ(Ω). Finally, consider the operator T:O→Odefined by agreeing that v:= Tu is the (unique) solution of ∆v−a(x, u)v=fin Ω,Tr v=g, v ∈Lp s+1/p(Ω).(3.56)
386 M. Dindoˇ s, M. Mitrea In order to see that vis well defined, it is sufficient to observe that for any u∈Owe have a(x, u(x)) ∈Lr(Ω) for some r>n/2. This, however, is a consequence of 0≤a(x, u)≤k(x) exp(A|u|γ)≤k(x) exp(A|h|γ).(3.57) Indeed, since h∈Xp, one can select Asmall enough so that exp(A|h|γ)∈ Lq(Ω). Thus, the desired conclusion follows by choosing qsufficiently large, so that 1/q +1/t<2/n. Also v=Tu∈Oby (3.54). Finally, Tis a continuous and compact map in the Lτ(Ω) topology for any τ∈[2,∞). Continuity follows from the fact that if ui→upointwise on Ω and all ui∈O, then also a(., ui(.)) →a(., u(.)) pointwise. As (3.57) gives a common majorant in Lr(Ω) (for some fixed r>n/2), we see that a(., ui(.)) →a(., u(.)) in Lr(Ω). With this at hand, Theorem 2.1 readily finishes the proof of the continuity of T. The proof of the compactness of Tgoes exactly as in Theorem 3.1. Hence, by Schauder’s fixed-point theorem, the map Thas a fixed point u∈Owhich, in turn, is the solution of the Poisson problem we seek. Moreover, usatisfies (3.52). There remains to show how to dispense off the extra hypothesis f∈ L∞(Ω). The idea is to approximate fin Lp s+1/p−2(Ω) by a sequence of functions fj∈L∞(Ω). Granted what we have proved so far, at each step j, we then solve ∆uj−a(x, uj)uj=fjin Ω,Tr uj=g, uj∈Lp s+1/p(Ω).(3.58) By (3.52), (uj)jis a bounded sequence in Xp. Going further, this entails that Vj(x):=a(x, uj(x)) are uniformly bounded in Lr(Ω), for some r>n/2. Since we have that uj=Tg(f+Vjuj) (here Tghas the same meaning as at the end of Section 2), an application of Lemma 2.2 shows that the sequence (uj)j≥1is bounded in Lp s+1/p(Ω) and that there exists a subsequence which converges to some uin the Lp s+1/p(Ω) norm. In turn, this readily yields that usolves (3.38), as desired. Finally, the proof of uniqueness is essentially the same as before. 4. Neumann boundary conditions Building on the sharp linear theory from [9], [30], in this section we analyze the semilinear Poisson problem with nonlinear Neumann boundary conditions.
Semilinear Poisson Problems 387 Theorem 4.1. Let Ω⊂Mbe an arbitrary Lipschitz domain with outward unit conormal ν∈T∗M. Also, fix a nonnegative function a(x, u)∈ L∞(Ω ×R)and consider L=∆−V, where V∈Lr(Ω) for some r>n, V≥0and V>0on a set of positive measure in Ω. Then there exists ε=ε(Ω,a)∈(0,1] with the following significance. If s∈(0,1), p∈(1,∞)satisfy either one of the three conditions in (2.2) (or, respectively, (2.10) if n=2) and if q:= (1 −1/p)−1,λ∈R,0<δ<1, then the Neumann Poisson problem (NP) Lu −a(x, u)u=f∈Lq 1 q−s−1,0(Ω), ∂νu+λ|u|δ=g∈Bq,q −s(∂Ω), u∈Lq 1−s+1 q (Ω), (4.1) has at least a solution which satisfies (4.2) ∀M>0∃K>0 so that fLq 1 q−s−1,0(Ω) +gBq,q −s(∂Ω) ≤M=⇒uLq 1 q−s+1(Ω) ≤K. In the case of linear boundary conditions, i.e., when λ=0, and when the nonlinearity b(x, u):=a(x, u)ualso satisfies 0≤∂ub(x, u)∈L∞(Ω× R), the solution is also unique. Furthermore, when ∂Ω∈C1, we can take any s∈(0,1) and p∈ (1,∞). Proof: If u∈Lq 1−s+1 q (Ω), then Tr u∈Bq,q 1−s(∂Ω) and recall from [33] that |·| δ:Bq,q 1−s(∂Ω) −→ Bq/δ,q/δ (1−s)δ(∂Ω)(4.3) is well-defined and bounded. Furthermore, by allowing an arbitrary small defect of smoothness (for the target space), this operator also becomes continuous and compact; cf. [32, Remark 5, p. 377]. Next, with uas above, we let v:= Tf,g(u) be the unique solution of the linear Poisson problem Lv −a(x, u)v=f∈Lp 1 q−s−1,0(Ω), ∂νv=g−λ|u|δ∈Bq,q −s(∂Ω), v∈Lq 1−s+1 q (Ω). (4.4)
388 M. Dindoˇ s, M. Mitrea That the latter is well posed, granted the current hypotheses, is guaranteed by the results in [30]. We aim at showing that Tf,g :Lq 1−s+1 q (Ω) −→ Lq 1−s+1 q (Ω)(4.5) is a continuous, compact mapping. First, let uj→u0in Lq 1−s+1 q (Ω) and set vj:= Tf,g(uj), v0:= Tf,g(u0). We will show that vj−→ v0in Lq 1−s+1 q (Ω).(4.6) Clearly, it suffices to prove the convergence in (4.6) for a subsequence (still denoted (vj)j≥1). Since, by the estimates of the linear theory and (4.3), there exists C=CaL∞,Ω,s,q,λ >0 such that vjLq 1−s+1 q (Ω) ≤CfLq 1 q−s−1(Ω) +gBq,q −s(∂Ω) +ujδ Lq 1 q−s−1(Ω),(4.7) an easy application of Rellich’s selection lemma, in concert with the uniqueness in the linear theory, gives that for each small ε>0 (and after possibly restricting to a subsequence), vj−→ v0in Lq 1−s+1 q−ε(Ω).(4.8) Next, observe that vj−vk∈Lq 1−s+1 q (Ω) satisfies (4.9) L(vj−vk)=a(x, uj)vj−a(x, uk)vk−→ 0inLq 1−s+1 q (Ω), and ∂ν(vj−vk)−→ 0inBq,q −s(Ω). Hence, once again by virtue of the estimates in the linear theory, vj−vkLq 1−s+1 q (Ω) −→ 0.(4.10) Clearly, this and (4.8) yield (4.6), hence the operator in (4.5) is continuous. Turning attention to the compactness of Tf,g in (4.5), assume that uj∈Lq 1−s+1 q (Ω) is an arbitrary bounded sequence and set vj:= Tf,g(uj), fj:= f+a(x, uj)vj∈Lq 1 q−s−1,0(Ω), gj:= g−λ|uj|δ∈Bq,q −s(∂Ω). (4.11)
Semilinear Poisson Problems 389 Since a(x, uj)∈L∞,vjis a bounded sequence in Lq 1−s+1 q (Ω) (cf. (4.7)) and the embedding L∞(Ω) ·Lq 1−s+1 q (Ω) !→Lq 1 q−s−1,0(Ω)(4.12) is compact, there is no loss of generality (cf. also the discussion pertaining to (4.3)) assuming that (4.13) fjconverges to some f0in Lq 1 q−s−1,0(Ω), and gjconverges to some g0in Bq,q −s(∂Ω). Granted this and observing that Lvj=fj∈Lq 1 q−s−1,0(Ω), ∂νvj=gj∈Bq,q −s(∂Ω), vj∈Lq 1−s+1 q (Ω), (4.14) it follows from the well-posedness of the linear problem that vj→v0in Lq 1−s+1 q (Ω), where v0is the unique solution of (4.14) with fj,gjreplaced, respectively, by f0and g0. This proves that the operator in (4.5) is also compact. Note that if we let BRstand for the closed ball (centered at the origin) of radius Rin the space Lq 1−s+1 q (Ω) then, by (4.7), the operator Tf,g :BR→BRis well-defined, continuous and compact, provided Ris large enough. At this stage, Schauder’s fixed-point theorem applies and takes care of the solvability of the Neumann Poisson problem (4.1), as well as, the accompanying estimate (4.2). Finally, granted that λ= 0 and the extra condition 0 ≤∂u[a(x, u)u]∈ L∞(Ω ×R), uniqueness for the problem (4.1) can be readily reduced to its linear version, i.e., when a(x, u) is independent of u. The reasoning when ∂Ω∈C1is similar and this finishes the proof. 5. Nonlinearities with sublinear growth We retain our standard hypotheses on Mand Ω made in Theorem 3.1. As usual, set L=∆−V. The main result of this section is the following.
390 M. Dindoˇ s, M. Mitrea Theorem 5.1. Let N(x, u)be a Carath´eodory function such that |N(x, u)|≤k1(x)+k2(x)|u(x)|δ,for some 0≤δ<1,(5.1) where 0≤kj∈Lqj(Ω) with qj>n/2,j=1,2. Also, assume that V satisfies (2.1). Then, if s,psatisfy either one of the three conditions in (2.2) and if (2.3) holds, the Poisson problem with Dirichlet boundary condition Lu −N(x, u)=f∈Lp s+1/p−2(Ω), Tr u=g∈Bp,p s(∂Ω),u∈Lp s+1/p(Ω), (5.2) has at least one solution. On the other hand, if q:= (1−1/p)−1and 0≤V∈Lr(M)with r>n, V>0on a set of positive measure in Ω, then the Poisson problem with Neumann boundary condition Lu −N(x, u)=f∈Lq 1/q−1−s,0(Ω), ∂νu=g∈Bq,q −s(∂Ω),u∈Lq 1−s+1/q(Ω), (5.3) has at least one solution. Similar results hold in dimension n=2granted that (2.10) is used in lieu of (2.2) and q1,q 2>(1 −|1 2p−s 2|)−1in the case of Dirichlet boundary conditions. Finally, when ∂Ω∈C1, one can simply take 1<p<∞and 0<s<1 in place of (2.2),(2.10). Proof: We shall only deal with the case of (5.2), since the case of Neumann boundary conditions is similar. To this effect, consider first the case when sp < n −1, and define the map T:Lnp/(n−1−sp)(Ω) → Lnp/(n−1−sp)(Ω) by taking v:= Tu to be the unique solution of Lv =f+N(x, u)inΩ, Tr v=g∈Bp,p s(∂Ω),v∈Lp s+1/p(Ω), (5.4) for each u∈Lnp/(n−1−sp)(Ω). Since by (5.1) and (the proof of) Lemma 2.2, N(x, u(x)) belongs to Lp s+1/p−2(Ω), Theorem 2.1 gives us that T is well defined. Furthermore, there exists C>0 such that (5.5) TuLnp/(n−1−sp)(Ω) ≤C(fLp s+1/p−2(Ω) +N(x, u)Lp s+1/p−2(Ω) +gBp,p s(∂Ω)). The estimate (5.1) also gives us N(x, u(x))Lp s+1/p−2(Ω) ≤K(1 + uδ Lnp/(n−1−sp)(Ω))(5.6)
Semilinear Poisson Problems 391 for some K=K(k1,k 2)>0 independent of u. In particular, if we take R>0 big enough such that C(fLp s+1/p−2(Ω) +gBp,p s(∂Ω) +K(1 + Rδ)) ≤R,(5.7) then Tmaps the ball {h∈Lnp/(n−1−sp)(Ω); hLnp/(n−1−sp)(Ω) ≤R} into itself. Next, the fact that Tis continuous and compact is seen essentially as in Theorem 3.1. Hence, by Schauder’s fixed-point theorem, the map T has a fixed point Tu =u. Since Tu solves (5.4) we also have that u=Tu ∈Lp s+1/p(Ω) and Tr u=Tr(Tu)=g. This concludes the proof of the case sp < n −1. The remaining cases discussed in the statement of the theorem are dealt with similarly and are somewhat easier; we omit the straightforward details. Example 5.1. Our first example illustrating Theorems 3.1, 4.1 is the boundary problem (DP)± ∆u±|u|qu=f∈Lp s+1 p−2(Ω), Tr u=g∈Bp,p s(∂Ω), u∈Lp s+1 p (Ω). (5.8) Theorem 3.1 gives that for the choice of the negative sign (5.8) is solvable for all p,ssatisfying (2.2) and qsuch that 0≤q< np n−1−sp ·min 2 n,1−n−1−sp np if sp < n −1,(5.9) and q>0ifsp ≥n−1, granted that n≥4. Explicit conditions, modeled upon (3.8)–(3.10), can be also given for n=2,3. A case which is not directly amenable to the analysis we have developed so far corresponds to the choice of a positive sign in (5.8). In this situation, a partial answer can be obtained by relying on Theorem 2.1 and proceeding much as in [32, Chapter 6] (parenthetically, it should be pointed out that this approach works only for small data). Finally, the range −1<q<0 is covered by Theorem 4.1 for either choice of the sign. Furthermore, for (DP)−one can also establish uniqueness in this range. This can be proved by adapting the argument used in [8, Example 4.5]. Next we discuss a two dimensional curvature equation (cf. also [7], [8] for a different context).
392 M. Dindoˇ s, M. Mitrea Example 5.2. Let Ω ⊂Mbe a connected Lipschitz domain on a two dimensional compact manifold M, equipped with a Riemannian metric g, whose Gauss curvature is k(x). The problem to be addressed is that of conformally altering gto a new metric ˜gin Ω with a prescribed Gaussian curvature ˜ k(x)≤0 on Ω, and such that ˜g|∂Ω=g|∂Ω. A well known formula, whose proof can be found in, e.g., [36, Appendix C], states that if gand ˜gare conformally related, i.e., ˜g=e2ug,(5.10) then the curvatures ˜ kand ksatisfy ˜ k(x)=e−2u(−∆u+k(x)),(5.11) where ∆ = ∆gis the Laplace-Beltrami operator associated with the original metric g. Thus, in view of (5.10)–(5.11), matters come down to solving the nonlinear PDE ∆u=k(x)−˜ k(x)e2u,u|∂Ω=0.(5.12) Next we study conditions guaranteeing that the nonlinear Dirichlet problem just formulated satisfies the assumptions of Theorem 3.2. First, observe that we can rewrite (5.12) as ∆u−−˜ k(x)e2u−1 uu=k(x)−˜ k(x),(5.13) which, in the notation employed in Theorem 3.2, translates into a(x, u):=−˜ k(x)e2u−1 u,and f(x):=k(x)−˜ k(x).(5.14) Clearly, a(x, u)≥0 since, by assumption, ˜ k(x)≤0. Also, both (3.39) and (3.40) hold provided ˜ k∈Lr(Ω) for some r>(1 + 1 2p−s 2)−1. Hence, if we assume that the original metric tensor gsatisfies (1.1), as well as, g∈Lp s+1 p (Ω),(5.15) then k∈Lp s+1 p−2(Ω), since it has the same smoothness as ∇2g. Thus, granted these conditions, the existence part of Theorem 3.2 is applicable. In fact, the uniqueness condition in Theorem 3.2 also holds, since b(x, u):=−˜ k(x)(e2u−1), and therefore ∂ ∂ub(x, u)=−2˜ k(x)e2u≥0.(5.16)
Semilinear Poisson Problems 393 The conclusion is that for any s,psatisfying (2.10), and sp ≥1, we can uniquely extend gsatisfying (5.15) and (1.1) conformally inside Ω to a new metric ˜gwith a prescribed curvature ˜ k≤0, ˜ k∈Lr(Ω), r>(1 + 1 2p−s 2)−1. A discussion of the case when k,˜ kare locally bounded in R2and entire solutions (with prescribed asymptotic behavior at ∞) are sought, is contained in [22]. 6. Other types of estimates and function spaces In this section we consider the regularity of the solution of Poisson type problems in terms of the so called nontangential maximal operator. Recall that for a function u∈L∞ loc(Ω), the latter is defined by u∗(x) := sup y∈γ(x) |u(x)|,for each x∈∂Ω.(6.1) Here γ(x) is the nontangential approach region with vertex at the boundary point x; cf. [27], [7]. In what follows, we make the assumption that ∪x∈∂Ωγ(x)=Ω. An issue which arises naturally in this context is that of providing an intrinsic description of the space {∆u;u∗∈Lp(∂Ω)}.(6.2) While at the present time this question remains open, our next definition identifies a rich linear subspace of (6.2). Specifically, let Ω be a Lipschitz subdomain of the Riemannian manifold Mand fix n/2<ρ≤∞,1≤q≤∞. We set (6.3) Lq ρ(Ω) := f= j ujvj;u∗ j∈Lq(∂Ω),v j∈Lρ(Ω), j u∗ jLq(∂Ω)vjLρ(Ω) <∞ and equip it with the norm (6.4) fLp ρ(Ω) := fL1(Ω) + inf j u∗ jLq(∂Ω)vjLρ(Ω);f= j ujvja.e. on Ω .
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Semilinear Poisson Problems 403 Martin Dindoˇs: Department of Mathematics Cornell University 310 Malott Hall Ithaca, NY, 14853 U.S.A. E-mail address:[email protected] Marius Mitrea: Department of Mathematics University of Missouri at Columbia Columbia, MO 65211 U.S.A. E-mail address:[email protected] Rebut el 17 de setembre de 2001.