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The Dirichlet Problem by Variational Methods THE DIRICHLET PROBLEM BY VARIATIONAL METHODS WOLFGANG ARENDT AND DANIEL DANERS Abstract. Let Ω ⊂ RN be an bounded open set and ϕ ∈ C(∂Ω). Assume that ϕ has an extension Φ ∈ C(Ω)¯ such that ∆Φ ∈ H−1(Ω). Then by the Riesz representation theorem there exists a unique 1 ′ u ∈ H0 (Ω) such that − ∆u =∆Φ in D(Ω) . We show that u + Φ coincides with the Perron solution of the Dirichlet problem ∆h = 0, h|∂Ω = ϕ. This extends recent results by Hildebrandt [Math. Nachr. 278 (2005), 141–144] and Simader [Math. Nachr. 279 (2006), 415– 430], and also gives a possible answer to Hadamard’s objection against Dirichlet’s principle. 1. The main result and its consequences Let Ω be a bounded open set in RN with boundary ∂Ω. Let ϕ ∈ C(∂Ω). We consider the Dirichlet problem D(ϕ, Ω) h ∈H(Ω) ∩ C(Ω)¯ , h|∂Ω = ϕ, where H(Ω) := {u ∈ C2(Ω): ∆u =0} denotes the space of all harmonic functions. It follows from the maximum principle that D(ϕ, Ω) has at most one solution. We say that Ω is Dirichlet regular if for each ϕ ∈ C(∂Ω) there exists a solution h of D(ϕ, Ω). Such a solution will be called a classical solution in what follows. If Ω is not Dirichlet regular, then there always exists a generalised solution of D(ϕ, Ω) namely the Perron solution hϕ (see Section 2 for the definition). Moreover, if D(ϕ, Ω) has a classical solution h, then h = hϕ. There is an elaborate theory describing the points z ∈ ∂Ω for which limx→z hϕ(x) = ϕ(z) for all ϕ ∈ C(∂Ω), those are called the regular points (this is equivalent to the existence of a barrier at z, see e.g. Kellogg [12, Section XI.17]). Our aim is to express that hϕ = ϕ on ∂Ω in a weak sense instead by 1 2 pointwise convergence. We denote by H (Ω) := {u ∈ L (Ω): Dju ∈ 2 1 L (Ω), j =1,...,N} the first Sobolev space and by H0 (Ω) the closure Date: June 29, 2007. 2000 Mathematics Subject Classification. 35J05, 31B05. Appeared in Bulletin of the London Mathematical Society, 40(2008), 51–56. doi:10.1112/blms/bdm091. 1 2 WOLFGANGARENDTANDDANIELDANERS of the test functions D(Ω) in H1(Ω). Finally, denote by D(Ω)′ the space of all distributions on Ω. Given h ∈H(Ω) we say that h = ϕ on ∂Ω weakly ¯ 1 if ϕ has an extension Φ ∈ C(Ω) such that u := h − Φ ∈ H0 (Ω). This implies that −∆u = ∆Φ in the sense of distributions, that is, (1) h∆Φ,vi := Φ∆vdx = ∇u∇vdx ZΩ ZΩ for all v ∈D(Ω). As a consequence 1/2 (2) |h∆Φ,vi| ≤ c |∇v|2 dx ZΩ 2 1/2 for all v ∈ D(Ω) where c = ( Ω |∇u| dx) . By virtue of Poincar´e’s 2 1/2 R 1 inequality, ( Ω |∇v| dx) defines an equivalent norm on H0 (Ω). Thus 1 (2) means thatR ∆Φ has a continuous extension from D(Ω) to H0 (Ω). −1 1 ′ We keep the notation for the extension ∆Φ ∈ H (Ω) := H0 (Ω) . Our main result is the following. Theorem 1.1. Let ϕ ∈ C(∂Ω) and assume that ϕ has an extension ¯ −1 1 Φ ∈ C(Ω) such that ∆Φ ∈ H (Ω). Let u ∈ H0 (Ω) be the unique solution of Poisson’s equation (3) − ∆u = ∆Φ in D(Ω)′. Then u +Φ= hϕ is the Perron solution of the Dirichlet problem. 1 As seen before (3) has a unique solution u ∈ H0 (Ω). It follows that ∆(u + Φ) = 0 in the sense of distributions and hence u + Φ ∈ H(Ω), see [7, Chapter II §3, Proposition 1] or [13, Appendix 34, Theorem 14]. Our main point is to prove that u +Φ = hϕ which will be done in 1 Section 2. The Riesz representation theorem also says that u ∈ H0 (Ω) is the unique minimiser of 1 2 1 2 1 |∇u| dx −h∆Φ,ui = min |∇v| dx −h∆Φ,vi: v ∈ H0 (Ω) 2 ZΩ n2 ZΩ o 1 (see [5, Th´eor`eme V.6]). Hence if Φ ∈ H (Ω), then hϕ is actually the solution of Dirichlet’s principle, which can now be formulated as follows. Corollary 1.2. Assume that ϕ has an extension Φ ∈ C(Ω)¯ ∩ H1(Ω). Then hϕ is the unique minimiser of 2 1 1 (4) min |∇w| dx: w ∈ H (Ω), w − Φ ∈ H0 (Ω) . nZΩ o 1 Proof. Substitute v ∈ H0 (Ω) in (4) by w = v + Φ. THE DIRICHLET PROBLEM BY VARIATIONAL METHODS 3 Hildebrandt [10, Theorem 1] shows that the minimiser h of (4) sat- isfies lim h(x)= ϕ(z) x→z for all regular points z ∈ ∂Ω. Thus, if Ω is Dirichlet regular, it follows that h = hϕ, which is also proved by Simader [15, Theorem 1.6] or [7, Proposition II.7.10]. However, even if Ω is Dirichlet regular, not every ϕ ∈ C(∂Ω) has an extension Φ ∈ C(Ω)¯ ∩ H1(Ω). This follows from Hadamard’s famous example [9] on the unit disc D of R2. Let ∞ ϕ(eiθ)= 2−n cos(22nθ). Xn=1 Then the classical solution of D(ϕ, D) is given by ∞ iθ 22n −n 2n hϕ(re )= r 2 cos(2 θ) Xn=1 (see e.g [6, page 179–180]), and the energy of hϕ is 2 |∇hϕ| dx = ∞, ZD 1 hence hϕ 6∈ H (Ω). As a consequence of Theorem 1.1, for this ϕ there 1 exists no extension Φ ∈ C(Ω)¯ ∩H (Ω) such that Φ|∂D = ϕ. Indeed, then 1 (3) would imply that also hϕ =Φ+ u ∈ H (Ω). We refer to Maz’ya and Shaposhnikova [14, §123] for the interesting history of Hadamard’s example. On the other hand, the condition that ϕ has an extension Φ ∈ C(Ω)¯ such that ∆Φ ∈ H−1(Ω) is weaker than Φ ∈ H1(Ω). Indeed, if D(ϕ, Ω) has a classical solution h ∈ H(Ω) ∩ C(Ω),¯ then ∆Φ = ∆h ∈ H−1(Ω) since ∆h = 0. Remark 1.3. Let Φ ∈ C(Ω).¯ The following assertions are equivalent. (i) ∆Φ ∈ H−1(Ω); 1 (ii) Φ ∈ Hloc(Ω) and there exists c> 0 such that 1/2 ∇Φ∇vdx ≤ c |∇v|2 dx ZΩ ZΩ for all v ∈D (Ω). −1 ∞ In fact, if ∆Φ ∈ H (Ω), then u+Φ = hϕ ∈ C (Ω) where ϕ = Φ|∂Ω and 1 1 u ∈ H0 (Ω) solves (3). Thus Φ ∈ Hloc(Ω). Now (2) implies the estimate. Conversely, (ii) implies (i) since ( |∇v|2 dx)1/2 is an equivalent norm on 1 1 H0 (Ω). However, as we saw aboveR (ii) does not imply that Φ ∈ H (Ω). We note two further consequences for Poisson’s equation. Let C0(Ω) := {u ∈ C(Ω):¯ u|∂Ω =0}. −1 Corollary 1.4. Let v ∈ C0(Ω) such that ∆v ∈ H (Ω). Then v ∈ 1 H0 (Ω). 4 WOLFGANGARENDTANDDANIELDANERS 1 Proof. Let Φ := v so that ϕ = Φ|∂Ω = 0. Let u ∈ H0 (Ω) be the solution of (3). Then by Theorem 1.1 we get u + v = hϕ = 0. The following result extends [2, Lemma 2.2]. Corollary 1.5. Assume that Ω is Dirichlet regular. Let f ∈ Lp(Ω) with N/2 < p ≤ ∞ if N ≥ 2, and let f be a bounded Borel measure on Ω if N = 1. Then there exists a unique solution u ∈ C0(Ω) of the Poisson equation −∆u = f in D(Ω)′. 1 q N 1 Proof. Since H0 (Ω) ⊂ L (Ω) whenever q < N−2 if N ≥ 2 and H0 (Ω) ⊂ L∞(Ω) in the case N = 1 we have Lp(Ω) ⊂ H−1(Ω) if p > N/2 and M(Ω) ⊂ H−1(Ω) if N = 1, where M(Ω) denotes the space of all bounded signed Borel measures on Ω. It follows from the Riesz 1 representation theorem that there exists a unique u ∈ H0 (Ω) such that ∇u∇vdx = fvdx ZΩ ZΩ 1 for all v ∈ H0 (Ω), that is, −∆u = f in D(Ω)′. Let Φ(x)= Ω f(y)E(x − y) dy if N ≥ 2 and Φ(x)= Ω E(x − y)df(y) if N = 1, whereR E is the Newtonian potential. ThenR Φ ∈ C(Ω).¯ This ′ p RN R follows from the fact that E ∈ Lloc( ) if N ≥ 2 and E ∈ C( ) if ′ N = 1. Moreover ∆Φ = f in D(Ω) . Let ϕ = Φ|∂Ω. It follows from Theorem 1.1 that hϕ =Φ+ u. Since Ω is Dirichlet regular, hϕ ∈ C(Ω)¯ and hϕ|∂Ω = ϕ. Thus u ∈ C0(Ω). In order to prove uniqueness, let ′ u ∈ C0(Ω) such that −∆u = f in D(Ω) . Then h = u + Φ ∈ C(Ω)¯ is a classical solution of D(ϕ, Ω). So uniqueness follows from the uniqueness of the classical solution of the Dirichlet problem D(ϕ, Ω). We conclude this section commenting on weak solutions of the Dirich- let problem. Remark 1.6. Let ϕ ∈ C(∂Ω). We call h ∈ H(Ω) a weak solution of D(ϕ, Ω) if ϕ has an extension Φ ∈ C(Ω)¯ such that ∆Φ ∈ H−1(Ω) and 1 h − Φ ∈ H0 (Ω). (a) It is not obvious that weak solutions are unique. Theorem 1.1 gives a positive answer: since h = hϕ and since the Perron solution hϕ is unique there is at most one solution. (b) If Ω is Dirichlet regular, then ϕ ∈ C(∂Ω) has an extension Φ ∈ −1 C(Ω)¯ with ∆Φ ∈ H (Ω), namely the Perron solution hϕ.
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