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The Dirichlet Problem for Certain Discrete Structures

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The Dirichlet Problem for Certain Discrete Structures The Dirichlet problem for certain discrete structures Abtin Daghighi U.U.D.M. Project Report 2005:4 Examensarbete i matematik, 20 poäng Handledare och examinator: Christer Kiselman April 2005 Department of Mathematics Uppsala University The Dirichlet problem for certain discrete structures Abtin Daghighi Abstract The paper reviews some of the early ideas behind the development of the theory of discrete harmonic functions. The connection to random walks is pointed out. Then a current and more general construction using weight functions is described. Discrete analogues of the Laplace operator are defined for Zn and a discrete planar hexagonal structure H. Discrete ana- logues of the Dirichlet problem and Poisson’s equation are formulated and existence and uniqueness of bounded solutions is proved for the finite case and also for a certain case of infinite sets in Zn. Explicit discrete analogues of the Green function are presented for Z3 and Z[i] equipped with a certain discrete calculus, which is also briefly presented. 1 Contents 1 Some results from the continuous case 3 1.1 The Dirichlet principle . 3 1.2 The Laplace equation and the maximum principle . 4 1.3 The method of Perron . 6 2 Introduction to the discrete case 7 3 The Dirichlet problem for subsets of Zn 8 3.1 Separation of variables . 12 4 Connection to martingales in two dimensions 14 5 A discrete analogue of Poisson’s equation for Z3 14 6 A discrete calculus for Z[i] 17 6.1 A discrete Green function for Z[i].................. 20 7 A more general Laplace operator and an analogue to the method of Perron 24 8 A hexagonal planar structure 27 9 References 28 10 Notation 29 2 1 Some results from the continuous case In this section some of the theory of partial differential equations which is used in connection with the Dirichlet problem will be presented. In some cases the ideas behind these results have been used to formulate and prove discrete analogues for functions defined on discrete structures, which is the main topic of this paper. Therefore it might be useful to first review some of the theory of this section to better understand the background of the chapters that follow. 1.1 The Dirichlet principle The Dirichlet problem for Laplace’s equation on Ω ⊂ Rn consists of finding a function u which satisfies the following conditions ( ∆u(x) = 0, x ∈ Ω; (∗) u(x) = g(x), x ∈ ∂Ω. The inhomogeneous Laplace equation is known as Poisson’s equation. The Dirichlet problem for Poisson’s equation is to find a function u ∈ C2(Ω), such that for a given function, f ∈ C0(Ω), the following conditions hold ( ∆u(x) = f(x), x ∈ Ω; (∗∗) u(x) = 0, x ∈ ∂Ω. Note that by adding a solution of the Dirichlet problem for the Laplace equa- tion on the same set one gets a solution for an inhomogeneous continuous bound- ary condition. The problem (∗) can be considered as a minimization problem according to the following principle. The Dirichlet principle. Let S = {w ∈ C2(Ω), w = g on ∂Ω} and let 1 Z I(w) = |∇w|2dx. 2 Ω A function u ∈ S is a solution to the Dirichlet problem for Laplace’s equation on Ω, (∗), if and only if I(u) = min I(w). w∈S 3 Proof. Suppose u is a solution of (*). Then for any w ∈ S, Z Z Z 0 = ∆u(u − w)dx = − |∇u|2dx + ∇u∇wdx ≤ Ω Ω Ω Z 1 Z − |∇u|2dx + |∇u|2 + |∇w|2 dx = Ω 2 Ω 1 Z 1 Z − |∇u|2dx + |∇w|2dx. 2 Ω 2 Ω Therefore Z Z |∇u|2dx ≤ |∇w|2dx, Ω Ω and since w ∈ S was arbitrary I(u) = min I(w). w∈S Next suppose that w ∈ S minimizes I. Let v ∈ C2 be such that v ≡ 0 for x ∈ ∂Ω. Then u + v ∈ S for all . Define J() ≡ I(u + v). Since u minimizes I, J must have minimun for = 0. Now Z J() = |∇u|2 + 2∇u∇v + 2|∇v|2 dx, Ω implies Z J 0() = ∇u∇v + 2|∇v|2dx, Ω so that Z Z ∂u Z J 0(0) = − (∆u)vdx + vdσ(x) = − (∆u)vdx Ω ∂Ω ∂ν Ω Now J 0(0) = 0 implies Z 0 = (∆u)vdx, Ω for all v ∈ C2(Ω) such that v = 0 for x ∈ ∂Ω, thus ∆u = 0. This completes the proof. 1.2 The Laplace equation and the maximum principle What follows is the derivation of a fundamental solution for the Laplace equation. Make the ansatz of a radial symmetric (around the point ξ ∈ Rn) harmonic function, v(r) ∈ C2(Ω), r = |x − ξ|, as a solution for (∗). The chain rule gives n − 1 ∆v(r) = v00(r) + v0(r) = 0, r 4 which implies 0 1−n v (r) = Ar Ar2−n v(r) = 2−n , n 6= 2 v(r) = A ln r, n = 2 Introducing the ball, B(ξ, ρ) ⊂ Ω, with surface, S(ξ, ρ), and using the Green identity Z Z ∂u ∂v (v∆u − u∆v) dx = v − u dx, Ω ∂Ω ∂n ∂n gives Z Z ∂u ∂v Z ∂u ∂v (v∆u − u∆v) dx = v − u dS + v − u dS = Ω ∂Ω ∂n ∂n S(ξ,ρ) ∂n ∂n Z ∂u ∂v Z Z v − u dS − v(ρ) ∆udx − v0(ρ) udS ∂Ω ∂n ∂n B(ξ,ρ) S(ξ,ρ) Z ∂u ∂v → v − u dS + Aωnu(ξ), ∂Ω ∂n ∂n dv(ρ) 0 0 where in the second equality the following is used: dn =n ˆ · v (r) = −v (ρ), and 2πn/2 n where ωn = Γ(n/2) , i.e., the surface area of the unit sphere in R , A ∈ R. So by choosing the constant A ∈ R to be 1 one gets ωn Z ∂u(x) ∂v Z u(ξ) = − v(x, ξ) − u(x) dS + v(x, ξ)∆u(x)dx. ∂Ω ∂n ∂n Ω ∞ In particular, replacing u(x) by φ(x) ∈ C0 (Ω) shows that ∆v(x, ξ) = δ(ξ), in the distribution sense, which makes it a fundamental solution of the Laplace operator. Further if there exists a fundamental solution say, w(x, ξ), such that w(x, ξ) = 0 for x ∈ ∂Ω, ξ ∈ Ω, then replacing v by w in the last equation yields a solution to (∗) and also to (∗∗), namely Z ∂w Z u(ξ) = g(x) (x, ξ)dS + w(x, ξ)f(x)dx. ∂Ω ∂n Ω A function f satisfying ∆f(x) ≥ 0 for all x ∈ Ω is called subharmonic on Ω. We shall now show a version of a result known as the maximum principle for subharmonic functions. This will be done in two steps. In the first step consider a function g ∈ C2(Ω) ∩ C0(Ω) satisfying ∆g > 0 in ◦ ∂2g Ω. If g assumes a maximum at ξ ∈ Ω then 2 ≤ 0 at ξ for all k , which implies ∂xk ∆g ≤ 0. This is a contradiction to the assumption ∆g > 0 in Ω. Thus max g = max g. Ω ∂Ω The next step is to consider u ∈ C2(Ω) ∩ C0(Ω) satisfying ∆u ≥ 0 in Ω. Let > 0, and g = |x|2. Then ∆g > 0, and so the result of the first step applies to 5 u + g. Hence one gets max(u + g) = max(u + g), Ω ∂Ω which implies max u ≤ max(u + g) = max(u + g) ≤ max u + max g. Ω Ω ∂Ω ∂Ω ∂Ω Letting → 0 gives max u ≤ max u. Ω ∂Ω This is the required result. 1.3 The method of Perron The method of Perron for findning solutions for (∗) uses the space of subharmonic functions over Ω, which is denoted by SH(Ω) = {u ∈ C2(Ω); ∆u(x) ≥ 0, x ∈ Ω}. Now for a given g ∈ C0(∂Ω), define 0 SHg(Ω) = {u ∈ C (Ω) ∩ SH(Ω); u(x) ≤ g(x), x ∈ ∂Ω}. Further define wg = supu∈SHg(Ω) u(x). Since the results from the continuous case are only given to provide a preliminary background and are not the main topic of this paper, only a brief summary of the method of Perron will be given. Several statements will thus be given without proof. Define ( uB(x) = u(x), x ∈ Ω, x∈ / B(ξ, r), ∆uB(x) = 0, x ∈ B(ξ, r), 0 where u ∈ C (Ω), B(ξ, r) ⊂ Ω, r > 0, ξ ∈ Ω. Such a uB will have the following property u(x) ≤ uB(x), x ∈ Ω, uB ∈ SH(Ω). The existence and uniqueness of such a uB is stated here without proof; cf. John (1982). The next step is to show that wg is harmonic in Ω. To do this, start by constructing a sequence of functions that are harmonic in B(ξ, r) and lie between fixed bounds. Let xk, k = 1, 2,..., be a sequence in B(ξ, r0) ⊂ Ω, r0 < r. Then j j k k there exists functions, uk ∈ SHg(Ω), such that limj→∞ uk(x ) = wg(x ), since wg j j j is defined as the supremum. Define u (x) = max(u1(x), . , uj(x)) ∈ SHg(Ω). j j Now u (x) ≥ uk(x) for x ∈ Ω, j ≥ k. Then also j k k (∗ ∗ ∗) limj→∞ u (x ) = wg(x ), for all k. Replacing if necessary uj by max(inf g, uj) provides a sequence uj(x) such j j j that inf g(x) ≤ u (x) ≤ sup g(x), x ∈ Ω. Then replacing u by uB, guarantees that each function of the series is harmonic in B(ξ, r), and the limit is the same 6 as the original sequence.
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