SOLUTION OF THE DIRICHLET PROBLEM WITH A VARIATIONAL METHOD. 1. Dirichlet integral

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1 SOLUTION OF THE DIRICHLET PROBLEM WITH A VARIATIONAL METHOD CRISTIAN E. GUTIÉRREZ FEBRUARY 3, 29. Dirichlet integral Let f C( ) with open and bounded. Let H = {u C ( ) : u = f on } and D(u) = Du(x) 2 dx. The objective is to prove that with minimizers of D(u) over H one can solve the Dirichlet problem in. We assume that f satisfies the following property: there exists v C ( ) such that v = f on. This is not a restriction to solve the Dirichlet problem with this approach because if f C( ) then by the Wierstrass approximation theorem there exist polynomials f k in R n such that f k converge uniformly to f on. If we can solve the Dirichlet problem with data u k = f k on, then by the maximum principle u k u uniformly in for some u and therefore u is harmonic in and has boundary values f. If u C 2 () C ( ) satisfies u = in and u = f on, then D(u) D(v), v H. That is, the Dirichlet integral is minimized by the solution of the Dirichlet problem. This follows writing g = v u with v H D(v) = D(g + u) = D(g) + 2D(g, u) + D(u) = D(g) + D(u) + 2 gd ν u dσ 2 = D(g) + D(u) D(u) g u dx from the first Green formula. There are continuous functions f C( ) such that the solution u of the Dirichlet problem with data f satisfies D(u) = +. An

2 2 C. E. GUTIÉRREZ FEBRUARY 3, 29 example is the function u(r, θ) = r n! cos(n!θ) k= that is harmonic in the unit disk, n 2 it has boundary values f (θ) = cos(n!θ) k=, and D(u) = +. n 2 We have that D(λu + v) = λ 2 D(u) + 2λD(u, v) + D(v) for all λ, and so D(u)D(v) D(u, v). Therefore u D = D(u) /2 defines a quasi norm in H, that is, D satisfies the triangle inequality and λu D = λ u D. 2. Poincaré inequality Let H = {u C ( ) : u = f on } and H = {u C ( ) : u = on }. Lemma. There exists a constant C >, depending only on the domain, such that for all w H. w(x) 2 dx C Dw(x) 2 dx, Proof. Since is bounded, Q = [, a] [, a]. Let w(x, y) = w(x, y) for (x, y) and w(x, y) = for (x, y) Q \. We assume that the intersection of with each vertical line is finite union of open intervals. Then the function w(x, ) is Lipschitz and therefore absolutely continuous. So we can write w(x, y) = y w y (x, ξ) dξ. Then squaring and using Cauchy-Schwartz we get w(x, y) 2 (y + a) y and integrating this inequality in x yields a Now integrating in y yields a a w(x, y) 2 dx 2a w y (x, ξ) 2 dξ 2a a a w(x, y) 2 dxdy 4a 2 a a w y (x, ξ) 2 dξ dx. a w y (x, ξ) 2 dξ, w y (x, ξ) 2 dξ dx. From the definition of w and since w y (x, y) = when (x, y) Q \, the lemma then follows with C = 4a 2.

3 SOLUTION OF THE DP WITH A VARIATIONAL METHOD February 3, 29 3 Remark 2. It is clear that if [a, b] [c, d], then the estimate holds with C = (b a)(d c). In higher dimensions, one obtains in the same way that if R with R an n-dimensional interval, then the lemma holds with C = R. 3. Solution to the Dirichlet problem By our assumption the set H, and therefore inf H D(u) = L <. So there exists a sequence (possibly not unique) v k H such that D(v k ) L as k. We call v k a minimizing sequence, and the objective is to construct with v k a harmonic function φ in such that φ = f on. Lemma 3. For each w H, we have lim D(v k, w) =, k where v k is the minimizing sequence. Moreover, if D(w k ) M with w k H, then D(v k, w k ) as k. Proof. We have v k + ɛw H for all ɛ and so L D(v k + ɛw) = D(v k ) + 2ɛD(v k, w) + ɛ 2 D(w), and the minimum of the right hand side is attained when ɛ = D(v k, w) D(w) yields L D(v k ) D2 (v k, w). D(w) Therefore and the lemma follows. D(v k, w) (D(v k ) L) /2 D(w) /2, which Lemma 4. The minimizing sequence v k is a Cauchy sequence in the norm L 2 () + D. Proof. Let w = v k v m. We have D(v k ) = D(v m + w) = D(v m ) + 2D(v m, w) + D(w). So D(w) D(v k ) D(v m ) + 2 D(v m, w) and therefore from the previous lemma, w D as k, m. From the Poincaré inequality, w L 2 () C w D and the lemma follows. Given x and B ρ (x) we let φ k (x, ρ) = v k (z) dz. B ρ (x)

4 4 C. E. GUTIÉRREZ FEBRUARY 3, 29 Lemma 5. Let K be closed K, and fix ρ < dist(k, c ). Then φ k (x, ρ) are continuous and converge uniformly for x K to a function φ(x, ρ) for each ρ. Proof. The functions φ k (x, ρ) are clearly continuous. We write φ k (x, ρ) φ m (x, ρ) = (v k (z) v m (z)) dz B ρ (x) /2 v k (z) v m (z) 2 dz B ρ (x) v /2 k v m L 2 (), B ρ (x) so the sequence φ k (x, ρ) is uniformly Cauchy and the lemma follows. Lemma 6. The function φ(x, ρ) is independent of ρ, i.e., φ(x, ρ) = φ(x). Proof. Fix x and let ρ < ρ 2 with B ρi (x). Suppose we are in dimension two and let ( log ρ + ( x z 2 )), x z ρ 2π ρ 2 2 ρ 2 ρ 2 ( 2 w(x) = x z log + ( )) x z 2, ρ 2π ρ 2 2 ρ 2 < x z ρ 2 2 ρ 2 < x z. We have We have that w H and ( (x z) ), x z ρ 2π ρ 2 ρ 2 ( 2 ) Dw(x) = (x z), ρ 2π ρ 2 x z 2 < x z ρ 2 2 ρ 2 < x z. πρ 2 w(x) = πρ 2 2, x z < ρ πρ 2 2, ρ < x z < ρ 2 ρ 2 < x z. To apply the first Green formula we remove the places where the Laplacian of w is discontinuous. We set ɛ = \ ( {z : ρ ɛ < x z < ρ + ɛ} {z : ρ 2 ɛ < x z < ρ 2 + ɛ} ).

5 SOLUTION OF THE DP WITH A VARIATIONAL METHOD February 3, 29 5 By the first Green formula applied in ɛ we have Dv k Dw dx + v k w dx = v k (z) η w(x) dσ(z). ɛ ɛ ɛ From the form of the gradient of w, the right hand side of the last identity tends to zero as ɛ, and so we obtain = D(v k, w) + v k w dx = D(v k, w) + v k (z) dz v k (z) dz, B ρ (x) B ρ2 (x) and letting k the lemma follows. Theorem 7. The function φ is harmonic in and φ = f on. Proof. We claim that if,, then In fact, we have and integrating in x φ k (x, a) dx = = φ k (x, a) dx = φ k (x, a) = v k (z) dz = B a (x) = φ k (x, b) dx. v k (z + x)dz dx = v k (u)du dz = B b (x +z) φ k (z, b) dz, which proves the claim. Letting k we obtain φ(x) dx = v k (z + x) dz, φ(x) dx, and since φ is continuous, letting b yields φ(x) dx = φ(x ), v k (z + x)dx dz φ k (x + z, b) dz that is, φ satisfies the mean value property in and the first part of the theorem is then proved. We shall prove in dimension two that lim φ(x) = f (x ), x x,x x,

6 6 C. E. GUTIÉRREZ FEBRUARY 3, 29 under the following assumption on : there exists R > such that {z : z x = ρ} for all x and for all ρ R. Let x and consider the ball B h (x ), and let x B h/2 (x ). Let σ = dist(x, ). There exists x such that x x = σ. We have σ h/2, x x < h, and B 3σ/2 (x ). We write f (x ) φ(x) f (x ) f (x ) + f (x ) v k (x) + v k(x) v k (z) dz + = I + II + III + IV. v k (z) dz φ(x) Since f is continuous, I is small for h small. II is small since v k C( ) and v k = f on. III is also small for h small since σ h/2. We estimate IV. Consider the circle C ρ (x ) for ρ 3σ/2. If 3σ/2 < R, where R appears in the condition on, then C ρ (x ) intersects at some point x ρ. Let w = v k v m. We have w H () and we extend w to be zero outside c and we still call this extension w. Let h(θ) = w(x + ρ(cos θ, sin θ)), x C ρ (x ), and x + ρ(cos, sin ) = x ρ, x + ρ(cos θ, sin θ ) = x. We have Then that is, w(x ) = θ h (θ) dθ = w(x ) θ θ Dw(x + ρ(cos θ, sin θ)) ( sin θ, cos θ)ρ dθ. Dw(x + ρ(cos θ, sin θ)) ρ dθ ( θ ρ θ /2 Dw(x + ρ(cos θ, sin θ)) 2 dθ ( 2π /2 ρ(2π) /2 Dw(x + ρ(cos θ, sin θ)) dθ) 2, 2π w(x + ρ(cos θ, sin θ )) 2 2π ρ 2 Dw(x + ρ(cos θ, sin θ)) 2 dθ. Integrating this inequality for θ 2π yields 2π 2π w(x + ρ(cos θ, sin θ )) 2 dθ 4π 2 ρ 2 and now integrating for ρ 3σ/2 we obtain w(z) 2 dz 9π 2 σ 2 Dw(z) 2 dz. B 3σ/2 (x ) B 3σ/2 (x ) ) /2 Dw(x + ρ(cos θ, sin θ)) 2 dθ,

7 SOLUTION OF THE DP WITH A VARIATIONAL METHOD February 3, 29 7 Therefore ( ) /2 (v k (z) v m (z)) dz v k (z) v m (z) 2 dz ( ) /2 3 w(z) 2 dz B 3σ/2 (x ) 2 ( ) /2 π Dw(z) 2 dz B 3σ/2 (x ) 2 ( π D(v k v m )(z) 2 dz which tends to zero as k, m. Lettting m we obtain v k (z) dz φ(x) < ɛ for all k sufficiently large and so IV is also small. ) /2 Department of Mathematics, Temple University, Philadelphia, PA 922 address: gutierre@temple.edu