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1 the Dirichlet Problem 2 the Poisson Kernel

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1 the Dirichlet Problem 2 the Poisson Kernel Math 857 Fall 2015 1 The Dirichlet Problem Before continuing to Fourier integrals, we consider first an application of 2 Fourier series. Let Ω ⊆ R be open and connected (region). Recall from complex analysis that a twice differentiable function U :Ω ! R is called harmonic, if uxx + uyy = 0 on Ω. Take Ω = D, the open unit disk, and consider the following question. If harmonic U in D is given, does there exist the boundary function of U on T? If it does exist, to what extent does it determine U? Problem. (Dirichlet problem) Given a continuous function f on T, does there exist a harmonic function U defined on D such that f is the boundary function of U? If yes, is U unique? We will see that the answer to both questions is essentially yes (after we define what we mean by the limit of U). 2 The Poisson kernel To analyze the previously described situation, we define a very useful peri- odic function, the Poisson kernel. For t 2 [0; 1] (can take R for simplicity) and 0 ≤ r < 1 define 1 − r2 P (t) = : r 1 − 2r cos 2πt + r2 Lemma 1. For 0 ≤ r < 1 and t 2 R X jnj 2πint Pr(t) = r e n2Z where the series on the right converges uniformly and absolutely. Proof. We have with z = re2πit that for jzj < 1 1 1 + z X = (1 + z)(1 + z + z2 + :::) = 1 + 2 rne2πint: 1 − z n=1 1 Take real parts on both sides to get 1 + z < = P (t): 1 − z r Finally multiply and divide (1 − z)=(1 + z) by 1 − z and separate the resulting expression into real and imaginary part. Evidently for 0 ≤ r < 1 and n 2 Z we have jrjnje2πintj = rjnj which is absolutely convergent. Since the summands on the right do not depend on x, the Weierstrass M-test gives uniform convergence of the series. Lemma 2. We have for 0 ≤ r < 1 and n 2 Z jnj Pbr(n) = r : Proof. Since the series converges uniformly, we may change integration and summation in the definition of the Fourier coefficient. Proposition 1. For 0 ≤ r < 1 and θ 2 R we have Z 1=2 Pr(θ)dθ = 1; −1=2 Pr(θ) > 0 for all θ; Pr(−θ) = Pr(θ); 1 P (θ) < P (δ) if 0 < δ < jθj ≤ : r r 2 and for each δ > 0 lim Pr(θ) = 0 r!1− uniformly in θ for π ≥ jθj ≥ δ. Proof. If r < 1 is fixed, then the series representation of the Poisson kernel converges uniformly in θ, hence the integration can be performed termwise, which gives the first claim. The second claim follows from 1 − r2 P (θ) = r j1 − re2πiθj2 and r < 1, the third claim follows since cos θ is even. 2 For the fourth claim we observe that 2π(1 − r2)2r sin 2πθ P 0(θ) = − < 0 r (1 − 2r cos 2πθ + r2)2 for θ 2 [δ; 1=2], and finally, the limit relation holds pointwise for θ 6= 0, and by the previous statement, uniformly in the stated region. The Dirichlet problem has a solution for the unit disk. Theorem 1. Assume that f : @D ! R is continuous. (It is crucial that f is real-valued.) Then there exists a continuous function u : D ! R such that u = f on @D and u is harmonic in D. Moreover, u is unique and can be defined by Z 1=2 2πiθ 2πit u(re ) = Pr(θ − t)f(e )dt −1=2 for 0 ≤ r < 1 and 0 ≤ θ ≤ 1. We see that u is the convolution of Pr and f. In terms of Fourier series, P 2πint if ane = f(t), then 2πit X jnj 2πint u(re ) = anr e ; n 2πit which means that Ur(t) defined by Ur(t) = u(re ) has Fourier coefficients jnj Ubr(n) = r an. Proof. Define u by the integral formula in the open unit disk and define u = f on the unit circle. We need then to show that u is continuous on the closed disk and harmonic on D. To show that u is harmonic on D, we note " # Z 1=2 1 + re2πi(θ−t) u(re2πiθ) = < f(e2πit)dt 2πi(θ−t) −1=2 1 − re "Z 1=2 2πit 2πiθ # e + re 2πit = < 2πit 2πiθ f(e )dt −1=2 e − re Write the fraction as (e2πit + z)=(e2πit − z) where z = re2πiθ,and note that Z 1=2 2πit e + z 2πit g(z) = 2πit f(e )dt −1=2 e − z is analytic in the open unit disk and satisfies <g = u. Hence, u is harmonic in D. The main difficulty lies in establishing continuity on the unit circle. 3 Lemma 3. Let α 2 [−1=2; 1=2] and " > 0. There exists 1 > ρ > 0 and an arc A of the unit circle with center e2πiα such that for r > ρ and any e2πiθ 2 A the inequality ju(re2πiθ) − f(e2πiα)j < " holds. The lemma implies continuity of u on the boundary. We use the same technique that we saw when proving density of the trigonometric polynomi- als in L2. Proof of the lemma. With a rotation we may assume that α = 0. We note that Z 1=2 2πiθ 2πit u(re ) − f(1) = Pr(θ − t)(f(e ) − f(1))dt; −1=2 and we split the integration into two parts, over jtj < δ and δ ≤ jtj ≤ 1=2, where δ remains to be chosen. The first condition for δ is that jf(e2πit) − f(1)j < " for jtj < δ. This is possible since f is continuous (even uniformly continuous), and allows us to estimate the first integral as Z < " Pr(θ − t)dt < " jtj<δ by extending the integration over the whole range. (This estimate is inde- pendent of θ and r.) For the second integral f(e2πit) and f(1) are far away from each other. We use the trivial estimate 2πit jf(e ) − f(1)j ≤ 2kfkL1(T): We therefore need to use properties of Pr to make the second integral small when θ is close to zero. Assume that jθj ≤ δ=2 and jtj ≥ δ. Then jt−θj ≥ δ=2. From the previous proposition we can find ρ < 1 so that Pr(η) < "=(2kfk1) 4 for all jηj ≥ δ=2 and ρ ≤ r < 1. Inserting this into the second integral with η = t − θ gives that the integral is < ". Replace " by "=2 throughout to get the final claim. To show uniqueness, note that for another solution v we have that u−v = 0 on the boundary of the unit disk, hence the maximum principle implies that u − v is the zero function. We note that since 1 + re2πiθ P (θ) = < ; r 1 − re2πiθ any real-valued function that is continuous on the closed unit disk and har- monic in the open unit disk may be written as the real part of Z 1=2 2πit e + z 2πit f(z) = 2πit u(e )dt: −1=2 e − z 5.
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