1 The Dirichlet Problem. 2 The Poisson kernel. Math 857 Fall 2015

Transcription

1 Math 857 Fall The Dirichlet Problem Before continuing to Fourier integrals, we consider first an application of Fourier series. Let Ω R 2 be open and connected (region). Recall from complex analysis that a twice differentiable function U : Ω R is called harmonic, if u xx + u yy = 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 periodic function, the Poisson kernel. For t [0, 1] (can take R for simplicity) and 0 r < 1 define P r (t) = Lemma 1. For 0 r < 1 and t R 1 r 2 1 2r cos 2πt + r 2. P r (t) = n Z r n e 2πint where the series on the right converges uniformly and absolutely. Proof. We have with z = re 2πit that for z < z 1 z = (1 + z)(1 + z + z2 +...) = r n e 2πint. n=1 1

2 Take real parts on both sides to get R 1 + z 1 z = P r(t). 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 Z we have r n e 2πint = r n 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 Z P r (n) = r n. 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 R we have and for each δ > 0 P r (θ)dθ = 1, P r (θ) > 0 for all θ, P r ( θ) = P r (θ), uniformly in θ for π θ δ. P r (θ) < P r (δ) if 0 < δ < θ 1 2. lim P r(θ) = 0 r 1 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 P r (θ) = 1 r 2 1 re 2πiθ 2 and r < 1, the third claim follows since cos θ is even. 2

3 For the fourth claim we observe that P r(θ) = 2π(1 r2 )2r sin 2πθ (1 2r cos 2πθ + r 2 ) 2 < 0 for θ [δ, 1/2], and finally, the limit relation holds pointwise for θ 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 u(re 2πiθ ) = for 0 r < 1 and 0 θ 1. P r (θ t)f(e 2πit )dt We see that u is the convolution of P r and f. In terms of Fourier series, if a n e 2πint = f(t), then u(re 2πit ) = n a n r n e 2πint, which means that U r (t) defined by U r (t) = u(re 2πit ) has Fourier coefficients Û r (n) = r n a n. 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 u(re 2πiθ ) = = R R [ [ 1 + re 2πi(θ t) 1 re 2πi(θ t) ] f(e 2πit )dt e 2πit + re 2πiθ e 2πit re 2πiθ f(e2πit )dt Write the fraction as (e 2πit + z)/(e 2πit z) where z = re 2πiθ,and note that e 2πit + z g(z) = e 2πit z f(e2πit )dt is analytic in the open unit disk and satisfies Rg = u. Hence, u is harmonic in D. The main difficulty lies in establishing continuity on the unit circle. 3 ]

4 Lemma 3. Let α [, 1/2] and ε > 0. There exists 1 > ρ > 0 and an arc A of the unit circle with center e 2πiα such that for r > ρ and any e 2πiθ A the inequality holds. u(re 2πiθ ) f(e 2πiα ) < ε The lemma implies continuity of u on the boundary. We use the same technique that we saw when proving density of the trigonometric polynomials in L 2. Proof of the lemma. With a rotation we may assume that α = 0. We note that u(re 2πiθ ) f(1) = P r (θ t)(f(e 2πit ) f(1))dt, and we split the integration into two parts, over t < δ and δ t 1/2, where δ remains to be chosen. The first condition for δ is that f(e 2πit ) f(1) < ε for t < δ. This is possible since f is continuous (even uniformly continuous), and allows us to estimate the first integral as < ε P r (θ t)dt < ε t <δ by extending the integration over the whole range. (This estimate is independent of θ and r.) For the second integral f(e 2πit ) and f(1) are far away from each other. We use the trivial estimate f(e 2πit ) f(1) 2 f L (T). We therefore need to use properties of P r to make the second integral small when θ is close to zero. Assume that θ δ/2 and t δ. Then t θ δ/2. From the previous proposition we can find ρ < 1 so that P r (η) < ε/(2 f ) 4

5 for all η δ/2 and ρ r < 1. Inserting this into the second integral with η = t θ gives that the integral is < ε. Replace ε by ε/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 + re 2πiθ P r (θ) = R 1 re 2πiθ, any real-valued function that is continuous on the closed unit disk and harmonic in the open unit disk may be written as the real part of f(z) = e 2πit + z e 2πit z u(e2πit )dt. 5