30 Poisson Integral Formulas

Transcription

1 3 Poisson Integral Formulas RecallCauchyIntegralTheorem: foranyholomorphicfunctionf definedinadomaind C, f(z) 1 f(ζ) dζ, z (a;r) D. i ζ z (a;r) By taking z aandaparametric functionφ : [,] C,θ a+e iθ for thecircle (a;r), we obtain f(a) 1 f(a+re iθ )dθ. (119) which is called the mean value theorem. It means that the central value of f is equal to the average value of f along the boundary. Question: Instead of a, for any point z (a;r), can the value of f(z) be equal to the average value of f along (a;r)? Poisson Integral Formula for holomorphic functions Theorem 3.1 Let f be a continuous function on the closed ball (a;r) and be holomorphic in (a;r). Then 84 f(z) 1 f(a+re iθ R 2 z a 2 ) Re iθ (z a) 2dθ, z (a;r). Proof: Assume a and R 1. To show: Recall that Φ z (ζ) ζ z 1 zζ f(z) 1 f(e iθ ) 1 z 2 e iθ z 2dθ, z (1). is a one-to-one and onto analytic function Φ z : (1) (1) satisfying Φ z (z). Then its inverse Φ 1 z f Φ 1 z () f(z). 84 In particular, when z a, this formula becomes f(a) 1 exists and it satisfies Φ 1 z () z so that f(a+re iθ )dθ, the mean value theorem. 182

2 Applying (119) to f Φ 1 z : f(z) f Φ 1 z () 1 f Φ 1 z (eiθ )dθ 1 i (1) f Φ 1 z (w) dw. (12) w Recall the formula of changing integral variable in Proposition 23.1: g(w)dw g C ψ 1 (C) ψ(z)ψ (z)dz. Let ψ Φ z which satisfies Φ 1 z ( (1)) (1). Then we get 1 f Φ 1 z (w) dw 1 (f Φ 1 z ) Φ z(ζ) Φ z (ζ)dζ i (1) w i Φ z (ζ) 1 f(ζ) Φ z (ζ)dζ i (1) Φ z (ζ) Φ 1 z ( (1)) 1 i f(e iθ )Φ z (eiθ )ie iθ dθ. (121) Φ z (e iθ ) In the last equality we have used the standard parametric function [,] (1),θ e iθ. Now we continue to calculate: so that Φ z (ζ) ζ z 1 zζ, Φ z(e iθ ) eiθ z 1 ze iθ Φ (1 zζ)+(ζ z)z z (ζ) (1 zζ) 2 1 z 2 (1 zζ) 2, Φ z (eiθ )ie iθ Φ z (e iθ ) i(1 z 2 ) e iθ z 2. 1 z 2 (1 ze iθ ) 2 1 zeiθ e iθ z ieiθ Substituting this into (12) and (121), we are done. Remarks i(1 z 2 ) ( z +e iθ )(e iθ z) i(1 z 2 ) (e iθ z)(e iθ z) 1. Write K(ζ,z) ζ 2 z 2 ζ z 2. Then the formula in Theorem 3.1 can be written as f(z) 1 f(ζ)k(ζ a,z a)dθ, z (a;r) where ζ Re iθ, which is called the Poisson integral formula. 183

3 2. When a, we have 3. If we take f 1, we find the formula f(z) 1 f(ζ)k(ζ, z)dθ, z (R). 1 1 K(ζ, z)dθ. Therefore, the Poisson integral formula means that a function valued can be obtained by an average of the values of the function. 4. Let us compare Cauchy integral formula and the Poisson integral formula. In Cauchy integral formula f(z) 1 f(ζ) i (R) ζ z dζ, the integral kernel is f(ζ) which is holomorphic in z. While in the Poisson integral ζ z formula, the integral kernel is K(ζ,z) ζ 2 z 2 which is not holomorphic in z. ζ z 2 In some sense, Cauchy integral formula is important to study holomorphic functions, for example, it implies that every holomorphic function must be analytic; but we cannot use Poisson integral formula to get the same conclusion. On the other hand, Poisson integral formula is important to study an important class of real-valued functions, which is called harmonic function, see below. Another equivalence for harmonic function is: a locally defined function is harmonic if and only if it is the real part of some holomorphic function. We ll discuss more on harmonic function in next lecture. Poisson integral formula for harmonic functions Writing f u + iv and separating the real part, we get u(z) 1 u(re iθ )K(Re iθ,z)dθ, z (R). Here we used the fact that K(ζ,z) is real-valued. Notice that we cannot get this formula from Cauchy integral formula. 184

4 Take ζ Re iθ and z re iψ. Then K(ζ,z) Re iθ re iψ 2 [R(cosθ +isinθ) r(cosψ +isinψ)][r(cosθ isinθ) r(cosψ isinψ)] [(Rcosθ rcosψ)+i(rsinθ rsinψ)][(rcosθ rcosψ) i(rsinθ rsinψ)] [R(cosθ rcosψ)] 2 +[Rsinθ rsinψ)] 2 R 2 2R r cos(ψ θ)+r 2 Then we have Theorem 3.2 Let f u + iv be a continuous function on the closed ball (R) and be holomorphic in (R). Then u(z) 1 u(re iθ ) R 2 2 R r cos(ψ θ)+r 2dθ, z reiψ (R). Reproducing holomorphic functions We notice that K(ζ,z) ζ 2 z 2 1 ( ζ +z ζ z 2 2 ζ z + ζ +z ) ( ) ζ +z Re. ζ z ζ z If u is the real part of a holomorphic function f which is defined on (R) and continuous on (R), by removing the real part from the formula u(z) : 1 ( ) ζ +z u(ζ)re dθ z (R), (ζ Re iθ ), ζ z we can define another function û(z) : 1 u(ζ) ζ +z ζ z dθ z (R), (ζ Reiθ ) which is a holomorphic function defined in (R) such that its real part Re(û)(z) u(z). 185

5 In other words, from a holomorphic function f, we obtain a real-valued function u; from the real-valued function u, we recovered a holomorphic function û. f u+iv u? û u Because f and û have the same real part, by Cauchy-Riemann Equations, the difference between the imaginary parts of f and û is a constant. If we write c f û, then c f() û() f() 1 u(re iθ )dθ f() u() iim(f()). Hence f(z) û(z)+i Im(f()). [Example] Let u(z) be a harmonic function on the half upper plane C + {z Im(z) > } and continue to the boundary R. Prove the analogy formula: for any point z C +, we have u(z) 1 π + y (t x) 2 +y 2u(t)dt. Recall the formula for harmonic function u(z) on the unit disk u(z) 1 1 z 2 )dθ 1 1 z 2 e iθ z 2u(eiθ ζ z 2u(ζ)dζ iζ. Here ζ e iθ. By using the Cayley transform φ(z) z i z+i, mapped C+ onto (1). Notice u φ 1 is defined on (1). Then by the integral formula, writing φ(z) a and φ(t) ζ, we have u(z) u φ 1 (a) 1 1 a 2 (ζ) dζ ζ a 2u φ 1 iζ π (1) 1 φ(z) 2 (t) φ(t) φ(z) 2u(t)φ iφ(t) dt R 1 z i z+i 2 R t i z i t+i z+i R + 2u(t) 2dt t 2 +1 (1) (by Theorem 31.1) (ζ φ(t), a φ(z)) ( φ(z) z i z +i ) [(y +1) 2 (y 1) 2 ](t 2 +1) u(t) 2[(t x) 2 +y 2 ] y (t x) 2 +y 2u(t)dt t 2 +1 dt

6 Here we have used φ (t) iφ(t) 2i (t+i) 2 t+i i(t i) 2 t 2 +1, 1 z i 2 z +i x+(y +1)i 2 x+(y 1)i 2 x2 +(y +1) 2 x 2 (y 1) 2 4y z +i 2 z +i 2 z +i 2, and t i x+iy i t+i x+iy +i 2 (t i)(x+iy +i) (t+i)(x+iy i) 2 t+i 2 z +i 2 4y2 +4(t x) 2 (t 2 +1) z +i