3 Basic boundary value problems for analytic function in the upper half plane

Transcription

1 3 Basc boundary value problems for analytc functon n the upper half plane 3. Posson representaton formulas for the half plane Let f be an analytc functon of z throughout the half plane Imz > 0, contnuous such that f satsfes the growth property where M(R, f) = max z =R f(z). 0 Imz lm M(R, f) = 0, (3.) R For any fxed pont z above the real axs let C R denote the upper half of the postvely orented crcle of radus R centered at the orgn, where R > z. Then, accordng to the Cauchy ntegral formula, f(z) = 2π C R f(s)ds s z + 2π R R f(t) t z (3.2) We fnd that the frst of these ntegrals approaches 0 as R tends to consequently, f(z) = f(t), (3.3) 2π t z where Imz > 0 and the ntegral s understood as a Cauchy prncpal value ntegral. Representaton (3.3) s the Cauchy ntegral formula for the half plane Imz > 0. When the pont z les below the real axs, the rght-hand sde of equaton (3.2) s zero; hence ntegral (3.3) s zero for such a pont. Consequently, when z s above the real axs, we have the formula f(z) = 2π ( t z + c )f(t), (3.4) t z where c s an arbtrary constant and Imz > 0. In the two cases c = and c = ths formula reduces, respectvely, to f(z) = yf(t) π t z 2, (3.5) where y > 0 and to 58

2 f(z) = (t x)f(t), (3.6) π t z 2 where y > 0. Here z = x + y s used. If f(z) = u(x, y) + v(x, y), t follows from formulas (3.5) and (3.6) that the harmonc functons u and v are represented n the half plane y > 0 n terms of the boundary values of u by the formulas where y > 0, u(x, y) = π = π yu(t, 0) t z 2 yu(t, 0) (t x) 2 + y2 (3.7) v(x, y) = (x t)u(t, 0) (3.8) π (t x) 2 + y2 where y > 0 Formula (3.7) s known as the Posson ntegral formula for the upper half plane and f(z) = u(t, 0) π t z + c 0 as the Schwarz ntegral formula, where c 0 s an arbtrary real constant. The constant c 0 can be determned e.g. by Imf() = c. Then and c = Im π f(z) = π u(t, 0) t + t c 0 = π ( u(t, 0) t z 3.2 Schwarz problem for the half plane t u(t, 0) t c 0 t ) + c. t 2 + Let F denote a real-valued functon of x that s bounded for all x and contnuous except for at most a fnte number of fnte jumps. When y ε and x ε, where ε s any postve constant, the ntegral 59

3 I(x, y) = F (t) (t x) 2 + y 2 converges unformly wth respect to x and y, as do the ntegrals of the partal dervatves of the ntegrand wth respect to x and y. Each of these ntegrals s the sum of a fnte number of mproper or defnte ntegrals over ntervals where F s contnuous; hence the ntegrand of each component ntegral s a contnuous functon of t, x and y when y ε. Moreover, each partal dervatve of I(x, y) s represented by the ntegral of the correspondng dervatve of the ntegrand whenever y > 0. We wrte U(x, y) = yi(x, y)/π. Thus U s the Posson ntegral transform of F, suggested by equaton (3.7). U(x, y) = π yf (t) (t x) 2 + y2, (3.9) where y > 0. Except for the factor /π, the kernel here s y/ t z 2. It s the magnary part of the functon /(t z) whch s analytc n z when y > 0. It follows that the kernel s harmonc, and so t satsfes Laplace s equaton n x and y. Because the order of dfferentaton and ntegraton can be nterchanged, functon (3.9) then satsfes that Laplace equaton. Consequently, U s harmonc when y > 0. To prove that lm U(x, y) = F (x) (3.0) y 0 y>0 for each fxed x at whch F s contnuous, we substtute t = x+y tan r n formula (3.9) and wrte where y > 0. If U(x, y) = π π/2 π/2 G(x, y, r) = F (x + y tan r) F (x) F (x + y tan r)dr, (3.) 60

4 and α s some small postve constant, then π[u(x, y) F (x)] = where π/2 π/2 I = I 2 = I 3 = G(x, y, r)dr = I (y) + I 2 (y) + I 3 (y), (3.2) π/2+α π/2 π/2 α π/2+α π/2 π/2 α G(x, y, r)dr G(x, y, r)dr G(x, y, r)dr. If M denotes an upper bound for F (x), then G(x, y, r) 2M. For a gven postve number ε we select α so that 6Mα < ε; then and I (y) 2Mα < ε/3 I 3 (y) 2Mα < ε/3 We next show that correspondng to ε there s a postve number δ such that I 2 (y) < ε/3, when 0 < y < δ. Snce F s contnuous at x, there s a postve number γ such that G(x, y, r) < ε 3π, when 0 < y tan r < γ. Note that the maxmum value of tan r as r ranges between π/2+ α and π/2 α s tan(π/2 α) = cot α. Hence f we wrte δ = γ tan α, t follows that 6

5 I 2 (y) < ε 3π (π 2α) < ε 3, when 0 < y < δ. We have thus shown that I (y) + I 2 (y) + I 3 (y) < ε, when 0 < y < δ. Conon (3.0) now follows from ths result and equaton (3.2). On the bass of (3.9) then for any real constant c the functon f(z) = π F (t) + c t z therefore solves the Schwarz problem for analytc functons for the half plane y > 0 wth the boundary conon lm z t Ref(z) = F (t), t R. 0<Imz It s evdent from the form (3.) of formula (3.9) that U(x, y) M n the half plane where M s an upper bound of F (x) ; that s, U s bounded. We note that U(x, y) = F 0 when F (x) = F 0, where F 0 s a constant. Accordng to formula (3.8) of the precedng secton, under certan conons on F the functon V (x, y) = π (x t)f (t) (t x) 2 + y2 + c, (3.3) where c s an arbtrary real constant and y > 0, s a harmonc conjugate of the functon U gven by formula (3.9). Actually, formula (3.3) furnshes a harmonc conjugate of U f F s everywhere contnuous, except for at most a fnte number of fnte jumps, and f F satsfes the growth property x k F (x) < M for some k > 0. For under those conons we fnd that U and V satsfy the Cauchy-Remann equatons when y > 0. 62

6 Remark 9 The boundary behavour (3.0) can be proved also n the followng way. Consder for z H We assert, that where t 0 R. f(z) = π F (t) t z. lm Ref(z) = F (t 0 ), z t 0 Proof Denotng u = Ref, then u(z) = π yf (t) t z 2. From (3.7) appled to u(t, y) t follows Hence, π y t z 2 =. Let u(z) F (t 0 ) = π y F (t) F (t 0) t z 2. t 0 x < 2 δ and 0 < y < 2 δ. By the contnuty of F at the pont t 0 for any ε > 0 there exsts a δ = δ(ε, t 0 ) > 0 such that for t t 0 < δ. Decomposng F (t) F (t 0 ) < ε, = t0 δ t0 +δ + + t 0 δ t 0 +δ 63

7 and observng the estmates t0+δ y and we have Hence, t 0 δ F (t) F (t 0 ) t0+δ t z 2 y ( t0 δ y2m = 2My t 0 δ ε t z ε 2 ) ( + y F (t) F (t 0) t 0 +δ t z 2 ( t0 δ ( δ + δ + t 0 +δ ) ) t z 2 ) ( t t 0 x ) 2 = 4My δ t 0 x 8My δ u(z) F (t 0 ) ε + 8My. δ t z 2 = ε lm u(z) = F (t 0 ). z t Drchlet problem for the half plane Theorem 3 Let w be an analytc functon n H, and a functon G L 2 (R; C) C(R; C) be gven. Then w(z) = 2π (3.4) t z s the unquely gven soluton to the Drchlet problem w = G on R f and only f for z H = 0. (3.5) 2π t z 64

8 Proof The proof of ths theorem conssts of two parts. )The suffcency of (3.5) follows at once from subtractng (3.5) from (3.4) leadng to w(z) = 2π t z = 2π t z 2π t z = π y G(t) (t x) 2 + y 2 Thus for z t 0 and t 0 R, lm z t0 w(z) = G(t 0 ) follows. 2)The formula (3.5) s shown to be necessary. Let w be a soluton to the Drchlet problem. Then w s an analytc functon n H havng contnuous boundary values where t 0 R. Consder for z / R the functon lm w(z) = G(t 0 ), (3.6) z t 0 and w(z) = 2π t z where z H. From w( z) = 2π t z, w(z) w( z) = π G(t) (3.7) (t x) 2 + y 2 and the propertes of the Posson kernel [see formula (3.7)] lm (w(z) w( z)) = G(t 0 ) (3.8) z t 0 65

9 follows. The formula (3.7) tends to G(t 0 ) when y tends to 0 and x tends to t 0. Then for the valy of where t 0 R, follows lm w(z) = G(t 0 ), z t 0 lm z t 0 w( z) = 0. Moreover, from t s seen w(ζ) = 2π G( ζ t) t ζ ζ w(ζ) ( ) ( G( ζ t) 2 2 2π = 2π ζ 2 ( ) ( G(t) 2 2 lm ζ Imζ<0 w(ζ) = 0., t ζ ζ 2 ) 2 t ζ ζ 2 Thus w(ζ) = 2π t ζ s an analytc functon n Imζ < 0 wth vanshng boundary values on R. As also w(ζ) = 0 lm ζ Imζ<0 then w vanshes dentcally n Imζ < 0. Ths n fact was requred to prove. ) 2 66

10 3.4 Neumann problem for the half plane Let w z = 0 n H and y w = w = G on R, w() = c, where G L 2 (R; C) C(R; C), satsfyng G(t) log( + t 2 ) L (R; C) and for z H 2π t z = 0.e. 2π R t z = 0, and c C. Then the Drchlet problem w = G on R s solvable by, see 3.3, f and only f w (z) = π = 2π = 2π yg(t) t z 2 G(t)( t z t z ) t z 2π t z = 0 for z H. Integraton leads to w(z) = c G(t) log t z 2 + G(t) log( + t 2 ). 2π 2π From w z (z) = 2π t z = 0 w s seen to be analytc. Moreover, 67

11 w (z) = 2π t z = π Summng up we have the followng theorem. G(t) t z 2, w() = c. Theorem 4 The Neumann problem w z = 0 n H, y w = G on R, w() = c wth G L 2 (R; C) C(R; C) and ( + t 2 )G(t) L (R; C) s solvable f and only f π for z H. Then the soluton s w(z) = c + 2π G(t) t z = Robn boundary value problem G(t) log t t z 2. At frst a partcular case s studed, see [3]. Specal Robn problem: Fnd an analytc functon n the upper half plane H, satsfyng the boundary conon w + ν w = γ (3.9) on R for gven γ L 2 (R; C) C(R; C). Note that n our case for the boundary conon (3.9) we have Thus (3.9) can be wrtten as ν w = y w = w. w + w = γ on R. As we know from Secton 3.3 w(z) + w (z) = 2π t z = π (3.20) t z 2 68

12 f and only f 2π t z = 0, for z H. The soluton to the dfferental equaton (3.20) s w(z) = C 0 e z π z e (z s) Ims ds. (3.2) t s 2 Remark 0 As the rght-hand sde of (3.20) s analytc the ntegral n (3.2) s analytc n H. Verfcaton of (3.2). By dfferentatng (3.2) w (z) = C 0 e z + π z e (z s) Ims t s 2 ds π t z 2 follows. Ths gves together wth (3.2) w(z) + w (z) = π In order to determne C 0 consder y t z 2. hence w() = C 0 e, C 0 = ew(), w(z) = w()e +z π z e (z s) Ims t s 2 ds. General Robn problem: Fnd an analytc functon n the upper half plane H, satsfyng the boundary conon αw + ν w = γ on R, 69

13 where α C(R; C) and w(0) = C 0. Ths boundary conon on R s αw + w = γ, whch s w αw = γ on R. (3.22) If γ = 0, then We can verfy ths for t R: w(t) = Ce t 0 α(ζ)dζ w (t) = Cα(t)e t 0 α(ζ)dζ = α(t)w(t). If γ 0, then by the method of varyng the constant for t R w(t) = C(t)e t 0 α(ζ)dζ and w (t) = C (t)e t 0 α(ζ)dζ + α(t)c(t)e t 0 α(ζ)dζ so that on R w (t) α(t)w(t) = C (t)e t 0 α(ζ)dζ + C(t)α(t)e t 0 α(ζ)dζ αc(t)e t 0 α(ζ)dζ =. Thus C (t) = e t 0 α(ζ)dζ. By ntegraton t C(t) = γ(s)e s 0 α(ζ)dζ ds + C 0 0 follows wth arbtrary C 0 C. Hence, on R 70

14 t w(t) = [ γ(s)e s 0 α(ζ)dζ ds + C 0 ]e t 0 α(ζ)dζ 0 = C 0 e t 0 α(ζ)dζ t 0 γ(s)e t s α(ζ)dζ ds. These are the Drchlet boundary values for the analytc functon w. The development of Secton 3.3 leads to the next result. Theorem 5 The Robn problem w z = 0 n H, αw + ν w = γ on R, w(0) = C 0 for α, γ L 2 (R; C) C(R; C), w(0) = C 0, C 0 C, s unquely solvable f and only f 2π {C 0 e t t 0 α(s)ds γ(σ)e t σ α(s)ds dσ} 0 t z = 0, where z H. The soluton s gven by w(z) = π {C 0 e t t 0 α(s)ds γ(σ)e t 0 σ α(s)ds dσ} t z 2. A specal case can be treated n another way. If α(t) are the boundary values of a functon α analytc n H, then s analytc n H so that w αw w αw = γ on R s a Drchlet problem for an analytc functon on H. Thus w (z) α(z)w(z) = 2π = π s an analytc soluton n H f and only f 7 t z (3.23) t z 2

15 2π t z = 0, where z H. If γ = 0, then w(z) = Ce z α(ζ)dζ. If γ 0,then the soluton to (3.28) s of the form w(z) = C(z)e z Therefore w = C e z α(ζ)dζ + αce z α(ζ)dζ, α(ζ)dζ. w αw = C e z α(ζ)dζ + αce z α(ζ)dζ αce z α(ζ)dζ.e. = π t z 2, C = π t z 2e z α(ζ)dζ. By ntegraton and C(z) = C 0 π z Ims s t s 2 e α(ζ)dζ ds w(z) = C 0 e z α(ζ)dζ z π follow. Obvously w() = C 0. Ths can be easly checked by w = C 0 αe z α(ζ)dζ π Ims z t s 2e s α(ζ)dζ ds y t z 2 72

16 from whch π z = αw π α(z)ims t s 2 e y t z 2 z s α(ζ)dζ ds follows. w αw = π y t z 2 Theorem 6 The Robn problem w z = 0 n H, αw + ν w = γ on R, w() = C 0 for α O(H) C(H; C), γ L 2 (R; C) C(R; C) and C 0 C s unquely solvable f and only f 2π t z = 0 for z H. The soluton s gven by w(z) = C 0 e z α(ζ)dζ z π Here O(H) denotes the set of analytc functons n H. z t s 2e s α(ζ)dζ ds. (3.24) 3.6 Neumann problem for harmonc functons n the upper half plane The Neumann problem for the Posson equaton s solved n Secton 2.4. Here ths problem s treated agan on the bass of the Posson formula for the Laplace equaton. Let G(x) be contnuous for all real x, except for at most a fnte number of fnte jumps, and xg(x) L 2 (R; C). For each fxed real number t the functon log z t s harmonc n the half plane Imz > 0. Consequently, the functon U(x, y) = log z t G(t) + U 0 π = 2π log[(t x) 2 + y 2 ]G(t) + U 0, (3.25) 73

17 where y > 0 and U 0 s a real constant, s harmonc n that half plane. Formula (3.25) s gven wth the Posson formula (3.9) n mnd; for t follows from formula (3.25) that U y (x, y) = π where y > 0. In vew of equatons (3.9) and (3.0), then, yg(t) (t x) 2 + y2, (3.26) lm U y(x, y) = G(x), y > 0, (3.27) y 0 at each pont x where G s contnuous. Integral formula (3.26) therefore solves the Neumann problem for the half plane y > 0 wth the boundary conon y U = G. But we have not presented conons on G that are suffcent to ensure that the harmonc functon U s bounded as z ncreases. When G s an even functon, formula (3.25) can be wrtten as U(x, y) = 2π 0 log (t x)2 + y 2 (t + x) 2 + y 2 G(t) + U 0, (3.28) where x > 0, y > 0. Ths represents a functon whch s harmonc n partcular n the frst quadrant x > 0, y > 0 and whch satsfes the boundary conons U(0, y) = U 0, y > 0 lm U y (x, y) = G(x), x > 0. y 0 y>0 74