34 Chapter 6 Conformal Mappings Solutions to Exercises 6.. An analytic function fz is conformal where f z. If fz = z + e z, then f z =e z z + z. We have f z = z z += z =. Thus f is conformal at all z. 5. The function fz =z + z is analytic at all z, and then f z = z.so f z = z = z = z = ±. Thus fz is conformal at all z, ±. 9. The function sin z is entire. Its derivative cos z is nonzero except for z = +k. Hence fz = sin z is conformal at all z + k. In particular, fz is conformal at z =,+ i a, i. At these points, fz rotates by an angle arg cos z and scales by a factor of cos z. For z =, we have cos =. Thus f rotates by an angle arg = and scales by a factor of. Thus at z =, sin z acts like the identity map z z. For z = + ia, we have cos + ia= cosia = cosh a. Thus f rotates by an angle arg cosh a = and scales by a factor of cosh a. For z = i, we have cos i = cosh. Thus f rotates by an angle arg cosh = and scales by a factor of cosh. 3. The lines x = a and y = b are clearly orthogonal and they intersect at the point a + ib. Their images by a mapping fz are two curves that intersect at the point fa + ib. The image curves will be orthogonal at the point fa + ib iffz is conformal at a + ib. Hence it is enough to check that fz is analytic and f z atz = a + ib, in order to conclude that the image curves are orthogonal at fa + ib. In the case of fz = e z, the image curves will always be orthogonal, because f z for all z. Indeed, the image of the line x = a or z = a + iy, <y< is the circle w = e a+iy = e a e iy, with center at and radius e a. The image of the line y = b or z = x + ib, <x< is the ray at angle b, w = e x e ib. The ray and the circle intersect at right angle at the point e a+ib. 7. a We have J z = z + = z + = Jz. z z b Fix R> and let S R = {z : z = R, Arg z}. Write the image of a point z = Re iθ on S R as w = u + iv. Then w = JRe iθ = Re iθ + R e iθ = Rcos θ + i sin θ++ R cos θ i sin θ = R + cos θ + i R sin θ. R R So u = R + cos θ and v = R sin θ; R R u v R + R = cos θ and R R = sin θ.
Section 6. Basic Properties 35 As θ varies from to, w traces the upper part of the ellipse [ u ] R + + R [ v ] R = cos θ + sin θ =. R. Let Ω be the slit plane, C \{z : z }, fz = Log z, and z n = +i n n n =,,... Note that f z = z for all z Ω. Also, is straightforward to check that f is one-to-one on Ω. However, as we now show, the sequence {z n } converges to a point on the boundary, namely, but fz n does not converge. Indeed, Log z n = Log +i n =ln n +i n n + i Arg +i n n = ln + n + iα n. If n =k is even, z k approaches from the upper half-plane. So Log z k tends to ln+i = i, as k. If n =k + is odd, z k+ approaches from the lower half-plane. So Log z k+ tends to ln i = i, ask. So Log z n does not converge as n.
36 Chapter 6 Conformal Mappings Solutions to Exercises 6.. Consider the linear fractional transformation LFT φz =i z +z. a We have φ =, φ = i, φi =i i +i =. b Let L denote the line through z =andz =, and L the line through z = and z 3 = i. We know that the image of a line by a LFT is either a line or a circle. So to determine whether φ[l j ] is a line or a circle, it suffices to check whether the images of three points on L j are colinear. Another way to determine the image of a line by a LFT is to check whether the image is bounded or unbounded. If it is unbounded, then it is necessarily a line. For example, the point z = is on L. Since φ =, we conclude that φ[l ] is a line through the points φz = and φz =i; that is, φ[l ] is the imaginary axis in the w-plane. Since L is perpendicular to L and φ is conformal at z =, the point of intersection, φ[l ] and φ[l ] must be orthogonal at φ[], their point of intersection. Now φ[l ] goes through the points i and. In order for φ[l ] to be perpendicular to the imaginary axis, φ[l ], it must be a circle that goes through the points and i. To get a third point on this circle, take z = i on L. Then φ i =i +i i =. 5. The inverse of the LFT w = φz =i z +z = i iz +z = iz + i z + is given by, where a = b = i, c = d =: Let us check the images of w j : z = ψw = z i z + i. 9. Suppose c and consider the equation φz =z ψw =ψ = = z ; ψw =ψi ==z ; ψw 3 =ψ = i = z 3. or az + b cz + d = z z d c. This equation is equivalent to cz +d az b =, which has at most two solutions. Thus in this case we have at most two distinct fixed points. If c =, the LFT takes the form φz =αz +β, where α. This linear function is either the identity α = or has one fixed point, which is obtained by solving z = αz + β, z = β α. In all case, the LFT has at most two fixed points. 3. The circle in the first figure is centered at + i and has radius. To map it to the circle shown in the second figure, use the translation f z =z i.
Section 6. Linear Fractional Transformations 37 To map the inside of the unit circle to its outside, and its outside to its inside, as shown in the mapping from the second to the third figure, use an inversion f w = w. Note that if z =, then f z = /z =, so f maps the unit circle to the unit circle. Since it takes boundary to boundary, and to, it will act as we claimed. Finally, to map the unit disk to the upper half-plane, use the LFT f 3 w =i w +w see Example a. The desired mapping is w = f 3 f f z =i w = i w +w + w = i w w + = iz i z i + = iz i. z i 7. The first mapping is f z = sin z see Example 4, Sec..6.. The second mapping is the LFT f w = i w i+w see Example b. Thus, w = f f z = i w = i sin z i + w i + sin z.. Consider the LFT f z =i z +z. We know from Example a that f takes the unit disk onto the upper half-plane. What does it do to the upper semi-disk? The lower boundary of the semi-disk, the interval [, ] is perpendicular to the upper semi-circle at the point. Since f is conformal at z =, the images of the interval [, ] and semi circle intersect at right angle. Since they both go through the point, which is mapped to, we conclude that these images are perpendicular lines. Using f = i and f i =, we conclude that [, ] is mapped to the upper half of the imaginary axis, and the semi circle is mapped to the right half of the real axis. Finally, testing the image of an interior point, say i/, we find that f i = 4 5 + 3 i, which is a point in the first quadrant. With this we conclude that 5 the upper semi-disk is mapped to the first quadrant, since boundary is mapped to boundary and interior points to interior points. To go from the first quadrant to a horizontal strip, we use a logarithmic mapping, because the logarithm maps the punctured plane onto a fundamental strip of width see Sec..7. For our purpose, it is easy to check that Log z will work: The semi-line, infty is mapped to the line, and the semi-line, z = iy, <y<, is mapped to Log iy =lny + i, which describes the horizontal line shown in the figure. The composition mapping is w = Log i z +z. 5. a Take <α< and let φ α z = z α αz. Since α is real, we have φ αz = z α αz. We know from Example 3, Sec. 3.7, that φ α maps the unit disk onto itself, taking the unit circle onto itself. As it is explained in Example 4, in order to center C, it is enough to choose α so that φ α a = φ α b. Equivalently, a α αa = b α αb ; a α αb = b α αa; a + bα + abα + a + b = α +ab a + b α +=.
38 Chapter 6 Conformal Mappings Note that a + b. Solving this quadratic equation in α, we obtain with roots +ab α = +ab +ab a + b + and α = +ab a + b a + b. a + b We next proceed to show that α > and<α <. So we must choose α in constructing φ α. b We have a <, <b<and a <b<. Since b>, if a<, then multiplying b by a, we get b>a b. If <a<, then b>a b, because the right side is negative, while the left side is positive. In all cases, + ab>a+ b>or+ab > a + b. +ab a+b Consequently, > ; hence the discriminant of the quadratic equation in α in part a is positive and so we have two distinct roots, α and α. c From the inequality + ab > a + b>, we conclude that +ab a+b > and so α >. The product of α α =, as can be checked directly or by using a well-known property that the product of the roots of the quadratic equation ax + bx + c =is c a. So since <α, we conclude that <α <, in order to have the equality α α =. d Now arguing exactly as we did in Example 4, we conclude that φz = z α +ab αz with α = a+b +ab a+b will map C onto C, C onto a circle centered at the origin with radius r = φb, and the region between C and C onto the annular region bounded by φ[c ] and the unit circle. 9. From the figure, the first mapping is a translation: f z =z 3 i. The second mapping is a linear fractional transformation that maps the shaded half plane to the disk of radius and center at the origin. This mapping is the inversion f w = w. To see this, recall that the mapping is conformal except at. So it will map perpendicular lines and circles to the same. The image of a circle centered at the origin, with radius R, is another circle centered at the origin with radius /R. Thus the circle in the second figure is mapped onto the circle with center at the origin and radius. The region inside the circle is mapped outside the region of the image circle and vice versa. In particular, the shaded half-plane is mapped inside the circle with radius. We must show that the image of the lower half-plane, which is also mapped inside the circle of radius, is the the smaller disk shown in the 3rd figure. To this end, we check the image of the horizontal boundary of this lower half-plane, which is the line through 3 i. This line is mapped to a circle the image of the line is contained in a bounded region, so it has to be a circle and not a line. Moreover, this circle makes an angle with the real axis the real axis is mapped to the real axis by an inversion. Since f = and f 3 i= 3i, we conclude that the image of the lower clear half-plane is the clear smaller disk as shown in the 3rd figure. To construct the last mapping of the sequence, we will appeal to the result of Exercise 5. Let us prepare the ground for the application of this result, by rotating the picture in the 3rd figure by, and then scaling by. This amounts to multiplying by e or i. So we introduce the mapping f 3 w = e w. This will map the outer circle in the 3rd figure to the unit circle and the inner circle to a circle of radius 6 and center at 6. In the notation of Exercise 5, we have a = and b = 3. According to Exercise 5, the mapping that will center the inner circle is φ αw 3 = w3 α αw 3, where α is the smaller of the roots of α α += α 6α +=. 3
Section 6. Linear Fractional Transformations 39 Thus α =3. Composing all the mappings together, we obtain: fz = w i 3 α = w α αw 3 α i = iw α +iαw = i w w α +i α w = i αw w + iα = i αz 3 i αz i 3α z 3 = i+iα z + i 3+α = 3 z i 33 z i = 3 z + i4 3 z i. 33. a We can reduce this problem to the one that is solved in Exercise 3 by first mapping the unit disk onto the upper half-space by using the linear fractional transformation φz =i z +z Example a. If f is a one-to-one analytic mappings of the upper half-plane onto the unit disk, then f φ is a one-to-one mapping of the unit disk onto itself. By Exercise 3, f φ = cφ α, where c =, α <, and φ α is the linear transformation given by 6. Consequently, f = cφ α φ, where φ is the inverse of φ. By Proposition with a = i, b = i, c = d =, we find φ z = z i z i = i z i+z. This is the mapping in Example b. To get the explicit form of f, we compute fz = cφ α φ z =c φ z α αφ z i z i+z = c α α i z i+z i α z + α = c i α+z + α, where α < and c =. It is not difficult to show that every such mapping takes the upper halfplane to the unit disk. Thus these are all the one-to-one analytic mappings of the upper half-plane to the unit disk. b Let ψz = i z i+z be the mapping given by Example b. Then ψ maps the upper half-plane onto the unit disk. If g is a one-to-one analytic mappings of the upper half-plane onto itself, then ψ g is a one-to-one analytic mappings of the upper half-plane onto the unit disk. By part a, we have i α z + α ψ gz =fz =c i α+z + α, for some unimodular c and constant α, with α <. Thus gz =ψ fz. To get the explicit formula of the LFT g, we recall that ψ z =i z +z.
4 Chapter 6 Conformal Mappings So gz = ψ fz =i fz +fz = i c i α z+α i α+z+α +c i α z+α i α+z+α i α c + cα+z + α + c + cα = i i α + c cα+z + α c cα iγ cγ +zβ + cβ = i iγ + cγ +zβ cβ, where β =+α and γ = α, c =, and α <. Conversely, any linear fractional transformation g of this form can be written as gz =ψ fz just reverse the preceding steps, and so, by the mapping properties of f and ψ, g maps the upper half-space one-to-one and onto itself. 37. For z S, z = x + iy, we have Ax + y +Bx + Cy + D =, where B + C 4AD >. Let Then where x and y satisfy f[s] = fz = z = x + iy = x iy x + y. { u + iv : u = x x + y,v= y } x + y, Ax + y +Bx + Cy + D =, where B + C 4AD >. We want to prove that u and v satisfy a similar identity with possibly different values of the coefficients. Let us consider such a relationship with u and v: Eu + v +Fu+ Gv + H = E x + y x + y + Fx x + y Gy x + y + H = = E x + y + Fx x + y Gy x + y + Hx + y x + y x + y Hx + y +Fx Gy + E. Choose H = A, F = B, G = C, and E = D. Then, for all x, y, in S, the last expression is equal to. Hence u, v satisfy Du + v +Bu Cv + A =. Moreover, B + C 4AD >. 4. a Consider a linear fractional transformation fz = az + b cz + d
Section 6. Linear Fractional Transformations 4 and suppose that a,c, and ad bc. We will not treat the easier cases in which a = or c =orad bc =. We have fz = a z + b a c z + d = a b a + d c c c z + d c = A + B z + C, where A, B, and C are constants with A and B. Let T z =z + C, φz = z and T z =Bz + A. Then f = T φ T, where T and T are linear transformations, and φ is an inversion. b Since linear mappings and inversions map lines and circles to the same, it follows that a linear fractional transformation does the same, because it can be written as a composition of linear mappings and an inversion.
4 Chapter 6 Conformal Mappings Solutions to Exercises 6.3. Transform the problem to the upper half-plane as in Example, by using the linear fractional transformation φz =i z +z. The LFT φ takes the unit disk onto the upper-half plane, the upper semi-circle onto the positive real axis, and the lower semi-circle onto the negative real axis. Thus the problem in the upper half of the w-plane becomes: U =, Uα = 7 if α>, Uα = 5 if α<. The solution in the w-plane is Uw =a Arg w+b, where a and b must be chosen in according to the boundary conditions: a Arg w+b = 7, when Arg = and a Arg w+b = 5, when Arg =. Thus b = 7 and a =. Hence Uw = Arg w + 7 and so uz =Uφz = Arg i z +z +7. 5. The mapping f z =z 4 takes the shaded region in Fig. 4 onto the upper semi-disk of radius. The upper semi-disk is mapped onto the first quadrant by f w =i w. +w Thus the mapping w = φz =i z4 +z 4 takes the shaded region in Fig. 4 onto the first quadrant. It is also a conformal mapping, being the composition of such mappings. We now determine the image of the boundary. The rays at angle and /4 are mapped onto the interval [, ] by f. The quarter of a circle is mapped to the upper semi-circle by f. The mapping f takes the interval [, ] onto the upper half of the imaginary axis, and the semi-circle onto the right half of the real axis. Thus the problem in the first quadrant of the w-plane becomes: U =, Uα = if α>, Uiα =ifα>. The solution in the w-plane is Uw =a Arg w+b, where a and b must be chosen in according to the boundary conditions: a Arg w+b =, when Arg = and a Arg w+b =, when Arg =. Thus b = and a =. Hence Uw = Arg w + and so uz =Uφz = i Arg z4 +z 4 +. 9. To solve the given problem, we can proceed as in Example 6 and make the necessary changes. A much quicker way is based is to use the solution in Example 6 and superposition, as follows. Let u z denote the solution in Example 6. Let u z =. It is clear that u is harmonic for all z. Thus u u is harmonic in the shaded region of Fig. 8. On the real axis, we have u z u z = =. On the upper semi-circle, we have u z u z = =. Thus u is the solution, where u is the solution in Example 6. 3. This problem is very similar to Example 3. The first step is to map the region to an annular region bounded by concentric circles. This can be done by using the linear fractional transformation φz = 4z 4 z.
Section 6.3 Solving Dirichlet Problems with Conformal Mappings 43 To derive this mapping, use the result of Exercise 7, Sec. 6., with a =,b = 8 7, and α = 4. Then φz =φ 4 z = z 4 4 = 4z 4 z. Using φ, we map the outer circle to the unit circle and the inner circle to the circle of radius 4 and center at. As in Example 3, the solution is ln w +ln4 ln w +ln4 ln w Uw = + = = ln/4 ln 4 ln 4. Thus the solution is uz = ln 4 ln 4z 4 z. 7. a The solution of the Dirichlet Problem in Figure 5a is obtained by applying the Poisson formula 3, with fs =s if <s<, fs = ifs< and fs =ifs>: ux + iy = y s x s + y ds + y x s + y ds + y ds x s + y = I + I + I 3. We compute each integral separately: I = y x s x s + y ds + y Similarly, we find Thus = y ln x s + y + xy x x s + y ds ds s x + y = y [ ln x + + y ln x + y ] + x y = y ln x + + y x + y + x du s= u + = y ln x + + y x + y + x s x tan y = y ln x + + y x + y + x [ tan +x y s x y Let u = s x. y s= I = y x s + y ds = + y +x tan ; y I 3 = y ds x s + y = y x tan. y + tan x y ds + ux + iy = y ln x + + y x + y + +x +x tan + +x x tan. y y ].
44 Chapter 6 Conformal Mappings b To solve the problem in Fig. 5b, we reduce to the problem in Fig. 5a by using the mapping sin z. This yields the solution ux + iy = Im sin z + Re sin z + + Re sin z + + sin x cosh y + ln Re sin z+ + Im sin z Re sin z + Im sin z + Re sin z tan Im sin z tan Re sin z Im sin z cos x sinh y = ln sin x cosh y + + cos x sinh y sin x cosh y + cos x sinh y + sin x cosh y + sin x cosh y + tan cos x sinh y, tan sin x cosh y cos x sinh y where we have used the fact that Re sin z = sin x cosh y and Im sin z = cos x sinh y. 5. To solve the Dirichlet problem in the upper half-plane with boundary function fx = x 4 +, we appeal to the Poisson formula, which gives: for y> and <x<, ux, y = y +s 4 s x + y ds. We evaluate the integral by using the residue techniques of Sec. 5.3. Let fz = +z 4 z x + y. By Proposition, Sec. 5.3, we have ux, y = y +s 4 s x + y ds =i y Res f,z j =iy Res f,z j, j j where the sum extends over the residues of f in the upper half-plane. We have z 4 = z 4 = e i z = e i 4,z = e i 3 4,z3 = e i 5 4,z4 = e i 7 4. Only z and z are in the upper half-plane. At these points, f has simple poles and the residues there are computed with the help of Proposition ii and iii, Sec. 5.: We have Res f,z = 4z 3 z=z z x + y z=z = 4e i 3 4 e i 4 x + y = R ; Res f,z = z=z z=z = 4z 3 z x + y 4e i 9 4 e i 3 4 x + y = 4e i 4 e i 3 4 x + y = R. The equation z x + y = has two roots z = x ± iy, as you can check. Only x + iy is in the upper half-plane, since y>. The residue of f at x + iy is: Res f,x + iy = z 4 = + z=x+iy z x z=x+iy x + iy 4 + iy = R 3.
Section 6.3 Solving Dirichlet Problems with Conformal Mappings 45 Thus, ux, y =iy 4e i 3 4 e i 4 x + y + 4e i 4 e i 3 4 x + y + x + iy 4 +. iy This answer should be real-valued, because the Poisson integral involves real-valued functions only. This is not obvious, but the integral can be simplified using Mathematica to yield an expression that is obviously real-valued. 9. We appeal to the Poisson formula the second version in 3, which gives: for y > and <x<, ux, y = y sin ax s This integral follows from Example, Sec. 5.4, because Thus sin ax s sin ax s s + y ds = s + y ds. sin ax cos as cos ax sin as s = + y s + y s + y. = sin ax = sin ax sin ax cos as s + y ds cos as s ds cos ax + y cos as s + y ds, cos ax sin as s + y ds = { }}{ sin as s + y ds where one of the integrals is because the integrand is an odd function. We now appeal to Example, Sec. 5.4 do the cases a> and a< separately: ux, y = y sin ax y sin ax ds = cos as s + y y e a y = e a y sin ax. 33. The conformal mapping that takes the first quadrant to the upper half-plane is w = φz =z. It also maps the interval [, ] in the z-plane to the interval [, ] in the w-plane. To determine the boundary values in the w-plane, we compose the inverse of φ with the boundary values in the z-plane. This gives the value for all points on the real axis outside the interval [, ], and for the w in the interval [, ], the boundary value is w = w. We thus obtain the boundary values in the w-plane: fw =w if <w< and otherwise. The solution in the w-plane is obtained by applying the Poisson integral function. We will use coordinates α, β in the w-plane. Thus Uα, β = β fs α s + β ds = β s α s + β ds = β [ α s α ] β tan + β lnβ +s α = α [ tan α β + tan α β ] + β ln β + α β + α, where the integral in s was evaluated by similar methods as in Exercise 7. The solution in the z-plane is obtained by replacing α by Re φz = Rez = x y and β by Im φz = Imz =xy. Thus ux, y = x y [ tan x y + tan x y ] + xy xy xy ln xy + x y xy +x y
46 Chapter 6 Conformal Mappings Exercises 6.4. The outside angles at w, w, and w 3 are θ =, θ =, and θ 3 =, respectively. By 7, we have z z + = i z. Applying, we find f = A z z + zdz+ B = A zdz+ B z z + = A zdz+ B = A z + B. i z i Using f± =, we obtain B =. Using f =, we obtain A = i. Thus fz = z. 5. The outside angles at w and w are θ = and θ =, respectively. By 7, we have z z + = i z. Applying, we find f = A z z + dz + B = ia z dz + B = ia [ z ] z + sin z + B, where in evaluating the last integral we used integration by parts. Using f =, we obtain ia + B =. Using f =, we obtain ia fz = [ z ] z + sin z. + B =. Thus B = and A = i 4.So 9. The solution of this problem is identical to the solution of Example 7, where instead of the angle you should use an arbitrary angle α. Thus the streamlines are the images by the mapping 4 fz =z α of the streamlines in the upper half-plane. The streamlines are thus parametrized by w =x + iy α, where y is arbitrary but fixed positive number, and <x<. The graph of the streamlines is like the one shown in Fig.5.
Section 6.5 Green s Functions 47 Solutions to Exercises 6.5. According to, Green s function for the shaded region is φζ φz Gz, ζ =ln φzφζ, where φ is a one-to-one conformal mapping of the region Ω onto the unit disk. Instead of doing the computations from scratch, we use our knowledge of Green s function for the upper-half plane, from Example. The rotation ρz =e i z takes Ω onto the upper-half plane. In Cartesian coordinates, this rotation is expressed as follows: ρx + iy =ix + iy = y + ix. Thus rho will map the point x, y to the point y, x and the point s, t to the point t, s. Applying this rotation then Green s function from Example, we obtain Green s function for Ω: Gz, ζ = ln y + t +x s y + t +x + s = ln x s +y t x + s +y t, where z = x + iy, ζ = s + it, x, s >. 5. As in Exercise, we will use Green s function for the upper half-plane, but first we must map Ω onto the upper half-plane using a one-to-one conformal mapping. It is straightforward to check that fz =e z takes Ω onto the upper-half plane. In Cartesian coordinates: fx + iy =e x cos y + ie x sin y. Thus f takes the point x, y to the point e x cos y, e x sin y and the point s, t to the point e s cos t, e s sin t. Applying f then Green s function from Example, we obtain Green s function for Ω: Gz, ζ = ln e x cos y e s cos t +e x sin y e s sin t e x cos y e s cos t +e x sin y + e s sin t, where z = x + iy, ζ = s + it, <x,s<, <y,t<. 9. a Green s function in the first quadrant, Ω, is derived in Exercise 3: For z and ζ in Ω, Gz, ζ =ln z ζ z + ζ z ζ z + ζ. Write z = x + iy, ζ = s + it, where x, y, s, and t are positive. Then Gz, ζ = x s ln +y t x + s +y + t x s + y t x + s + y + t = ln x s +y t + ln x + s +y + t ln x s +y + t ln x + s + y + t. The boundary of Ω consists of two half-lines: the positive real axis, and the upper part of the imaginary axis. On the imaginary axis, the normal derivative is minus the derivative with respect
48 Chapter 6 Conformal Mappings to s. Thus, on the imaginary axis, n Gz, ζ = s Gz, ζ s= = s ln x s +y t s ln x + s +y + t + s ln x s +y + t + s ln x + s + y + t s= x s = x s +y t x + s x + s +y + t x s x s +y + t + x + s s= x + s + y + t x = x +y t x x +y + t x x +y + t + x x + y + t x = x +y t x x +y + t. By a similar argument, we find that, on the real axis, n Gz, ζ = t Gz, ζ t= = y x s + y y x + s + y. From 9, we have uz = uζ Gz, ζds. Γ n We have to figure our the symbols in this integral. For the part of the boundary that is on the imaginary axis, ds = dt, and uζ =gt. For the part of the boundary that is on the real axis, uζ =fs, and ds = ds bad notation: on the left, ds stands for element of arc length; on the right ds stands for element of integration with respect to the s variables. Using the values of the normal derivative, we get fs ds uz = y + x x s + y x + s + y gt x +y t x +y + t dt. b Consider the special case in which gt =. Call the solution in this u. From part a, we have u z = y fs x s + y x + s + y ds, z = x + iy. We will show how to derive this solution by reducing the problem to a Dirichlet problem in the upper half-plane. Indeed, consider the Dirichlet problem in the upper half-plane with boundary values ux, = fx ifx> and ux, = f x ifx<. Thus the boundary function is the odd extension of fx to the entire real line. We will use the same notation for the odd extension as for the function f. Then, by the Poisson integral formula on the real line, ux, y = y fs x s ds, y >, <x<. + y
Section 6.5 Green s Functions 49 To determine the values of u on the upper part of the imaginary axis, we set x = and get u, y= y fs ds =, s + y because f is odd. So u is harmonic in the upper half-plane; is equal to on the imaginary axis; and is equal to fx on the x-axis. Therefore, its restriction to the first quadrant solves the Dirichlet problem with boundary values on the imaginary axis and fx on the positive real axis. Thus u agrees with u in the first quadrant. Because f is odd, we can write ux, y = y = y = y = y = y fs x s + y ds fs x s + y ds + y f s x + s + y ds + y fs x + s + y ds + y fs fs x s + y ds fs x s + y ds fs x s + y ds x + s + y + x s + y which is precisely the formula for u that we obtained earlier by using Green s function in the first quadrant. c The case in Figure 9 in which fs = is very similar to the case that we treated in b. You just have to interchange x and y, s and t. The solution in this case is u x, y = x gt y + t + x + dt, y t + x which is again the formula that you obtain by setting f = in part a. d The Dirichlet problem in Fig. 9 can be written as the sum of two Dirichlet problems: the problem in which g =, and the problem in which f =. The first problem is solved in b the solution is u, and the second problem is solved in c the solution is u. It is easy to straightforward to verify that u = u + u is the solution of the problem in Fig. 9. Adding the two formulas that we have for u and u, we obtain the formula for u that we derived in a. The point of this problem is that we were able to solve the Dirichlet problem in the first quadrant by reducing the problem to two problem in the upper half-plane and then using the Poisson integral formula on the real line. ds,
5 Chapter 6 Conformal Mappings Solutions to Exercises 6.6. To determine a Neumann function for an unbounded region, we will apply Proposition 3. For that purpose, we must describe a one-to-one conformal mapping of the region Ω onto the upper half-plane. It is clear that a translation by, followed by a rotation by an angle of will do the job here. So φz =e i z = iz and the Neumann function is Nz, ζ = ln φz φζ +ln φz φζ = ln iz iζ +ln iz iζ = ln iz ζ +ln iz + i iζ + i = ln z ζ +ln z + ζ = ln x s +y t + ln y + s +x + t, where z = x + iy, ζ = s + it, x, s >, and <y,t<. 5. We apply Proposition 3, with φz = sin z see Example 4, Sec..6. We also use the fact that sinx + iy = sin x cosh y + i cos x sinh y see 6, Sec..6. Then Nz, ζ = ln φz φζ +ln φz φζ = ln sin z sin ζ +ln sin z sin ζ = ln sin x cosh y sin s cosh t + cos x sinh y cos s sinh t + ln cos x sinh y + sin s cosh t + sin x cosh y cos s sinh t, where z = x + iy, ζ = s + it, <x,s<, and <y,t.