CHAP. 16 Laurent Series. Residue integration PROOF We enclose each of the singular points z, in a circle Cj with radius small enough that those k circles and C are all separated (Fig. 370). Then f(z) is analytic in the multiply connected domain D bounded by C and C], ■ ■ ■ , Ck and on the entire boundary of D. From Cauchy's integral theorem we thus have (7) f(z) dz f(z) dz f(z) dz + ■■■ +

0). Ans. 2iri(—jjj — jjg + \tt1) = tt(it2 — \)l = 30.221; by the residue theorem. PRO B LEM 5 ET 16.3 1. Verify the calculations in Example 3 and find the other residues. 2. Verify the calculations in Example 4 and find the other residue. 3-121 RESIDUES Find all the singular points and the corresponding residues. (Show the details of your work.) 1 cos 7 4 + z< sin z _2 _6 7. cotz 9. (zz - iy 11. tanz 8. sec z 1/3 10. 12. 1 - 1 13. CAS PROJECT. Residue at a Pole. Write a program for calculating the residue at a pole of any order. Use it for solving Probs. 3-8. 14-251 RESIDUE INTEGRATION Evaluate (counterclockwise). (Show the details.) sin ttz 14. _4 dz, C: \z -i=2 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. : dz, C: z = dz sinh \ttz C:\z-W 1.4 tan 77Z dz, C: \z\ = 1 tan ttz dz, C: \z\ = 2 dz, C: \z\ = 4.5 c tosz coth zdz, C:\zl 1 c c COS ttz cosh : c * -3« tan ttz c dz, C: 1 - 4z + 6z2 : (l2 + |)(2 - z) 30z2, - 23z + 5 c (2z - l)2(3z - 1) dz, C: \z - i\ = 1.5 «fe, C: \z\ = 1 z + i«1 = 1 - e'1") = — 2i 2/ Since F is rational in cos 0 and sin 0, Eq. (2) shows that F is now a rational function of Z, say, /(/). Since dzldd = ielß, we have rfö = dz/iz and the given integral lakes the form (3) J = j> Rz) — Jc « and, as 0 ranges from 0 to 2it in (1), the variable z = eie ranges counterclockwise once around the unit circle \z\ = 1. (Review Sec. 13.5 if necessary.) EXAMPLE 1 An Integral of the Type (1) j2.TT f dB Show bv the present method that —/=- = 2n. J0 V2-cose Solution. We use cos 0 = \(:. + 1/r) aiid tld = dz/iz. Then the integral becomes dz/iz X dz V'2 - (z - V2 - - V2 + I) We see that the integrand lias a simple pole at Zj = V'2 I 1 outside the unit circle C, so that it is of no interest here, and another simple pole at :2 = V2 - 1 (where - - V'2 +1=0) inside C with residue [by (3), Sec. 16-3] 1 Res -; 2=22 (z - V2 - - V'2 I- 1) -4—1 z - V2 - 1 J s-vf-i 1 _ 2 ' Answer: 27n'(- 2//)( —1/2) = 2tt. (Here -2// is the factor in front of the last integral.) SEC. 16.4 Residue Integration of Real Integrals 719 As another large class, let us consider real integrals of the form (4) / _ f(x) dx. Such an integral, whose interval of integration is not finite is called an improper integral, and it has the meaning r" r° rb (5 ) f(x) dx = lim fix) dx + lim fix) dx. If both limits exist, we may couple the two independent passages to — =° and °o, and write (5) f fix) dx = lim f /(*) dx. The limit in (5) is called the Cauchy principal value of the integral. It is written cc pr. v. f{x) dx. It may exist even if the limits in (5') do not. Example: lim R-^cc J x dx = lim , _K R-.00 \ 2 R2 R = 0, but lim x d: We assume that the function fix) in (4) is a real rational function whose denominator is different from zero for all real x and is of degree at least two units higher than the degree of the numerator. Then the limits in (5 ) exist, and we may start from (5). We consider the corresponding contour integral (5*) m dZ. around a path C in Fig. 371. Since fix) is rational, /(z) has finitely many poles in the upper half-plane, and if we choose R large enough, then C encloses all these poles. By the residue theorem we then obtain °°, the value of the integral over the semicircle S approaches zero. If we set z = Reie, then S is represented by R = const, and as z ranges along .V, the variable 0 ranges from 0 to tt. Since, by assumption, the degree of the denominator of /(z) is at least two units higher than the degree of the numerator, we have l.fU)| < (|z| = R>R0) for sufficiently large constants k and ^0. By the ML-inequality in Sec. 14.1, dz k kTT < —w ttR = - R R (R > R0). Hence, as R approaches infinity, the value of the integral over S approaches zero, and (5) and (6) yield the result (7) J f(x) dx = 2m 2 Res /(z) where we sum over all the residues of /(z) at the poles of f{z) in the upper half-plane. EXAMPLE 2 An Improper Integral from 0 to Using (7), show that 00 f dx I 1 +-v4 2V2 y Fig. 372. Example 2 Solution. Indeed, /(;) = 1/(1 + -4) has four simple poles at the points (make a sketch) 7ri/4 _ 3ir»/4 _ „-37ri/4 _ -rri/l zi = e , z2 - e , z3 - e , z4 - e The first two of these poles lie in the upper half-plane (Fig. 372). From (4) in the last section we find the residues SEC. 16.4 Residue Integration of Real Integrals 721 Res f(z) = Res f{z) = Z=Z'i (1 - ;V 4;'" 1 1 1 (1 +:4)' . Z-Z2 .4?. ~ 4 Z-Z2 -9iri/4 _ J_ -iri/4 4 (Here we used e71 1 and e 2rn = 1.) By (1) in Sec. 13.6 and (7) in this section. 2m 00 ■Loo 1 + x 2-rri TT TT IT —— • 2i - sin — = TT sin — = —1= . 4 4 4 V2 .Since 1/(1 + a) is an even function, we thus obtain, as asserted. OG _3 1 + 2V2 Fourier Integrals The method of evaluating (4) by creating a closed contour (Fig. 371) and "blowing it up" extends to integrals (8) 00 cos sx dx and J fix) si sin sx dx (s real) as they occur in connection with the Fourier integral (Sec. 11.7). If f(x) is a rational function satisfying the assumption on the degree as for (4), we may consider the corresponding integral 0) — DC where we sum the residues of f(z)elsz at its poles in the upper half-plane. Equating the real and the imaginary parts on both sides of (9), we have (10) oo [ /(*) cos sx dx = -277 2 Im Res [fiz)eisz\, DC [ f(x) sin sx dx = 2tt 2 Re Res [/(z)ete]. •*_r*"i is > 0) To establish (9), we must show [as for (4)] that the value of the integral over the semicircle S in Fig. 371 approaches 0 as R —» °°. Now s > 0 and 5 lies in the upper half-plane _y i£ 0. Hence = 1 • e~sy SI (s > 0, y § 0). From this we obtain the inequality = |/(z)| |eisz| S |/(z)| (s > 0, v S 0). This reduces our present problem to that for (4). Continuing as before gives (9) and (10). ■ 722 EXAMPLE 3 An Application of (10) Show that COS sx tt k2 + x2 k O i dx = 0 (s > 0, k > 0). Solution. In fact, elszl{k2 + z2) has only one pole in the upper half-plane, namely, a simple pole at z = ik, and from (4) in Sec. 16.3 we obtain Res e-ks z=ik Hk Thus i dx = 2m k + x< Since e"* = cos sac + i sin sx, this yields the above results [see also (15) in Sec. 11.7.] Another Kind of Improper Integral We consider an improper integral (11) \f(x)dx whose integrand becomes infinite at a point a in the interval of integration, lim = cc. By definition, this integral (11) means (12) f f(x) dx = lim [ f(x) dx + lim f f(x) dx where both e and 77 approach zero independently and through positive values. It may happen that neither of these two limits exists if e and 17 go to 0 independently, but the limit J fix) dx + J fix) dx (13) lim exists. This is called the Cauchy principal value of the integral. It is written pr. v. I fix) dx. ja For example, 1 dx lim r* dx r1 d 1 dx = 0; the principal value exists, although the integral itself has no meaning. In the case of simple poles on the real axis we shall obtain a formula for the principal value of an integral from -00 to ». This formula will result from the following theorem. SEC. 16.4 Residue Integration of Real Integrals 723 THEOREM 1 Simple Poles on the Real Axis If f(z) has a simple pole at z = a on the real axis, then (Fig. 373) L-2 dz. = Tri Res f(z). a — r a a + r x Fig. 373. Theorem 1 PROOF By the definition of a simple pole (Sec. 16.2) the integrand /(-) has for 0 < \z — a\ < R the Laurent series f(z) = + g(z), bx = Res f(z). Here g(z) is analytic on the semicircle of integration (Fig. 373) C2: z = a + re1", 0 S 0 ^ tt and for all z between C2 and the x-axis, and thus bounded on C2, say, \g(z)\ — M. By integration, | f(z) dz=\ \ ire'0 dd + f g(z) dz = iHTTi + \ g(z) dz. Jc2 Jo re ->c.}. Jc2 The second integral on the right cannot exceed Mtrr in absolute value, by the ML-inequality (Sec. 14.1), and ML = M-nr-> 0 as r^> 0. ■ Figure 374 shows the idea of applying Theorem 1 to obtain the principal value of the integral of a rational function f(x) from — oo to °°. For sufficiently large R the integral over the entire contour in Fig. 374 has the value J given by 2m limes the sum of the residues of f(z) at the singularities in the upper half-plane. We assume that f{x) satisfies the degree -R a—r a a+rR Fig. 374. Application of Theorem 1 724 condition imposed in connection with (4). Then the value of the integral over the large semicircle S approaches 0 as R —» °°. For r —» 0 the integral over C2 (clockwise!) approaches the value K = - iri Res f(z) z — a by Theorem 1. Together this shows that the principal value P of the integral from — oc to oo plus K equals J; hence P = J — K = J + iri Resz^0 f(z). If /(z) has several simple poles on the real axis, then K will be -m times the sum of the corresponding residues. Hence the desired formula is oc (14) pr. v. [ fix) dx = 2m 2 Res /(z) + m 2 Res /(z) where the first sum extends over all poles in the upper half-plane and the second over all poles on the real axis, the latter being simple by assumption. EXAMPLE 4 Poles on the Real Axis Find the principal value Solution. Since a2 - 3.v + 2 = {x - 1)(je - 2), the integrand f(x). considered for complex z. has simple poles at I _ (z - 2)(z2 + 1) J z-i 1 " 2 ' I _ (z - 1)(;2 + 1) J z=2 1 5 ' 1 (z2 - 3z + 2)(z + 0 J s=i 1 3 - i 6 + 2i 20 ' and at z = —i in the lower half-plane, which is of no interest here. From (14) we get the answer More integrals of the kind considered in this section arc included in the problem set. Try also your CAS, which may sometimes give you false results on complex integrals. pr. v oo f -i—- J_oc (x2 - 3x + 2 Wf + 1) z = 1, Res /(z) = z = 2, Res /(?) = 2 = 2 Z = i. Res /(z) = 2=i SEC. 16.4 Residue Integration of Real Integrals 725 PJ$J3BLEM S ET 1 6.4 1-8 INTEGRALS INVOLVING COSINE AND SINE Evaluate ihe following integrals. (Show the details of your work.) f2" d.6 C dO ' J0 7 + 6 cos 6 ' J0 2 + cos 0 f2" dO r2" di " L 37 - 12 cos 0 ' L 8-2 )0 8 - 2 sin 6 r de sin J0 5 - 4 sin 6 ' J„ 5-4 J-2 77 o 13 cos 8 12 cos 26 řfr/ir. cos 20 = — 2 1 + 4 cos e o 17 - 8 cos 0 9-22 IMPROPER INTEGRALS: INFINITE INTERVAL OF INTEGRATION Evaluate (showing the details): 1 dx + 1 C dx /' dx U' J.oe a^T f 13. —s-? J-^ (a + 4)2 15. o d J-=a 1 + X8 10. 12. 14. f03 X J (a-2 - 2a + 5)2 x + 16 CO r dx 16. —=-5— J-cc (a2 + 1)(a-2 + 9) ,,r (a - 2a + 2)2 7? & 18. 20. 22. r - J-c a4 /cc cos a -oc Í — dx dx f sin a sin 3a a + 1 <7.Y í/.Y cos 4a 5a2 + 4 c/v 23-271 IMPROPER INTEGRALS: POLES ON THE REAL AXIS Find the Cauchy principal value (showing details): x + 2 23. -t;- c/a- 24. a + a I r a2 ■L a4 - 1 CC _ f a + 5 J_cc a3 — a • J , a4 + 3.y 27. | - — x a dx 28. TEAM PROJECT. Comments on Real Integrals. (a) Formula (10) follows from (9). Give the details. _ 2 (b) Use of auxiliary results. Integrating e around the boundary C of the rectangle with vertices —a, a, a + ib, —a + ib. letting a —» oo, and using c e x dx show that f vtt (This integral is needed in heat conduction in Sec. 12.6.) (c) Inspection. Solve Probs. 15 and 21 without calculation. 29. CAS EXPERIMENT. Check your CAS. Find out to what extent your CAS can evaluate integrals of the form (1), (4), and (8) correctly. Do this by comparing the results of direct integration (which may come out false) with those of using residues. 30. CAS EXPERIMENT. Simple Poles on the Real Axis. Experiment with integrals J%c /(a) dx, fix) = \ix ~ «i)(a - a2) ■ ■ ■ (x - 1. Conjecture that the principal value of these integrals is 0. Try to prove this for a special k, say, k = 3. For general k. 726 CHAP. 16 Laurent Series. Residue Integration 3EBXSOBW2BS3ffiBEqpE£STIONS AND PROBLEMS 1. Laurent scries generalize Taylor series. Explain the details. 2. Can a function have several Laurent scries with the same center? Explain. If your answer is yes, give examples. 3. What is the principal part of a Laurent scries'.' Its significance? cosh 5z 21. ., . , C: z 22. z2 + 4 ' 4zs + Iz , C: Z + 1 = 25. 26. 27. 28. 4zz cos; 4. What is a pole? An essential singularity? Give 24. — examples. 5. What is Picard's theorem? Why did it occur in this chapter? 6. What is the Riemann sphere? The extended complex plane? Its significance? 7. Is ellz2 analytic or singular at infinity? cosh z? (z — 4)3? Explain. 8. What is the residue? Why is it important? 9. State formulas for residues from memory. 10. State some further methods for calculating residues. cos z 23. cot 8-, C: |z| = Ü.2 2 , sin: I , C: z ■ . C: z = 1 2 + 1 _2 — 2? : + 9 - 9z 15 z + 9 ,3 - 9z . C: ~x2 + f , C:\z- 3 C: 29-35 REAL INTEGRALS details) ,.2 11. What is residue integration? To what kind of complex integrals does it apply? 12. By what idea can we apply residue integration to real evaluate by the methods of this chapter (showing the integrals from to x? Give simple examples. 13. What is a zero of an analytic function? How arc zeros classified? 14. What are improper integrals? Cauchy principal values? Give examples. 15. Can the residue at a singular point be 0'.' At a simple pole? dO 29. 30. r71_cw_ J0 25 - 24 , 24 cos 6 r" do ■In k + co . k> 1 16. What is a meromorphic function? An entire function? | Give examples. -V 17-28 COMPLEX INTEGRALS Integrate counterclockwise around C. (Show the details.) 1 tan z 17. —r- , C: 18. 19. 20. sin 2z , C: Iz = 1 10z 2z + i iz + C: z - 2i\ = 3 , C: |z -1=3 ■La •f - 1 2 sin 9 cos 6 de 33 34 (7.v dx 0 (1 1 - 2x2 + 4x" dx iz + 2 36. Obtain the answer to Prob. IX in Sec. 16.4 from the present Prob. 35. Summary of Chapter 16 727 JU IY1JY1 M IV I Xjr v jf\m I L- IV, „...!.,.*r Laurent Series. Residue Integration A Laurent series is a series of the form (I) Hz) = 2 ««(2 - So)" + X (7—7 1n »7=0 n^l Vi ~0' or, more briefly written [but this means the same as (1 )!| (Sec. 16.1) (1*) f(z) = 2 an(z - ZoT 1 2777 m ... __\n+l a< C U M)> where n = 0, ±1, ±2, • • • . This series converges in an open annulus (ring) A with center z0. In A the function f(z) is analytic. At points not in A it may have singularities. The first series in (1) is a power series. In a given annulus, a Laurent series of/(-) is unique, but f(z) may have different Laurent series in different annuli with the same center. Of particular importance is the Laurent series (1) that converges in a neighborhood of z0 except at z0 itself, say, for 0 < \z - z0\ < R (R > 0, suitable). The series (or finite sum) of the negative powers in this Laurent series is called the principal part of f(z) at zo- The coefficient b1 of l/(z — z,0) in this series is called the residue of f(z) at Zo and is given by [see (1) and (1*)] (2) by = Res f(z) 1 /(£*) dz*. Thus f(z*) dz* = 2m Res f(z). bx can be used for integration as shown in (2) because it can be found from 1 (3) Res /(-) (in - 1)! s-*o \d m-i Lb — 2o) J(z)\ (Sec. 16.3), provided f(z) has at ;0 a pole of order m; by definition this means that that principal part has l/(- — Zo)m as its highest negative power. Thus for a simple pole (in = 1), Res f(z) = lim {z - Zo)f(z); p(z) P(Zo) also. Res z-z„ q(z) q (z0) If the principal part is an infinite series, the singularity of f(z) at s() is called an essential singularity (Sec. 16.2). Section 16.2 also discusses the extended complex plane, that is, the complex plane with an improper point °° ("infinity") attached. Residue integration may also be used to evaluate certain classes of complicated real integrals (Sec. 16.4). If a complex function w = f(z) is defined in a domain D of the --plane, then to each point in D there corresponds a point in the w-plane. In this way we obtain a mapping of D onto the range of values of f(z) in the w-plane. We shall see that if f(z) is an analytic function, then the mapping given by w - f(z) is conformal (angle-preserving), except at points where the derivative / (z) is zero. Conformality appeared early in history in connection with constructing maps of the globe, which can be conformal (can give directions correctly) or "equiareal" (give areas correctly, except for a scale factor), but cannot have both properties, as can be proved (see [GR8] in App. 1). Conformality is the most important geometric property of analytic functions and gives the possibility of a geometric approach to complex analysis. Indeed, just as in calculus we use curves of real functions v = f(x) for studying "geometric" properties of functions, in complex analysis we can use conformal mappings for obtaining a deeper understanding of properties of functions, notably of those discussed in Chap. 13. Indeed, we shall first define the concepts of conformal mapping and then consider mappings by those elementary analytic functions in Chap. 13. This is one purpose of this chapter. A second purpose, more important to the engineer and physicist, is the use of conformal mapping in connection with potential problems. In fact, in this chapter and in the next one we shall see that conformal mapping yields a standard method for solving boundary value problems in (two-dimensional) potential theory by transforming a complicated region into a simpler one. Corresponding applications will concern problems from electrostatics, heat flow, and fluid flow. In the last section (17.5) we explain the concept of a Ricmann surface, which fits well into the present discussion of "geometric" ideas. Prerequisite: Chap. 13. Sections that may be omitted in a shorter course: 17.3 and 17.5 References and Answers to Problems: App. 1 Part D, App. 2. 728 SEC. 17.1 Geometry of Analytic Functions: Conformal Mapping 17.1 Geometry of Analytic Functions: Conformal Mapping A complex function (1) w = f{z) = u(x, y) + iv(x, y) (z = x + iy) of a complex variable z gives a mapping of its domain of definition D in the complex z-plane into the complex w-plane or onto its range of values in that plane.1 For any point z0 in D the point w0 = f(z0) is called the image of z0 with respect to /. More generally, for the points of a curve C in D the image points form the image of C; similarly for other point sets in D. Also, instead of the mapping by a function w = f(z) we shall say more briefly the mapping w = f(z). EXAMPLE 1 Mapping w =/(z) = z2 Using polar forms z = re'H and w = Re"1', we have w = z2 = r2e2'0. Comparing moduli and arguments gives R = 2 and = 26. Hence circles r = r0 are mapped onto circles R = r02 and rays 0 = d0 onto rays 4> = 260. Figure 375 shows this for the region 1 s |z| g 3/2. 77/6 £9^ ir/3, which is mapped onto the region l § |vf| S 9/4, rr/3 § 6 S 2tt/3. In Cartesian coordinates we have z = Jc + iy and « = Re (z2) = x2 - >'2, v = Im (z2) = 2xy. Hence vertical lines x = c = const arc mapped onto u = c2 - y2, v = 2cy. From this we can eliminate y. We obtain y2 c2 - u and v2 = 4c2y2. Together, v2 = 4c2(c2 - «) (Fig. 376). These parabolas open to the left. Similarly, horizontal lines y = k = const are mapped onto parabolas opening to the right, v2 = 4k2(k2 + u) (Fig. 376). ■ lThe general terminology is as follows. A mapping of a set A into a set B is called surjective or a mapping of A onto B if every clement of B is the image of at least one clement of A. It is called injective or one-to-one if different elements of A have different images in B. Finally, it is called bijective if it is both surjective and injective. 730 CHAP. 17 Conformal Mapping THEOREM 1 Conformal Mapping A mapping w = f(z) is called conformal if it preserves angles between oriented curves in magnitude as well as in sense. Figure 377 shows what this means. The angle a (0 ^ a ^ tt) between two intersecting curves Cx and C2 is defined to be the angle between their oriented tangents at the intersection point z0. And conformality means that the images Cx* and C2* of Cj and C2 make the same angle as the curves themselves in both magnitude and direction. Conformality of Mapping by Analytic Functions The mapping w = f(z) by an analytic function f is conformed, except at critical points, that is, points at which the derivative f' is zero. PROOF w = z2 has a critical point at z = 0, where f'(z) = 1z (see Fig. 375), so that conformality fails. The idea of proof is to consider a curve 0 and the angles are doubled (2) C: z{t) = x(t) + iy(t) in the domain of f(z) and to show that w = f(z) rotates all tangents at a point z0 (where f\z0) + 0) through the same angle. Now 'z(t) = dzldt = x(t) + iy(t) is tangent to C in (2) because this is the limit of (jj — Zo)/&t (which has the direction of the secant Zi — Zn (2-plane) (u)-plane) Fig. 377. Curves C, and C2 and their respective images C|* and C2* under a conformal mapping w = f(z) SEC. 17.1 Geometry of Analytic Functions: Conformal Mapping 731 in Fig. 378) as Zi approaches z0 along C. The image C* of C is w = f(z(t)). By the chain rule, w = f'(z(t))'z(t). Hence the tangent direction of C* is given by the argument (use (9) in Sec. 13.2) (3) arg w = arg /' + arg 'z where arg z gives the tangent direction of C. This shows that the mapping rotates all directions at a point z0 in the domain of analyticity of / through the same angle arg /'(z0). which exists as long as f (z0) # 0. But this means conformably, as Fig. 377 illustrates for an angle a between two curves, whose images C{* and C2* make the same angle (because of the rotation). ■ Fig. 378. Secant and tangent of the curve C In the remainder of this section and in the next ones we shall consider various conformal mappings that are of practical interest, for instance, in modeling potential problems. EXAMPLE 2 Conformality of w = z" The mapping w = z", n = 2, 3, • • ■ , is conformal. except at z = 0. where w' = «z"_1 = 0. For n = 2 this is shown in Fig. 375; we see that at 0 the angles arc doubled. For general n the angles at 0 are multiplied by a factor n under the mapping. Hence the sector 0? os irln is mapped by onto the upper half-plane V § 0 (Fig. 379). ■ x u Fig. 379. Mapping by w = z" EXAMPLE 3 Mapping w — z + 1/z. Joukowski Airfoil In terms of polar coordinates this mapping is 1 w = it + iv = r(cos 0 + i sin 0) + — (cos 8 - / sin 0). By separating the real and imaginary parts we thus obtain u = a cos 0. v = h sin 6 where a = r + — , b = r — — . r r Hence circles \;\ = r = const f- 1 are mapped onto ellipses x2/a2 - y2lb2 = 1. The circle r = 1 is mapped onto the segment — 2 ^ u ^ 2 of the «-axis. See Fig. 380. 732 y " Fig. 380. Example 3 Now the derivative of w is i fj + Da - i) » - 1 ~ ~~2 ~ 2 Z z which is 0 at z = ± 1. These are the points at which the mapping is not conformal. The two circles in Fig. 381 pass through z, = — 1. The larger is mapped onto a Joukowski airfoil. The dashed circle passes through both - 1 and 1 and is mapped onto a curved segment. Another interesting application of w = z + 1/?. (the flow around a cylinder) will be considered in Sec. 18.4. I y v Fig. 381. Joukowski airfoil EXAMPLE 4 Conformality of w = ez From (10) in Sec. 13.5 wc have \ez\ = ex and Arg z = y. Hence ez maps a vertical straight line x = x0 — const onto the circle \w\ = e 0 and a horizontal straight line y = y'o - cons/ onto the ray arg w = y0. The rectangle in Fig. 382 is mapped onto a region bounded by circles and rays as shown. The fundamental region —at< Arg z = it of ez in the z-plane is mapped objectively and conformally onto the entire iv-plane without the origin w = 0 (because ez = 0 for no z). Figure 383 shows that the upper half 0 < y § 7T of the fundamental region is mapped onto the upper half-plane 0 < arg w S= it, the left half being mapped inside the unit disk |w| £ 1 and the right half outside (why?). v 3 - o v/ // / / 2 f r A* \ x -3 -2 -1 0 Fig. 382. Mapping by w = ez 3 " y 0 x (z-plane) -1 0 1 (m-plane) Fig. 383. Mapping by w = ez SEC. 17.1 Geometry of Analytic Functions: Conformal Mapping 733 EXAMPLE 5 Principle of Inverse Mapping. Mapping w = Ln z / V) of w f(z) is obtained by interchanging the roles of the Principle. The mapping by the inverse : z-plane and the w-plane in the mapping by w = /(z). Now the principal value w = /(;) = Ln z of the natural logarithm has the inverse z = f (w) = ew. From Example 4 (with the notations z and w interchanged!) we know that f (w) = ew maps the fundamental region of the exponential function onto the z-plane without z = 0 (because ew i 0 for every w). Hence w = /(z) = Ln z maps the z-plane without the origin and cut along the negative real axis (where ti = ImLn z jumps by 2tt) conformally onto the horizontal strip — tt < v S= 77 of the h-plane, where w = it + iv. Since the mapping w = Ln z + 2tri differs from w = Ln z by the translation 2tt/ (vertically upward), this function maps the z-planc (cut as before and 0 omitted) onto the strip w< v £ 3tt. Similarly for each of the infinitely many mappings w = ln z = Ln z — 2nm (n = 0, 1, 2, • • •)• The corresponding horizontal strips of width 2t7 (images of the z-plane under these mappings) together cover the whole w-plane without overlapping. I Magnification Ratio. By the definition of the derivative we have ' fid ~ Kzo) (4) lim z-»z(1 l/'feo)l Therefore, the mapping w = f(z) magnifies (or shortens) the lengths of short lines by approximately the factor |/'(z0)|- The image of a small figure conforms to the original figure in the sense that it has approximately the same shape. However, since f'(z) varies from point to point, a large figure may have an image whose shape is quite different from that of the original figure. More on the Condition f(z) + 0. From (4) in Sec. 13.4 and the Cauchy-Riemann equations we obtain (5') \f'(z) that is, 2 _ du dv — + i — dx dx (5) 1 du y \ dx ) + du du dx dy dv dv dx dv du dv dx dv d{u, v) d(x, y) du dv dy dx This determinant is the so-called Jacobian (Sec. 10.3) of the transformation w = f(z) written in real form u = u(x, v), v = v(x, y). Hence f'(Zo) + 0 implies that the Jacobian is not 0 at za- This condition is sufficient that the mapping w = f(z) in a sufficiently small neighborhood of Zo is one-to-one or injective (different points have different images). See Ret". [GR41 in App. 1. "P'ROfili'EWI SEX 17 1 1. Verify all calculations in Example 1. 2. Why do the images of the curves \z\ — const and arg z = const under a mapping by an analytic function f(z) intersect at right angles, except at points at which f\z) = 0? 3. Does the mapping w = z = x — iy preserve angles in size as well as in sense? [4^6] MAPPING OF CURVES Find and sketch or graph the image of the given curves under the given mapping. 4. x = 1, 2, 3, 4.v = 1, 2, 3, 4; w = zz 5. Curves as in Prob. 4, w = iz (Rotation) 6. |-| = 1/3, 1/2, 1, 2, 3; Arg z = 0, ±tt/4, ±77/2, ±3tt/2, ±77; w = l/z. CHAP. 17 Conformal Mapping 7-15 MAPPING OF REGIONS 17-23 FAILURE OF CONFORMALITY Find and sketch or graph the image of the given region under the given mapping. 7. -tt/4 < Argz < tt/4, I,-J < 1/2. w = z3 8. x g 1, w = 1/z 9. [z| > 1, w = 3z 10. Im z > 0, w = 1 - z 11. a" g 0, y S 0, |z| g 4; w = z2 12. -1 S * S 1, -ir < v < tt; »• = 3 by one, and only one, linear fractional transformation w = / (z). This mapping is given implicitly by the equation w — wx ^ w2 — W3 _ z — Z\ _ Z2 ~ Z3 w - vv3 w2 - z - ZZ z2 - Z] (If one of these points is the point °°, the quotient of the two differences containing this point must be replaced by 1.) PROOF Equation (2) is of the form F(w) = G(z) with linear fractional F and G. Hence w = F-1(G(z)) = f(z), where F_1 is the inverse of F and is linear fractional (see (4) in Sec. 17.2) and so is the composite F_1(G(z)) (by Prob. 21), that is, w = f(z) is linear fractional. Now if in (2) we set w = w\, w2, w3 on the left and z = Zi, Za> 011 me tight-we sec that F(wi) = 0, F(w2) = 1, F(w3) = o= G(zi) = 0, G(z2) = 1, G(zz) = oo. From the first column, FCwx) = G(zi), thus = F_1(G(zx)) - f(zx). Similarly, w2 = f(z2), w3 = /(z3). This proves the existence of the desired linear fractional transformation. To prove uniqueness, let w = g(z) be a linear fractional transformation, which also maps Zj onto Wj,j = 1, 2, 3. Thus vv, = g(Zj)- Hence £-1(h^) = Zj, where Wj = /(z,). Together, g_1(.f(Zj)) = Zj, a mapping with the three fixed points -L, z2, z3- By Theorem 2 in Sec. 17.2, this is the identity mapping, g~1(f(z)) = z for all z. Thus f(z) = g(z) for all Z, the uniqueness. The last statement of Theorem 1 follows from the General Remark in Sec. 17.2. ■ Mapping of Standard Domains by Theorem 1 Using Theorem 1, we can now find linear fractional transformations according to the following Principle. Prescribe three boundary points Z\, z2» Z3 of the domain D in the z-plane. Choose their images wv w2, w3 on the boundary of the image D* of D in the vv-plane. Obtain the mapping from (2). Make sure that D is mapped onto D*, not onto its complement. In the latter case, interchange two vv-points. (Why does this help?) EXAMPLE 1 Mapping of a Half-Plane onto a Disk (Fig. 385) Find the linear fractional transformation (1) that maps z\ = -1. c2 = Q< '3 = ' onto w1 = — 1. u2 = w3 = 1, respectively. Solution. From (2) we obtain w - (-1) -» - 1 z - (-1) 0 - I w - 1 ' -i - (-1) ~ z - 1 ' 0 - (-1) thus THEOREM 1 SEC. 17.3 Special Linear Fractional Transformations v x = 0 Fig. 385. Linear fractional transformation in Example 1 Let us show that we can determine the specific properties of such a mapping without much calculation. For - = x we have w - (x - (')/(—ix + 1). thus \w\ = 1, so that the a-axis maps onto the unit circle. Since z = /' gives w = 0, the upper half-plane maps onto the interior of that circle and the lower half-plane onto the exterior. Z = 0, /'. oo go onto w = —i, 0, i, so that the positive imaginary axis maps onto the segment 5: « = 0, — lSugl, The vertical lines x = const map onto circles (by Theorem I, Sec. 17.2) through w = /' (the image of z = oc) and perpendicular to \w\ = 1 (by conformality; see Fig. 385). Similarly, the horizontal lines y = const map onto circles through w = i and perpendicular to 5 (by conformality). Figure 385 gives these circles for v — 0, and for y < 0 they lie outside the unit disk shown. EXAMPLE 2 Occurrence of Determine the linear fractional transformation that maps Z\ = 0, z-z = 1. z% = x onto wx = — 1, n2 = — i. w3 = 1. respectively. Solution. From (2) we obtain the desired mapping This is sometimes called the Cayley transformation?' In this case, (2) gave at first the quotient (1 - x)/(z — °°), which we had to replace by 1. EXAMPLE 3 Mapping of a Disk onto a Half-Plane Find the linear fractional transformation that maps Zi = -1, z2 = '• *3 = ' onto wx = 0, vv2 = i. w3 = », respectively, such that the unit disk is mapped onto the right half-plane. (Sketch disk and half-plane.) Solution. From (2) we obtain, after replacing (/ - =c)/(u< - oo) by 1, Z L Mapping half-planes onto half-planes is another task of practical interest. For instance, we may wish to map the upper half-plane y § 0 onto the upper half-plane V & 0. Then the x-axis must be mapped onto the w-axis. 2ARTHUR CAYLEY (1821-1895), English mathematician and professor at Cambridge, is known for his important work in algebra, matrix theory, and differential equations. 740 CHAP. 17 Conformal Mapping EXAMPLE 4 Mapping of a Half-Plane onto a Half-Plane Find the linear fractional transformation that maps Z\ = - 2, ;2 = 0, 7.3 = 2 onto w1 = 00, w2 = 1/4, ws = 3/8, respectively. Solution. You may verify that (2) gives the mapping function z -i 1 2z + 4 What is the image of the jc-axis? Of the v-axis? Mappings of disks onto disks is a third class of practical problems. We may readily verify that the unit disk in the /.-plane is mapped onto the unit disk in the w-plane by the following function, which maps Zo onto the center w — 0. ? — Zo 1 1 , (3) vv =-- , c = z.0, |z0| < 1. cz - 1 To see this, take 1/1= 1, obtaining, with c = z0 as 'n (3), \z — Zq\ = \z — c\ = \zz - cz.\ = |l - cz\ = \cz - 1|. Hence |h»| = \z — z0\l\cz — l| = 1 from (3), so that \z\ = 1 maps onto \w\ = 1, as claimed, with z0 going onto 0, as the numerator in (3) shows. Formula (3) is illustrated by the following example. Another interesting case will be given in Prob. 10 of Sec. 18.2. EXAMPLE 5 Mapping of the Unit Disk onto the Unit Disk Taking za = \ in (3), we obtain (verify!) 2s - 1 _ w -— (Kg- 386). ■ Fig. 386. Mapping in Example 5 741 EXAMPLE 6 Mapping of an Angular Region onto the Unit Disk Certain mapping problems can be solved by combining linear fractional transformations with others. For instance, to map the angular region D: -tt/6 ^ arg z " - 77/6 (Fig. 387) onto the unit disk \w\ 3= 1. we may map D by Z = z3 onto the right Z-half-plane and then the latter onto the disk \w\ ^ 1 by Z - 1 z3 ~ 1 w = i - , combined w = i —s- . Z + 1 z3 + 1 (z-plane) (Z-plane) (zu-plane) Fig. 387. Mapping in Example 6 This is the end of our discussion of linear fractional transformations. In the next section we turn to conformal mappings by other analytic functions (sine, cosine, etc.). 1. Derive the mapping in Example 2 from (2). 2. (Inverse) Find the inverse of the mapping in Example 1. Show that under that inverse the lines x = const are the images of circles in the w-plane with centers on the line v = 1. 3. Verify the formula (3) for disks. 4. Derive the mapping in Example 4 from (2). Find its inverse and prove by calculation that it has the same fixed points as the mapping itself. Is this surprising? 5. (Inverse) If w = /(z) is any transformation that has an inverse, prove the (trivial!) fact that / and its inverse have the same fixed points. 6. CAS EXPERIMENT. Linear Fractional Transformations (LFTs). (a) Graph typical regions (squares, disks, etc.) and their images under the LFTs in Examples 1-5. (b) Make an experimental study of the continuous dependence of LFTs on their coefficients. For instance, change the LFT in Example 4 continuously and graph the changing image of a fixed region (applying animation if available). 7-15 LFTs FROM THREE POINTS AND THEIR IMAGES Find the LFT that maps the given three points onto the three given points in the respective order. 7. -1,0, 1 onto -0.6 - 0.8/, -1, -0.6, + O.82 8. 0, 1, 2 onto 1, |, I 9. 21 -2i, 4 onto -4 + 2i, -4 - 2i, 0 10. i, —1, 1 onto —1, —«', i 11. 0, 1, 00 onto =0, 1, 0 12. 0, — i, i onto -1, 0, =c 13. 2i, i, 0 onto §2i, =° 14. 0, 2i, -2i onto -1, 0, 00 15. -1,0, 1 onto 0. 1, -1 16. Find all LFTs w(z) that map the x-axis onto the «-axis. 17. Find a LFT that maps |z| £ 1 onto |w| S 1 so that z = ill is mapped onto w = 0. Sketch the images of the lines x = const and y = const. 18. Find an analytic function that maps the second quadrant of the z-plane onto the interior of the unit circle in the w-plane. 19. Find an analytic function w = f (z) that maps the region 0 S arg z = tt/4 onto the unit disk |vv| 1. 20. (Composite) Show that the composite of two LFTs is a LFT. 742 CHAP. 17 Conformal Mapping 17.4 Conformal Mapping by Other Functions So far we have discussed the mapping by zn, ez (Sec. 17.1) and linear fractional transformations (Sees. 17.2, 17.3), and we shall now turn to the mapping by trigonometric and hyperbolic analytic functions. (z-plane) tie-plane) Fig. 388. Mapping w = u + iv = sin z Sine Function. Figure 388 shows the mapping by (1) w = u + iv = sin z = sin x cosh y + i cos x sinh y Hence (Sec. 13.6). (2) u — sin x cosh y, cos x sinh y. Since sin z is periodic with period 2tt, the mapping is certainly not one-to-one if we consider it in the full z-plane. We restrict z to the vertical strip S: —\tt x S in Fig. 388. Since f'(z) = cos z = 0 at z = i^77' me mapping 1S nor conformal at these two critical points. We claim that the rectangular net of straight lines x = const and y = const in Fig. 388 is mapped onto a net in the w'-plane consisting of hyperbolas (the images of the vertical lines x = const) and ellipses (the images of the horizontal lines y = const) intersecting the hyperbolas at right angles (conformality!). Corresponding calculations are simple. From (2) and the relations sin2 x + cos2 x = 1 and cosh2 y — sinh2 y = 1 we obtain sin2 x cos2 x cosh2 y — sinh2 v = 1 (Hyperbolas) -^--1--v?— = sin2 x + cos2 x = 1 (Ellipses). cosh y sinh y Exceptions are the vertical lines x = ±2?r, which are "folded" onto u ' —1 and u s 1 ([) = 0), respectively. Figure 389 illustrates this further. The upper and lower sides of the rectangle are mapped onto semi-ellipses and the vertical sides onto —cosh 1 S u § —1 and 1 ^ u i cosh 1 (v = 0), respectively. An application to a potential problem will be given in Prob. 5 of Sec. 18.2. SEC. 17.4 Conformal Mapping by Other Functions Fig. 189. Mapping by w = sin z Cosine Function. The mapping w = cos z could be discussed independently, but since (3) w = cos z = sin (z + 5 77), we see at once that this is the same mapping as sin z preceded by a translation to the right through g7T units. Hyperbolic Sine. Since (4) w = sinhz = —i sin (iz), the mapping is a counterclockwise rotation Z = iz through \tt (i.e., 90°), followed by the sine mapping Z* = sin Z, followed by a clockwise 90°-rotation w = ~iZ*. Hyperbolic Cosine. This function (5) w = cosh z = cos (iz) defines a mapping that is a rotation Z = iz followed by the mapping w = cos Z. Figure 390 shows the mapping of a semi-infinite strip onto a half-plane by w — cosh z. Since cosh 0=1, the point z = 0 is mapped onto w = 1. For real z = x = 0, cosh z is real and increases with increasing x in a monotone fashion, starting from 1. Hence the positive .v-axis is mapped onto the portion u § 1 of the m-axis. For pure imaginary z = iy we have cosh iy = cos y. Hence the left boundary of the strip is mapped onto the segment I = m = — 1 of the m-axis, the point z = iri corresponding to w = cosh iir — cos tt = —I. On the upper boundary of the strip, y = 77, and since sin tt = 0 and cos rr = — 1, it follows that this part of the boundary is mapped onto the portion u = — 1 of the m-axis. Hence the boundary of the strip is mapped onto the m-axis. It is not difficult to see that the interior of the strip is mapped onto the upper half of the w-plane, and the mapping is one-to-one. This mapping in Fig. 390 has applications in potential theory, as we shall see in Prob. 12 of Sec. 18.3. ao- v A* -—-—Qui i 1 ■■ -0— - -10 1 Fig. 190. Mapping by w = cosh z 744 CHAP. 17 Conformal Mapping Tangent Function. Figure 391 shows the mapping of a vertical infinite strip onto the unit circle by w = tan z, accomplished in three steps as suggested by the representation (Sec. 13.6) sin; _ (elz - e~iz)/i _ (e2iz - \)li vv — tan z — — j, _jz — o;z , , cos z e + e e + 1 Hence if we set Z = e2lz and use Hi = —i, we have Z - 1 (6) w = tan z = -iW, W = - 1 , Z = e2;\ We now see that vv = tan z is a linear fractional transformation preceded by an exponential mapping (see Sec. 17.1) and followed by a clockwise rotation through an angle\tt (90°). The strip is S: —\tt < x < \tt, and we show that it is mapped onto the unit disk in the vv-plane. Since Z = e2iz = e-2y,2il\ we see from (10) in Sec. 13.5 that |Z| = e~2y, Arg Z = 2x. Hence the vertical lines x = —irIA, 0, tt/4 are mapped onto the rays Arg Z = -tt/2, 0, tt/2, respectively. Hence S is mapped onto the right Z-half-plane. Also |Z] = e~2y < I if y > 0 and |Z| > 1 if y < 0. Hence the upper half of S is mapped inside the unit circle |Z| = 1 and the lower half of S outside |Z| = 1, as shown in Fig. 391. Now comes the linear fractional transformation in (6), which we denote by g(Z): (7) W=g(Z}=^\. For real Z this is real. Hence the real Z-axis is mapped onto the real W-axis. Furthermore, the imaginary Z-axis is mapped onto the unit circle \W\ = 1 because for pure imaginary Z = iY we get from (7) \8(iY)\ = iY - 1 iY + 1 = 1. The right Z-half-plane is mapped inside this unit circle \ W\ = 1, not outside, because Z = 1 has its image g(l) = 0 inside that circle. Finally, the unit circle |Z| = 1 is mapped iz-plane) (Z-plane) (W'-plane) Fig. 391. Mapping by w = tan z ((c-plane) SEC. 17.4 Conformal Mapping by Other Functions onto the imaginary IF-axis, because this circle is Z = e14', so that (7) gives a pure imaginary expression, namely, et4> _ 1 j+IS _ -i.h/2 ■ ( . /2) g(e"h = - = - = -——- gV ; e'4' + 1 e1*'2 + e-i4><2 cos (0/2) ' From the W-plane we get to the vv-plane simply by a clockwise rotation through tt/2; see (6). Together we have shown that w = tan z maps S: —tt/4 < Re z < tt/4 onto the unit disk \w\ = 1, with the four quarters of S mapped as indicated in Fig. 391. This mapping is conformal and one-to-one. 1-7 CONFORMAL MAPPING w = e1 Find and sketch the image of the given region under w = e~. 1. 0SjS2,-jtS v ■< tt 2. -lSxgO,OSyS ir/2 3. -0.5 < x < 0.5, 3tt/4 < y < 5tt/4 4. -3 < x < 3, 7r/4 < y < 3tt/4 5. 0 < x < 1, 0 < y < tt 6. * < 0, -ir/2 < v < tt/2 7. x arbitrary, 0 S y S 2tt 8. CAS EXPERIMENT. Conformal Mapping. If your CAS can do conformal mapping, use it to solve Prob. 5. Then increase y beyond tt, say, to 507ror IOO77-. State what you expected. See what you get as the image. Explain. 9-12 CONFORMAL MAPPING w = sin z Find and sketch or graph the image of the given region under w = sin z. 9. OSxStt, OSjSI 10. 0 < x < 77/6, y arbitrary 11. 0 < x < 2tt, 1 < y < 5 12. - 77/4 < x < tt/4, 0 < y < 3 13. Determine all points at which w = sin z is not conformal. 14. Find and sketch or graph the images of the lines x = 0, ±tt/6, ±7r/3, ±77/2 under the mapping w = sinz. 15. Find an analytic function that maps the region R bounded by the positive x- and y-axes and the hyperbola xy = tt/2 in the first quadrant onto the upper half-plane. Hint. First map the region onto a horizontal strip. 16. Describe the mapping w = cosh z in terms of the mapping w = sin z and rotations and translations. 17. Find all points at which the mapping w = cosh ttz is not conformal. 18-221 CONFORMAL MAPPING w = cos z Find and sketch or graph the image of the given region under w = cos z, 18. 0 < x < tt/2, 0 < y < 2 19. 0 < x < it, 0 < y < 1 20. -1 S x S 1, 0 S y S 1 21. 77 < x < 2t7, y < 0 22. 0 < x < 2tt, 1/2 < y < 1 23. Find the images of the lines y mapping w = cos z. con.sY under the 24. Show that w = Ln z + 1 maps (he upper half-plane onto the horizontal strip 0 ^ im >v S 77 as shown in the figure. (-) -1 0 1 (z-plane) in Ó M H*(~) -6— 0 (it'-plane) Problem 24 25. Find and sketch the image of R: 2 ^ \z\ ^ 3, tt/4 s 6 tt/2 under the mapping vv = Ln z. CHAP. 17 Conformal Mapping Riemann Surfaces. Optional Riemann surfaces are surfaces on which multivalued relations, such as w = Vz or w = In z, become single-valued, that is, functions in the usual sense. Wc explain the idea, which is simple—but ingenious, one of the greatest in complex analysis. The mapping given by (1) w = u + iv = z2 (Sec. 17.1) is conformal, except at z = 0, where w' = 2z = 0. At z = 0, angles are doubled under the mapping. Thus the right --half-plane (including the positive y-axis) is mapped onto the full n-plane, cut along the negative half of the H-axis; this mapping is one-to-one. Similarly for the left z-half-plane (including the negative y-axis). Hence the image of the full z-plane under w = z2 "covers the vt'-plane twice" in the sense that every w + 0 is the image of two z-points; if Z\ is one, the other is — Zj. For example, z = i and — i are both mapped onto w = — 1. Now comes the crucial idea. We place those two copies of the cut vt'-plane upon each other so that the upper sheet is the image of the right half z-plane R and the lower sheet is the image of the left half z-plane L. Wc join the two sheets crosswise along the cuts (along the negative m-axis) so that if z moves from R to L, its image can move from the upper to the lower sheet. The two origins are fastened together because w = 0 is the image of just one z-point, z = 0. The surface obtained is called a Riemann surface (Fig. 392a). w = 0 is called a "winding point" or branch point, w = z2 maps the full z-plane onto this surface in a one-to-one manner. By interchanging the roles of the variables z and w it follows that the double-valued relation (2) w = Vz (Sec. 13.2) becomes single-valued on the Riemann surface in Fig. 392a, that is, a function in the usual sense. We can let the upper sheet correspond to the principal value of Vz. Its image is the right vt'-half-plane. The other sheet is then mapped onto the left w-half-plane. I— h— (a) Riemann surface of \z (b) Riemann surface of Nz Fig. 392. Riemann surfaces Similarly, the triple-valued relation vv = ^z becomes single-valued on the three-sheeted Riemann surface in Fig. 392b, which also has a branch point at z = 0. Chapter 17 Review Questions and Problems 747 The infinitely many-valued natural logarithm (Sec. 13.7) w -\nz= Lnz + 2«77; (« = 0, ±1, ±2, • • •) becomes single-valued on a Riemann surface consisting of infinitely many sheets, w = Ln z corresponds to one of them. This sheet is cut along the negative x-axis and the upper edge of the slit is joined to the lower edge of the next sheet, which corresponds to the argument 77 < 6 g 377, that is, to w — Ln z + 277/, The principal value Ln z maps its sheet onto the horizontal strip — 77 < v = it. The function W = Ln z + 2m maps its sheet onto the neighboring strip 77 < v = 3 77, and so on. The mapping of the points z, 0 of the Riemann surface onto the points of the vv-plane is one-to-one. See also Example 5 in Sec. 17.1. 1. Consider w = Vz. Find the path of the image point w of a point z that moves twice around the unit circle, starting from the initial position z = 1. n — 2. Show that the Riemann surface of w = Vz consists of n sheets and has a branch point at z — 0. 3. Make a sketch, similar to Fig. 392, of the Riemann surface of vz. 4. Show that the Riemann surface of w has branch points at z = 1 and z = -- V(z- -2) 2 and consists of two sheets that may be cut along the line segment from 1 to 2 and joined crosswise. Hint. Introduce polar coordinates z — 1 = r-^e1"1, z — 2 = r2e0'2. 5-101 RIEMANN SURFACES Find the branch points and the number of sheets of the Riemann surface. 5 7 9. e V3z + 5 5 + ^2z Vz + i 6. V(l - z2)(4 - z2) 8. ln (3z - 40 10. V? a^EEÖ^^SbESraffil^BSE^^ßKSTIONS AND PROBLEMS 1. How did we define the angle of intersection of two oriented curves, and what does it mean to say that a mapping is conformal? 2. At what points is a mapping w = f(z) by an analytic function not conformal? Give examples. 3. What happens to angles at z0 tinder a mapping w = f(z) iff'(Zo) = 0.f"(zo) = 0, f"'(z0) * 0? 4. What do "surjective," "injective," and "bijective" mean? 5. What mapping gave the loukowski airfoil? 6. What are linear fractional transformations (LFTs)? Why are they important in connection with the extended complex plane? 7. Why did we require that ad — be ¥= 0 for a LFT? 8. What are fixed points of a mapping? Give examples. 9. Can you remember mapping properties of w = sin z? 10. What is a Riemann surface? Why was it introduced? Explain the simplest example. 11-16 MAPPING w Find and sketch the imaee of the given curve or region under w = z2. 11. y = -1, y = 1 12. xy = -4 13. \z\ = 4.5, |arg z| < tt/4 14. 0 < y < 2 15. \ < x < 1 16. Im z > 0 17-22 J MAPPING iv = 1/z Find and sketch the image of the given curve or region under w = 1/z. 17. x = -1 18. y = 1 748 CHAP. 17 Conformal Mapping 19. 21. |arg z\ < tt/4 2Ü. 22. < \, y < 0 < 1, x < 0, y > 0 23-28 FAILURE OF CONFORMALITY Where is the mapping by the given function not conformal? (Give reason.) 23. 5z7 + 7z5 24. cosh 2z 25. sin 2z + cos 2z 26. cos ttz2 27. cxp (z4 + z2) 28. z + 1/z (z * 0) 29-341 LINEAR FRACTIONAL TRANSFORMATIONS (LFTs) Find the LFT that maps 29. 0, 1,2 onto 0, i, 2i, respectively 30. -1, 1, 2 onto 0, 2, 3/2, respectively 31. 1, —1, —/onto 1, —1, i, respectively 32. (' onto 1 - i, 2, 0, respectively 33. 0, «=, —2 onto 0, 1, so, respectively 34. 0, i, 2/ onto 0, °°, 2i 35-40 Fixed Points. Find all fixed points of z + 2 z + 1 3z + 2 35. w = 36. w 37. vr 39. w z - 1 (2 + i) z + 1 38. w = 2iz - 1 z + 2i iz + 5 5z + (" 40. w 41-45 GIVEN REGIONS Find an analytic function vv = f(z) that maps: 41. The infinite strip Q < y < rr/3 onto the tipper half-plane v > 0. 42. The interior of the unit circle |z| = 1 onto the exterior of the circle |vv + l| = 5. 43. The region x > 0, y > 0, xy < k onto the strip 0 < v < 1. 44. The semi-disk lz| < 1, x > 0 onto the exterior of the unit circle |w| = 1. 45. The sector 0 < arg z < tt/3 onto the region u < 1. ?^HH3«3iR¥ OF CHAPTER 17 Conformal Mapping A complex function w = f(z) gives a mapping of its domain of definition in the complex z-plane onto its range of values in the complex vv-plane. If/(z) is analytic, this mapping is conformal, that is, angle-preserving: the images of any two intersecting curves make the same angle of intersection, in both magnitude and sense, as the curves themselves (Sec. 17.1). Exceptions are the points at which /'(z) = 0 ("critical points," e.g. z = 0 for w = z2). For mapping properties of ez, cos z, sin z, etc. see Sees. 17.1 and 17.4. Linear fractional transformations, also called Möbius transformations az + b (1) w =-- (Sees. 17.2, 17.3) cz + a (ad — be ¥= 0) map the extended complex plane (Sec. 17.2) onto itself. They solve the problems of mapping half-planes onto half-planes or disks, and disks onto disks or half-planes. Prescribing the images of three points determines (1) uniquely. Riemann surfaces (Sec. 17.5) consist of several sheets connected at certain points called branch points. On them, multivalued relations become single-valued, that is, functions in the usual sense. Examples. For w = Vz we need two sheets (with branch point 0) since this relation is doubly-valued. For w = In z wc need infinitely many sheets since this relation is infinitely many-valued (see Sec. 13.7). CHÁPTE r1 8 Complex Analysis and Potential Theory Laplaces's equation V2(l> = 0 is one of the most important PDEs in engineering mathematics, because it occurs in gravitation (Sees. 9.7, 12.10), electrostatics (Sec. 9.7), steady-state heat conduction (Sec. 12.5), incompressible fluid flow, etc. The theory of solutions of this equation is called potential theory (although "potential" is also used in a more general sense in connection with gradients: see Sec. 9.7). In the "two-dimensional case" when depends only on two Cartesian coordinates x and v, Laplace's equation becomes V2$ = + are closely related to complex analytic functions + iXV. This relation is the main reason for the importance of complex analysis in physics and engineering. (We use the notation 0 + i since u + iv will be needed in conformal mapping.) In this chapter we shall consider this connection and its consequences in detail and illustrate it by modeling typical examples from electrostatics (Sees. 18.1, 18.2), heat conduction (Sec. 18.3), and hydrodynamics (Sec. 18.4). This will lead to boundary value problems, some of which involving functions whose mapping properties we have studied in Chap. 17. Further relating to that chapter, in Sec. 18.2 we explain conformal mapping as a method in potential theory. In Sec. 18.5 we derive the important Poisson formula for potentials in a circular disk. Finally, in Sec. 18.6 we show that results on analytic functions can be used to characterize general properties of harmonic functions (solutions of Laplace's equation whose second partial derivatives are continuous). Prerequisite: Chaps. 13, 14, 17. References and Answers to Problems: App. 1 Part D, App. 2. 749 750 CHAP. 18 Complex Analysis and Potential Theory 18.1 Electrostatic Fields The electrical force of attraction or repulsion between charged particles is governed by Coulomb's law. This force is the gradient of a function , called the electrostatic potential. At any points free of charges. $ is a solution of Laplace's equation V2 = 0. The surfaces = const are called equipotential surfaces. At each point P at which the gradient of <1> is not the zero vector, it is perpendicular to the surface $ = const through P; that is, the electrical force has the direction perpendicular to the equipotential surface. (See also Sees. 9.7 and 12.10.) The problems wc shall discuss in this entire chapter are two-dimensional (for the reason just given in the chapter opening), that is, they model physical systems that lie in three-dimensional space (of course!), but are such that the potential 4> is independent of one of the space coordinates, so that depends only on two coordinates, which we call a" and y. Then Laplace's equation becomes Equipotential surfaces now appear as equipotential lines (curves) in the xv-plane. Let us illustrate these ideas by a few simple basic examples. EX A MP L E 1 Potential Between Parallel Plates Find the potential of the field between two parallel conducting plates extending to infinity (Fig. 393), which are kept at potentials X and 2. respectively. Solution. From the shape of the plates it follows that depends only on x. and Laplace's equation becomes " = 0. By integrating twice we obtain = ax + b. where the constants a and b are determined by the given boundary values of

(.t) = |(4>2 _ + |(*2 1 The equipotential surfaces are parallel planes. I y Fig. 393. Potential in Example 1 Fig. 394. Potential in Example 2 SEC. 18.1 Electrostatic Fields 751 EXAMPLE 2 Potential Between Coaxial Cylinders Find the potential between two coaxial conducting cylinders extending to infinity on both ends (Fig. 394) and kept at potentials i and 2- respectively. Solution. Here depends only on r = V.v2 I y2. for reasons of symmetry, and Laplace's equation r2un. + ruT + um = 0 [(5), Sec. 12.9] with uge = 0 and u = 4> becomes r' = 0. By separating variables and intearatina we obtain " In $' = -In ;- + 2, . LI = - a In r - b and a and b are determined by the given values of 0? on the cylinders. Although no infinitely extended conductors exist, the field in our idealized conductor will approximate the field in a long finite conductor in that part which is far away from the two ends of the cylinders. EXAMPLE 3 Potential in an Angular Region Fig. 395. Potential in Example 3 Find the potential between the conducting plates in Fig. 395, which are kept at potentials i (the lower platei and 2, and make an angle a, where () < a ^ tt. (In the figure we have a = 120" = 2tt/3.) Solution, a = Arg z (z = x + iy # 0) is constant on rays 0 = const. It is harmonic since it is the imaginary part of an analytic function. Ln ; (Sec. 13.7). Hence the solution is 2. Thus a = (2 + l) + ~ (*2 2 a i)0. Complex Potential Let y) be harmonic in some domain D and ^(x, y) a harmonic conjugate of in D. (See Sec. 13.4, where we wrote u and v, now needed in conformal mapping from the next section on; hence the change to cfJ and M*1.) Then (2) F(z) = *(x, y) + W(x, y) is an analytic function of z = x + iy. This function F is called the complex potential corresponding to the real potential . Recall from Sec. 13.4 that for given = const in the .vy-plane at right angles [except where F (z) = 0], Hence they have the direction of the electrical force and, therefore, are called lines of force. They are the paths of moving charged particles (electrons in an electron microscope, etc.). EXAMPLE 4 Complex Potential In Example 1. a conjugate is 1f = ay. It follows that the complex potential is F(z) = az + b = cu" + b + iay. 752 CHAP. 18 Complex Analysis and Potential Theory and the lines of force arc horizontal straight lines y = const parallel to the v-axis. EXAMPLE 5 Complex Potential In Example 2 we have = a In r i b = a In |z| I- b. A conjugate is ^ = a Arg z. Hence the complex potential is F(z) = oLn; + b and the lines of force arc straight lines through the origin. F(z) may also be interpreted as the complex potential of a source line (a wire perpendicular to the icy-plane) whose trace in the xy-plane is the origin. EXAMPLE 6 Complex Potential In Example 3 we get F(z) by noting that i Ln z = i In \z\ - Arg.-, multiplying this by -b, and adding a: F(z) = a- ibLaz = a + b Arg - - ib ln \z\. We sec from this that the lines of force are concentric circles \z\ = const. Can you sketch them? Superposition More complicated potentials can often be obtained by superposition. EXAMPLE 7 Potential of a Pair of Source Lines (a Pair of Charged Wires) Determine the potential of a pair of oppositely charged source lines of the same strength at the points z = c and z = -c on the real axis. Solution. From Examples 2 and 5 it follows that the potential of each of the source lines is <1>! = K In |z - c\ and 2 = ~K ln |z + c\. respectively. Here the real constant K measures the strength (amount of charge). These are the real parts of the complex potentials Fxiz) = K Ln (z - c) and F2(z) = -KLniz F c). Hence the complex potential of the combination of the two source lines is (3) F(z) - Fjjiz) + F2(z) = K [Ln (z - c) - Ln (z + c)]. The equipotcntial lines are the curves 4> = Re F(z) = K In thus Z + c z + c These are circles, as you may show by direct calculation. The lines of force are 1r = 1m F(z) - ATArg (z - c) Arg (z + c)] = const. We write this briefly (Fig. 3%) ^ = K(81 - 6Z) = const. Now By - 92 is the angle between the line segments from z to c and —c (Fig. 396). Hence the lines of force are the curves along each of which the line segment S: -cSiSc appears under a constant angle. These curves are the totality of circular arcs over S, as is (or should be) known from elementary geometry. Hence the lines of force are circles. Figure 397 shows some of them together with some equipotcntial lines. In addition to the interpretation as the potential of two source lines, this potential could also be thought of as the potential between two circular cylinders whose axes are parallel but do not coincide, or as the potential between two equal cylinders that lie outside each other, or as the potential between a cylinder and a plane wall. Explain this, using Fig. 397. The idea of the complex potential as just explained is the key to a close relation of potential theory to complex analysis and will recur in heat How and fluid flow. SEC. 18.1 Electrostatic Fields 753 8.-9 Fig. 396. Arguments in Example 7 Fig. 397. Equipotential lines and lines of force (dashed) in Example 7 POTENTIAL 13-15 Find and sketch the potential. Find the complex potential: 1. Between parallel plates at x = -3 and 3, potentials 140 V and 260 V, respectively 2. Between parallel plates at x — —4 and 10. potentials 4.4 kV and 10 kV. respectively 3. Between the axes (potential 110 V) and the hyperbola xy = 1 (potential 60 V) 4. Between parallel plates at y = x and x + k. potentials 0 and 100 V, respectively 5-81 COAXIAL CYLINDERS Find the potential between two infinite coaxial cylinders of radii t\ and r2 having potentials U1 and U2, respectively. Find the complex potential. i = 0.5, r2 1, = 2.0, U1 110 V, U2 = 110V 't '"t = 1, r2 = 10, Ui = 100 V, U2 = 1 r2 = 4, [/, = 200 V, U2 = 0 kV 0.1, 10. 17, = 150 V. U2 = 50 V 9. Show that = 9/ir = (l/ir) arctan (y/x) is harmonic in the upper half-plane and satisfies the boundary condition 0, and ihe corresponding complex potential is F(z) = —(Utt) Ln z. 10. Map the upper half ;-planc onto the unit disk \w\ ^ 1 so that 0,», — 1 are mapped onto 1,- i, respectively. What are the boundary conditions on |vv| = 1 resulting from Ihe potential in Prob. 9? What is the potential at w = 0? 11. Verify by calculation that the equipotential lines in Example 7 are circles. 12. CAS EXPERIMENT. Complex Potentials. Graph the equipotential lines and lines of force in (a)-(d) (four graphs, Re F(z) and Im F(z) on the same axes). Then explore further complex potentials of your choice with the purpose of discovering configurations that might be of practical interest. (a) F(z) = z2 (b) F(z.) = iz2 (c) F(z) = 1/z (d) F(z) = ilz 13. POTENTIALS FOR OTHER CONFIGURATIONS Show that F(z) = arccos z (defined in Problem Set 13.7) gives the potential in Figs. 398 and 399. Fig. 398. Slit 14. 15. Fig. 399. Other apertures Find the real and complex potentials in the sector -tt/6 Sfls 77/6 between the boundary 8 = ±17-/6 (kept at 0) and the curve x3 - 3xy2 = I, kept at 110 V. Find the potential in the first quadrant of the xy-plane between the axes (having potential 220 V) and the hyperbola xy = 1 (having potential 110 V). 754 CHAP. 18 Complex Analysis and Potential Theory 18.2 Use of Conformal Mapping. Modeling Complex potentials relate potential theory closely to complex analysis, as we have just seen. Another close relation results from the use of conformal mapping in modeling and solving boundary value problems for the Laplace equation, that is, in finding a solution of the equation in some domain assuming given values on the boundary ("Dirichlet problem"; see also Sec. 12.5). Then conformal mapping is used to map a given domain onto one for which the solution is known or can be found more easily. This solution is then mapped back to the given domain. This is the idea. That it works is due to the fact that harmonic functions remain harmonic under conformal mapping: THEOREM 1 Harmonic Functions Under Conformal Mapping Let * be harmonic in a domain D in the w-plane. Suppose that w = u + iv = f(z) is analytic in a domain D in the z-plane and maps D conformally onto D*. Then the function (1) y) = <&*(u(jc, y), v(x, y)) is harmonic in D. PROOF The composite of analytic functions is analytic, as follows from the chain rule. Hence, taking a harmonic conjugate W*(u, v) of *, as defined in Sec. 13.4, and forming the analytic function F*(w) = <&*(u, v) + i^*(u, v), we conclude that F(z) = F*(f{z)) is analytic in D. Hence its real part (x, y) = Re F(z) is harmonic in D. This completes the proof. We mention without proof that if D* is simply connected (Sec. 14.2), then a harmonic conjugate of * exists. Another proof of Theorem 1 without the use of a harmonic conjugate is given in App. 4. ■ EXAMPLE 1 Potential Between Noncoaxial Cylinders Model the electrostatic potential between the cylinders Cx: \z\ 1 and C2: \z — 2/51 = 2/5 in Fig. 400. Then give the solution for the case that Cx is grounded, C'j = 0 V, and C2 has the potential U2 - 110 V. Solution. We map the unit disk \z\ = I onto the unit disk \w\ = I in such a way that C2 is mapped onto some cylinder C2*: |tv| r0. By (3), Sec. 17.3. a linear fractional transformation mapping the unit disk onto the unit disk is z - b 11) (a) z-plane (b) w-plane Fig. 400. Example 1 SEC. 18.2 Use of Conformal Mapping. Modeling where we have chosen b = zn real without restriction. z0 is of no immediate help here because centers of circles do not map onto centers of the images, in general. However, we now have two free constants b and r0 and shall succeed by imposing two reasonable conditions, namely, that 0 and 4/5 (Fig. 400) should be mapped onto r0 and -r0, respectively. This gives by (2) 0 - b 4/5 b 4/5 - ;•„ r„ = -z-—- = b. and with this. -r0 = 4fc/5 _ , = _ ^ . a quadratic equation in r0 with solutions ra = 2 (no good because r0 < 1) and r0 = 1/2. Hence our mapping function (2) with b = 1/2 becomes that in Example 5 of Sec. 17.3, 2z - 1 (3) w = ,f(z) =-- . From Example 5 in Sec. 18.1, writing w for z we have as the complex potential in the iv-plane the function F*(w) = a Ln w + k and from this the real potential **(«, v) = Re F*(w) - a In |w| + k. This is our model. Wc now determine a and k from the boundary conditions. If \w\ = 1, then * = a In 1 + k — 0, hence it = 0. If \w\ = r0 = 1/2, then * a In (1/2) = 110. hence a = 110/ln (1/2) = -158.7. Substitution of (3) now gives the desired solution in the given domain in the z-plane 2z - I F(z) = F*(f(z)) = a Ln-- . The real potential is U. v) = Re F(z) 2z - 1 -158.7. Can we "see" this result? Well, *(u, v) in the right half-plane of the w-plane. Example 3 in Sec. 18.1 with a = -ft, = -3000, and U2 = 3000 [with v) instead of 4>(x, y)J yields 6000 v («, v) = - (p, if = arctan — . 77 u On the positive half of the imaginary axis (q> = tt/2), this equals 3000 and on the negative half -3000, as it should be. d>* is the real part of the complex potential 6000 i F (w) =---Ln w. Slep 3. We substitute the mapping function into F* to get the complex potential F(z.) in Fig. 401a in the form 6000 i I + z F{z) = /••*(/(.-)) =---l.n ;- . 77 1 - c The real part of this is the potential we wanted to determine: 6000 1 + z 6000 1 + z y) = Re F(z) = - Im Ln---= - Arg 77 1 — " 77 1 — Z As in Example 1 we conclude that the equipotential lines (.r, y) = const are circular arcs because they correspond to Arg |(1 _ z)] = const, hence to Arg w = const. Also, Arg w = const are rays from 0 to «, the images of z = -1 and z = 1, respectively. Hence the equipotential lines all have -1 and 1 (the points where the boundary potential jumps) as their endpoints (Fig. 401a). The lines of force are circular arcs. too. and since they must be orthogonal to the equipotential lines, their centers can be obtained as intersections of tangents to the unit circle with the x-axis, (Explain!) Further examples can easily be constructed. Just take any mapping w = f(z) in Chap. 17, a domain D in the z-plane, its image D* in the vv-plane, and a potential 4>* in D*. Then (1) gives a potential in D. Make up some examples of your own, involving, for instance, linear fractional transformations. Basic Comment on Modeling We formulated the examples in this section as models on the electrostatic potential. It is quite important to realize that this is accidental. We could equally well have phrased everything in terms of (time-independent) heat flow; then instead of voltages we would have had temperatures, the equipotential lines would have become isotherms (= lines of constant temperature), and the lines of the electrical force would have become lines along which heat flows from higher to lower temperatures (more on this in the next section). Or we could have talked about fluid flow; then the electrostatic lines of force would have become streamlines (more on this in Sec. 18.4). What we again see here is the unifying power of mathematics: different phenomena and systems from different areas in physics having the same types of model can be treated by the same mathematical methods. What differs from area to area is just the kinds of problems that are of practical interest. SEC. 18.3 Heat Problems 757 1. Verify Theorem 1 for 3>*(«, y) = u2 - v2. w = /(') = e" and any domain />. 2. Verily Theorem 1 for *(«, f) = uv, w = /(ľ) = ez, and D: je š 0, 0 S y S tt. Sketch D and D*. 3. Carry out all steps of the second proof of Theorem 1 (given in App. 4) in detail. 4. Derive (3) from (2). 5. Let D* be Ihe image of the rectangle D: 0 š x S \ 7t, 0 S y S 1 under w = sin z, and in D and its boundary values. 6. What happens in Prob. 5 if you replace the potential by the conjugate tt>* = 2ml? Skelch or graph some of the equipotential lines 4> = const. 7. CAS PROJECT. Graphing Potential Fields. (a) Graph equipotential lines in Probs. 1 and 2. (b) Graph equipotential lines if the complex potential is F(z) = iz\ F(z) = e\ F(z) = iez, F(z) = eiz. (c) Graph equipotential surfaces corresponding to F(z) = In z as cylinders in space. 8. TEAM PROJECT. Noncoaxial Cylinders. Find the potential between the cylinders C,: \z\ = 1 (potential U1 = 0) and C2: \z - c\ = c (t/2 = 110 V), where 0 < c < |. Sketch or graph the equipotential curves and their orthogonal trajectories for c = 0.1, 0.2, 0.3, 0.4. Try to think of the further extension Q: \z\ = 1, C2: \z - c| = p + c. 9. Find Ihe potential 4> in the region /? in the first quadrant of the z-plane bounded by the axes (having potential £/q) and the hyperbola y = 1/jc (having potential 0) in two ways, (i) directly, (ii) by mapping R onto a suitable infinite strip. 10. (Extension of Example 2) Find the linear fractional transformation z = g(Z) that maps \Z\ S 1 onto \z\ = 1 with Z = i/2 being mapped onto z = 0. Show that Zi = 0.6 + 0.8/ is mapped onto z = — 1 and Z2 = —0.6 + 0.8/ onto z = 1, so that the equipotential lines of Example 2 look in |Z| § 1 as shown in Fig. 402. Y -3 kV _o_^/ J \ 1 / 2 I 3 kV Fig. 402. Problem 10 11. The equipotential lines in Prob. 10 are circles. Why? 12. Show that in Example 2 the y-axis is mapped onto the unit circle in the w-plane. 13. Find the complex and real potentials in the upper half-plane with boundary values 0 if x < 4 and 10 kV if x > 4 on the .r-axis. 14. (Angular region) Applying a suitable conformal mapping, obtain from Fig. 401b the potential in the angular region -577- < Arg z < \tt such that = -3 kV if Arg z = ~\tt and = 3 kV if Arg z = \rr. 15. At z = ±1 in Fig. 401a the tangents to the equipotential lines shown make equal angles (tt/6). Why? 18.3 Heat Problems Laplace's equation also governs heat flow problems that are steady, that is, time-independent. Indeed, heat conduction in a body of homogeneous material is modeled by the heat equation Tt = c2V2T where the function T is temperature, T, = dT/dt, t is time, and c2 is a positive constant (depending on the material of the body; see Sec. 12.5). Hence if a problem is steady, so that Tt = 0, and two-dimensional, then the heat equation reduces to the two-dimensional Laplace equation (1) ^T =r»+ Tyy = 0, so that the problem can be treated by our present methods. 758 CHAP. 18 Complex Analysis and Potential Theory T(x, y) is called the heat potential. It is the real part of the complex heat potential F(z) = T(x, y) + W{x, y). The curves T(x, y) = const are called isotherms (= lines of constant temperature) and the curves ^{x, y) = const heat flow lines, because along them, heat flows from higher to lower temperatures. It follows that all the examples considered so far (Sees. 18.1. 18.2) can now be reinterpreted as problems on heat flow. The electrostatic equipotential lines $(.*% >') = const now become isotherms T(x, y) = const, and the lines of electrical force become lines of heat flow, as in the following two problems. EXAMPLE 1 Temperature Between Parallel Plates Hind the temperature between two parallel plates x = 0 and x = d in Fig. 403 having temperatures 0 and 100°C. respectively. Solution. As in Example I of Sec. 18.1 we conclude that 7(.v, y) ax + b. From the boundary conditions. b = 0 and a = IOO/<) =- — .v [°C]. a The corresponding complex potential is h'(z) = (I00W):. Heat flows horizontally in the negative .v-direclion along the lines y = const. EXAMPLE 2 Temperature Distribution Between a Wire and a Cylinder Find the temperature field around a long thin wire of radius i\ = 1 mm that is electrically heated to Tx = 500"F and is surrounded by a circular cylinder of radius r2 = 100 mm, which is kept at temperature 72 = 60°F by cooling it with air. See Fig. 404. (The wire is at the origin of the coordinate system.) Solution. T depends only on r, for reasons of symmetry. Hence, as in Sec. 18.1 (Example 2), T(x, y) = a In r + b. The boundary conditions are 11 = 500 = a In 1 + b, T2 = 60 = a In 100 + />. Hence b = 500 (since In 1 = 0) and a = (60 - &)/ln 100 = -95.54. The answer is T(x. y) = 500 - 95.54 In r [°F]. The isotherms are concentric circles. Heat flows from the wire radially outward to the cylinder. Sketch T as a function of r. Does it look physically reasonable? Fig. 403. Example 1 T= 60"f Fig. 404. Example 2 nsulated Fig. 405. Example 3 SEC. 18.3 Heat Problems Mathematically the calculations remain the same in the transition to another field of application. Physically, new problems may arise, with boundary conditions that would make no sense physically or would be of no practical interest. This is illustrated by the next two examples. EXAMPLE 3 A Mixed Boundary Value Problem Find the temperature distribution in the region in Fig. 405 (cross section of a solid quarter-cylinder), whose vertical portion of the boundary is at 20°C, the horizontal portion at 50°C, and the circular portion is insulated. Solution. The insulated portion of the boundary must be a heat flow line, since by the insulation, heat is prevented from crossing such a curve, hence heat must flow along the curve. Thus the isotherms must meet such a curve at right angles. Since T is constant along an isotherm, this means that (2) ST cm = 0 along an insulated portion of the boundary. Here dT/dn is the normal derivative of T, that is, the directional derivative (Sec. 9.7) in the direction normal (perpendicular) to the insulated boundary. Such a problem in which 7"is prescribed on one portion of the boundary and dT/dn on the other portion is called a mixed boundary value problem. In our case, the normal direction to the insulated circular boundary curve is the radial direction toward the origin. Hence (2) becomes dT/Br = 0, meaning that along this curve the solution must not depend on r. Now Arg z = 9 satisfies (1), as well as this condition, and is constant (0 and it/2) on the straight portions of the boundary. Hence the solution is of the form T{x, y) = a0+ b. The boundary conditions yield a • tt/2 + b = 20 and a • 0 ■+ b = 50. This gives 60 T(x, y) = 50 e, 0 = arctan The isotherms are portions of rays 0 Fig. 405) to the y-axis. const. Heat flows from the .v-axis along circles r = const (dashed in y \ X \ T = 0'C -1 ^--Insulated 1 r=20°C (a) 2-plane Insulated (b) to-plane Fig. 406. Example 4 EXAMPLE 4 Another Mixed Boundary Value Problem in Heat Conduction Find the temperature field in the upper half-plane when the x-axis is kept at T = 0°C for a < — 1, is insulated for — 1 < .v 1, and is kept at T = 20CC for x > 1 (Fig. 406a). Solution. We map the half-plane in Fig. 406a onto the vertical strip in Fig. 406b, find the temperature T*(u, v) there, and map it back to get the temperature T(x, y) in the half-plane. The idea of using that strip is suggested by Fig. 388 in Sec. 17.4 with the roles of z = x + iy and w = u - iv interchanged. The figure shows that z = sin w maps our present strip onto our half-plane in Fig. 406a. Hence the inverse function w = f(z) = arcsin z 760 CHAP. 18 Complex Analysis and Potential Theory maps that half-plane unto the strip in the ii-plane. This is the mapping function that we need according to Theorem 1 in See. 18.2. The insulated segment I < x < 1 on the x-axis maps onto the segment — tt/2 < u < tt/2 on the m-axis. The rest of the x-axis maps onto the two vertical boundary portions u = —tt/2 and tt/2, v > 0. of the strip. This gives the transformed boundary conditions in Fig. 406b for T*(u, u). where on the insulated horizontal boundary, i)T*/i)n dT*/i)v = 0 because v is a coordinate normal to that segment. Similarly to Example 1 we obtain 20 r*(m, v) = 10 + — u 77 which satisfies all the boundary conditions. This is the real part of the complex potential F*(w) = 10 + (20/Tr)H\ Hence the complex potential in the --plane is 20 F(z) = F*(f(z)) = 10 + — arcsin z 77 and T(x. y) = Re F(z) is the solution. The isotherms are u = const in the strip and the hyperbolas in the z-plane, perpendicular to which heat flows along the dashed ellipses from the 20°-portion to the cooler O'-portion of the boundary, a physically very reasonable result. This section and the last one show the usefulness of conformal mappings and complex potentials. The latter will also play a role in the next section on fluid flow. 1. CAS PROJECT. Isotherms. Graph isotherms and lines of heat flow in Examples 2-4. Can you sec from the graphs where the heat flow is very rapid? 2. Find the temperature and the complex potential in an infinite plate with edges y = x — 2 and y = x + 2 kept at -10°C and 20°C, respectively. 3. Find the temperature between two parallel plates y = 0 and y = d kept at temperatures 0CC and 100PC, respectively, (i) Proceed directly, (ii) Use Example 1 and a suitable mapping. 4. Find the temperature 7" in the sector 0 Arg z = ir/3, \z\ ^ 1 if T = 20°C on the x-axis, T = 50°C on y = V3 x, and the curved portion is insulated. 5. Find the temperature in Fig. 405 if T = -20CC on the y-axis, T = 100°C on the x-axis, and the circular portion of the boundary is insulated as before. 6. Interpret Prob. 10 in Sec. 18.2 as a heat flow problem (wiui boundary temperatures, say, 20°C and 300°C). Along what curves does the heat flow? 7. Find the temperature and the complex potential in the first quadrant of the z-plane if the y-axis is kept at 100=C, the segment 0 < x < 1 of the x-axis is insulated and the portion x > 1 of the x-axis is kept at 200=C. Hint. Use Example 4. 8. TEAM PROJECT. Piecewise Constant Boundary Temperatures, (a) A basic building block is shown in Fig. 407. Find the corresponding temperature and complex potential in the upper half-plane. (b) Conformal mapping. What temperature in the first quadrant of the z-plane is obtained from (a) by the mapping w = a + z2 and what are the transformed boundary conditions? (c) Superposition. Find the temperature T* and the complex potential F* in the upper half-plane satisfying the boundary condition in Fig, 408. (d) Semi-infinite strip. Applying w = cosh z to (c), obtain the solution of the boundary value problem in Fig. 409. T:" = T? « T" = T Fig. 407. Team Project 8(a) 5T* = 0 T* = TQ T* = 0 " Fig. 408. Team Project 8(c) y T=0 7t o T = T, o 0 6 T = 0 * Fig. 409. Team Project 8(d) SEC. 18.4 Fluid Flow 761 9-14 j TEMPERATURE DISTRIBUTIONS IN PLATES Find the temperature T(x, y) in the given thin metal plate whose faces are insulated and whose edges are kepi al the indicated temperatures or arc insulated as shown. 11. 12. T= 100°C 0 T=100°C 9. 10. 60° T = 0°C ^Insulated T = 0°C 13. T= 50=C nsulated 14. y o Ö o 1 o 1 ._ T= lOOxf T = 0 C x 18.4 Fluid Flow Laplace's equation also plays a basic role in hydrodynamics, in steady nonviscous fluid flow under physical conditions discussed later in this section. In order that methods of complex analysis can be applied, our problems will be two-dimensional, so that the velocity vector V by which the motion of the fluid can be given depends only on two space variables x and y and the motion is the same in all planes parallel to the jcy-plane. Then we can use for the velocity vector V a complex function (1) V = V, + iV2 giving the magnitude \ V\ and direction Arg V of the velocity at each point z = x + iy. Here Vx and V2 are the components of the velocity in the x and y directions. V is tangential to the path of the moving particles, called a streamline of the motion (Fig. 410). We show that under suitable assumptions (explained in detail following the examples), for a given flow there exists an analytic function (2) F(z) = (*, y) + i¥(jc,.v), called the complex potential of the flow, such that the streamlines are given by "V(x, y) = const, and the velocity vector or, briefly, the velocity is given by (3) V = Vj + iV2 = F' (z) y X Fig. 410. Velocity 762 CHAP. 18 Complex Analysis and Potential Theory where the bar denotes the complex conjugate. is called the stream function. The function <£> is called the velocity potential. The curves $>(x, y) = const are called equipotential lines. The velocity vector V is the gradient of O; by definition, this means that (4) V, = d d.v V9 = 3 Indeed, for F = + i*, Eq. (4) in Sec. 13.4 is F' = y by the second Cauchy-Riemann equation. Together we obtain (3): J7?!) = *, - i¥x = *« + i% = Vt + iV2 = V. Furthermore, since F(z) is analytic, and ^ satisfy Laplace's equation (5) V2cJ> = civ2 + 0, V2^ - a2^ 0. Whereas in electrostatics the boundaries (conducting plates) are equipotential lines, in fluid flow the boundaries across which fluid cannot flow must be streamlines. Hence in fluid flow the stream function is of particular importance. Before discussing the conditions for the validity of the statements involving (2)-(5), let us consider two flows of practical interest, so that we first see what is going on from a practical point of view. Further (lows are included in the problem set. EXAMPLE 1 Flow Around a Corner The complex potential F(z) = z2 = x2 y2 + 2ixy models a flow with Equipotential lines <1> = x2 - y2 = const (Hyperbolas) Streamlines *P = 2xy = const (Hyperbolas). From (3) we obtain the velocity vector V --21= 2(x - iy), that is, V1 = 2x, V2 = -2y. The speed (magnitude of the velocity) is |V| = \'v2 + V22 = 2\i x2 + y2. The flow may be interpreted as the flow in a channel bounded by the positive coordinates axes and a hyperbola, say, xy = I (Fig. 411). We note that the speed along a streamline S has a minimum at the point P where the cross section of the channel is large. y I Fig. 411. Flow around a corner (Example 1) SEC. 18.4 Fluid Flow EXAMPLE 2 Flow Around a Cylinder Consider the complex potential 1 F(z) = *(*, y) + i*(.v. y) = i + - . Using the polar form z — re1"■ we obtain F(z) = re'0 + ~ e~ie = (r + j j cos 0 + i |r - j\ sin 0. Hence the streamlines are xl'(x, y) = (r - — j sin 9 const. In particular, 1?(x, y) = 0 gives r — Mr = 0 or sin 0 = 0. Hence this streamline consists of the unit circle (r = 1/r gives r = 1) and the x-axis (0=0 and 0 = it). For large |;| the term 1/z in is small in absolute value, so that for these z the flow is nearly uniform and parallel to the .v-axis. Hence we can interpret this as a How around a long circular cylinder of unit radius that is perpendicular to the ;-plane and intersects it in the unit circle |-| = I and whose axis corresponds to z = 0. The flow has two stagnation points (that is. point'- at which the velocity V is zero), at ; = ±1. This follows from (3) and F'(z) =1--j ' hcncc ~2 ~ 1 = °- (see fi§- 412-j " y Fig. 412. Flow around a cylinder (Example 2) Assumptions and Theory Underlying (2)-(5) THEOREM 1 Complex Potential of a Flow If the domain of flow is simply connected and the flow is irrotational and incompressible, then the statements involving (2)-(5) hold. In particular, then the flow has a complex potential F(z), which is an analytic function. (Explanation of terms below.) PROOF We prove this theorem, along with a discussion of basic concepts related to fluid flow. (a) First Assumption: Irrotational. Let C be any smooth curve in the z-plane given by z(s) - x(s) + iy(s), where s is the arc length of C. Let the real variable Vt be the component of the velocity Vtangent to C (Fig. 413). Then the value of the real line integral (6) 1 V. ds 764 CHAP. 18 Complex Analysis and Potential Theory Fig. 41 J. Tangential component of the velocity with respect to a curve C taken along C in the sense of increasing s is called the circulation of the fluid along C, a name that will be motivated as we proceed in this proof. Dividing the circulation by the length of C, we obtain the mean velocity1 of the flow along the curve C. Now Vt = \V\ cos a (Fig. 413). Hence Vt is the dot product (Sec. 9.2) of V and the tangent vector dzlds of C (Sec. 17.1); thus in (6), Vt ds dx + Vo dy ds ds The circulation (6) along C now becomes ds = VY dx + V2 dy. (7) f Vt ds = f (V, dx + V2 dy). As the next idea, let C be a closed curve satisfying the assumption as in Green' s theorem (Sec. 10.4), and let C be the boundary of a simply connected domain D. Suppose further that V has continuous partial derivatives in a domain containing D and C. Then we can use Green's theorem to represent the circulation around C by a double integral, (8) The integrand of this double integral is called the vorticity of the flow. The vorticity divided by 2 is called the rotation (9) 0(x, y) 1 / dV2 2 \ dx dV1 dy 1 f Definitions:---- f(x) dx b — a J mean value of / on the interval a s x ^ b, 1 f — I f(s) ds = mean value of / on C (L = length ol C), jj.li*. y) dx dy = mean value of / on D (A = area of D). SEC. 18.4 Fluid Flow 765 We assume the flow to be irrotational, that is, w(x, y) = 0 throughout the flow; thus, dV2 aVi (10) —--L = 0. ox dy To understand the physical meaning of vorticity and rotation, take for C in (8) a circle. Let r be the radius of C. Then the circulation divided by the length 2irr of C is the mean velocity of the fluid along C. Hence by dividing this by r we obtain the mean angular velocity . (b) Second Assumption: Incompressible. Our second assumption is that the fluid is incompressible. (Fluids include liquids, which are incompressible, and gases, such as air, which are compressible.) Then dV, dV2 (li) —- + —- = o dx dy in every region that is free of sources or sinks, that is, points at which fluid is produced or disappears, respectively. The expression in (11) is called the divergence of V and is denoted by div V. (See also (7) in Sec. 9.8.) (c) Complex Velocity Potential. If the domain D of the flow is simply connected (Sec. 14.2) and the flow is irrotational, then (10) implies that the line integral (7) is independent of path in D (by Theorem 3 in Sec. 10.2, where F1 = V,, F2 = V2, F3 = 0, and z is the third coordinate in space and has nothing to do with our present z). Hence if we integrate from a fixed point (a, b) in D to a variable point (x, y) in D, the integral becomes a function of the point (x, y), say, •): (12) $0, y) = (V, dx + V2 dy). (a. b) We claim that the flow has a velocity potential <$>, which is given by (12). To prove this, all we have to do is to show that (4) holds. Now since the integral (7) is independent of path, V1 dx + V2 dy is exact (Sec. 10.2), namely, the differential of a* V1 dx + V2 d\ = - dx + -dy. 3.x dy From this we see that V, = d<$>/dx and V2 = d

. Then the other equation in (5) holds. Also, since the second partial derivatives of and XP are continuous, we see that the complex function /•(;:> = «l>( v. vi + /M'l.v. y) is analytic in D. Since the curves xP(x, y) = const are perpendicular to the equipotential curves (x, y) = const (except where F'(z) = 0), we conclude that yV{x, y) = const are the streamlines. Hence ^ is the stream function and F(z) is the complex potential of the flow. This completes the proof of Theorem 1 as well as our discussion of the important role of complex analysis in compressible fluid flow. ■ 11-151 FLOW PATTERNS: STREAMLINES, COMPLEX POTENTIAL These problems should encourage you lo experiment with various functions F(z), many of which model interesting How patterns. 1. (Parallel flow) Show that F(z) = -iKz (K positive real) describes a uniform flow upward, which can be interpreted as a uniform flow between two parallel lines (parallel planes in three-dimensional space). Sec Fig. 414. Find the velocity vector, the streamlines, and the equipotential lines. y I X Fig. 414. Parallel flow in Problem 1 2. (Conformal mapping) Obtain the flow in Example 1 from that in Prob. 1 by a suitable conformal mapping. 3. Find the complex potential of a uniform flow parallel to the x-axis in the positive x-direction. 4. What happens lo the ilow in Prob. 1 if you replace z by ze~'" with constant a, e.g., a = tt/4? 5. What is the complex potential of an upward parallel flow in the direction of y = 2x? 6. (Extension of Example 1) Sketch or graph the flow in Example I on the whole upper half-plane. Show that you can interpret it as as flow against a horizontal wall (the x-axis). 7. What F(z) would be suitable in Example 1 if the angle of the corner were tt/3? 8. Sketch or graph the streamlines and equipotential lines of F(z) = iz3. Find V. Find all points at which V is parallel to the x-axis. 9. Find and graph the streamlines of F(z) = z.2 + 2z. Interpret the flow. 10. Show that F(z) = iz2 models a flow around a corner. Sketch the streamlines and equipotential lines. Find V. 11. (Potential F(z) = 1/z) Show that the streamlines of F(z) = 1/c are circles through the origin. 12. (Cylinder) What happens in Example 2 if you replace Z by -2? Sketch and interpret the resulting flow in the first quadrant. 13. Change F(z) in Example 2 slightly to obtain a flow around a cylinder of radius r0 that gives the flow in Example 2 if r0 —» 1. 14. (Aperture) Show that F(z) = arccosh z gives confoeal hyperbolas as streamlines, with foci at z = ±1, and the flow may be interpreted as a flow through an aperture (Fig. 415). 15. (Elliptical cylinder) Show that F(z) = arccos - gives confoeal ellipses as streamlines, with foci at z = ±1, and that the flow circulates around an elliptic cylinder or a plate (the segment from -1 to 1 in Fig. 416). Fig. 415. Flow through an aperture in Problem 14 767 Fig. 416. Flow around a plate in Problem 15 Fig. 417. Point source y I Fig. 418. Vortex flow 16. TEAM PROJIXT. Role of the Natural Logarithm in Modeling Flows, (a) Basic flows: Source and sink. Show thai F(z) = (c/2ir) In z with constant positive real c gives a flow directed radially outward (Fig. 417). so that F models a point source at ; — 0 (that is. a source line x = 0, y = 0 in space) at which fluid is produced, c is called the strength or discharge of the source. If c is negative real, show that the How is directed radially inward, so that F models a sink at z = 0, a point at which fluid disappears. Note that r = 0 is the singular point of F(z). (b) Basic Hows: Vortex. Show that F(z.) = -(KHItt) In z with positive real K gives a How circulating counterclockwise around z = 0 (Fig. 418). z = 0 is called a vortex. Note that each time we travel around the vortex, the potential increases by K. (c) Addition of flows. Show that addition of the velocity vectors of two flows gives a flow whose complex potential is obtained by adding the complex potentials of those flows. (d) Source and sink combined. Find the complex poteniials of a flow with a source of strength 1 at z = —a and of a flow with a sink of strength 1 at z = a. Add both and sketch or graph the streamlines. Show lhat for small \a\ these lines look similar lo those in Prob. 11. (e) Flow with circulation around a cylinder. Add the potential in (b) to that in Example 2. Show that this gives a How for which the cylinder wall |z\ = 1 is a streamline. Find the speed and show thai the stagnation points are \K I -K'2 4tt V l67^2 if K = 0 they are at ± 1; as K increases they move up on the unit circle until they unite at z = i(K = 4tt, sec Fig. 419), and if K > 4irthey lie on the imaginary axis (one lies in the field of flow and the other one lies inside the cylinder and has no physical meaning). Fig. 419. Flow around a cylinder without circulation (K = 0) and with circulation CHAP. 18 Complex Analysis and Potential Theory 18.5 Poisson's Integral Formula for Potentials So far in this chapter we have seen that complex analysis offers powerful methods for modeling and solving two-dimensional potential problems based on conformal mappings and complex potentials. A further method results from complex integration. As a most important result it yields Poisson's integral formula (5) for potentials in a standard domain (a circular disk) and from (5) a useful series (7) for these potentials. Hence we can solve problems for disks and then map solutions conformally onto other domains. Poisson's formula will follow from Cauchy's integral formula (Sec. 14.3) 1 £ F(z*) 2tti jc z z Here C is the circle z* = Rela (counterclockwise, O^aS 2tt), and we assume that F(z*) is analytic in a domain containing C and its full interior. Since dz* = iRela da = iz* da, we obtain from (1) (2) F(z) = -1- f F(z*) —^— da {z* = Re™, z = reie). Now comes a little trick. If instead of z inside C we take a Z outside C, the integrals (1) and (2) are zero by Cauchy's integral theorem (Sec. 14.2). We choose Z = z*z*/z = R2lz, which is outside C because |Z| = R2/\z\ = R2/r > R. From (2) we thus have z and by straightforward simplification of the last expression on the right, 1 f2- z 0 = — F(z*)--— da. 2tt j0 z - z* We subtract this from (2) and use the following formula that you can verify by direct calculation (zz* cancels): z* z z*z* - zz (3) ---= -. z* - z z-z* {z* - z)(z* - z) We then have (4) F(z) »77 J0 F(z*) z*z* (z* - z)(z* - z) da. From the polar representations of z and z* we see that the quotient in the integrand is real and equal to R2 - r2 R2 - r2 (Re'a - relU)(Re-la - re'10) R2 - 2Rr cos (6 - a) + f SEC. 18.5 Poisson's Integral Formula for Potentials 769 We now write F(z) = 0(r, 6) + i"*¥(r, 0) and take the real part on both sides of (4). Then we obtain Poisson's integral formula2 1 r2v R2 - r2 (5) (*, a)-----—-j da. 277 J0 « _ 2i?r cos (0 - a) + r This formula represents the harmonic function <3*> in the disk \z\ = R in terms of its values <&(/?, a) on the boundary (the circle) \z\ = R. Formula (5) is still valid if the boundary function <&(R, a) is merely piecewise continuous (as is practically often the case; see Fig. 401 in Sec. 18.2 for an example). Then (5) gives a function harmonic in the open disk, and on the circle |z| = R equal to the given boundary function, except at points where the latter is discontinuous. A proof can be found in Ref. [Dl] in App. 1. Series for Potentials in Disks From (5) we may obtain an important series development of <$> in terms of simple harmonic functions. We remember that the quotient in the integrand of (5) was derived from (3). We claim that the right side of (3) is the real part of _ 7*7 _|_ ;* + z _ (z* + z)(z* - z) z*-z ~ (z* - z)(f * - z) Indeed, the last denominator is real and so is z*z* — zz in the numerator, whereas ~~Z*Z + zz* = 2i Im (zz*') in the numerator is pure imaginary. This verifies our claim. Now by the use of the geometric series we obtain (develop the denominator) (6) z* + z + (z/z*) z* - z 1 - (z/z*) Since z = re10 and z* = Re'", we have - / \n~ I 7 \ Re n— 1 " rn . . . — = Re \z* _ _ Rn cos (nO — na). On the right, cos (nQ — na) = cos nd cos na + sin nB sin na. Hence from (6) we obtain Re = 1 (6*) 2 2 Re n=l = 1 + 22 R (cos nO cos na + sin nd sin na). 2SIMEON DENIS POISSON (1781-1840), French mathematician and physicist, professor in Paris from 1809. His work includes potential theory, partial differential equations (Poisson equation, Sec. 12.1), and probability (Sec. 24.7). 770 CHAP. 18 Complex Analysis and Potential Theory This expression is equal to the quotient in (5), as we have mentioned before, and by inserting it into (5) and integrating term by term with respect to a from 0 to 2tt we obtain (7) (R, a) da, an = — f (7?, a); see Sec. 11.1. Now for r = R the series (7) becomes the Fourier series of (7v, a) on the boundary can be represented by a Fourier series. EXAMPLE 1 Dirichlet Problem for the Unit Disk Find the electrostatic potential (r, 0) in the unit disk r < 1 having the boundary values q'/tt if — tt < a < 0 4>(1, a) alir if 0 < a < tt (Fig. 420). Solution. Since <1>(1, a) is even, bn = 0, and from (8) we obtain a0 = \ and a t a • da [a fa — cos na da + — cos na j tt jn tt — TT if 2 (cos hit — 1). Hence, a„ -4/(7! 7r ) if » is odd, an - 0 if n = 2, 4, • • •, and the potential is r cos 9 + cos 39 h —r cos 50 -I 32 52 Figure 421 shows the unit disk and some of the cquipotential lines (curves = const). Fig. 420. Boundary values in Example 1 Fig. 421. Potential in Example 1 SEC. 18.6 General Properties of Harmonic Functions 771 1. Verify (3). 2. Show that every term in (7) is a harmonic function in the disk r < R. 3. Give the details of the derivation of the series (7) from the Poisson formula (5). 4-13 HARMONIC FUNCTIONS IN A DISK Using (7), find the potential <£(?-, 0) in the unit disk r < 1 having the given boundary values $(1, 6). Using the sum of the first few terms of the series, compute some values of and sketch a figure of the equipolenlial lines. 4. 4X1, 6) = sin 26 5. *(1, 6) 6. (l, 0) 11. (l, 9) 12. (l, 0) 0) tt 2 sin2 0 cos2 56 0if -tt < 9 if 0 < 8 < 2tt sin3 26 cos4 9 92 if - 77 < 0 < 77 1 if -|77 < 0 < |t7 0 if 3-77 < 0<§77 13. $(1. 0) = 0if -^77 < 0 < <1>(1, 6) = 77 - 0 il'577 < 8 77. 14. TEAM PROJECT. Potential in a Disk, (a) Mean value property. Show that the value of a harmonic function + i"^. 15. CAS EXPERIMENT. Series (7). Write a program for scries developments (7). Experiment on accuracy by computing values from partial sums and comparing them with values that you obtain from your CAS graph. Do this (a) for Example 1 and Fig. 421, (b) for in Prob. 8 (which is discontinuous on the boundary!), (c) for a $ of your choice with continuous boundary values, (d) for with discontinuous boundary values. 18.6 General Properties of Harmonic Functions General properties of harmonic functions can often be obtained from properties of analytic functions in a simple fashion. Specifically, important mean value properties of harmonic functions follow readily from those of analytic functions. The details are as follows. THEOREM 1 Mean Value Property of Analytic Functions Lei f(z) be analytic in a simply connected domain D. Then the value of F(z) at a point Zo in D is equal to the mean value ofF(z) on any circle in D with center at Zo- PROOF In Cauchy's integral formula (Sec. 14.3) 1 f F(z) (1) F(z0) = —(x, y) be harmonic in a simply connected domain D. Then the value of tf>(x, y) at a point (x0, y0) in D is equal to the mean value of<&(x, y) on any circle in D with center at (x0, y0). This value is also equal to the mean value of <$>{x, y) on any circular disk in D with center (x0, y0). [Sec footnote 1 in Sec. 18.4.] PROOF The first part of the theorem follows from (2) by taking the real parts on both sides, 1 c27T OOo v0) = Re F(x0 + iv0) = — ®(xo + >"cos a< v0 + r sin a) da. 2-7T -»o The second part of the theorem follows by integrating this formula over r from 0 to r0 (the radius of the disk) and dividing by r02/2, (3) 1 trr0 J0 J0 j j d5(x0 + r cos a, y0 + r sin a) r da dr. The right side is the indicated mean value (integral divided by the area of the region of integration). Returning to analytic functions, we state and prove another famous consequence of Cauchy's integral formula. The proof is indirect and shows quite a nice idea of applying the ML-inequality. (A bounded region is a region that lies entirely in some circle about the origin.) THEOREM 3 Maximum Modulus Theorem for Analytic Functions Let F(z) be analytic and nonconstant in a domain containing a bounded region R and its boundary. Then the absolute value \F(z)\ cannot have a maximum at an interior point of R. Consequently, the maximum of\F(z)\ is taken on the boundary of R. If F(z) + 0 in R, the same is true with respect to the minimum of\F{z)\. PROOF We assume that \F(z)\ has a maximum at an interior point z0 of R and show that this leads to a contradiction. Let |F(z0)| = M be this maximum. Since F(z) is not constant, \F(z)\ is not constant, as follows from Example 3 in Sec. 13.4. Consequently, we can find a circle C of radius r with center at z0 such that the interior of C is in R and \F(z)\ is smaller than M at some point P of C. Since |F(z) is continuous, it will be smaller than M on an arc C, of C that contains P (see Fig. 422), say, |F(z)| g M - k (k > 0) for all z on Ct. SEC. 18.6 General Properties of Harmonic Functions Let Ct have the length Lx. Then the complementary arc C2 of C has the length 2irr — L\. We now apply the ML-inequality (Sec. 14.1) to (1) and note that \z — z0| = r. We then obtain (using straightforward calculation in the second line of the formula) M = \F(Zo 1 I 277 1: F(z.) dz 1 i F(z) 1 IM — k\ 1 I M\ ~ - Lx + — — (2i 7T \ r / 277 \ r J Lj) = M 2irr < M that is, M < M, which is impossible. Hence our assumption is false and the first statement is proved. Next we prove the second statement. If F(z) + 0 in then l/F(z) is analytic in 7a\ From the statement already proved it follows that the maximum of l/|F(z)| lies on the boundary of R. But this maximum corresponds to the minimum of \F(z)\. This completes the proof. ■ This theorem has several fundamental consequences for harmonic functions, as follows. THEOREM 4 Harmonic Functions Let (x, y) be harmonic in a domain containing a simply connected bounded region R and its boundary curve C. Then: (I) (Maximum principle) lf<£>(x, y) is not constant, it has neither a maximum nor a minimum in R. Consequently, the maximum and the minimum are taken on the boundary of R. (II) lf<$>(x, y) is constant on C, then (x, y) is a constant. (III) If h(x, y) is harmonic in R and on C and if h(x, y) = <&(a', y) on C, then h(x, y) = <£>(x, y) everywhere in R. PROOF (I) Let ty(x, >) be a conjugate harmonic function of (x, y) in R. Then the complex function F(z) = ^(x, y) + ixV(x, y) is analytic in R, and so is G(z) = eF(z). Its absolute value is (0(2)1 = erh f(z) = e*te' y\ From Theorem 3 it follows that \G(z)\ cannot have a maximum at an interior point of R. Since is a monotone increasing function of the real variable the statement about the CHAP. 18 Complex Analysis and Potential Theory maximum of follows. From this, the statement about the minimum follows by replacing by -. (II) By (I) the function (.v, y) takes its maximum and its minimum on C. Thus, if (.i\ y) must be a constant. (III) If h and 4> are harmonic in R and on C, then h — is also harmonic in R and on C, and by assumption, h — <5 = 0 everywhere on C. By (II) we thus have h — <6 = 0 everywhere in /?, and (III) is proved. ■ The last statement of Theorem 4 is very important. It means that a harmonic function is uniquely determined in R by its values on the boundary of R. Usually, (jr, y) = dMxn, y0), where (jc0, y0) is on the boundary and (x, y) is in R. Under these assumptions the maximum principle (I) is still applicable. The problem of determining (.v, y) when the boundary values are given is called the Dirichlet problem for the Laplace equation in two variables, as we know. From (III) we thus have, as a highlight of our discussion. THEOREM 5 Uniqueness Theorem for the Dirichlet Problem If for a given region and given boundary values the Dirichlet problem for the Laplace equation in two variables has a solution, the solution is unique. PROBLEM SET 18.6 1. Integrate \z\2 around the unit circle. Does your result contradict Theorem 1 ? 2-4 | VERIFY THEOREM 1 for the given F(z), ;0, and circle of radius 1. 2. (z + I)3, 3. (z 4. lOz* -2)2 -« . = 0 _ l 0 _ 3 to 5-7 VERIFY THEOREM 2 for the given M.v. >•), (Xq, >'o) and circle of radius 1. 5. (x - 2)(y - 2), (4, -4) 6. x .2 V2, (3, 8) (1, 1) 8. Derive Theorem 2 from Poisson's integral formula. 9. CAS EXPERIMENT. Graphing Potentials. Graph the potentials in Probs. 5 and 7 and for three other functions of your choice as surfaces over a rectangle or a disk in the .vy-plane. Find the locations of maxima and minima by inspecting these graphs. 10. TEAM PROJECT. Maximum Modulus of Analytic Functions, (a) Verify Theorem 3 for (i) F(z) = z2 and the square 4 S jr S 6, 2 S y S 4, (ii) Flz) and any bounded domain, (iii) F(z) = sin z and the unit disk. (b) F(x) = cos.r (x real) has a maximum 1 at 0. How does it follow that this cannot be a maximum of \F(z)\ = |cos z\ in a domain containing z = 0? (c) F(z) = 1 + |z|2 is not zero in the disk \z\ ~: 4 and has a minimum at an interior point. Does this contradict Theorem 3? (d) If F(z) is analytic and not constant in the closed unit disk D: \z\ = 1 and \F(z)\ = c = const on the unit circle, show that F(z) must have a zero in D. Can you extend this to an arbitrary simple closed curve? Chapter 18 Review Questions and Problems 11-13 MAXIMUM MODULUS Find the location and size of the maximum of \F(z)\ in the unit disk \z\ = 1. 11. F(z) = z2 - 1 12. /•'(-) = az + b {a, b complex) 13. F(z) = cos 2z 14. Verify the maximum principle for (x, y) and the rectangle a b, 0 15. (Conjugate) Do and a harmonic conjugate XV of in a region R have their maximum at the same point of R7 16. (Conformal mapping) Find the location of the maximum of * = eu cos v in 7?*: |w| § 1, c 6 0, where w = it + z'p. Find the region R that is mapped onto R* by vv = f(z) = z2. Find the potential in R resulting from $* and the location (xla yx) of the maximum. Is (ttj, ff) the image of (x^ v^)? If so, is this just by chance? 3E3Hra^»EE&EffHEffiBB^aKSTIONS AND PROBLEMS 1. Why can potential problems be modeled and solved by complex analysis? For what dimensions? 2. What is a harmonic function? A harmonic conjugate? 3. Give a few examples of potential problems considered in this chapter. 4. What is a complex potential? What does it give physically? 5. How can conformal mapping be used in connection with the Dirichlet problem? 6. What heat problems reduce to potential problems? Give a few examples. 7. Write a short essay on potential theory in fluid flow from memory. 8. What is a mixed boundary value problem? Where did it occur? 9. Stale Poisson's formula and its derivation from Cauchy's formula. 10. State the maximum modulus theorem and mean value theorems for harmonic functions. 11. Find the potential and complex potential between the plates y = x and y = x + 10 kept at 10 V and 110 V, respectively. 12. Find the potential between the cylinders \z\ = 1 cm having potential 0 and \z\ = 10 cm having potential 20 kV. ' 13. Find the complex potential in Prob. 12. 14. Find the equipotential line U = 0 V between the cylinders |z| = 0.25 cm and \z\ = 4 cm kept at —220 V and 220 V, respectively. (Guess first.) 15. Find the potential between the cylinders \z\ = 10 cm and \z\ = 100 cm kept at the potentials 10 kV and 0. respectively. 16. Find the potential in the angular region between the plates Arg z = tt/6, kept at 8 kV, and Arg z = 7i73, kept at 6 kV. 17. Find the equipotential lines of F(z) = i Ln z. 18. Find and sketch the equipotential lines of F(z) = (1 + j)/z. 19. What is the complex potential in the upper half-plane if the negative half of the jr-axis has potential 1 kV and the positive half is grounded? 20. Find the potential on the ray y = x, x > 0, and on the positive half of the x-axis if the positive half of the y-axis is at 1200 V and the negative half is grounded. 21. Interpret Prob. 20 as a problem in heat conduction. 22. Find the temperature in the upper half-plane if the portion x > 2 of the x-axis is kept at 50°C and the other portion at 0°C. 23. Show that the isotherms of F(z) = -iz2 + z are hyperbolas. 24. If the region between two concentric cylinders of radii 2 cm and 10 cm contains water and the outer cylinder is kepi at 20°C, to what temperature must we heat the inner cylinder in order to have 30°C at distance 5 cm from the axis? 25. Whal are the streamlines of F(z) = ilz?. 26. What is the complex potential of a flow around a cylinder of radius 4 without circulation' 27. Find the complex potential of a source at z = 5. What are the streamlines? 28. Find the temperature in the unit disk \z\ — 1 in the form of an infinite series if the left semicircle of \z\ = 1 has the temperature of 50°C and the right semicircle has the temperature 0°C. 29. Same task as in Prob. 28 if the upper semicircle is at 40°C and the lower at 0"C. 30. Find a scries for the potential in the unit disk with boundary values $(1, 0} = 62 (-v < $ < it). 776 CHAP. 18 Complex Analysis and Potential Theory Potential theory is the theory of solutions of Laplace's equation (1) V2 = 0. Solutions whose second partial derivatives are continuous are called harmonic functions. Equation (1) is the most important PDE in physics, where it is of interest in two and three dimensions. It appears in electrostatics (Sec. 18.1), steady-state heat problems (Sec. 18.3), fluid flow (Sec. 18.4), gravity, etc. Whereas the three-dimensional case requires other methods (see Chap. 12), two-dimensional potential theory can be handled by complex analysis, since the real and imaginary parts of an analytic function are harmonic (Sec. 13.4). They remain harmonic under conformal mapping (Sec. 18.2), so that conformal mapping becomes a powerful tool in solving boundary value problems for (1), as is illustrated in this chapter. With a real potential $ in (1) we can associate a complex potential Then both families of curves = const and yV = const have a physical meaning. In electrostatics, they are equipotcntial lines and lines of electrical force (Sec. 18.1). In heat problems, they are isotherms (curves of constant temperature) and lines of heat flow (Sec. 18.3). In fluid flow, they are equipotential lines of the velocity potential and streamlines (Sec. 18.4). For the disk, the solution of the Dirichlet problem is given by the Poisson formula (Sec. 18.5) or by a series that on the boundary circle becomes the Fourier series of the given boundary values (Sec. 18.5). Harmonic functions, like analytic functions, have a number of general properties; particularly important are the mean value property and the maximum modulus property (Sec. 18.6), which implies the uniqueness of the solution of the Dirichlet problem (Theorem 5 in Sec. 18.6). (2) F{z) =