MTH302 Unit 6 Harmonic FunctionsTransforming Circles: ExampleFind the image of the circle C(0, 2) under theMobius transformation T(z) = ( 1+z)/(z-1).T(z) = (z-1+2)/(z-1)=1 + 2/(z-1)Since C(0,2) does not pass through 1, theimage is a circle.T is a composite of (1) a translation by -1,(2) An inversion, (3) a magnification by 2 and(4) translation by 1C(0,2) is mapped onto C(-1,2)by the translation by -1 with equationx2 + y2 + 2x 3= 0 (1)In general form A(x2 + y2) + Bx + Cy + D = 0A = 1, B = 2, C = 0 and D = -3The circle (1) is mapped under inversion to the circlewith equationA + Bu Cv +D(u2 + v2) = 0,i.e., 1 + 2u 3(u2+v2) = 01 + 2u 3(u2+v2) = 0i.e., (u-1/3)2 + v2 = 4/9 = (2/3)2A circle centred at (1/3,0) of radius 2/3, C(1/3, 2/3).Under magnification by 2, the circle C(1/3, 2/3) ismapped onto the circle C(2/3, 4/3).Finally under translation by 1, this circle is mappedontoC(5/3, 4/3).Harmonic maps: Definition(Defnnition 1.1.)Let D be a domain in C. Suppose u : D ? R is a realvalued function. (Here we are taking the complexplane C as R2 . We say the function u is harmonic onD if u(x, y) has continuous partial derivatives uptoand including second and satisfies the followingLaplace equations.Ø2uØx2 + ØØx2u2 = 0The relationship is stated as the following theorem.Theorem(Theorem 1.2.)Suppose D is a domain in C and f : D ? C isanalytic. Then the real and imaginary parts of f isharmonic on D, i.e., if f = u + i v, then u and v areharmonic on D.Proof. Recall from Corollary 1.10 Unit 6MTH301, that if f is analytic at z = a + i b, then uand v has continuous partial derivatives of allorders at (a, b). Thus they both have continuouspartial derivatives upto and including the secondorder. We have actually shown that both u and vsatisfy the Laplace equations.Definition(Definition 1.3.) Suppose u : D ? R is a harmonicfunction. If v: D ? R is a harmonic function on Dsuch that the function f = u + iv is analytic on D,then v is called a harmonic conjugate of u.Theorem(Theorem 1.4.)Let D be a simply connected domain in C.If u : D ? R is harmonic on D, then it has aharmonic conjugate function v on D.If v1 is another harmonic conjugate of u, then v v1 is a constant function on D.Proof. The scheme of the proof is to produce ananalytic function f on D with u as its real part.Observe that if f exists, then by the Cauchy-RiemannTheorem (Theorem 2.2 Unit 4 MTH301),f ?(a + bi) = ØØux(a, b) iØØuy(a, b)(Cauchy Riemann equations.) So we see that thederivative of f, if it exists is entirely determined by u.All we need do is to find a primitive for the expression.ØuØx(a, b) iØØuy(a, b)Let g : D ? C be defined byg(x + iy) = ØØux (x, y) i ØØuy (x, y).Since u is harmonic, all partial derivatives of u upto and includingthe second order are continuous on D. Therefore, the partialderivatives of real and imaginary parts of g exist and arecontinuous on D. In particular, the real and imaginary parts of gsatisfies the Cauchy-Riemann equations on D, as ØØxØuØx (x, y) = on DØØu y (x, y) ØØy because u satisfies the Laplace equation,and ØØyØuØx (x, y) =on D,ØØuy (x, y) ØØx because the mixed second partial derivatives are equalIt follows by Theorem 2.3 Unit 4 MTH301, that gis analytic on D. Since D is simply connected (forexample if D is star-shaped), then by ageneralization of the Cauchys Integral Theorem tosimply connected domain, g has a primitive say f, such that f = g on D. Supposef = U + i V. Thenf ?(x + yi) g(x, y)= .ØUØx (x, y) ØØux(x, y) i ØØUy (x, y) ØØux(x, y) = 0Thus we have,Ø(U u) on D .Øx (x, y) = 0 and Ø(UØy- u)(x, y) = 0Note that U u has continuous partial derivatives and obviouslysatisfies the Cauchy Riemann equations and so U u is analytic onD. Plainly h = U u satisfies h = 0 on D. Therefore, since D is adomain in C (is open and path connected), by Corollary 3.13 Unit 4MTH301, h is a constant function. Since U u is real, U = u + cfor some real number c. Hence F = f c is an analytic functionsuch that Re F = U c = u and the Im F = V is the harmonicconjugate of u on D.Suppose V1 is another harmonic conjugate of u. Then F1 = u + iV1is analytic on D. ThenF F1 = i(V V1) is analytic on D. As a complex function (F F1) = 0 on D. Thus as above, by Corollary 3.13 Unit 4 MTH301,F F1 is a constant function. Alternatively, we can use the OpenMapping Theorem of Unit 3 to deduce that F F1 is a constantfunction. Hence Real part of F F1 is 0 and V V1 is a constant onD.Example(Example 1.5.) Let D be the punctured diskD'(0,2) and letu(x, y) = ln(x2 + y2) , ((x, y) ? D).Note that D is not simply connected. Then u isharmonic on D but u has no harmonic conjugateon D.Proof. It is easy to verify that u is harmonic on D.Suppose u has a harmonic conjugate v on D.Define g :[0, 2?] ? R byg(t) = v(cos(t), sin(t)), for t in[0, 2?].Then g is continuous and differentiable on [0, 2?] withg(0) = g(2?).By the Chain Rule,g?(t) = Ø Øv x((cos(t), sin(t))(- sin(t)) + Ø Øv y((cos(t), sin(t))(cos(t)).Now u and v satisfies the Cauchy Riemann equations, and,ØuØx (x, y) = x22+xy2 and ØØuy (x, y) = x22+yy2g?(t) = -ØØuy ((cos(t), sin(t))(- sin(t)) + ØØu x ((cos(t), sin(t)) cos(t)= 2 sin(t)cos 2(t) + sin2(t)(- sin(t)) + cos22(tco ) +s(si t)n2(t) cos(t) = 2for all t in (0, 2?).Therefore, integrating we getg(t) = g(0) + 2t, for t in [0, 2?].It follows that g(2?) = g(0) + 4? ? g(0). Thiscontradicts g(0) = g(2?) and so u cannot have aharmonic conjugate.Examples: finding harmonic conjugate(Example 1.6.) We can actually construct a harmonic conjugatedirectly.Find the harmonic conjugate of u(x, y) = x3 3xy2 + y on C.u is harmonic on C and C is simply connected.If a conjugate harmonic function v exists and f = u + iv is analyticon C, then by the Cauchy Riemann equationsØ Øv y(x, y) = ØØux(x, y) and Ø Øv x(x, y) = -ØØuy(x, y) . (1)Choose (a, b) in C, we can define for (x, y) in C,v(x, y) = -¶ ax ØØuy(t, b)dt + ¶by ØØu x (x, t)dt.Now ØØu x (x, y) = 3x 2 3y 2 , ØØuy(x, y) = -6xy + 1. (2)Therefore, taking a = b = 0, we have¶0y ØØux(x, t)dt = ¶0y(3×2 3t2)dt = 3x2y y3, ¶0x ØØuy(t, 0)dt = ¶1dt = x, 0x and sov(x, y) = -x + 3x2y y3is a harmonic conjugate of u.We may do this in steps. From (1) and (2) we getØ Øv y(x, y) = 3×2 3y2 - (3)Ø Øv x(x, y) = 6xy 1 (4)From (3) by integrating with respect to y and keeping xconstant, we getv(x, y) = 3x 2y y3 + g(x). (5)Then from (5) we get Ø Øv x(x, y) = 6xy + g?(x) (6) Therefore, equating this with (4)we musy have g'(x) = -1 .Hence g(x) = x + c for some constant c. Thusv(x, y) = 3x 2y y3 + g(x) = 3x2y y3 x + c.(Example 1.7.) Let u(x, y) = x2 y2 and v(x,y) = 2
xy. Then v is aharmonic conjugate of u on C but u is not a harmonic conjugate of von C.Proof. Notice that u and v are respectively the real and imaginaryparts of the function f (z) = z2 , which is obviously analytic on C.Hence v is a harmonic conjugate of u. Now we have Now we examine the Cauchy Riemann equations for g(x, y) = v(x,y) +i u(x, y).The equations areØv , and ,Øx(x, y) = ØØuy (x, y) Ø Øv y(x, y) = -ØØux(x, y)which gives 2y = 2y and 2x = -2x.Hence y = 0 and x = 0 and g is differentiable at (0,0) only and nowhereelse. Therefore, g is nowhere analytic. That means u cannot be ahamonic conjugate of v.Theorem(Theorem 1.8. )Suppose D is a domain in C.Suppose u:D ? R is a harmonic function on D.If u is non-constant, then u has no absolutemaximum nor absolute minimum points on D.Proof. First we show that there does not exist an open disk in D onwhich u is constant.Suppose there is such a disk D(z0, r0) in D such that u(z) = C for allu in D(z0, r0).Take any ? in D. Then since D is path connected, there exists apath ? in D joining z0 to ?.Since D is open, for each point z in ?*, there exists a disk D(z, r(z))for some r(z) > 0 such that D(z, r(z)) ? D.The collection of these disks together with D(z0, r0 ) is an opencover for ?*.Since ?? is compact, it has a finite subcover D(z0, r0 ), D(z1, r1),D(z2, r2), , D(zn, rn), zk ? ??, such that?? ? D(z0, r0 )? D(z1, r1)?D(z2, r2)? ?D(zn, rn) = H ? D.Then H is open and path connected since ??? D(zk, rk) ?? for k =0, 1, , n and ? is a path. H contains both z0 and ?.By Theorem 1.4, u has harmonic conjugate functions vk on D(zk, rk) suchthat f k = u + vk is analytic on D(zk, rk), k = 0, 1, , n. Since ?? is pathconnected, we can order these disks in such a way thatD(zk, rk)?( D(z0, r)? D(z1, r1)?D(z2, r2)? D(zk-1, rk-1)) ?? for k =1, , n. Since u is constant on D(z0, r0), f 0 is constant on D(z0, r0).Let f 0 = c on D(z0, r0). Then Re c = C.Since D(z1, r1 )? D(z0, r) ?? , by Theorem 1.3 Unit 5, f 1 is constant onD(z1, r1 ) and so the Re f 1 = u is constant on D(z1, r1 ) and is equal to C.By setting Im f 1 = Im c, then f 1 is also the constant function c.Repeat this argument for f 2 , since D(z2, r2)?( D(z0, r)? D(z1, r1)) ?? ,deducing that Re f 2 = C on D(z2, r2) and we can take f 2 to be theconstant function c.In this way we deduce that u is constant on H and so is constant on ?* .Therefore u(?) = C.Since ? is arbitrary, this shows that u is constant on the whole of D.This is the required contradiction.u is an open mapping:Take a point ? in D. Since D is open, there exists an open disk D(?,r) for some r > 0 such that D(?, r) ? D.Then u is non-constant on D(?, r). By Theorem 1.4, there is ananalytic function f = u|D(,r) + iv on D(?, r).Since u is non-constant, f is non-constant on D(?, r). It follows bythe Open Mapping Theorem (Theorem 2.2 unit 3) that f (D(?, r)) isopen in C.That means there is ? > 0 such thatD( f (?), ? ) ? f (D(?, r)). - (1)Now the interval (u(?) ?, u(?) +?), satisfies(u(?) ?, u(?) +?)×v(?) = { (x, v(?)): x ? (u(?) ?, u(?) +?)}? D( f (?), ? ).Then (1) says that(u(?) ?, u(?) +?) ? u(D(?, r))) ? u(D).Therefor, u(D) is a neighbourhood of u(?) for each ? in D and so u(D)is open.Therefore, u is an open mapping on D. Therefore,it has no local maximum nor local minimumpoints. It follows that u has no absolute maximumnor absolute minimum points on D.Theorem(Theorem 1.9. )Suppose D is a domain in C and u:D ? R is aharmonic function on D. If u is non-constant, thenu is an open mapping.(Corollary 1.10. )If u : D ? R is harmonic on a domain D and uattains its maximum or minimum value at somepoint in D, then u is a constant function.Theorem(Theorem 1.11.)Suppose D is a bounded domain in C and D itsclosure.If u is a continuous real value function on Dharmonic on D, then u attains its absolutemaximum and minimum on the boundary ?D.Proof. Since D is closed and bounded, by the HeineBorel Theorem (Theorem 2.19 Unit 2 MTH301), Dis compact.By the Extreme Value Theorem (Theorem 1.10 Unit3 MTH301) u must attain its absolute maxmum andminimum on D .If u attains its maximum and minimum on theboundary ?D, we have nothing to prove.If u attains its absolute maximum or minimum in D,then by Corollary 1.10, u is constant on D andconsequently, by continuity u is constant on D .Therefore, it attains its absolute maximum andminimum on the boundary ?D.(Corollary 1.12. )Suppose D is a bounded domain in C and D itsclosure. Suppose u is a continuous real valuefunction on D harmonic on D.If u is non-constant, then u attains its absolutemaximum and minimum on the boundary ?D andnowhere else.(Theorem 1.13.)Suppose D is a bounded domain in C and D its closure.If ? and ? are continuous real value functions definedon D and are harmonic on D and if ? and ? agree onthe boundary ?D, then ? is identically equal to ? on D .Proof. If ? and ? are harmonic on D, then theirdifference ? ? is also harmonic on D. By Theorem1.11, ? ? attains its absolute maximum andminimum on the boundary ?D.But ? ? = 0 on the boundary ?D as ? and ? agree onthe boundary ?D and so ? ? = 0 on D becuaeabsolute maximum and absolute minimum are equal to0.. It follows that ? is identically equal to ? on D .(Theorem 1.14.)Suppose that an analytic functionw = f (z) = u(x, y) + i v(x, y) (B)maps a domain Dz in the z-plane onto a domain Dwin the w-plane.If h(x,y) is a harmonic function defined on Dw , thenH(x,y) = h(u(x,y), v(x, y)) - (C)is harmonic in Dz.Proof. We first prove the simpler case when Dw issimply connected.Suppose now Dw is simply connected. Then byTheorem 1.4, h(x, y) has a harmonic conjugate g(x, y)so that the functionF (z) = F(x + i y) = h(x, y) + i g(x, y) (1)is analytic on Dw.By the Chain Rule F ) f is analytic on Dz since f mapsDz onto Dw.The real part of F ) f is given byh(u(x,y), v(x, y)) = H(x, y)on Dz and so H is harmonic by Theorem 1.2.If Dw is not simply connected, we use the local toglobal approach.Take any z0 in Dz. Let w0 = f (z0). Since Dw is openany point w0 has an open disk D(w0, ?) for some ? >0 such that D(w0, ?) ? Dw .Since D(w0, ?) is simply connected by the simplercase, there is a function F of the form (1) analytic onD(w0, ?). Since f is contnuous, there is a ? > 0 suchthat f ( D(z0, ?)) ? D(w0, ?).It follows that the composition F ) f is analytic onD(z0, ?)). Therefore, h(u(x,y), v(x, y)) = H(x, y) isharmonic in D(z0, ?)). Since this is true for any z0 inDz , h(u(x,y), v(x, y)) = H(x, y) is harmonic on Dz.Example(Example 1.14.)Let Dz = {(x,y): x > 0, y > 0}. Then Dz consists ofpoints in the first quadrant.Let Dw = {(u, v): v > 0} = points in the upper halfplane.Let f (z) = z2. Then f is analytic on C. Inparticular, f maps Dz onto Dw .NowDw = {(u, v): v > 0} ={ r ei ? : r > 0, 0 0, y > 0} = { r ei ? : r > 0, 0