MTH302 Unit 6 Fluid Flow & Julia SetsDefinitionThe fluid velocity V will be a vector field depending on the coorrdinates x and y.That is V is a vector field on R2 considered as the complex plane C.The components of the velocity V along the coordinate axes are Vx and Vy.Think of V as the complex velocity functionV(x,y) = Vx(x,y) + i Vy(x,y), (D)regarding complex numbers as vectors. Under suitable condition asteady(irrotational) fluid flow, there is a velocity potential function ? (dependentupon the presure) such thatV = grad ?,that is, Vx = ØØ&x and Vy = ØØ&y. (E)Under steady state condition in a domain free from sources and sinks, thevolume of fluid in the domain remains constant in time. This means as muchfluid flows into during any time as flows out. This requires that the potentialfunction must satisfy the Laplace equation,.Ø2&Øx2 +Ø2&Øy2 = 0Assuming all partial derivatives including the secondorder are continuous, ? is a harmonic function in thedomain. Assuming that the domain is simplyconnected, by Theorem 1.4, ? has a harmonicconjugate ? such that?(x, y) = ?(x, y) + i ?(x,y)is analytic in the domain D.? is called a stream function. ? is called the complexvelocity potential, the level curves ?(x, y) = c arecalled equipotentials and the level curves ?(x, y) = care called the streamlines.The two families of curves are orthogonal, i.e., theequipotentials and the streamlines are orthogonalto each orther.In particular a boundary across which fluid cannotflow is a streamline.Thus the fluid velocity vector at a point is alwaystangent to the streamline passing through thatpoint.If we follow the progress of a moving particle ofthe fluid, we find that its path is the streamline.Now by the Cauchy-Rieman equations,,ddz =Ø&Øx iØ&Øy so thatddz =Øx + iØy = V,or? = V - (F) Ø&Ø&i,e,, the conjugate of the derivative of the complexpotential function is the velocity function.Therefore, the speed or the magnitude of thevelocity vector V is given byV = ? = ? - (G)Example(Example 2.1.) Uniform Flow across the upper plane.Suppose the complex potential function is given by?(z) = 5z.Then ?(x, y) = 5x and ?(x, y) = 5y, where ?(x, y) = ?(x, y) + i ?(x,y).Therefore, the steamlines ?(x, y) = c are horizontal lines y = c/5 .The velocity at any point is given by V = ? = 5.Note that then the x-axis, y = 0 is a stream line. OxV(x, y) y(5,0)Fig 1.?(x,y) is the rate of flow across any contour drawnfrom the origin to the point (x, y) as shown inFig.1.The flow is uniform and is to the right as thevelocity function is (5,0).We can interpret this as the uniform flow in theupper half plane bounded by the x-axis which is astreamline.The function ?(x,y)= 5y is harmonic in the upperhalf plane H and is 0 on the boundary y = 0.A harmonic function is not always determined by the values on theboundary of the domain.We can take any other harmonic function that vanishes on thex-axis. For instanceg(x, y) = sin(y)ex .is harmonic on H and vanishes on the boundary but the streamlineg(x, y) = 0 not only consists of the x axis but also all the horizontallines y = n?, n = 1,2, . In this case the complex potentialfunction is?1(z)=ez .The velocity potential function ?1 is the real part of ?1(z) and isequal to cos(y)ex .Then the velocity function is given byV1(x, y) = ?(x, y) = ex cos(y) ie x sin(y).Thus, the fluid in this case flows to the right along the x-axis and tothe left along y = ? .Example(Eample 2.2. )We shall use the anlytic function f (z) = z2 to transform theflow in the half plane to a flow in the first quadrant.Let Dz = {(x,y): x > 0, y > 0} = points in the first quadrant.Let Dw = {(u, v): v > 0} = points in the upper half plane.Let f (z) = z2. f maps Dz onto Dw. (See Example 1.15),The complex potential function for the uniform flow to theright in the upper half plane in Example 2.1 is ?(z) = 5z.The complex potential function for the flow in the firstquadrant is then given by ?( f (z)) = 5z2 . It isF(x, y) = 5(x2 y2) + i 10xy.Therefore, the stream function for the flow in the first quadrantis given by?(x, y)= 10x y.The stream function is harmonic in the first quadrant andobviously vanishes on the boundary of the first quadrant.The streamlines ?(x, y)= 10x y = c, are branches of therectangular hyperbolas10x y = cin the first quadrant. By using (F) the velocity of the fluid V isgiven byV(z) = F?(z) = 10z = 10x i10yand the speed of the fluid is then given by|V(z)| = 10 x2 + y2 .Note that the speed of the particle is directly proportional to itsdistance from the origin. Note that V(z) always pointdownward. Therefore, the streamlines come in downwardparallel to the y-axis and are forced to turn a corner near theorigin and proceed away from the origin but parallel to thex-axis as shown in Fig. 2.y OxFig 2.Example(Example 2.3.) Flow pass an obstacle.Consider planar uniform flow pass a circurlar obstacle.Represent the circlular obstclae by a circle of radius 1.By symmetry we may consider the domain as the region above thehalf plane and exterior to the unit circle.Its boundary is the upper unit circle and the parts of the x-axisexterior to the unit circle.We need to find an analytic function mapping this domain to theupper half plane. Such a function is provided by f (z) = z + 1/z.Let Dw = {(u, v): v > 0} = points in the upper half plane and let Dz bepoints in the upper half plane above the unit circle.Take the complex potential function for the uniform flow to the rightin the upper half plane Dw as in Example 2.1 to be ?(z) = 5z. Thenthe complex potential function for the flow in the given region Dz isgiven byF(z) = ?( f (z)) = 5( z + 1/z).The velocity function,V(z) = F?(z) = 5(1 1 . - (1)z2 ) = 5(1 1z2 )From the above expression, we note thatV(z) = V(z)which means the velocity function also represent the flow inthe lower region in thelower half plane below the unit circle.The stream function is given by the following i n polarco-ordinate,)(r, ) = 5(r 1r ) sin(). - (2)From (1) , we see that the velocity function approaches 5 as ztends to ?.The streamlines,5(r 1r ) sin() = care clearly symemtric about the y = axis and also about thex-axis (taking the lower region together).The streamline 5(r 1r ) sin() = 0 consists of the unit circle andpart of the x axis exterior to the circle (see Fig 3).yx 1Fig. 3 ExampleFind the electrostatic potential ? in the semidiskwith the boundary values as shown below, -1? = 01? = 0i?=1?=1 To find a harmonic function on the semi disk, wefirst map the semidisk conformally to the whole ofthe first quadrant by a map T then define an obviousharmonic map ? on the first quadrant. Then byTheorem 1.14 Unit 6 , ? = ? T is the requiredpotential function.To find the conformal map T we only need to specifythe images of 3 points.We want T(-1) = 0, T(1) = ? and T(i) = i.Then T maps the real segments [-1, 1] to the positiveu-axis and the semicircular arc to the positivev-axis.Let T(z) = az + b ,(ad ? bc).cz + d T(1) = a + bc + d = ?if and only if a + b ? 0, c + d = 0 - (1) and T(-1 ) = -a + b if and only if a = b and c ? d (2)-c + d = 0and T(i ) = ai + b if and only ifci + d = iai + b = -c + di - (3)(1) and (2) implies that a,b,c,d are not 0.Hence we have from (1) c = -d, a = b ,(3) gives a(i+1) = d(1+i) so that a = d.Hence T(z) = dz + d .-dz + d =1 + z1
zNote that T(2i ) = 3 5 + 4 5i is in the first quadrant and so since i/2is in the interior of the semidisk, T maps the interior of thesemi disk onto the interior of the first quadrant. T maps thez-plane to the w-plane.Define the harmonic function ? on the w-plane by)(w) = 2Arg(w) = 2 tan-1(uy) if u ! 0 .2 if u = 0It can be easily verified that ? is harmonic in the first quadrantof the w-plane minus the origin, ? = 0 on the positive u-axisand 1 on the positive vertical axis.Th e re fo re, & (z) = ) (T (z)) = 2 A rg 1 1 + z z is th ereq u ired p oten tial fu n ction ,? = 0? = 1vuDefinitionIf we are given a function f : C ? C, and startingfrom a given point z0 we can form the iteratedsequence z1 = f (z0), z2 = f (z1), , zn+1= f (zn), .This sequence is called the orbit of z0 under f .There are two questions we can ask:(1) Does zn ? ? ?(2) If it does not diverge to ?, what does it do?We shall assume that f is an analytic functionthroughout.(Definition 2.4.)A point z0 in C is said to be a fixed point of f iff (z0) = z0.For example the function f (z) = z2 has 0 and 1 asfixed points.(Theorem 2.5.)Suppose z0 is a fixed point of an analytic functionf and | f (z0) | 0 suchthat for all z in D(z0, r), the orbit sequence startingat z converges to z0, i.e.,,nd?lim f n(z) = z0where f n(z) denote the iterated mapf n(z) = f (f n-1(z)) , f 1(z) = f (z).Proof. Let 0 0 suchthat for all z in D(z0, r), | f (z) | 1,(3) indifferent, if | f (z0) | = 1 ,(4) super-attracting if | f (z0) | = 0.(Defnition 2.6. ) If z0 is an attracting fixed point ofan anlytic function f , the basin of attraction of z 0under f is the set {z: f n(z) ? z0 }.(Definition 2.7.) The filled Julia set for a polynomialfunction f (z) is defined to be the set of points thatlauch bounded orbits through iteration of f ; the Juliaset is the boundary of the filled Julia set.(Example 2.8.)The Julia set for the function f (z) = z2 is the unitcircle and the filled Julia set is the closed unitdisk. This is because f n(z) ? 0 if |z| 1 and the orbit{ f n(z) } stays bounded if |z| =1 .(Definition 2.9.)The set of functions {Pc : c ? C} , where Pc(z) =z2 + c} is called the family of basic quadraticfunctions.(Theorem 2.10.)Let rc = 1 2 + 1 4 + |c| . Then for |z0| > rc , ( Pc n(z0) )is an increasing sequence and Pc n(z0) d ? .Proof.Note that P c(z) = |z2 + c| m |z|2 |c|. Nowconsider the quadratic equationx2 |c| = x.The positive solution of this equation isrc = .12+ 14 + |c|Now we claim that if x > rc + ?, ? > 0, thenx2 |c| > (1+ ?)x. (1)To see this, note that if x > rc + ?, then 1x rc + |c ||c| x2 |c| = x x x> x (rc + ) rc+ (2)Now(rc + ) rc|c+| = rc 2 + 2rrcc++2 |c| = rc +r2c+rc+ 2because rc 2 |c| = rc,mrc + + rc + 2rc + since rc ? 1,? 1 + ?.From (2) then x2 |c| > (1+ ?)x for x > rc + ?.Thus starting with |z0| > rc + ?, by (1)Pc(z0) = |z0 2 + c| m |z0|2 |c| > (1 + )|z0| > rc + .Applying (1) again,P .2c(z0) = |Pc(z0)2 + c| m |Pc(z0)|2 |c| > (1 + )|Pc(z0)| > |Pc(z0)|Observe that if z satisfies |z| > rc + ? so does |Pc(z)|.So we can apply (1) to the terms of the sequence ( Pc n(z0) )Hence,Pnc+1(z0) = |Pc n(z0)2 + c| m |Pc n(z0)|2 |c| > (1 + )|Pc n(z0)|> |Pc n(z0)| - (3)Hence ( Pc n(z0) ) is an increasing sequence.Moreover by (3),P .nc(z0) m (1 + )n|z0|Since, (1+ ?)n ? ? and |z0| > 0, by the Comparison Test,P .nc(z0) d ?Definition(Definition 2.11.)For c ? C, the escape set Ec is the set{z : Pc n(z) d ? as n d ?}.The complement of the escape set Ec is the filledJulia set or Keep set Kc.Definition 2.12. A subset A of C is said to becompletely invariant under a function f if andonly if z ? A ? f (z) ? A.Theorem(Theorem 2.13.) For each c in C, the escape set Ecand the keep set or filled Julia set Kc have thefollowing properties.(i) Ec ? {z : |z| > rc} and Kc ? {z : |z| ? rc};(ii) Ec is open and Kc is closed;(iii) Ec ? C and Kc ??;(iv) Ec and Kc are each completely invariant under Pc;(v) Ec and Kc are each symmetric under rotation by ?about the origin 0;(vi) Ec is (path) connected and Kc has no holes in it.Proof. Part (i)By Theorem 2.11, Ec ? {z : |z| > rc} andThe complement of Ec = Kc ? complement of {z : |z| > rc} ={z : |z| ?rc}.For part (ii), sufficient to prove that Ec is open since Kc is thecomplement of Ec.Take z0 in Ec, then since Pc n(z0) d ?, there exists a integer N suchthat n > N implies that Pc n(z0) > rc.Take a fixed m > N such that Pc m(z0) > rc.By continuity there exists a ? > 0 such thatfor all z in D(z0, ?), Pc m(z) > rc.( Take = Pc m(z0) rc > 0 and invoke definition of continuity.)Therefore, by Theorem 2.11, for z in D(z0, ?), . Pc n(z) d ?Hence D(z0, ?) ? Ec.Hence Ec is a neighbourhood of each of its points and so is open.(Part iii). Since Pc is a polynomial, it always has afixed point as Pc(z) z is also a plynomial and alwayshas a root in C. Therefore,Kc ?? and so Ec ? CPart (iv) is trivial as plainly f (Ec) ? Ec and f (Kc) ? Kc.Part (v) follows from the fact that Pc is an evenfunction, Pc(z) = Pc(-z) for all z in C.For Part (vi),By part (i) Ec ? {z : |z| > rc}.Since A= {z : |z| > rc} is path connected, to show thatEcis path connected we only have to show that any zin Ec can be joined by a path in Ec to a point in A.We shall prove this by contradiction.Suppose there exists a point ? in Ec that cannot be joined by apath in Ec to any point in A.Consider all those points can be joined to ? by a path in Ec, B ={z ? Ec : z can be joined by a path in Ec to ?}.We shall show that B must intersect A.Then B ? ? since B contains ? and B ? Ec.B is open:Take z in B, then there is a path in Ec joining z to ?. By part (ii)Ec is open, so there is a disk D(z, r), for some r > 0 such that D(z,r) ? Ec . Since D(z, r) is path connected any ? in D(z, r) can bejoined by a path in Ec to z and then to ?.Therfore, D(z, r) ? B. This means B is open.B is plainly path connected since every point can be joined by apath in Ec (hence in B) to ?. Note that B is bounded since Bcannot contain any points in A and so B ? {z : |z| ? rc}.?B?Ec = ?:If there exists a point ? in ?B?Ec, then since Ec is open,there exists an open disk D(?, ?) for some ? > 0 suchthat D(?, ?) ? Ec .But D(?, ?) is path connected and D(?, ?)?B ??(because ? is a boundary point of B) and so ? can bejoined by a path in D(?, ?) ? Ec to a point in B, whichin turn can be joined by a path in Ec to ?. Therfore, ? isin B = Int B, contradicting that ? ? ?B.Hence ?B?Ec = ?.Therefore, ?B ? Complement of Ec = Kc . This meansfor any ? in ?B and for any integer n in P,P .nc() [ rcSince B is open and path connected, by the MaximumModulus Principle (Theorem 2.3 Unit 2), themaximum modulus of each of the function P on thencclosure of B, occurs in a boundary point. Hence foreach n in P and for all z in B,P .nc(z) [ rcThis means the sequence (Pc n(z)) for each z in B isbounded and so B ? C Ec , contradicting that B ? Ec.Hence this shows that Ec is (path) connected.Kccannot contain any holes:Suppose Kc has a hole that means there exists acircle C(z0, R) such that C(z, R) ? Kc and thereexists a point ? in D(z0, R) such that ? ? Kc.Therefore, ? ? Ec.But we have jsut proved that Ec is path connectedand so there exists a path in Ec joining ? to somepoint in {z : |z| > rc}. But C(z, R) ? {z : |z| ? rc}.That means such a path must intersect the circleC(z, R) contradicting C(z, R) ? complement of Ec.Hence Kc cannot contain any holes.