MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 10 pps

MAPPINGS 313 Fig. 12.12 Conformal mapping 12.4.1 Conformal Mappings To see an interesting property of analytic functions we differentiate w = f ( 12.115) at zo, where the modulus and the arguments. of the derivative are given as I $ I and a, respectively. We now use polar coordinates to write the modulus 20 lim At-0 and the argument (Fig. 12.12) as a = lim arg[Azu] - lim arg[Az] Az-0 Ar-0 (12.116) ( 12.1 17) (12.118) (12.119) Since this function, f (z) , maps a curve c, in the z-plane into another curve c, in the w-plane, from the arguments [&. (12.119)] it is seen that if the slope of c, at is 00, then the slope of cw at wo is a+&. For a pair of curves intersecting at a the angle between their tangents in the w- and z-planes will be equal, that is, 4'2 - 41 = (02 + a) - (0, + a), ( 12.120) = e2 -el. (12.121) Hence analytic functions preserve angles between the curves they map (Fig. 12.12). For this reason they are also called conformal mappings or transfor- mations. 314 COMPLEX VARIABLES AND FUNCTIONS fig. 12.13 Two plates with hyperbolic cross sections 12.4.2 Electrostatics and Conformal Mappings Conformal mappings are very useful in electrostatic and laminar (irrotational) flow problems, where the Laplace equation must be solved. Even though the method is restricted to cases with one translational symmetry, it allows one to solve analytically some complex boundary value problems. Example 12.10. Conformal mappings and electrostatics: Let us con- sider two conductors held at potentials Vi and V2 with hyperbolic cross sections x2 - 9’ = c1 and x2 - y2 = c2. (12.122) We want to find the equipotentials and the electric field lines. In the complex z-plane the problem can be shown as in Figure 12.13. We use the conformal mapping (12.123) =x2-$+2(2Xy) ( 12.124) 2 W=Z to map these hyperbolae to the straight lines u = c1 and u = c2 (12.125) in the w-plane (Fig. 12.14). The problem is now reduced to finding the equipotentials and the electric field lines between two infinitely long MAPPINGS 315 fig. 12.14 Equipotentials and electric field lines in the w-plane parallel plates held at potentials Vl and .V2, where the electric field lines are given by the family of lines 2, = c3 (12.126) and the equipotentials are given by the lines perpendicular to these as u = ci. (12.127) Because the problem is in the z-plane, we make the inverse transforma- tion to obtain the electric field lines as (v =) 2zy = cj (12.128) and the equipotentials as (u =) z2 - y2 = cz. (12.129) In three dimensions, to find the equipotential surfaces these curves must be extended along the direction of the normal to the plane of the paper. Example 12.11. Electrostatics and conformal mappings: We now find the equipotentials and the electric field lines inside two conductors with semicircular cross sections separated by an insulator and held at poten- tials +V, and -V& respectively (Fig. 12.15). The equation of a circle 316 COMPLEX VARIABLES AND FUNCTIONS + z-plane Fig. 12.15 Two conductors with semicircular cross sections in the z-plane is given as 22+y2= 1. We use the conformal mapping w (2) = In (-) l+z 1-2 ' (12.130) (12.131) to map these semicircles into straight lines in the w-plane (Fig. 12.16). Using Equation (12.131) we write (12.132) l+z+iy 1-2-zy u + iu = In 1 1 - 22 - y2 + 2iy 1 -22+22+y2 =In[ and express the argument of the In function as Reia: u+zv=InR+icu. Now the u function is found as U=Ly 2Y 1 - (22fy2)' = tan-' (12.133) (12.134) (12.135) MAPPINGS 317 4 U v = -d2 6 Fig. 12.16 Two semicircular conductors in the w-plane From the limits and (12.136) we see that the two semicircles in the z-plane are mapped to two straight lines given as Equipotential surfaces in the w-plane cam now be written easily as av,, V(v) = ?I. 7r (12.139) Using Equation (12.135) we transform this into the z-plane to find the equipotentials as - - R (12.140) (I 2.14 1) Because this problem has translational symmetry perpendicular to the plane of the paper, equipotential surfaces in three dimensions can be 318 COMPLEX VARIABLES AND FUNCTIONS found by extending these curves in that direction. The solution to this problem has been found rather easily and in closed form. Compare this with the separation of variables method, where the solution is given in terms of the Legendre polynomials as an infinite series. However, applications of conformal mapping are limited to problems with one translational symmetry, where the problem can be reduced to two di- mensions. Even though there are tables of conformal mappings, it is not always easy as in this case to find an analytic expression for the needed mapping. 12.4.3 For laminar (irrotational) and frictionless flow, conservation of mass is given as Fluid Mechanics and Conformal Mappings g + 7. (p-i') = 0, (12.142) where p(?",) and -i'(?",t) represent the density and the velocity of a fluid element. For stationary flow we write aP - = 0, at (12.143) thus Equation (12.142) becomes 3. (p-i') = 0. (12.144) Also, a lot of realistic situations can be approximated by the incompressible fluid equation of state, that is, p = const. (12.145) This further reduces Equation (12.144) to V.3=O0. (12.146) This equation alone is not sufficient to determine the velocity field *(?",t). If the flow is irrotational, it will also satisfy ?XT=O, (12.147) thus the two equations V-3=0 (12.148) and Vx-i'=O (12.149) MAPPINGS 319 Fig. 12.17 Flow around a wall of height h completely specify the kinematics of laminar, frictionless flow of incompress- ible fluids. These equations are also the expressions of linear and angular momentum conservations for the fluid elements. Fluid elements in laminar flow follow streamlines, where the velocity g(?",t) at a given point is tan- gent to the streamline at that point. Equations (12.148) and (12.149) are the sarme as Maxwell's equations in electrostatics. Following the definition of electrostatic potential, we use Equa- tion (12.149) to define a velocity potential as (12.150) ++ 21 (T ,t) = Tkq7,t). Substituting this into Equation (12.148) we obtain the Laplace equation V2@(7, t) = 0. (12.151) We should note that even though a(?", t) is known as the velocity potential it is very different from the electrostatic potential. Example 12.12. Flow around an obstacle of height h: Let us consider laminar flow around an infinitely long and thin obstacle of height h. Since the problem has translational symmetry, we can show it in two dimen- sions as in Figure 12.17, where we search for a solution of the Laplace equation in the region R. Even though the velocity potential satisfies the Laplace equation like the electrostatic potential, we have to be careful with the boundary conditions. In electrostatics, electric field lines are perpendicular to the equipotentials; hence the test particles can only move perpendicular to the conducting surfaces. In the laminar flow case, motion perpendicular 320 COMPLEX VARIABLES AND FUNCTIONS to the surfaces is not allowed because fluid elements follow the contours of the bounding surfaces. For points far away from the obstacle, we take the flow lines as parallel to the z-axis. As we approach the obstacle, the flow lines follow the contours of the surface. For points away from the obstacle, we set v, = 1. (12.152) We now look for a transformation that maps the region R in the z-plane to the upper half of the w-plane. Naturally, the lower boundary of the region R in Figure 12.17 will be mapped to the real axis of the w-plane. We now construct this transformation in three steps: We first use w1= z 2 (12.153) to map the region R to the entire wl-plane. Here the obstacle is between 0 and -h2. As our second step, we translate the obstacle to the interval between 0 and h2 by ~2 = z2 + h2. (12.154) Finally we map the w2-plane to the upper half of the w-plane by w = 6. (12.155) The complete transformation from the z-plane to the w-plane can be written as (Fig. 12.18) w = JW. The Laplace equation can now be easily solved in the upper half of the w-plane, and the streamlines are obtained as 21 = cj. Curves perpendicular to these will give the velocity equipotentials as U= bj. Finally transforming back to the z-plane we find the streamlines as the curves cj = Irn[J2TP], and the velocity of the fluid elements that are tangents to the streamlines (Fig. 12.19) are given as MAPPINGS 321 wz= w , + h2 I fig. 12.18 Transition from the z-plane to the w-plane 322 COMPLEX VARIABLES AND FUNCTIONS h T’ 4 z-plane fig. 12.20 upper half of the w-plane Schwara-Christoffel transformation maps the inside of a polygon to the 12.4.4 Schwarz-Christoffel Transformations We have seen that analytic transformations are also conformal mappings, which preserve angles. We now introduce the Schwarz-Christoffel transfor- mations, where the transformation is not analytic at an isolated number of points. Schwarz-Christoffel transformations map the inside of a polygon in the z-plane to the upper half of the w-plane (Fig. 12.20). To construct the Schwarz-Christoffel transformations let us consider the function (12.156) [...]... Conducting circular cylinder parallel to infinite metallic plate 333 This Page Intentionally Left Blank 13 COMPLEX INTEGRALS and SERIES In this chapter we first introduce the complex integral theorems Using analytic continuation we discuss how these theorenis can be used to evaluate some frequently encountered definite integrals In conjunction with our discussion of definite integrals, we also introduce... u(2, = sin z cosh y + x 2 - y2 + 4xy y) is a harmonic function and find its conjugate 12.3 Show that u(z, = sin 2z/(cosh 2 + cos 2z) y) y can be the real part of an analytic function f ( z ) Find its imaginary part and express f (z ) explicitly as a function of z 12.4 Using cylindrical coordinates and the method of separation of variables find the equipotentials and the electric field lines inside... CLASSIFICATION OF SINGULAR POINTS 13.4 347 CLASSIFICATION OF SINGULAR POINTS Using Laurent the series we can classify singular points of a function Definition I Isolated singular point: If a function is not analytic at a but analytic at every other point in some neighborhood of a, then is called an isolated singular point Definition I1 Essential singular point: In the Laurent series of a function: C 00 f(z) = an(Z-*)... the equipotentials and the electric field lines and plot 12.13 Map the real w-axis into the triangular region shown in Figure 12.31, in the limit a s 25 f 03 and 23 -f -03 12.14 Find the equipotentials and the electric field lines for a conducting and parallel to a grounded infinite concircular cylinder held at potential VO ducting plane (Fig 12.32) Hint: Use the transformation z = a tanhiw/2 PROBLEMS... n 7 (13.62) n= M if for n < - I l m < 0, a, = 0 and a-lrnl then # 0, is called a singular point of order m Definition I11 Essential singular point: If m is infinity, then singular point a is called an essential Definition IV Simple Pole: In Definition 11, if m = 1, then zo is called a simple pole Definition V Entire function:When a function is analytic in the entire z-plane it is called an entire function... (’ d (13.20) where z’ is a point on the contour and z is a point inside the contour C (Fig 13.4) We can now write Equation (13.20) as = - Since the inequality Iz the binomial formula k j6, i f ( 2 )dz’ [(z’ a)- ( z - zo)] f 1 < Iz’ - - (2’) dz’ (13.21) is satisfied in and on C , we can use 00 (13.22) t o write (13.23) Interchanging the integral and the summation signs we find (13.24) which gives us the... the singularities o f f(z ) = tanh L 10 Show that the transformation F 331 332 COMPLEX VARIABLES AND FUNCTIONS t v=o fig 12.30 Rectangular region surrounded by metallic plates or maps the 21 =const lines into circles in the z-plane 12.11 Use the transformation given in Problem 12 .10 to find the equipotentials and the electric field lines for the electrostatics problem of two infinite parallel cylindrical... conditions are satisfied in and on the closed path C, the right-hand side of Equation (13.5) is zero, thus proving the theorem 11 Cauchy Integral Theorem If f ( z ) is analytic in and on a closed path C in a simply connected domain (Fig 13.2) and if ~0 is a point inside the path C, then we can write the integral (13.6) COMPLEX IN JEGRAL THEOREMS 337 Fig 13.2 Contour for the Cauchy integral theorem Proof... contribution of Un(n 1 and about (13.54)... th'e path in Figure 13.2 and use that in Figure 3.3, where we can use the Cauchy-Goursat theorem t o write (13.7) This integral must be evaluated in the liniit as the radius of the path Co goes to zero Integrals over L1 and L2 cancel each other Also noting that the integral over Co is taken clockwise, we write (13.8) where both integrals are now taken counterclockwise The integral on the left-hand side . lines u = c1 and u = c2 (12.125) in the w-plane (Fig. 12.14). The problem is now reduced to finding the equipotentials and the electric field lines between two infinitely long MAPPINGS. coordinates and the method of separation of variables find the equipotentials and the electric field lines inside two conductors with semi-circular cross sections separated by an insulator and. given in Problem 12 .10 to find the equipoten- tials and the electric field lines for the electrostatics problem of two infinite parallel cylindrical conductors, each of radius R and separated